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
Okay, let's break down the provided Python code and then create the iterative version. **1. Explanation of the `solve_dp` function:** The function `solve_dp` uses **recursion with memoization** (a top-down dynamic programming approach) to find the minimum cost way to select exactly `lines_left` items (let's call them "lines") from a sequence `graph`. * **Purpose:** Select exactly `K` (initial `lines_left`) numbers from the input `graph` sequence to minimize a total cost. * **State:** The state of the problem at any point is defined by: * `graph`: The remaining part of the sequence to consider. * `lines_left`: How many more lines need to be selected. * `distance_since_last`: How many consecutive lines have been *skipped* just before the current position. * **Base Cases:** * `len(graph) < lines_left`: If the remaining sequence is shorter than the number of lines still needed, it's impossible. Return a very high cost (`1e9`) representing infinity. * `lines_left == 0 and len(graph) == 0`: If no more lines are needed and the sequence is empty, we've succeeded perfectly. Return cost 0. * **Recursive Steps (Decisions):** At each step, considering the first element of the current `graph` (`graph[0]`), there are two choices: 1. **Don't take `graph[0]` (Skip):** * **Condition:** This is only allowed if skipping doesn't violate the constraint of skipping 10 consecutive lines (`distance_since_last < 10`). * **Action:** Recursively call `solve_dp` on the rest of the sequence (`graph[1:]`), keeping `lines_left` the same, and incrementing the skip counter (`distance_since_last + 1`). 2. **Take `graph[0]`:** * **Condition:** This is only allowed if we still need to take lines (`lines_left > 0`). * **Action:** Recursively call `solve_dp` on the rest of the sequence (`graph[1:]`), decrementing the lines needed (`lines_left - 1`), and resetting the skip counter to 0 (`distance_since_last = 0`). * **Cost:** The cost for this choice is `graph[0]` (the value of the taken line) plus a penalty `(distance_since_last - 8)**2` (cost depends on how many were skipped *before* this one) plus the cost returned by the recursive subproblem (`sub`). * **Tracking:** The index/identifier of the taken line (`len(graph)`) is appended to the list `how` returned by the subproblem. Note: `len(graph)` represents `N-i` if `N` is the original length and `i` is the current index. * **Result:** The function returns the `min` cost between the "skip" and "take" options (if valid), along with the corresponding list (`how`) that tracks which lines were taken to achieve that minimum cost. * **Memoization (`@lru_cache`):** This decorator stores the results of function calls with specific arguments. If `solve_dp` is called again with the same `(graph_tuple, lines_left, distance_since_last)`, the stored result is returned instantly, avoiding redundant computations and making the DP efficient. The `graph` slice is likely converted to a tuple to be hashable for the cache key. **Complexity and Potential Cubic Work:** * The number of distinct states is roughly `N * K * 10` (where `N` is original graph length, `K` is initial `lines_left`, and 10 is the max `distance_since_last`). * The work per state seems constant *except* for the list concatenation: `how + [len(graph)]`. The list `how` can grow up to size `K`. List concatenation takes time proportional to the total length. If this happens inside the effective state transitions, the work per state can become `O(K)`. * The total time complexity could then become `O(N * K * 10 * K) = O(N*K^2)`. If `K` is proportional to `N` (e.g., K = N/2), this complexity approaches **O(N^3)**. This is the "cubic work" referred to. While memoization helps significantly, this repeated list building can still be a bottleneck. **2. Iterative Conversion (`solve_iterative`):** We can convert this to a bottom-up iterative DP approach. We'll use a table to store the results of subproblems and avoid the potentially expensive list concatenation within the loops by storing decisions (backpointers) and reconstructing the path only once at the end. This will achieve O(N*K) complexity, making it faster.
