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
