Algorithms

void memoryFill( void* target, char* pattern, size_t ptrnlen, size_t numRepeats )
{  
   char* trgt = (char*)target;  //for ease in pointer manipulation.   
   ...
   //
   // If we KNEW that memcpy copied from the base of the src array to the base
   // of the target array, we could write an EXTREMELY fast algorithm:
   //    memcpy( trgt, pattern, ptrnlen );                     //put one copy at the base.
   //    memcpy( trgt+ptrnlen, trgt, ptrnlen*(numRepeats-1) ); //destructive copy the rest
   // Conversely, if we KNEW that memcpy copied from the end of the src array 
   // to the end of the target array, we could write:
   //    memcpy( trgt+ptrnlen*(numRepeats-1) , pattern, ptrnlen ); //put one copy at the end.
   //    memcpy( trgt, trgt+ptrnlen, ptrnlen*(numRepeats-1) );     //destructive copy the rest
   // However, the C standard distinctly says that memcpy can do it's copying
   // in whatever order it wants.  We CANNOT rely on any specific order.
   //
   ...
}

smyers wrote:

You don't need an algorithm to check the difference in size between two files. You can just read each file one line at a time and count the bytes of each or if the files are small get a byte array of the files and then comapare the sizes of the arrays.

Juraj Borza wrote:

The program would then evaluate how much the generated image look like the original (like fitness function in GA), I mean, evaluate many solutions, find the best ones, combine/mutate them in some way to produce better results in the next generation.

Ennis Ray Lynch, Jr. wrote:

Suppose I have an image of a region of earth of arbitrary size and I have Lat/Long from two points on the map is there a simple formula for converting a given pixel to a Lat/Long that takes into account the curvature of Earth? Assume perfect lens and that the map is of such a scale and size that a given pixel distance will be the same across the image.