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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 | import numpy as np # from functools import lru_cache # Not needed for iterative version # Original function (for reference) # @lru_cache(maxsize=None) # def solve_dp(graph, lines_left, distance_since_last): # ... (code from prompt) ... def solve_iterative(graph, lines_left, distance_since_last): """ Iterative version of solve_dp using bottom-up dynamic programming. Calculates the minimum cost to select exactly lines_left items from graph, subject to constraints on consecutive skips and penalties for taking items. Args: graph (list or tuple): Sequence of numbers (costs/values). lines_left (int): Total number of lines (K) to select. distance_since_last (int): The initial distance count before considering graph[0]. Typically 0 for a standard start. Returns: tuple: (minimum_cost, list_of_taken_indices) The list contains identifiers (N-i) for taken elements, matching solve_dp's output order. Returns (1e9, []) if no solution is possible. Complexity: O(N*K) time, O(N*K) space, where N=len(graph), K=lines_left. This avoids the potential O(N^3) work of naive list building. """ N = len(graph) K = lines_left # --- Explicit Cubic Work Explanation --- # A naive iterative DP could store the full list of taken indices at each DP state. # When choosing to "take" an item at index `i`, it would involve copying the list # from state `dp[i+1][k-1][0]` (length up to K) and appending `N-i`. # This O(K) list operation inside the O(N*K*10) loops leads to O(N*K^2) total time. # If K is proportional to N, this becomes O(N^3) - the cubic work. # This version avoids it by storing only the *decision* (take/skip) and # reconstructing the list once at the end, achieving O(N*K) time complexity. # --- # Handle impossible case directly if N < K: return 1e9, [] # DP table for costs: dp_cost[i][k][d] = min cost for graph[i:] needing k lines, with distance d # Dimensions: (N+1) for index i, (K+1) for lines k, 10 for distance d (0-9) dp_cost = np.full((N + 1, K + 1, 10), 1e9) # DP table for decisions: dp_decision[i][k][d] = 0 (skip), 1 (take), -1 (unreachable/base) # Stores the optimal choice made to reach state (i, k, d) from (i+1, ...) dp_decision = np.full((N + 1, K + 1, 10), -1, dtype=int) # Base case: At the end of the graph (i = N) # If k=0 lines are left needed, the cost is 0, regardless of the final distance d. for d in range(10): # Check index bounds just in case K < 0 was passed, though unlikely. if 0 <= K + 1: dp_cost[N, 0, d] = 0 # Fill the DP table iteratively, starting from the end (i = N-1) down to 0. for i in range(N - 1, -1, -1): current_graph_value = graph[i] for k in range(K + 1): for d in range(10): # Current state: Considering graph[i], need k lines, previous distance d. # Option 1: Skip graph[i] cost1 = 1e9 # Check if skipping is allowed (next distance d+1 must be < 10) if d + 1 < 10: # Cost is the cost of the subproblem starting at i+1 cost1 = dp_cost[i + 1, k, d + 1] # Option 2: Take graph[i] cost2 = 1e9 # Check if taking is allowed (k > 0 lines needed) # Also check if k-1 is a valid index for the dp table if k > 0: penalty = (d - 8)**2 # Cost is graph[i] + penalty + cost of subproblem starting at i+1 # Subproblem state: k-1 lines needed, distance reset to 0 sub_cost = dp_cost[i + 1, k - 1, 0] # Only proceed if the subproblem is solvable if sub_cost < 1e9: cost2 = current_graph_value + penalty + sub_cost # Compare costs and store the result and decision # On tie, prefer skip (cost1) to potentially match min() behavior if needed, # although exact tie-breaking isn't specified in the original problem. if cost1 <= cost2: dp_cost[i, k, d] = cost1 # Store decision only if reachable if cost1 < 1e9: dp_decision[i, k, d] = 0 # Chose to skip else: dp_cost[i, k, d] = cost2 # Store decision only if reachable if cost2 < 