The XOR trick is only cool in its undefined-behavior form:
a^=b^=a^=b;
Which allegedly saves you 0.5 seconds of typing in competitive programming competitions from 20 years ago and is known to work reliably (on MinGW under Windows XP).
Bonus authenticity: use `a^=a` to zero a register in a single x86 instruction (and makes a real difference for compiler toolchains 30+ years old).
For real now, a very useful application of XOR is its relation to the Nim game [0], which comes in very handy if you need to save your village from an ancient disgruntled Chinese emperor.
Xor swap trick has perfect profile for underhanded C contests. It generally works until a specific condition triggers its failure. The condition is "the arguments are aliases", so for example XOR_SWAP(a[i], a[j]) when i=j.
You can use this same property of xor to make a double-linked list using just one pointer per item, which is xor of the previous and next item addresses!
> given a list where every value appears exactly twice except one, XOR all the values together and the duplicates cancel out, leaving the unique element
For some reason this reminds me of the Fourier transform. I wonder if it can be performed with XOR tricks and no complicated arithmetic?
There's a more general formulation, which is that every value but one must appear even numbers of times, and the one must appear some odd number of times.
How about every value appears 0 mod n times except one which appears 1 mod n times? :)
Solution: xor is just addition mod 2. Write the numbers in base n and do digit-wise addition mod n (ie without carry). Very intuitive way to see the xor trick.
a^=b^=a^=b;
Which allegedly saves you 0.5 seconds of typing in competitive programming competitions from 20 years ago and is known to work reliably (on MinGW under Windows XP).
Bonus authenticity: use `a^=a` to zero a register in a single x86 instruction (and makes a real difference for compiler toolchains 30+ years old).
For real now, a very useful application of XOR is its relation to the Nim game [0], which comes in very handy if you need to save your village from an ancient disgruntled Chinese emperor.
[0] https://en.wikipedia.org/wiki/Nim
For some reason this reminds me of the Fourier transform. I wonder if it can be performed with XOR tricks and no complicated arithmetic?
Solution: xor is just addition mod 2. Write the numbers in base n and do digit-wise addition mod n (ie without carry). Very intuitive way to see the xor trick.