Pascal's triangle: Difference between revisions
→{{header|jq}}: Using recurse/1
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[1,3,3,1]
[1,4,6,4,1]</lang>
'''Using recurse/1'''
Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function.
<lang jq>def pascal(n):
if n <= 0 then empty
else []
| recurse( if length >= n then empty
else . as $in
| reduce range(0;length-1) as $i
([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1]
end)
end;</lang>
=={{header|K}}==
| |||