Misunderstanding about ZA size and shape

Assume SVL = 512 bits, and the element type is float32.

Misunderstanding ❌: ZA has size $SVL×SVL$, so I would have $512*512 = 2^{18}$ bits, which means my ZA register can hold $2^{13}$ floats.

Correct ✅: The ZA matrix is square at the “byte” level, not at the “bit” level.

ZA is not a bitmap. It is a “rectangle” with a height of $SVL/8$ rows, and each row is $SVL/8$ bytes wide.

      <--------- Width = 512 bits (64 Bytes) --------->
    +--------------------------------------------------+
 R  | Byte 0 | Byte 1 | ...                  | Byte 63 |
 o  |--------+--------+----------------------+---------|
 w  | Byte 0 | Byte 1 | ...                  | Byte 63 |
 s  |--------+--------+----------------------+---------|
    | ...    | ...    | ...                  | ...     |
(64)|--------+--------+----------------------+---------|
    | Byte 0 | Byte 1 | ...                  | Byte 63 |
    +--------------------------------------------------+

So the actual size of the ZA matrix is:

$64*512 \text{ bits} = 2^{15} \text{ bits} = 2^{12} \text{ bytes}$

About tile size and shape

In SME: one tile is always a square of N x N elements.

The value of N depends on how many such elements fit in one Z register.

Therefore, the number of tiles changes with the element size.