Cryptography-Digest Digest #446, Volume #14 Sat, 26 May 01 04:13:01 EDT
Contents:
Re: OAP-L3: "The absurd weakness." (Anthony Stephen Szopa)
Re: RSA's new Factoring Challenges: $200,000 prize. ("Tapio")
James Felling: Sorry to break your bubble (Anthony Stephen Szopa)
Re: Kernaugh maps (try #2) (Christian Wittenhorst)
----------------------------------------------------------------------------
From: Anthony Stephen Szopa <[EMAIL PROTECTED]>
Crossposted-To: alt.hacker,talk.politics.crypto
Subject: Re: OAP-L3: "The absurd weakness."
Date: Sat, 26 May 2001 00:37:18 -0700
James Felling wrote:
>
> > <sniping my posting>
>
> >
> > You are obviously trying a little too hard to be a little too smooth.
> >
> > If you want to claim that I took your advice and that of others on
> > this point regarding the scramble process, go ahead. But I am sure
> > you will agree that OAP-L3 is basically thoroughly based on
> > permutation manipulation. With this in mind you must agree that I
> > most probably am quite well versed in the simplest most fundamental
> > permutation transformations. If you think your claim of advising me
> > on the scramble process is believable then anyone who believes this
> > must be rather gullible. The transformation of permutations is
> > lesson #1 in linear algebra. But if you say it is so then I have
> > nothing else to say about this.
>
> If you learned your permutation theory in linear algebre, that explains alot.
>
> >
> >
> > There were design choices made. And there is a very good reason I
> > chose 14. I mentioned it earlier. One reason is that the number
> > lends itself to easy implementation and ease of use for a user.
> > Also it provides a certain consistency within the overall program.
>
> OK. I was just curious as to why you a composite number and that specific one.
> Ease of use and programing simplicity are reasonable, I guess.
>
> >
> >
> > "there is a cap upon how much randomness this permutation can
> > contribute" Can you prove this or is this something we should all
> > realize on its face?
>
> This is something that is obvious on its face. Given you have n cards, you
> cannot arange then in more than n! possible ways. thus there is a cap on the
> randomness contributed.
>
> > Seems to me what you are saying is analogous
> > to saying that shuffling a deck of cards by splitting the deck in
> > two halves then interleaving the two halves card by card from each
> > half will certainly limit the outcome of the overall sequence of the
> > deck as compared to some other shuffle.
>
> A said nothing of the sort.
>
> > Are you serious? Are you
> > claiming that there is no way that one can arrive at the same final
> > outcome sequence of a deck of cards if two different methods of
> > shuffling them are used?
>
> Given certian side conditions yes. Ok. you are doing a single shuffle. if you
> grab n cards starting at card m out of the middle and put them on top, there is
> no way that that can be duplicated by a single riffle shuffle. Given more
> shuffles you will eventuall reach all possible arengements of the deck. using
> either method. however. the riffle shuffle will tend to fill the space more
> quickly than the n card block shuffle.
>
> >
> >
> > Your arguments are ridiculous. "It will fill that space fairly
> > slowly" You realize that the space for three MixFiles is about
> > 1E66,000,000 and 1E22,000,000 for one MixFile.
>
> Yes. If you are to have optimally random daty you want to fill that space as
> uniformly and agressively as possible. Like the shuffling example above. The
> more effective your elemental ops are at shuffling the large "decks' being used,
> the faster an arbitrary permutation is reached, and the more uniformly random
> the data is. If I had a huge deck of cards, I would want it shuffeld alot, and
> very effectively, much more so than if I had a smaller one. would you trust a
> deck of a million cards that had been riffle shuffled 7 times.( probably not).
> Would you trust a standard 52 card deck that had been( probably).
>
> >
> >
> > Each final output sequence occupies just one place in this entire
> > space. I claim I can fill one of these spaces faster with a
> > sequence of 14 numbers than you can using a sequence of 105
> > numbers. Yep. While your method has to search for each number in
> > 105, my process just makes two passes counting off in sequence
> > either 1, or 2, or 3 or 4, etc. numbers and copying them to the
> > final output file. Now think about what you just said and you will
> > see that I am right.
