在递归代码中断功能中实现记忆

Question

我似乎无法弄清楚是什么破坏了我的代码。

我写了一个取金字塔的代码：

[[1],
 [2,3],
 [4,5,6],
 [7,8,9,10]]

从头开始（金字塔[0][0]）计算移动到下面的项目，或者递归到下面的项目并向右移动可能达到的最大总和

在此示例中，输出应为 20。

这是没有记忆的代码，工作正常：

def max_trail(pyramid,row=0,col=0):
    if row == (len(pyramid)-1):
        return pyramid[row][col]
    else:
        return pyramid[row][col] + max(max_trail(pyramid ,row+1 ,col),
                                       max_trail(pyramid ,row+1, col+1))

但是当我尝试应用记忆时，过程中出现了问题。我错过了什么？

这是损坏的代码：

def max_trail_mem(pyramid,row=0,col=0,mem=None):
    if mem==None:
        mem={}
    key = ((row),(col))
    if row == (len(pyramid)-1):
        if key not in mem:
            value = pyramid[row][col]
            mem[key]=value
            return mem[key]
        return mem[key]
    else:
        key = (row+1),(col)
        if key not in mem:
            mem[(row+1),(col)] = max_trail_mem(pyramid ,row+1 ,col,mem)
        key = (row+1),(col+1)
        if key not in mem:
            mem[(row+1),(col+1)]=max_trail_mem(pyramid ,row+1, col+1,mem)
    return max(mem[(row+1),(col)],mem[(row+1),(col+1)])

这让我的学生生活少了几个小时。任何帮助将不胜感激！

Answer 1

您上次 return 在 max(... 之前忘记了 pyramid[row][col] +。添加它会为您的示例提供 20（请参阅代码的最后一行）：

def max_trail_mem(pyramid,row=0,col=0,mem=None):
    if mem==None:
        mem={}
    key = ((row),(col))
    if row == (len(pyramid)-1):
        if key not in mem:
            value = pyramid[row][col]
            mem[key]=value
            return mem[key]
        return mem[key]
    else:
        key = (row+1),(col)
        if key not in mem:
            mem[(row+1),(col)] = max_trail_mem(pyramid ,row+1 ,col,mem)
        key = (row+1),(col+1)
        if key not in mem:
            mem[(row+1),(col+1)] = max_trail_mem(pyramid ,row+1, col+1,mem)
    return pyramid[row][col] + max(mem[(row+1),(col)],mem[(row+1),(col+1)])

Answer 2

from timeit import timeit
import math

from repoze.lru import CacheMaker
cache_maker=CacheMaker()


def max_trail(pyramid,row=0,col=0):
    if row == (len(pyramid)-1):
        return pyramid[row][col]
    else:

        mt1 = max_trail(pyramid ,row+1 ,col)
        mt2 = max_trail(pyramid ,row+1, col+1)
        return pyramid[row][col] + max(mt1, mt2)

@cache_maker.lrucache(maxsize='1000', name='pyramid')
def max_trail_with_memoization(pyramid,row=0,col=0):
    if row == (len(pyramid)-1):
        return pyramid[row][col]
    else:

        mt1 = max_trail(pyramid ,row+1 ,col)
        mt2 = max_trail(pyramid ,row+1, col+1)
        return pyramid[row][col] + max(mt1, mt2)

# Build pyramid
pyramid = ()
c = 0
for i in range(20):
    row = ()
    for j in range(i):
        c += 1
        row += (c,)
    if row:
        pyramid += (tuple(row),)

# Repetitions to time
number = 20

# Time it
print('without memoization:  ', timeit('f(t)', 'from __main__ import max_trail as f, pyramid as t', number=number))
print('with memoization      ', timeit('f(t)', 'from __main__ import max_trail_with_memoization as f, pyramid as t', number=number))




print max_trail(pyramid)

Returns:

without memoization:   3.9645819664001465
with memoization:      0.18628692626953125

在递归代码中断功能中实现记忆

Implementing memoization in recursive code break functionality

python

recursion

memoization