CuTe Layout Algebra

Introduction

CuTe layout algebra is extremely important for understanding and applying CUTLASS for accelerated computing. Despite the fact that CuTe has a documentation for its layout algebra, it cannot be understood completely without first understanding its mathematical foundations. I tried to create some proofs for the CuTe layout algebra on my own and realized that it was a huge amount of work. Gratefully, Jay Shah has created a paper “A Note on the Algebra of CuTe Layouts” that completes the CuTe layout algebra mathematical foundations that I wanted to create.

As my proofreading, I found Jay Shah’s paper mostly error-free, except for a few very minor oversights and typos. However, it does skip some details without which the paper is a little bit hard to understand. In this article, based on Jay Shah’s paper, I would like to provide more details to the proofs and explanations of the CuTe layout algebra. Most of the definitions and annotations will follow Jay Shah’s paper.

This article can be read as a complement to Jay Shah’s paper, but it’s also standalone for understanding the CuTe layout algebra.

Layout Algebra Preliminaries

Definition 2.1: Layout

A layout $L$ is a pair of positive integer tuples $\mathbf{S}$ and $\mathbf{D}$ of matching dimensions. We call $\mathbf{S}$ the shape and $\mathbf{D}$ the stride. We write $L = \mathbf{S}:\mathbf{D}$.

A flattened layout means that there is no internal parentheses in the shape and stride. For example, $L = (5, 2, 2):(16, 80, 4)$ is a flattened layout, whereas $L = (5, (2, 2)):(16, (80, 4))$ is not. Flattening a layout will not change the semantics and operations of the layout.

Definition 2.2: Layout Size, Length, and Mode

Let $\alpha \geq 0$ be an integer and $L = \mathbf{S}:\mathbf{D} = (M_0, M_1, \ldots, M_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ be a layout. Then:

Concatenation

Given two layouts $L = \mathbf{S}:\mathbf{D}$ and $L^{\prime} = \mathbf{S}^{\prime}:\mathbf{D}^{\prime}$, let $\mathbf{S}^{\prime\prime}$ and $\mathbf{D}^{\prime\prime}$ be the shape and stride tuples given by (the flattening of) $(\mathbf{S}, \mathbf{S}^{\prime})$ and $(\mathbf{D}, \mathbf{D}^{\prime})$ respectively. Then the concatenation of $L$ and $L^{\prime}$ is given by the layout

$$(L, L^{\prime}) = \mathbf{S}^{\prime\prime}:\mathbf{D}^{\prime\prime}$$

and we say that $(L, L^{\prime})$ is decomposed by $L$ and $L^{\prime}$.

Inductively, given layouts $L_0, L_1, \ldots, L_N$, we can then form the concatenation $(L_0, L_1, \ldots, L_N)$. Conversely, given $L$ a layout, $L$ is maximally decomposed by its modes.

Isomorphism

Let $\mathbf{S} = (M_0, M_1, \ldots, M_{\alpha})$ and $\mathbf{D} = (d_0, d_1, \ldots, d_{\alpha})$ be the respective shape and stride tuples of $L = \mathbf{S}:\mathbf{D}$. Let $M = M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha}$ be the size of $L$ and let $[0, M) \subset \mathbb{N}$ be the subset of the natural numbers given by ${0, 1, 2, \ldots, M - 1}$. Then we have an isomorphism

$$\begin{align}\iota: [0, M) \cong [0, M_0) \times [0, M_1) \times \ldots \times [0, M_{\alpha}) \\\end{align}$$

Given any $x \in [0, M)$, the isomorphism $\iota$ maps $x$ to the tuple

$$\begin{align}x \mapsto \left(x \mod M_0, \left\lfloor \frac{x}{M_0} \right\rfloor \mod M_1, \ldots, \left\lfloor \frac{x}{M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \mod M_{\alpha}\right)\end{align}$$

The isomorphism mapping is bijective. In our case, given any tuple $(x_0, x_1, \ldots, x_{\alpha}) \in [0, M_0) \times [0, M_1) \times \ldots \times [0, M_{\alpha})$, the isomorphism inverse maps the tuple to the integer

$$\begin{align}\left(x_0, x_1, \ldots, x_{\alpha}\right) \mapsto x_0 + x_1 \cdot M_0 + x_2 \cdot M_0 \cdot M_1 + \ldots + x_{\alpha} \cdot M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}\end{align}$$

It’s straightforward to verify the above isomorphism mapping is valid and proof the above isomorphism mapping is bijective (by contradiction).

One could imagine the isomorphism as a mapping between a one-dimensional coordinate and a multi-dimensional coordinate.

Definition 2.3: Layout Function

Given a layout $L$, its layout function is the function $f_L: [0, M) \to \mathbb{N}$ is defined to be the composite

$$\begin{align}[0, M) \cong [0, M_0) \times [0, M_1) \times \ldots \times [0, M_{\alpha}) \subset \mathbb{N}^{\times (\alpha + 1)} \xrightarrow{\cdot d_0, \cdot d_1, \ldots, \cdot d_{\alpha}} \mathbb{N}^{\times (\alpha + 1)} \xrightarrow{+} \mathbb{N}\end{align}$$

In other words, $f_L$ is the composition of the multilinear function

$$\begin{align}[0, M_0) \times [0, M_1) \times \ldots \times [0, M_{\alpha}) \to \mathbb{N} \\(x_0, x_1, \ldots, x_{\alpha}) \mapsto x_0 \cdot d_0 + x_1 \cdot d_1 + \ldots + x_{\alpha} \cdot d_{\alpha}\end{align}$$

determined by the stride, with the isomorphism $\iota$, determined by the shape.