>
> I can do it in a single pass. simple and easy assuming I use a generic
> permutation method. I do 1 loop containing 105 assigns. easy.
>
> But that is not the claim I am making. The claim I am making is that your mix a
> mixfile is a poor shuffle over the space it exists in.( 105 unit permutations)
> It has a fixed point, and a tendency toward preserving the order of the first
> few digits that is well worse than chance. It will never reach certian points on
> the space ( It cannot reverse the order of the 105 items no matter how many
> times it is itterated.( for example).) Even if you evaluate it over a reduced
> subset( permutations of 105 items where the first item is fixed and the
> remaining 104 are shuffled arbitrarially it will take an inordinately long time
> filling that space.( It takes a long time to reach certian corners of the space
> -- far longer than the numbers you bandy about would suggest.) Further, points
> near the fixed point tent to remain stable as well the odds of the first n sets
> being unchanged after m randomly chosen shuffles is ((15-n)/14)^m. this means
> that after 16 arbitrary shuffles the odds of the first two elements being in the
> same place are better than 50%. Does this sound like a good primitive to you? I
> am somewhat iffy as to how well it does its job. The fact that there are better
> methods that just beg to be used, methods without fixed points, and without such
> islands of stablity that are simple to code and fix makes me wonder how much
> time you actually put into your design of this program.
>
> >
> >
> > "and unevenly"? Not so. If the user inputs a random 14 number
> > sequence then each random 14 number sequence has an equal chance
> > of being input as any other 14 number sequence.
>
> true, but that does not mean that the permutation fills the space of possible
> results evenly. There are arangements of data that are more likely( like the
> first two items being unchanged) that happen with far greater likelyhood than in
> a better arangment.
>
> >
> >
> > "This means that certian permutative structures will be more common
> > thatn others."
>
> yep.
>
> > If the user inputs a random 14 number sequence,
> > after one running of the process there will be 14! = 87,000,000,000
> > possible outcomes. Run the process twice using another random 14
> > number sequence the second time and there will be (14!)^2 possible
> > outcomes. Are you saying that as this process is repeated one per
> > second in the time remaining in your lifetime that the number of
> > possible outcomes will be less than (14!)^(years remaining in your
> > life * 365 * 24 * 60 * 60)?
>
> Yes. it will infact be less than (105!). Further I claim that you should know
> this if you actually understand what you are doing.
>
> >
> >
> > And are you saying that you know of a method to somehow determine
> > what the number of times this process was run and what the random
> > sequences used were in each running of the process by just examining
> > the final sequence output?
>
> No
>
> >
> >
> > Or are you simply saying that you can predict how many times the
> > process is run and what the input is and what the final output
> > sequence of the MixFile will be before the user even generates the
> > random number sequences and runs the processes?
>
> No I am claiming that your method is suboptimal. That there are better ways of
> doing things than those you have chosen
>
> >
> >
> > You ARE really good:
> >
> > "the methods used have slow difusion, require alot of work and have
> > weaknesses such as fixed points, poor mixing, and other such flaws."
> >
> > You must admit that all of these statements are generalities which
> > completely miss the point of the purpose of OAP-L3. Primarily you
> > have not quantified any of these points. And this is crucial in any
> > serious critique of OAP-L3. I believe that you have intentionally
> > neglected to mention these because your intention is to look good
> > while feebly attempting to trash OAP-L3. To have flagrantly ignored
> > any quantification to support your statements proves your intentions
> > beyond any doubt.
>
> Yep. my weak mind batters febly against your wonderful algorithim. Maybe you
> should take a look at what is being said, and not smoke screen us all.
>
> >
> >
> > Your flaws are figments of your imagination. Do you refute that
> > running the process results in the possible outcomes such as: (14!)^
> > (number of times the process is run using random input)?
>
> Yes.
>
> > And if not
> > is it significant?
>
> Not terribly, but why should I accept a compromise vs. other methods that do mix
> efficiently?
>
> > And why do you not mention the variety of
> > processes available to a user. This one process is not recommended
> > to be used exclusively. The software is to be used in its entirety.
>
> Of course. You asked me to point out what methods and means are flawed, and now
> that I have pointed at one you tell me I am being too narrow in my view of your
> program. make up your mind sir.
