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Euler 11

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Jul 31st, 2018
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  1. # PROBLEM DESCRIPTION:
  2. # In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
  3. # 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
  4. # 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
  5. # 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
  6. # 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
  7. # 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
  8. # 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
  9. # 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
  10. # 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
  11. # 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
  12. # 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
  13. # 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
  14. # 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
  15. # 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
  16. # 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
  17. # 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
  18. # 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
  19. # 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
  20. # 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
  21. # 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
  22. # 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
  23. # The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
  24. # What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
  25.  
  26. # SOLUTION DESCRIPTION:
  27. # First, we use the Q quoting mechanism with the `to` adverb with value `END`. This will start quoting very literally
  28. #   on the following line. We can treat this expression as a value, splitting it on newlines and skipping empty segments
  29. #   (due to the :skip-empty adverb, which is shorthand for :skip-empty(True) or skip-empty => True)
  30. #   Then, we split each of the strings in the resulting list on the space character. To make these Strs into Ints, we
  31. #   call .deepmap, which will recurse down as far as it can go, applying the WhateverCode, *.Int, to each individual
  32. #   element.
  33. # Since we have a row-major grid, getting the horizontal lines of 4 numbers is easy with .rotor. We call .rotor on each
  34. #   row, passing it the Pair, 4 => -3, which indicates that the method should grab chunks of 4, rewinding by 3 each
  35. #   time (forward 4, back 3 -> net movement of 1), so it grabs all sets of 4 consecutive elements.
  36. #   Then, we call .map on each row (actually each list of sets of 4), telling the method to reduce each element with the
  37. #   infix:<*> routine (operator *). Note that we refer to the routine as a noun with the & sigil.
  38. #   The resulting list (still nested) is flattened out and passed to the max subroutine.
  39. # Vertical lines of 4 numbers are more tricky because our grid is row-major. You should be able to understand most of
  40. #   this code. Here are some comments to help:
  41. #     ^16 (read: up to sixteen) is the same as 0..^16 is the same as 0..15.
  42. #     You can take multiple elements of an array by passing a list to the postcircumfix [ ] operator.
  43. #     Circumfix [ ] takes an operator and makes a reduction function from it. [*], then, creates a function that reduces
  44. #       by applying * between all the elements. Though it doesn't matter for *, this reduction metaoperator also
  45. #       respects the associativity of the operator, so you can think of it as putting the operator between all the
  46. #       elements of the list passed in:
  47. #       2 ** 3 ** 4 == 2 ** (3 ** 4) == [**] 2, 3 ,4
  48. #     The array indexing operator, postcirumfix [ ], can also be hyper-called. We use this to quickly access the element
  49. #     in the same position in each row.
  50. # We then take the top-left -> bottom-right diagonal lines and the top-right -> bottom-left diagonal lines.
  51. # Finally, we find the max of the maxes from each section.
  52.  
  53. my @grid = Q:to<END>.split("\n", :skip-emtpy)>>.split(' ').deepmap(*.Int);
  54. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
  55. 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
  56. 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
  57. 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
  58. 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
  59. 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
  60. 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
  61. 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
  62. 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
  63. 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
  64. 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
  65. 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
  66. 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
  67. 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
  68. 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
  69. 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
  70. 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
  71. 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
  72. 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
  73. 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
  74. END
  75.  
  76. my $horizontal-max = max @grid>>.rotor(4 => -3)>>.map(*.reduce(&infix:<*>)).flat;
  77.  
  78. my $vertical-max = 0;
  79. for ^16 -> $first-row
  80. {
  81.     my @nums = @grid[$first-row..$first-row + 3];
  82.     for ^20 -> $column
  83.     {
  84.         my $product = [*] @nums>>[$column];
  85.         $vertical-max = $product if $product > $vertical-max;
  86.     }
  87. }
  88.  
  89. my $diagonal1-max = 0;
  90. for ^16 -> $row
  91. {
  92.     for ^16 -> $col
  93.     {
  94.         my $product = [*]
  95.                 @grid[$row; $col],
  96.                 @grid[$row + 1; $col + 1],
  97.                 @grid[$row + 2; $col + 2],
  98.                 @grid[$row + 3; $col + 3];
  99.         $diagonal1-max = $product if $product > $diagonal1-max;
  100.     }
  101. }
  102.  
  103. my $diagonal2-max = 0;
  104. for ^16 -> $row
  105. {
  106.     for 3..^20 -> $col
  107.     {
  108.         my $product = [*]
  109.                 @grid[$row; $col],
  110.                 @grid[$row + 1; $col - 1],
  111.                 @grid[$row + 2; $col - 2],
  112.                 @grid[$row + 3; $col - 3];
  113.         $diagonal2-max = $product if $product > $diagonal2-max;
  114.     }
  115. }
  116.  
  117. say max $horizontal-max, $vertical-max, $diagonal1-max, $diagonal2-max;
  118.  
  119. # ANSWER: 70600674
  120. # TIME:   0.11025621 s
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