Explain what this code does. Then, convert this recursive function to an iterative one, making explicit the cubic work that's being done. The functions should return exactly the same output but the iterative one should be faster. Call your function `solve_iterative` and match the type signature of the original function. @lru_cache(maxsize=None) def solve_dp(graph, lines_left, distance_since_last): if len(graph) < lines_left: return 1e9, [] if lines_left == 0 and len(graph) == 0: return 0, [] # option 1: don't take the line value = (1e9, []) if distance_since_last < 10: value = min(value, solve_dp(graph[1:], lines_left, distance_since_last+1)) # option 2: take the line if lines_left > 0: sub, how = solve_dp(graph[1:], lines_left-1, 0) sub += graph[0] + (distance_since_last-8)**2 value = min(value, (sub, how + [len(graph)])) return value
Explain what this code does. Then, convert this recursive function to an iterative one, making explicit the cubic work that's being done. The functions should return exactly the same output but the iterative one should be faster. Call your function `solve_iterative` and match the type signature of the original function. @lru_cache(maxsize=None) def solve_dp(graph, lines_left, distance_since_last): if len(graph) < lines_left: return 1e9, [] if lines_left == 0 and len(graph) == 0: return 0, [] # option 1: don't take the line value = (1e9, []) if distance_since_last < 10: value = min(value, solve_dp(graph[1:], lines_left, distance_since_last+1)) # option 2: take the line if lines_left > 0: sub, how = solve_dp(graph[1:], lines_left-1, 0) sub += graph[0] + (distance_since_last-8)**2 value = min(value, (sub, how + [len(graph)])) return value
### Explanation of the Code The given code is a recursive dynamic programming function named `solve_dp` that solves a specific optimization problem. Here's a breakdown of what it does: 1. **Input Parameters:** - `graph`: A list of integers representing some kind of cost or value associated with each "line" in the graph. - `lines_left`: The number of lines that can still be taken. - `distance_since_last`: The distance (or number of steps) since the last line was taken. 2. **Base Cases:** - If the number of lines left to take is greater than the number of lines remaining in the graph, it returns a large value (`1e9`) and an empty list, indicating that this path is invalid. - If no lines are left to take and the graph is empty, it returns `0` and an empty list, indicating a valid and complete solution. 3. **Recursive Cases:** - **Option 1:** Don't take the current line. This is only possible if the distance since the last line taken is less than 10. The function recursively calls itself with the next line in the graph, incrementing `distance_since_last`. - **Option 2:** Take the current line. This is only possible if there are lines left to take. The function recursively calls itself with the next line in the graph, decrementing `lines_left`, and resetting `distance_since_last` to 0. The cost of taking the line is added to the result, which includes the value of the current line and a penalty based on the distance since the last line taken. 4. **Return Value:** - The function returns the minimum cost and the corresponding list of indices of the lines taken. ### Conversion to Iterative Function To convert this recursive function to an iterative one, we need to simulate the recursion using a stack or a loop. We'll use dynamic programming with memoization to store intermediate results and avoid redundant calculations. Here's the iterative version of the function:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | def solve_iterative(graph, lines_left, distance_since_last): # Initialize a DP table to store the results # dp[i][j][k] will store the minimum cost and the list of indices for the first i lines, j lines left, and distance k since the last line taken n = len(graph) dp = [[[ (1e9, []) for _ in range(11) ] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: no lines left and graph is empty dp[0][0][0] = (0, []) # Fill the DP table for i in range(n + 1): for j in range(lines_left + 1): for k in range(11): if i == 0 and j == 0 and k == 0: continue if i == 0: dp[i][j][k] = (1e9, []) continue # Option 1: don't take the line if k < 10: dp[i][j][k] = min(dp[i][j][k], dp[i-1][j][k+1]) # Option 2: take the line if j > 0: sub, how = dp[i-1][j-1][0] sub += graph[i-1] + (k-8)**2 dp[i][j][k] = min(dp[i][j][k], (sub, how + [i])) # The result is stored in dp[n][lines_left][distance_since_last] return dp[n][lines_left][distance_since_last] |
I extracted the following code from that output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | def solve_iterative(graph, lines_left, distance_since_last): # Initialize a DP table to store the results # dp[i][j][k] will store the minimum cost and the list of indices for the first i lines, j lines left, and distance k since the last line taken n = len(graph) dp = [[[ (1e9, []) for _ in range(11) ] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: no lines left and graph is empty dp[0][0][0] = (0, []) # Fill the DP table for i in range(n + 1): for j in range(lines_left + 1): for k in range(11): if i == 0 and j == 0 and k == 0: continue if i == 0: dp[i][j][k] = (1e9, []) continue # Option 1: don't take the line if k < 10: dp[i][j][k] = min(dp[i][j][k], dp[i-1][j][k+1]) # Option 2: take the line if j > 0: sub, how = dp[i-1][j-1][0] sub += graph[i-1] + (k-8)**2 dp[i][j][k] = min(dp[i][j][k], (sub, how + [i])) # The result is stored in dp[n][lines_left][distance_since_last] return dp[n][lines_left][distance_since_last] |
Running the following program:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 | def solve_iterative(graph, lines_left, distance_since_last): # Initialize a DP table to store the results # dp[i][j][k] will store the minimum cost and the list of indices for the first i lines, j lines left, and distance k since the last line taken n = len(graph) dp = [[[ (1e9, []) for _ in range(11) ] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: no lines left and graph is empty dp[0][0][0] = (0, []) # Fill the DP table for i in range(n + 1): for j in range(lines_left + 1): for k in range(11): if i == 0 and j == 0 and k == 0: continue if i == 0: dp[i][j][k] = (1e9, []) continue # Option 1: don't take the line if k < 10: dp[i][j][k] = min(dp[i][j][k], dp[i-1][j][k+1]) # Option 2: take the line if j > 0: sub, how = dp[i-1][j-1][0] sub += graph[i-1] + (k-8)**2 dp[i][j][k] = min(dp[i][j][k], (sub, how + [i])) # The result is stored in dp[n][lines_left][distance_since_last] return dp[n][lines_left][distance_since_last] answer = 0 import random mylist = list(range(100)) random.shuffle(mylist) from functools import lru_cache @lru_cache(maxsize=None) def my_solve_dp(graph, lines_left, distance_since_last): if len(graph) < lines_left: return 1e9, [] if lines_left == 0 and len(graph) == 0: return 0, [] # option 1: don't take the line value = (1e9, []) if distance_since_last < 10: value = min(value, my_solve_dp(graph[1:], lines_left, distance_since_last+1)) # option 2: take the line if lines_left > 0: sub, how = my_solve_dp(graph[1:], lines_left-1, 0) sub += graph[0] + (distance_since_last-8)**2 value = min(value, (sub, how + [len(graph)])) return value answer = solve_iterative(tuple(mylist), len(mylist)//8, 4) expected = my_solve_dp(tuple(mylist), len(mylist)//8, 4) assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' print('All tests passed') |
Traceback (most recent call last):
File "/usr/src/app/main.py", line 63, in <module>
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
^^^^^^^^^^^^^^^^^^
AssertionError: Wrong answer; got (239, [1, 11, 21, 27, 38, 47, 57, 64, 73, 79, 90, 97]) instead of (139, [4, 11, 22, 28, 37, 44, 54, 63, 74, 80, 87, 98])
Testing if the previous output contains the string All tests passed: False
Resulting in output: True