Rational precision proposal (Raul Miller)
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Message: 1
Date: Sun, 15 May 2022 16:12:42 +0200
From: Jan-Pieter Jacobs
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] newbie question
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actually, J lets you approximate p
81828459045235360287471352662497757147554933
> NB. ^
>
> NB. view in fixed width please.
>
> https://www.jsoftware.com/help/dictionary/dx009.htm
>
>
> > Date: Sun, 15 May 2022 10:28:30 +0200
> > From: yt
>
iew in fixed width please.
https://www.jsoftware.com/help/dictionary/dx009.htm
Date: Sun, 15 May 2022 10:28:30 +0200
From: yt
To:programm...@jsoftware.com
Subject: Re: [Jprogramming] newbie question
Message-ID:
Content-Type: text/plain; charset=UTF-8; format=flowed
Dear All,
i come back
I actually made a video for Pi day showing this technique and the various
adjustments you need to make to get pi in an extended form from its original
float representation.
https://youtu.be/vyILnD0e2IE
Cheers, bob
> On May 15, 2022, at 07:12, Jan-Pieter Jacobs
> wrote:
>
> actually, J lets
actually, J lets you approximate pi to as many digits as you'd want, but
you have to know the trick:
pix =. ([: <.@:o. 10^x:)
_10 ]\ ": pix 100
3141592653
5897932384
6264338327
9502884197
1693993751
0582097494
4592307816
4062862089
9862803482
5342117067
9
gives the first 101 digits of pi (inclu
J has a few different ways of representing numbers, including:
- Machine integers (limited to 64 or 32 bit); this is what you get if you type
a number like 123
- Machine floating-point numbers (limited precision, but can have decimal
points in them); this is what you get if you type a number l
Dear All,
i come back to start in J
in jijx>tour>overview
3p5 NB. Pi (3 * Pi ^ 5)
918.059
may be it is not the better way to find Pi :
1p1 NB. Pi (1 * Pi ^ 1)
1p1
3.14159
1p1x
|ill-formed number
| 1p1x
| ^
but this is ok for :
^/ 2 3 4
Good luck.
And, not also that (z);t 4 should not behave any different from z;t 4
Take care,
--
Raul
On Sat, Oct 16, 2021 at 10:48 AM Richard Donovan wrote:
>
> Thanks for a quick reply Raul. I have printed off the parsing rules for later
> study and I use your workaround!
>
> Richard
>
> >
Thanks for a quick reply Raul. I have printed off the parsing rules for later
study and I use your workaround!
Richard
> On 16 Oct 2021, at 15:30, Raul Miller wrote:
>
> What you are seeing here is that J needs to understand what names mean
> before it can act on those names.
>
> A concise
What you are seeing here is that J needs to understand what names mean
before it can act on those names.
A concise summary of the issues is visible here:
https://www.jsoftware.com/help/dictionary/dicte.htm
In 'z; t 4' the evaluation of t 4 corresponds to the second line of
the parsing table which
Hi. I write a function t, execute it, then execute it again.
On the second execution, I want to display the intermediate result then the
final result
I have coded it as below
The intermediate result is displayed as the intermediate result of the FIRST
execution
I was surprised as I though
At long last -
I've only just managed to find some further solutions!
I generate triples of P5s,
such that P5(a) + P5(b) = P5(c) for triple (a,b,c)
and check for which of them P5(d) = P5(a) + P5(c) for an integer d:
|:Ptriples 30
4 7 12 19
7 23 22 22
8 24 25 29
Here are some timings on my la
Larger pentagonal numbers are further apart.
Here, after 1827553 pentagonal numbers, they're too far apart for adjacent
numbers to be closer together.
(Note that this also limits the search space--though being slightly liberal
with bounds and testing blocks at a time might work to keep interprete
Yes, Pascal found the same two pentagonal numbers I found:
*pn=.3 :'y*(1-~3*y)%2'*
* p=:pn>:i.5000x*
Paschal's solution: P(2166)-P(1019)
*2166 1019{p*
*7042750 1560090*
My solution:
*]n=.m#p2*
*1560090 7042750*
The question is, do these two pentagonal numbers have the smallest D?
Skip
Quick correction: the answer given by Pascal is not 1019 but P(2166)-P(1019),
and my email should use that instead of 1019 wherever it is mentioned.
