3 Day 3: Testing Triangles

Read the description of today’s puzzle. Here is my input.

3.1 How many of the listed triangles are possible?

This turned out to be the easiest pair of puzzles to date. Maybe I’m getting better at this.

The puzzle tells us that for a triangle to be valid, the sum of any two sides must be greater than the remaining side. If you think about it for a bit, you’ll see that the two smallest sides are the only ones you need to add.

There’s nothing left to do except convert each line of the input into a triangle—that is, a list containing three numbers—and map the triangle-valid? function onto the resulting list of triangles, converting each one into a simple boolean #t or #f value.

Racket’s count function counts values list that pass a given set of criteria. The first argument is the name of the function we want to use to tell it which values to count: if that function returns anything besides #f for a given value, that value is included in the count. And since (identity x) simply returns x, passing identity to count amounts to saying, "count all the non-false values in this list".

3.2 How many triangles are possible when reading the list vertically?

You’ve got to be kidding me. You’re asking me to rearrange a list of lists into a new list of lists, in a language where lists are basically everything?

Please, don’t throw me in that briar patch!

Ok, so: all we need is a new string->triangles function that looks at the lines three at a time, and pulls from the first, second, and third columns of each. It was in the course of writing this function that I learned of Racket’s for*/list function: it works exactly like nested for loops work in other languages, but without the redundancy.

3.3 Testing Day 3