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Economy / Services / Re: [BOUNTY] 200$ for the first to provide purely mathematic expression of XOR

on: January 23, 2014, 04:35:34 PM

I'm with BurtW on this. BTW.. way more than $200 worth of effort in this thread already

Economy / Services / Re: [BOUNTY] 200$ for the first to provide purely mathematic expression of XOR

on: January 22, 2014, 11:35:35 PM

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))

Economy / Services / Re: [BOUNTY] 200$ for the first to provide purely mathematic expression of XOR

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)

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