Sorry, I forgot to spell it out.
tap achieves the same as perm, though I think perm is an explicit recursive or
looping
function.
tap is probably one of Roger’s whizzbangs; tap is an acronym for something
like Table all
permutations.
The version I happened to use yesterday is
i.@! A. i.
but I love the more compact
A.&i.~!
In either case, my expression involving tap could just as well be
1,. 2 + perm 7
which generates 5040 permutations of 2,3,...8 and pretends 1 in “fixed”
position.
Cheers,
Mike
Sent from my iPad
> On 15 Mar 2021, at 02:35, Skip Cave <[email protected]> wrote:
>
> Here is the J Programming thread about Rotational Duplicates:
> http://jsoftware.com/pipermail/programming/2020-May/055839.html
>
> Mike Day - What is the "tap" verb in your equation?
>
> $s3=.+/"1[3,\"1|:n,2{.n=.|:1,.2+tap 7
> 5040 8
>
>
> Skip Cave
> Cave Consulting LLC
>
>
>> On Sun, Mar 14, 2021 at 8:01 PM Skip Cave <[email protected]> wrote:
>>
>> Mike,
>> You are right. I forgot about rotational duplicates.
>> I had a similar problem awhile back, and Roger showed us several ways to
>> remove rotational duplicates.
>> Here is one of Roger's verbs:
>>
>> rdr=.3 :'~.(y i."_1<./"1 y)|."_1 y' NB. Remove rotational duplicates.
>>
>> $n=.>:perm 8
>>
>> 40320 8
>>
>> $rdr n
>>
>> 5040 8
>>
>> So there are only 5,040 unique vertex permutations.
>>
>> Skip Cave
>> Cave Consulting LLC
>>
>>
>> On Sun, Mar 14, 2021 at 6:55 PM 'Mike Day' via Chat <[email protected]>
>> wrote:
>>
>>> As these are supposed to be regular polygons, shouldn’t you identify
>>> 123...8 with
>>> 234...81 etc, in which case there are !7 rather than !8 different
>>> octagons.
>>>
>>> Also, we only need consider the minimum triplet sum in each figure,
>>> since, by definition, all other triplet sums in that octagon are greater
>>> or equal.
>>>
>>> So:
>>> $s3=.+/"1[3,\"1|:n,2{.n=.|:1,.2+tap 7
>>> 5040 8
>>>
>>> +/m=.*./"1[11<"1 s3. NB. 8 times smaller because of equivalence under
>>> rotation...
>>> 40
>>>
>>> +/11<<./"1 s3. NB. same comparing minimum triplet-sums
>>> 40
>>>
>>> It is fairly straightforward to develop a constructive function to
>>> generate polygons with n
>>> sides for which all k-tuplets of consecutive corners sum to more than
>>> some value. It’s a lot slower than using perm or tap for octagons, but
>>> would be competitive for somewhat larger polygons.
>>>
>>> Since +/>: i. 8 = 36, the average triplet sum is 13.5 = 108%3 , we know
>>> that there is no
>>> such octagon with all triplet sums exceeding 13.
>>>
>>> Mike
>>>
>>>
>>> Sent from my iPad
>>>
>>>> On 14 Mar 2021, at 22:00, Skip Cave <[email protected]> wrote:
>>>>
>>>> I found this problem on Quora. I had fun solving it in J. Are there more
>>>> efficient ways?
>>>>
>>>> The first 8 natural numbers (1 to 8) are positioned on the vertices of
>>> an
>>>> octagon. Is it possible to position the numbers such that the sum of
>>> any 3
>>>> adjacent vertices is greater than 11? Greater than 13?
>>>>
>>>> Using the J programming language, brute force approach:
>>>>
>>>> Wikipedia: "J (programming language)"
>>>>
>>>> J primitives: NuVoc - J Wiki <https://code.jsoftware.com/wiki/NuVoc>
>>>>
>>>> Generate all permutations of the integers 1–8, & store them in n. Count
>>> the
>>>> permutations & display the count:
>>>>
>>>> $n=.|:>:perm 8
>>>>
>>>> 8 40320
>>>>
>>>> So n contains the 40,320 possible permutations of the integers 1–8,
>>>> representing all the possible permutations of the numbered vertices of
>>> an
>>>> octagon.
>>>>
>>>> Now sum consecutive sets of three integers from each set of 8 integers
>>> with
>>>> wraparound, which will create sets of 8 sums, and store the sums in s3.
>>>> Count the sums & display the count:
>>>>
>>>> $s3=.+/"1[3,\"1|:n,2{.n
>>>>
>>>> 40320 8
>>>>
>>>> So s3 has 40,320 sets of 8 sums, where each set of 8 sums represents
>>> the 8
>>>> sums of 3 adjacent vertices on each unique octagon.
>>>>
>>>> Now mark all the octagons have all their 3 consecutive vertex sums
>>> greater
>>>> than 11, sum the marks, and list the sum:
>>>>
>>>> +/m=.*./"1[11<"1 s3
>>>>
>>>> 320
>>>>
>>>> So 320s of the 40,320 possible octagons have all their 3 consecutive
>>> vertex
>>>> sums greater than 11:
>>>>
>>>> Use the mark vector to extract & display the first & last few sets of
>>>> octagon vertices whose 3 consecutive vertex sums are greater than 11:
>>>>
>>>> {m#|:n
>>>>
>>>> │1 4 7 2 5 6 3 8│1 4 7 2 6 5 3 8│1 5 6 2 4 7 3 8│1 5 6 2 7 3 4 8│1 5 6
>>> 2 7
>>>> 4 3 8│1 5 6 2 8 3 4 7│1 5 6 3 7 2 4 8│1 5 6 3 8 2 4 7│1 5 6 4 2 7 3 8│1
>>> 5 6
>>>> 4 7 2 3 8│….
>>>>
>>>> …..│8 3 5 6 2 7 4 1│8 3 6 4 2 7 5 1│8 3 6 5 1 7 4 2│8 3 6 5 2 7 4 1│8 3
>>> 7 2
>>>> 4 6 5 1│8 3 7 4 2 6 5 1│8 4 2 7 3 5 6 1│8 4 2 7 3 6 5 1│8 4 3 7 2 5 6
>>> 1│8 4
>>>> 3 7 2 6 5 1│
>>>>
>>>> How many octagons have all their 3 consecutive vertex sums greater than
>>> 13?
>>>>
>>>> +/*./"1[13<"1 s3
>>>>
>>>> 0
>>>>
>>>> There are no octagons whose 3 consecutive vertex sums are all greater
>>> than
>>>> 13.
>>>>
>>>> How many octagons have all their 3 consecutive vertex sums greater than
>>> 10?
>>>>
>>>> +/*./"1[10<"1 s3
>>>>
>>>> 3456
>>>>
>>>> So 3,456 octagons of the 40,320 possible octagons have all their 3
>>>> consecutive vertex sums greater than 10.
>>>>
>>>> Skip Cave
>>>> Cave Consulting LLC
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>
>>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
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