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    • Evil-Knievel (OP)
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    January 22, 2014, 09:02:53 PM
    Last edit: April 17, 2016, 09:23:14 PM by Evil-Knievel
     #1
    This message was too old and has been purged

  • dirvine
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    January 22, 2014, 09:14:01 PM
     #2
    NUM_SIZE=32

    A XOR B

    for n=0 to n=32; 2^n * (((A/2^n) % 2) + (B/2^n) %2) %2)

    Probably easier for me though to have written a for loop.



    13YSCv8SLrBw27AdyRQY8adfsGa56viQcJ

  • Evil-Knievel (OP)
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    January 22, 2014, 09:16:32 PM
    Last edit: April 17, 2016, 09:23:08 PM by Evil-Knievel
     #3
    This message was too old and has been purged

  • dirvine
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    January 22, 2014, 09:26:17 PM
     #4
    Quote from: Evil-Knievel on January 22, 2014, 09:16:32 PM
    Quote from: dirvine on January 22, 2014, 09:14:01 PM
    NUM_SIZE=32

    A XOR B

    for n=0 to n=32; 2^n * (((A/2^n) % 2) + (B/2^n) %2) %2)

    Probably easier for me though to have written a for loop.



    13YSCv8SLrBw27AdyRQY8adfsGa56viQcJ

    Hey, thanks for your answer and your intrest. However a purely arithmetic solution is search. The for loop actually is more than just maths which prevents it from putting the formula into a CAS to process it any further. Also only modulus 2^32-1 is allowed ... unfortunately there is a mod 2 in your equation.

    But this idea is pretty good, I hope you come up with a pure mathematical/algebraic representation soon.  Smiley

    Yes I spotted the mod statement and also signed numbers too after posting. I will have a wee think now :-) cheers for the question it's pretty cool.

  • niniyo
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    January 22, 2014, 09:32:27 PM
     #5
    How is this?

    Sum from n = 1 to 32:
      (((a mod 2^n) 2^(n-1)) * 2^(n-1)) + (((b mod 2^n) 2^(n-1)) * 2^(n-1))

    I don't quite understand what you mean about only being able to use modulus 2^32-1?  Isn't 2^32-1 the maximum number, so anything mod that would just give the same number back?

  • Evil-Knievel (OP)
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    January 22, 2014, 09:49:49 PM
    Last edit: April 17, 2016, 09:22:55 PM by Evil-Knievel
     #6
    This message was too old and has been purged

  • niniyo
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    January 22, 2014, 10:11:35 PM
     #7
    OK how about this:

    function mymod (value, modpow) =>
      ; returns the 32-bit result of value mod 2^modpow
      ((value * 2^(32-modpow) MOD 2^32) 2^(32-modpow));


    ; note that the above is a mathematical expression, and i'm just going to use it like a macro to avoid repeating the same expression all the time

    function myxor (a, b) =>
      ; return the 32-bit result of a bitwise xor b
      sum from n = 1 to 32:
        mymod(mymod(a,n) 2^(n-1) * 2^(n-1) + mymod(a,n) 2^(n-1) * 2^(n-1), n)

    EDIT: I used the expression ((n 2^n) * 2^n) to zero out the lowest n bits, because I'm assuming these are integer operations so a division results in the truncated integer result.  Is that suitable?

  • BurtW
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    January 22, 2014, 10:22:21 PM
    Last edit: January 22, 2014, 10:56:32 PM by BurtW
     #8
    Code:
    #include <iostream>
    #include <stdio.h>

    using namespace std;

    int main (int argc, char *argv[])
    {
        unsigned long a = 0x01234567ul;
        unsigned long b = 0x87654321ul;
        unsigned long x = 0;
        unsigned long d = 1;
        unsigned long n, r;

        for (n=0; n<32; n++) {
            r = a + b;
            r *= 2147483648ul;
            r= (2147483648ul/d);
            x += r;
            a= 2;
            b= 2;
            d *= 2;
        }
        printf("x=0x%08lX xor is 0x%08lX\n", x, 0x01234567ul ^ 0x87654321ul);
    }


    C:\Work\xor\Debug>xor
    x=0x86460646 xor is 0x86460646


    Don't know if this is what you are looking for but it did work.

    PS:  I kind of like that snip of code...
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  • niniyo
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    January 22, 2014, 10:26:41 PM
     #9
    OP - can you clarify whether integer operations are supported?  Like, 5 2 = 2?  Or, since it's purely mathematical, would that be 2.5?  If so, can we use the floor() function?

