fix: P2D

[axiles]

The previous implementation could give an incorrect result when the first node was not a root node.

Example:

P2D 0 2 2

correctly returns:

0 1 0 0 2 1

However

P2D 1 1

returns

1 0 0 1

Therefore, it says that no permutation is necessary to have the depth-first traversal.

The reason is that, in the matrix (-1+d)↑⍤0 1⊢⌽z, a zero can either mean the index of a node or an element used to fill the empty positions.

In the last example, (-1+d)↑⍤0 1⊢⌽z is:

1 0  <- path of the node is 1 0
1 0  <- path of the node is 1

Grade Up says that rows of this matrix are already sorted.

A quick fix is just to add one to the matrix z. Therefore, we have instead (-1+d)↑⍤0 1⊢⌽1+z) which is in our example:

2 1
2 0

And we have the expected result:

>> P2D 1 1

1 0     1 0

Notes:

• There are no issue when the first node is a root node.
• There is no issue when ⎕IO is equal to 1. Actually, the value we need to add to z is 1 - ⎕IO, if we wanted a solution that works for any value of ⎕IO.

Edit: Cosmetic

[arcfide]

Thanks for this! I'll have to think about whether I like this fix or would like to keep it the way it is or even try a different approach. At the very least, it's very good to have a note about this limitation in the current code. I'm keeping this open for a bit while I get some other stuff done, but I plan to come back to this and make sure it gets a proper seeing to.

[axiles]

Another possibility (even though probably less efficient since it involves nested arrays) is to split to matrix into lines:

P2D←{z←⍪⍳≢⍵ ⋄ d←⍵≠,z ⋄ _←{p⊣d+←⍵≠p←⍺[z,←⍵]}⍣≡⍨⍵ ⋄ d(⍋(-1+d)↑¨↓⌽z)}

If we go that route, we can have something even shorter (but less efficient) using ∪ instead of using d:

P2D←{z←⍪⍳≢⍵ ⋄ d←⍵≠,z ⋄ _←{p⊣d+←⍵≠p←⍺[z,←⍵]}⍣≡⍨⍵ ⋄ d(⍋∪¨↓⌽z)}