AoC_2016_2_Reibello

#AoC 2016 Puzzle 2a
input_file_object = open("aoc16_input_2a.txt")
input_as_string = input_file_object.read()
input_file_object.close()

keypad = ((1,2,3),(4,5,6),(7,8,9))

start_y = 1
start_x = 1

keycode = []

input_by_line = input_as_string.split("\n")

for sequence in input_by_line:
    print(start_x)
    print(start_y)
    for char in sequence:
        if char == 'U':
            if start_y > 0:
                start_y -= 1
        elif char == 'D':
            if start_y < 2:
                start_y += 1
        elif char == 'L':
            if start_x > 0:
                start_x -= 1
        elif char == 'R':
            if start_x < 2:
                start_x += 1
    keycode.append(keypad[start_y][start_x])

print(keycode)

#Time for 1b

keypad = (('','','1','',''),('','2','3','4',''),('5','6','7','8','9'),('','A','B','C',''),('','','D','',''))

start_y = 2
start_x = 0

keycode = []

for sequence in input_by_line:
    print(start_x)
    print(start_y)
    for char in sequence:
        if char == 'U':
            if start_y > 0 and keypad[start_y - 1][start_x] != '':
                start_y -= 1
        elif char == 'D':
            if start_y < 4 and keypad[start_y + 1][start_x] != '':
                start_y += 1
        elif char == 'L':
            if start_x > 0 and keypad[start_y][start_x -1] != '':
                start_x -= 1
        elif char == 'R':
            if start_x < 4 and keypad[start_y][start_x +1] != '':
                start_x += 1
    keycode.append(keypad[start_y][start_x])
print(keycode)