from:"jonas . thornvall"

Den onsdag 13 juli 2016 kl. 00:24:23 UTC+2 skrev Ian:
> On Tue, Jul 12, 2016 at 11:35 AM,   wrote:
> >
> > No it is only compressible down to a limit given by the algorithm.
> 
> Then your algorithm does not compress random data as you claimed.
> 
> For some input, determine the limiting output that it ultimately
> compresses down to. Take that output and feed it through your
> algorithm as if it were the original input. If the data are to be
> considered random, then this input is just as probable as the
> original. What output does the algorithm now create? If it just
> returns the input unchanged, then how do you discern the original
> input from this input when decompressing? If it returns a different
> output of the same size, then repeat the process with the new output.
> Now there are *two* outputs of that size that can't be repeated. There
> are only finitely many possible outputs of that size, so eventually
> you're going to have to get to one that either repeats an output -- in
> which case your algorithm produces the same output for two different
> inputs and is therefore incorrect -- or you will get to an input that
> produces an output *larger* in size than the original.

The later sounds reasonable that is start toggle between states.
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Re: Compression of random binary data

Den onsdag 13 juli 2016 kl. 00:24:23 UTC+2 skrev Ian:
> On Tue, Jul 12, 2016 at 11:35 AM,   wrote:
> >
> > No it is only compressible down to a limit given by the algorithm.
> 
> Then your algorithm does not compress random data as you claimed.
> 
> For some input, determine the limiting output that it ultimately
> compresses down to. Take that output and feed it through your
> algorithm as if it were the original input. If the data are to be
> considered random, then this input is just as probable as the
> original. What output does the algorithm now create? If it just
> returns the input unchanged, then how do you discern the original
> input from this input when decompressing? If it returns a different
> output of the same size, then repeat the process with the new output.
> Now there are *two* outputs of that size that can't be repeated. There
> are only finitely many possible outputs of that size, so eventually
> you're going to have to get to one that either repeats an output -- in
> which case your algorithm produces the same output for two different
> inputs and is therefore incorrect -- or you will get to an input that
> produces an output *larger* in size than the original.

The dataset must have a certain size that is the only requirment, of course you 
can not compress something into nothing, at least the arithmetic ruleset need 
to be encoded but what would be the point to just compress something less than 
a couple of bytes. And of course the number of rounds you applied the algorithm 
must be stored. But that is no problem for small datasets.

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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 21:40:36 UTC+2 skrev jonas.t...@gmail.com:
> Den tisdag 12 juli 2016 kl. 20:20:52 UTC+2 skrev Michael Torrie:
> > On 07/12/2016 11:46 AM, jonas.thornv...@gmail.com wrote:
> > > Well the algorithm start with looking up a suitable folding structure
> > > "close enough to the number", then it works down the folding
> > > structure finding the fold where the difference or sum between the
> > > folds closest to zero.
> > > 
> > > You do the same prinicple with the remainder until zero is achieved.
> > > 
> > > So our first fold can either be bigger or smaller, and it seek a
> > > configuration for the fold that close in max on the actual random
> > > number. The second fold could be a fold that depending upon our first
> > > fold was bigger or smaller than number either will add or subtract
> > > lower layers of the fold.
> > > 
> > > There will come out a difference that need to be folded, the process
> > > is repeated until there is nothing to fold.
> > > 
> > > It is basicly a search algorithm looking for suitable folding
> > > structures.
> > 
> > Better patent it quickly then.  And you will win a noble prize for math
> > if you could do what you say you could.
> 
> I must stress when i say number here i really mean +10 decimal digit. So 
> i basicly search in on big numbers that i compress. So i divide the dataset 
> into suitable sizes for compression.

And the dataset chunks that comes out from the process can also be treated like 
a new datafile, so the compression is iterative down to a limit. 
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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 20:20:52 UTC+2 skrev Michael Torrie:
> On 07/12/2016 11:46 AM, jonas.thornv...@gmail.com wrote:
> > Well the algorithm start with looking up a suitable folding structure
> > "close enough to the number", then it works down the folding
> > structure finding the fold where the difference or sum between the
> > folds closest to zero.
> > 
> > You do the same prinicple with the remainder until zero is achieved.
> > 
> > So our first fold can either be bigger or smaller, and it seek a
> > configuration for the fold that close in max on the actual random
> > number. The second fold could be a fold that depending upon our first
> > fold was bigger or smaller than number either will add or subtract
> > lower layers of the fold.
> > 
> > There will come out a difference that need to be folded, the process
> > is repeated until there is nothing to fold.
> > 
> > It is basicly a search algorithm looking for suitable folding
> > structures.
> 
> Better patent it quickly then.  And you will win a noble prize for math
> if you could do what you say you could.

I must stress when i say number here i really mean +10 decimal digit. So i 
basicly search in on big numbers that i compress. So i divide the dataset into 
suitable sizes for compression.
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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 20:20:52 UTC+2 skrev Michael Torrie:
> On 07/12/2016 11:46 AM, jonas.thornv...@gmail.com wrote:
> > Well the algorithm start with looking up a suitable folding structure
> > "close enough to the number", then it works down the folding
> > structure finding the fold where the difference or sum between the
> > folds closest to zero.
> > 
> > You do the same prinicple with the remainder until zero is achieved.
> > 
> > So our first fold can either be bigger or smaller, and it seek a
> > configuration for the fold that close in max on the actual random
> > number. The second fold could be a fold that depending upon our first
> > fold was bigger or smaller than number either will add or subtract
> > lower layers of the fold.
> > 
> > There will come out a difference that need to be folded, the process
> > is repeated until there is nothing to fold.
> > 
> > It is basicly a search algorithm looking for suitable folding
> > structures.
> 
> Better patent it quickly then.  And you will win a noble prize for math
> if you could do what you say you could.

I doubt it i never got anyone before for my ideas.
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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 17:19:54 UTC+2 skrev Steven D'Aprano:
> On Wed, 13 Jul 2016 12:29 am, jonas.thornv...@gmail.com wrote:
> 
> > Den tisdag 12 juli 2016 kl. 05:01:20 UTC+2 skrev Lawrence D’Oliveiro:
> >> On Tuesday, July 12, 2016 at 5:52:27 AM UTC+12, jonas.t...@gmail.com
> >> wrote:
> >> 
> >> > What kind of statistic law or mathematical conjecture  or is it even a
> >> > physical law is violated by compression of random binary data?
> >> 
> >> Try compressing already-compressed data.
> >> 
> >> Does that answer your question?
> > 
> > Yes that is my question, and also a claim i can do it.
> 
> Can you also make a perpetual motion machine, square the circle, and find an
> exact rational fraction equal to pi?
> 
> 
> What gets me is the people who *say* that they can compress already
> compressed data. We know they can't, because if they could, they could
> compress it again and again and again and again until there was only a
> single bit, AND STILL REVERSE IT, using no external storage. Your lookup
> tables are part of the compressed data. If the "compressed file" plus the
> lookup table is bigger than the original file, then you haven't really
> compressed anything. You've just moved some of it from the file into a
> lookup table.
> 
> So why do people claim that they can compress already compressed data? Who
> are they fooling? Themselves?
> 
> 
> 
> -- 
> Steve

Well the algorithm start with looking up a suitable folding structure "close 
enough to the number", then it works down the folding structure finding the 
fold where the difference or sum between the folds closest to zero.

You do the same prinicple with the remainder until zero is achieved.

So our first fold can either be bigger or smaller, and it seek a configuration 
for the fold that close in max on the actual random number. The second fold 
could be a fold that depending upon our first fold was bigger or smaller than 
number either will add or subtract lower layers of the fold. 

There will come out a difference that need to be folded, the process is 
repeated until there is nothing to fold.

It is basicly a search algorithm looking for suitable folding structures.
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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 17:12:01 UTC+2 skrev Steven D'Aprano:
> On Wed, 13 Jul 2016 12:24 am, jonas.thornv...@gmail.com wrote:
> 
> > Den måndag 11 juli 2016 kl. 20:38:51 UTC+2 skrev Steven D'Aprano:
> >> On Tue, 12 Jul 2016 03:52 am, jonas.thornv...@gmail.com wrote:
> >> 
> >> > What kind of statistic law or mathematical conjecture  or is it even a
> >> > physical law is violated by compression of random binary data?
> >> 
> >> The pigeon hole principle. If you have 100 pigeon holes, and 101 pigeons,
> >> then clearly at least one pigeon hole must have two pigeons in it.
> [...]
> > But it seem your reasoning is based upon interpretation of the actual
> > digits, bits and bytes value.
> 
> Not at all. If you think that, you've misread my example. There's no
> interpretation of the bytes: they are just 8-bit numbers from 0 to 255. You
> cannot losslessly compress all 256 of them to just four 2-bit numbers.
> 
> 
> > There could be different interpretation 
> > worlds of course you would have to chose one using digits, An 
> > interpretationworld here could be reading out different word lengths of
> > the dataset and maybe a lookup table.
> 
> Any lookup table you have counts as part of the compressed data.
> 
> 
> > But it could also be arithmetic rules that magically recreate a number
> > from a number of folds or difference of folds.
> 
> Oh, sure, if you believe in magic, anything is possible. Just close your
> eyes, click your heels together, and wish really, really hard.
> 
> Suppose I could compress ANY random data, no matter what, down to 10% of the
> original size. Okay, let's start with a million bits of data. Compress it
> down to 100,000 bits.
> 
> But I believe that I can compress *anything*, any random collection of data.
> Okay, let me compress it again. Now I have 10,000 bits.
> 
> Compress it again. Now I have 1,000 bits.
> 
> Compress it again. Now I have 100 bits.
> 
> Compress it again. Now I have 10 bits.
> 
> Compress it again. Now I have 1 bit, either a 0 or a 1.
> 
> 
> Can you not see how absurd this is? I have claimed that I can take *any*
> random set of data, and by compressing it again and again and again,
> compress it down to ONE BIT, either a 0 or a 1, WITHOUT LOSS. Somehow I
> have to take that 0 bit and uncompress it back to the Complete Works Of
> William Shakespeare, and *also* uncompress it back to the recent Deadpool
> movie, AND uncompress it back to last year's Ant Man movie, AND uncompress
> it back to some funny picture of a cat.
> 
> How can I possibly know which of the billions and billions of different
> files this 0 bit represents?
> 
> If you pass me a 0 bit, and say "uncompress this", and I get The Lord Of The
> Rings novels, and then you pass me another 0 bit, and I uncompress it and
> get The Hobbit, well, how did I tell the two bits apart? They're both zero.
> 
> 
> 
> The alternative is to say, it doesn't matter how clever you are, you can't
> compress *everything*. There are some things that simply won't compress.
> Eventually you get something no longer compresses. If you could compress
> EVERYTHING, then you could compress the compressed data, and compress the
> compressed-compressed data, and so on, until you've got only a single bit.
> And that is ridiculous.
> 
> 
> 
> -- 
> Steven
> “Cheer up,” they said, “things could be worse.” So I cheered up, and sure
> enough, things got worse.

No it is only compressible down to a limit given by the algorithm.
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Re: Compression of random binary data

Den tisdag 12 juli 2016 kl. 05:01:20 UTC+2 skrev Lawrence D’Oliveiro:
> On Tuesday, July 12, 2016 at 5:52:27 AM UTC+12, jonas.t...@gmail.com wrote:
> 
> > What kind of statistic law or mathematical conjecture  or is it even a
> > physical law is violated by compression of random binary data? 
> 
> Try compressing already-compressed data.
> 
> Does that answer your question?

Yes that is my question, and also a claim i can do it.
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Re: Compression of random binary data

Den måndag 11 juli 2016 kl. 20:38:51 UTC+2 skrev Steven D'Aprano:
> On Tue, 12 Jul 2016 03:52 am, jonas.thornv...@gmail.com wrote:
> 
> > What kind of statistic law or mathematical conjecture  or is it even a
> > physical law is violated by compression of random binary data?
> 
> The pigeon hole principle. If you have 100 pigeon holes, and 101 pigeons,
> then clearly at least one pigeon hole must have two pigeons in it.
> 
> To keep the numbers small and manageable, let's say we are going to compress
> one byte at a time. Now a byte has eight bits, so there are exactly 256
> possible bytes:
> 
>  
>  0001
>  0010
> ...
>  1110
>  
> 
> Now, suppose I claim that I can LOSSLESSLY (that is, reversibly) compress
> any random byte to just two bits. The lossless part is important: its not
> hard to compress random data by irreversibly throwing some of it away, and
> there's no violation there.
> 
> So I claim that you can give me any random byte, I will compress it to just
> two bits:
> 
> 00
> 01
> 10
> 11
> 
> and then be able to decompress it back again to give you the original byte
> once more.
> 
> Obviously I'm trying to pull a fast one! There's no way I can do this. I can
> squeeze 256 pigeons into just four pigeon holes, but only by doubling them
> up. Suppose I compress these three bytes to 00:
> 
>  
> 0110 1001
> 1100 0110
> 
> Now when I go to uncompress 00, what should I return? There is no way for me
> to know which of the three was the original value.
> 
> (If I'm cunning, I'll have sneakily stored some data *elsewhere*, say, in
> the file name, or in a database, so that you need this extra hidden data to
> uncompress the 00 back to a full byte. But then I'm not compressing eight
> bits down to two. I'm compressing eight bits down to two bits plus
> who-knows-how-many-bits of hidden data.)
> 
> So the pigeon hole principle tells us one of two things:
> 
> (1) If you compress random data, then it must be lossy; I can compress eight
> bits to two, but then I can't uncompress it back again, at least not
> without throwing away some data.
> 
> (2) Or, if the compression is lossless, then some data must be expanded
> rather than compressed. If you pick data at random, some of it will be
> expanded.
> 
> Suppose I have a compression algorithm that infallibly and reversibly
> compresses as follows:
> 
>   <--> 00
>  0001 <--> 01
>  0010 <--> 10
>  0011 <--> 11
> 
> That part is fine. But what will my algorithm do with the other 252 bytes?
> At *best* it will leave them untouched:
> 
>  0100 <-->  0100
> ...
>   <-->  
> 
> which is no compression at all, but at worst it will actually expand them
> and make them bigger. (After all, it's likely that my compression format
> has at least a bit of overhead.)
> 
> In practice, compression algorithms are designed to look for particular
> kinds of order or structure in the data, and compress that. That's fine for
> the sorts of non-random data we care about: pictures are rarely pictures of
> static, text files are rarely random collections of bits. But if you do
> throw a random set of bits at a lossless compression algorithm, it will at
> best not compress it at all, and at worst actually make the file bigger.
> 
> 
> > What is to say that you can not do it if the symbolic representation is
> > richer than the symbolic represenatation of the dataset.
> > 
> > Isn't it a fact that the set of squareroots actually depict numbers in a
> > shorter way than their actual representation.
> 
> Sure. But you need to know what √2 means. It *represents* the number
> 1.41421356237... but doesn't compress it. There's nothing you can do to the
> symbol √2 that will uncompress back to the infinite series of digits. All
> you can do is look it up somewhere to see what the digits are.
> 
> > Now the inpretator or program must know the rules. And i have very good
> > rules to make it happen.
> 
> Right. How much information is in the rules? More than you save with
> the "compression". Consider:
> 
> 1.41421356237 compressed down to √2, that's 13 characters down to 2. Great!
> But to *uncompress*, you need to store a rule:
> 
> √2=1.41421356237 
> 
> and that's *sixteen* characters. So your "compression" is:
> 
> original: 13
> compressed to: 2
> plus rule: 16
> 
> means you have "compressed" 13 characters to 18.
> 
> Now, this is still worth doing if you need to repeat the √2 many times, so
> long as you don't have to repeat the rule. That's useful. But it's not
> compression. It's more like keeping an index to a database, or a scrap of
> paper with the title of a book written on it:
> 
> "See Lord Of The Rings, by J.R.R. Tolkien"
> 
> That's a lot smaller than the actual book: eight words, instead of who knows
> how many tens of thousands. But you can't call it compression: you can't
> sit down with the scrap of paper *and nothing else* and uncompress it back
> to the entire LOTR trilogy.
> 
> 
> 
>

Re: Compression of random binary data

2016-07-11 Thread jonas . thornvall

Den måndag 11 juli 2016 kl. 20:24:37 UTC+2 skrev jonas.t...@gmail.com:
> Den måndag 11 juli 2016 kl. 20:09:39 UTC+2 skrev Waffle:
> > On 11 July 2016 at 20:52,   wrote:
> > > What kind of statistic law or mathematical conjecture  or is it even a 
> > > physical law is violated by compression of random binary data?
> > >
> > > I only know that Shanon theorised it could not be done, but were there 
> > > any proof?
> > 
> > Compression relies on some items in the dataset being more frequent
> > than others, if you have some dataset that is completely random it
> > would be hard to compress as most items have very similar number of
> > occurrances.
> > 
> > > What is to say that you can not do it if the symbolic representation is 
> > > richer than the symbolic represenatation of the dataset.
> > >
> > > Isn't it a fact that the set of squareroots actually depict numbers in a 
> > > shorter way than their actual representation.
> > 
> > A square root may be smaller numerically than a number but it
> > definitely is not smaller in terms of entropy.
> > 
> > lets try to compress the number 2 for instance using square roots.
> > sqrt(2) = 1.4142
> > the square root actually takes more space in this case even tho it is
> > a smaller number. so having the square root would have negative
> > compression in this case.
> > with some rounding back and forth we can probably get around the fact
> > that sqrt(2) would take an infinite amout of memory to accurately
> > represent but that neccesarily means restricting the values we are
> > possible of encoding.
> > 
> > for sqrt(2) to not have worse space consumprion than the number 2
> > itself we basically have to trow away precision so sqrt(2) ~= 1
> > now i challenge you to get that 2 back out of that 1..
> 
> Well who it to say different kind of numbers isn't treated differently, i 
> mean all numbers isn't squares. All numbers isn't naturals.