Computing the value of a layout function $f_L$ at a point $x \in [0, M)$ can be decomposed into computing the sum of the values of the layout function at multiple points. This is sometimes useful for computing the value of the layout function at a point handily.

Given a layout $L = (M_0, M_1, \ldots, M_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ and $x \in [0, M)$,

$$\begin{align}x \mapsto \left(x_0, x_1, \ldots, x_{\alpha}\right) \mapsto x_0 \cdot d_0 + x_1 \cdot d_1 + \ldots + x_{\alpha} \cdot d_{\alpha}\end{align}$$

We also have

$$\begin{align}x^{\prime}_{0} &\mapsto \left(x_0, 0, 0, \ldots, 0\right) \mapsto x_0 \cdot d_0 \\x^{\prime}_{1} &\mapsto \left(0, x_1, 0, \ldots, 0\right) \mapsto x_1 \cdot d_1 \\&\vdots \\x^{\prime}_{\alpha} &\mapsto \left(0, 0, 0, \ldots, x_{\alpha}\right) \mapsto x_{\alpha} \cdot d_{\alpha}\end{align}$$

Therefore, we have

$$\begin{align}f_L(x) = f_L(x^{\prime}_{0}) + f_L(x^{\prime}_{1}) + \ldots + f_L(x^{\prime}_{\alpha})\end{align}$$

where

$$\begin{align}x^{\prime}_{0} &= x \mod M_0 \\x^{\prime}_{1} &= \left\lfloor \frac{x}{M_0} \right\rfloor \mod M_1 \cdot M_0 \\&\vdots \\x^{\prime}_{\alpha} &= \left\lfloor \frac{x}{M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \mod M_{\alpha} \cdot M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}\end{align}$$

For example, given a layout $L = (3, 2):(2, 3)$ and $x = 5$, we have

$$\begin{align}f_L(5) &= f_L(5 \mod 3) + f_L(\left\lfloor \frac{5}{3} \right\rfloor \mod 2 \cdot 3) \\&= f_L(2) + f_L(3) \\&= 2 \cdot 2 + \left\lfloor \frac{3}{3} \right\rfloor \cdot 3 \\&= 4 + 3 \\&= 7\end{align}$$

Extension of Layout Function

Based on the definition of layout function, the extension of the layout function $f_L$ is the function, $\widehat{f}_L: \mathbb{N} \to \mathbb{N}$, defined by replacing $M_{\alpha}$ with $\infty$ in the definition of $f_L$, i.e., the composite

$$\begin{align}\mathbb{N} \cong [0, M_0) \times [0, M_1) \times \ldots \times [0, M_{\alpha - 1}) \times \mathbb{N} \subset \mathbb{N}^{\times (\alpha + 1)} \xrightarrow{\cdot d_0, \cdot d_1, \ldots, \cdot d_{\alpha}} \mathbb{N}^{\times (\alpha + 1)} \xrightarrow{+} \mathbb{N}\end{align}$$

where the extension of the isomorphism $\iota$, $\widehat{\iota}$, is given by

$$\begin{align}x \mapsto \left(x \mod M_0, \left\lfloor \frac{x}{M_0} \right\rfloor \mod M_1, \ldots, \left\lfloor \frac{x}{M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 2}} \right\rfloor \mod M_{\alpha - 1}, \left\lfloor \frac{x}{M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor\right)\end{align}$$

The extension of the isomorphism mapping is also bijective. The inverse mapping of the extension of the isomorphism is also given by

$$\begin{align}\left(x_0, x_1, \ldots, x_{\alpha - 1}, x_{\alpha}\right) \mapsto x_0 + x_1 \cdot M_0 + x_2 \cdot M_0 \cdot M_1 + \ldots  + x_{\alpha} \cdot M_0 \cdot M_1 \cdot \ldots \cdot M_{\alpha - 1}\end{align}$$

One could imagine the extension of the isomorphism defines the last dimension of the shape to be a “batch” dimension and the batch size can be infinite.

Complementation

Definition 2.4: Sorted Layout

Let $A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ be a layout. We say that $A$ is sorted if $d_0 \leq d_1 \leq \ldots \leq d_{\alpha}$ and for every $i < j$, if $d_i = d_j$, then $N_i \leq N_j$.

Note that sorting a layout, or more generally, changing the order of modes of a layout, will change the semantics and operations of the layout.

