Back on the laptop, running J903 rather than J701 which crashed on the
less compendious iPad.
Here's a somewhat speedier version of Bob Theriault's approach. (I
wasn't seeking to speed it up,
but it wouldn't complete in reasonable time without a tweak or two.)
rt =: 3 : 0
N =. y
n =. >: i. N
NB. +/ @:(*./"1)@:(= <.)@:(3 %~ (* (0 -.~ _&q:))) n NB. RT's Original
answer for the list up to n^n
+/ @:(3 (0*/"1@:=|) (*{:"2@:(__&q:))) n NB. more space-efficient equivalent
)
I'll repeat the definition of qprob here for convenience:
qprob =: 3 : 0
N =. y
t1 =. N <.@% 3. NB. Floor (N % 3)
t2 =. N <.@^ %3 NB. Floor (N ^ 1%3)
t3 =. 3 <.@%~ N <.@^ %3 NB. Floor ( (N ^ 1%3) % 3)
(;~ +/) t1, t2, -t3
)
And here are the comparisons:
(<@rt, qprob)"0 ] 10 100 10000 100000 NB. Theriault's and my results
+-----+-----+------------+
|5 |5 |3 2 0 | NB. No type 3s <: 10 (added comment!)
+-----+-----+------------+
|36 |36 |33 4 _1 |
+-----+-----+------------+
|3347 |3347 |3333 21 _7 |
+-----+-----+------------+
|33364|33364|33333 46 _15|
+-----+-----+------------+
Mike
On 21/03/2021 11:25, 'Mike Day' via Programming wrote:
OK...
here’s what I should have sent earlier:
1. The required number includes all nn = n raised to n for which n are
multiples of 3.
2. It also includes all nn for which n are already powers of 3
3. But we need to avoid double-counting nn for which both 1 & 2 apply.
So:
qprob =: 3 : 0
N =. y
t1 =. N <.@% 3. NB. Floor (N % 3)
t2 =. N <.@^ %3 NB. Floor (N ^ 1%3)
t3 =. 3 <.@%~ N <.@^ %3 NB. Floor ( (N ^ 1%3) % 3)
(;~ +/) t1, t2, -t3
)
qprob 1000
+---+---------+
|340|333 10 _3|
+---+---------+
qprob 10000
+----+----------+
|3347|3333 21 _7|
+----+----------+
qprob 100000
+-----+------------+
|33364|33333 46 _15|
+-----+------------+
Consistent with Bob’s results for 1000 and 10000. The iPad’s J session crashed
checking
his method for 100000; not his fault!
Sorry, this looks crummy on the iPad.
Mike
Sent from my iPad
On 21 Mar 2021, at 08:41, 'robert therriault' via Programming
<programm...@jsoftware.com> wrote:
n=. >: i. 1000 NB. set up list from 1 to 1000
_&q: 1000 NB. returns the prime exponents of 1000
3 0 3
(* _&q:) 1000 NB. returns the prime exponents of 1000^1000
3000 0 3000
(= <.)@:(3 %~ (* _&q:)) 1000 NB. divides the exponents by 3 then returns 1
for integer result
1 1 1
(*./"1)@:(= <.)@:(3 %~ (* _&q:)) 1000 NB. if all exponents are divisible by
3 then it is a cube
1
+/ @:(*./"1)@:(= <.)@:(3 %~ (* _&q:)) 1000 NB. sum up the number of cubes -
returns 1 since 1000^1000 is a cube
1
+/ @:(*./"1)@:(= <.)@:(3 %~ (* _&q:)) n NB. answer for the list up to n^n
340
as it turns out the 340 are these
17 20 $ >: I. *./"1@:(= <.)@:(3 %~ (* _&q:)) n
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