A crippled rook can move on a chessboard in the following way: from a square, it can move to an adjacent square sharing a common side, and every two consecutive moves must be at right angles (i.e., the rook makes a 90∘ turn at every move).
A cycle is a sequence of squares which start and end at the same square, and traces out a valid path that the crippled rook can move according to the rules above. A non-intersecting cycle consists of pairwise distinct squares, with the sole exception of the starting and ending square.
What is the length of the longest possible cyclic, non-intersecting route of a crippled rook on a 15×15 chessboard?
Details and assumptions
The length of the route is the number of squares that the rook travels on.
To find the length of the longest possible cyclic, non-intersecting route of a crippled rook on a 15×15 chessboard, we can approach this problem using a strategy known as backtracking. Backtracking is a technique used to systematically explore all possible solutions to a problem by incrementally building a solution and undoing steps when they lead to a dead end.
Here's how we can solve this problem step by step:
1. Define a recursive function, let's call it "findLongestRoute", that takes the following parameters:
- The current position on the chessboard (represented by coordinates x and y)
- The number of moves made so far
- A set of visited squares to keep track of where the rook has been already
2. Implement the base cases for the function:
- If the current position is already visited, return the number of moves made so far (the length of the route).
- If the number of moves made so far is equal to the number of squares on the chessboard (225 in this case), return 0 (no more moves can be made).
3. Initialize a variable "maxRouteLength" to keep track of the longest route length found so far. Set it to 0 initially.
4. Iterate over the four possible directions that the rook can move (up, down, left, right) and explore each direction recursively:
- Calculate the new position by adding the corresponding offsets to the current position.
- If the new position is within the bounds of the chessboard (0 ≤ x < 15 and 0 ≤ y < 15), call the "findLongestRoute" function recursively with the new position, incremented number of moves, and updated set of visited squares.
- Update "maxRouteLength" with the maximum value between the current "maxRouteLength" and the return value from the recursive call.
- Undo the move by removing the new position from the set of visited squares.
5. Return the final value of "maxRouteLength", which will be the length of the longest possible cyclic, non-intersecting route.
Using this approach, the implementation will involve a depth-first search of the possible routes on the chessboard, exploring all possible paths from the starting square and keeping track of the maximum route length found. As the function recursively backtracks, it explores different possible routes until it exhausts all options.
Note: The implementation to find the longest route for a 15×15 chessboard can take some time to complete, as the number of possible routes is very large. It is recommended to optimize the solution by using memoization (storing previously computed results) to avoid redundant computations.