looks right

One 1 per element.

ohhhhhhhhhhhhhhh

⍥

well it's not 3 but =⍨≡¨ is fun using monadic ≡

×≢¨

oops no that doesn't work for length 0 things

Hint: This CMC is connected to the lesson learned from the "decrement first" CMC.

well there's ≡⍨¨ but that's not particularly interesting

Should I give it away? Or do you want more hints?

I don't mind it being given away, but if others want to try and get it that's fine

OK, here's another hint: It uses two functions, of which one is an arithmetic function.

Oh well, here goes…

3

2

1

!⍷⍨

anyone ready for nother one?

I'm going to bed soon, but sure, feel free to post it. You may want to start with **CMC:** so it is easier to spot.

@Adám using ⍷⍨! nice

      a
1 1 1 1
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
3 3 3 3
4 4 4 4
4 4 4 4
4 4 4 4

A A A A
1 1 1 1
1 1 1 1
1 1 1 1
B B B B
2 2 2 2
2 2 2 2
2 2 2 2
C C C C
3 3 3 3
3 3 3 3
3 3 3 3

@nathanrogers What is the expected result for those a and b?

@nathanrogers What is b here?

b is 4 4⍴4/⎕A

a in the base case is 4 4⍴4/⍳4

and you can increase the size of a for part 1 with {(⍵,4)⍴⍵/⍳4} where ⍵ is divisible by 4

all the 1's are "nested between rows A and B, all 2's nested between rows B and C... etc

Ah, so instead of a simple perfect shuffle, it is a distributed perfect shuffle.

Right

Think about ⎕XML arguments

@nathanrogers by elements you mean major cells?

Sure, too late to edit

@nathanrogers Will a always be the shorter one?

should have given better names... bleh, B is the groups, A is the nested values, so for part 1 A will always be of equal length for each row of B

Must work for any A where ⍵ is divisible by 4, and constant B

      B Solution A 4
A A A A
1 1 1 1
B B B B
2 2 2 2
C C C C
3 3 3 3
D D D D
4 4 4 4

So 1's are grouped with A, 2's are grouped with B, 3's with C, etc

      B Solution A 8
A A A A
1 1 1 1
1 1 1 1
B B B B
2 2 2 2
2 2 2 2
C C C C
3 3 3 3
3 3 3 3
D D D D
4 4 4 4
4 4 4 4

etc for all A (4×⍳∞) with ⎕IO←1

@nathanrogers Where did the letters come from?

B←4 4⍴4/⎕A

X( I'm not good at this

Should have typed it out first

Yes.

@nathanrogers Only matrices or any rank arrays?

only matrices

@nathanrogers I think {c[⍋⍋×(1+⍵÷⍥≢⍺)|⍳≢c←⍺⍪⍵;]} does it.

your ⎕IO?

yep, works for ⎕IO←0

⎕IO←0

@nathanrogers {c⌷⍨⊂⍋⍋×(1+⍵÷⍥≢⍺)|⍳≢c←⍺⍪⍵} for arbitrary rank arrays.

sol←{(⍺⍪⍵)[⍋n,((≢⍵)÷≢⍺)/(n←⍳≢⍺);]}

this is my solution for ⎕IO independent

Hold on, I have an idea for a neater solution.

@nathanrogers {,[⍳2]⍺⍪⍤1 2⊢4(⍵÷⍥≢⍺)4⍴⍵} any ⎕IO

whoah, I'm gonna have to look at these

Hold on, I have an idea for a neater solution.

:O

Nah, turned out ugly: {⊃⍪/(↓⍺)⍪¨⍵⊂[0]⍨⍵(⊣⍴1↑⍨÷)⍥≢⍺}

@nathanrogers Tacit version of my second solution, but with swapped arguments: ,[⍳2]⊢⍪⍤1 2⊣⍴⍨4,÷⍥≢,4⍨

wow neat

Costs one byte to swap args.

I'm sure you can modify my solution to handle that.

min you just modify the replicate of the rightmost ⍳

sol←{(⍺⍪⍵)[⍋n,((MODIFY THIS PART)/(n←⍳≢⍺);]}

is basically the solution

      xmlInput←(rowxml⍪cellsxml)[⍋(⍳≢rn),(2×rcc)/⍳≢rn;]         ⍝ multiply the row cell count by 2, this accounts for both cell tags and value tags to be passed to ⎕XML

This line in APL2XL was written by Morten, and I thought it was pretty dope. The 2× is because cells xml has 2 nested elements per row (values inside cells inside row), and you count up the cells in each row, you have how many pairs will exist inside that row