Thun/docs/Advent_of_Code_2017_December_2nd.ipynb

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.

In [1]:

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:

In [2]:

define('maxmin [max] [min] cleave')

In [3]:

J('[1 2 3] maxmin')

3 1

Then F just does that then subtracts the min from the max:

F == maxmin -

So:

In [4]:

define('AoC2017.2 [maxmin - +] step_zero')

In [5]:

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?

In [6]:

J('[5 9 2 8] sort reverse')

[9 8 5 2]

In [7]:

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

In [8]:

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

In [9]:

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.

In [10]:

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] tailrec

n [...] p [0 >] [roll> popop] [roll< popop uncons [G] nullary] tailrec

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.

In [11]:

define('find-result [0 >] [roll> popop] [roll< popop uncons [G] nullary] tailrec')

In [12]:

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.

In [13]:

define('prep-row sort reverse 0 tuck')

Now we can define our program.

In [14]:

define('AoC20017.2.extra [prep-row find-result +] step_zero')

In [15]:

J('''

[[5 9 2 8]
 [9 4 7 3]
 [3 8 6 5]] AoC20017.2.extra

''')

9.0