# Python 为什么将 DP 添加到我的递归中会使其停止正常工作

Can someone please tell me why the two codes give different outputs. Aren't they the same code with two different implementations (one is brute force and the other one is using Dynamic Programming).

Program Description: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

``````def minPathSum_dp(grid):
def dfs(i, j, pathSum):
if (i, j) in dp:
return dp[(i, j)]

if i == len(grid)-1 and j == len(grid[i])-1:
return pathSum + grid[i][j]

if i == len(grid)-1:
return dfs(i, j+1, pathSum + grid[i][j])

if j == len(grid[i])-1:
return dfs(i+1, j, pathSum + grid[i][j])

path1 = dfs(i+1, j, pathSum + grid[i][j])
path2 = dfs(i, j+1, pathSum + grid[i][j])
dp[(i, j)] = min(path1, path2)
return dp[(i, j)]

dp = {}
return dfs(0, 0, 0)

def minPathSum_bf(grid):
def dfs(i, j, pathSum):
if i == len(grid)-1 and j == len(grid[i])-1:
return pathSum + grid[i][j]

if i == len(grid)-1:
return dfs(i, j+1, pathSum + grid[i][j])

if j == len(grid[i])-1:
return dfs(i+1, j, pathSum + grid[i][j])

path1 = dfs(i+1, j, pathSum + grid[i][j])
path2 = dfs(i, j+1, pathSum + grid[i][j])
return min(path1, path2)

return dfs(0, 0, 0)

grid = [[1,4,8,6,2,2,1,7],[4,7,3,1,4,5,5,1],[8,8,2,1,1,8,0,1],[8,9,2,9,8,0,8,9],[5,7,5,7,1,8,5,5],[7,0,9,4,5,6,5,6],[4,9,9,7,9,1,9,0]]

UPDATED: WORKING SOLUTION

def minPathSum(self, grid: List[List[int]]) -> int:
def dfs(i, j):
if (i, j) in dp:
return dp[(i, j)]

if i == len(grid)-1 and j == len(grid[i])-1:
return grid[i][j]

if i == len(grid)-1:
dp[(i, j)] = grid[i][j] + dfs(i, j+1)
return dp[(i, j)]

if j == len(grid[i])-1:
dp[(i, j)] = grid[i][j] + dfs(i+1, j)
return dp[(i, j)]

path1 = grid[i][j] + dfs(i+1, j)
path2 = grid[i][j] + dfs(i, j+1)
dp[(i, j)] = min(path1, path2)
return dp[(i, j)]

dp = {}
return dfs(0, 0)

print(minPathSum_dp(grid))
print(minPathSum_bf(grid))
``````

## 推荐答案

(A quick terminology note - what you’re doing here seems to me to be closer to memoization than dynamic programming, since you’re computing things top-down rather than bottom-up.)

Your recursive function has three parameters. The first two are “where am I right now?” The third is “what is the sum along the path that’s taken me here so far?” So, for example, calling `dfs(row, col, 137)` gives you the best cost you can reach if you are at (row, col) and the cost of the current path is 137, and calling `dfs(row, col, 42)` gives you the cost if your current path has cost 42.

However, your DP/memoization table is only keyed on the first two parameters. This means that the value you’re writing down at position (row, col) would need to be the answer for both `dfs(row, col, 137)` and `dfs(row, col, 42)`. But that’s clearly not going to work.

You could technically fix this by having your DP/memoization table write down answers keyed by all three parameters. However, that’s not going to make things run quickly, since you’re unlikely to end up with two or more recursive calls being made to the same position in the grid with the same prefix sum.

The more proper way to fix this is to change your recursive strategy so that you don’t need that third parameter of the path cost up to your current point. By having that parameter, each recursive call has to be aware of the specific call chain that got it there. Instead, see if you can find a way to make the recursive function just take in a (row, col) pair and return the best cost from that point to the destination. Once you’ve gotten that working, you can add in memoization and it’ll work just fine.