433 lines
9.4 KiB
ReStructuredText
433 lines
9.4 KiB
ReStructuredText
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Advent of Code 2017
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===================
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December 2nd
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------------
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For each row, determine the difference between the largest value and the
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smallest value; the checksum is the sum of all of these differences.
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For example, given the following spreadsheet:
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::
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5 1 9 5
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7 5 3
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2 4 6 8
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- The first row's largest and smallest values are 9 and 1, and their
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difference is 8.
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- The second row's largest and smallest values are 7 and 3, and their
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difference is 4.
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- The third row's difference is 6.
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In this example, the spreadsheet's checksum would be 8 + 4 + 6 = 18.
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.. code:: ipython2
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from notebook_preamble import J, V, define
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I'll assume the input is a Joy sequence of sequences of integers.
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::
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[[5 1 9 5]
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[7 5 3]
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[2 4 6 8]]
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So, obviously, the initial form will be a ``step`` function:
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::
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AoC2017.2 == 0 swap [F +] step
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This function ``F`` must get the ``max`` and ``min`` of a row of numbers
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and subtract. We can define a helper function ``maxmin`` which does
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this:
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.. code:: ipython2
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define('maxmin == [max] [min] cleave')
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.. code:: ipython2
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J('[1 2 3] maxmin')
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.. parsed-literal::
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3 1
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Then ``F`` just does that then subtracts the min from the max:
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::
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F == maxmin -
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So:
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.. code:: ipython2
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define('AoC2017.2 == [maxmin - +] step_zero')
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.. code:: ipython2
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J('''
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[[5 1 9 5]
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[7 5 3]
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[2 4 6 8]] AoC2017.2
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''')
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.. parsed-literal::
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18
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...find the only two numbers in each row where one evenly divides the
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other - that is, where the result of the division operation is a whole
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number. They would like you to find those numbers on each line, divide
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them, and add up each line's result.
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For example, given the following spreadsheet:
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::
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5 9 2 8
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9 4 7 3
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3 8 6 5
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- In the first row, the only two numbers that evenly divide are 8 and
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2; the result of this division is 4.
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- In the second row, the two numbers are 9 and 3; the result is 3.
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- In the third row, the result is 2.
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In this example, the sum of the results would be 4 + 3 + 2 = 9.
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What is the sum of each row's result in your puzzle input?
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.. code:: ipython2
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J('[5 9 2 8] sort reverse')
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.. parsed-literal::
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[9 8 5 2]
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.. code:: ipython2
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J('[9 8 5 2] uncons [swap [divmod] cons] dupdip')
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.. parsed-literal::
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[8 5 2] [9 divmod] [8 5 2]
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::
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[9 8 5 2] uncons [swap [divmod] cons F] dupdip G
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[8 5 2] [9 divmod] F [8 5 2] G
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.. code:: ipython2
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V('[8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip')
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.. parsed-literal::
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. [8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip
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[8 5 2] . [9 divmod] [uncons swap] dip dup [i not] dip
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[8 5 2] [9 divmod] . [uncons swap] dip dup [i not] dip
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[8 5 2] [9 divmod] [uncons swap] . dip dup [i not] dip
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[8 5 2] . uncons swap [9 divmod] dup [i not] dip
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8 [5 2] . swap [9 divmod] dup [i not] dip
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[5 2] 8 . [9 divmod] dup [i not] dip
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[5 2] 8 [9 divmod] . dup [i not] dip
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[5 2] 8 [9 divmod] [9 divmod] . [i not] dip
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[5 2] 8 [9 divmod] [9 divmod] [i not] . dip
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[5 2] 8 [9 divmod] . i not [9 divmod]
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[5 2] 8 . 9 divmod not [9 divmod]
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[5 2] 8 9 . divmod not [9 divmod]
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[5 2] 1 1 . not [9 divmod]
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[5 2] 1 False . [9 divmod]
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[5 2] 1 False [9 divmod] .
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Tricky
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------
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Let's think.
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Given a *sorted* sequence (from highest to lowest) we want to \* for
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head, tail in sequence \* for term in tail: \* check if the head % term
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== 0 \* if so compute head / term and terminate loop \* else continue
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So we want a ``loop`` I think
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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::
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[a b c d] True [Q] loop
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[a b c d] Q [Q] loop
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``Q`` should either leave the result and False, or the ``rest`` and
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True.
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::
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[a b c d] Q
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-----------------
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result 0
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[a b c d] Q
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-----------------
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[b c d] 1
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This suggests that ``Q`` should start with:
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::
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[a b c d] uncons dup roll<
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[b c d] [b c d] a
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Now we just have to ``pop`` it if we don't need it.
