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 provided recursive function `solve_dp` is a dynamic programming solution designed to determine the optimal way to select a certain number of lines (`lines_left`) from a given `graph`. The goal is to minimize a computed cost while adhering to specific constraints related to the distance between selected lines. Here's a step-by-step breakdown of what the function does: 1. **Base Cases:** - **Insufficient Lines:** If the number of remaining lines in the `graph` is less than the `lines_left`, it's impossible to select the required number of lines. The function returns a very large cost (`1e9`) and an empty list to indicate an invalid path. - **Exact Match:** If there are no lines left to select (`lines_left == 0`) and the `graph` is fully traversed (`len(graph) == 0`), the cost is `0`, and the path is an empty list, indicating a successful selection. 2. **Recursive Cases:** - **Option 1: Do Not Take the Current Line** - This option considers skipping the current line in the `graph`. - It only proceeds if the `distance_since_last` taken line is less than `10`, ensuring that lines are not too far apart. - The function recursively calls itself with the remaining graph (`graph[1:]`), the same number of `lines_left`, and increments the `distance_since_last` by `1`. - **Option 2: Take the Current Line** - This option considers selecting the current line. - It only proceeds if there are lines left to take (`lines_left > 0`). - The function recursively calls itself with the remaining graph (`graph[1:]`), decrements the `lines_left` by `1`, and resets the `distance_since_last` to `0` since a line has been taken. - The cost is updated by adding the value of the current line (`graph[0]`) and a penalty based on the distance since the last taken line (`(distance_since_last - 8)^2`). - The position of the taken line (`len(graph)`) is appended to the path. - **Choosing the Minimum Cost:** - After evaluating both options, the function selects the one with the minimum cost and returns the corresponding cost and path. 3. **Memoization:** - The `@lru_cache(maxsize=None)` decorator is used to cache the results of subproblems, avoiding redundant computations and enhancing performance. ### Iterative Version: `solve_iterative` To optimize the recursive approach and eliminate the overhead of recursive calls and slicing, we can convert the function to an iterative dynamic programming (DP) solution. This iterative version explicitly handles the cubic complexity by iterating through all possible positions, remaining lines, and distances since the last taken line. Here's the step-by-step approach for the iterative solution: 1. **DP Table Setup:** - Create a 3D DP table where `dp[i][j][k]` represents the minimum cost and corresponding path when considering the first `i` lines of the graph, having `j` lines left to take, and a distance of `k` since the last taken line. 2. **Initialization:** - Initialize the DP table with infinite costs and empty paths. - Set the base case where no lines are left to take and no lines are considered; the cost is `0`, and the path is empty. 3. **State Transition:** - Iterate through each position in the graph. - For each position, iterate through all possible numbers of lines left to take. - For each combination, consider both options: - **Not Taking the Current Line:** Update the distance since the last taken line and carry forward the previous cost and path. - **Taking the Current Line:** Reset the distance since the last taken line, update the cost with the current line's value and the distance penalty, and append the current position to the path. 4. **Result Extraction:** - After filling the DP table, extract the minimum cost and corresponding path from the last position with the required number of lines taken. Below is the iterative implementation matching the original function's signature:
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 | from typing import List, Tuple def solve_iterative(graph: List[int], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) INF = float('inf') # Initialize DP table # dp[i][j][k] = (cost, path) # i: current position in graph (0 to n) # j: lines left to take (0 to lines_left) # k: distance since last taken line (0 to distance_limit) distance_limit = 10 # As per the condition distance_since_last <10 dp = [[[ (INF, []) for _ in range(distance_limit +1)] for _ in range(lines_left +1)] for _ in range(n +1)] # Base case: at position 0, with no lines left, distance_since_last can be anything for k in range(distance_limit +1): dp[0][0][k] = (0, []) for i in range(n): for j in range(lines_left +1): for k in range(distance_limit +1): current_cost, current_path = dp[i][j][k] if current_cost == INF: continue # Option 1: Do not take the current line if k < distance_limit: new_k = k +1 if new_k <= distance_limit: new_cost, new_path = dp[i+1][j][new_k] if current_cost < new_cost: dp[i+1][j][new_k] = (current_cost, current_path.copy()) # Option 2: Take the current line if j < lines_left: new_j = j +1 new_k = 0 # Calculate new cost penalty = (k -8)**2 total_cost = current_cost + graph[i] + penalty existing_cost, existing_path = dp[i+1][j+1][new_k] if total_cost < existing_cost: dp[i+1][j+1][new_k] = (total_cost, current_path + [i]) # Find the minimum cost among all possible distance_since_last at the end min_cost = INF min_path = [] for k in range(distance_limit +1): cost, path = dp[n][lines_left][k] if cost < min_cost: min_cost = cost min_path = path return min_cost, min_path |
1 2 3 4 5 6 7 | if __name__ == "__main__": graph = [5, 3, 6, 2, 7, 4, 5, 3, 6, 2] lines_left = 3 distance_since_last = 0 cost, path = solve_iterative(graph, lines_left, distance_since_last) print(f"Minimum Cost: {cost}") print(f"Path: {path}") |
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 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 | from typing import List, Tuple def solve_iterative(graph: List[int], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) INF = float('inf') # Initialize DP table # dp[i][j][k] = (cost, path) # i: current position in graph (0 to n) # j: lines left to take (0 to lines_left) # k: distance since last taken line (0 to distance_limit) distance_limit = 10 # As per the condition distance_since_last <10 dp = [[[ (INF, []) for _ in range(distance_limit +1)] for _ in range(lines_left +1)] for _ in range(n +1)] # Base case: at position 0, with no lines left, distance_since_last can be anything for k in range(distance_limit +1): dp[0][0][k] = (0, []) for i in range(n): for j in range(lines_left +1): for k in range(distance_limit +1): current_cost, current_path = dp[i][j][k] if current_cost == INF: continue # Option 1: Do not take the current line if k < distance_limit: new_k = k +1 if new_k <= distance_limit: new_cost, new_path = dp[i+1][j][new_k] if current_cost < new_cost: dp[i+1][j][new_k] = (current_cost, current_path.copy()) # Option 2: Take the current line if j < lines_left: new_j = j +1 new_k = 0 # Calculate new cost penalty = (k -8)**2 total_cost = current_cost + graph[i] + penalty existing_cost, existing_path = dp[i+1][j+1][new_k] if total_cost < existing_cost: dp[i+1][j+1][new_k] = (total_cost, current_path + [i]) # Find the minimum cost among all possible distance_since_last at the end min_cost = INF min_path = [] for k in range(distance_limit +1): cost, path = dp[n][lines_left][k] if cost < min_cost: min_cost = cost min_path = path return min_cost, min_path |
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 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 | from typing import List, Tuple def solve_iterative(graph: List[int], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) INF = float('inf') # Initialize DP table # dp[i][j][k] = (cost, path) # i: current position in graph (0 to n) # j: lines left to take (0 to lines_left) # k: distance since last taken line (0 to distance_limit) distance_limit = 10 # As per the condition distance_since_last <10 dp = [[[ (INF, []) for _ in range(distance_limit +1)] for _ in range(lines_left +1)] for _ in range(n +1)] # Base case: at position 0, with no lines left, distance_since_last can be anything for k in range(distance_limit +1): dp[0][0][k] = (0, []) for i in range(n): for j in range(lines_left +1): for k in range(distance_limit +1): current_cost, current_path = dp[i][j][k] if current_cost == INF: continue # Option 1: Do not take the current line if k < distance_limit: new_k = k +1 if new_k <= distance_limit: new_cost, new_path = dp[i+1][j][new_k] if current_cost < new_cost: dp[i+1][j][new_k] = (current_cost, current_path.copy()) # Option 2: Take the current line if j < lines_left: new_j = j +1 new_k = 0 # Calculate new cost penalty = (k -8)**2 total_cost = current_cost + graph[i] + penalty existing_cost, existing_path = dp[i+1][j+1][new_k] if total_cost < existing_cost: dp[i+1][j+1][new_k] = (total_cost, current_path + [i]) # Find the minimum cost among all possible distance_since_last at the end min_cost = INF min_path = [] for k in range(distance_limit +1): cost, path = dp[n][lines_left][k] if cost < min_cost: min_cost = cost min_path = path return min_cost, min_path 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 86, in <module>
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
AssertionError: Wrong answer; got (210, [3, 13, 20, 26, 33, 40, 48, 56, 65, 72, 83, 91]) instead of (211, [9, 17, 28, 35, 44, 52, 60, 67, 74, 80, 87, 97])
Testing if the previous output contains the string All tests passed: False
Resulting in output: True
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 provided `solve_dp` function is a recursive dynamic programming (DP) solution designed to process a sequence of elements (`graph`) and make optimal decisions based on two parameters: 1. **`lines_left`**: The number of lines (or selections) remaining that can be chosen from the `graph`. 2. **`distance_since_last`**: The number of steps (or elements) since the last line was taken. The goal of the function is to minimize a certain cost associated with selecting lines from the `graph`. The function returns a tuple consisting of the minimum total cost and a list of indices corresponding to the selected lines. Here's a step-by-step breakdown of what the function does: 1. **Base Cases**: - **Insufficient Elements**: If the number of remaining elements in `graph` is less than `lines_left`, it's impossible to select the required number of lines. In this case, it returns a large cost (`1e9`) and an empty list. - **Exact Match**: If there are no lines left to select (`lines_left == 0`) and no elements left in `graph`, the cost is `0`, and the list of selected indices is empty. 2. **Decision Making**: - **Option 1: Do Not Take the Current Line**: - If the `distance_since_last` is less than `10`, the function considers not taking the current line. It recursively calls itself with the remaining `graph` (excluding the first element), the same `lines_left`, and increments `distance_since_last` by `1`. - The cost and selection from this option are stored in `value`. - **Option 2: Take the Current Line**: - If there are lines left to take (`lines_left > 0`), the function considers taking the current line. It recursively calls itself with the remaining `graph` (excluding the first element), decrements `lines_left` by `1`, and resets `distance_since_last` to `0`. - It then adds the cost of the current line (`graph[0]`) and a penalty term `(distance_since_last - 8)^2` to the subtotal cost. - The current index (`len(graph)`) is appended to the list of selected indices. - This option's cost and selection are compared with `value`, and the minimum is retained. 3. **Memoization**: - The function uses `@lru_cache(maxsize=None)` to memoize results of subproblems, ensuring that each unique state (`graph`, `lines_left`, `distance_since_last`) is computed only once. This significantly optimizes the recursive solution by avoiding redundant computations. ### Iterative Conversion To optimize the recursive DP solution further, we can convert it into an iterative DP approach. The recursive solution, even with memoization, can have overhead due to function calls and managing the cache. An iterative approach uses explicit loops and DP tables, which can be more efficient, especially for larger inputs. Below is the `solve_iterative` function that mirrors the behavior of `solve_dp` but operates iteratively. It constructs a DP table to store intermediate results, ensuring that each state is computed only once without the overhead of recursion and memoization.
