Solution for Advent of Code 2016 - Day 1: No Time for a Taxicab

Link to the puzzle text

Part 1

In this puzzle we are given a grid and a list of instruction to walk on this grid. The instructions are either "left turn", "right turn" or "walk straight for n tiles". We are asked to find the tile after we follow the instructions.

We implement the current position as a complex number with the real part being the North-South-axis and the imaginary part the East-West-axis. So a number x + iy means a position of [x,y]. For the current orientation we also have a complex number. Rotating right and left is now simply mulitplying the orientation by either 1i or -1i, while walking for n tiles is n*orientation. For the answer we are asked to return the distance between the origin and the final tile. This is abs(tile.real) + abs(tile.imaginary). 

pos = 0 + 0j
facing = 1 + 0j

for instruction in instructions:
    direction, length = instruction[:1], int(instruction[1:])
    if direction == "R":
        facing *= 1j
    else:
        facing *= -1j

    pos += length * facing

print(int(abs(pos.real) + abs(pos.imag)))

Part 2

For part 2, we are also asked to find the first tile visited twice. 

For this, we keep a set of visited tiles and check if the current tile has been visited after each move. The move itself has to be modified from a length * facing to walking one step at a time for length steps, so we don't skip over previously visited tiles.

for i in range(length):
    pos += facing

    if pos in pos_visited:
        print("PART 2")
        print(int(abs(pos.real) + abs(pos.imag)))

    pos_visited.append(pos)

To reduce the output, I also added a helper variable to only print the output for the first time a tile is visited twice.

Link to my solutions