# Advent of Code 2017 ## December 2nd For each row, determine the difference between the largest value and the smallest value; the checksum is the sum of all of these differences. For example, given the following spreadsheet: 5 1 9 5 7 5 3 2 4 6 8 * The first row's largest and smallest values are 9 and 1, and their difference is 8. * The second row's largest and smallest values are 7 and 3, and their difference is 4. * The third row's difference is 6. In this example, the spreadsheet's checksum would be 8 + 4 + 6 = 18. ```python from notebook_preamble import J, V, define ``` I'll assume the input is a Joy sequence of sequences of integers. [[5 1 9 5] [7 5 3] [2 4 6 8]] So, obviously, the initial form will be a `step` function: AoC2017.2 == 0 swap [F +] step This function `F` must get the `max` and `min` of a row of numbers and subtract. We can define a helper function `maxmin` which does this: ```python define('maxmin == [max] [min] cleave') ``` ```python J('[1 2 3] maxmin') ``` 3 1 Then `F` just does that then subtracts the min from the max: F == maxmin - So: ```python define('AoC2017.2 == [maxmin - +] step_zero') ``` ```python J(''' [[5 1 9 5] [7 5 3] [2 4 6 8]] AoC2017.2 ''') ``` 18 ...find the only two numbers in each row where one evenly divides the other - that is, where the result of the division operation is a whole number. They would like you to find those numbers on each line, divide them, and add up each line's result. For example, given the following spreadsheet: 5 9 2 8 9 4 7 3 3 8 6 5 * In the first row, the only two numbers that evenly divide are 8 and 2; the result of this division is 4. * In the second row, the two numbers are 9 and 3; the result is 3. * In the third row, the result is 2. In this example, the sum of the results would be 4 + 3 + 2 = 9. What is the sum of each row's result in your puzzle input? ```python J('[5 9 2 8] sort reverse') ``` [9 8 5 2] ```python J('[9 8 5 2] uncons [swap [divmod] cons] dupdip') ``` [8 5 2] [9 divmod] [8 5 2] [9 8 5 2] uncons [swap [divmod] cons F] dupdip G [8 5 2] [9 divmod] F [8 5 2] G ```python V('[8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip') ``` . [8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip [8 5 2] . [9 divmod] [uncons swap] dip dup [i not] dip [8 5 2] [9 divmod] . [uncons swap] dip dup [i not] dip [8 5 2] [9 divmod] [uncons swap] . dip dup [i not] dip [8 5 2] . uncons swap [9 divmod] dup [i not] dip 8 [5 2] . swap [9 divmod] dup [i not] dip [5 2] 8 . [9 divmod] dup [i not] dip [5 2] 8 [9 divmod] . dup [i not] dip [5 2] 8 [9 divmod] [9 divmod] . [i not] dip [5 2] 8 [9 divmod] [9 divmod] [i not] . dip [5 2] 8 [9 divmod] . i not [9 divmod] [5 2] 8 . 9 divmod not [9 divmod] [5 2] 8 9 . divmod not [9 divmod] [5 2] 1 1 . not [9 divmod] [5 2] 1 False . [9 divmod] [5 2] 1 False [9 divmod] . ## Tricky Let's think. Given a *sorted* sequence (from highest to lowest) we want to * for head, tail in sequence * for term in tail: * check if the head % term == 0 * if so compute head / term and terminate loop * else continue ### So we want a `loop` I think [a b c d] True [Q] loop [a b c d] Q [Q] loop `Q` should either leave the result and False, or the `rest` and True. [a b c d] Q ----------------- result 0 [a b c d] Q ----------------- [b c d] 1 This suggests that `Q` should start with: [a b c d] uncons dup roll< [b c d] [b c d] a Now we just have to `pop` it if we don't need it. [b c d] [b c d] a [P] [T] [cons] app2 popdd [E] primrec [b c d] [b c d] [a P] [a T] [E] primrec ------------------- w/ Q == [% not] [T] [F] primrec [a b c d] uncons a [b c d] tuck [b c d] a [b c d] uncons [b c d] a b [c d] roll> [b c d] [c d] a b Q [b c d] [c d] a b [% not] [T] [F] primrec [b c d] [c d] a b T [b c d] [c d] a b / roll> popop 0 [b c d] [c d] a b F Q [b c