What does (2#x)#1,x#0 do in q ?

In q (the programming language used with kdb+), the expression (2#x)#1,x#0 constructs a unit matrix (identity matrix) of a given size.

Let’s break it down step-by-step.
1. Understanding the components

    # is the take function when used in a binary context:
        a#b takes a elements from b (or repeats b if a is larger than the count of b).
    x is assumed to be an integer (the desired size of the identity matrix).

2. Inner part: 1,x#0

    x#0 creates a list of x zeros.
    Example: if x=3, then 3#0 is 0 0 0.

    1,x#0 prepends a 1 to that list of zeros.
    Example: 1,3#0 gives 1 0 0 0.

So 1,x#0 is a list of length x+1, starting with 1 followed by x zeros.
3. Outer part: (2#x)#

    2#x creates a list of two xs.
    Example: 2#3 is 3 3.

    So (3 3)#1 0 0 0 means: create a 3×3 matrix by taking elements from the list 1 0 0 0 repeatedly, row by row.

4. How the repetition works

The list 1 0 0 0 is:
First row: take first 3 elements → 1 0 0
Second row: continue from 4th element → 0 (list wraps around because we need 3 elements) → next elements: 1 0 (so second row: 0 1 0)
Wait—let’s check carefully because this is the key to the identity matrix.

Actually, let’s trace for x=3:

List: 1 0 0 0 (length 4)
We are taking 3×3=9 elements cyclically:

Row-major order:

    Row 1: elements 1–3: 1 0 0
    Row 2: elements 4, then wrap to 1, 2: 0 1 0
    Row 3: next element after Row 2’s last (position 3 in list) → increment: element 4 (0), then 1 (1), then 2 (0) → 0 1 0? Hmm, this doesn’t seem right. Let’s place carefully.

Better to check actual q output:
q

q) x:3
q) 1,x#0
1 0 0 0
q) (2#x)#1,x#0
1 0 0
0 1 0
0 0 1

Yes! It works. Let’s see the data fill pattern:

We have list L = 1 0 0 0 (length 4).

Reshape to 3×3 takes items in order:

Positions in L (0-based index for reasoning):
0: 1
1: 0
2: 0
3: 0
4 (wrap 0): 1
5 (wrap 1): 0
6 (wrap 2): 0
7 (wrap 3): 0
8 (wrap 0): 1

That doesn’t seem to align. Let’s do it step by step:

Index mapping for 3×3:

Element (i,j) gets index = 3*i + j in the flat result, taken from L cyclically.

Flat result length 9:
Take from L repeatedly:
1(0) 0(1) 0(2)
0(3) 1(4→0) 0(5→1)
0(6→2) 0(7→3) 1(8→0)

So indeed:
Row0: 1 0 0
Row1: 0 1 0
Row2: 0 0 1

Yes! The wrap-around causes the diagonal.
5. General case

For any x, (x,x)#1,x#0 creates an x by x identity matrix because the list is 1 followed by x zeros, so the 1 appears at positions 0, x+1, 2*(x+1), … but after wrapping modulo (x+1), that 1 appears exactly at diagonal positions (where row index = col index in terms of flat index mod (x+1) = 0).

Actually, easier: The pattern repeats every x+1 elements, and we are taking x elements per row, so each row shifts the starting point by one relative to the list, placing the 1 one step to the right each row → identity matrix.
6. Summary

(2#x)#1,x#0 creates an x-by-x identity matrix.

Example:
q

q) x:4
q) (2#x)#1,x#0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1