Pascal Triangle Recursive Program optimization in C++

Question

I have built recursive function to compute Pascal's triangle values.

Is there a way to optimize it?

Short reminder about Pascal's triangle: C(n, k) = C(n-1, k-1) + C(n-1, k) My code is:

int Pascal(int n, int k) {
if (k == 0) return 1;
if (n == 0) return 0;
return Pascal(n - 1, k - 1) + Pascal(n - 1, k);
}

The inefficiency I see is that it stores some values twice. Example: C(6,2) = C(5,1) + C(5,2) C(6,2) = C(4,0) + C(4,1) + C(4,1) + C(4,2) it will call C(4,1) twice

Any idea how to optimize this function?

Thanks

Answer 1

The following routine will compute the n-choose-k, using the recursive definition and memoization. The routine is extremely fast and accurate:

inline unsigned long long n_choose_k(const unsigned long long& n,
                                     const unsigned long long& k)
{
   if (n  < k) return 0;
   if (0 == n) return 0;
   if (0 == k) return 1;
   if (n == k) return 1;
   if (1 == k) return n;

   typedef unsigned long long value_type;

   class n_choose_k_impl
   {
   public:

      n_choose_k_impl(value_type* table,const value_type& dimension)
      : table_(table),
      dimension_(dimension / 2)
      {}

      inline value_type& lookup(const value_type& n, const value_type& k)
      {
         const std::size_t difference = static_cast<std::size_t>(n - k);
         return table_[static_cast<std::size_t>((dimension_ * n) + ((k < difference) ? k : difference))];
      }

      inline value_type compute(const value_type& n, const value_type& k)
      {
         // n-Choose-k = (n-1)-Choose-(k-1) + (n-1)-Choose-k
         if ((0 == k) || (k == n))
            return 1;
         value_type v1 = lookup(n - 1,k - 1);
         if (0 == v1)
            v1 = lookup(n - 1,k - 1) = compute(n - 1,k - 1);
         value_type v2 = lookup(n - 1,k);
         if (0 == v2)
            v2 = lookup(n - 1,k) = compute(n - 1,k);
         return v1 + v2;
      }

      value_type* table_;
      const value_type dimension_;
   };

   static const std::size_t static_table_dim = 100;
   static const std::size_t static_table_size = static_cast<std::size_t>((static_table_dim * static_table_dim) / 2);
   static value_type static_table[static_table_size];
   static bool static_table_initialized = false;

   if (!static_table_initialized && (n <= static_table_dim))
   {
      std::fill_n(static_table,static_table_size,0);
      static_table_initialized = true;
   }

   const std::size_t table_size = static_cast<std::size_t>(n * (n / 2) + (n & 1));

   unsigned long long dimension = static_table_dim;
   value_type* table = 0;

   if (table_size <= static_table_size)
      table = static_table;
   else
   {
      dimension = n;
      table = new value_type[table_size];
      std::fill_n(table,table_size,0LL);
   }

   value_type result = n_choose_k_impl(table,dimension).compute(n,k);

   if (table != static_table)
      delete [] table;

   return result;
}

Answer 2

Keep a table of previously returned results (indexed by their n and k values); the technique used there is memoization. You can also change the recursion to an iteration and use dynamic programming to fill in an array containing the triangle for n and k values smaller than the one you are trying to evaluate, then just get one element from it.