>
> >
> >
> > Here is my last shot at a fixed target. "So how would YOU fix that
> > nasty fixed point issue?" (really pathetic innuendo: "nasty.")
>
> Ok, so the official position is that is a flaw, patch around it and dont worry
> about it. Acceptable, but, maybe fixing it inpalce would be good.
>
> >
>
> >
> >
> > I just told you. You use a random combination of the several other
> > methods.
> >
> > Thanks for revealing some of what you are. Don't take that personal.
> > Much of what we are is not of our own doing.
> >
> > You may confound some with your leaky arguments but I guess we get
> > what we pay for when we read one of your posts.
>
> I feel the same about your postings sir.
Sorry to break your bubble
Tell me, do you have very many stupid people who pay you money for
expounding such logic as you have demonstrated in your past three
or four posts? Wait! Don't answer that right now. First read the
following.
As you will see from looking at the first 105 permutations that the
first 5 digits are: 0 1 2 3 4. No matter how many times you run
your 105! process these first five digits of the group of 105
permutations will always be the same.
Now if you are aware of the way OAP-L3 works, you will know that
this will result in very very poor random digit output: basically
unusable.
You will also find further redundancy in the sixth digit. Over a
group of just 105 permutations this is unacceptable with OAP-L3.
You can also find redundancy in other digit locations both singly
and in smaller groups of two and three digits. These last
redundancies by themselves are not significant but when combined
with the redundancies in the first 5 or 6 digits in such a small group
as 105 permutations they do add to the overall redundancy of
your suggested "fix." Your process will simply never work
with OAP-L3.
Now what do you have to say?
The only way out of this predicament is to increase the number from
105 to 210 permutations but you will find the same problem. Only
when you start to approach a significant percentage of the entire
3,628,800 possible permutations will you be able to even begin to
approach any usable random digit output.
Thus your recommended "fix" is practicably unusable. What user is
going to generate perhaps a million true random numbers to gain
1,000,000! and key them all in? If they could do this then why not
just use the random numbers themselves and forget about OAP-L3?
I could pursue this analysis further but your suggestion of my
"design flaws" has turned out to be nothing more than your myth and
your suggestion to correct these fantastic "flaws" has turned out to
be a fraud.
I sure hope everyone is listening.
Array number: 0 0 1 2 3 4 5 6 7 8 9
Array number: 1 0 1 2 3 4 5 6 7 9 8
Array number: 2 0 1 2 3 4 5 6 8 7 9
Array number: 3 0 1 2 3 4 5 6 8 9 7
Array number: 4 0 1 2 3 4 5 6 9 7 8
Array number: 5 0 1 2 3 4 5 6 9 8 7
Array number: 6 0 1 2 3 4 5 7 6 8 9
Array number: 7 0 1 2 3 4 5 7 6 9 8
Array number: 8 0 1 2 3 4 5 7 8 6 9
Array number: 9 0 1 2 3 4 5 7 8 9 6
Array number: 10 0 1 2 3 4 5 7 9 6 8
Array number: 11 0 1 2 3 4 5 7 9 8 6
Array number: 12 0 1 2 3 4 5 8 6 7 9
Array number: 13 0 1 2 3 4 5 8 6 9 7
Array number: 14 0 1 2 3 4 5 8 7 6 9
Array number: 15 0 1 2 3 4 5 8 7 9 6
Array number: 16 0 1 2 3 4 5 8 9 6 7
Array number: 17 0 1 2 3 4 5 8 9 7 6
Array number: 18 0 1 2 3 4 5 9 6 7 8
