## Re: [PrimeNumbers] Re: Boolean algebra solver?

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• ... Unfortunately, people have been trying this approach, unsuccessfully so far, for quite a few years. If you make progress, it will be *very* interesting!
Message 1 of 13 , May 31, 2004
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On Mon, 2004-05-31 at 12:49, Ron wrote:

> Ha - that would be really neat Décio, if I had actually
> discovered a new or novel method of doing something, but
> considering my lack of background in math, the technique is
> probably as old as the hills and I just don't know the correct
> name for it.

Unfortunately, people have been trying this approach, unsuccessfully so
far, for quite a few years. If you make progress, it will be *very*
interesting!

> I guess what I've done is to transform the factoring problem into
> a boolean satisfiability problem (SAT), but I thought factoring
> was already known (or at least widely believed to be) a very
> difficult problem - I guess "NP-Hard" is the way to describe it(?)
> Since it should be trivial to transform the SAT equations (if
> that's what they are) back into the product of two binary
> polynomials, I would assume the transformation of these particular
> equations is "isomorphic," if I understand the meaning of that
> term correctly.

You've shown that factoring is no harder than SAT, something that is
moderately well-known.

Factoring is clearly in NP. If you guess the factors of an integer, a
polynomial time algorithm (multiplication takes (log n)^2 or better)
will verify the guess. SAT is also in NP.

What is not yet known is what complexity class factoring belongs to.
It's possible it may be NP-complete or NP-hard. It's possible that it's
in P. Until very recently primality testing wasn't known to be in P.

Paul
• ... Hash: SHA1 I see. Thank you Paul, as well as Décio for saving me countless hours of unnecessarily wasted effort. As I understand it then, the real problem
Message 2 of 13 , Jun 1, 2004
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--- In primenumbers@yahoogroups.com, Paul Leyland <pcl@w...> wrote:
> What is not yet known is what complexity class factoring
> belongs to.

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I see. Thank you Paul, as well as Décio for saving me countless
hours of unnecessarily wasted effort. As I understand it then, the
real problem in this particular case would be to "reverse" the
equations I mentioned so that the a's and b's are solved for in
terms of the bits of the composite number to be factored rather
than the way I show them? In other words, what I need (for
example) is to convert equations like these:

n1 = a1 ^ 1
n2 = a1 ^ b1
(Where n1 and n2 are binary bits of the composite number to be
factored)

Into something like this:

a1 = n1 ^ 1
b1 = n1 ^ n2 ^ 1

I realize this is probably impossible in general, but if it *were*
possible it would show that factoring is in P, right?

Thanks,

Ron

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• Sorry, what is the purpose of n1 = a1 ^ 1 and a1 = n1 ^ 1 ? Aren t these just n1 = a1 and a1 = n1 ? It would be helpful to explain these, or have
Message 3 of 13 , Jun 2, 2004
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Sorry, what is the purpose of n1 = a1 ^ 1 and a1 = n1 ^ 1 ?
Aren't these just n1 = a1 and a1 = n1 ?

thought out more before posting.

Thanks.

Milton L. Brown
miltbrown@...

----- Original Message -----
From: "Ron" <Yaho6Hb3c@...>
Sent: Tuesday, June 01, 2004 8:13 PM
Subject: [PrimeNumbers] Re: Boolean algebra solver?

--- In primenumbers@yahoogroups.com, Paul Leyland <pcl@w...> wrote:
> What is not yet known is what complexity class factoring
> belongs to.

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I see. Thank you Paul, as well as Décio for saving me countless
hours of unnecessarily wasted effort. As I understand it then, the
real problem in this particular case would be to "reverse" the
equations I mentioned so that the a's and b's are solved for in
terms of the bits of the composite number to be factored rather
than the way I show them? In other words, what I need (for
example) is to convert equations like these:

n1 = a1 ^ 1
n2 = a1 ^ b1
(Where n1 and n2 are binary bits of the composite number to be
factored)

Into something like this:

a1 = n1 ^ 1
b1 = n1 ^ n2 ^ 1

I realize this is probably impossible in general, but if it *were*
possible it would show that factoring is in P, right?

Thanks,

Ron

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• ... No they are not. When he uses the ^ operator, he means the xor (exclusive or) operator. If you don t know about the xor operator, here s a quick tut: 0^0
Message 4 of 13 , Jun 2, 2004
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Milton Brown wrote:
>
> Sorry, what is the purpose of n1 = a1 ^ 1 and a1 = n1 ^ 1 ?
> Aren't these just n1 = a1 and a1 = n1 ?

No they are not. When he uses the ^ operator, he means the xor (exclusive
or) operator. If you don't know about the xor operator, here's a quick
tut:
0^0 = 0
0^1 = 1
1^0 = 1
1^1 = 0
So, what n1 = a1 ^ 1 basically means is if a1 is 0 then n1 is 1, and vice
versa. HTH.

-David C.

