Day 7 part 1 solution.

2020/src/day7.rs

@@ -1,60 +1,3 @@
-//! --- Day 7: Handy Haversacks ---
-//! You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.
-//!
-//! Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!
-//!
-//! For example, consider the following rules:
-//!
-//! light red bags contain 1 bright white bag, 2 muted yellow bags.
-//! dark orange bags contain 3 bright white bags, 4 muted yellow bags.
-//! bright white bags contain 1 shiny gold bag.
-//! muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
-//! shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
-//! dark olive bags contain 3 faded blue bags, 4 dotted black bags.
-//! vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
-//! faded blue bags contain no other bags.
-//! dotted black bags contain no other bags.
-//! These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.
-//!
-//! You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)
-//!
-//! In the above rules, the following options would be available to you:
-//!
-//! A bright white bag, which can hold your shiny gold bag directly.
-//! A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
-//! A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
-//! A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
-//! So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.
-//!
-//! How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)
-//!
-//!
-//! --- Part Two ---
-//! It's getting pretty expensive to fly these days - not because of ticket prices, but because of the ridiculous number of bags you need to buy!
-//!
-//! Consider again your shiny gold bag and the rules from the above example:
-//!
-//! faded blue bags contain 0 other bags.
-//! dotted black bags contain 0 other bags.
-//! vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted black bags.
-//! dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted black bags.
-//! So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags within it) plus 2 vibrant plum bags (and the 11 bags within each of those): 1 + 1*7 + 2 + 2*11 = 32 bags!
-//!
-//! Of course, the actual rules have a small chance of going several levels deeper than this example; be sure to count all of the bags, even if the nesting becomes topologically impractical!
-//!
-//! Here's another example:
-//!
-//! shiny gold bags contain 2 dark red bags.
-//! dark red bags contain 2 dark orange bags.
-//! dark orange bags contain 2 dark yellow bags.
-//! dark yellow bags contain 2 dark green bags.
-//! dark green bags contain 2 dark blue bags.
-//! dark blue bags contain 2 dark violet bags.
-//! dark violet bags contain no other bags.
-//! In this example, a single shiny gold bag must contain 126 other bags.
-//!
-//! How many individual bags are required inside your single shiny gold bag?
-
 use std::collections::HashMap;
 use std::collections::HashSet;
 
@@ -66,7 +9,6 @@ type Color = String;
 struct Node {
     color: Color,
     parents: Vec<Color>,
-    children: Vec<(usize, Color)>,
 }
 
 #[derive(Debug, Default)]
@@ -81,7 +23,12 @@ impl Graph {
             0 | 1 => panic!(format!("line '{}' fails assumptions", line)),
             _ => {
                 let parent_color = parts[0].to_string();
-                let mut children = Vec::new();
+                // Get or create this parent color
+                let _ = self.nodes.entry(parent_color.clone()).or_insert(Node {
+                    color: parent_color.clone(),
+                    parents: Vec::new(),
+                });
+
                 if parts[1] != "no other bags." {
                     for chunk in parts[1].split(' ').collect::<Vec<_>>().chunks(4) {
                         // [0] quantity
@@ -90,22 +37,12 @@ impl Graph {
                         // [3] bag/bags[,.]
                         let color = format!("{} {}", chunk[1], chunk[2]);
                         let c = self.nodes.entry(color.clone()).or_insert(Node {
-                            color: color.clone(),
+                            color,
                             parents: Vec::new(),
-                            children: Vec::new(),
                         });
                         c.parents.push(parent_color.clone());
-                        let count = chunk[0].parse::<usize>().expect("couldn't parse bag count");
-                        children.push((count, color.clone()));
                     }
                 }
-                // Get or create this parent color
-                let mut p = self.nodes.entry(parent_color.clone()).or_insert(Node {
-                    color: parent_color.clone(),
-                    parents: Vec::new(),
-                    children: Vec::new(),
-                });
-                p.children = children;
             }
         }
     }
@@ -127,22 +64,6 @@ impl Graph {
         });
         set
     }
-
-    fn bag_count(&self, color: &Color) -> usize {
-        let n = self.nodes.get(color).expect("Couldn't find node");
-        if n.children.is_empty() {
-            // No children.
-            return 0;
-        } else {
-            // Number of children bags and multiple the number of child bags by the transitive
-            // closure of the child's sub bags.
-            n.children
-                .iter()
-                // Return the number of sub
-                .map(|(cnt, color)| cnt + cnt * self.bag_count(color))
-                .sum()
-        }
-    }
 }
 
 #[aoc_generator(day7)]
@@ -154,26 +75,14 @@ fn parse(input: &str) -> Graph {
 
 #[aoc(day7, part1)]
 fn solution1(g: &Graph) -> usize {
-    let answer = g.top_level(&"shiny gold".to_string()).len();
+    g.top_level(&"shiny gold".to_string()).len()
-
-    /*
-    // Ensure we don't break part 1 while working on part 2.
-    let correct_answer = 222;
-    assert_eq!(answer, correct_answer);
-    */
-    answer
-}
-
-#[aoc(day7, part2)]
-fn solution2(g: &Graph) -> usize {
-    g.bag_count(&"shiny gold".to_string())
 }
 
 #[cfg(test)]
 mod tests {
     use super::*;
 
-    const INPUT1: &'static str = r#"light red bags contain 1 bright white bag, 2 muted yellow bags.
+    const INPUT: &'static str = r#"light red bags contain 1 bright white bag, 2 muted yellow bags.
 dark orange bags contain 3 bright white bags, 4 muted yellow bags.
 bright white bags contain 1 shiny gold bag.
 muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
@@ -185,20 +94,6 @@ dotted black bags contain no other bags."#;
 
     #[test]
     fn part1() {
-        assert_eq!(solution1(&parse(INPUT1)), 4);
+        assert_eq!(solution1(&parse(INPUT)), 4);
-    }
-
-    const INPUT2: &'static str = r#"shiny gold bags contain 2 dark red bags.
-dark red bags contain 2 dark orange bags.
-dark orange bags contain 2 dark yellow bags.
-dark yellow bags contain 2 dark green bags.
-dark green bags contain 2 dark blue bags.
-dark blue bags contain 2 dark violet bags.
-dark violet bags contain no other bags."#;
-
-    #[test]
-    fn part2() {
-        assert_eq!(solution2(&parse(INPUT1)), 32);
-        assert_eq!(solution2(&parse(INPUT2)), 126);
     }
 }