d] [c d] a b pop swap uncons ... Q [b c d] [c d] a swap uncons ... Q [b c d] a [c d] uncons ... Q [b c d] a c [d] roll> Q [b c d] [d] a c Q Q == [% not] [/ roll> popop 0] [pop swap uncons roll>] primrec uncons tuck uncons roll> Q ```python J('[8 5 3 2] 9 [swap] [% not] [cons] app2 popdd') ``` [8 5 3 2] [9 swap] [9 % not] ------------------- [a b c d] uncons a [b c d] tuck [b c d] a [b c d] [not] [popop 1] [Q] ifte [b c d] a [] popop 1 [b c d] 1 [b c d] a [b c d] Q a [...] Q --------------- result 0 a [...] Q --------------- 1 w/ Q == [first % not] [first / 0] [rest [not] [popop 1]] [ifte] a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte] a [b c d] first % not a b % not a%b not bool(a%b) a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte] a [b c d] first / 0 a b / 0 a/b 0 a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte] a [b c d] rest [not] [popop 1] [Q] ifte a [c d] [not] [popop 1] [Q] ifte a [c d] [not] [popop 1] [Q] ifte a [c d] [not] [popop 1] [Q] ifte a [c d] not a [] popop 1 1 a [c d] Q uncons tuck [first % not] [first / 0] [rest [not] [popop 1]] [ifte] ### I finally sat down with a piece of paper and blocked it out. First, I made a function `G` that expects a number and a sequence of candidates and return the result or zero: n [...] G --------------- result n [...] G --------------- 0 It's a recursive function that conditionally executes the recursive part of its recursive branch [Pg] [E] [R1 [Pi] [T]] [ifte] genrec The recursive branch is the else-part of the inner `ifte`: G == [Pg] [E] [R1 [Pi] [T]] [ifte] genrec == [Pg] [E] [R1 [Pi] [T] [G] ifte] ifte But this is in hindsight. Going forward I derived: G == [first % not] [first /] [rest [not] [popop 0]] [ifte] genrec The predicate detects if the `n` can be evenly divided by the `first` item in the list. If so, the then-part returns the result. Otherwise, we have: n [m ...] rest [not] [popop 0] [G] ifte n [...] [not] [popop 0] [G] ifte This `ifte` guards against empty sequences and returns zero in that case, otherwise it executes `G`. ```python define('G == [first % not] [first /] [rest [not] [popop 0]] [ifte] genrec') ``` Now we need a word that uses `G` on each (head, tail) pair of a sequence until it finds a (non-zero) result. It's going to be designed to work on a stack that has some candidate `n`, a sequence of possible divisors, and a result that is zero to signal to continue (a non-zero value implies that it is the discovered result): n [...] p find-result --------------------------- result It applies `G` using `nullary` because if it fails with one candidate it needs the list to get the next one (the list is otherwise consumed by `G`.) find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec n [...] p [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec The base-case is trivial, return the (non-zero) result. The recursive branch... n [...] p roll< popop uncons [G] nullary find-result [...] p n popop uncons [G] nullary find-result [...] uncons [G] nullary find-result m [..] [G] nullary find-result m [..] p find-result The puzzle states that the input is well-formed, meaning that we can expect a result before the row sequence empties and so do not need to guard the `uncons`. ```python define('find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec') ``` ```python J('[11 9 8 7 3 2] 0 tuck find-result') ``` 3.0 In order to get the thing started, we need to `sort` the list in descending order, then prime the `find-result` function with a dummy candidate value and zero ("continue") flag. ```python define('prep-row == sort reverse 0 tuck') ``` Now we can define our program. ```python define('AoC20017.2.extra == [prep-row find-result +] step_zero') ``` ```python J(''' [[5 9 2 8] [9 4 7 3] [3 8 6 5]] AoC20017.2.extra ''') ``` 9.0