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 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 | from typing import List, Tuple import sys def solve_iterative(graph: List[float], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) # Define maximum possible distance MAX_DISTANCE = 10 # as per the condition distance_since_last < 10 # Initialize DP table # dp[i][j][k] = (cost, list_of_indices) # i: current index in graph (0 to n) # j: lines left (0 to lines_left) # k: distance since last taken (0 to MAX_DISTANCE) # Initialize with infinity cost and empty list dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: at the end of the graph for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif j > 0 and n < j: dp[n][j][k] = (1e9, []) # Iterate backwards for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line option1_cost = 1e9 option1_indices = [] if k < 10: next_cost, next_indices = dp[i+1][j][k+1] option1_cost = next_cost option1_indices = next_indices.copy() # Option 2: Take the line option2_cost = 1e9 option2_indices = [] if j > 0: next_cost, next_indices = dp[i+1][j-1][0] temp_cost = next_cost + graph[i] + (k - 8) ** 2 if temp_cost < option2_cost: option2_cost = temp_cost option2_indices = next_indices.copy() option2_indices.append(i) # Choose the minimum if option1_cost < option2_cost: dp[i][j][k] = (option1_cost, option1_indices) else: dp[i][j][k] = (option2_cost, option2_indices) # The initial state depends on the starting distance_since_last # However, in recursive version, the initial distance_since_last is given # So we need to adjust the base cases and iteration accordingly # To handle arbitrary starting distance_since_last, we need to iterate on k accordingly # For simplicity, assume distance_since_last is part of the state # Since our current dp table does not account for starting distance, adjust it # Let's redefine the DP to handle arbitrary starting distance # Redefine DP with the correct base cases dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base cases for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif n < j: dp[n][j][k] = (1e9, []) # Fill the DP table for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line if k < MAX_DISTANCE: cost1, indices1 = dp[i+1][j][k+1] else: cost1, indices1 = 1e9, [] # Option 2: Take the line if j > 0: cost2_prev, indices2_prev = dp[i+1][j-1][0] cost2 = cost2_prev + graph[i] + (k - 8) ** 2 indices2 = indices2_prev.copy() indices2.append(i) else: cost2 = 1e9 indices2 = [] # Choose the minimum if cost1 < cost2: dp[i][j][k] = (cost1, indices1.copy()) else: dp[i][j][k] = (cost2, indices2.copy()) # The initial state is at index 0, with 'lines_left' lines left, and 'distance_since_last' final_cost, final_indices = dp[0][lines_left][distance_since_last] return final_cost, final_indices |
1 2 3 4 5 6 7 | if __name__ == "__main__": graph = [5.0, 3.2, 4.1, 2.3, 6.5] lines_left = 2 distance_since_last = 0 cost, selected_indices = solve_iterative(graph, lines_left, distance_since_last) print(f"Minimum Cost: {cost}") print(f"Selected Indices: {selected_indices}") |
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 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 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 | from typing import List, Tuple import sys def solve_iterative(graph: List[float], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) # Define maximum possible distance MAX_DISTANCE = 10 # as per the condition distance_since_last < 10 # Initialize DP table # dp[i][j][k] = (cost, list_of_indices) # i: current index in graph (0 to n) # j: lines left (0 to lines_left) # k: distance since last taken (0 to MAX_DISTANCE) # Initialize with infinity cost and empty list dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: at the end of the graph for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif j > 0 and n < j: dp[n][j][k] = (1e9, []) # Iterate backwards for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line option1_cost = 1e9 option1_indices = [] if k < 10: next_cost, next_indices = dp[i+1][j][k+1] option1_cost = next_cost option1_indices = next_indices.copy() # Option 2: Take the line option2_cost = 1e9 option2_indices = [] if j > 0: next_cost, next_indices = dp[i+1][j-1][0] temp_cost = next_cost + graph[i] + (k - 8) ** 2 if temp_cost < option2_cost: option2_cost = temp_cost option2_indices = next_indices.copy() option2_indices.append(i) # Choose the minimum if option1_cost < option2_cost: dp[i][j][k] = (option1_cost, option1_indices) else: dp[i][j][k] = (option2_cost, option2_indices) # The initial state depends on the starting distance_since_last # However, in recursive