Recursive change-making algorithm

问题:

Given a target amount and a list of coin denominations, my code is supposed to find the fewest coins needed to reach the target amount.

Examples:

• C(78, [1, 5, 10, 25, 50]) = 6

  • we can make 78 from 3x25 + 3x1, so 6 coins are required

• C(48, [1, 7, 24, 42]) = 2

  • 48 = 2x24, so 2 coins are sufficient

• C(35, [1, 3, 16, 30, 50]) = 3

  • we can make 35 from 2x16 + 1x3, so 3 coins suffice

I made the code with for loops, but how do I make it recursive?

def C(i, coins, cdict = None):
    if cdict == None:
        cdict = {}
    if i <= 0:
        cdict[i] = 0
        return cdict[i]
    elif i in cdict:
        return cdict[i]
    elif i in coins:
        cdict[i] = 1
        return cdict[i]
    else:
        min = 0
        for cj in coins:
            result = C(i - cj, coins)
            if result != 0:
                if min == 0 or (result + 1) < min:
                    min = 1 + result
        cdict[i] = min
        return cdict[i]

回答1:

It's the change-making problem. Here's the standard recursive solution, V is the list of coins and C the target amount of money:

def min_change(V, C):
    def min_coins(i, aC):
        if aC == 0:
            return 0
        elif i == -1 or aC < 0:
            return float('inf')
        else:
            return min(min_coins(i-1, aC), 1 + min_coins(i, aC-V[i]))
    return min_coins(len(V)-1, C)

And this is an optimized version, using dynamic programming:

def min_change(V, C):
    m, n = len(V)+1, C+1
    table = [[0] * n for x in xrange(m)]
    for j in xrange(1, n):
        table[0][j] = float('inf')
    for i in xrange(1, m):
        for j in xrange(1, n):
            aC = table[i][j - V[i-1]] if j - V[i-1] >= 0 else float('inf')
            table[i][j] = min(table[i-1][j], 1 + aC)
    return table[m-1][n-1]

Both implementations will always return the optimal solution, but the second one will be much faster for large inputs. Notice that the greedy algorithm suggested in other answers gives an optimal solution only for certain combinations of currency - for instance, it works for the American coins.

回答2:

Try using a greedy algorithm (largest coins first). Remove the coin from the list after you apply to the total and call the function again from within.

回答3:

So the best place to look for dynamic programming implementation is: interactive python

However, after a bit of testing I found that it is not the fastest solution. The good thing about it is that one run is enough to find values for all change values up needed amount.

My favourite is still to use recursion with caching:

from collections import defaultdict
from functools import lru_cache


@lru_cache(maxsize=1024)
def greedy(coin_list, change):
    if change in coin_list:
        return defaultdict(int, [(change, 1)])
    else:
        min_coins = change
        for c in [coin for coin in coin_list if coin < change]:
            result_min1 = greedy(coin_list, change - c)
            num_coins = 1 + sum(result_min1.values())
            if num_coins <= min_coins:
                min_coins = num_coins
                result = defaultdict(int, [(c, 1)])
                for c1, n in result_min1.items():
                    result[c1] += n
        return result