Module day25

Let It Snow

There are two parts to solving this problem.

The first is converting the row and column to a zero-based index. Using the example of the 12th code at row 4 column 2:

       | 1   2   3   4   5   6
    ---+---+---+---+---+---+---+
     1 |  1   3   6  10  15  21
     2 |  2   5   9  14  20
     3 |  4   8  13  19
     4 |  7  12  18
     5 | 11  17
     6 | 16

First we observe that the numbers on the top row are the triangular numbers that can be calculated with the formula (n * (n + 1)) / 2 for the nth number.

Starting at the chosen number 12 and moving diagonally upwards to the right we intersect the top row at column column + row - 1 = 2 + 4 - 1 = 5. This gives the triangular number 5 * (5 + 1) / 2 = 15. Then we count backward by row elements to get the one less zero-based index 15 - 4 = 11.

The second part is realizing that the description of the code generation is modular exponentiation. The exponent of the first code is zero, which is the reason for using a zero-based index.

Functions

parse

part1

part2

Type Aliases

Input 🔒