  • OP_CHECKSIG
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    January 22, 2014, 10:27:37 PM
     #10
    Here's javascript code that works...

    function xor32bit(a,b){
       var sum=0;
       var x,shift;
       var s32 = Math.pow(2,32);
       var s31 = Math.pow(2,31);

       for (n=0;n<32;n++){
          shift = Math.pow(2,n);
          // move current bits to b0 and add
          x= Math.floor(a/shift)+Math.floor(b/shift);
          // move to most significant bit to clear out left bits
          x*= s31;
          // mod to keep in 32 bits range
          x=x%s32;
          // divide to put back in right it position
          x=Math.floor(x/Math.pow(2,(31-n)));
          // add to sum
          sum+=x;
       }
       return sum;
    }

    console.log(xor32bit(0x12345678,0).toString(16));
    console.log(xor32bit(0x12345678,0xffffffff).toString(16));

    Equation is sum for n 0..31 (((a/2^n+b/2^n)x2^31)mod 2^32)/2^(31-n)

  • Evil-Knievel (OP)
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    January 22, 2014, 10:44:00 PM
    Last edit: April 17, 2016, 09:22:36 PM by Evil-Knievel
     #11
    This message was too old and has been purged

  • niniyo
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    January 22, 2014, 10:47:35 PM
     #12
    OK here is my final submission which should be purely mathematical and doesn't require rounded integer arithmetic operations.  It uses a floor() function.

    Code:
    ; note that the following two helper functions are mathematical expressions, and i'm just going to use them like macros to avoid repeating the same expression all the time
    function mymod (value, modpow) =>
      ; returns the 32-bit result of value mod 2^modpow
      ((value * 2^(32-modpow) MOD 2^32) 2^(32-modpow));

    function clearbits (value, n) =>
      ; returns the 32-bit value with the lowest n bits cleared
      floor(value 2^n) * 2^n;

    ; main function for calculating the result.  returns the 32-bit result of a bitwise-xor b.
    function myxor (a, b) =>
      sum from n = 1 to 32:
        ; add the value of the nth bit of the xor result
        mymod(clearbits(mymod(a,n), n-1) + clearbits(mymod(a,n), n-1), n)

  • BurtW
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    January 22, 2014, 10:48:56 PM
     #13
    Quote from: OP_CHECKSIG on January 22, 2014, 10:27:37 PM
    Here's javascript code that works...

    function xor32bit(a,b){
       var sum=0;
       var x,shift;
       var s32 = Math.pow(2,32);
       var s31 = Math.pow(2,31);

       for (n=0;n<32;n++){
          shift = Math.pow(2,n);
          // move current bits to b0 and add
          x= Math.floor(a/shift)+Math.floor(b/shift);
          // move to most significant bit to clear out left bits
          x*= s31;
          // mod to keep in 32 bits range
          x=x%s32;
          // divide to put back in right it position
          x=Math.floor(x/Math.pow(2,(31-n)));
          // add to sum
          sum+=x;
       }
       return sum;
    }

    console.log(xor32bit(0x12345678,0).toString(16));
    console.log(xor32bit(0x12345678,0xffffffff).toString(16));

    Equation is sum for n 0..31 (((a/2^n+b/2^n)x2^31)mod 2^32)/2^(31-n)


    Basically what I did up above in my post but it looks like he is not looking for an algorithm...
    Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!

  • OP_CHECKSIG
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    January 22, 2014, 11:35:35 PM
     #14
    Expand the sum to long form?
    Code:
    x = ((((a+b)x2^31)mod 2^32)/2^(31))+((((a/2+b/2)x2^31)mod 2^32)/2^(30))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(29))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(28))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(27))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(26))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(25))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(24))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(23))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(22))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(21))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(20))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(19))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(18))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(17))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(16))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(15))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(14))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(13))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(12))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(11))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(10))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(9))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(8))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(7))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(6))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(5))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(4))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(3))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(2))+
    ((((a/(2^2)+b/(2^2))x2^31)mod 2^32)/2^(1))+((((a/(2^3)+b/(2^3))x2^31)mod 2^32)/2^(0))

  • BurtW
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    January 23, 2014, 12:02:29 AM
    Last edit: January 23, 2014, 01:09:02 AM by BurtW
     #15
    The above post is NOT correct but this one is, and it is just the unwinding of my loop in post #8 above so it is the same solution.

