In today's post we quickly review the Longest Common Subsequence and a hashing function
Let c[i, j] be the length of the longest common subsequence of the prefix  of X[1..i] and Y[1..j]. Recursion:
c[i, j] = { 0 if i =  0 or j = 0
          = { c[i-1, j-1] + 1 if i,j > 0 and xi = yj
          = { max( c[i,j-1], c[i-1,j] ) otherwise
This way we utilize the solutions we computed. We compute the table c[i,j] bottom-up and also store the value b[i,j] that captures whether c[i,j-1], c[i-1,j] or c[i-1,j-1] + 1is optimum.  We therefore find the longest common subsequence by walking backwards on the path from the last cell of b[i,j] to the first.

Hashing function
template<class T> size_t Hash<T>::operator()(const T& obj)
{
  size_t len = sizeof(obj);
  size_t res = 0;
  const char* ptr = reinterpret_cast<const char*>(&obj);
  while (len--) res = (res << 1) ^ *ptr++;
  return res;
}

Void memcpy ( char* src, char* dest, size_t len)
{
While (len--){
      *dest++ = *src++;
}
}

void GaussianAverage(char* pSrc, char* pDest, int i, int j, int row, int col)
{
  // parameter validation
  int val = pSrc[i * col + j];
  int numNeighbors = 0;
  for (int m = i - 2; m <= i+2; m++)
        for (int n = j- 2; n <= j+2; n++)
           if ( m >= 0 && n >= 0 && m < row && n < col )
           {
                     val +=  pSrc[m * col + n]; 
                     numNeighbors++;
            }
 val = val / (numNeighbors + 1);
 pDest[m * col + n]   = val;
}