The old algorithm was to go through all the variations and try to find a guess that matches the current state of the table. Assuming that the currently examined guess is the secret, will we get the same answers for the guesses that are already on the table as the answers are actually on the table? If yes, then the current guess can be the secret, and it is just as good a guess as any other guesses.
A more complex approach can implement the min-max algorithm (https://en.wikipedia.org/wiki/Minimax). This algorithm does not simply get the next possible guess but also looks at all the possible guesses and selects the one that shortens the outcome of the game the most. If there is a guess that can be followed by three more guesses in the worst case, and there is another for which this number is only two, then min-max will choose the latter. It is a good exercise for the interested readers...