>
> It would be helpful to explain these, or have your mathematics
> thought out more before posting.
>
> Thanks.
>
> Milton L. Brown
• ... Hash: SHA1 Justin asked me for a real life example of how this technique can be used to find factors, so at the risk of boring everyone else, I thought I d
Message 5 of 13 , Jun 3, 2004
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Justin asked me for a real life example of how this technique
can be used to find factors, so at the risk of boring everyone
else, I thought I'd post it here on the forum. Here is a sample
Maple code snippet ('#' begins a comment) that illustrates the
problem:

n:= [0,0,1,1,1,1]: # factor binary 001111 = decimal 15 = 3*5

c1:=a1*b1:
c2:=(a2+a1*b1+b2)*c1+1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1):
c3:=(a2+a1*b1+b2)*c1*(1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1)):
c4:=1+(a2*b1*a1*b2+1)*(a2*b1*c2+1)*(a1*b2*c2+1):
c5:=1+(a2*b2*c3+1)*(a2*b2*c4+1)*(c3*c4+1):

eqn := {a1+b1 = 1,
a2+a1*b1+b2+c1 = 1,
a2*b1+a1*b2+c2 = 1,
a2*b2+c3+c4 = 0,
c5 = 0};

solve(eqn,{a1,a2,b1,b2}); # <= Maple can't solve this

# But the solution is:

a2:=0: a1:=1: # a0:=1, so that a= "011" binary = 3 decimal
b2:=1: b1:=0: # a0:=1, so that b= "101" binary = 5 decimal

# Where a=3 and b=5 are the factors of 15.

Ron Dotson
6/03/2004

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• ... Actually one can solve the system eqn with Maple (in two ways): First way: (note in this case one has free variables b1,b2,c1,c4 so we allow them to be
Message 6 of 13 , Jun 3, 2004
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On Thu, 3 Jun 2004, Ron wrote:

>
> n:= [0,0,1,1,1,1]:  # factor binary 001111 = decimal 15 = 3*5
>
> c1:=a1*b1:
> c2:=(a2+a1*b1+b2)*c1+1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1):
> c3:=(a2+a1*b1+b2)*c1*(1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1)):
> c4:=1+(a2*b1*a1*b2+1)*(a2*b1*c2+1)*(a1*b2*c2+1):
> c5:=1+(a2*b2*c3+1)*(a2*b2*c4+1)*(c3*c4+1):
>
> eqn := {a1+b1 = 1,
> a2+a1*b1+b2+c1 = 1,
> a2*b1+a1*b2+c2 = 1,
> a2*b2+c3+c4 = 0,
> c5 = 0};
>
> solve(eqn,{a1,a2,b1,b2});  # <= Maple can't solve this

Actually one can solve the system eqn with Maple (in two ways):

First way: (note in this case one has free variables b1,b2,c1,c4 so we
allow them to be any of 0 or 1 and see what works)

Sol:=solve(eqn);
2
Sol := {c5 = 0, a1 = -b1 + 1, a2 = b1 - b1 - b2 - c1 + 1,

3 2
c2 = -b1 + b1 + 2 b2 b1 + b1 c1 - b1 - b2 + 1,

2 2
c3 = -b2 b1 + b2 b1 + b2 + b2 c1 - b2 - c4, b1 = b1,

b2 = b2, c1 = c1, c4 = c4}

> for b1 from 0 to 1 do
> for b2 from 0 to 1 do
> for c1 from 0 to 1 do
> for c4 from 0 to 1 do
> a:=eval(a2*4+a1*2+1,Sol);
> b:=eval(b2*4+b1*2+1,Sol);
> if a*b = 15 then print(a,b); fi;
> od;
> od;
> od;
> od:

Second way: Us msolve to solve modulo 2. It gives 5 solutions, but it can
pick the one that works: (It may just be luck that this works?) Here's
how:

> eqn := {a1+b1 = 1,
> a2+a1*b1+b2+c1 = 1,
> a2*b1+a1*b2+c2 = 1,
> a2*b2+c3+c4 = 0,
> c5 = 0}:
> Sol:=msolve(eqn,2);

Sol := {c5 = 0, b1 = 1, a1 = 0, c2 = 1, c1 = 1, a2 = 0, b2 = 0,

c3 = c4}, {c5 = 0, b1 = 1, a1 = 0, c2 = 1, c1 = 0, b2 = 1,

a2 = 0, c3 = c4}, {c5 = 0, b1 = 1, a1 = 0, a2 = 1, c1 = 0,

b2 = 0, c3 = c4, c2 = 0}, {c5 = 0, b1 = 1, a1 = 0, a2 = 1,

c1 = 1, b2 = 1, c2 = 0, c3 = c4 + 1}, {c5 = 0, a1 = 1, c2 = 1,

c1 = 1, a2 = 0, b2 = 0, c3 = c4, b1 = 0}

> for i from 1 to nops([Sol]) do
> a:=eval(a2*4+a1*2+1,Sol[i]);
> b:=eval(b2*4+b1*2+1,Sol[i]);
> if a*b = 15 then print(a,b); fi
> od:

5, 3

I'm not sure why we don't also get 3,5 this way.

