I recently started playing Flow Free Game.
Connect matching colors with pipe to creat
I suspect that no polynomial-time algorithm is guaranteed to solve every instance of this problem. But since one of the requirements is that every square must be covered by pipe, a similar approach to what both people and computers use for solving Sudoku should work well here:
When picking a square to branch on, it's generally a good idea to pick a square with as few allowed colours as possible.
[EDIT: It's important to avoid the possibility of forming invalid "loops" of pipe. One way to do this is by maintaining, for each allowed colour i of each square x, 2 bits of information: whether the square x is connected by a path of definite i-coloured tiles to the first i-coloured endpoint, and the same thing for the second i-coloured endpoint. Then when recursing, don't ever pick a square that has two neighbours with the same bit set (or with neither bit set) for any allowed colour.]
You actually don't need to use any logical deductions at all, but the more and better deductions you use, the faster the program will run as they will (possibly dramatically) reduce the amount of recursion. Some useful deductions include:
More advanced deductions based on path connectivity might help further -- e.g. if you can determine that every path connecting some pair of connectors must pass through a particular square, you can immediately assign that colour to the square.
This simple approach infers a complete solution without any recursion in your 5x5 example: the squares at (5, 2), (5, 3), (4, 3) and (4, 4) are forced to be orange; (4, 5) is forced to be green; (5, 5) is also forced to be green by virtue of the fact that no other colour could get to this square and then back again; now the orange path ending at (4, 4) has nowhere to go except to complete the orange path at (3, 4). Also (3, 1) is forced to be red; (3, 2) is forced to be yellow, which in turn forces (2, 1) and then (2, 2) to be red, which finally forces the yellow path to finish at (3, 3). The red pipe at (2, 2) forces (1, 2) to be blue, and the red and blue paths wind up being completely determined, "forcing each other" as they go.