But it could be all numbers are foldable. Both the integer parts and the real 
parts.And be expressed by the folding differences.
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Re: Compression of random binary data

2016-07-11 Thread jonas . thornvall

Den måndag 11 juli 2016 kl. 20:09:39 UTC+2 skrev Waffle:
> On 11 July 2016 at 20:52,   wrote:
> > What kind of statistic law or mathematical conjecture  or is it even a 
> > physical law is violated by compression of random binary data?
> >
> > I only know that Shanon theorised it could not be done, but were there any 
> > proof?
> 
> Compression relies on some items in the dataset being more frequent
> than others, if you have some dataset that is completely random it
> would be hard to compress as most items have very similar number of
> occurrances.
> 
> > What is to say that you can not do it if the symbolic representation is 
> > richer than the symbolic represenatation of the dataset.
> >
> > Isn't it a fact that the set of squareroots actually depict numbers in a 
> > shorter way than their actual representation.
> 
> A square root may be smaller numerically than a number but it
> definitely is not smaller in terms of entropy.
> 
> lets try to compress the number 2 for instance using square roots.
> sqrt(2) = 1.4142
> the square root actually takes more space in this case even tho it is
> a smaller number. so having the square root would have negative
> compression in this case.
> with some rounding back and forth we can probably get around the fact
> that sqrt(2) would take an infinite amout of memory to accurately
> represent but that neccesarily means restricting the values we are
> possible of encoding.
> 
> for sqrt(2) to not have worse space consumprion than the number 2
> itself we basically have to trow away precision so sqrt(2) ~= 1
> now i challenge you to get that 2 back out of that 1..

Well who it to say different kind of numbers isn't treated differently, i mean 
all numbers isn't squares. All numbers isn't naturals.
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Compression of random binary data

2016-07-11 Thread jonas . thornvall

What kind of statistic law or mathematical conjecture  or is it even a physical 
law is violated by compression of random binary data? 

I only know that Shanon theorised it could not be done, but were there any 
proof? 

What is to say that you can not do it if the symbolic representation is richer 
than the symbolic represenatation of the dataset. 

Isn't it a fact that the set of squareroots actually depict numbers in a 
shorter way than their actual representation. 

Now the inpretator or program must know the rules. And i have very good rules 
to make it happen. 
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Re: Compact connected regular graphs

2016-03-06 Thread jonas . thornvall

Den söndag 6 mars 2016 kl. 12:01:02 UTC+1 skrev Mark Lawrence:
> On 06/03/2016 10:39, jonas.thornv...@gmail.com wrote:
> > How come all graphs using 42 links and more then 84 nodes have compact  
> > regular connected solutions?
> >
> > Did turn off animation for more than a couple of thousands of  links so the 
> > animation is off when searching but script give message and numerical 
> > results on search.
> >
> > http://jt.node365.se/nodes15.html
> >
> > Do anyone know?
> >
> >
> > http://jt.node365.se/Compact.rtf
> >
> 
> Where is the Python question here?
> 
> -- 
> My fellow Pythonistas, ask not what our language can do for you, ask
> what you can do for our language.
> 
> Mark Lawrence

It is the Python that squeeze the information out of the graphs.
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Compact connected regular graphs

2016-03-06 Thread jonas . thornvall

How come all graphs using 42 links and more then 84 nodes have compact  regular 
connected solutions?

Did turn off animation for more than a couple of thousands of  links so the 
animation is off when searching but script give message and numerical results 
on search.

http://jt.node365.se/nodes15.html

Do anyone know?


http://jt.node365.se/Compact.rtf
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Re: Condition fullfilled to early but only "sometimes"

2016-03-01 Thread jonas . thornvall

Den tisdag 1 mars 2016 kl. 08:49:57 UTC+1 skrev Ian:
> It's not at all clear what the problem is from your description. What
> is it that you expect the code to do? What is it doing instead that
> violates your expectation? Why are you asking for Javascript help on a
> Python mailing list?
> 
> On Mon, Feb 29, 2016 at 10:40 PM,   wrote:
> > I've been looking at the code for two days so i am a bit crosseyed.
> > If anyone could help me set the condition so the loop catch the last 
> > pair/pairs, it is kind of weird that it succeed sometimes and break to 
> > early and report fail others.
> >
> > I would be very greatful if anyone can see why it break to early and return 
> > fail. Although it is clearly a succes looking at the last remaining pair 
> > none connected pair.It should report fail for the network permutations that 
> > stall due to link exhausting, but not stall creating last pair.
> >
> > It just a minor bug due to some condition that i just not get.
> > Have a go 5 munutes maybe someone catch the faulthy condition.
> >
> > http://jt.node365.se/mydebug1.html
> > --
> > https://mail.python.org/mailman/listinfo/python-list

Here is a script to draw networks, drag and drop nodes. So the function will go 
here, right now only node deep one work.

http://jt.node365.se/nodes11.html
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Re: Condition fullfilled to early but only "sometimes"

2016-03-01 Thread jonas . thornvall

Den tisdag 1 mars 2016 kl. 08:49:57 UTC+1 skrev Ian:
> It's not at all clear what the problem is from your description. What
> is it that you expect the code to do? What is it doing instead that
> violates your expectation? Why are you asking for Javascript help on a
> Python mailing list?
> 
> On Mon, Feb 29, 2016 at 10:40 PM,   wrote:
> > I've been looking at the code for two days so i am a bit crosseyed.
> > If anyone could help me set the condition so the loop catch the last 
> > pair/pairs, it is kind of weird that it succeed sometimes and break to 
> > early and report fail others.
> >
> > I would be very greatful if anyone can see why it break to early and return 
> > fail. Although it is clearly a succes looking at the last remaining pair 
> > none connected pair.It should report fail for the network permutations that 
> > stall due to link exhausting, but not stall creating last pair.
> >
> > It just a minor bug due to some condition that i just not get.
> > Have a go 5 munutes maybe someone catch the faulthy condition.
> >
> > http://jt.node365.se/mydebug1.html
> > --
> > https://mail.python.org/mailman/listinfo/python-list

Here is example output and as you can see there is no reason to not create the 
last pair, the link is not a copy it should be created but condition somehow 
wrong *sometimes*.

Hurray failed generate network ***not quite if single pair two nodes missing 
0'st NODE Links-> 7,3,5
1'st NODE Links-> 4,6,3
2'st NODE Links-> 5,9,3
3'st NODE Links-> 0,1,2
4'st NODE Links-> 1,8,5
5'st NODE Links-> 0,2,4
6'st NODE Links-> 1,7,9
7'st NODE Links-> 0,6,8
8'st NODE Links-> 4,7
9'st NODE Links-> 2,6
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Re: Condition fullfilled to early but only "sometimes"

2016-03-01 Thread jonas . thornvall

Den tisdag 1 mars 2016 kl. 08:49:57 UTC+1 skrev Ian:
> It's not at all clear what the problem is from your description. What
> is it that you expect the code to do? What is it doing instead that
> violates your expectation? Why are you asking for Javascript help on a
> Python mailing list?
> 
> On Mon, Feb 29, 2016 at 10:40 PM,   wrote:
> > I've been looking at the code for two days so i am a bit crosseyed.
> > If anyone could help me set the condition so the loop catch the last 
> > pair/pairs, it is kind of weird that it succeed sometimes and break to 
> > early and report fail others.
> >
> > I would be very greatful if anyone can see why it break to early and return 
> > fail. Although it is clearly a succes looking at the last remaining pair 
> > none connected pair.It should report fail for the network permutations that 
> > stall due to link exhausting, but not stall creating last pair.
> >
> > It just a minor bug due to some condition that i just not get.
> > Have a go 5 munutes maybe someone catch the faulthy condition.
> >
> > http://jt.node365.se/mydebug1.html
> > --
> > https://mail.python.org/mailman/listinfo/python-list

The program creates search for uniform networks, that is x nodes each with y 
links. Only a subset of permutations possible "easiest found out with pen and 
paper". So to the left is the node and to the right the nodes it links to.

When you run new network, sometimes it is full every node have x links, you 
create a uniform network. But often ir reports fail the links was exhausted and 
you stuck into a loop, that i fortunatily have the conditions to break.

But it *sometimes* break to early when there is still one or two pairs that 
would had worked. Not it is easy to make a fix for those, but it certainly 
would be more beautiful setting the correct condition.

function createLinks()
{
   var i = 0;
   var j = 0;

   while(i < nodes)
   {
// This see so that links already generated accounted for if one link than j
  j = arr[i].nodelinks.length;
  stupid = nodes - 1;
  temparr = new Array();
  while(j < links)
  {
 dublett = false;
 // Onlygenerate random values bigger than "i" else all links  exhausted
 if(i == 0)
 {
aLink = Math.floor(Math.random() * (nodes - 1));
aLink ++ ;
 }
 else
 {
aLink = Math.floor(Math.random() * (stupid - i)) + i + 1;
 }
 if (aLink == nodes)aLink -- ;

 if (temparr[0] == null)
 {
temparr[0] = aLink;
 }
 for(k = 0; k < arr[i].nodelinks.length; k ++ )
 {
if(aLink == arr[i].nodelinks[k])
{
   dublett = true;
}
 }
 inmylist = false;
 var t = 0;
 for(var m = 0; m < temparr.length; m ++ )
 {
if(temparr[m] == aLink)
{
   inmylist = true;
}
 }
 if (inmylist == false)
 {
temparr[temparr.length] = aLink;
 
 }
 else
 {
inmylist = false
 }
 scope = (nodes - 1) - i;
 if (temparr.length >= scope)
 {
myboolean = false;
return myboolean;
 }

 if(dublett == false && arr[aLink].nroflinks < links)
 {
arr[i].nodelinks[arr[i].nodelinks.length] = aLink;
arr[aLink].nodelinks[arr[aLink].nodelinks.length] = i;
arr[i].nroflinks ++ ;
arr[aLink].nroflinks ++ ;
j ++ ;
 }
  }
  i ++ ;
   }
   myboolean = true;
   return myboolean;
}
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Condition fullfilled to early but only "sometimes"

I've been looking at the code for two days so i am a bit crosseyed.
If anyone could help me set the condition so the loop catch the last 
pair/pairs, it is kind of weird that it succeed sometimes and break to early 
and report fail others.

I would be very greatful if anyone can see why it break to early and return 
fail. Although it is clearly a succes looking at the last remaining pair none 
connected pair.It should report fail for the network permutations that stall 
due to link exhausting, but not stall creating last pair.

It just a minor bug due to some condition that i just not get.
Have a go 5 munutes maybe someone catch the faulthy condition.

http://jt.node365.se/mydebug1.html
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Re: Loop awareness

Den måndag 29 februari 2016 kl. 21:32:51 UTC+1 skrev Ian:
> On Mon, Feb 29, 2016 at 1:07 PM,   wrote:
> > This program creates a uniform linktree of x nodes, and it knows when it 
> > get stuck in a loop generating random values.
> >
> > Because the networks random generated, only a subset of the permutations 
> > will generate a uniform network most get stuck in loops generating random 
> > values. But the program keep track the exhausted links and knows when no 
> > uniform network possible.
> >
> > Is this related to the halting problem?
> >
> > http://jt.node365.se/mydebug1.html
> 
> No, the halting problem is that you can't algorithmically determine
> whether an *arbitrary* program given *arbitrary* input will ever halt.
> When you narrow down the scope of the problem considerably, as you've
> done here, then it's no longer necessarily undecidable.

Here i told it to only report the unfiorm subset of working networks.



http://jt.node365.se/mydebug2.html
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Loop awareness

This program creates a uniform linktree of x nodes, and it knows when it get 
stuck in a loop generating random values.

Because the networks random generated, only a subset of the permutations will 
generate a uniform network most get stuck in loops generating random values. 
But the program keep track the exhausted links and knows when no uniform 
network possible.

Is this related to the halting problem?

http://jt.node365.se/mydebug1.html
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Re: General computer language, syntax question.

Den måndag 29 februari 2016 kl. 14:57:04 UTC+1 skrev jonas.t...@gmail.com:
> Den måndag 29 februari 2016 kl. 14:45:37 UTC+1 skrev jonas.t...@gmail.com:
> > I have a problem programming uniform networks, "x nodes with y links" that 
> > turn out to be really hairy to solve for me and i feel i really lack the 
> > programming features 
> > 
> > "Actually i program in javascript" but the problem seem general for all 
> > programming languages including Pyhton.
> > 
> > For a beautiful solution it would require "If in list/array return boolean" 
> > "If not in list/array return boolean". 
> > 
> > But there is no such feature Python nor Javascript, so instead i set 
> > boolean value "inlist" to false and loop thru, to see if it is already in 
> > list. If not it is added to list. 
> > 
> > So if the current node i generate links for is x and i try to generate a 
> > link to  node z, and z's links exhausted i will add it to exhausted list.
> > 
> > And if there is node-x exhausted entries in list, well then script should 
> > break because it will lock into a cycle. The cyclic lockup is due to only a 
> > subset of the networks on the form (links*deep)+1=nodes is possible.
> > 
> > So what i need is to know howto write "if list/array ***empty*** do 
> > {something}"
> > 
> > I sure know howto check if an array have 1 element 2,3,4 but how do you 
> > check for empty.
> > 
> > I think it is crazy that you can not do general comparissons of the type 
> > If in list/array
> > If not in list/array
> > 
> > But it is even more crazy that you can not check if list/array is empty, 
> > that is just madness.
> 
> I mean for for example Javascript
> if (typeof array[index] !== 'undefined' && array[index] !== null) {
> or this one
> if (array[index] != null) {
> 
> I mean how do they come up with such convoluted syntax, do they pull it out 
> of ass? Well one can always say it is needed because undefined not equal to 
> null.
> 
> But would not if (array[index]==empty) suffice and be alot clearer?

Sorry but would not if (array==empty) suffice and be alot clearer?
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Re: General computer language, syntax question.

Den måndag 29 februari 2016 kl. 14:45:37 UTC+1 skrev jonas.t...@gmail.com:
> I have a problem programming uniform networks, "x nodes with y links" that 
> turn out to be really hairy to solve for me and i feel i really lack the 
> programming features 
> 
> "Actually i program in javascript" but the problem seem general for all 
> programming languages including Pyhton.
> 
> For a beautiful solution it would require "If in list/array return boolean" 
> "If not in list/array return boolean". 
> 
> But there is no such feature Python nor Javascript, so instead i set boolean 
> value "inlist" to false and loop thru, to see if it is already in list. If 
> not it is added to list. 
> 
> So if the current node i generate links for is x and i try to generate a link 
> to  node z, and z's links exhausted i will add it to exhausted list.
> 
> And if there is node-x exhausted entries in list, well then script should 
> break because it will lock into a cycle. The cyclic lockup is due to only a 
> subset of the networks on the form (links*deep)+1=nodes is possible.
> 
> So what i need is to know howto write "if list/array ***empty*** do 
> {something}"
> 
> I sure know howto check if an array have 1 element 2,3,4 but how do you check 
> for empty.
> 
> I think it is crazy that you can not do general comparissons of the type 
> If in list/array
> If not in list/array
> 
> But it is even more crazy that you can not check if list/array is empty, that 
> is just madness.

I mean for for example Javascript
if (typeof array[index] !== 'undefined' && array[index] !== null) {
or this one
if (array[index] != null) {

I mean how do they come up with such convoluted syntax, do they pull it out of 
ass? Well one can always say it is needed because undefined not equal to null.

But would not if (array[index]==empty) suffice and be alot clearer?



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General computer language, syntax question.

I have a problem programming uniform networks, "x nodes with y links" that turn 
out to be really hairy to solve for me and i feel i really lack the programming 
features 

"Actually i program in javascript" but the problem seem general for all 
programming languages including Pyhton.

