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Re: [PrimeNumbers] Re: Boolean algebra solver?

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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 1 of 13 , Jun 2, 2004
      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 2 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 3 of 13 , Jun 3, 2004
          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 4 of 13 , Jun 4, 2004
            --- 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 5 of 13 , Jun 4, 2004
              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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