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Re: [PrimeNumbers] Re: Boolean algebra solver?

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  • Milton Brown
    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 1 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 ?

      It would be helpful to explain these, or have your mathematics
      thought out more before posting.

      Thanks.

      Milton L. Brown
      miltbrown at earthlink.net
      miltbrown@...

      ----- Original Message -----
      From: "Ron" <Yaho6Hb3c@...>
      To: <primenumbers@yahoogroups.com>
      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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    • David Cleaver
      ... 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 2 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
      • Ron
        ... 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 3 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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        • Edwin Clark
          ... 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 4 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/
            ---------------------------------------------------------
          • Ron
            ... 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 5 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
              to try it themselves, you can download the carry definition file
              (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("Reading Equations");
              read "equ.dat":
              print("Reading Carries");
              read "Carries.txt":
              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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            • Justin
              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 6 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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