For a beautiful solution it would require "If in list/array return boolean" "If 
not in list/array return boolean". 

But there is no such feature Python nor Javascript, so instead i set boolean 
value "inlist" to false and loop thru, to see if it is already in list. If not 
it is added to list. 

So if the current node i generate links for is x and i try to generate a link 
to  node z, and z's links exhausted i will add it to exhausted list.

And if there is node-x exhausted entries in list, well then script should break 
because it will lock into a cycle. The cyclic lockup is due to only a subset of 
the networks on the form (links*deep)+1=nodes is possible.

So what i need is to know howto write "if list/array ***empty*** do {something}"

I sure know howto check if an array have 1 element 2,3,4 but how do you check 
for empty.

I think it is crazy that you can not do general comparissons of the type 
If in list/array
If not in list/array

But it is even more crazy that you can not check if list/array is empty, that 
is just madness.
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Primality sieve challenge

2016-01-17 Thread jonas . thornvall

Maybe we could have a challenge finding the base with best reducion factor upto 
base 100 00? One probably do not want to store more vectors than that in 
memory, and i guess

Reducing composites in the integer field
http://jt.node365.se/composite.html


I've made my own kind of primality sieve that work by reducing lthe number 
field into composite "legs" and prime "egs". The legs that contain just 
composites in the numberfield will be thrown away sieved?

I asked the people at sci.math if there was a known upper limit for the 
reduction using this type of counters in the integer field, they said no.

I wonder if there is a name for this type of sieve/counter within information 
theory?

My idea is to find a base with very high percentage of composite legs, remove 
all start numbers that produce only composites in their legs, and then store 
the other egs that contain prime in an array.

That array will than serve the purpose modelling the new integer field, using 
the searched base with highest composite reduction.


The script seaching the bases take a few seconds, so there could be alot of 
improvements. I have a feeling that using bignumb bases as sieves will be out 
of range for us mere mortals, but who knows maybe for computer theorists.
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Re: Primality sieve challenge

2016-01-17 Thread jonas . thornvall

Den söndag 17 januari 2016 kl. 10:44:47 UTC+1 skrev jonas.t...@gmail.com:
> Maybe we could have a challenge finding the base with best reducion factor 
> upto base 100 00? One probably do not want to store more vectors than 
> that in memory, and i guess
> 
> Reducing composites in the integer field
> http://jt.node365.se/composite.html
> 
> 
> I've made my own kind of primality sieve that work by reducing lthe number 
> field into composite "legs" and prime "egs". The legs that contain just 
> composites in the numberfield will be thrown away sieved?
> 
> I asked the people at sci.math if there was a known upper limit for the 
> reduction using this type of counters in the integer field, they said no.
> 
> I wonder if there is a name for this type of sieve/counter within information 
> theory?
> 
> My idea is to find a base with very high percentage of composite legs, remove 
> all start numbers that produce only composites in their legs, and then store 
> the other egs that contain prime in an array.
> 
> That array will than serve the purpose modelling the new integer field, using 
> the searched base with highest composite reduction.
> 
> 
> The script seaching the bases take a few seconds, so there could be alot of 
> improvements. I have a feeling that using bignumb bases as sieves will be out 
> of range for us mere mortals, but who knows maybe for computer theorists.

At least i am impressed how well it reduce the composite integer field but of 
course you need a big array...

The math guys said just construct the base using the primes 2*3*5*7*11*13*17... 
and so on.

NEWBASE = 510510 --
BASE= 510510 Composite legs= 92.91551585669234% ===
One could say it is a bit clumsy but i think it is neat be able to peel of 
numbers that do not even need to be considered for primality test.
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Keen eyes

2016-01-16 Thread jonas . thornvall

This is not python just a short snippet of javascript that refuse tracing, i've 
staired blind upon it but since it does something weird with allocating memory 
i have no idea what is going on and the parrots and monkeys at 
comp.lang.javascript refuse to give a hint.

Something in those loops really wrong but it is no giant numbers.



function factor_it(i){
prime=true;
sqroot=Math.floor(Math.sqrt(i));
for (j=2;j

Re: Keen eyes

2016-01-16 Thread jonas . thornvall

Den lördag 16 januari 2016 kl. 23:30:48 UTC+1 skrev Chris Angelico:
> On Sun, Jan 17, 2016 at 9:23 AM,   wrote:
> > function factor_it(i){
> > prime=true;
> > sqroot=Math.floor(Math.sqrt(i));
> > for (j=2;j > {return prime}}
> > return prime;
> > }
> 
> A couple of potential problems here. The first thing that comes to
> mind is that floating point inaccuracy is going to bite you long
> before the numbers "seem huge" to someone who's thinking about 2**53.
> The second is an off-by-one error: a perfect square may come up as
> prime.
> 
> Check for those and see how it looks.
> 
> Also, check your double-use of the 'prime' variable, which also
> appears to be global here.
> 
> ChrisA

Thank you Chris, this is not really for factoring it is meant to create a 
composite sieve. Well all the legs just holding composites will be false.

Regarding the problem no number bigger than 100 is sent to the function, there 
is something weird with either the loop structure or the break.

What should happen...
j=1
i=j

1+10 break (because prime in leg)
j++; well j is 2 and so is i

2+10,2+20,2+30,2+40.2+50,2+60,2+70,2+80,2+90 ((condidion break loop i>100
j++ ->i=3
3+10 break (because prime in leg)
j++

And so on...
And as you can see the outer loop does this until j reaches ***base which is 
10***.
In all it is about 50 operations
So how can it allocate all memory?
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Re: Keen eyes

2016-01-16 Thread jonas . thornvall

Den lördag 16 januari 2016 kl. 23:30:48 UTC+1 skrev Chris Angelico:
> On Sun, Jan 17, 2016 at 9:23 AM,   wrote:
> > function factor_it(i){
> > prime=true;
> > sqroot=Math.floor(Math.sqrt(i));
> > for (j=2;j > {return prime}}
> > return prime;
> > }
> 
> A couple of potential problems here. The first thing that comes to
> mind is that floating point inaccuracy is going to bite you long
> before the numbers "seem huge" to someone who's thinking about 2**53.
> The second is an off-by-one error: a perfect square may come up as
> prime.
> 
> Check for those and see how it looks.
> 
> Also, check your double-use of the 'prime' variable, which also
> appears to be global here.
> 
> ChrisA

Thank you Chris your comment resolved it double use of j in two different 
functions and loops.
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 02:09:17 UTC+2 skrev Ben Bacarisse:
 Ian Kelly ian.g.ke...@gmail.com writes:
 
  On Mon, Jun 29, 2015 at 4:56 PM, Ian Kelly ian.g.ke...@gmail.com wrote:
  On Mon, Jun 29, 2015 at 4:39 PM,  jonas.thornv...@gmail.com wrote:
  http://jt.node365.se/baseconversion8.html
 snip
  By the way, I think you have a bug. I did my time testing by
  concatenating the default input number to itself several times. At
  5184 decimal digits, which was the last case I tried, the 58th output
  digit was 111, after which point the remaining 672 output digits
  are all 12665464, which seemed rather improbable.
 
 Yes, it's a bug.  Because it's off-topic, I won't say more here but I
 posted a simpler test case to comp.lang.javacript just now (1000
 converted to base 100).  I would not normally reply here, but I wanted
 to acknowledge your post as prompting me to look for other cases.
 Followup-to: set.
 
 -- 
 Ben.

Thank you, for finding the bug Ben, it turned out i did basicly same thing two 
times. One of the times was the more general case and that turned out to do the 
correct thing.

jt.node365.se/baseconversion9.html

It still bug out on very big numbers if base outside integer scope.
I am very keen on suggestions regarding the logic to make it faster.
I will read in base into an array holding bignumb, and then it should work for 
anybase and any number as far i can understand. 
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 11:43:55 UTC+2 skrev jonas.t...@gmail.com:
 Den tisdag 30 juni 2015 kl. 11:35:06 UTC+2 skrev jonas.t...@gmail.com:
  Den tisdag 30 juni 2015 kl. 11:08:01 UTC+2 skrev Christian Gollwitzer:
   Am 30.06.15 um 10:52 schrieb jonas.thornv...@gmail.com:
It still bug out on very big numbers if base outside integer scope.
I am very keen on suggestions regarding the logic to make it faster.
   
   Concerning the algorithmic complexity, it can't be faster than square 
   time in the number of digits N. Baseconversion needs to do a sequence of 
   division operations, where every operation gves you one digit in the new 
   base. The number of digits in the new base is proportional to the number 
   of digits in the old base (the ratio is log b1/log b2). Therefore it 
   will be O(N^2).
   
 Christian
  
  Any new digit will be found in SQRT(base) comparissons. 
  Averaged case will be in (SQRT(base)*(SQRT(base)+1))/2
 
 Well that averaging was balloney. How do i write that the digit will be found 
 in.
 Average values from 1 to SQRT(base)?

Regarding the time it seem to double the digits quadruple the time. And that is 
still linear or?
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 11:08:01 UTC+2 skrev Christian Gollwitzer:
 Am 30.06.15 um 10:52 schrieb jonas.thornv...@gmail.com:
  It still bug out on very big numbers if base outside integer scope.
  I am very keen on suggestions regarding the logic to make it faster.
 
 Concerning the algorithmic complexity, it can't be faster than square 
 time in the number of digits N. Baseconversion needs to do a sequence of 
 division operations, where every operation gves you one digit in the new 
 base. The number of digits in the new base is proportional to the number 
 of digits in the old base (the ratio is log b1/log b2). Therefore it 
 will be O(N^2).
 
   Christian

Any new digit will be found in SQRT(base) comparissons. 
Averaged case will be in (SQRT(base)*(SQRT(base)+1))/2
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 15:22:44 UTC+2 skrev jonas.t...@gmail.com:
 Den tisdag 30 juni 2015 kl. 11:43:55 UTC+2 skrev jonas.t...@gmail.com:
  Den tisdag 30 juni 2015 kl. 11:35:06 UTC+2 skrev jonas.t...@gmail.com:
   Den tisdag 30 juni 2015 kl. 11:08:01 UTC+2 skrev Christian Gollwitzer:
Am 30.06.15 um 10:52 schrieb jonas.thornv...@gmail.com:
 It still bug out on very big numbers if base outside integer scope.
 I am very keen on suggestions regarding the logic to make it faster.

Concerning the algorithmic complexity, it can't be faster than square 
time in the number of digits N. Baseconversion needs to do a sequence 
of 
division operations, where every operation gves you one digit in the 
new 
base. The number of digits in the new base is proportional to the 
number 
of digits in the old base (the ratio is log b1/log b2). Therefore it 
will be O(N^2).

Christian
   
   Any new digit will be found in SQRT(base) comparissons. 
   Averaged case will be in (SQRT(base)*(SQRT(base)+1))/2
  
  Well that averaging was balloney. How do i write that the digit will be 
  found in.
  Average values from 1 to SQRT(base)?
 
 Regarding the time it seem to double the digits quadruple the time. And that 
 is still linear or?

2x seem linear to me?
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 18:12:46 UTC+2 skrev Michael Torrie:
 Do you have some Python code to show us?

No i just thought you would find the digit search algorithm interesting.
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 01:11:51 UTC+2 skrev Ian:
 On Mon, Jun 29, 2015 at 4:56 PM, Ian Kelly ian.g.ke...@gmail.com wrote:
  On Mon, Jun 29, 2015 at 4:39 PM,  jonas.thornv...@gmail.com wrote:
  http://jt.node365.se/baseconversion8.html
 
  Back of the envelope mental calculation, that appears to be quadratic,
  not linear. Doubling the length of the input results in an approximate
  quadrupling of the time taken to produce the output.
 
  That noted, this is off-topic for this list, which is for discussion
  about Python. Please take this to somewhere else like
  comp.lang.javascript instead.
 
 By the way, I think you have a bug. I did my time testing by
 concatenating the default input number to itself several times. At
 5184 decimal digits, which was the last case I tried, the 58th output
 digit was 111, after which point the remaining 672 output digits
 are all 12665464, which seemed rather improbable.

It isn't i used the power to check with wolfram, but there is a plus 1 bug as 
ben noticed.

But not in this slower version.
http://jt.node365.se/baseconversion5.html
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 11:35:06 UTC+2 skrev jonas.t...@gmail.com:
 Den tisdag 30 juni 2015 kl. 11:08:01 UTC+2 skrev Christian Gollwitzer:
  Am 30.06.15 um 10:52 schrieb jonas.thornv...@gmail.com:
   It still bug out on very big numbers if base outside integer scope.
   I am very keen on suggestions regarding the logic to make it faster.
  
  Concerning the algorithmic complexity, it can't be faster than square 
  time in the number of digits N. Baseconversion needs to do a sequence of 
  division operations, where every operation gves you one digit in the new 
  base. The number of digits in the new base is proportional to the number 
  of digits in the old base (the ratio is log b1/log b2). Therefore it 
  will be O(N^2).
  
  Christian
 
 Any new digit will be found in SQRT(base) comparissons. 
 Averaged case will be in (SQRT(base)*(SQRT(base)+1))/2

Well that averaging was balloney. How do i write that the digit will be found 
in.
Average values from 1 to SQRT(base)?
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Re: Linear time baseconversion

Den tisdag 30 juni 2015 kl. 11:08:01 UTC+2 skrev Christian Gollwitzer:
 Am 30.06.15 um 10:52 schrieb jonas.thornv...@gmail.com:
  It still bug out on very big numbers if base outside integer scope.
  I am very keen on suggestions regarding the logic to make it faster.
 
 Concerning the algorithmic complexity, it can't be faster than square 
 time in the number of digits N. Baseconversion needs to do a sequence of 
 division operations, where every operation gves you one digit in the new 
 base. The number of digits in the new base is proportional to the number 
 of digits in the old base (the ratio is log b1/log b2). Therefore it 
 will be O(N^2).
 
   Christian

There is not a single division except for the 2 one, used in halfinterval 
search 
This algorithm only use multiplication and a modified GCD search algorithm.

http://jt.node365.se/baseconversion9.html
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Re: Linear time baseconversion

2015-06-29 Thread jonas . thornvall

Den tisdag 30 juni 2015 kl. 01:11:51 UTC+2 skrev Ian:
 On Mon, Jun 29, 2015 at 4:56 PM, Ian Kelly ian.g.ke...@gmail.com wrote:
  On Mon, Jun 29, 2015 at 4:39 PM,  jonas.thornv...@gmail.com wrote:
  http://jt.node365.se/baseconversion8.html
 
  Back of the envelope mental calculation, that appears to be quadratic,
  not linear. Doubling the length of the input results in an approximate
  quadrupling of the time taken to produce the output.
 
  That noted, this is off-topic for this list, which is for discussion
  about Python. Please take this to somewhere else like
  comp.lang.javascript instead.
 
 By the way, I think you have a bug. I did my time testing by
 concatenating the default input number to itself several times. At
 5184 decimal digits, which was the last case I tried, the 58th output
 digit was 111, after which point the remaining 672 output digits
 are all 12665464, which seemed rather improbable.

No there is no bug, try 12665465^77 wolfram alpha gives
7976781785088446084873306945009387424538813088491673402748035781202063831184317457635773873975198429536727051788368983080016275695338973787545408588270715883570324803120051806974411884509607864124408509087974736382639726555779113221450078243163340395784592348080359062163885323462144257596380766496966084237358222613649636595300081207683175410584139478853066960795599027150403706812116411354218686846750615496852415389467898730884554700066802269059177874281189887271741594702334892800692129299635716260471508809448693000376806594431400299072265625

Try that number you get a ONE followed by 77 ZEROS
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Linear time baseconversion

2015-06-29 Thread jonas . thornvall

http://jt.node365.se/baseconversion8.html 
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Python functionality in Javascript

2015-06-21 Thread jonas . thornvall

Check this out using a 8 digit base with a 100 digit number no problem.
http://jt.node365.se/baseconversion3.html
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Re: find all multiplicands and multipliers for a number

Den lördag 11 april 2015 kl. 09:35:22 UTC+2 skrev Steven D'Aprano:
 On Sat, 11 Apr 2015 04:08 pm, Paul Rubin wrote:
 
  Steven D'Aprano steve+comp.lang.pyt...@pearwood.info writes:
  It may be a bit slow for very large numbers. On my computer, this takes
  20 seconds:
  py pyprimes.factors.factorise(2**111+1)
  [3, 3, 1777, 3331, 17539, 25781083, 107775231312019L]
  
  This takes about 4 seconds on a Intel(R) Core(TM) i5-3230M CPU @ 2.60GHz
  laptop (64 bit linux):
 
 Nice for some who have fast machines, but on my old jalopy, your code takes
 110 seconds, more than five times slower than mine. It also returns the
 wrong result:
 
 [3, 3, 3, 3, 7, 19, 73, 87211, 262657, 1.4411518807585587e+17]
 
 Oops, you have a float in there, how did that happen?
 