Array number: 19 0 1 2 3 4 5 9 6 8 7
Array number: 20 0 1 2 3 4 5 9 7 6 8
Array number: 21 0 1 2 3 4 5 9 7 8 6
Array number: 22 0 1 2 3 4 5 9 8 6 7
Array number: 23 0 1 2 3 4 5 9 8 7 6
Array number: 24 0 1 2 3 4 6 5 7 8 9
Array number: 25 0 1 2 3 4 6 5 7 9 8
Array number: 26 0 1 2 3 4 6 5 8 7 9
Array number: 27 0 1 2 3 4 6 5 8 9 7
Array number: 28 0 1 2 3 4 6 5 9 7 8
Array number: 29 0 1 2 3 4 6 5 9 8 7
Array number: 30 0 1 2 3 4 6 7 5 8 9
Array number: 31 0 1 2 3 4 6 7 5 9 8
Array number: 32 0 1 2 3 4 6 7 8 5 9
Array number: 33 0 1 2 3 4 6 7 8 9 5
Array number: 34 0 1 2 3 4 6 7 9 5 8
Array number: 35 0 1 2 3 4 6 7 9 8 5
Array number: 36 0 1 2 3 4 6 8 5 7 9
Array number: 37 0 1 2 3 4 6 8 5 9 7
Array number: 38 0 1 2 3 4 6 8 7 5 9
Array number: 39 0 1 2 3 4 6 8 7 9 5
Array number: 40 0 1 2 3 4 6 8 9 5 7
Array number: 41 0 1 2 3 4 6 8 9 7 5
Array number: 42 0 1 2 3 4 6 9 5 7 8
Array number: 43 0 1 2 3 4 6 9 5 8 7
Array number: 44 0 1 2 3 4 6 9 7 5 8
Array number: 45 0 1 2 3 4 6 9 7 8 5
Array number: 46 0 1 2 3 4 6 9 8 5 7
Array number: 47 0 1 2 3 4 6 9 8 7 5
Array number: 48 0 1 2 3 4 7 5 6 8 9
Array number: 49 0 1 2 3 4 7 5 6 9 8
Array number: 50 0 1 2 3 4 7 5 8 6 9
Array number: 51 0 1 2 3 4 7 5 8 9 6
Array number: 52 0 1 2 3 4 7 5 9 6 8
Array number: 53 0 1 2 3 4 7 5 9 8 6
Array number: 54 0 1 2 3 4 7 6 5 8 9
Array number: 55 0 1 2 3 4 7 6 5 9 8
Array number: 56 0 1 2 3 4 7 6 8 5 9
Array number: 57 0 1 2 3 4 7 6 8 9 5
Array number: 58 0 1 2 3 4 7 6 9 5 8
Array number: 59 0 1 2 3 4 7 6 9 8 5
Array number: 60 0 1 2 3 4 7 8 5 6 9
Array number: 61 0 1 2 3 4 7 8 5 9 6
Array number: 62 0 1 2 3 4 7 8 6 5 9
Array number: 63 0 1 2 3 4 7 8 6 9 5
Array number: 64 0 1 2 3 4 7 8 9 5 6
Array number: 65 0 1 2 3 4 7 8 9 6 5
Array number: 66 0 1 2 3 4 7 9 5 6 8
Array number: 67 0 1 2 3 4 7 9 5 8 6
Array number: 68 0 1 2 3 4 7 9 6 5 8
Array number: 69 0 1 2 3 4 7 9 6 8 5
Array number: 70 0 1 2 3 4 7 9 8 5 6
Array number: 71 0 1 2 3 4 7 9 8 6 5
Array number: 72 0 1 2 3 4 8 5 6 7 9
Array number: 73 0 1 2 3 4 8 5 6 9 7
Array number: 74 0 1 2 3 4 8 5 7 6 9
Array number: 75 0 1 2 3 4 8 5 7 9 6
Array number: 76 0 1 2 3 4 8 5 9 6 7
Array number: 77 0 1 2 3 4 8 5 9 7 6
Array number: 78 0 1 2 3 4 8 6 5 7 9
Array number: 79 0 1 2 3 4 8 6 5 9 7
Array number: 80 0 1 2 3 4 8 6 7 5 9
Array number: 81 0 1 2 3 4 8 6 7 9 5
Array number: 82 0 1 2 3 4 8 6 9 5 7
Array number: 83 0 1 2 3 4 8 6 9 7 5
Array number: 84 0 1 2 3 4 8 7 5 6 9
Array number: 85 0 1 2 3 4 8 7 5 9 6
Array number: 86 0 1 2 3 4 8 7 6 5 9
Array number: 87 0 1 2 3 4 8 7 6 9 5
Array number: 88 0 1 2 3 4 8 7 9 5 6
Array number: 89 0 1 2 3 4 8 7 9 6 5
Array number: 90 0 1 2 3 4 8 9 5 6 7
Array number: 91 0 1 2 3 4 8 9 5 7 6
Array number: 92 0 1 2 3 4 8 9 6 5 7
Array number: 93 0 1 2 3 4 8 9 6 7 5
Array number: 94 0 1 2 3 4 8 9 7 5 6
Array number: 95 0 1 2 3 4 8 9 7 6 5
Array number: 96 0 1 2 3 4 9 5 6 7 8
Array number: 97 0 1 2 3 4 9 5 6 8 7
Array number: 98 0 1 2 3 4 9 5 7 6 8
Array number: 99 0 1 2 3 4 9 5 7 8 6
Array number: 100 0 1 2 3 4 9 5 8 6 7
Array number: 101 0 1 2 3 4 9 5 8 7 6
Array number: 102 0 1 2 3 4 9 6 5 7 8
Array number: 103 0 1 2 3 4 9 6 5 8 7
Array number: 104 0 1 2 3 4 9 6 7 5 8
Tell me, do you have very many stupid people who pay you money for
expounding such logic as you have demonstrated in your past three
or four posts? Wha'da ya say?
