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- # PROBLEM DESCRIPTION:
- # In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
- # 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
- # 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
- # 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
- # 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
- # 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
- # 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
- # 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
- # 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
- # 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
- # 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
- # 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
- # 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
- # 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
- # 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
- # 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
- # 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
- # 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
- # 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
- # 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
- # 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
- # The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
- # What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
- # SOLUTION DESCRIPTION:
- # First, we use the Q quoting mechanism with the `to` adverb with value `END`. This will start quoting very literally
- # on the following line. We can treat this expression as a value, splitting it on newlines and skipping empty segments
- # (due to the :skip-empty adverb, which is shorthand for :skip-empty(True) or skip-empty => True)
- # Then, we split each of the strings in the resulting list on the space character. To make these Strs into Ints, we
- # call .deepmap, which will recurse down as far as it can go, applying the WhateverCode, *.Int, to each individual
- # element.
- # Since we have a row-major grid, getting the horizontal lines of 4 numbers is easy with .rotor. We call .rotor on each
- # row, passing it the Pair, 4 => -3, which indicates that the method should grab chunks of 4, rewinding by 3 each
- # time (forward 4, back 3 -> net movement of 1), so it grabs all sets of 4 consecutive elements.
- # Then, we call .map on each row (actually each list of sets of 4), telling the method to reduce each element with the
- # infix:<*> routine (operator *). Note that we refer to the routine as a noun with the & sigil.
- # The resulting list (still nested) is flattened out and passed to the max subroutine.
- # Vertical lines of 4 numbers are more tricky because our grid is row-major. You should be able to understand most of
- # this code. Here are some comments to help:
- # ^16 (read: up to sixteen) is the same as 0..^16 is the same as 0..15.
- # You can take multiple elements of an array by passing a list to the postcircumfix [ ] operator.
- # Circumfix [ ] takes an operator and makes a reduction function from it. [*], then, creates a function that reduces
- # by applying * between all the elements. Though it doesn't matter for *, this reduction metaoperator also
- # respects the associativity of the operator, so you can think of it as putting the operator between all the
- # elements of the list passed in:
- # 2 ** 3 ** 4 == 2 ** (3 ** 4) == [**] 2, 3 ,4
- # The array indexing operator, postcirumfix [ ], can also be hyper-called. We use this to quickly access the element
- # in the same position in each row.
- # We then take the top-left -> bottom-right diagonal lines and the top-right -> bottom-left diagonal lines.
- # Finally, we find the max of the maxes from each section.
- my @grid = Q:to<END>.split("\n", :skip-emtpy)>>.split(' ').deepmap(*.Int);
- 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
- 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
- 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
- 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
- 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
- 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
- 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
- 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
- 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
- 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
- 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
- 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
- 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
- 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
- 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
- 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
- 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
- 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
- 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
- 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
- END
- my $horizontal-max = max @grid>>.rotor(4 => -3)>>.map(*.reduce(&infix:<*>)).flat;
- my $vertical-max = 0;
- for ^16 -> $first-row
- {
- my @nums = @grid[$first-row..$first-row + 3];
- for ^20 -> $column
- {
- my $product = [*] @nums>>[$column];
- $vertical-max = $product if $product > $vertical-max;
- }
- }
- my $diagonal1-max = 0;
- for ^16 -> $row
- {
- for ^16 -> $col
- {
- my $product = [*]
- @grid[$row; $col],
- @grid[$row + 1; $col + 1],
- @grid[$row + 2; $col + 2],
- @grid[$row + 3; $col + 3];
- $diagonal1-max = $product if $product > $diagonal1-max;
- }
- }
- my $diagonal2-max = 0;
- for ^16 -> $row
- {
- for 3..^20 -> $col
- {
- my $product = [*]
- @grid[$row; $col],
- @grid[$row + 1; $col - 1],
- @grid[$row + 2; $col - 2],
- @grid[$row + 3; $col - 3];
- $diagonal2-max = $product if $product > $diagonal2-max;
- }
- }
- say max $horizontal-max, $vertical-max, $diagonal1-max, $diagonal2-max;
- # ANSWER: 70600674
- # TIME: 0.11025621 s
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