 We can tell your code gives the wrong result. It claims that 7 is a factor
 of 2**111+1:
 
 py n = 2**111 + 1
 py itertools.islice(fac(n), 0, 5)
 itertools.islice object at 0xb7c8f914
 py list(itertools.islice(fac(n), 0, 5))
 [3, 3, 3, 3, 7]
 
 
 but that can't be right:
 
 py n % 7
 2L
 
 Seven in not a factor of 2**111 + 1.
 
 The reason your code gives wrong results is that you perform true division /
 rather than integer division // which means that you with n a float, which
 loses precision:
 
 py (n/9) % 3  # nine is a factor
 0.0
 py (n//9) % 3  # exact, without rounding
 1L
 
 
 
 -- 
 Steven

Love it.
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Re: Generarl programming question.

Den lördag 11 april 2015 kl. 17:26:03 UTC+2 skrev Steven D'Aprano:
 On Sun, 12 Apr 2015 01:00 am, jonas.thornv...@gmail.com wrote:
 
  If two functions crossreference eachother back and forth what happen with
  the local variables.
 
 Nothing. They are local to the function that creates them.
 
 
  Will there be a new instance of function holding the variables or do they
  get messed up?
 
 No to both of those. You have two functions, each with it's own locals.
 
 
 def spam():
 colour = red
 print(Inside spam: colour is:, colour)
 eggs()
 print(Inside spam after calling eggs: colour is:, colour)
 eggs()
 
 
 def eggs():
 colour = yellow
 print(Inside eggs: colour is:, colour)
 
 
 Calling spam() gives you this output:
 
 py spam()
 Inside spam: colour is: red
 Inside eggs: colour is: yellow
 Inside spam after calling eggs: colour is: red
 Inside eggs: colour is: yellow
 
 
 Even if the functions call each other (mutual recursion) each function's
 local variables remain local.
 
 
 
 -- 
 Steven

I don't think it matter butt eggs also calls spam, once more.
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Re: Generarl programming question.

Den lördag 11 april 2015 kl. 17:16:09 UTC+2 skrev Chris Angelico:
 On Sun, Apr 12, 2015 at 1:00 AM,  jonas.thornv...@gmail.com wrote:
  If two functions crossreference eachother back and forth what happen with 
  the local variables.
 
  Will there be a new instance of function holding the variables or do they 
  get messed up?
 
 You mean if one function calls another, and that function calls the
 first? That's called mutual recursion:
 
 def func1(x):
 if x % 2: return x + 1
 return func2(x - 2)
 
 def func2(x):
 if x % 3: return x + 2
 return func1(x - 3)
 
 The 'x' inside each function is completely separate, no matter how
 many times they get called. They're usually stored on something called
 a call stack - you put another sheet of paper on top of the stack
 every time you call a function, local variables are all written on
 that paper, and when you return from a function, you discard the top
 sheet and see what's underneath.
 
 For more information, search the web for the key terms in the above
 description, particularly the ones I put in quotes.
 
 If this isn't what you're talking about, the best way to clarify your
 question is probably to post a simple (even stupidly trivial, like the
 one above) example, and ask a question about that code. Someone'll
 doubtless help out!
 
 ChrisA

Thanks i was worried, i try to make a generic base choice algorithm that should 
work for anybase, and i just realised that the bignumb add would need to call 
the bignumb subtraction and viceversa. I thought there may be instances but i 
was not sure. 

But I have a feeling the code will be hard to debug.
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Generarl programming question.

If two functions crossreference eachother back and forth what happen with the 
local variables.

Will there be a new instance of function holding the variables or do they get 
messed up?
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Re: Best search algorithm to find condition within a range

2015-04-09 Thread jonas . thornvall

Den torsdag 9 april 2015 kl. 17:02:24 UTC+2 skrev Chris Angelico:
 On Fri, Apr 10, 2015 at 12:53 AM, Alain Ketterlin
 al...@dpt-info.u-strasbg.fr wrote:
  Ouch, you're right, I tried to stick with Marko's example and forgot the
  basics. I meant signed ints, but the removable condition should be
  something like:
 
  if ( x0  y0  zx )
  ...
 
  i.e., some condition that is never true (either false or undefined).
  Thanks for the correction.
 
 Ah, yep. And yes, that's absolutely right - the compiler's allowed to
 treat that as never true.
 
 ChrisA

I guess when working on highlevel arithmetic one does not need to think about 
such trivial? things, the expression is broken down into pairs fed into the 
functions and the parser handle the signs?
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Re: Best search algorithm to find condition within a range

2015-04-09 Thread jonas . thornvall

Den torsdag 9 april 2015 kl. 17:04:53 UTC+2 skrev Ian:
 On Thu, Apr 9, 2015 at 8:54 AM,  jonas.thornv...@gmail.com wrote:
  Aside from the base changer i've written a bignumb anybase generic 
  multiplication, addition and subtraction routine. My goal is to make the 
  arithmetic in the base of choice whatever size.
 
  And i think i can do it, i already have mult, add , subt working in anybase 
  but not bignumb.
 
  Does this mult work for arbitrary size?
  Or will it but out at some point?
 
 Dunno, I have no interest in verifying your code for you. Especially
 since this is a Python list and said code is not Python.
 
 If you want to build confidence that your code works, I suggest
 writing some unit tests for it. (I'd also suggest keeping it separate
 from your html in a .js file, which would make it a lot easier to test
 and reuse.)

Well with the condition i can get the bignumb arithmetic fully working, i am on 
my way of working bignumb arithmetic in anybase. 

I think this will be the first implementation ever to use arithmetic in the 
base of your choice. 

And i know for a fact (i've factored RSA in the 200 digit range on a 486 in a 
couple of hours 98-) 

So there is more to this than fancy overlay, it is about the digit 
representation at the binary level, and how the binary groups interpretated.

So in the end, even with none optimised type like Javascript integer it will 
save operations at high level as well as lowlevel.

That said highlevel and highbase arithmetic will work alot faster than binary.
And using modular arithmetic you may find out that some operations requiering 
exponential work like factoring will turn out to be done in polynomial time.

SCRIPT LANGUAGE=Javascript
/* MULTIPLY TWO VARIABLES */

//CONVERT A DECIMAL NUMBER INTO ANYBASE
function newbase(decNumber,base){
var out =[];
digits=1;
digmult=1;
while(digmult*base=decNumber){
digmult=digmult*base
digits++;
}
digsave=digmult;
while(decNumber0 || digits0){
loop=1;
digit=0;
   for (digit=0;digit=base;digit++) {
if((digit+1)*digmultdecNumber)break;
   }
out[digits]=digit;
digmult=digmult*digit;
decNumber=decNumber-digmult;
digsave=digsave/base;
digmult=digsave;
digits--;
}

return out;

}

function naiveMult(base, multOne, multTwo)
{  
   var total = [];
   var addResult = [];
   total[0] = 0;

   //Check which array bigger minimize multiplications
   if (multOne.lengthmultTwo.length){ mshort=multOne.length; 
mlong=multTwo.length;}
   else { mshort=multOne.length; mlong=multTwo.length;}

   for (var i = 0; i  mshort; i ++ )
   {
  multresult = [];
  remainder = 0;
  tempVal = 0;
  tempDig = 0;
  j = 0;

  if(i  0)
  {
 for(zero = 0; zero  i; zero ++ ) multresult[zero] = 0;
  }

  while(j  mlong)
  {
 intMult++;
   
 tempVal = (multOne[i] * multTwo[j]) + remainder;
 remainder = 0;

 if (tempVal = base)
 {
tempDig = tempVal % base;
remainder = (tempVal - tempDig) / base;
multresult[j + i] = tempDig;
 }
 else
 {
multresult[j + i] = tempVal;
 }
 j ++ ;
  }

  if(remainder != 0)
  {
 multresult[j + i] = remainder;
  }
  addResult=naiveAdd(base, total, multresult);
  total=addResult;
   }
   return total;
}

/* ADD TWO VARIABLES */

function naiveAdd(base, arrOne, arrTwo)
{
   
   shortArr=[];
   longArr=[];
   addResult = [];
   remainder = 0;
   //Find shortest and longest array
   if (arrOne.length = arrTwo.length) {shortArr = arrTwo; longArr = arrOne;}
else  {shortArr = arrOne;longArr = arrTwo;}
   for (i = 0; i  shortArr.length; i ++ )
   {
  intAdd ++ ;
  arrOne[i] = parseInt(arrOne[i]);
  arrTwo[i] = parseInt(arrTwo[i]);
  if (isNaN(arrOne[i])) arrOne[i] = 0;
  if (isNaN(arrTwo[i])) arrTwo[i] = 0;
  addResult[i] = arrOne[i] + arrTwo[i] + remainder;
 
  if (addResult[i] = base)
  {
 addResult[i] = addResult[i] - base;
 remainder = 1;
  }
  else
  {
 remainder = 0;
  }
   }
   //Copying values from the shorter string
   while(ilongArr.length) 
{longArr[i]=parseInt(longArr[i]);addResult[i]=longArr[i]+remainder; 
remainder=0; i++;}
   //If strings of equal lenght but there is a remainder;
   if (remainder == 1) addResult[i++] = 1;
   else addResult.splice(i, 1);
   return addResult;
}

/* MAIN */
function fetchValues()
{
intMult=0;
intAdd=0;
base=10;
x=1;
y=1;
document.calc.result.value=;
document.calc.stats.value=;
document.calc.outbase.value=;
strEval = document.calc.eval.value;
genArr = strEval.split(*);
//fONE=newbase(genArr[0],base);
//document.write(fONE[2]);
//document.calc.outbase.value +=fONE+ + \n;
//fTWO=newbase(genArr[1],base);
//document.calc.outbase.value +=fTWO+\n;

Re: Best search algorithm to find condition within a range

2015-04-09 Thread jonas . thornvall

Den torsdag 9 april 2015 kl. 16:08:48 UTC+2 skrev Chris Angelico:
 On Thu, Apr 9, 2015 at 11:57 PM, Alain Ketterlin
 al...@dpt-info.u-strasbg.fr wrote:
  Because, in:
 
  z = x+y; // all signed ints
  if ( z  x )
  ...
 
  either there was no overflow (and the condition is false), or there was,
  and the result is undefined, i.e., zx can be considered false also.
 
 Do you mean all unsigned ints here? Because if y is negative, the
 condition will be true without overflow. If you didn't, then I'm
 puzzled as to where the undefined behaviour is coming from.
 
 ChrisA

Aside from the base changer i've written a bignumb anybase generic 
multiplication, addition and subtraction routine. My goal is to make the 
arithmetic in the base of choice whatever size.

And i think i can do it, i already have mult, add , subt working in anybase but 
not bignumb.

Does this mult work for arbitrary size?
Or will it but out at some point?


SCRIPT LANGUAGE=Javascript
/* MULTIPLY TWO VARIABLES */

//CONVERT A DECIMAL NUMBER INTO ANYBASE
function newbase(decNumber,base){
var out =[];
digits=1;
digmult=1;
while(digmult*base=decNumber){
digmult=digmult*base
digits++;
}
digsave=digmult;
while(decNumber0 || digits0){
loop=1;
digit=0;
for (digit=0;digit=base;digit++) {
if((digit+1)*digmultdecNumber)break;
}
out[digits]=digit;
digmult=digmult*digit;
decNumber=decNumber-digmult;
digsave=digsave/base;
digmult=digsave;
digits--;
}

return out;

}

function naiveMult(base, multOne, multTwo)
{  
   var total = [];
   var addResult = [];
   total[0] = 0;

   //Check which array bigger minimize multiplications
   if (multOne.lengthmultTwo.length){ mshort=multOne.length; 
mlong=multTwo.length;}
   else { mshort=multOne.length; mlong=multTwo.length;}

   for (var i = 0; i  mshort; i ++ )
   {
  multresult = [];
  remainder = 0;
  tempVal = 0;
  tempDig = 0;
  j = 0;

  if(i  0)
  {
 for(zero = 0; zero  i; zero ++ ) multresult[zero] = 0;
  }

  while(j  mlong)
  {
 intMult++;
   
 tempVal = (multOne[i] * multTwo[j]) + remainder;
 remainder = 0;

 if (tempVal = base)
 {
tempDig = tempVal % base;
remainder = (tempVal - tempDig) / base;
multresult[j + i] = tempDig;
 }
 else
 {
multresult[j + i] = tempVal;
 }
 j ++ ;
  }

  if(remainder != 0)
  {
 multresult[j + i] = remainder;
  }
  addResult=naiveAdd(base, total, multresult);
  total=addResult; 
   }
   return total;
}

/* ADD TWO VARIABLES */

function naiveAdd(base, arrOne, arrTwo)
{
   
   shortArr=[];
   longArr=[];
   addResult = [];
   remainder = 0;
   //Find shortest and longest array
   if (arrOne.length = arrTwo.length) {shortArr = arrTwo; longArr = arrOne;}
else  {shortArr = arrOne;longArr = arrTwo;}
   for (i = 0; i  shortArr.length; i ++ )
   {
  intAdd ++ ;
  arrOne[i] = parseInt(arrOne[i]);
  arrTwo[i] = parseInt(arrTwo[i]);
  if (isNaN(arrOne[i])) arrOne[i] = 0;
  if (isNaN(arrTwo[i])) arrTwo[i] = 0;
  addResult[i] = arrOne[i] + arrTwo[i] + remainder;
 
  if (addResult[i] = base)
  {
 addResult[i] = addResult[i] - base;
 remainder = 1;
  }
  else
  {
 remainder = 0;
  }
   }
   //Copying values from the shorter string
   while(ilongArr.length) 
{longArr[i]=parseInt(longArr[i]);addResult[i]=longArr[i]+remainder; 
remainder=0; i++;}
   //If strings of equal lenght but there is a remainder;
   if (remainder == 1) addResult[i++] = 1;
   else addResult.splice(i, 1);
   return addResult;
}

/* MAIN */
function fetchValues()
{
intMult=0;
intAdd=0;
base=10;
x=1;
y=1;
document.calc.result.value=;
document.calc.stats.value=;
document.calc.outbase.value=;
strEval = document.calc.eval.value;
genArr = strEval.split(*);
//fONE=newbase(genArr[0],base);
//document.write(fONE[2]);
//document.calc.outbase.value +=fONE+ + \n;
//fTWO=newbase(genArr[1],base);
//document.calc.outbase.value +=fTWO+\n;

arrOne=genArr[0].split().reverse();
arrTwo=genArr[1].split().reverse();
base = document.calc.formbase.value;
out=naiveMult(base,arrOne,arrTwo);
document.calc.stats.value +=Multiplications = +intMult+\n;
document.calc.result.value += Product =  + out.reverse()+\n;
}
/SCRIPT
!DOCTYPE html
html lang=en
headmeta charset=utf-8//head
body bgcolor=gold onload=fetchValues()
H1Big numbers/H1P
FORM name=calc onSubmit=fetchValues(); return false;
BINPUT DEC-input type=submit name=calc value=Calc input type=text 
name=eval value=32432439*12 size=300br
BASE input type=text name=formbase value=10 size=10br
This calculationinput type=text name=outbase value= size=60br
FONT COLOR=white
RESULTbr
textarea name=result rows=20 cols=120/textareabr
STATUS/Bbr
textarea name=stats cols=120 rows=40

Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 03:00:12 UTC+2 skrev Gregory Ewing:
 Steven D'Aprano wrote:
 
  What digits would you use for base one-million?
 
 Interesting question. Unicode currently has about
 75,000 CJK characters, so we would need to find about
 12 more independently developed cultures with similar
 writing systems to get enough characters. I suggest
 revisiting the issue when interstellar travel and
 exploration have been better developed.
 
 -- 
 Greg

No need Greg it is an easy task creating visual interpretable symbols from low 
to high using lines in a point/pixel space. I could create the set in a day.

And if i use colour space it is even easier.

The hard part is making the point space visible and interpretabel without 
calculation and holding remainders in head. 

But then again when numbers have more than 5 digitplaces people tend to 
interpretate them digitwise anyway.

10010 is interpretated as a single digit by brain.
7582560 people mostly interpretate as 7 m 582 000 and 560.

You can tell how people group and think about numbers by having them reading up 
a very long telephone number.
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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 02:38:57 UTC+2 skrev Steven D'Aprano:
 On Wed, 8 Apr 2015 03:44 am, Ian Kelly wrote:
 
 
 to_base(2932903594368438384328325832983294832483258958495845849584958458435439543858588435856958650865490,
  429496729)
  [27626525, 286159541, 134919277, 305018215, 329341598, 48181777,
  79384857, 112868646, 221068759, 70871527, 416507001, 31]
 
 
 They're not exactly *digits* though, are they? Without an easy to use set of
 429496729 different symbols, the whole exercise is rather pointless. It's
 not more compact: 97 decimal digits, versus 121 characters in the list
 representation. 110 if you strip out the spaces between items. It's
 certainly not more memory efficient: the long int 293...490 takes 56 bytes,
 compared to 80 bytes for just the list, not including the memory used by
 its 12 int items. (Results may vary in other versions of Python.) You can't
 do arithmetic on it faster than Python's built-ins.
 
