The solutions based on (p.) are correct, but (p.) is insufficient. Roger's
work-around is very nice!
Den 5:01 mandag den 10. august 2015 skrev Roger Hui
<[email protected]>:
An advantage of solutions based on taking cube roots, on the expression
>.@%: in particular, is working correctly with large integers. For example:
g=: = (^&3 - ])@:(3&(>.@%:))
n=: */ 0 1 2 + 11^20x
n
304481639541418099575807073027324053193216441247734861355941206
g n
1
g n+1
0
In contrast:
p. (-n),2 3 1
┌─┬────────────────────────────────────────────────────────┐
│1│6.7275e20 _3.36375e20j5.82619e20 _3.36375e20j_5.82619e20│
└─┴────────────────────────────────────────────────────────┘
p. (-n+1),2 3 1
┌─┬────────────────────────────────────────────────────────┐
│1│6.7275e20 _3.36375e20j5.82619e20 _3.36375e20j_5.82619e20│
└─┴────────────────────────────────────────────────────────┘
(p. (-n),2 3 1) -:!.0 p.(-n+1),2 3 1
1
The last expression shows that a solution based on p. can not discriminate
between n and n+1.
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