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
>> > ----------------------------------------------------------------------
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>> ----------------------------------------------------------------------
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>>
>
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