version, the initial distance_since_last is given # So we need to adjust the base cases and iteration accordingly # To handle arbitrary starting distance_since_last, we need to iterate on k accordingly # For simplicity, assume distance_since_last is part of the state # Since our current dp table does not account for starting distance, adjust it # Let's redefine the DP to handle arbitrary starting distance # Redefine DP with the correct base cases dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base cases for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif n < j: dp[n][j][k] = (1e9, []) # Fill the DP table for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line if k < MAX_DISTANCE: cost1, indices1 = dp[i+1][j][k+1] else: cost1, indices1 = 1e9, [] # Option 2: Take the line if j > 0: cost2_prev, indices2_prev = dp[i+1][j-1][0] cost2 = cost2_prev + graph[i] + (k - 8) ** 2 indices2 = indices2_prev.copy() indices2.append(i) else: cost2 = 1e9 indices2 = [] # Choose the minimum if cost1 < cost2: dp[i][j][k] = (cost1, indices1.copy()) else: dp[i][j][k] = (cost2, indices2.copy()) # The initial state is at index 0, with 'lines_left' lines left, and 'distance_since_last' final_cost, final_indices = dp[0][lines_left][distance_since_last] return final_cost, final_indices |
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 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 | from typing import List, Tuple import sys def solve_iterative(graph: List[float], lines_left: int, distance_since_last: int) -> Tuple[float, List[int]]: n = len(graph) # Define maximum possible distance MAX_DISTANCE = 10 # as per the condition distance_since_last < 10 # Initialize DP table # dp[i][j][k] = (cost, list_of_indices) # i: current index in graph (0 to n) # j: lines left (0 to lines_left) # k: distance since last taken (0 to MAX_DISTANCE) # Initialize with infinity cost and empty list dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base case: at the end of the graph for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif j > 0 and n < j: dp[n][j][k] = (1e9, []) # Iterate backwards for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line option1_cost = 1e9 option1_indices = [] if k < 10: next_cost, next_indices = dp[i+1][j][k+1] option1_cost = next_cost option1_indices = next_indices.copy() # Option 2: Take the line option2_cost = 1e9 option2_indices = [] if j > 0: next_cost, next_indices = dp[i+1][j-1][0] temp_cost = next_cost + graph[i] + (k - 8) ** 2 if temp_cost < option2_cost: option2_cost = temp_cost option2_indices = next_indices.copy() option2_indices.append(i) # Choose the minimum if option1_cost < option2_cost: dp[i][j][k] = (option1_cost, option1_indices) else: dp[i][j][k] = (option2_cost, option2_indices) # The initial state depends on the starting distance_since_last # However, in recursive version, the initial distance_since_last is given # So we need to adjust the base cases and iteration accordingly # To handle arbitrary starting distance_since_last, we need to iterate on k accordingly # For simplicity, assume distance_since_last is part of the state # Since our current dp table does not account for starting distance, adjust it # Let's redefine the DP to handle arbitrary starting distance # Redefine DP with the correct base cases dp = [[[ (1e9, []) for _ in range(MAX_DISTANCE + 1)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Base cases for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): if j == 0 and n == 0: dp[n][j][k] = (0, []) elif n < j: dp[n][j][k] = (1e9, []) # Fill the DP table for i in range(n-1, -1, -1): for j in range(lines_left + 1): for k in range(MAX_DISTANCE + 1): # Option 1: Don't take the line if k < MAX_DISTANCE: cost1, indices1 = dp[i+1][j][k+1] else: cost1, indices1 = 1e9, [] # Option 2: Take the line if j > 0: cost2_prev, indices2_prev = dp[i+1][j-1][0] cost2 = cost2_prev + graph[i] + (k - 8) ** 2 indices2 = indices2_prev.copy() indices2.append(i) else: cost2 = 1e9 indices2 = [] # Choose the minimum if cost1 < cost2: dp[i][j][k] = (cost1, indices1.copy()) else: dp[i][j][k] = (cost2, indices2.copy()) # The initial state is at index 0, with 'lines_left' lines left, and 'distance_since_last' final_cost, final_indices = dp[0][lines_left][distance_since_last] return final_cost, final_indices 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 134, in <module>
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
AssertionError: Wrong answer; got (1000000000.0, []) instead of (184, [3, 13, 23, 32, 40, 50, 60, 68, 76, 82, 89, 99])
Testing if the previous output contains the string All tests passed: False
Resulting in output: True