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::
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[b c d] [b c d] a [P] [T] [cons] app2 popdd [E] primrec
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[b c d] [b c d] [a P] [a T] [E] primrec
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--------------
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::
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w/ Q == [% not] [T] [F] primrec
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[a b c d] uncons
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a [b c d] tuck
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[b c d] a [b c d] uncons
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[b c d] a b [c d] roll>
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[b c d] [c d] a b Q
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[b c d] [c d] a b [% not] [T] [F] primrec
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[b c d] [c d] a b T
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[b c d] [c d] a b / roll> popop 0
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[b c d] [c d] a b F Q
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[b c d] [c d] a b pop swap uncons ... Q
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[b c d] [c d] a swap uncons ... Q
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[b c d] a [c d] uncons ... Q
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[b c d] a c [d] roll> Q
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[b c d] [d] a c Q
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Q == [% not] [/ roll> popop 0] [pop swap uncons roll>] primrec
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uncons tuck uncons roll> Q
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.. code:: ipython2
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J('[8 5 3 2] 9 [swap] [% not] [cons] app2 popdd')
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.. parsed-literal::
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[8 5 3 2] [9 swap] [9 % not]
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--------------
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::
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[a b c d] uncons
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a [b c d] tuck
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[b c d] a [b c d] [not] [popop 1] [Q] ifte
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[b c d] a [] popop 1
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[b c d] 1
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[b c d] a [b c d] Q
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a [...] Q
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---------------
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result 0
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a [...] Q
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---------------
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1
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w/ Q == [first % not] [first / 0] [rest [not] [popop 1]] [ifte]
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a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]
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a [b c d] first % not
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a b % not
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a%b not
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bool(a%b)
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a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]
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a [b c d] first / 0
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a b / 0
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a/b 0
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a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]
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a [b c d] rest [not] [popop 1] [Q] ifte
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a [c d] [not] [popop 1] [Q] ifte
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a [c d] [not] [popop 1] [Q] ifte
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a [c d] [not] [popop 1] [Q] ifte
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a [c d] not
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a [] popop 1
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1
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a [c d] Q
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uncons tuck [first % not] [first / 0] [rest [not] [popop 1]] [ifte]
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I finally sat down with a piece of paper and blocked it out.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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First, I made a function ``G`` that expects a number and a sequence of
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candidates and return the result or zero:
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::
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n [...] G
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---------------
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result
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n [...] G
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---------------
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0
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It's a recursive function that conditionally executes the recursive part
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of its recursive branch
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::
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[Pg] [E] [R1 [Pi] [T]] [ifte] genrec
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The recursive branch is the else-part of the inner ``ifte``:
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::
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G == [Pg] [E] [R1 [Pi] [T]] [ifte] genrec
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== [Pg] [E] [R1 [Pi] [T] [G] ifte] ifte
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But this is in hindsight. Going forward I derived:
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::
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G == [first % not]
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[first /]
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[rest [not] [popop 0]]
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[ifte] genrec
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The predicate detects if the ``n`` can be evenly divided by the
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``first`` item in the list. If so, the then-part returns the result.
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Otherwise, we have:
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::
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n [m ...] rest [not] [popop 0] [G] ifte
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n [...] [not] [popop 0] [G] ifte
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This ``ifte`` guards against empty sequences and returns zero in that
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case, otherwise it executes ``G``.
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.. code:: ipython2
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define('G == [first % not] [first /] [rest [not] [popop 0]] [ifte] genrec')
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Now we need a word that uses ``G`` on each (head, tail) pair of a
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sequence until it finds a (non-zero) result. It's going to be designed
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to work on a stack that has some candidate ``n``, a sequence of possible
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divisors, and a result that is zero to signal to continue (a non-zero
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value implies that it is the discovered result):
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::
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n [...] p find-result
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---------------------------
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result
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It applies ``G`` using ``nullary`` because if it fails with one
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candidate it needs the list to get the next one (the list is otherwise
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consumed by ``G``.)
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::
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find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec
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n [...] p [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec
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The base-case is trivial, return the (non-zero) result. The recursive
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branch...
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::
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n [...] p roll< popop uncons [G] nullary find-result
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[...] p n popop uncons [G] nullary find-result
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[...] uncons [G] nullary find-result
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m [..] [G] nullary find-result
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m [..] p find-result
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The puzzle states that the input is well-formed, meaning that we can
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expect a result before the row sequence empties and so do not need to
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guard the ``uncons``.
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.. code:: ipython2
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define('find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec')
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.. code:: ipython2
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J('[11 9 8 7 3 2] 0 tuck find-result')
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.. parsed-literal::
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3.0
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In order to get the thing started, we need to ``sort`` the list in
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descending order, then prime the ``find-result`` function with a dummy
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candidate value and zero ("continue") flag.
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.. code:: ipython2
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define('prep-row == sort reverse 0 tuck')
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Now we can define our program.
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.. code:: ipython2
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define('AoC20017.2.extra == [prep-row find-result +] step_zero')
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.. code:: ipython2
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J('''
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[[5 9 2 8]
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[9 4 7 3]
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[3 8 6 5]] AoC20017.2.extra
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''')
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.. parsed-literal::
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9.0
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