Find Consecutive Location from Python List of List

Question

I have a list that contains 3 lists.

locations = [[(1,1), (3,3), (6,7)], [(5,2), (3,4), (2,3)], [(3,7), (3,5), (7,7)]]

The first list contains, the location of alphabet A

The second list contains, the location of alphabet B

The third list contains, the location of alphabet C

My task is to find them together where they sit close to each other. It's a word puzzle problem. Thus, alphabet B could sit up, down, right, or left of the A. The C has to be the same direction as the B to A.

In this case, the solution will be a new list with their locations. The answer is -

puzzle_solved_location = [(3,3), (3,4), (3,5)]

I was thinking about using the nested list. But there will be some problems where some words with 10 or 15 letters will be given. I need one program that will solve any word puzzle with any number of letters.

Currently, I'm trying a nested loop but it won't work for longer length word and the codes need to be edited each time for new words. The code is given below-

A= [(1,1), (3,3), (6,7)]
B= [(5,2), (3,4), (2,3)]
C= [(3,7), (3,5), (7,7)]


for row_A,column_A in A:
    for row_B,column_B in B:
        if (row_A+1==row_B or row_A-1==row_B or row_A==row_B)
 and (column_A+1==column_B or column_A-1 ==column_B or column_A==column_B):
            for row_C,column_C in C:
                x_co, y_co = row_B-row_A, column_B-column_A
                if row_B+x_co ==row_C and column_B+y_co ==column_C:
                    print (row_A, column_A, row_B, column_B, row_C, column_C)

The result gives the location of the solved puzzle.

>> 3 3 3 4 3 5

Is there any other way to do that without writing if statement for each alphabet?

Answer 1

You can use a list comprehension to iterate through the 4 possible directions and the possible starting points (of A 's) to see which ones would result in finding the corresponding points in the rest of the locations:

[[(ax + dx * i, ay + dy * i) for i in range(len(locations))] for dx, dy in ((1, 0), (0, 1), (-1, 0), (0, -1)) for ax, ay in locations[0] if all((ax + dx * i, ay + dy * i) in c for i, c in enumerate(locations[1:], 1))]

This returns:

[[(3, 3), (3, 4), (3, 5)]]

Note that this returns a list of lists of tuples because there could potentially be more than one solution.