How to speed up a recursive algorithm

Question

I'm trying to solve the Hackerrank challenge Game of Stones, a (shortened) problem statement of which is copied below.

I came up with the following solution:

# The lines below are for the Hackerrank submission
# T = int(raw_input().strip())
# ns = [int(raw_input().strip()) for _ in range(T)]

T = 8
ns = [1, 2, 3, 4, 5, 6, 7, 10]

legal_moves = [2, 3, 5]

def which_player_wins(n):
    if n <= 1:
        return "Second"               # First player loses
    elif n in legal_moves:
        return "First"                # First player wins immediately
    else:
        next_ns = map(lambda x: n - x, legal_moves)
        next_ns = filter(lambda x: x >= 0, next_ns)
        next_n_rewards = map(which_player_wins, next_ns)      # Reward for opponent
        if any(map(lambda x: x=="Second", next_n_rewards)):            # Opponent enters a losing position
            return "First"
        else:
            return "Second"

for n in ns:
    print which_player_wins(n)

The algorithm is essentially the minimax algorithm as it looks one move ahead and then recursively calls the same function. The problem is that in Hackerrank, it is terminating due to timeout:

Indeed, I've noticed that evaluating which_player_wins(40) already takes ~2 seconds. Any ideas for a faster solution which would not time out?

Answer 1

It seems from the problem description that you are able to keep the intermediate and final results from each calculation to use in later calculations. If that is so, a recursive algorithm is not optimal. Instead, use a dynamic programming style.

In other words, keep a global array that stores the results from previous determinations of wins and losses. When you determine for a new value of n, rather than recursing all the way to the end, use the previous determinations.

For example, when you get to n=10, you see that the first player removing 3 stones leaves 7 stones, which you have already seen as a win for the second player. Therefore, 10 stones is a win for the first player.

I believe that your recursive algorithm would calculate all over again the result for n=7 rather than using the previous work.

Answer 2

Your solution would work (though very slowly) if you had infinite recursion levels available. But since you're not re-using already calculated results you are very quickly hitting Python's recursion limit. One good solution, as suggested above, is to re-write the code using a non-recursive algorithm.

Alternatively, you could keep the code you have, but get it to re-use any result that you've already seen before. This is called memoization, and an easy way to implement that is to add a memoization decorator to the part of the code that computes the next move. Here's a way to use memoization to speed up your recursive algorithm, which I think is what you've specifically asked for:

1. Convert all of the code after the first else into a function, next_move(n) that returns either 'First' or 'Second'.

2. Add a memoize(f) function which will take next_move(n) and avoid the recursion call if the result for n has already been calculated.

3. Add the decorator line @memoize immediately before the next_move definition.

Resulting code:

T = 8
ns = [1, 2, 3, 4, 5, 6, 7, 10]

legal_moves = [2, 3, 5]

def memoize(f):
    memo = {}
    def helper(x):
        if x not in memo:
            memo[x] = f(x)
        return memo[x]
    return helper

@memoize
def next_move(n):
    next_ns = map(lambda x: n - x, legal_moves)
    next_ns = filter(lambda x: x >= 0, next_ns)
    next_n_rewards = map(which_player_wins, next_ns)  # Reward for opponent
    if any(map(lambda x: x == "Second", next_n_rewards)):  # Opponent enters a losing position
        return "First"
    else:
        return "Second"

def which_player_wins(n):
    if n <= 1:
        return "Second"               # First player loses
    elif n in legal_moves:
        return "First"                # First player wins immediately
    else:
        return next_move(n)

for n in ns:
    print which_player_wins(n)

This hugely speeds up the calculation as well as reducing the levels of recursion required. On my computer n=100 is solved in 0.8 ms.