 Besides, it isn't clear to me whether Jonas wants to convert decimal
 293...490 *into* base 429496729 as you have done, or *base 429496729*
 293...490 into decimal.
 
 
 
 -- 
 Steven

Well Steven you just draw a line under connecting them all, and now you are 
allowed to call whatever combination you write down a single digit.

And you have just created 429496729 unique symbols ;), in a pencil stroke.
I know at least one intergalactic poster that will be impressed ;)
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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 09:16:24 UTC+2 skrev Ian:
 On Tue, Apr 7, 2015 at 4:35 PM,  jonas.thornv...@gmail.com wrote:
  I am not sure you guys realised, that althoug the size of the factors to 
  muliply expands according to base^(exp+1) for each digitplace the number of 
  comparissons needed to reach the digit place (multiple of base^exp+1) is 
  constant with my approach/method.
 
 No it isn't. You do one comparison on every iteration of your while
 loop, and you do one iteration for every digit. How is that constant?

Ok the number of comparissons is linear for each digitplace not exponential.
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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 09:16:24 UTC+2 skrev Ian:
 On Tue, Apr 7, 2015 at 4:35 PM,  jonas.thornv...@gmail.com wrote:
  I am not sure you guys realised, that althoug the size of the factors to 
  muliply expands according to base^(exp+1) for each digitplace the number of 
  comparissons needed to reach the digit place (multiple of base^exp+1) is 
  constant with my approach/method.
 
 No it isn't. You do one comparison on every iteration of your while
 loop, and you do one iteration for every digit. How is that constant?

The for loop should be replaced with a version of binary search making 
comparisson instead of adding to digit. It will be alot faster with a base 
using 32-bits.

But the point is the number of operations is linear for each digitplace with my 
approach/method, you will multiply bigger numbers. But the numbers of 
comparissons and multiplications is linear over the digitspace.
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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 15:40:46 UTC+2 skrev Mel Wilson:
 On Tue, 07 Apr 2015 23:19:49 -0700, jonas.thornvall wrote:
 
  And you have just created 429496729 unique symbols ;), in a pencil
  stroke.
 
 No.  You did that, when you said base 429496729.  Representing the 
 symbols in a computer is no problem, any Python long int can do that.  To 
 display the symbols, you can use PIL to make up glyphs out of coloured 
 pixels 864x864.  You can keep your glyphs in GIFs.  Where you keep them 
 is up to you.  Keeping them in Tumbolia and computing them out as 
 required will work well.

Well many interpretate a numeral digit as a single unique symbol representing a 
number written out in one pensttroke. I just pointed out that using handwriting 
or underscore a compination can be considered a numerical digit in itself.

There is no need for inventing a new set of characters representing 32-bit 
numbers. You will not be able to learn them by heart anyway, unless they build 
on a interpretation system binaries, decimals. 

Of course you could rather easily create a new set building on Points and lines 
and their interpretation.

Well just saying the Babylonians and Sumerians already did this, even the old 
greek had some sort of system with 120 unique digits. But what would be the 
need for such system when we so easy can create a unique permutation 
representing any number, using our 10 digits.

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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 19:34:39 UTC+2 skrev Mel Wilson:
 On Wed, 08 Apr 2015 07:56:05 -0700, jonas.thornvall wrote:
 
  There is no need for inventing a new set of characters representing
  32-bit numbers. You will not be able to learn them by heart anyway,
  unless they build on a interpretation system binaries, decimals.
 
 See Jorge Luis Borges, _Funes the Memorious_.  Gotta keep up with the 
 literature.

One thing is true though the bigger the chunks the less operations doing add.
So arithmetic may turn out to be obsolete and replaced with search in 
lookuptables.

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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 21:28:34 UTC+2 skrev jonas.t...@gmail.com:
 Den onsdag 8 april 2015 kl. 19:34:39 UTC+2 skrev Mel Wilson:
  On Wed, 08 Apr 2015 07:56:05 -0700, jonas.thornvall wrote:
  
   There is no need for inventing a new set of characters representing
   32-bit numbers. You will not be able to learn them by heart anyway,
   unless they build on a interpretation system binaries, decimals.
  
  See Jorge Luis Borges, _Funes the Memorious_.  Gotta keep up with the 
  literature.
 
 One thing is true though the bigger the chunks the less operations doing add.
 So arithmetic may turn out to be obsolete and replaced with search in 
 lookuptables.

When doing by hand and working within the digitspace of a single decimal digit 
like 3+4 the operation is implicit. You do not really add anything you use a 
lookup table.

But if you take an expression like 193+169, most people perform the arithmetic. 
Unless they do not happen to be your favourit numbers and you learnt the sum by 
heart.

Top down or bottom up is basicly the choices.
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Best search algorithm to find condition within a range



I want todo faster baseconversion for very big bases like base 1 000 000, so 
instead of adding up digits i search it. 

I need the fastest algorithm to find the relation to a decimal number. 
Digmult is an instance of base at a digitplace (base^x) what i try to find is 
the digit for the below condition is true and the loop break. 


* 
for (digit=0;digit=base;digit++) { 
   if((digit+1)*digmultdecNumber)break; 
} 
* 

So i am looking for the digit where following condition true. 

if((digit)*digmultdecNumber) AND if((digit+1)*digmultdecNumber) then BREAK; 

One could start at half base searching, but then i Think i've read that using 
1/3 closing in faster? 

I Think also i remember that if the search space so big that at least 22 or 23 
guesses, needed.A random Oracle may even faster? 

Just pick up a number and get lucky, is it any truth to that? 

Below the actual algorithm. 



SCRIPT LANGUAGE=Javascript 
//CONVERT A DECIMAL NUMBER INTO ANYBASE 
function newbase(decNumber,base){ 
digits=1; 
digmult=1; 
while(digmult*base=decNumber){ 
digmult=digmult*base 
digits++; 
} 
digsave=digmult; 
while(decNumber0 || digits0){ 
loop=1; 
digit=0; 
   for (digit=0;digit=base;digit++) { 
if((digit+1)*digmultdecNumber)break; 
   } 
out[digits]=digit; 
digmult=digmult*digit; 
decNumber=decNumber-digmult; 
digsave=digsave/base; 
digmult=digsave; 
digits--; 
} 
return out; 
} 

var out= []; 
base=256; 
number=854544; 
out=newbase(number,base); 
out.reverse(); 
document.write(Number = ,out,BR); 
/script 

 












 
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 15:30:36 UTC+2 skrev Dave Angel:
 On 04/07/2015 05:44 AM, jonas.thornv...@gmail.com wrote:
 
 
  I want todo faster baseconversion for very big bases like base 1 000 000, 
  so instead of adding up digits i search it.
 
 For this and most of the following statements:  I can almost guess what 
 you're trying to say.  However, I cannot.  No idea why you're adding up 
 digits, that sounds like casting out nines.  And in base-N, that would 
 be casting out (N-1)'s.
 
 What's the it you're trying to search?
 
 How do you know the baseconversion is the bottleneck, if you haven't 
 written any Python code yet?
 
 
 
  I need the fastest algorithm to find the relation to a decimal number.
 
 What relation would that be?  Between what and what?
 
  Digmult is an instance of base at a digitplace (base^x) what i try to find 
  is the digit for the below condition is true and the loop break.
 
 
 You haven't defined a class Base yet.  In fact, I don't see any Python 
 code in the whole message.
 
 
  *
  for (digit=0;digit=base;digit++) {
  if((digit+1)*digmultdecNumber)break;
  }
  *
 
 
 
  So i am looking for the digit where following condition true.
 
  if((digit)*digmultdecNumber) AND if((digit+1)*digmultdecNumber) then 
  BREAK;
 
 You could try integer divide.  That's just something like
   digit = decNumber // digmult
 But if you think hard enough you'd realize that
 
 
 
  One could start at half base searching, but then i Think i've read that 
  using 1/3 closing in faster?
 
  I Think also i remember that if the search space so big that at least 22 or 
  23 guesses, needed.A random Oracle may even faster?
 
  Just pick up a number and get lucky, is it any truth to that?
 
  Below the actual algorithm.
 
 
 
  SCRIPT LANGUAGE=Javascript
  //CONVERT A DECIMAL NUMBER INTO ANYBASE
  function newbase(decNumber,base){
  digits=1;
  digmult=1;
  while(digmult*base=decNumber){
   digmult=digmult*base
   digits++;
  }
  digsave=digmult;
  while(decNumber0 || digits0){
   loop=1;
   digit=0;
  for (digit=0;digit=base;digit++) {
   if((digit+1)*digmultdecNumber)break;
  }
   out[digits]=digit;
   digmult=digmult*digit;
   decNumber=decNumber-digmult;
   digsave=digsave/base;
   digmult=digsave;
   digits--;
   }
  return out;
  }
 
  var out= [];
  base=256;
  number=854544;
  out=newbase(number,base);
  out.reverse();
  document.write(Number = ,out,BR);
  /script
 
 
 If that code were in Python, I could be more motivated to critique it. 
 The whole algorithm could be much simpler.  But perhaps there is some 
 limitation of javascript that's crippling the code.
 
 How would you do it if you were converting the base by hand?  I 
 certainly wouldn't be doing any trial and error.  For each pass, I'd 
 calculate quotient and remainder, where remainder is the digit, and 
 quotient is the next value you work on.
 
 
 -- 
 DaveA

I am doing it just like i would do it by hand finding the biggest digit first. 
To do that i need to know nearest base^exp that is less than the actual number. 
Add up the digit (multiply) it to the nearest smaller multiple. Subtract that 
number (base^exp*multiple). 

Divide / Scale down the exponent with base. And record the digit.
And start looking for next digit doing same manipulation until remainder = 0.

And that is what i am doing.
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 16:30:15 UTC+2 skrev Denis McMahon:
 On Tue, 07 Apr 2015 09:29:59 -0400, Dave Angel wrote:
 
  On 04/07/2015 05:44 AM, jonas.thornv...@gmail.com wrote:
 
  I want todo faster baseconversion for very big bases like base 1 000
  000, so instead of adding up digits i search it.
 
  How do you know the baseconversion is the bottleneck, if you haven't
  written any Python code yet?
 
 He doesn't. He doesn't comprehend that as far as a computer is concerned 
 an integer has no specific 'base', it's only when presented in a form for 
 humans to read that it gets base information added in the representation.
 
 He's making these and other similar errors in the javascript groups too.
 
 I suspect he's one of those people that spends his time thinking up 
 elaborate solutions that he has no idea how to implement as a response to 
 dreamt up non existent problems.
 
 -- 
 Denis McMahon, denismfmcma...@gmail.com

Bullshit declare two integers in any language one 7 and one 4 and then write 
x=7+4; if you find a programming language where that does not yield 11 tell me.

Integers are internally assumed to be base 10 otherwise you could not calculate 
without giving the base.

All operations on integers addition, subtraction, multiplication and division 
assume base 10.
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 16:32:56 UTC+2 skrev Ian:
 On Tue, Apr 7, 2015 at 3:44 AM,  jonas.thornv...@gmail.com wrote:
 
 
  I want todo faster baseconversion for very big bases like base 1 000 000, 
  so instead of adding up digits i search it.
 
  I need the fastest algorithm to find the relation to a decimal number.
  Digmult is an instance of base at a digitplace (base^x) what i try to find 
  is the digit for the below condition is true and the loop break.
 
 
  *
  for (digit=0;digit=base;digit++) {
 if((digit+1)*digmultdecNumber)break;
  }
  *
 
  So i am looking for the digit where following condition true.
 
  if((digit)*digmultdecNumber) AND if((digit+1)*digmultdecNumber) then 
  BREAK;
 
 I'm not sure that I understand what it is that you're trying to
 accomplish. Are you trying to find the digits without using division
 because division is slow? If that's the case, then let me show you
 something.
 
 $ python -m timeit -s n = 1523837293 n // 100
 100 loops, best of 3: 0.286 usec per loop
 
 $ python -m timeit -s n = 1523 n * 100; (n+1) * 100
 100 loops, best of 3: 0.455 usec per loop
 
 In Python, one addition and two multiplications are already slower
 than a single division. The only way you're going to beat division by
 using trial multiplication is if the first digit that you try is
 always correct. To do that, you would need an oracle feeding your
 search algorithm, and then you might as well just use the oracle.
 
  One could start at half base searching, but then i Think i've read that 
  using 1/3 closing in faster?
 
 Do you mean binary search? That would be an improvement over the
 linear search algorithm you've shown. Whether a trinary search might
 be faster would depend on the distribution of the numbers you expect.
 If they're evenly distributed, it will be slower.
 
  I Think also i remember that if the search space so big that at least 22 or 
  23 guesses, needed.A random Oracle may even faster?
 
  Just pick up a number and get lucky, is it any truth to that?
 
 On average, a random Oracle with a search space of 100 will need
 100 guesses.

Well of course you use same principles like a binary search setting min and 
max, closing in on the digit. In this case the searched numbers   base^exp  
and number base^exp+1.

But since the search is within large bases upto 32-bit space, so base 
4294967295 is the biggest allowed. I need to find the nearest less exp in base 
for each (lets call them pseudo digits). But as you see there it will take time 
to add them up. So better doing a binary search, you know min-max half 
(iteration). You can do the same for a random oracle min max within range, and 
if the number of tries in general over 22 i think a random oracle do it better 
than a binary search. 

It was a long time since i did this, but i do know there is a threshold where 
searching min max with the oracle will be faster than the binary search.
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 18:34:32 UTC+2 skrev Dave Angel:
On 04/07/2015 11:40 AM, jonas.thornv...@gmail.com wrote:
Den tisdag 7 april 2015 kl. 16:32:56 UTC+2 skrev Ian:
On Tue, Apr 7, 2015 at 3:44 AM, jonas.thornv...@gmail.com wrote:

I want todo faster baseconversion for very big bases like base 1 000 000,
so instead of adding up digits i search it.

I need the fastest algorithm to find the relation to a decimal number.
Digmult is an instance of base at a digitplace (base^x) what i try to
find is the digit for the below condition is true and the loop break.

*
for (digit=0;digit=base;digit++) {
if((digit+1)*digmultdecNumber)break;
}
*

snip

One could start at half base searching, but then i Think i've read that
using 1/3 closing in faster?

Do you mean binary search? That would be an improvement over the
linear search algorithm you've shown. Whether a trinary search might
be faster would depend on the distribution of the numbers you expect.
If they're evenly distributed, it will be slower.

I Think also i remember that if the search space so big that at least 22
or 23 guesses, needed.A random Oracle may even faster?

Just pick up a number and get lucky, is it any truth to that?

On average, a random Oracle with a search space of 100 will need
100 guesses.

Well of course you use same principles like a binary search setting min and
max, closing in on the digit. In this case the searched numbers base^exp
and number base^exp+1.

But since the search is within large bases upto 32-bit space, so base
4294967295 is the biggest allowed. I need to find the nearest less exp in
base for each (lets call them pseudo digits). But as you see there it will
take time to add them up. So better doing a binary search, you know min-max
half (iteration). You can do the same for a random oracle min max within
range, and if the number of tries in general over 22 i think a random
oracle do it better than a binary search.

It was a long time since i did this, but i do know there is a threshold
where searching min max with the oracle will be faster than the binary
search.

Once again, there's no point in doing a search, when a simple integer
divide can give you the exact answer. And there's probably no point in
going left to right when right to left would yield a tiny, fast program.

I haven't seen one line of Python from you yet, so perhaps you're just
yanking our chain. I'm not here to optimize Javascript code.

Using only Python 3.4 and builtin functions, this function can be
implemented straightforwardly in 7 lines, assuming number is nonnegative
integer, and base is positive integer. It definitely could be done
smaller, but then the code might be more confusing.

--
DaveA

So you can tell me the first (higest) digit of the integer
2932903594368438384328325832983294832483258958495845849584958458435439543858588435856958650865490

Using base 429496729?

How long time did it take to find it?
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 17:00:53 UTC+2 skrev MRAB:
 On 2015-04-07 15:36, jonas.thornv...@gmail.com wrote:
  Den tisdag 7 april 2015 kl. 16:30:15 UTC+2 skrev Denis McMahon:
  On Tue, 07 Apr 2015 09:29:59 -0400, Dave Angel wrote:
 
  On 04/07/2015 05:44 AM, jonas.thornv...@gmail.com wrote:
 
  I want todo faster baseconversion for very big bases like base
  1 000 000, so instead of adding up digits i search it.
 
