Fast

I believe I don’t have to describe what primes are, what are their properties and what not. This post is more a tribute to geek-ness of 2 friends-and-colleagues (@lucabox) that have fun thinking of algorithms to solve stupid (or less stupid), and always useless problems ;-) . Optimus Prime :-P - yeah yeah, a stupid joke Briefing This code is based on the assumption that we want to generate very very large primes, so it uses unsigned long long to store the values, instead of classical unsigned int. Live with that. ...

Someone says this is an old lame trick. I think it’s simple and clever use of XOR. How/Why does it work? It’s built around the properties of the XOR ^ operator, who has the following properties: A ^ B = B ^ A (commutative) A ^ 0 = A A ^ 1 = ~A A ^ A = 0 So, you can see how it get’s applied here: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 #include <stdio .h> int main(void) { unsigned int a, b; // ... populate somehow "a" and "b"... printf("a = %d - b = %d\n", a, b); a ^= b; // store in "a" the value of "a XOR b" b ^= a; // store in "b" the value of "a XOR b XOR b" = "a XOR 0" = "a" a ^= b; // store in "a" the velue of "a XOR b XOR a" = "b XOR 0" = "b" printf("a = %d - b = %d\n", a, b); } Neat.

Searching a string inside another string is a very easy task. Given a string A and a string B, start a loop over A from left to right and, for every letter, start an internal loop to see if there is a match with the letters of B. A ball of string Pseudo code would look something like this: 1 2 3 4 5 6 7 function NaiveSearch(string s[1..n], string sub[1..m]) for i from 1 to n-m+1 for j from 1 to m if s[i+j-1] ≠ sub[j] jump to next iteration of outer loop return i return not found Time Complexity? O(n * m), where n is the size of A and m is the size of B. ...