For example, suppose we have a layout $A = (2, 4):(4, 1)$ and a layout $B = (4, 2):(1, 4)$. We could see that $B$ is the sorted version of $A$. We could compute the layout function of $A$ and $B$ as follows using lookup tables:

$$\begin{align}f_A(0) &= f_A(0, 0) = 0 \cdot 4 + 0 \cdot 1 = 0 \\f_A(1) &= f_A(1, 0) = 1 \cdot 4 + 0 \cdot 1 = 4 \\f_A(2) &= f_A(0, 1) = 0 \cdot 4 + 1 \cdot 1 = 1 \\f_A(3) &= f_A(1, 1) = 1 \cdot 4 + 1 \cdot 1 = 5 \\f_A(4) &= f_A(0, 2) = 0 \cdot 4 + 2 \cdot 1 = 2 \\f_A(5) &= f_A(1, 2) = 1 \cdot 4 + 2 \cdot 1 = 6 \\f_A(6) &= f_A(0, 3) = 0 \cdot 4 + 3 \cdot 1 = 3 \\f_A(7) &= f_A(1, 3) = 1 \cdot 4 + 3 \cdot 1 = 7\end{align}$$

$$\begin{align}f_B(0) &= f_B(0, 0) = 0 \cdot 1 + 0 \cdot 4 = 0 \\f_B(1) &= f_B(1, 0) = 1 \cdot 1 + 0 \cdot 4 = 1 \\f_B(2) &= f_B(2, 0) = 2 \cdot 1 + 0 \cdot 4 = 2 \\f_B(3) &= f_B(3, 0) = 3 \cdot 1 + 0 \cdot 4 = 3 \\f_B(4) &= f_B(0, 1) = 0 \cdot 1 + 1 \cdot 4 = 4 \\f_B(5) &= f_B(1, 1) = 1 \cdot 1 + 1 \cdot 4 = 5 \\f_B(6) &= f_B(2, 1) = 2 \cdot 1 + 1 \cdot 4 = 6 \\f_B(7) &= f_B(3, 1) = 3 \cdot 1 + 1 \cdot 4 = 7\end{align}$$

We could see that the layout $B$ is typically referred as the column-major layout, and the layout $A$ is typically referred as the row-major layout. They are completely different layouts.

More generally, the sorted layout is a just like the “generalization” of the column-major layout.

Definition 2.5: Admission for Complementation

Let $A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ be a layout and $M$ be a positive integer. If $A$ is not sorted then replace $A$ with its sorted version. We say that the pair $\{A, M\}$ is admissible for complementation (or simply admissible) if:

That $\{A, M\}$ is admissible for complementation also implies:

Definition 2.6: Complementation

Let $A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ be a layout and $M$ be a positive integer. If $\{A, M\}$ is admissible for complementation, then if $A$ is not sorted, replace $A$ with its sorted version. The complement of $\{A, M\}$ is defined to be the layout

$$\begin{align}\text{complement}(A, M) = \left(d_0, \frac{d_1}{N_0 d_0}, \frac{d_2}{N_1 d_1}, \ldots, \frac{d_{\alpha}}{N_{\alpha - 1} d_{\alpha - 1}}, \frac{M}{N_{\alpha} d_{\alpha}} \right): \left(1, N_0 d_0, N_1 d_1, \ldots, N_{\alpha} d_{\alpha}\right)\end{align}$$

Note that the size of the complement of $\{A, M\}$, $\text{size}(\text{complement}(A, M))$, is $\frac{M}{\text{size}(A)} = \frac{M}{N_0 \cdot N_1 \cdot \ldots \cdot N_{\alpha}}$.

By definition, the complement of $\{A, M\}$ is insensitive to the order of the modes of $A$, since it will always be sorted before complementation.

The complement of $\{A, M\}$ is strictly increasing. This might not be very obvious, so we will show a proof.

Proof

Suppose $B = \text{complement}(A, M)$, to show that the layout function $f_{B}$, whose domain is a set of natural numbers, is strictly increasing, we need to show that for every two adjacent natural numbers $x$ and $x + 1$, $0 \leq x < x + 1 < \text{size}(B)$, we have $f_{B}(x) < f_{B}(x + 1)$.

Because of the isomorphism, suppose the mapping of $x$ is as follows:

$$\begin{align}x &\mapsto \left(x_0, x_1, \ldots, x_{\alpha}, x_{\alpha + 1}\right) \\\end{align}$$

By definition of the layout function $f_{B}$, we have

$$\begin{align}f_{B}(x) &= x_0 + x_1 \cdot N_0 d_0 + x_2 \cdot N_1 d_1 + \ldots + x_{\alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{\alpha + 1} \cdot N_{\alpha} d_{\alpha} \\\end{align}$$

The mapping of $x + 1$ can have many different cases.

In the simplest case,

$$\begin{align}x + 1 &\mapsto \left(x_0 + 1, x_1, \ldots, x_{\alpha}, x_{\alpha + 1}\right) \\\end{align}$$

Then we have

$$\begin{align}f_{B}(x + 1) &= x_0 + 1 + x_1 \cdot N_0 d_0 + x_2 \cdot N_1 d_1 + \ldots + x_{\alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{\alpha + 1} \cdot N_{\alpha} d_{\alpha} \\&= f_{B}(x) + 1 \\&> f_{B}(x)\end{align}$$

In a more complicated case, where $x_0 = d_0 - 1$ and $x_1 < \frac{d_1}{N_0 d_0} - 1$, we have

$$\begin{align}x + 1 &\mapsto \left(0, x_1 + 1, \ldots, x_{\alpha}, x_{\alpha + 1}\right) \\\end{align}$$

