BFS is the right instinct here since you're looking for shortest path on an unweighted graph.
Model the problem as a graph where each cell is a node and edges represent valid knight moves for a given (a,b). For each (a,b), run BFS from (0,0) to find the shortest path to (n-1,n-1). Optimize by precomputing move offsets and handling unreachable cases.
Pro tip: Mention that BFS is optimal for unweighted graphs, but also discuss potential optimizations like bidirectional BFS or A* with Manhattan distance heuristic to reduce search space, especially for large n.
Clarify that n is the board size, (a,b) are move parameters with 1 ≤ a,b ≤ n-1, and moves are the 8 knight-like jumps. Note that each (a,b) is independent and we need the minimum moves for each.
Represent each cell as a node. For a fixed (a,b), generate all valid moves from a cell. Since all moves have equal cost, BFS from (0,0) gives the shortest path to (n-1,n-1).
Use a queue and a distance array (or visited set) to avoid revisiting cells. Precompute the 8 move offsets for the given (a,b) to avoid redundant calculations.
If BFS exhausts all reachable cells without reaching (n-1,n-1), return -1. Also consider special cases like n=1 (start equals target, 0 moves) and when a or b equals 0 (not allowed by constraints).
For each (a,b), BFS takes O(n^2) time and space. With O(n^2) pairs, total O(n^4) time. Discuss possible optimizations like bidirectional BFS or precomputing reachability for symmetric pairs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as a graph where each position is a node, and edges represent valid reversals that move the token from one position to another. Then run BFS from each starting position to compute shortest distances to all targets, skipping forbidden positions. Optimize by precomputing all possible moves for each position and using multi-source BFS if multiple starts share the same graph.
Pro tip: Clarify constraints first (n, k, number of queries) to choose the right algorithm; for large n, precompute moves and use BFS with early termination, and consider bidirectional BFS if only a few targets are needed.
Determine all valid reversals: for a token at position i, a segment of length k containing i can start at s where max(1, i-k+1) ≤ s ≤ min(i, n-k+1). The token moves to s + (s+k-1 - i) = 2s + k - 1 - i. Ensure the new position is not forbidden.
Create an adjacency list where each allowed position i has edges to all reachable positions j via a valid reversal, excluding forbidden positions. This graph is undirected because reversing the same segment again returns the token.
For each starting position p (if not forbidden), perform BFS to compute the minimum number of moves to every other allowed position. If p is forbidden, all distances are -1. Use a queue and distance array initialized to -1.
If there are many starting positions, precompute all-pairs shortest paths using BFS from each node (O(n*(n+m))). If n is large, consider that the graph may have structure (e.g., moves depend only on parity or modulo) and derive a formula or use bidirectional BFS for specific targets.
Check if k > n (no moves possible), if k = 1 (no movement), if the start is forbidden, and if the target is forbidden. Also consider disconnected components and unreachable targets.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.