If you have two encryption software methods I suggest you consider a
corollary of Occum's Razor: choose the one that is simplest and
easiest to understand in its entirety over the one you cannot
comprehend with certainty to the fullest.
(Damn, that felt good. Maybe I'm ready for Bruce?)
------------------------------
From: "Tapio" <[EMAIL PROTECTED]>
Crossposted-To: sci.math
Subject: Re: RSA's new Factoring Challenges: $200,000 prize.
Date: Sat, 26 May 2001 10:41:16 +0300
"Steve Lord" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
>
> On 24 May 2001, SCOTT19U.ZIP_GUY wrote:
>
> > [EMAIL PROTECTED] (Peter Trei) wrote in
> > <[EMAIL PROTECTED]>:
> >
> > >RSA Security, has revamped its Factoring Challenges.
> They request a "Description of Effort". I think it would be possible to
> submit a factorization without mentioning how you obtained it, but they
> would be curious. In addition, if someone in the NSA _did_ violate their
> NDA, then as soon as the name of the winner was released, they would be in
> DEEP trouble and would most likely lose their job. I don't quite think
> it's worth it.
So:
1) Ref. and specify exactly RSA Security problem so they know what you are
taking about
2) Publish your solutions here in sci.math.
3) If they are really interested in your solution they pay you more than
promised. It becomes cheaper.... ;-)
"Anyway the wind blows" (Bohemian...)
Tapio
> Steve L
>
>
------------------------------
From: Anthony Stephen Szopa <[EMAIL PROTECTED]>
Crossposted-To: alt.hacker,talk.politics.crypto
Subject: James Felling: Sorry to break your bubble
Date: Sat, 26 May 2001 00:42:58 -0700
James Felling: Sorry to break your bubble
Reference: OAP-L3: "The absurd weakness."
Tell me, do you have very many stupid people who pay you money for
expounding such logic as you have demonstrated in your past three
or four posts? Wait! Don't answer that right now. First read the
following.
As you will see from looking at the first 105 permutations that the
first 5 digits are: 0 1 2 3 4. No matter how many times you run
your 105! process these first five digits of the group of 105
permutations will always be the same.
Now if you are aware of the way OAP-L3 works, you will know that
this will result in very very poor random digit output: basically
unusable.
You will also find further redundancy in the sixth digit. Over a
group of just 105 permutations this is unacceptable with OAP-L3.
You can also find redundancy in other digit locations both singly
and in smaller groups of two and three digits. These last
redundancies by themselves are not significant but when combined
with the redundancies in the first 5 or 6 digits in such a small
group as 105 permutations they do add to the overall redundancy of
your suggested "fix." Your process will simply never work
with OAP-L3.
Now what do you have to say?
The only way out of this predicament is to increase the number from
105 to 210 permutations but you will find the same problem. Only
when you start to approach a significant percentage of the entire
3,628,800 possible permutations will you be able to even begin to
approach any usable random digit output.
Thus your recommended "fix" is practicably unusable. What user is
going to generate perhaps a million true random numbers to gain
1,000,000! and key them all in? If they could do this then why not
just use the random numbers themselves and forget about OAP-L3?