Then we have

$$\begin{align}f_{B}(x + 1) &= 0 + (x_1 + 1) \cdot N_0 d_0 + x_2 \cdot N_1 d_1 + \ldots + x_{\alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{\alpha + 1} \cdot N_{\alpha} d_{\alpha} \\&= f_{B}(x) - x_0 + N_0 d_0 \\&= f_{B}(x) - (d_0 - 1) + N_0 d_0 \\&= f_{B}(x) + 1 + (N_0 - 1) d_0 \\&> f_{B}(x)\end{align}$$

Because $N_0 \geq 1$, we have $(N_0 - 1) d_0 \geq 0$, so we have

$$\begin{align}f_{B}(x + 1) &> f_{B}(x)\end{align}$$

In general, when $x_0 = d_0 - 1$, for some $k \in [1, \alpha - 1]$, $x_i = \frac{d_i}{N_{i - 1} d_{i - 1}} - 1$ for every $i \in [1, k]$, $x_{k + 1} < \frac{d_{k + 1}}{N_k d_k} - 1$, we have

$$\begin{align}x + 1 &\mapsto \left(0, 0, \ldots, 0, x_{k + 1} + 1, \ldots, x_{\alpha}, x_{\alpha + 1}\right) \\\end{align}$$

Then we have

$$\begin{align}f_{B}(x + 1) &= 0 + 0 \cdot N_0 d_0 + \ldots + 0 \cdot N_{k - 1} d_{k - 1} + (x_{k + 1} + 1) \cdot N_k d_k + \ldots + x_{\alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{\alpha + 1} \cdot N_{\alpha} d_{\alpha} \\&= f_{B}(x) - x_0 - \left(\sum_{i = 1}^{k} x_i \cdot N_{i - 1} d_{i - 1}\right) + N_k d_k \\&= f_{B}(x) - (d_0 - 1) - \left(\sum_{i = 1}^{k} \left(\frac{d_i}{N_{i - 1} d_{i - 1}} - 1\right) \cdot N_{i - 1} d_{i - 1}\right) + N_k d_k \\&= f_{B}(x) - (d_0 - 1) - \left(\sum_{i = 1}^{k} \left(d_i - N_{i - 1} d_{i - 1}\right)\right) + N_k d_k \\&= f_{B}(x) - (d_0 - 1) + \sum_{i = 1}^{k} N_{i - 1} d_{i - 1} - \sum_{i = 1}^{k} d_i + N_k d_k \\&=  f_{B}(x) + \sum_{i = 0}^{k} \left(N_{i} - 1\right) d_{i} + 1 \\\end{align}$$

Because $N_{i} \geq 1$ for every $i$, we have $\left(N_{i} - 1\right) d_{i} \geq 0$ for every $i$, so we have

$$\begin{align}f_{B}(x + 1) &> f_{B}(x)\end{align}$$

This concludes the proof. $\square$

Similarly, we could also prove that the extension of the complement of $\{A, M\}$ is strictly increasing.

Proposition 2.7

Let $\{A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha}), M\}$ be admissible for complementation and $B = \text{complement}(A, M)$. Let $C = (A, B)$ be the concatenated layout. Then the size of $C$ is $M$ and $f_C: [0, M) \to \mathbb{N}$ restricts to a bijection $[0, M) \cong [0, M)$.

Proof

Because $\text{size}(A) = \prod_{i = 0}^{\alpha} N_i$ and $\text{size}(B) = \frac{M}{\prod_{i = 0}^{\alpha} N_i}$, we have $\text{size}(C) = \text{size}(A) \cdot \text{size}(B) = M$. Thus the domain of $f_C$ is $[0, M)$.

Note that the image of $f_C$ is the same as that of $f_{C^{\prime}}$ for any permutation $C^{\prime}$ of $C$.

To see this, suppose we have the following layout $C$ and its permutation $C^{\prime}$ in which only one pair of the modes is permuted.

$$\begin{align}C &=  \left(N_0, N_1, \ldots, N_{i}, \ldots, N_{j}, \ldots, N_{\alpha} \right): \left(d_0, d_1, \ldots, d_{i}, \ldots, d_{j}, \ldots, d_{\alpha}\right) \\C^{\prime} &=  \left(N_0, N_1, \ldots, N_{j}, \ldots, N_{i}, \ldots, N_{\alpha} \right): \left(d_0, d_1, \ldots, d_{j}, \ldots, d_{i}, \ldots, d_{\alpha}\right)\end{align}$$

The domains of $f_C$ and $f_C^{\prime}$ are both $[0, M)$. For any $x_C \in [0, M)$, we have

$$\begin{align}x_C &\mapsto \left(x_0, x_1, \ldots, x_{i}, \ldots, x_{j}, \ldots, x_{\alpha}\right) \\x_{C^{\prime}} &\mapsto \left(x_0, x_1, \ldots, x_{j}, \ldots, x_{i}, \ldots, x_{\alpha}\right)\end{align}$$

and $x_C$ and $x_{C^{\prime}}$ are bijective.

Because by definition, $f_C(x_C) = f_{C^{\prime}}(x_{C^{\prime}})$, the image of $f_C$ is the same as that of $f_{C^{\prime}}$.