  How do you know the baseconversion is the bottleneck, if you
  haven't written any Python code yet?
 
  He doesn't. He doesn't comprehend that as far as a computer is
  concerned an integer has no specific 'base', it's only when
  presented in a form for humans to read that it gets base
  information added in the representation.
 
  He's making these and other similar errors in the javascript groups
  too.
 
  I suspect he's one of those people that spends his time thinking
  up elaborate solutions that he has no idea how to implement as a
  response to dreamt up non existent problems.
 
  Bullshit declare two integers in any language one 7 and one 4 and
  then write x=7+4; if you find a programming language where that does
  not yield 11 tell me.
 
  Integers are internally assumed to be base 10 otherwise you could not
  calculate without giving the base.
 
  All operations on integers addition, subtraction, multiplication and
  division assume base 10.
 
 Sorry to say this, but that's nonsense.
 
 It doesn't matter what base it's working in internally; usually it's
 base 2 (binary), because that's simpler to implement.
 
 It's only when you're converting from or to text that you need specify
 a base. Humans prefer base 10, so they've make that the default.

No that is not what i am saying, i am saying if you do operations on two 
integers the machine will assume base 10.
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 16:30:15 UTC+2 skrev Denis McMahon:
 On Tue, 07 Apr 2015 09:29:59 -0400, Dave Angel wrote:
 
  On 04/07/2015 05:44 AM, jonas.thornv...@gmail.com wrote:
 
  I want todo faster baseconversion for very big bases like base 1 000
  000, so instead of adding up digits i search it.
 
  How do you know the baseconversion is the bottleneck, if you haven't
  written any Python code yet?
 
 He doesn't. He doesn't comprehend that as far as a computer is concerned 
 an integer has no specific 'base', it's only when presented in a form for 
 humans to read that it gets base information added in the representation.
 
 He's making these and other similar errors in the javascript groups too.
 
 I suspect he's one of those people that spends his time thinking up 
 elaborate solutions that he has no idea how to implement as a response to 
 dreamt up non existent problems.
 
 -- 
 Denis McMahon, denismfmcma...@gmail.com

Well Denis if it hasn't a base then how can you add integers, the integer 
operations assume base 10. 
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 17:07:36 UTC+2 skrev Ian:
 On Tue, Apr 7, 2015 at 8:36 AM,  jonas.thornv...@gmail.com wrote:
  Den tisdag 7 april 2015 kl. 16:30:15 UTC+2 skrev Denis McMahon:
  On Tue, 07 Apr 2015 09:29:59 -0400, Dave Angel wrote:
 
   On 04/07/2015 05:44 AM, jonas.thornv...@gmail.com wrote:
 
   I want todo faster baseconversion for very big bases like base 1 000
   000, so instead of adding up digits i search it.
 
   How do you know the baseconversion is the bottleneck, if you haven't
   written any Python code yet?
 
  He doesn't. He doesn't comprehend that as far as a computer is concerned
  an integer has no specific 'base', it's only when presented in a form for
  humans to read that it gets base information added in the representation.
 
  He's making these and other similar errors in the javascript groups too.
 
  I suspect he's one of those people that spends his time thinking up
  elaborate solutions that he has no idea how to implement as a response to
  dreamt up non existent problems.
 
  --
  Denis McMahon, denismfmcma...@gmail.com
 
  Bullshit declare two integers in any language one 7 and one 4 and then 
  write x=7+4; if you find a programming language where that does not yield 
  11 tell me.
 
  Integers are internally assumed to be base 10 otherwise you could not 
  calculate without giving the base.
 
  All operations on integers addition, subtraction, multiplication and 
  division assume base 10.
 
 You're conflating the internal representation of the integer with the
 formatting that is done to display the integer. When you do
 print(x), the computer doesn't just dump the internal representation
 of x onto the display. It formats x as character data and displays
 *that*. For integers, the vast majority of programming languages will
 do the formatting as base 10 by default, since that is the format
 preferred by most humans.

No i don't i say the operations assume base ten other wise INT A=7,B=4 
Calculate C=A+B would not yield 11 as an answer. The programming languages 
assume integer operations are done in base 10, at least the one where you can 
not specify in what base the integer arithmetic is done. And there probably is 
such languages, Python maybe?
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Re: Best search algorithm to find condition within a range

Den tisdag 7 april 2015 kl. 21:27:20 UTC+2 skrev Ben Bacarisse:
 Ian Kelly ian.g.ke...@gmail.com writes:
 
  On Tue, Apr 7, 2015 at 12:55 PM, Terry Reedy tjre...@udel.edu wrote:
  On 4/7/2015 1:44 PM, Ian Kelly wrote:
 
  def to_base(number, base):
 
  ... digits = []
  ... while number  0:
  ... digits.append(number % base)
  ... number //= base
  ... return digits or [0]
  ...
 
 
  to_base(2932903594368438384328325832983294832483258958495845849584958458435439543858588435856958650865490,
  429496729)
 
  [27626525, 286159541, 134919277, 305018215, 329341598, 48181777,
  79384857, 112868646, 221068759, 70871527, 416507001, 31]
  About 15 microseconds.
 
 
  % and probably // call divmod internally and toss one of the results.
  Slightly faster (5.7 versus 6.1 microseconds on my machine) is
 
  Not on my box.
 
  $ python3 -m timeit -s n = 100; x = 42 n % x; n // x
  1000 loops, best of 3: 0.105 usec per loop
  $ python3 -m timeit -s n = 100; x = 42 divmod(n,x)
  1000 loops, best of 3: 0.124 usec per loop
 
 I get similar results, but the times switch over when n is large enough
 to become a bignum.
 
 -- 
 Ben.

I am not sure you guys realised, that althoug the size of the factors to 
muliply expands according to base^(exp+1) for each digitplace the number of 
comparissons needed to reach the digit place (multiple of base^exp+1) is 
constant with my approach/method.
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Re: Best search algorithm to find condition within a range

Den onsdag 8 april 2015 kl. 00:57:27 UTC+2 skrev Steven D'Aprano:
 On Tue, 7 Apr 2015 07:44 pm, jonas.thornv...@gmail.com wrote:
 
 
  I want todo faster baseconversion for very big bases like base 1 000 000,
  so instead of adding up digits i search it.
 
 What digits would you use for base one-million?
 
 Base 2 uses 0 1.
 Base 3 uses 0 1 2.
 Base 10 uses 0 1 2 3 4 5 6 7 8 9.
 Base 16 uses 0 1 2 3 4 5 6 7 8 9 A B C D E F.
 
 Base one million uses what?
 
 How would you write down 12345 in base one-million?
 
 
 
 -- 
 Steven

Well digit places have to be comma separated i think i told you, in low bases 
we have unique digits so there is no problems write out numbers without 
separaation of digitplaces. But when the base is higher than the number of 
digits you have a choice to make. Either you invent a new numeral(number digit.

I think you realise that BB = 11,11

So your example would of course be ,12345

I hope you see A,B,C,D,E,F is only replacement for 10,11,12,13,14,15
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Re: More or less code in python?

Den söndag 15 mars 2015 kl. 20:01:36 UTC+1 skrev Paul Rubin:
 jonas.thornv...@gmail.com writes:
  I though it would be interesting doing comparissons in timing adding
  massive digits in different bases. Especially in Python.
 
 Python has built-in bignums.  Try print 2**500.

I know.
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Re: More or less code in python?

Den söndag 15 mars 2015 kl. 20:01:36 UTC+1 skrev Paul Rubin:
 jonas.thornv...@gmail.com writes:
  I though it would be interesting doing comparissons in timing adding
  massive digits in different bases. Especially in Python.
 
 Python has built-in bignums.  Try print 2**500.

I will try implement the common operators + - * / pow sqrt operators mod floor 
and and a generic parser taking arguments in any base.
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Re: More or less code in python?

Den söndag 15 mars 2015 kl. 19:32:02 UTC+1 skrev Joel Goldstick:
 On Sun, Mar 15, 2015 at 2:07 PM,  jonas.thornv...@gmail.com wrote:
  SCRIPT LANGUAGE=Javascript
 
  function naiveAdd(base,arrOne,arrTwo) {
  if (arrOne.length=arrTwo.length) {length=arrOne.length;} else 
  {length=arrTwo.length;}
  out=;
  remainder=0;
  for (i=0;ilength;i++){
  one=arrOne[i];
  two=arrTwo[i];
  one=parseInt(one);
  two=parseInt(two);
  if (isNaN(one)) one = 0;
  if (isNaN(two)) two = 0;
  sum=one+two+remainder;
 
  if (sum=base) { sum=sum-base; remainder=1;} else {remainder=0;}
  out=,+sum+out;
  }
  if (remainder==1) out=remainder+out;
  return out;
  }
  base=16;
  strOne=2,2;
  strTwo=13,13;
  arrOne=strOne.split(,);
  arrTwo=strTwo.split(,);
  arrOne.reverse();
  arrTwo.reverse();
  naiveAdd(base,arrOne,arrTwo);
  document.write(Sum = ,out,BR);
  /SCRIPT
  --
  https://mail.python.org/mailman/listinfo/python-list
 
 I'm guessing less, if your question is whether this code would be
 fewer lines in python.  However, it would be nice if you formatted it
 so that it is readable, and gave a description of what you think the
 code is doing, or should be doing.  Its an odd question in that python
 generally runs on a server and javascript in a browser
 
 -- 
 Joel Goldstick
 http://joelgoldstick.com

I though it would be interesting doing comparissons in timing adding massive 
digits in different bases. Especially in Python.
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More or less code in python?

SCRIPT LANGUAGE=Javascript 

function naiveAdd(base,arrOne,arrTwo) { 
if (arrOne.length=arrTwo.length) {length=arrOne.length;} else 
{length=arrTwo.length;} 
out=; 
remainder=0; 
for (i=0;ilength;i++){ 
one=arrOne[i]; 
two=arrTwo[i]; 
one=parseInt(one); 
two=parseInt(two); 
if (isNaN(one)) one = 0; 
if (isNaN(two)) two = 0; 
sum=one+two+remainder; 

if (sum=base) { sum=sum-base; remainder=1;} else {remainder=0;} 
out=,+sum+out; 
} 
if (remainder==1) out=remainder+out; 
return out; 
} 
base=16; 
strOne=2,2; 
strTwo=13,13; 
arrOne=strOne.split(,); 
arrTwo=strTwo.split(,); 
arrOne.reverse(); 
arrTwo.reverse(); 
naiveAdd(base,arrOne,arrTwo); 
document.write(Sum = ,out,BR); 
/SCRIPT
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Re: More or less code in python?

Den söndag 15 mars 2015 kl. 19:32:02 UTC+1 skrev Joel Goldstick:
 On Sun, Mar 15, 2015 at 2:07 PM,  jonas.thornv...@gmail.com wrote:
  SCRIPT LANGUAGE=Javascript
 
  function naiveAdd(base,arrOne,arrTwo) {
  if (arrOne.length=arrTwo.length) {length=arrOne.length;} else 
  {length=arrTwo.length;}
  out=;
  remainder=0;
  for (i=0;ilength;i++){
  one=arrOne[i];
  two=arrTwo[i];
  one=parseInt(one);
  two=parseInt(two);
  if (isNaN(one)) one = 0;
  if (isNaN(two)) two = 0;
  sum=one+two+remainder;
 
  if (sum=base) { sum=sum-base; remainder=1;} else {remainder=0;}
  out=,+sum+out;
  }
  if (remainder==1) out=remainder+out;
  return out;
  }
  base=16;
  strOne=2,2;
  strTwo=13,13;
  arrOne=strOne.split(,);
  arrTwo=strTwo.split(,);
  arrOne.reverse();
  arrTwo.reverse();
  naiveAdd(base,arrOne,arrTwo);
  document.write(Sum = ,out,BR);
  /SCRIPT
  --
  https://mail.python.org/mailman/listinfo/python-list
 
 I'm guessing less, if your question is whether this code would be
 fewer lines in python.  However, it would be nice if you formatted it
 so that it is readable, and gave a description of what you think the
 code is doing, or should be doing.  Its an odd question in that python
 generally runs on a server and javascript in a browser
 
 -- 
 Joel Goldstick
 http://joelgoldstick.com

It add two postive numbers working in any base you prefer. There is no 
restriction on the size of operands, it dependens upon the memory of your 
computer or string limitations in the implementation of ECMASCRIPT.(If there 
is).

Basicly the function doing addition in Python shell, but you can chose base you 
work in.
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Re: Brython (Python in the browser)

2013-12-27 Thread jonas . thornvall

Den fredagen den 27:e december 2013 kl. 07:14:35 UTC+1 skrev Pierre Quentel:
 Hi,
 
 
 
 Ever wanted to use Python instead of Javascript for web client programming ? 
 Take a look at Brython, an implementation of Python 3 in the browser, with an 
 interface with DOM elements and events
 
 
 
 Its use is very simple :
 
 - load the Javascript library brython.js : script src=/path/to/brython.js
 
 - embed Python code inside a tag script type=text/python
 
 - run the Python script on page load : body onload=brython()
 
 
 
 The Python code is translated into Javascript and executed on the fly
 
 
 
 Brython supports the DOM API, HTML5, SVG, with some syntaxic sugar to make 
 the interface more concise (a la jQuery) ; interaction with Javascript 
 libraries is very straightforward. The Brython site provides documentation 
 and many examples
 
 
 
 After 1 year of intense development, Brython now covers most of the Python3 
 syntax and can run most of the modules of the Python3.3 standard distribution 
 unmodified, including complex packages like unittest. The team aims at 
 covering 100% of all of Python that makes sense in a browser environment
 
 
 
 Home page : http://www.brython.info
 
 Development site : https://bitbucket.org/olemis/brython/src
 
 Downloads : https://bitbucket.org/olemis/brython/downloads
 
 Community : https://groups.google.com/forum/?fromgroups=#!forum/brython

I am not sure i understand the concept correct it is not serverside, i have to 
install the libraries on my local computer, and it is accessible via 
javascript, not on its own in browsers?

In the future will it always be a library that has to be downloaded and 
installed?

Could in not be implemented like an activeX plugin as flashplayer, when you 
come to a Brython side you have a message to download plugin?

If i understand correct, now one can implement Brython script in JAVASCRIPT 
invoked HTML on a server, but it will not work for people surfing the web 
because the browsers lack support for Brython libraries?

Is Brython thought to be a webapplication or can it read/write to files?
So what is its future?

1. A standard incorporated and implemented by browsers?
2. A plugin like flashplayer?
3. A standalone library to be downloaded and installed on the local computer to 
run python scripts one make?
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Line based graphic in canvas (vectorgraphic?) scale incorrectly on HDMI monitor.

2013-12-21 Thread jonas . thornvall

Is there a way to make linebased graphic used in canvas scale correct on any 
monitor?

I run in 1920*1080 on a Philips TV used as monitor does it matter, lines tend 
to get longer vertical then horizontal?

Strange is i really do not see it writing out recangles.







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Re: Line based graphic in canvas (vectorgraphic?) scale incorrectly on HDMI monitor.

2013-12-21 Thread jonas . thornvall

Den lördagen den 21:e december 2013 kl. 20:03:17 UTC+1 skrev Ned Batchelder:
 On 12/21/13 1:30 PM, jonas.thornv...@gmail.com wrote:
 
  Is there a way to make linebased graphic used in canvas scale correct on 
  any monitor?
 
 
 
  I run in 1920*1080 on a Philips TV used as monitor does it matter, lines 
  tend to get longer vertical then horizontal?
 
 
 
  Strange is i really do not see it writing out recangles.
 
 
 
 I'm not sure how we can help without seeing any code.  There's no 
 
 information here that we can use to make a concrete suggestion.
 
 
 
 -- 
 
 Ned Batchelder, http://nedbatchelder.com
A vertical and horisontal line of same length will be plottet out like this.
There is really no code to show yet, i just write out some lines and noted that 
writing out a verital line of length 400 is longer then a horisontal line of 
same length e.g.

line(100,100,500,100); is shorter
line(300,100,300,500); is longer

__
|
|
|
|
|
|
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Re: Line based graphic in canvas (vectorgraphic?) scale incorrectly on HDMI monitor.

2013-12-21 Thread jonas . thornvall

Den lördagen den 21:e december 2013 kl. 20:56:54 UTC+1 skrev Ned Batchelder:
 On 12/21/13 2:12 PM, jonas.thornv...@gmail.com wrote:
 
  Den lï¿½rdagen den 21:e december 2013 kl. 20:03:17 UTC+1 skrev Ned 
  Batchelder:
 
  On 12/21/13 1:30 PM, jonas.thornv...@gmail.com wrote:
 
 
 
  Is there a way to make linebased graphic used in canvas scale correct on 
  any monitor?
 
