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  • \$\begingroup\$ (∘.=⍨⍳) -> ∘.=⍨∘⍳ \$\endgroup\$ Commented Jan 8, 2018 at 17:30
  • \$\begingroup\$ @ngn I have this golf on stand by, I was waiting to see if I can golf some more bytes before I put this in because editing posts is really pain using mobile \$\endgroup\$ Commented Jan 8, 2018 at 18:15
  • \$\begingroup\$ I took the liberty to edit it for you. I too suspect there might be a better answer - it reminds me of but I can't quite figure out how it could work... \$\endgroup\$ Commented Jan 8, 2018 at 19:30
  • \$\begingroup\$ matrix division was fruitless but I found an interesting circular function: {÷¯1+4○⍵*.5}⍪⍳∘.=⍳ \$\endgroup\$ Commented Jan 9, 2018 at 20:41
  • \$\begingroup\$ @ngn thanks. I used a tacit version of your solution for the same count \$\endgroup\$ Commented Jan 10, 2018 at 13:03