    Code:
        x = (a/0x00000001 + b/0x00000001) * 0x80000000 0x80000000 +
            (a/0x00000002 + b/0x00000002) * 0x80000000 0x40000000 +
            (a/0x00000004 + b/0x00000004) * 0x80000000 0x20000000 +
            (a/0x00000008 + b/0x00000008) * 0x80000000 0x10000000 +
            (a/0x00000010 + b/0x00000010) * 0x80000000 0x08000000 +
            (a/0x00000020 + b/0x00000020) * 0x80000000 0x04000000 +
            (a/0x00000040 + b/0x00000040) * 0x80000000 0x02000000 +
            (a/0x00000080 + b/0x00000080) * 0x80000000 0x01000000 +
            (a/0x00000100 + b/0x00000100) * 0x80000000 0x00800000 +
            (a/0x00000200 + b/0x00000200) * 0x80000000 0x00400000 +
            (a/0x00000400 + b/0x00000400) * 0x80000000 0x00200000 +
            (a/0x00000800 + b/0x00000800) * 0x80000000 0x00100000 +
            (a/0x00001000 + b/0x00001000) * 0x80000000 0x00080000 +
            (a/0x00002000 + b/0x00002000) * 0x80000000 0x00040000 +
            (a/0x00004000 + b/0x00004000) * 0x80000000 0x00020000 +
            (a/0x00008000 + b/0x00008000) * 0x80000000 0x00010000 +
            (a/0x00010000 + b/0x00010000) * 0x80000000 0x00008000 +
            (a/0x00020000 + b/0x00020000) * 0x80000000 0x00004000 +
            (a/0x00040000 + b/0x00040000) * 0x80000000 0x00002000 +
            (a/0x00080000 + b/0x00080000) * 0x80000000 0x00001000 +
            (a/0x00100000 + b/0x00100000) * 0x80000000 0x00000800 +
            (a/0x00200000 + b/0x00200000) * 0x80000000 0x00000400 +
            (a/0x00400000 + b/0x00400000) * 0x80000000 0x00000200 +
            (a/0x00800000 + b/0x00800000) * 0x80000000 0x00000100 +
            (a/0x01000000 + b/0x01000000) * 0x80000000 0x00000080 +
            (a/0x02000000 + b/0x02000000) * 0x80000000 0x00000040 +
            (a/0x04000000 + b/0x04000000) * 0x80000000 0x00000020 +
            (a/0x08000000 + b/0x08000000) * 0x80000000 0x00000010 +
            (a/0x10000000 + b/0x10000000) * 0x80000000 0x00000008 +
            (a/0x20000000 + b/0x20000000) * 0x80000000 0x00000004 +
            (a/0x40000000 + b/0x40000000) * 0x80000000 0x00000002 +
            (a/0x80000000 + b/0x80000000) * 0x80000000 0x00000001 ;

    Tested and it works on C++.  You may need to add some masking and truncating operations in there - depending on exactly how the programming language you use works.  However, the basic idea is sound.

    Did I win?
    Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!

  • gogodr
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    January 23, 2014, 12:55:12 AM
    Last edit: January 23, 2014, 01:07:23 AM by gogodr
     #16
    uhm...I have a very simple solution here...

    entry 1:
    XOR:
    f(x,y) = 0^(0^((1+x)mod(1+y)));

    entry 2:
    XOR:
    f(x,y) = 0^(0^(x-y)); ,x>=y
    f(x,y) = 0^(0^(y-x)); ,x<y


    you know 0^0 == 1 , you can use that to convert logic into mathematics.
    f(5,5) == 0
    f(5,6) == 1
    f(6,0) == 1
    f(0,0) == 0

    edit: I know 0^0 should be indeterminate but it is a convention.

  • BurtW
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    January 23, 2014, 01:04:38 AM
     #17
    Quote from: gogodr on January 23, 2014, 12:55:12 AM
    uhm...I have a very simple solution here...

    entry 1:
    XOR:
    f(x,y) = 0^(0^((1+x)mod(1+y)));

    entry 2:
    XOR:
    f(x,y) = 0^(0^(x-y)); ,x>=y
    f(x,y) = 0^(0^(y-x)); ,x<y


    you know 0^0 == 1 , you can use that to convert logic into mathematics.
    f(5,5) == 0
    f(5,6) == 1
    f(6,0) == 1
    f(0,0) == 0


    He wants the bitwise XOR of the values:

    XOR(5,5) = 0
    XOR(5,6) = 3
    XOR(6,0) = 6
    XOR(0,0) = 0

    So your formula does not give the correct results, however, mine does Wink
    Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!

  • niniyo
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    January 23, 2014, 01:15:15 AM
     #18
    Quote from: BurtW on January 23, 2014, 12:02:29 AM
    You may need to add some masking and truncating operations in there - depending on exactly how the programming language you use works.  However, the basic idea is sound.

    Did I win?

