Pascal Triangle Recursive Program optimization in C++
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 store开发者_运维技巧s 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
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;
}
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.
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