For any permutation $C^{\prime}$ of $C$, it can be obtained by permuting one pair of the modes of $C$ at a time and each time the image of $f_C$ is the same as that of $f_{C^{\prime}}$. Therefore, the image of $f_C$ is the same as that of $f_{C^{\prime}}$ for any permutation $C^{\prime}$ of $C$.

When computing the image of $f_C$ we may sort $C$. Without loss of generality, suppose $A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$ is already sorted. After sorting $C$, the sorted $C^{\prime}$ could only be as follows:

$$\begin{align}C^{\prime} &= \left(d_0, N_0, \frac{d_1}{N_0 d_0}, N_1, \frac{d_2}{N_1 d_1}, N_2, \ldots, \frac{d_{\alpha}}{N_{\alpha - 1} d_{\alpha - 1}}, N_{\alpha}, \frac{M}{N_{\alpha} d_{\alpha}} \right): \left(1, d_0, N_0 d_0, d_1, N_1 d_1, d_2, \ldots, N_{\alpha - 1} d_{\alpha - 1}, d_{\alpha}, N_{\alpha} d_{\alpha}\right)\end{align}$$

Because $d_i \leq N_i d_i$ and $N_i d_i \leq d_{i + 1}$ for every $i$, when $N_i = 1$, $N_i \leq \frac{d_{i + 1}}{N_i d_i}$, when $N_i d_i = d_{i + 1}$, $\frac{d_{i + 1}}{N_i d_i} \leq N_{i + 1}$, thus $C^{\prime}$ is sorted and any permutation of $C^{\prime}$ will make it not sorted.

Then we may rewrite

$$\begin{align}C^{\prime} &= \left(r_0, r_1, r_2, \ldots, r_{\beta} \right): \left(1, r_0, r_0 r_1, \ldots, r_0 r_1 \ldots r_{\beta - 1}\right)\end{align}$$

where $\beta = 2 \alpha + 1$ and the maximum value that $f_{C^{\prime}}$ attains is computed as follows:

$$\begin{align}f_{C^{\prime}}(M - 1) &= f_{C^{\prime}}(r_0 - 1, r_1 - 1, r_2 - 1, \ldots, r_{\beta - 1} - 1, r_{\beta} - 1) \\&= (r_0 - 1) + (r_1 - 1) \cdot r_0 + (r_2 - 1) \cdot r_0 r_1 + \ldots + (r_{\beta - 1} - 1) \cdot r_0 r_1 \ldots r_{\beta - 2} + (r_{\beta} - 1) \cdot r_0 r_1 \ldots r_{\beta - 1} \\&= r_0 - 1 + r_0 r_1  - r_0 + r_0 r_1 r_2 - r_0 r_1 + \ldots + r_0 r_1 \ldots r_{\beta - 1} - r_0 r_1 \ldots r_{\beta - 2} + r_0 r_1 \ldots r_{\beta} - r_0 r_1 \ldots r_{\beta - 1} \\&= r_0 r_1 \ldots r_{\beta} - 1 \\&= M - 1\end{align}$$

Then in this case, to establish the bijectivity assertion, it’s sufficient to just show $f_{C^{\prime}}(x)$ is injective, i.e., for any $x, y \in [0, M)$, if $f_{C^{\prime}}(x) = f_{C^{\prime}}(y)$, then $x = y$.

Suppose the isomorphism mapping of $x$ and $y$ are as follows:

$$\begin{align}x &\mapsto \left(x_0, x_1, \ldots, x_{\beta}\right) \\y &\mapsto \left(y_0, y_1, \ldots, y_{\beta}\right)\end{align}$$

Because $f_{C^{\prime}}(x) = f_{C^{\prime}}(y)$, we have

$$\begin{align}x_0 + x_1 \cdot r_0 + x_2 \cdot r_0 r_1 + \ldots + x_{\beta} \cdot r_0 r_1 \ldots r_{\beta - 1} = y_0 + y_1 \cdot r_0 + y_2 \cdot r_0 r_1 + \ldots + y_{\beta} \cdot r_0 r_1 \ldots r_{\beta - 1}\end{align}$$

We will use strong induction to show that $x_i = y_i$ for every $i \in [0, \beta]$.

Because $f_{C^{\prime}}(x) \mod r_0 =  f_{C^{\prime}}(y) \mod r_0$, we have $x_0 = y_0$.

Now suppose by the strong induction that given $i \in (0, \beta]$, for all $j < i$, we have $x_j = y_j$. we have

$$\begin{align}x_i \cdot r_0 r_1 \ldots r_{i - 1} + x_{i + 1} \cdot r_0 r_1 \ldots r_{i} + \ldots + x_{\beta} \cdot r_0 r_1 \ldots r_{\beta - 1} = y_i \cdot r_0 r_1 \ldots r_{i - 1} + y_{i + 1} \cdot r_0 r_1 \ldots r_{i} + \ldots + y_{\beta} \cdot r_0 r_1 \ldots r_{\beta - 1}\end{align}$$

Because $x_i \in [0, r_i)$ and $y_i \in [0, r_i)$, taking this equation modulo $r_0 r_1 \ldots r_{i}$ and dividing by $r_0 r_1 \ldots r_{i - 1}$, we have $x_i = y_i$.

