Created
June 11, 2016 09:30
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Chapter 5 : Problem 1
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#include <iostream> | |
using namespace std; | |
/* Code to generate a LUT for parity */ | |
short computeParity( short num ) | |
{ | |
short parity = 0; | |
while (num > 0) | |
{ | |
parity ^= (x&1); | |
x >>= 1; | |
} | |
return parity; | |
} | |
/* | |
The while loop in the compute parity can be optimized by just unsetting the last possible 1 bit set | |
That way even though the complexity is still at O(N) where N is the size of bits in short integer type~ 16 we | |
still optimize to only loop the number of times a bit is set: | |
while (num > 0) | |
{ | |
parity ^= 1; | |
x &= (x - 1) // This unsets the last bit | |
} | |
*/ | |
int main (int argc, char *argv[]) | |
{ | |
parityLUT[0x10000]; | |
//First make sure you populate the LUT for parity. We could provide a static table. But I am lazy. | |
for (short i = 0; i <= 0xFFFF; i++) | |
{ | |
parityLUT[i] = computeParity(i); // Compute the parity of all values from 0 to 0xFFFF | |
} | |
// Now that we have ourself the table we can calculate the parity of the actual 64 bit number. | |
unsigned long num; | |
cin<<num; | |
// We now need to break down the 64 bit long into smaller 16 bit parts | |
unsigned short finalParity = 0; | |
finalParity = parityLUT[ (num & 0xFFFF)] ^ parityLUT[(num >> 16) & 0xFFFF] ^ parityLUT[(num >> 32) & 0xFFFF] ^ parityLUT[(num >> 56)]; | |
cout<<"The final parity value is " + finalParity; | |
return 0; | |
} |
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