I could pursue this analysis further but your suggestion of my
"design flaws" has turned out to be nothing more than your myth and
your suggestion to correct these fantastic "flaws" has turned out to
be a fraud.
I sure hope everyone is listening.
Array number: 0 0 1 2 3 4 5 6 7 8 9
Array number: 1 0 1 2 3 4 5 6 7 9 8
Array number: 2 0 1 2 3 4 5 6 8 7 9
Array number: 3 0 1 2 3 4 5 6 8 9 7
Array number: 4 0 1 2 3 4 5 6 9 7 8
Array number: 5 0 1 2 3 4 5 6 9 8 7
Array number: 6 0 1 2 3 4 5 7 6 8 9
Array number: 7 0 1 2 3 4 5 7 6 9 8
Array number: 8 0 1 2 3 4 5 7 8 6 9
Array number: 9 0 1 2 3 4 5 7 8 9 6
Array number: 10 0 1 2 3 4 5 7 9 6 8
Array number: 11 0 1 2 3 4 5 7 9 8 6
Array number: 12 0 1 2 3 4 5 8 6 7 9
Array number: 13 0 1 2 3 4 5 8 6 9 7
Array number: 14 0 1 2 3 4 5 8 7 6 9
Array number: 15 0 1 2 3 4 5 8 7 9 6
Array number: 16 0 1 2 3 4 5 8 9 6 7
Array number: 17 0 1 2 3 4 5 8 9 7 6
Array number: 18 0 1 2 3 4 5 9 6 7 8
Array number: 19 0 1 2 3 4 5 9 6 8 7
Array number: 20 0 1 2 3 4 5 9 7 6 8
Array number: 21 0 1 2 3 4 5 9 7 8 6
Array number: 22 0 1 2 3 4 5 9 8 6 7
Array number: 23 0 1 2 3 4 5 9 8 7 6
Array number: 24 0 1 2 3 4 6 5 7 8 9
Array number: 25 0 1 2 3 4 6 5 7 9 8
Array number: 26 0 1 2 3 4 6 5 8 7 9
Array number: 27 0 1 2 3 4 6 5 8 9 7
Array number: 28 0 1 2 3 4 6 5 9 7 8
Array number: 29 0 1 2 3 4 6 5 9 8 7
Array number: 30 0 1 2 3 4 6 7 5 8 9
Array number: 31 0 1 2 3 4 6 7 5 9 8
Array number: 32 0 1 2 3 4 6 7 8 5 9
Array number: 33 0 1 2 3 4 6 7 8 9 5
Array number: 34 0 1 2 3 4 6 7 9 5 8
Array number: 35 0 1 2 3 4 6 7 9 8 5
Array number: 36 0 1 2 3 4 6 8 5 7 9
Array number: 37 0 1 2 3 4 6 8 5 9 7
Array number: 38 0 1 2 3 4 6 8 7 5 9
Array number: 39 0 1 2 3 4 6 8 7 9 5
Array number: 40 0 1 2 3 4 6 8 9 5 7
Array number: 41 0 1 2 3 4 6 8 9 7 5
Array number: 42 0 1 2 3 4 6 9 5 7 8
Array number: 43 0 1 2 3 4 6 9 5 8 7
Array number: 44 0 1 2 3 4 6 9 7 5 8
Array number: 45 0 1 2 3 4 6 9 7 8 5
Array number: 46 0 1 2 3 4 6 9 8 5 7
Array number: 47 0 1 2 3 4 6 9 8 7 5
Array number: 48 0 1 2 3 4 7 5 6 8 9
Array number: 49 0 1 2 3 4 7 5 6 9 8
Array number: 50 0 1 2 3 4 7 5 8 6 9
Array number: 51 0 1 2 3 4 7 5 8 9 6
Array number: 52 0 1 2 3 4 7 5 9 6 8
Array number: 53 0 1 2 3 4 7 5 9 8 6
Array number: 54 0 1 2 3 4 7 6 5 8 9
Array number: 55 0 1 2 3 4 7 6 5 9 8
Array number: 56 0 1 2 3 4 7 6 8 5 9
Array number: 57 0 1 2 3 4 7 6 8 9 5
Array number: 58 0 1 2 3 4 7 6 9 5 8
Array number: 59 0 1 2 3 4 7 6 9 8 5
Array number: 60 0 1 2 3 4 7 8 5 6 9
Array number: 61 0 1 2 3 4 7 8 5 9 6
Array number: 62 0 1 2 3 4 7 8 6 5 9