Because $(x_0, x_1, \ldots, x_{\beta}) = (y_0, y_1, \ldots, y_{\beta})$, and the isomorphism mapping is bijective, we have $x = y$.

Therefore $f_{C^{\prime}}: [0, M) \to \mathbb{N}$ restricts to a bijection $[0, M) \cong [0, M)$. So does $f_C$.

This concludes the proof. $\square$

Corollary 2.8 Complementation Disjointness

The Corollary 2.8 explains what it means of taking a complement of a layout.

In the setting of Proposition 2.7, let $I = [0, \text{size}(A)) = [0, N_0 N_1 \ldots N_{\alpha})$ be the domain of $f_A$. Then

$$\begin{align}f_A(I) \cap \widehat{f}_B(I) = \{0\}\end{align}$$

In other words, $\widehat{f}_A$ and $\widehat{f}_B$ have disjoint image when restricted to the domain of $f_A$, apart from 0.

Note that in the corollary, $f_A$ and $\widehat{f}_A$ are actually interchangeable, because the function domain is restricted to the domain of $f_A$.

Proof

Let $J = [0, \text{size}(B)) = [0, \frac{M}{N_0 N_1 \ldots N_{\alpha}})$ be the domain of $f_B$. Then by Proposition 2.7, we have

$$\begin{align}f_A(I) \cap f_B(J) = \{0\}\end{align}$$

To understand this, for any $x_A \in I$ and any $x_B \in J$, because of the isomorphism, we have

$$\begin{align}x_A &\mapsto \left(x_{A, 0}, x_{A, 1}, \ldots, x_{A, \alpha} \right) \\x_B &\mapsto \left(x_{B, 0}, x_{B, 1}, \ldots, x_{B, \alpha}, x_{B, \alpha + 1} \right)\end{align}$$

Then we have

$$\begin{align}f_A(x_A) &= x_{A, 0} + x_{A, 1} \cdot N_0 + x_{A, 2} \cdot N_0 N_1 + \ldots + x_{A, \alpha} \cdot N_0 N_1 \ldots N_{\alpha - 1} \\f_B(x_B) &= x_{B, 0} + x_{B, 1} \cdot N_0 d_0 + x_{B, 2} \cdot N_1 d_1 + \ldots + x_{B, \alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{B, \alpha + 1} \cdot N_{\alpha} d_{\alpha}\end{align}$$

We orchestrate new coordinates for layout $C$ as follows:

$$\begin{align}x^{\prime}_A &\mapsto \left(0, x_{A, 0}, 0, x_{A, 1}, 0, x_{A, 2}, \ldots, 0, x_{A, \alpha}, 0 \right) \\x^{\prime}_B &\mapsto \left(x_{B, 0}, 0, x_{B, 1}, 0, x_{B, 2}, \ldots, x_{B, \alpha}, 0, x_{B, \alpha + 1} \right)\end{align}$$

Then we have

$$\begin{align}f_C(x^{\prime}_A) &= x_{A, 0} + x_{A, 1} \cdot N_0 + x_{A, 2} \cdot N_0 N_1 + \ldots + x_{A, \alpha} \cdot N_0 N_1 \ldots N_{\alpha - 1} \\&= f_A(x_A) \\f_C(x^{\prime}_B) &= x_{B, 0} + x_{B, 1} \cdot N_0 d_0 + x_{B, 2} \cdot N_1 d_1 + \ldots + x_{B, \alpha} \cdot N_{\alpha - 1} d_{\alpha - 1} + x_{B, \alpha + 1} \cdot N_{\alpha} d_{\alpha} \\&= f_B(x_B)\end{align}$$

By the Proposition 2.7, we have $f_C: [0, M) \to \mathbb{N}$ restricts to a bijection $[0, M) \cong [0, M)$. If $x^{\prime}_A \neq x^{\prime}_B$, then $f_C(x^{\prime}_A) \neq f_C(x^{\prime}_B)$, and $f_A(x_A) \neq f_B(x_B)$.

Obviously, other than $(0, 0, \ldots, 0)$, for any values of $x_{A, 0}, x_{A, 1}, \ldots, x_{A, \alpha}$ and $x_{B, 0}, x_{B, 1}, \ldots, x_{B, \alpha}, x_{B, \alpha + 1}$, $\left(0, x_{A, 0}, 0, x_{A, 1}, 0, x_{A, 2}, \ldots, 0, x_{A, \alpha}, 0 \right) \neq \left(x_{B, 0}, 0, x_{B, 1}, 0, x_{B, 2}, \ldots, x_{B, \alpha}, 0, x_{B, \alpha + 1} \right)$, $x^{\prime}_A \neq x^{\prime}_B$, $f_C(x^{\prime}_A) \neq f_C(x^{\prime}_B)$, and $f_A(x_A) \neq f_B(x_B)$.

This means, for any $x \in I$ that $x \neq 0$, there is no $y \in J$ such that $f_A(x) = f_B(y)$.

When $x = 0$, we have $f_A(x) = f_B(x) = 0$. Thus we could claim that

$$\begin{align}f_A(I) \cap f_B(J) = \{0\}\end{align}$$

In the Definition 2.6: Complementation, we have shown that the complement of $\{A, M\}$, $f_B$, as well as its extension $\widehat{f}_B$, are strictly increasing.

