在python中递归实现“最小硬币数”

Question

这个问题和这里问的一样。

给定一个硬币列表，它们的值（c1，c2，c3，... cj，...），以及总和i。找出硬币总数为i的最小数量（我们可以根据需要使用一种类型的硬币），或者报告不可能以总和为S的方式选择硬币。

我昨天刚刚被介绍到动态编程，我试图为它编写一个代码。

# Optimal substructure: C[i] = 1 + min_j(C[i-cj])
cdict = {}
def C(i, coins):

    if i <= 0:
        return 0

    if i in cdict:
        return cdict[i]
    else:
        answer = 1 + min([C(i - cj, coins) for cj in coins])
        cdict[i] = answer
        return answer

这里，C[i]是货币量“i”的最优解。可用的硬币有{c1，c2，…，cj，…}对于这个程序，我增加了递归限制，以避免最大递归深度超过错误。但是，这个程序只给出了一些正确的答案，当一个解决方案不可能时，它并不表明这一点。

我的代码有什么问题，如何更正？

Answer 1

这里有一个有趣的方法。有点像黑客，但这就是为什么它很有趣。

    import math

    def find_change(coins, value):
        coins = sorted(coins, reverse=True)
        coin_dict = {}
        for c in coins:
            if value % c == 0:
                coin_dict[c] = value / c
                return coin_dict
            else:
                coin_dict[c] = math.trunc(value/ float(c))
                value -= (c * coin_dict[c])

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}

下面是带有边缘大小写保护的注释的相同解决方案

    import math
    def find_change(coins, value):
        '''
        :param coins: List of the value of each coin [25, 10, 5, 1]
        :param value: the value you want to find the change for ie; 69 cents
        :return: a change dictionary where the key is the coin, and the value is how many times it is used in finding the minimum change
        '''
        change_dict = {}  # CREATE OUR CHANGE DICT, THIS IS A DICT OF THE COINS WE ARE RETURNING, A COIN PURSE
        coins = sorted(coins, reverse=True)  # SORT COINS SO WE START LOOP WITH BIGGEST COIN VALUE
        for c in coins:
            for d in coins:     # THIS LOOP WAS ADDED BY A SMART STUDENT:  IE IN THE CASE OF IF THERE IS A 7cent COIN AND YOU ARE LOOKING FOR CHANGE FOR 14 CENTS, WITHOUT THIS D FOR LOOP IT WILL RETURN 10: 1, 1: 4
                if (d != 1) & (value % d == 0):
                    change_dict[d] = value / d
                    return change_dict
            if value % c == 0:      # IF THE VALUE DIVIDED BY THE COIN HAS NO REMAINDER,  # ie, if there is no remainder, all the neccessary change has been given  # PLACE THE NUMBER OF TIMES THAT COIN IS USED IN THE change_dict  # YOU ARE FINISHED NOW RETURN THE change_dict
                change_dict[c] = value / c
                return change_dict
            else:
                change_dict[c] = math.trunc(value/ float(c))        # PLACE THAT NUMBER INTO OUR coin_dict  # DIVIDE THE VALUE BY THE COIN, THEN GET JUST THE WHOLE NUMBER  # IE 69 / 25.0 = 2.76  # math.trunc(2.76) == 2    # AND THAT IS HOW MANY TIMES IT WILL EVENLY GO INTO THE VALUE,
                amount = (c * change_dict[c])  # NOW TAKE THE NUMBER OF COINS YOU HAVE IN YOUR  UPDATE THE VALUE BY SUBTRACTING THE c * TIME NUMBER OF TIMES IT WAS USED    # AMOUNT IS HOW MUCH CHANGE HAS BEEN PUT INTO THE CHANGE DICT ON THIS LOOP  # FOR THE CASE OF 69, YOU GIVE 2 25CENT COINS, SO 2 * 25 = 50, 19 = 69 - 50
                value = value - amount              # NOW, UPDATE YOUR VALUE, SO THE NEXT TIME IT GOES INTO THIS LOOP, IT WILL BE LOOKING FOR THE MIN CHANGE FOR 19 CENTS...

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}
    edge_case_coins = [1, 7, 10, 25]
    edge_case_answer = find_change(coins, 14)
    print edge_case_answer
    [OUT]: {7: 2}

Answer 2

正如注释所说，当i

cdict = {}
def C(i, coins):

    if i == 0:
        return 0

   if i < 0:
        return 1e100    # Return infinity in ideally

    if i in cdict:
        return cdict[i]
    else:
        answer = 1 + min([C(i - cj, coins) for cj in coins])
        cdict[i] = answer
    return answer

现在，只要函数返回1e100，就意味着解决方案是不可能的。

例如：

$ python2 coins.py 13555 1 5 9
1507  coins
$ python2 coins.py 139 1 5 9
19  coins
$ python2 coins.py 139  5 9
19  coins
$ python2 coins.py 13977  5 9
1553  coins
$ python2 coins.py 13977   9
1553  coins
$ python2 coins.py 139772   9
1e+100  coins

与用法：

python2 coins.py <amount> <coin1> <coin2> ...