Array number: 63 0 1 2 3 4 7 8 6 9 5
Array number: 64 0 1 2 3 4 7 8 9 5 6
Array number: 65 0 1 2 3 4 7 8 9 6 5
Array number: 66 0 1 2 3 4 7 9 5 6 8
Array number: 67 0 1 2 3 4 7 9 5 8 6
Array number: 68 0 1 2 3 4 7 9 6 5 8
Array number: 69 0 1 2 3 4 7 9 6 8 5
Array number: 70 0 1 2 3 4 7 9 8 5 6
Array number: 71 0 1 2 3 4 7 9 8 6 5
Array number: 72 0 1 2 3 4 8 5 6 7 9
Array number: 73 0 1 2 3 4 8 5 6 9 7
Array number: 74 0 1 2 3 4 8 5 7 6 9
Array number: 75 0 1 2 3 4 8 5 7 9 6
Array number: 76 0 1 2 3 4 8 5 9 6 7
Array number: 77 0 1 2 3 4 8 5 9 7 6
Array number: 78 0 1 2 3 4 8 6 5 7 9
Array number: 79 0 1 2 3 4 8 6 5 9 7
Array number: 80 0 1 2 3 4 8 6 7 5 9
Array number: 81 0 1 2 3 4 8 6 7 9 5
Array number: 82 0 1 2 3 4 8 6 9 5 7
Array number: 83 0 1 2 3 4 8 6 9 7 5
Array number: 84 0 1 2 3 4 8 7 5 6 9
Array number: 85 0 1 2 3 4 8 7 5 9 6
Array number: 86 0 1 2 3 4 8 7 6 5 9
Array number: 87 0 1 2 3 4 8 7 6 9 5
Array number: 88 0 1 2 3 4 8 7 9 5 6
Array number: 89 0 1 2 3 4 8 7 9 6 5
Array number: 90 0 1 2 3 4 8 9 5 6 7
Array number: 91 0 1 2 3 4 8 9 5 7 6
Array number: 92 0 1 2 3 4 8 9 6 5 7
Array number: 93 0 1 2 3 4 8 9 6 7 5
Array number: 94 0 1 2 3 4 8 9 7 5 6
Array number: 95 0 1 2 3 4 8 9 7 6 5
Array number: 96 0 1 2 3 4 9 5 6 7 8
Array number: 97 0 1 2 3 4 9 5 6 8 7
Array number: 98 0 1 2 3 4 9 5 7 6 8
Array number: 99 0 1 2 3 4 9 5 7 8 6
Array number: 100 0 1 2 3 4 9 5 8 6 7
Array number: 101 0 1 2 3 4 9 5 8 7 6
Array number: 102 0 1 2 3 4 9 6 5 7 8
Array number: 103 0 1 2 3 4 9 6 5 8 7
Array number: 104 0 1 2 3 4 9 6 7 5 8
Tell me, do you have very many stupid people who pay you money for
expounding such logic as you have demonstrated in your past three
or four posts? Wha'da ya say?
If you have two encryption software methods I suggest you consider a
corollary of Occum's Razor: choose the one that is simplest and
easiest to understand in its entirety over the one you cannot
comprehend with certainty to the fullest.
(Damn, that felt good. Maybe I'm ready for Bruce?)
------------------------------
From: Christian Wittenhorst <[EMAIL PROTECTED]>
Subject: Re: Kernaugh maps (try #2)
Date: Sat, 26 May 2001 10:07:34 +0200
On Wed, 16 May 2001 23:38:31 -0400, "Jeffrey Walton"
<[EMAIL PROTECTED]> wrote:
>So, my final equation would be f = a'c + a'b'd' + ab'c + ab'd
There is simpler form: f = a'c + a'b'd' + b'c + ab'd
(The two 1's at the right bottom can be joined with the the two 1's at
the left bottom to form a 2x2 group)
Yours,
wiwi
------------------------------
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