In addition, by the extension of the isomorphism, we have

$$\begin{align}\text{size}(B) \mapsto \left(0, 0, \ldots, 0, \frac{M}{N_{\alpha} d_{\alpha}} \right) \\\end{align}$$

Then we have

$$\begin{align}\widehat{f}_{B}(\text{size}(B)) &= 0 + 0 \cdot 1 + 0 \cdot N_0 d_0 + \ldots + 0 \cdot N_{\alpha - 1} d_{\alpha - 1} + \frac{M}{N_{\alpha} d_{\alpha}} \cdot N_{\alpha} d_{\alpha} \\&= M\end{align}$$

The largest value attained by $f_A$ is at $N_0 N_1 \ldots N_{\alpha} - 1$, and $f_A(N_0 N_1 \ldots N_{\alpha} - 1) = (N_0 - 1) d_0 + (N_1 - 1) d_1 + \ldots + (N_{\alpha} - 1) d_{\alpha}$.

Because $(N_0 - 1) d_0 < N_0 d_0$ and $N_i d_i \leq d_{i + 1}$ for every $i \in [0, \alpha - 1]$, $N_{\alpha} d_{\alpha} \leq M$, we have

$$\begin{align}f_A(N_0 N_1 \ldots N_{\alpha} - 1) &= (N_0 - 1) d_0 + (N_1 - 1) d_1 + \ldots + (N_{\alpha} - 1) d_{\alpha} \\&< N_0 d_0 + N_1 d_1 - d_1 + N_2 d_2 - d_2 + \ldots + N_{\alpha} d_{\alpha} - d_{\alpha} \\&\leq d_1 + N_1 d_1 - d_1 + N_2 d_2 - d_2 + \ldots + N_{\alpha} d_{\alpha} - d_{\alpha} \\&= N_1 d_1 + N_2 d_2 - d_2 + \ldots + N_{\alpha} d_{\alpha} - d_{\alpha} \\&\leq d_2 + N_2 d_2 - d_2 + \ldots + N_{\alpha} d_{\alpha} - d_{\alpha} \\&\vdots \\&\leq d_{\alpha} + N_{\alpha} d_{\alpha} - d_{\alpha} \\&= N_{\alpha} d_{\alpha} \\&\leq M\end{align}$$

Thus $f_A(N_0 N_1 \ldots N_{\alpha} - 1) < \widehat{f}_{B}(\text{size}(B))$.

In the case of $I \cap J = I$, i.e., $\text{size}(A) \leq \text{size}(B)$. Then we have

$$\begin{align}f_A(I) \cap f_B(I) = \{0\}\end{align}$$

Because in this case, $f_B(I) = \widehat{f}_B(I)$, we have

$$\begin{align}f_A(I) \cap \widehat{f}_B(I) = \{0\}\end{align}$$

In the other case of $I \cap J = J$, i.e., $\text{size}(A) \geq \text{size}(B)$. Because the largest value attained by $f_A$ is $f_A(N_0 N_1 \ldots N_{\alpha} - 1)$, and $f_A(N_0 N_1 \ldots N_{\alpha} - 1) < \widehat{f}_{B}(\text{size}(B))$, for any $x \in I/J$, we have $f_A(x) < \widehat{f}_{B}(\text{size}(B))$.

Thus,

$$\begin{align}f_A(I) \cap \widehat{f}_B(I/J) = {\emptyset}\end{align}$$

Therefore,

$$\begin{align}f_A(I) \cap \widehat{f}_B(I) &= f_A(I) \cap \left(\widehat{f}_B(I) \cup \widehat{f}_B(I/J)\right) \\&= f_A(I) \cap \left(f_B(I) \cup \widehat{f}_B(I/J)\right) \\&= \left(f_A(I) \cap f_B(I)\right) \cup \left(f_A(I) \cap \widehat{f}_B(I/J)\right) \\&= \{0\} \cup {\emptyset} \\&= \{0\}\end{align}$$

Taken together, we have

$$\begin{align}f_A(I) \cap \widehat{f}_B(I) = \{0\}\end{align}$$

This concludes the proof. $\square$

A short note on the original proof of Corollary 2.8 in the paper is that Jay Shah claimed $f_A(I \cap J) \cap f_B(I \cap J) = \{0\}$, which is insufficient to show the proof. The sufficient statement should be $f_A(I) \cap f_B(J) = \{0\}$.

Remark 2.9 Complementation Disjointness, Ordering, and Boundedness

The complement $B$ of a layout $A$ with respect to an integer $M$ should satisfy three properties:

The property 1 and 2 have been proved in the Corollary 2.8 and Definition 2.6. We will show a proof of the property 3.

Proof

By Definition 2.6, we have $\text{size}(B) = \frac{M}{\text{size}(A)}$.

Because cosize is insensitive to the ordering of the layout, without loss of generality, we sorted $A$ so that $A = (N_0, N_1, \ldots, N_{\alpha}):(d_0, d_1, \ldots, d_{\alpha})$.