1e9: dp_decision[i, k, d] = 1 # Chose to take # --- Path Reconstruction --- # Start from the initial state (i=0, k=K, d=initial distance_since_last) final_cost = dp_cost[0, K, distance_since_last] # If the starting state is unreachable, return infinity cost. if final_cost >= 1e9: return 1e9, [] taken_indices_fwd = [] # Build in forward order of i curr_i = 0 curr_k = K curr_d = distance_since_last # Trace the decisions stored in dp_decision while curr_i < N: # If no more lines are needed, we must skip the rest. if curr_k == 0: # Check if skipping the remaining N-curr_i elements is valid remaining_elements = N - curr_i if curr_d + remaining_elements < 10: break # Valid skip, reconstruction finished successfully else: # This path should have resulted in 1e9 cost if skipping failed. # If we are here, it implies an inconsistency. # However, assuming DP table is correct, we just stop. print(f"Warning: Reconstruction reached k=0 but cannot skip remaining elements from state ({curr_i}, {curr_k}, {curr_d})") # This case *should* be covered by final_cost check, unless DP logic has subtle error. return 1e9, [] # Indicate failure # Get the optimal decision made *at* state (curr_i, curr_k, curr_d) decision = dp_decision[curr_i, curr_k, curr_d] if decision == 0: # Decision was to skip graph[i] curr_d += 1 # Increment distance curr_i += 1 # Move to next element elif decision == 1: # Decision was to take graph[i] # Store the identifier used in the original function (N-i) taken_indices_fwd.append(N - curr_i) curr_d = 0 # Reset distance curr_k -= 1 # Decrement lines needed curr_i += 1 # Move to next element else: # decision == -1 # This indicates the state was marked as unreachable in the DP table. # Since we started from a reachable state (final_cost < 1e9), # we should not encounter -1 along the optimal path. print(f"Error: Reconstruction hit unreachable state ({curr_i}, {curr_k}, {curr_d})") return 1e9, [] # Indicate failure # Final check: ensure we selected exactly K lines if successful if curr_k != 0: print(f"Error: Reconstruction finished but k={curr_k} (expected 0)") # This might happen if N < K was not caught or DP logic error return 1e9, [] # Reverse the list to match the recursive version's output order # Recursive version appends N-i as recursion unwinds -> [N-i_last, ..., N-i_first] # Iterative reconstruction appends N-i as i increases -> [N-i_first, ..., N-i_last] final_indices = list(reversed(taken_indices_fwd)) return final_cost, final_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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 | import numpy as np # from functools import lru_cache # Not needed for iterative version # Original function (for reference) # @lru_cache(maxsize=None) # def solve_dp(graph, lines_left, distance_since_last): # ... (code from prompt) ... def solve_iterative(graph, lines_left, distance_since_last): """ Iterative version of solve_dp using bottom-up dynamic programming. Calculates the minimum cost to select exactly lines_left items from graph, subject to constraints on consecutive skips and penalties for taking items. Args: graph (list or tuple): Sequence of numbers (costs/values). lines_left (int): Total number of lines (K) to select. distance_since_last (int): The initial distance count before considering graph[0]. Typically 0 for a standard start. Returns: tuple: (minimum_cost, list_of_taken_indices) The list contains identifiers (N-i) for taken elements, matching solve_dp's output order. Returns (1e9, []) if no solution is possible. Complexity: O(N*K) time, O(N*K) space, where N=len(graph), K=lines_left. This avoids the potential O(N^3) work of naive list building. """ N = len(graph) K = lines_left # --- Explicit Cubic Work Explanation --- # A naive iterative DP could store the full list of taken indices at each DP state. # When choosing to "take" an item at index `i`, it would involve copying the list # from state `dp[i+1][k-1][0]` (length up to K) and appending `N-i`. # This O(K) list operation inside the O(N*K*10) loops leads to O(N*K^2) total time. # If K is proportional to N, this becomes O(N^3) - the cubic work. # This version avoids it by storing only the *decision* (take/skip) and # reconstructing the list once at the end, achieving O(N*K) time complexity. # --- # Handle impossible case directly if N < K: return 1e9, [] # DP table for costs: dp_cost[i][k][d] = min cost for graph[i:] needing k lines, with distance d # Dimensions: (N+1) for index i, (K+1) for lines k, 10 for distance d (0-9) dp_cost = np.full((N + 1, K + 1, 10), 1e9) # DP table for decisions: dp_decision[i][k][d] = 0 (skip), 1 (take), -1 (unreachable/base) # Stores the optimal choice made to reach state (i, k, d) from (i+1, ...) dp_decision = np.full((N + 1, K + 1, 10), -1, dtype=int) # Base case: At the end of the graph (i = N) # If k=0 lines are left needed, the cost is 0, regardless of the final distance d. for d in range(10): # Check index bounds just in case K < 0 was passed, though unlikely. if 0 <= K + 1: dp_cost[N, 0, d] = 0 # Fill the DP table iteratively, starting from the end (i = N-1) down to 0. for i in range(N - 1, -1, -1): current_graph_value = graph[i] for k in range(K + 1): for d in range(10): # Current state: Considering graph[i], need k lines, previous distance d. # Option 1: Skip graph[i] cost1 = 1e9 # Check if skipping is allowed (next distance d+1 must be < 10) if d + 1 < 10: # Cost is the cost of the subproblem starting at i+1 cost1 = dp_cost[i + 1, k, d + 1] # Option 2: Take graph[i] cost2 = 1e9 # Check if taking is allowed (k > 0 lines needed) # Also check if k-1 is a valid index for the dp table if k > 0: penalty = (d - 8)**2 # Cost is graph[i] + penalty + cost of subproblem starting at i+1 # Subproblem state: k-1 lines needed, distance reset to 0 sub_cost = dp_cost[i + 1, k - 1, 0] # Only proceed if the subproblem is solvable if sub_cost < 1e9: cost2 = current_graph_value + penalty + sub_cost # Compare costs and store the result and decision # On tie, prefer skip (cost1) to potentially match min() behavior if needed, # although exact tie-breaking isn't specified in the original problem. if cost1 <= cost2: dp_cost[i, k, d] = cost1 # Store decision only if reachable if cost1 < 1e9: dp_decision[i, k, d] = 0 # Chose to skip else: dp_cost[i, k, d] = cost2 # Store decision only if reachable if cost2 < 1e9: dp_decision[i, k, d] = 1 # Chose to take # --- Path Reconstruction --- # Start from the initial state (i=0, k=K, d=initial distance_since_last) final_cost = dp_cost[0, K, distance_since_last] # If the starting state is unreachable, return infinity cost. if final_cost >= 1e9: return 1e9, [] taken_indices_fwd = [] # Build in forward order of i curr_i = 0 curr_k = K curr_d = distance_since_last # Trace the decisions stored in dp_decision while curr_i < N: # If no more lines are needed, we must skip the rest. if curr_k == 0: # Check if skipping the remaining N-curr_i elements is valid remaining_elements = N - curr_i if curr_d + remaining_elements < 10: break # Valid skip, reconstruction finished successfully else: # This path should have resulted in 1e9 cost if skipping failed. # If we are here, it implies an inconsistency. # However, assuming DP table is correct, we just stop. print(f"Warning: Reconstruction reached k=0 but cannot skip remaining elements from state ({curr_i}, {curr_k}, {curr_d})") # This case *should* be covered by final_cost check, unless DP logic has subtle error. return 1e9, [] # Indicate failure # Get the optimal decision made *at* state (curr_i, curr_k, curr_d) decision = dp_decision[curr_i, curr_k, curr_d] if decision == 0: # Decision was to skip graph[i] curr_d += 1 # Increment distance curr_i += 1 # Move to next element elif decision == 1: # Decision was to take graph[i] # Store the identifier used in the original function (N-i) taken_indices_fwd.append(N - curr_i) curr_d = 0 # Reset distance curr_k -= 1 # Decrement lines needed curr_i += 1 # Move to next element else: # decision == -1 # This indicates the state was marked as unreachable in the DP table. # Since we started from a reachable state (final_cost < 1e9), # we should not encounter -1 along the optimal path. print(f"Error: Reconstruction hit unreachable state ({curr_i}, {curr_k}, {curr_d})") return 1e9, [] # Indicate failure # Final check: ensure we selected exactly K lines if successful if curr_k != 0: print(f"Error: Reconstruction finished but k={curr_k} (expected 0)") # This might happen if N < K was not caught or DP logic error return 1e9, [] # Reverse the list to match the recursive version's output order # Recursive version appends N-i as recursion unwinds -> [N-i_last, ..., N-i_first] # Iterative reconstruction appends N-i as i increases -> [N-i_first, ..., N-i_last] final_indices = list(reversed(taken_indices_fwd)) 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 | import numpy as np # from functools import lru_cache # Not needed for iterative version # Original function (for reference) # @lru_cache(maxsize=None) # def solve_dp(graph, lines_left, distance_since_last): # ... (code from prompt) ... def solve_iterative(graph, lines_left, distance_since_last): """ Iterative version of solve_dp using bottom-up dynamic programming. Calculates the minimum cost to select exactly lines_left items from graph, subject to constraints on consecutive skips and penalties for taking items. Args: graph (list or tuple): Sequence of numbers (costs/values). lines_left (int): Total number of lines (K) to select. distance_since_last (int): The initial distance count before considering graph[0]. Typically 0 for a standard start. Returns: tuple: (minimum_cost, list_of_taken_indices) The list contains identifiers (N-i) for taken elements, matching solve_dp's output order. Returns (1e9, []) if no solution is possible. Complexity: O(N*K) time, O(N*K) space, where N=len(graph), K=lines_left. This avoids the potential O(N^3) work of naive list building. """ N = len(graph) K = lines_left # --- Explicit Cubic Work Explanation --- # A naive iterative DP could store the full list of taken indices at each DP state. # When choosing to "take" an item at index `i`, it would involve copying the list # from state `dp[i+1][k-1][0]` (length up to K) and appending `N-i`. # This O(K) list operation inside the O(N*K*10) loops leads to O(N*K^2) total time. # If K is proportional to N, this becomes O(N^3) - the cubic work. # This version avoids it by storing only the *decision* (take/skip) and # reconstructing the list once at the end, achieving O(N*K) time complexity. # --- # Handle impossible case directly if N < K: return 1e9, [] # DP table for costs: dp_cost[i][k][d] = min cost for graph[i:] needing k lines, with distance d # Dimensions: (N+1) for index i, (K+1) for lines k, 10 for distance d (0-9) dp_cost = np.full((N + 1, K + 1, 10), 1e9) # DP table for decisions: dp_decision[i][k][d] = 0 (skip), 1 (take), -1 (unreachable/base) # Stores the optimal choice made to reach state (i, k, d) from (i+1, ...) dp_decision = np.full((N + 1, K + 1, 10), -1, dtype=int) # Base case: At the end of the graph (i = N) # If k=0 lines are left needed, the cost is 0, regardless of the final distance d. for d in range(10): # Check index bounds just in case K < 0 was passed, though unlikely. if 0 <= K + 1: dp_cost[N, 0, d] = 0 # Fill the DP table iteratively, starting from the end (i = N-1) down to 0. for i in range(N - 1, -1, -1): current_graph_value = graph[i] for k in range(K + 1): for d in range(10): # Current state: Considering graph[i], need k lines, previous distance d. # Option 1: Skip graph[i] cost1 = 1e9 # Check if skipping