 
 
 
 
 
 
  I run in 1920*1080 on a Philips TV used as monitor does it matter, lines 
  tend to get longer vertical then horizontal?
 
 
 
 
 
 
 
  Strange is i really do not see it writing out recangles.
 
 
 
 
 
 
 
  I'm not sure how we can help without seeing any code.  There's no
 
 
 
  information here that we can use to make a concrete suggestion.
 
 
 
 
 
 
 
  --
 
 
 
  Ned Batchelder, http://nedbatchelder.com
 
  A vertical and horisontal line of same length will be plottet out like this.
 
  There is really no code to show yet, i just write out some lines and noted 
  that writing out a verital line of length 400 is longer then a horisontal 
  line of same length e.g.
 
 
 
  line(100,100,500,100); is shorter
 
  line(300,100,300,500); is longer
 
 
 
  __
 
   |
 
   |
 
   |
 
   |
 
   |
 
   |
 
 
 
 
 
 Sounds like you don't have square pixels.  Either use a monitor which 
 
 does, or adjust your coordinates to take the shape of the pixel into 
 
 account, or try different resolution settings to see if one of them does 
 
 have square pixels.  Also, the settings on the monitor matter also, you 
 
 may be able to adjust them to correct the shape of the pixel.
 
 
 
 -- 
 
 Ned Batchelder, http://nedbatchelder.com

It is weird because it seem to be compensated writing out rectangles, but not 
for linegraphics.
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Install Tkinter for Windows 7 64-bit

2013-11-11 Thread jonas . thornvall

I have installed Python 3.3, and i want to add a library with some basic 
functions like canvas and basic geomteric objects, fonts etc. Preferably 
something similar to the Javascript canvas.

I've looked for graphic packages, and from what i can see something called 
Tkinter may be what i look for.

However i have no luck install it, i installed something called Visual studio 
in hope it would incorporate Tkinter, but it did not and it also do not seem to 
work with 3.3 only 3.1 and below.

Next i found ActiveStateTCL that from what i could see incorporated Tkinter 
library?

But i have no luck runn the Tkinter example file i downloaded in idel, it still 
says no module called Tkinter.


===
Traceback (most recent call last):
  File D:\Python33\test2.py, line 16, in module
from Tkinter import Tk, Canvas, Frame, BOTH
ImportError: No module named 'Tkinter'
===

Maybe i should use some other module, i just need a canvas and the basic 
geometric objects functions line,rectangle,circle and some font.

JT
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Re: Install Tkinter for Windows 7 64-bit

2013-11-11 Thread jonas . thornvall

Here is the example file i have tried.

#!/usr/bin/python
# -*- coding: utf-8 -*-


ZetCode Tkinter tutorial

This program draws three
rectangles filled with different
colors.

author: Jan Bodar
last modified: January 2011
website: www.zetcode.com


from Tkinter import Tk, Canvas, Frame, BOTH


class Example(Frame):
  
def __init__(self, parent):
Frame.__init__(self, parent)   
 
self.parent = parent
self.initUI()

def initUI(self):
  
self.parent.title(Colors)
self.pack(fill=BOTH, expand=1)

canvas = Canvas(self)
canvas.create_rectangle(30, 10, 120, 80, 
outline=#fb0, fill=#fb0)
canvas.create_rectangle(150, 10, 240, 80, 
outline=#f50, fill=#f50)
canvas.create_rectangle(270, 10, 370, 80, 
outline=#05f, fill=#05f)
canvas.pack(fill=BOTH, expand=1)



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Re: Install Tkinter for Windows 7 64-bit

2013-11-11 Thread jonas . thornvall

Den måndagen den 11:e november 2013 kl. 17:43:12 UTC+1 skrev Chris “Kwpolska” 
Warrick:
 On Mon, Nov 11, 2013 at 5:38 PM,  jonas.thornv...@gmail.com wrote:
 
  But i have no luck runn the Tkinter example file i downloaded in idel, it 
  still says no module called Tkinter.
 
 IDLE*
 
  ===
 
  Traceback (most recent call last):
 
File D:\Python33\test2.py, line 16, in module
 
  from Tkinter import Tk, Canvas, Frame, BOTH
 
  ImportError: No module named 'Tkinter'
 
  ===
 
 
 
 In Python 3, 'Tkinter' was renamed to 'tkinter' (both sans quotes).
 
 Please change the file to reflect that (as the traceback points out,
 
 line 16).
 
 
 
 -- 
 
 Chris “Kwpolska” Warrick http://kwpolska.tk
 
 PGP: 5EAAEA16
 
 stop html mail | always bottom-post | only UTF-8 makes sense

Thanks!
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-08 Thread jonas . thornvall

Den fredagen den 8:e november 2013 kl. 03:43:17 UTC+1 skrev zipher:
  I am not sure if it is just stupidness or laziness that prevent you from 
  seeing that 4^8=65536.
 
 
 
  I can see that 4^8 = 65536. Now how are you going to render 65537? You
 
  claimed that you could render *any* number efficiently. What you've
 
  proven is that a small subset of numbers can be rendered efficiently.
 
 
 
 I think the idea would be to find the prime factorization for a given
 
 number, which has been proven to be available (and unique) for any and
 
 every number.  Most numbers can compress given this technique.  Prime
 
 numbers, of course, would not be compressed.
 
 
 
 -- 
 
 MarkJ
 
 Tacoma, Washington

3^2-2^2=5 
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-07 Thread jonas . thornvall

Den torsdagen den 7:e november 2013 kl. 23:26:45 UTC+1 skrev Chris Angelico:
 On Fri, Nov 8, 2013 at 5:59 AM, Mark Janssen dreamingforw...@gmail.com 
 wrote:
 
  I think the idea is that you could take any arbitrary input sequence,
 
  view it as a large number, and then find what exponential equation can
 
  produce that result.  The equation becomes the compression.
 
 
 
 Interesting idea, but I don't see how this has anything to do with
 
 compressing arbitrary data. We already have a compact notation for
 
 representing a large number as 2^24*x+2^16*y+2^8*z+q, so I don't see
 
 how this will be any better, other than eliding NULs.
 
 
 
 ChrisA

I guess what matter is how fast an algorithm can encode and decode a big 
number, at least if you want to use it for very big sets of random data, or 
losless video compression?
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-07 Thread jonas . thornvall

Den fredagen den 8:e november 2013 kl. 03:17:36 UTC+1 skrev Chris Angelico:
 On Fri, Nov 8, 2013 at 1:05 PM,  jonas.thornv...@gmail.com wrote:
 
  I guess what matter is how fast an algorithm can encode and decode a big 
  number, at least if you want to use it for very big sets of random data, or 
  losless video compression?
 
 
 
 I don't care how fast. I care about the laws of physics :) You can't
 
 stuff more data into less space without losing some of it.
 
 
 
 Also, please lose Google Groups, or check out what other people have
 
 said about making it less obnoxious.
 
 
 
 ChrisA

Please, you are he obnoxious, so fuck off or go learn about reformulation of 
problems. Every number has an infinite number of arithmetical solutions. So 
every number do has a shortest arithmetical encoding. And that is not the hard 
part to figure out, the hard part is to find a generic arithmetic encoding.

I am not sure if it is just stupidness or laziness that prevent you from seeing 
that 4^8=65536.
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-04 Thread jonas . thornvall

Den lördagen den 2:e november 2013 kl. 22:31:09 UTC+1 skrev Tim Roberts:
 jonas.thornv...@gmail.com wrote:
 
 
 
 Well then i have news for you.
 
 
 
 Well, then, why don't you share it?
 
 
 
 Let me try to get you to understand WHY what you say is impossible.  Let's
 
 say you do have a function f(x) that can produce a compressed output y for
 
 any given x, such that y is always smaller than x.  If that were true, then
 
 I could call f() recursively:
 
 f(f(...f(f(f(f(f(x)...))
 
 and eventually the result get down to a single bit.  I hope it is clear
 
 that there's no way to restore a single bit back into different source
 
 texts.
 
 
 
 Here's another way to look at it.  If f(x) is smaller than x for every x,
 
 that means there MUST me multiple values of x that produce the same f(x).
 
 Do you see?  If x is three bits and f(x) is two bits, that means there are
 
 8 possible values for x but only 4 values for f(x).  So, given an f(x), you
 
 cannot tell which value of x it came from.  You have lost information.
 
 -- 
 
 Tim Roberts, t...@probo.com
 
 Providenza  Boekelheide, Inc.

Well let me try to explain why it is working and i have implemented one.
I only need to refresh my memory it was almost 15 years ago.
This is not the solution but this is why it is working.
65536=256^2=16^4=***4^8***=2^16

Yes i am aware that 256 is a single byte 8 bits, but the approach is valid 
anyway.
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-04 Thread jonas . thornvall

Den måndagen den 4:e november 2013 kl. 14:53:28 UTC+1 skrev 
jonas.t...@gmail.com:
 Den lördagen den 2:e november 2013 kl. 22:31:09 UTC+1 skrev Tim Roberts:
 
  jonas.thornv...@gmail.com wrote:
 
  
 
  
 
  
 
  Well then i have news for you.
 
  
 
  
 
  
 
  Well, then, why don't you share it?
 
  
 
  
 
  
 
  Let me try to get you to understand WHY what you say is impossible.  Let's
 
  
 
  say you do have a function f(x) that can produce a compressed output y for
 
  
 
  any given x, such that y is always smaller than x.  If that were true, then
 
  
 
  I could call f() recursively:
 
  
 
  f(f(...f(f(f(f(f(x)...))
 
  
 
  and eventually the result get down to a single bit.  I hope it is clear
 
  
 
  that there's no way to restore a single bit back into different source
 
  
 
  texts.
 
  
 
  
 
  
 
  Here's another way to look at it.  If f(x) is smaller than x for every x,
 
  
 
  that means there MUST me multiple values of x that produce the same f(x).
 
  
 
  Do you see?  If x is three bits and f(x) is two bits, that means there are
 
  
 
  8 possible values for x but only 4 values for f(x).  So, given an f(x), you
 
  
 
  cannot tell which value of x it came from.  You have lost information.
 
  
 
  -- 
 
  
 
  Tim Roberts, t...@probo.com
 
  
 
  Providenza  Boekelheide, Inc.
 
 
 
 Well let me try to explain why it is working and i have implemented one.
 
 I only need to refresh my memory it was almost 15 years ago.
 
 This is not the solution but this is why it is working.
 
 65536=256^2=16^4=***4^8***=2^16
 
 
 
 Yes i am aware that 256 is a single byte 8 bits, but the approach is valid 
 anyway.

You see if the program behave intelligent some small operations can perform 
wonder on a number.
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Re: Algorithm that makes maximum compression of completly diffused data.

2013-11-04 Thread jonas . thornvall

Den måndagen den 4:e november 2013 kl. 15:27:19 UTC+1 skrev Dave Angel:
 On Mon, 4 Nov 2013 05:53:28 -0800 (PST), jonas.thornv...@gmail.com 
 
 wrote:
 
  Den lördagen den 2:e november 2013 kl. 22:31:09 UTC+1 skrev Tim 
 
 Roberts:
 
   Here's another way to look at it.  If f(x) is smaller than x for 
 
 every x,
 
 
 
 
 
 
 
 
 
   that means there MUST me multiple values of x that produce the 
 
 same f(x).
 
 
 
 
 
 
 
 
 
   Do you see?  If x is three bits and f(x) is two bits, that means 
 
 there are
 
 
 
 
 
 
 
 
 
   8 possible values for x but only 4 values for f(x).  So, given an 
 
 f(x), y=
 
 
 
 
 
 
 
 
 
   cannot tell which value of x it came from.  You have lost 
 
 information.
 
 
 
 
 
 
 
 
 
  Well let me try to explain why it is working and i have implemented 
 
 one.
 
  I only need to refresh my memory it was almost 15 years ago.
 
  This is not the solution but this is why it is working.
 
  65536=256^2=16^4=***4^8***=2^16
 
 
 
 
 
  Yes i am aware that 256 is a single byte 8 bits, but the approach 
 
 is valid =
 
  anyway.
 
 
 
 And e ^ (I * pi) == -1
 
 So what. ?
 

e is an approximation... and your idea is not general for any n.
 
 
 Better file that patent, before the patent office realizes the 
 
 analogy to the perpetual motion machine.
 
 
 
 -- 
 
 DaveA

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Re: First day beginner to python, add to counter after nested loop

2013-11-03 Thread jonas . thornvall

Den lördagen den 2:e november 2013 kl. 21:19:44 UTC+1 skrev Tim Roberts:
 jonas.thornv...@gmail.com wrote:
 
 
 
 I certainly do not like the old bracket style it was a catastrophe, but 
 
 in honesty the gui editor of python should have what i propose, a parser
 
 that indent automaticly at loops, functions and end.
 
 
 
 Many editors do that.  Vim, which is what I use, certainly does.
 
 
 
 I promise you it will save millions of hours of bug searching all over 
 
 world in a month.
 
 
 
 I suspect you meant dozens rather than millions...
 
 
 
 Look, both schemes have their pitfalls.  With an end requirement, it's
 
 easy to have code where the indenting doesn't match the actual nesting, and
 
 that causes human confusion.  Without the end requirement, it's not hard
 
 to type code where you forget to dedent.  Those are just two manifestations
 
 of the exact same problem.  Neither scheme is provably superior to the
 
 other.  It's just a choice that a language designer has to make.
 
 
 
 I happen to like Python's choice.  You'll get used to it.
 
 -- 
 
 Tim Roberts, t...@probo.com
 
 Providenza  Boekelheide, Inc.

What does Vim stand for Voyager interstellar mission?
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Re: First day beginner to python, add to counter after nested loop

2013-11-03 Thread jonas . thornvall

Den lördagen den 2:e november 2013 kl. 21:19:44 UTC+1 skrev Tim Roberts:
 jonas.thornv...@gmail.com wrote:
 
 
 
 I certainly do not like the old bracket style it was a catastrophe, but 
 
 in honesty the gui editor of python should have what i propose, a parser
 
 that indent automaticly at loops, functions and end.
 
 
 
 Many editors do that.  Vim, which is what I use, certainly does.
 
 
 
 I promise you it will save millions of hours of bug searching all over 
 
 world in a month.
 
 
 
 I suspect you meant dozens rather than millions...
 
 
 
 Look, both schemes have their pitfalls.  With an end requirement, it's
 
 easy to have code where the indenting doesn't match the actual nesting, and
 
 that causes human confusion.  Without the end requirement, it's not hard
 
 to type code where you forget to dedent.  Those are just two manifestations
 
 of the exact same problem.  Neither scheme is provably superior to the
 
 other.  It's just a choice that a language designer has to make.
 
 
 
 I happen to like Python's choice.  You'll get used to it.
 
 -- 
 
 Tim Roberts, t...@probo.com
 
 Providenza  Boekelheide, Inc.

Was there a VIM discussion group?
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 jonas.thornv...@gmail.com wrote:
 
 
 
 Why did Python not implement end... The end is really not necessary for
 
 the programming language it can be excluded, but it is a courtesy to
 
 the programmer and could easily be transformed to indents automaticly,
 
 that is removed before the compiliation/interpretation of code.  
 
 
 
 You only say that because your brain has been poisoned by languages that
 
 require some kind of end.  It's not necessary, and it's extra typing. 99%
 
 of programmers do the indentation anyway, to make the program easy to read,
 
 so why not just make it part of the syntax?  That way, you don't
 
 accidentally have the indentation not match the syntax.
 
 -- 
 
 Tim Roberts, t...@probo.com
 
 Providenza  Boekelheide, Inc.

It maybe common practice in program languages, but to me it is slightly 
confusing to have the while/for loop on the same indent level, as the regular 
statements.
Because when the while loop ends there is no other identification then that the 
indent stopped, and to me it is easy to interpret that terms that actually 
straight under the loop belongs to it.

But of course it is a matter of adjust the way i look at the code.
Thanks for help guys.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 09:52:16 UTC+1 skrev 
jonas.t...@gmail.com:
 Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 
  jonas.thornv...@gmail.com wrote:
 
  
 
  
 
  
 
  Why did Python not implement end... The end is really not necessary for
 
  
 
  the programming language it can be excluded, but it is a courtesy to
 
  
 
  the programmer and could easily be transformed to indents automaticly,
 
  
 
  that is removed before the compiliation/interpretation of code.  
 
  
 
  
 
  
 
  You only say that because your brain has been poisoned by languages that
 
  
 
  require some kind of end.  It's not necessary, and it's extra typing. 99%
 
  
 
  of programmers do the indentation anyway, to make the program easy to read,
 
  
 
  so why not just make it part of the syntax?  That way, you don't
 
  
 
  accidentally have the indentation not match the syntax.
 
  
 
  -- 
 
  
 
  Tim Roberts, t...@probo.com
 
  
 
  Providenza  Boekelheide, Inc.
 