--Edwin Clark

>
> # But the solution is:
>
> a2:=0: a1:=1:  # a0:=1, so that a= "011" binary = 3 decimal
> b2:=1: b1:=0:  # a0:=1, so that b= "101" binary = 5 decimal
>
> # Where a=3 and b=5 are the factors of 15.
>
> Ron Dotson
> 6/03/2004
>
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>
>
>
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--
---------------------------------------------------------
W. Edwin Clark, Math Dept, University of South Florida
http://www.math.usf.edu/~eclark/
---------------------------------------------------------
• ... Hash: SHA1 Very perceptive Edwin. I didn t know we had any Maple users in this group. You are correct that if you copy/paste the following lines into a
Message 7 of 13 , Jun 4, 2004
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--- In primenumbers@yahoogroups.com, Edwin Clark <eclark@m...> wrote:
>
> Actually one can solve the system eqn with Maple (in two ways):
>

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Very perceptive Edwin. I didn't know we had any Maple users
in this group. You are correct that if you copy/paste the
following lines into a Maple worksheet:

c1:=a1*b1:
c2:=(a2+a1*b1+b2)*c1+1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1):
c3:=(a2+a1*b1+b2)*c1*(1+(a2*a1*b1+1)*(a2*b2+1)*(a1*b1*b2+1)):
c4:=1+(a2*b1*a1*b2+1)*(a2*b1*c2+1)*(a1*b2*c2+1):
c5:=1+(a2*b2*c3+1)*(a2*b2*c4+1)*(c3*c4+1):
eqn := {a1+b1 = 1,
a2+a1*b1+b2+c1 = 1,
a2*b1+a1*b2+c2 = 1,
a2*b2+c3+c4 = 0,
c5 = 0}:
Sol:=msolve(eqn,2);

Maple provides the solutions immediately:
Sol := {a1 = 0, b1 = 1, a2 = 1, b2 = 0},
{a1 = 1, b1 = 0, a2 = 0, b2 = 1}

I should have used "msolve()" in my example instead of
"solve()." *However*, if you use the actual RSA-640 carry
definitions and equations (all 135MB of them!) instead of the
equations in this simple example, Maple's "msolve()" also fails.
If you don't include the carry definitions it fails immediately,
and if you include them it runs for awhile (hours) and then
terminates with an "Out of Memory" error while trying to expand
the carry definitions (if I recall correctly). If anyone wants
(Carries.txt) and the equation file (equ.dat) as a single gzipped
tar file from:
http://snipurl.com/6uqu/rsa640.tar.gz

or as a Windows ZIP file from:
http://snipurl.com/6uqu/rsa640.zip

(They are both about 10MB in size)

To have Maple try to solve them, paste the following
into a Maple worksheet and execute it from the same directory
you unzipped "equ.dat" and "Carries.txt" into:

restart:
print("Solving");
Sol:=msolve(equ,2);

Maybe it will work on a computer that's faster than
mine and has more RAM, but I doubt it. I have
an Athlon 1.1 GHz AMD processor with 1 GB RAM.

Ron Dotson
6/04/2004

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• Hi Ron, R Justin asked me for a real life example of how this technique R can be used to find factors, so at the risk of boring everyone R else, I thought
Message 8 of 13 , Jun 4, 2004
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Hi Ron,

R> Justin asked me for a real life example of how this technique
R> can be used to find factors, so at the risk of boring everyone
R> else, I thought I'd post it here on the forum. Here is a sample
R> Maple code snippet ('#' begins a comment) that illustrates the
R> problem:

Thanks Ron. I think I've finally understood the idea of the technique. I
also understand long-hand multiplication at long last ;)

R> n:= [0,0,1,1,1,1]: # factor binary 001111 = decimal 15 = 3*5

R> eqn := {a1+b1 = 1,
R> a2+a1*b1+b2+c1 = 1,
R> a2*b1+a1*b2+c2 = 1,
R> a2*b2+c3+c4 = 0,
R> c5 = 0};

As you know, these equations in the Z(2) field are equivalent to boolean
expressions. So they can be transformed the following way to eliminate the
equality:

lhs == 1 -> lhs
lhs == 0 -> NOT(lhs) -> lhs+1 (equivalent to NOT using only xor)

now all the ex-equations can be combined to one with AND (ie multiplication
in Z(2)) to form one expression:

(a1+b1) (a2+a1*b1+b2+c1) (a2*b1+a1*b2+c2) (a2*b2+c3+c4+1) (c5+1)

This can be reduced algebraically and transformed to boolean:

(a1 && b2 && ! a2 && ! b1) || (a2 && b1 && ! a1 && ! b2)

Which in effect is indeed your solution:

R> # But the solution is:

R> a2:=0: a1:=1: # a0:=1, so that a= "011" binary = 3 decimal
R> b2:=1: b1:=0: # a0:=1, so that b= "101" binary = 5 decimal

R> # Where a=3 and b=5 are the factors of 15.

I use Mathematica, and am not familiar with Maple, but I suppose it is
just as capable of these transformations.

I'm not sure about the practicability though. It already took a few seconds
to reduce your example expressions (albeit with somewhat straightforward
and probably not really efficient reduction algorithm).

But then, what do I know? ;P

--
Justin
ICQ 37456745
AIM jasticle
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