递归记忆化解决方案"count changes"

Question

我正在尝试通过记忆解决 "Counting Change" 问题。

Consider the following problem: How many different ways can we make change of .00, given half-dollars, quarters, dimes, nickels, and pennies? More generally, can we write a function to compute the number of ways to change any given amount of money using any set of currency denominations?

以及递归的直观解决方案。

使用n种硬币改变金额a的方法数等于

使用除第一种硬币以外的所有硬币来改变 a 的方法数，加上
使用所有n种硬币来改变较小金额a - d的方法的数量，其中d是第一种硬币的面额。

#+BEGIN_SRC python :results output
# cache = {} # add cache 
def count_change(a, kinds=(50, 25, 10, 5, 1)):
    """Return the number of ways to change amount a using coin kinds."""
    if a == 0:
        return 1
    if a < 0 or len(kinds) == 0:
        return 0
    d = kinds[0]　# d for digit
    return count_change(a, kinds[1:]) + count_change(a - d, kinds) 
print(count_change(100))
#+END_SRC
#+RESULTS:
: 292

我尽量发挥记忆力，

Signature: count_change(a, kinds=(50, 25, 10, 5, 1))
Source:   
def count_change(a, kinds=(50, 25, 10, 5, 1)):
    """Return the number of ways to change amount a using coin kinds."""
    if a == 0:
        return 1
    if a < 0 or len(kinds) == 0:
        return 0
    d = kinds[0]
    cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
    return cache[a]

它适用于像

这样的小数字

In [17]: count_change(120)
Out[17]: 494

研究大数

In [18]: count_change(11000)                        
---------------------------------------------------------------------------
RecursionError                            Traceback (most recent call last)
<ipython-input-18-52ba30c71509> in <module>
----> 1 count_change(11000)

/tmp/ipython_edit_h0rppahk/ipython_edit_uxh2u429.py in count_change(a, kinds)
      9         return 0
     10     d = kinds[0]
---> 11     cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
     12     return cache[a]

... last 1 frames repeated, from the frame below ...

/tmp/ipython_edit_h0rppahk/ipython_edit_uxh2u429.py in count_change(a, kinds)
      9         return 0
     10     d = kinds[0]
---> 11     cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
     12     return cache[a]

RecursionError: maximum recursion depth exceeded in comparison

记忆解决方案有什么问题？

Answer 1

在memoized版本中，count_change函数必须考虑到您可以使用的最高硬币指数当你进行递归调用时，这样你就可以使用已经计算出的值...

def count_change(n, k, kinds):
    if n < 0:
        return 0
    if (n, k) in cache:
        return cache[n,k]
    if k == 0:
        v = 1
    else:
        v = count_change(n-kinds[k], k, kinds) + count_change(n, k-1, kinds)
    cache[n,k] = v
    return v

你可以试试:

cache = {}
count_change(120,4, [1, 5, 10, 25, 50])

给出 494

而 :

cache = {}
count_change(11000,4, [1, 5, 10, 25, 50])

输出：9930221951

递归记忆化解决方案"count changes"

Recursive memoization solutio to solve "count changes"

python-3.x

algorithm

coin-change