Sorry for the noise,
Louis
> On 14 Jul 2019, at 01:58, Louis de Forcrand wrote:
>
> I’m with Skip here: how do y’all guarantee that your brute-fo
I’m with Skip here: how do y’all guarantee that your brute-force answers (which
search through a list of the first few pentagonal numbers) actually return the
pair with the smallest _difference_? How do you know there isn’t a pair outside
your search range, where each number in the pair is much
Another approach to Euler 44:
NB. make a pentagonal number generator verb:
*pn=.3 :'y*(1-~3*y)%2'*
NB. Generate the first 5000 pentagonal numbers and store them in p. Then
find all possible pair combinations of the first 5000 pentagonal numbers
and store the pairs in p2. Then store the two inte
Ahhh, the anti-base is clever. Now I get it.
3 4 #: 0 6 7
0 0
1 2
1 3
While the spare-array trick is great, I think this answer resonates as more
in keeping with an array-oriented 'style' that I'm trying to internalize.
It also succinctly does what I was trying to do with modulus/division hack
Sorry, that tacit form got butchered in copy/pasting through different
mediums and is actually
(|@-)/@(0&{)@(a {~ (b {~ I.@(a e.~ (|@:-)/"1@:(a {~ b=:4&$.@$.@(a e.~ (
wrote:
> Mike, thanks for thinking like I had a analytic insight, but I made a
> mistake.
>
> In my defense, I did know to divide
Mike, thanks for thinking like I had a analytic insight, but I made a
mistake.
In my defense, I did know to divide by 2, but at the time I was exploring
what forms of verbs could automatically be inverted (obverted?) to create a
possible "is_pentagonal" predicate and accidentally copied my truncat
Trying out Pascal Jasmin’s brute-force approach (beware his spoiler!) runs out
of memory on this iPad.
Some of the solvers’ forum messages, from 2005 and later, boast of times as
low as 0.2sec (Delphi)
I’ve just worked up something akin to Oleg Kolobchenko’s approach; it takes
about 12 sec on
It looks as if Daniel was solving the problem for pentagonal numbers * 2.
Clearly, if there’s a solution in true Pn for a pair of indices, m > n,
ie, (-/ Pn m,n) = +/ Pn m,n,
it will also be a solution for 2 * Pn .
I suppose there might be some half-integral solutions for 2 * Pn which don’t
your definition of pentagonal doesn't do the divide by 2 step
pentagonal =: [ -:@* ( 1 -~ 3 * ])
this combines (,:) the +/ and -/ tables into 2, then finds intersection (*./).
4$.$. returns index(es)
4 $. $. *./@:(e.~ -/~ ,: +/~) pentagonal >: i.20115
2166 1019
On Thursday, July 11, 201
Sorry - I see Louis has just posted a very similar comment, but I'll
persist!...
I usually use something more basic and more boring!
The idea is that you use the shape of your input as the basis for a
representation of the indices of the ravelled version of the array.
Here goes:
] toy =: 3 4
Hi Daniel,
If the ravel indices into the array M are stored in I, the multi-dimensional
indices are then
($M) #: I
So for your toy example a possible verb is
v=: $ #: , I.@:= _1:
Cheers,
Louis
> On 12 Jul 2019, at 02:11, Daniel Eklund wrote:
>
> Double thanks.
>
> One for that sparse arra
Double thanks.
One for that sparse array trick. That will go into my toolbox.
And secondly, for the work you've been putting into the youtube videos -- I
think they've been an invaluable teaching aid, and a welcome addition to
the relative paucity of J learning outside of the main website.
On
Also, I am not sure that I would call it an anti-pattern but you can simplify
your general example by using Compose (&)
( pentagonal 20+i.6) +/ pentagonal >: i. 5
1182 1190 1204 1224 1250
1304 1312 1326 1346 1372
1432 1440 1454 1474 1500
1566 1574 1588 1608 1634
1706 1714 1728 1748 1774
1852
Wow Daniel,
I am sincerely impressed at how you wrestled that one to the ground.
A trick I learned a while ago on these forums was the use of Sparse ($.)
toy e. _1
1 0 0 0
0 0 1 1
0 0 0 0
$. toy e. _1 NB. converts dense array to sparse form
0 0 │ 1
1 2 │ 1
1 3 │ 1
4 $. $. toy e. _1 NB.
Hi everyone,
I’m looking for some newbie help. I feel I’ve come so far but I’ve run
into something that is making me think I’m not really getting something
fundamental.
Rather than try to come up with a contrived example, I’ll just say outright
that I’m trying to solve one of the project Euler q
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