 
 
 It maybe common practice in program languages, but to me it is slightly 
 confusing to have the while/for loop on the same indent level, as the regular 
 statements.
 
 Because when the while loop ends there is no other identification then that 
 the indent stopped, and to me it is easy to interpret that terms that 
 actually straight under the loop belongs to it.
 
 
 
 But of course it is a matter of adjust the way i look at the code.
 
 Thanks for help guys.

To show you guys that i am not totally uneducable i actually followed your 
sugestions ;D. 
(Is there any support similar javascript canvas for drawing, and HTML for 
interactivity, textbox, buttons in python?). 

I have been programming some PHP long time ago, basicly only remember you had 
to have a server running to interact via HTML. 

#!/usr/bin/python
import math
# Function definition is here
def sq(number):
   
  square=1
  factor=2
  multip=exponent*exponent
  print(x,= , end=)
  while number=multip:
 while square=number:
factor+=1
square=factor*factor
 factor-=1  
 print(factor,^2+,sep=,end=)
 square=factor*factor
 number=number-(factor*factor)
 square=1
 factor=1
  print(number)

#Set exponent here  
exponent=3
print(Exp=x^,exponent,sep=)
#Set range of numbers x
for x in range (1,100):
  sq(x);
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 10:48:36 UTC+1 skrev 
jonas.t...@gmail.com:
 Den onsdagen den 30:e oktober 2013 kl. 09:52:16 UTC+1 skrev 
 jonas.t...@gmail.com:
 
  Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 
  
 
   jonas.thornv...@gmail.com wrote:
 
  
 
   
 
  
 
   
 
  
 
   
 
  
 
   Why did Python not implement end... The end is really not necessary for
 
  
 
   
 
  
 
   the programming language it can be excluded, but it is a courtesy to
 
  
 
   
 
  
 
   the programmer and could easily be transformed to indents automaticly,
 
  
 
   
 
  
 
   that is removed before the compiliation/interpretation of code.  
 
  
 
   
 
  
 
   
 
  
 
   
 
  
 
   You only say that because your brain has been poisoned by languages that
 
  
 
   
 
  
 
   require some kind of end.  It's not necessary, and it's extra typing. 
   99%
 
  
 
   
 
  
 
   of programmers do the indentation anyway, to make the program easy to 
   read,
 
  
 
   
 
  
 
   so why not just make it part of the syntax?  That way, you don't
 
  
 
   
 
  
 
   accidentally have the indentation not match the syntax.
 
  
 
   
 
  
 
   -- 
 
  
 
   
 
  
 
   Tim Roberts, t...@probo.com
 
  
 
   
 
  
 
   Providenza  Boekelheide, Inc.
 
  
 
  
 
  
 
  It maybe common practice in program languages, but to me it is slightly 
  confusing to have the while/for loop on the same indent level, as the 
  regular statements.
 
  
 
  Because when the while loop ends there is no other identification then that 
  the indent stopped, and to me it is easy to interpret that terms that 
  actually straight under the loop belongs to it.
 
  
 
  
 
  
 
  But of course it is a matter of adjust the way i look at the code.
 
  
 
  Thanks for help guys.
 
 
 
 To show you guys that i am not totally uneducable i actually followed your 
 sugestions ;D. 
 
 (Is there any support similar javascript canvas for drawing, and HTML for 
 interactivity, textbox, buttons in python?). 
 
 
 
 I have been programming some PHP long time ago, basicly only remember you had 
 to have a server running to interact via HTML. 
 
 
 
 #!/usr/bin/python
 
 import math
 
 # Function definition is here
 
 def sq(number):
 

 
   square=1
 
   factor=2
 
   multip=exponent*exponent
 
   print(x,= , end=)
 
   while number=multip:
 
  while square=number:
 
 factor+=1
 
 square=factor*factor
 
  factor-=1
 
  print(factor,^2+,sep=,end=)
 
  square=factor*factor
 
  number=number-(factor*factor)
 
  square=1
 
  factor=1
 
   print(number)
 
 
 
 #Set exponent here  
 
 exponent=3
 
 print(Exp=x^,exponent,sep=)
 
 #Set range of numbers x
 
 for x in range (1,100):
 
   sq(x);

Forgot change the static square that was written out, i am a bit undecuable 
afterall...

#!/usr/bin/python
import math
# Function definition is here
def sq(number):
   
  square=1
  factor=2
  multip=exponent*exponent
  print(x,= , end=)
  while number=multip:
 while square=number:
factor+=1
square=factor*factor
 factor-=1  
 print(factor,^,exponent,+,sep=,end=)
 square=factor*factor
 number=number-(factor*factor)
 square=1
 factor=1
  print(number)

#Set exponent here  
exponent=3
print(Exp=x^,exponent,sep=)
#Set range of numbers x
for x in range (1,100):
  sq(x);
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 jonas.thornv...@gmail.com wrote:
 
 
 
 Why did Python not implement end... The end is really not necessary for
 
 the programming language it can be excluded, but it is a courtesy to
 
 the programmer and could easily be transformed to indents automaticly,
 
 that is removed before the compiliation/interpretation of code.  
 
 
 
 You only say that because your brain has been poisoned by languages that
 
 require some kind of end.  It's not necessary, and it's extra typing. 99%
 
 of programmers do the indentation anyway, to make the program easy to read,
 
 so why not just make it part of the syntax?  That way, you don't
 
 accidentally have the indentation not match the syntax.
 
 -- 
 
 Tim Roberts, t...@probo.com
 
 Providenza  Boekelheide, Inc.

Well Tim ***one could argue*** why not do a (i think it is called parser) that 
react to loop, end and function. And lazy like me do not have to think 
about what is not part of program.

I certainly do not like the old bracket style it was a catastrophe, but in 
honesty the gui editor of python should have what i propose, a parser that 
indent automaticly at loops, functions and end. I promise you it will save 
millions of hours of bug searching all over world in a month.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 11:08:11 UTC+1 skrev 
jonas.t...@gmail.com:
 Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 
  jonas.thornv...@gmail.com wrote:
 
  
 
  
 
  
 
  Why did Python not implement end... The end is really not necessary for
 
  
 
  the programming language it can be excluded, but it is a courtesy to
 
  
 
  the programmer and could easily be transformed to indents automaticly,
 
  
 
  that is removed before the compiliation/interpretation of code.  
 
  
 
  
 
  
 
  You only say that because your brain has been poisoned by languages that
 
  
 
  require some kind of end.  It's not necessary, and it's extra typing. 99%
 
  
 
  of programmers do the indentation anyway, to make the program easy to read,
 
  
 
  so why not just make it part of the syntax?  That way, you don't
 
  
 
  accidentally have the indentation not match the syntax.
 
  
 
  -- 
 
  
 
  Tim Roberts, t...@probo.com
 
  
 
  Providenza  Boekelheide, Inc.
 
 
 
 Well Tim ***one could argue*** why not do a (i think it is called parser) 
 that react to loop, end and function. And lazy like me do not have to 
 think about what is not part of program.
 
 
 
 I certainly do not like the old bracket style it was a catastrophe, but in 
 honesty the gui editor of python should have what i propose, a parser that 
 indent automaticly at loops, functions and end. I promise you it will save 
 millions of hours of bug searching all over world in a month.

Instead of having going over indent manually, you just drop in an end it is so 
simple, no marking no meny indent unindent it is automaticly done. And that was 
the purpose of python to remove idiocies, like brackets and indents.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 11:00:30 UTC+1 skrev Mark Lawrence:
 On 30/10/2013 09:52, jonas.thornv...@gmail.com wrote:
 
 
 
 Please stop sending us double spaced crap.
 
 
 
 -- 
 
 Python is the second best programming language in the world.
 
 But the best has yet to be invented.  Christian Tismer
 
 
 
 Mark Lawrence
I am not sure what you want.
You want me to remove the empty line in function? I do it for it is easier to 
read for me.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 11:11:17 UTC+1 skrev 
jonas.t...@gmail.com:
 Den onsdagen den 30:e oktober 2013 kl. 11:08:11 UTC+1 skrev 
 jonas.t...@gmail.com:
 
  Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 
  
 
   jonas.thornv...@gmail.com wrote:
 
  
 
   
 
  
 
   
 
  
 
   
 
  
 
   Why did Python not implement end... The end is really not necessary for
 
  
 
   
 
  
 
   the programming language it can be excluded, but it is a courtesy to
 
  
 
   
 
  
 
   the programmer and could easily be transformed to indents automaticly,
 
  
 
   
 
  
 
   that is removed before the compiliation/interpretation of code.  
 
  
 
   
 
  
 
   
 
  
 
   
 
  
 
   You only say that because your brain has been poisoned by languages that
 
  
 
   
 
  
 
   require some kind of end.  It's not necessary, and it's extra typing. 
   99%
 
  
 
   
 
  
 
   of programmers do the indentation anyway, to make the program easy to 
   read,
 
  
 
   
 
  
 
   so why not just make it part of the syntax?  That way, you don't
 
  
 
   
 
  
 
   accidentally have the indentation not match the syntax.
 
  
 
   
 
  
 
   -- 
 
  
 
   
 
  
 
   Tim Roberts, t...@probo.com
 
  
 
   
 
  
 
   Providenza  Boekelheide, Inc.
 
  
 
  
 
  
 
  Well Tim ***one could argue*** why not do a (i think it is called parser) 
  that react to loop, end and function. And lazy like me do not have to 
  think about what is not part of program.
 
  
 
  
 
  
 
  I certainly do not like the old bracket style it was a catastrophe, but in 
  honesty the gui editor of python should have what i propose, a parser that 
  indent automaticly at loops, functions and end. I promise you it will save 
  millions of hours of bug searching all over world in a month.
 
 
 
 Instead of having going over indent manually, you just drop in an end it is 
 so simple, no marking no meny indent unindent it is automaticly done. And 
 that was the purpose of python to remove idiocies, like brackets and indents.

You could have it in the menu indent parser on off. If on you write out ends, 
and the text is automaticly indented, by using end or a reserved sign. It is 
quite simple i could program it in a day...
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 11:00:30 UTC+1 skrev Mark Lawrence:
 On 30/10/2013 09:52, jonas.thornv...@gmail.com wrote:
 
 
 
 Please stop sending us double spaced crap.
 
 
 
 -- 
 
 Python is the second best programming language in the world.
 
 But the best has yet to be invented.  Christian Tismer
 
 
 
 Mark Lawrence

I think it is not me it is probably google groups, well maybe they should 
consider changing linebreak sign as stored in database.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 11:11:17 UTC+1 skrev 
jonas.t...@gmail.com:
 Den onsdagen den 30:e oktober 2013 kl. 11:08:11 UTC+1 skrev 
 jonas.t...@gmail.com:
 
  Den onsdagen den 30:e oktober 2013 kl. 08:07:31 UTC+1 skrev Tim Roberts:
 
  
 
   jonas.thornv...@gmail.com wrote:
 
  
I suddenly realised i mixed code from a plain square system with the generic 
exponential modulus system.

Here is the code, what it does is encode numbers in an exponential modulus 
base. So it is basicly a numbersystem of my own you do not need to write out + 
^ and exponent because the numbersystem create unique values for every natural 
number.

Well without the + ^ it is hard to read but so are binary, ternary and 
hexadecimal. I think the requirment for a numbersystem is that it have a 
working arithmetic, and mine have.

So here is the working code, to write out exponential modulus number for and 
exponent.

#!/usr/bin/python
import math
# Function definition is here
def sq(number):
   
  exp=1
  factor=2
  multip=math.pow(2,exponent)
  print(x,= , end=)
  while number=multip:
 while exp=number:
factor+=1
exp=math.pow(factor,exponent)
 factor-=1  
 print(factor,^,exponent,+,sep=,end=)
 exp=math.pow(factor,exponent)
 number=number-exp
 exp=1
 factor=1
  print(number)

#Set exponent here  
exponent=2
print(Exp=x^,exponent,sep=)
#Set range of numbers x
for x in range (1,100):
  sq(x) 

 
   
 
  
 
   
 
  
 
   
 
  
 
   Why did Python not implement end... The end is really not necessary for
 
  
 
   
 
  
 
   the programming language it can be excluded, but it is a courtesy to
 
  
 
   
 
  
 
   the programmer and could easily be transformed to indents automaticly,
 
  
 
   
 
  
 
   that is removed before the compiliation/interpretation of code.  
 
  
 
   
 
  
 
   
 
  
 
   
 
  
 
   You only say that because your brain has been poisoned by languages that
 
  
 
   
 
  
 
   require some kind of end.  It's not necessary, and it's extra typing. 
   99%
 
  
 
   
 
  
 
   of programmers do the indentation anyway, to make the program easy to 
   read,
 
  
 
   
 
  
 
   so why not just make it part of the syntax?  That way, you don't
 
  
 
   
 
  
 
   accidentally have the indentation not match the syntax.
 
  
 
   
 
  
 
   -- 
 
  
 
   
 
  
 
   Tim Roberts, t...@probo.com
 
  
 
   
 
  
 
   Providenza  Boekelheide, Inc.
 
  
 
  
 
  
 
  Well Tim ***one could argue*** why not do a (i think it is called parser) 
  that react to loop, end and function. And lazy like me do not have to 
  think about what is not part of program.
 
  
 
  
 
  
 
  I certainly do not like the old bracket style it was a catastrophe, but in 
  honesty the gui editor of python should have what i propose, a parser that 
  indent automaticly at loops, functions and end. I promise you it will save 
  millions of hours of bug searching all over world in a month.
 
 
 
 Instead of having going over indent manually, you just drop in an end it is 
 so simple, no marking no meny indent unindent it is automaticly done. And 
 that was the purpose of python to remove idiocies, like brackets and indents.
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Re: First day beginner to python, add to counter after nested loop

Den onsdagen den 30:e oktober 2013 kl. 15:22:50 UTC+1 skrev Alister:
 On Wed, 30 Oct 2013 13:42:37 +0100, Antoon Pardon wrote:
 
 
 
  Op 30-10-13 13:17, Chris Angelico schreef:
 
  On Wed, Oct 30, 2013 at 11:01 PM, Antoon Pardon
 
  antoon.par...@rece.vub.ac.be wrote:
 
  I broadly agree with your post (I'm of the school of thought that
 
  braces are better than indentation for delimiting blocks), but I don't
 
  think this argument holds water. All you need to do is be consistent
 
  about tabs OR spaces (and I'd recommend tabs, since they're simpler and
 
  safer), and you'll never have this trouble.
 
  
 
  Easier said than done. First of all I can be as consistent as possible,
 
  I can't just take code from someone else and insert it because that
 
  other person may be consistenly doing it different from me.
 
 
 
 I disagree it is very easy.
 
 
 
 1) make sure you editor is set to inset 4 spaces rather than tab when 
 
 pressing the tab key. consistency in your own code is now not an issue.
 
 
 
 2) when importing code from someone else a simple search  replace of tab 
 
 with 4 spaces will instantly correct the formatting on code using tab 
 
 without breaking code that doesn't.
 
 
 
 
 
  
 
  Then if you are working on different machines, the settings of your
 
  editor may not always be the same so that you have tabs on one machine
 
  and spaces on an other, which causes problem when you move the code.
 
  
 
 that is fixed by setting your environment consistantly but step 2 above 
 
 will fix it if you forget.
 
 
 
  Also when you have an xterm, selecting a tab and pasting it into another
 
  it will turn the tab into spaces.
 
 
 
 Read pep 11  always use 4 spaces for indentation not tab.
 
 
 
  
 
  All these things usually can be ignored, they typically only show up
 
  when you print something and things aren't aligned as you expect but
 
  with python you are forced to correct those things immediately, forcing
 
  you to focus on white space layout issues instead of on the logic of the
 
  code.
 
  
 
  Also, the parser should tell you if you mix tabs and spaces, so that
 
  won't trip anything either.
 
  
 
  Maybe you mean something different than I understand but a program
 
  throwing a syntax error because there is a tab instead of a number of
 
  spaces or vice versa, is something I would understand as tripping.
 
 
 
 no more than failing to close a brace in a C like language
 
 indentation is the syntax of python you will grow to love it, like most 
 
 people I found it distracting at first even though i tended to indent 
 
 other code (inconsistently)to make it readable.
 
 
 
 
 
 
 
 
 
 -- 
 
 I am what you will be; I was what you are.

Alister i do not ask for changing the actual implementation with indents that 
the compiler/interpretator work with. What i ask for is some courtesy relative 
the programmers using IDLE, to incorporate a simple automatic parser that let 
them who like to write slopy formatted with end instead to do so. And the 
parser in editor automaticly go in and autoindent *function, loops, if and 
allow end that the editor autoindent to end of loop. It can not be that hard i 
have implemented my own python using this...
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Re: First day beginner to python, add to counter after nested loop