Squares With Three Sides

Advent of Code 2016 Day 3

Part 1

Three sums and three comparisons

I need to determine which three-sided shapes are valid triangles.

The proof for a valid triangle, per the instructions:

the sum of any two sides must be larger than the remaining side

This seems like the best - perhaps only? - approach:

 |\
 | \  
 |  \
A|   \ B
 |    \
 |_____\
    C

A + B > C?
A + C > B?
B + C > A?

My algorithm in JavaScript:

input.reduce((valids, triangle) => {
  let [A,B,C] = [...triangle.matchAll(/\d+/g)].map(el => +el[0])
  return valids += (
    A + B > C && 
    A + C > B &&
    B + C > A
  ) ? 1 : 0
}, 0)

It generated the correct answer for Part 1!

Part 2

Columns instead of rows

A fun twist that will require nested for loops instead of a single reduce().

My algorithm as pseudocode:

Extract the digits from each line
  Generate a 3-element array in place of the string

Set valid count as 0

For each 3-element array except the last two, skipping two each time
  For each element in the array
    Generate a 3-element array containing the numbers in the same position as the element in the array...from the current and next two arrays
    Increment valid count by 1 only if each pair of side lengths is greater than the non-included side

Return valid count

My algorithm in JavaScript:

let sides = input.map(
  line => [...line.matchAll(/\d+/g)].map(el => +el[0])
)
let valids = 0
for (let row = 0; row < sides.length - 2; row += 3) {
  for (let col = 0; col < 3; col++) {
    let [A,B,C] = sides.slice(row, row + 3).map(el => el[col])
    valids += (
      A + B > C && 
      A + C > B &&
      B + C > A
    ) ? 1 : 0
  }
}
return valids

It generated the correct answer for Part 2!

I did it!!

• I solved both parts!
• I leveraged my growing familiarity with regex, reduce() and several array manipulation techniques!
• With the exception of Day 11, I continue a 2-star streak since Day 22!