Python helper function for simple game

In a small game that I am developing, the screen is square of a given size, s, measured in pixels. The x- and y-coordinates go from (0,0) to (s-1, s-1). Boulders are rolling across the screen, and the player must avoid being hit by them. When a boulder rolls off one edge of the screen, it appears at the opposite edge of the screen, preserving the other coordinate. For example, if s = 100 and a boulder on the middle of the right edge of the screen, at coordinate (99, 50), moves to the right, it would reappear at coordinates (0, 50) (i.e. in the middle on the left edge of the screen). The screen is essentially the surface of a torus (doughnut shape) where you can imagine the right side has been joined to the left side, and the top of the screen to the bottom. he rolling boulders are simulated by updating their (x, y) coordinates in discrete steps. Each rolling boulder is considered to move along a trajectory defined by a sequence of discrete points: THE i ARE subscripts. (x0, y0), (x1, y1) ... (xi, yi) It moves along a straight line joining consecutive points in the sequence. The trajectory of a rolling boulder is parameterised by two integers a and b so that x(i+1), y(i+1) is x(i+1) = a(xi) - yi y(i+1) = xi + b(yi) If either coordinate returned by the calculation is either negative or larger than , then it is considered to wrap around to the other side of the screen, and carry on counting. For example, if s= 100 , then the coordinate (131, 257) would correspond to the point at (31,57). Also coordinate (-1, -3) would correspond to the coordinate (99, 97). Another example: For function def findPositions(a, b, s, queries): First line (a, b, s) Second Line (number of queries) Lines After (queries) If I read in the following: 2 3 100 3 1 0 3 0 1 4 1 1 3 I should get as positions 1 18 45 39 83 37 Explanation: The first query asks if a boulder starts from (1,0) where will it end up after 3 steps. Following the rules: (1,0) --> (2 X 1 - 0, 1 + 3 X 0) = (2,1) --> (2 X 2 -1, 2 + 3 X 1) = (3,5) --> (2 X 3 -5, 3 + 3 X 5) = (1, 18) So we output 1 18 on the first line Similarly, for the second query, find where a boulder that starts from (0,1) would be after 4 steps. The following trajectory: (0,1) --> (2(0) - 1,0 + 3(1)) = (-1,3) = (99,3) --> (198 -3,99+9) = (195,108) = (95,8) --> (2(95) - 8, 95+3(8)) = (182, 119) = (82,19) --> (164 -19, 82 + 57) = (145,139) = (45,39) The negative coordinate to a valid screen coordinate according to the given screen width of 100 (value of s). The third query: (1,1) --> (2 -1, 1+3) = (1,4) --> (2 -4, 1+ 12) = (-2,13) = (98, 13) --> (196 - 13, 98 + 39) = (183, 137) = (83, 37)