On 4/23/20, Brian Schott wrote:
> Harvey,
> Your explanations were very helpful. Especially of the second line of
> RSLTNT output.
> There is no listing of the notes that go with the example RSLTNT 5 3, are
> there? Your explanation just understands what notes are typical and the
> meaning of "int
I want to find all the integer factors of a positive integer,
which includes 1 and the integer.
Here's a brute force verb I wrote:
fac=:3 : 'a#~0=(a=.1+i.y)|y'
Test it:
fac each 40+i.5
│1 2 4 5 8 10 20 40│1 41│1 2 3 6 7 14 21 42│1 43│1 2 4 11 22 44│
However my fac verb is inefficient, slow, an
See https://code.jsoftware.com/wiki/Essays/Factorings
On Fri, Apr 24, 2020 at 11:29 AM Skip Cave wrote:
> I want to find all the integer factors of a positive integer,
> which includes 1 and the integer.
> Here's a brute force verb I wrote:
> fac=:3 : 'a#~0=(a=.1+i.y)|y'
>
> Test it:
>
> fac e
using builtin q:
I get correct results using
/:~ ~.@, ((*/)@#"1~ #:@:i.@(2&^)@#) 1,q: 4711
1 7 673 4711
/:~ ~.@, ((*/)@#"1~ #:@:i.@(2&^)@#) 1,q: 42
1 2 3 6 7 14 21 42
less brute force but it’s just a first attempt
Am 24.04.20 um 20:28 schrieb Skip Cave:
I want to find all the integer f
I have used the wrong terminology. I want to find all the positive integer
*divisors* of an integer n (not just primes) which includes 1 and the
integer.
Here's a brute force verb I wrote:
div=:3 : 'a#~0=(a=.1+i.y)|y'
Test it:
div each 40+i.5
│1 2 4 5 8 10 20 40│1 41│1 2 3 6 7 14 21 42│1 43│1 2
does counting suffice? then that would be φ which I’m sure
someone must have written code for already
Am 24.04.20 um 21:09 schrieb Skip Cave:
I have used the wrong terminology. I want to find all the positive integer
*divisors* of an integer n (not just primes) which includes 1 and the
integer.
Euler's phi function? That's
5 p: ] or 5&p:
the "totient"
Not what Skip is looking for, though?
On 24/04/2020 20:14, Hauke Rehr wrote:
does counting suffice? then that would be φ which I’m sure
someone must have written code for already
Am 24.04.20 um 21:09 schrieb Sk
Sorry, I must have confused things.
This is not at all the correct answer.
I thought it should be (- φ) n, but obviously it’s wrong.
(tested with some numbers)
Thanks for pointing to 5&p: for φ, though.
Am 24.04.20 um 21:31 schrieb 'Michael Day' via Programming:
Euler's phi function? That's
It occurred to me that I had forgotten two other ways that the
"rhythm" resultant of 3 2 1 3 1 2 3 could be used--namely, for volume
(dynamics) and for speed (tempo) of some music. The point is that any
resultant of "interference" of 2 or more rhythms (whether constant
"beats" or including another
I now have an answer for the number of results:
*/ >: +/"1 (=/~ ~.) q: 4711
4
*/ >: +/"1 (=/~ ~.) q: 42
8
I hesitate using =/~ (it’s actually deprecated)
but I wouldn’t know a better way in this case
Am 24.04.20 um 21:09 schrieb Skip Cave:
I have used the wrong terminology. I want to fi
The number of ways a prime p can participate in a list of divisors is 1+e
where e is its exponent in the prime factorization of x . __ q: x gives
you the primes and the corresponding exponents. Therefore:
# div 24
8
*/ 1 + {: __ q: 24
8
# div 64e8
135
*/ 1 + {: __ q: 64e8
135
On
I think this is near to the approved way:
(*/"1@:>:@(_& q:))24 360 4711
8 24 4
Mike
Sent from my iPad
> On 24 Apr 2020, at 21:13, Hauke Rehr wrote:
>
> I now have an answer for the number of results:
>
> */ >: +/"1 (=/~ ~.) q: 4711
> 4
> */ >: +/"1 (=/~ ~.) q: 42
> 8
>
> I hesitate u
I seem to recall there is a way to use the "cut" conjunction with a Boolean
left argument specifying partition boundaries but I cannot find where this
is mentioned on the J wiki.
Is there a way to do this?
Thanks,
Devon
--
Devon McCormick, CFA
Quantitative Consultant
I must have been thinking of "key", e.g.
1 0 1 0 wrote:
> I seem to recall there is a way to use the "cut" conjunction with a
> Boolean left argument specifying partition boundaries but I cannot find
> where this is mentioned on the J wiki.
>
> Is there a way to do this?
>
> Thanks,
>
> Devon
but in that case a boolean left argument will
give a partition with at most two elements
in general you need n different values
for partitioning into n sets this way
Am 24.04.20 um 23:45 schrieb Devon McCormick:
I must have been thinking of "key", e.g.
1 0 1 0 wrote:
I seem to recall the
Or not ...
1 0 1 0 0 +/;.1 ]i. 5
1 9
> On 2020Apr 24, at 14:45, Devon McCormick wrote:
>
> I must have been thinking of "key", e.g.
> 1 0 1 0 +---+---+
> |0 2|1 3|
> +---+---+
>
>
> On Fri, Apr 24, 2020 at 5:37 PM Devon McCormick wrote:
>
>> I seem to recall there is a way to use th
This is missing a few divisors. It seems Skip's original method would be
difficult to beat. It's O(n) in time and space. You could improve space
complexity by doing a simple loop, but as far as I can see there are no
shortcuts, and the best way is to just "brute force" it.
Sent from BlueMail
Numbers of divisors below, not the divisors themselves... the thread had split!
I was attempting to comment on Hauke Rehr’s post.
M
Sent from my iPad
> On 24 Apr 2020, at 23:52, 'Jon' via Programming
> wrote:
>
> This is missing a few divisors. It seems Skip's original method would be
> diff
I found this one in my bag of tricks:
allFactors=: */&>@{@((^ i.@>:)&.>/)@q:~&__
#,allFactors */p:i.10
1024
On Fri, Apr 24, 2020 at 7:15 PM 'Mike Day' via Programming <
[email protected]> wrote:
> Numbers of divisors below, not the divisors themselves... the thread had
> split! I w
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