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Pascal's triangle: Difference between revisions
→{{header|Haskell}}: using comprehensions
(→{{header|Haskell}}: using comprehensions) |
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Line 2,363:
pascal = iterate nextRow [1]</lang>
Alternatively, using list comprehensions:
<lang haskell>
pascal :: [[Integer]]
pascal =
(1 : [ 0 | _ <- head pascal])
: [zipWith (+) (0:row) row | row <- pascal]
</lang>
{{out}}<pre>
Pascal> take 5 <$> (take 5 $ triangle)
[[1,0,0,0,0],[1,1,0,0,0],[1,2,1,0,0],[1,3,3,1,0],[1,4,6,4,1]]
</pre>
With binomial coefficients:
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