Spiral matrix: Difference between revisions
→Functional Solutions: Added a pure functional variant (without generative mutation)
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Line 3,858:
print ' '.join('%3s' % x for x in row)</lang>
===Functional
{{works with|Python|2.6, 3.0}}
<lang python>import itertools
Line 3,879:
for row in spiral(5):
print(' '.join('%3s' % x for x in row))</lang>
Or, as an alternative to generative mutation:
{{Works with|Python|3.7}}
{{Trans|Haskell}}
<lang python>'''Spiral Matrix'''
# spiral :: Int -> [[Int]]
def spiral(n):
'''The rows of a spiral matrix of order N.
'''
def go(rows, cols, x):
return [list(range(x, x + cols))] + [
list(reversed(x)) for x
in zip(*go(cols, rows - 1, x + cols))
] if 0 < rows else [[]]
return go(n, n, 0)
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Spiral matrix of order 5, in wiki table markup'''
print(wikiTable(
spiral(5)
))
# FORMATTING ----------------------------------------------
# wikiTable :: [[a]] -> String
def wikiTable(rows):
'''Wiki markup for a no-frills tabulation of rows.'''
return '{| class="wikitable" style="' + (
'width:12em;height:12em;table-layout:fixed;"|-\n'
) + '\n|-\n'.join([
'| ' + ' || '.join([str(cell) for cell in row])
for row in rows
]) + '\n|}'
# MAIN ---
if __name__ == '__main__':
main()</lang>
{| class="wikitable" style="width:12em;height:12em;table-layout:fixed;"|-
| 0 || 1 || 2 || 3 || 4
|-
| 15 || 16 || 17 || 18 || 5
|-
| 14 || 23 || 24 || 19 || 6
|-
| 13 || 22 || 21 || 20 || 7
|-
| 12 || 11 || 10 || 9 || 8
|}
=== Simple solution ===
| |||