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day08: replace static foreach with workaround
author Jordi Gutiérrez Hermoso <>
date Tue, 16 Jan 2018 11:28:55 -0500
parents 049fb8e56025
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--- Day 7: Recursive Circus ---

Wandering further through the circuits of the computer, you come upon
a tower of programs that have gotten themselves into a bit of trouble.
A recursive algorithm has gotten out of hand, and now they're balanced
precariously in a large tower.

One program at the bottom supports the entire tower. It's holding a
large disc, and on the disc are balanced several more sub-towers. At
the bottom of these sub-towers, standing on the bottom disc, are other
programs, each holding their own disc, and so on. At the very tops of
these sub-sub-sub-...-towers, many programs stand simply keeping the
disc below them balanced but with no disc of their own.

You offer to help, but first you need to understand the structure of
these towers. You ask each program to yell out their name, their
weight, and (if they're holding a disc) the names of the programs
immediately above them balancing on that disc. You write this
information down (your puzzle input). Unfortunately, in their panic,
they don't do this in an orderly fashion; by the time you're done,
you're not sure which program gave which information.

For example, if your list is the following:

pbga (66)
xhth (57)
ebii (61)
havc (66)
ktlj (57)
fwft (72) -> ktlj, cntj, xhth
qoyq (66)
padx (45) -> pbga, havc, qoyq
tknk (41) -> ugml, padx, fwft
jptl (61)
ugml (68) -> gyxo, ebii, jptl
gyxo (61)
cntj (57)

...then you would be able to recreate the structure of the towers that
looks like this:

         ugml - ebii
       /      \     
      |         jptl
      |         pbga
     /        /
tknk --- padx - havc
     \        \
      |         qoyq
      |         ktlj
       \      /     
         fwft - cntj

In this example, tknk is at the bottom of the tower (the bottom
program), and is holding up ugml, padx, and fwft. Those programs are,
in turn, holding up other programs; in this example, none of those
programs are holding up any other programs, and are all the tops of
their own towers. (The actual tower balancing in front of you is much

Before you're ready to help them, you need to make sure your
information is correct. What is the name of the bottom program?

Your puzzle answer was xegshds.

--- Part Two ---

The programs explain the situation: they can't get down. Rather, they
could get down, if they weren't expending all of their energy trying
to keep the tower balanced. Apparently, one program has the wrong
weight, and until it's fixed, they're stuck here.

For any program holding a disc, each program standing on that disc
forms a sub-tower. Each of those sub-towers are supposed to be the
same weight, or the disc itself isn't balanced. The weight of a tower
is the sum of the weights of the programs in that tower.

In the example above, this means that for ugml's disc to be balanced,
gyxo, ebii, and jptl must all have the same weight, and they do: 61.

However, for tknk to be balanced, each of the programs standing on its
disc and all programs above it must each match. This means that the
following sums must all be the same:

    ugml + (gyxo + ebii + jptl) = 68 + (61 + 61 + 61) = 251
    padx + (pbga + havc + qoyq) = 45 + (66 + 66 + 66) = 243
    fwft + (ktlj + cntj + xhth) = 72 + (57 + 57 + 57) = 243

As you can see, tknk's disc is unbalanced: ugml's stack is heavier
than the other two. Even though the nodes above ugml are balanced,
ugml itself is too heavy: it needs to be 8 units lighter for its stack
to weigh 243 and keep the towers balanced. If this change were made,
its weight would be 60.

Given that exactly one program is the wrong weight, what would its
weight need to be to balance the entire tower?

Your puzzle answer was 299.

Both parts of this puzzle are complete! They provide two gold stars: **