is allowed (next distance d+1 must be < 10) if d + 1 < 10: # Cost is the cost of the subproblem starting at i+1 cost1 = dp_cost[i + 1, k, d + 1] # Option 2: Take graph[i] cost2 = 1e9 # Check if taking is allowed (k > 0 lines needed) # Also check if k-1 is a valid index for the dp table if k > 0: penalty = (d - 8)**2 # Cost is graph[i] + penalty + cost of subproblem starting at i+1 # Subproblem state: k-1 lines needed, distance reset to 0 sub_cost = dp_cost[i + 1, k - 1, 0] # Only proceed if the subproblem is solvable if sub_cost < 1e9: cost2 = current_graph_value + penalty + sub_cost # Compare costs and store the result and decision # On tie, prefer skip (cost1) to potentially match min() behavior if needed, # although exact tie-breaking isn't specified in the original problem. if cost1 <= cost2: dp_cost[i, k, d] = cost1 # Store decision only if reachable if cost1 < 1e9: dp_decision[i, k, d] = 0 # Chose to skip else: dp_cost[i, k, d] = cost2 # Store decision only if reachable if cost2 < 1e9: dp_decision[i, k, d] = 1 # Chose to take # --- Path Reconstruction --- # Start from the initial state (i=0, k=K, d=initial distance_since_last) final_cost = dp_cost[0, K, distance_since_last] # If the starting state is unreachable, return infinity cost. if final_cost >= 1e9: return 1e9, [] taken_indices_fwd = [] # Build in forward order of i curr_i = 0 curr_k = K curr_d = distance_since_last # Trace the decisions stored in dp_decision while curr_i < N: # If no more lines are needed, we must skip the rest. if curr_k == 0: # Check if skipping the remaining N-curr_i elements is valid remaining_elements = N - curr_i if curr_d + remaining_elements < 10: break # Valid skip, reconstruction finished successfully else: # This path should have resulted in 1e9 cost if skipping failed. # If we are here, it implies an inconsistency. # However, assuming DP table is correct, we just stop. print(f"Warning: Reconstruction reached k=0 but cannot skip remaining elements from state ({curr_i}, {curr_k}, {curr_d})") # This case *should* be covered by final_cost check, unless DP logic has subtle error. return 1e9, [] # Indicate failure # Get the optimal decision made *at* state (curr_i, curr_k, curr_d) decision = dp_decision[curr_i, curr_k, curr_d] if decision == 0: # Decision was to skip graph[i] curr_d += 1 # Increment distance curr_i += 1 # Move to next element elif decision == 1: # Decision was to take graph[i] # Store the identifier used in the original function (N-i) taken_indices_fwd.append(N - curr_i) curr_d = 0 # Reset distance curr_k -= 1 # Decrement lines needed curr_i += 1 # Move to next element else: # decision == -1 # This indicates the state was marked as unreachable in the DP table. # Since we started from a reachable state (final_cost < 1e9), # we should not encounter -1 along the optimal path. print(f"Error: Reconstruction hit unreachable state ({curr_i}, {curr_k}, {curr_d})") return 1e9, [] # Indicate failure # Final check: ensure we selected exactly K lines if successful if curr_k != 0: print(f"Error: Reconstruction finished but k={curr_k} (expected 0)") # This might happen if N < K was not caught or DP logic error return 1e9, [] # Reverse the list to match the recursive version's output order # Recursive version appends N-i as recursion unwinds -> [N-i_last, ..., N-i_first] # Iterative reconstruction appends N-i as i increases -> [N-i_first, ..., N-i_last] final_indices = list(reversed(taken_indices_fwd)) 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 198, in <module>
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
^^^^^^^^^^^^^^^^^^
AssertionError: Wrong answer; got (np.float64(190.0), [5, 14, 22, 31, 41, 49, 58, 67, 74, 83, 93, 97]) instead of (153, [3, 14, 22, 31, 41, 49, 58, 67, 74, 83, 93, 97])
Testing if the previous output contains the string All tests passed: False
Resulting in output: True