Answer 3

这是一个很好的算法问题，但是说实话，我不认为你的实现是正确的，或者可能是我不理解你的函数的输入/输出，为此我道歉。

这是您的实现的修改版本。

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]

这是我试图解决一个类似的问题，但这次返回一个硬币列表。我最初从一个递归算法开始，它接受一个总和和和一个硬币列表，如果找不到这样的配置，它可以返回一个最小硬币数的列表，或者返回一个无。

def get_min_coin_configuration(sum = None, coins = None):
if sum in coins: # if sum in coins, nothing to do but return.
    return [sum]
elif max(coins) > sum: # if the largest coin is greater then the sum, there's nothing we can do.
    return None
else: # check for each coin, keep track of the minimun configuration, then return it.
    min_length = None
    min_configuration = None
    for coin in coins:
        results = get_min_coin_configuration(sum = sum - coin, coins = coins)
        if results != None:
            if min_length == None or (1 + len(results)) < len(min_configuration):
                min_configuration = [coin] + results
                min_length = len(min_configuration)
    return min_configuration

现在让我们看看是否可以通过使用动态编程（我称之为缓存）来改进它。

def get_min_coin_configuration(sum = None, coins = None, cache = None):
if cache == None: # this is quite crucial if its in the definition its presistent ...
    cache = {}
if sum in cache:
    return cache[sum]
elif sum in coins: # if sum in coins, nothing to do but return.
    cache[sum] = [sum]
    return cache[sum]
elif max(coins) > sum: # if the largest coin is greater then the sum, there's nothing we can do.
    cache[sum] = None
    return cache[sum]
else: # check for each coin, keep track of the minimun configuration, then return it.
    min_length = None
    min_configuration = None
    for coin in coins:
        results = get_min_coin_configuration(sum = sum - coin, coins = coins, cache = cache)
        if results != None:
            if min_length == None or (1 + len(results)) < len(min_configuration):
                min_configuration = [coin] + results
                min_length = len(min_configuration)
    cache[sum] = min_configuration
    return cache[sum]

现在让我们运行一些测试。

assert all([ get_min_coin_configuration(**test[0]) == test[1] for test in
[({'sum':25,  'coins':[1, 5, 10]}, [5, 10, 10]),
 ({'sum':153, 'coins':[1, 5, 10, 50]}, [1, 1, 1, 50, 50, 50]),
 ({'sum':100, 'coins':[1, 5, 10, 25]}, [25, 25, 25, 25]),
 ({'sum':123, 'coins':[5, 10, 25]}, None),
 ({'sum':100, 'coins':[1,5,25,100]}, [100])] ])

假设这些测试不够健壮，您也可以这样做。

import random
random_sum = random.randint(10**3, 10**4)
result = get_min_coin_configuration(sum = random_sum, coins = random.sample(range(10**3), 200))
assert sum(result) == random_sum

有可能没有这样的硬币组合等于我们的随机总和，但我相信这不太可能。。。

我确信有更好的实现，我试图强调可读性而不是性能。祝你好运

更新的以前的代码有一个小错误，它假设检查最小硬币而不是最大硬币，重新编写符合pep8的算法，并在找不到组合时返回[]，而不是无。

def get_min_coin_configuration(total_sum, coins, cache=None):  # shadowing python built-ins is frowned upon.
    # assert(all(c > 0 for c in coins)) Assuming all coins are > 0
    if cache is None:  # initialize cache.
        cache = {}
    if total_sum in cache:  # check cache, for previously discovered solution.
        return cache[total_sum]
    elif total_sum in coins:  # check if total_sum is one of the coins.
        cache[total_sum] = [total_sum]
        return [total_sum]
    elif min(coins) > total_sum:  # check feasibility, if min(coins) > total_sum
        cache[total_sum] = []     # no combination of coins will yield solution as per our assumption (all +).
        return []
    else:
        min_configuration = []  # default solution if none found.
        for coin in coins:  # iterate over all coins, check which one will yield the smallest combination.
            results = get_min_coin_configuration(total_sum - coin, coins, cache=cache)  # recursively search.
            if results and (not min_configuration or (1 + len(results)) < len(min_configuration)):  # check if better.
                min_configuration = [coin] + results
        cache[total_sum] = min_configuration  # save this solution, for future calculations.
    return cache[total_sum]



assert all([ get_min_coin_configuration(**test[0]) == test[1] for test in
             [({'total_sum':25,  'coins':[1, 5, 10]}, [5, 10, 10]),
              ({'total_sum':153, 'coins':[1, 5, 10, 50]}, [1, 1, 1, 50, 50, 50]),
              ({'total_sum':100, 'coins':[1, 5, 10, 25]}, [25, 25, 25, 25]),
              ({'total_sum':123, 'coins':[5, 10, 25]}, []),
              ({'total_sum':100, 'coins':[1,5,25,100]}, [100])] ])