    No.  The whole point was that it has to be a purely mathematical solution.  Relying on a programming language to truncate values and handle overflows is not purely mathematical.  You're supposed to think in terms of mathematical operations, not machine operations.

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    January 23, 2014, 01:26:38 AM
     #19
    Here I added the one allowed mod operation.  Per this explaination of what is allowed:

    Quote from: Evil-Knievel on January 22, 2014, 09:49:49 PM
    That means, that if you create an overflow and you temporarily use more than 32 bits, you can strip it down to 32bit my using a "mod 2^32-1" operation  Wink

    The allowed mod operation is actually mod 2^32 not 2^32-1 as suggested in the OP - OP should be corrected.

    This is pure integer math, not relying on the "machine" to do the truncation of the multiply.

    As far as I can tell this single equation should satisfy the requirements of the puzzle.

    Code:
        x = (((a/0x00000001 + b/0x00000001) * 0x80000000) % 0x100000000) 0x80000000 +
    (((a/0x00000002 + b/0x00000002) * 0x80000000) % 0x100000000) 0x40000000 +
    (((a/0x00000004 + b/0x00000004) * 0x80000000) % 0x100000000) 0x20000000 +
    (((a/0x00000008 + b/0x00000008) * 0x80000000) % 0x100000000) 0x10000000 +
    (((a/0x00000010 + b/0x00000010) * 0x80000000) % 0x100000000) 0x08000000 +
    (((a/0x00000020 + b/0x00000020) * 0x80000000) % 0x100000000) 0x04000000 +
    (((a/0x00000040 + b/0x00000040) * 0x80000000) % 0x100000000) 0x02000000 +
    (((a/0x00000080 + b/0x00000080) * 0x80000000) % 0x100000000) 0x01000000 +
    (((a/0x00000100 + b/0x00000100) * 0x80000000) % 0x100000000) 0x00800000 +
    (((a/0x00000200 + b/0x00000200) * 0x80000000) % 0x100000000) 0x00400000 +
    (((a/0x00000400 + b/0x00000400) * 0x80000000) % 0x100000000) 0x00200000 +
    (((a/0x00000800 + b/0x00000800) * 0x80000000) % 0x100000000) 0x00100000 +
    (((a/0x00001000 + b/0x00001000) * 0x80000000) % 0x100000000) 0x00080000 +
    (((a/0x00002000 + b/0x00002000) * 0x80000000) % 0x100000000) 0x00040000 +
    (((a/0x00004000 + b/0x00004000) * 0x80000000) % 0x100000000) 0x00020000 +
    (((a/0x00008000 + b/0x00008000) * 0x80000000) % 0x100000000) 0x00010000 +
    (((a/0x00010000 + b/0x00010000) * 0x80000000) % 0x100000000) 0x00008000 +
    (((a/0x00020000 + b/0x00020000) * 0x80000000) % 0x100000000) 0x00004000 +
    (((a/0x00040000 + b/0x00040000) * 0x80000000) % 0x100000000) 0x00002000 +
    (((a/0x00080000 + b/0x00080000) * 0x80000000) % 0x100000000) 0x00001000 +
    (((a/0x00100000 + b/0x00100000) * 0x80000000) % 0x100000000) 0x00000800 +
    (((a/0x00200000 + b/0x00200000) * 0x80000000) % 0x100000000) 0x00000400 +
    (((a/0x00400000 + b/0x00400000) * 0x80000000) % 0x100000000) 0x00000200 +
    (((a/0x00800000 + b/0x00800000) * 0x80000000) % 0x100000000) 0x00000100 +
    (((a/0x01000000 + b/0x01000000) * 0x80000000) % 0x100000000) 0x00000080 +
    (((a/0x02000000 + b/0x02000000) * 0x80000000) % 0x100000000) 0x00000040 +
    (((a/0x04000000 + b/0x04000000) * 0x80000000) % 0x100000000) 0x00000020 +
    (((a/0x08000000 + b/0x08000000) * 0x80000000) % 0x100000000) 0x00000010 +
    (((a/0x10000000 + b/0x10000000) * 0x80000000) % 0x100000000) 0x00000008 +
    (((a/0x20000000 + b/0x20000000) * 0x80000000) % 0x100000000) 0x00000004 +
    (((a/0x40000000 + b/0x40000000) * 0x80000000) % 0x100000000) 0x00000002 +
    (((a/0x80000000 + b/0x80000000) * 0x80000000) % 0x100000000) 0x00000001 ;
    Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!

  • empoweoqwj
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    January 23, 2014, 04:07:28 AM
     #20
    Surely if the solution exists, you could just google it and paste in the solution? I'm guessing it can't be done ....
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