By the definition of cosize, we have

$$\begin{align}\text{cosize}(B) &= f_B(\text{size}(B) - 1) + 1 \\&= f_B\left(d_0 - 1, \frac{d_1}{N_0 d_0} - 1, \ldots, \frac{d_{\alpha}}{N_{\alpha - 1} d_{\alpha - 1}} - 1, \frac{M}{N_{\alpha} d_{\alpha}} - 1\right) + 1 \\&= (d_0 - 1) + \left(\frac{d_1}{N_0 d_0} - 1\right) \cdot N_0 d_0 + \ldots + \left(\frac{d_{\alpha}}{N_{\alpha - 1} d_{\alpha - 1}} - 1\right) \cdot N_{\alpha - 1} d_{\alpha - 1} + \left(\frac{M}{N_{\alpha} d_{\alpha}} - 1\right) \cdot N_{\alpha} d_{\alpha} + 1 \\&= d_0 + d_1 + \ldots + d_{\alpha} - N_0 d_0 - N_1 d_1 - \ldots - N_{\alpha} d_{\alpha} + M \\&= M - \left(\left(N_0 - 1\right) d_0 + \left(N_1 - 1\right) d_1 + \ldots + \left(N_{\alpha} - 1\right) d_{\alpha}\right) \\&= M - f_A(\text{size}(A) - 1) \\&= M - \left( \text{cosize}(A) - 1 \right) \\\end{align}$$

To obtain the inequality $\text{cosize}(B) \leq \left\lfloor\frac{M}{\text{cosize}(A)}\right\rfloor \cdot \text{cosize}(A)$, we divide the above equation by $\text{cosize}(A)$.

$$\begin{align}\frac{\text{cosize}(B)}{\text{cosize}(A)} &= \frac{M - \left( \text{cosize}(A) - 1 \right)}{\text{cosize}(A)} \\&= \frac{M}{\text{cosize}(A)} - 1 + \frac{1}{\text{cosize}(A)} \\\end{align}$$

and we have to show that

$$\begin{align}\frac{M}{\text{cosize}(A)} - 1 + \frac{1}{\text{cosize}(A)} \leq \left\lfloor\frac{M}{\text{cosize}(A)}\right\rfloor\end{align}$$

In fact, for any $a, b \in \mathbb{N}$ and $a \geq 1$, we have

$$\begin{align}\frac{b}{a} - 1 + \frac{1}{a} \leq \left\lfloor\frac{b}{a}\right\rfloor\end{align}$$

To see this, suppose $\frac{b}{a} = \left\lfloor\frac{b}{a}\right\rfloor + c$, where $c = \frac{k}{a}$ and $k$ is an integer such that $0 \leq k < a$. Then we have

$\frac{1}{a} \leq c < 1$ and $1 \leq ac < a$. Then we want to show that

$$\begin{align}\frac{b}{a} - 1 + \frac{1}{a} \leq \frac{b}{a} - c  \\\end{align}$$

$$\begin{align}-a + 1 \leq -ac \\\end{align}$$

$$\begin{align}a - ac \geq 1 \\\end{align}$$

$$\begin{align}a - k \geq 1 \\\end{align}$$

Because $a$ and $k$ are both integers and $0 \leq k < a$, we have $a - k \geq 1$. Thus the inequality holds.

This concludes the proof. $\square$

Coalescence

Coalescence simplifies the layout and does not change the layout function.

Coalescence Rules

Considering a layout with just two integral modes, $A = (N_{0}, N_{1}):(d_{0}, d_{1})$, we have four cases to consider:

In the first case, obviously, $A = (N_{0}, 1):(d_{0}, d_{1}) = (N_{0}):(d_{0})$.

In the second case, also obviously, $A = (1, N_{1}):(d_{0}, d_{1}) = (N_{1}):(d_{1})$.

In the third case, we have $A = (N_{0}, N_{1}):(d_{0}, N_{0} d_{0}) = (N_{0} N_{1}):(d_{0})$.

In the fourth case, we could do nothing and $A$ remains the same.

There is one case that can often be misunderstood, that is $d_{0} = N_{1} d_{1}$. In this case, we have $A = (N_{0}, N_{1}):(N_{1} d_{1}, d_{1})$. At first glance, it seems that we could coalesce $A$ to $(N_{0} N_{1}):(d_{1})$. However, this is not correct, because it changes the layout function.

Composition

Definition 2.11 Left Divisibility

Let $M, d > 0$ be positive integers and let $M = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha}$ be a given factorization of $M$ by integers $M_{k} > 1$ for $k \in [0, \alpha]$. Replacing $M_{\alpha}$ by $\infty$, let

$$\begin{align}\widehat{M} = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty\end{align}$$

and consider $\infty$ to be divisible by every positive integer. We say that $M$ is left divisible by $d$ (implicitly, with respect to the given factorization) if there exists $0 \leq i \leq \alpha$ such that:

Here $i$ is necessarily unique if it exists. We could prove this by contradiction.

Proof

Suppose there exists two distinct $i$ and $j$ such that the three conditions are satisfied. Without loss of generality, suppose $i < j$.

There are two cases to consider.

In the case where $j < \alpha$, we will also have $i < \alpha$. Then we have

$$\begin{align}d &= c \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \\\end{align}$$

where $c$ is some positive integer such that $1 \leq c < M_{i}$.

Similarly,

$$\begin{align}d &= c^{\prime} \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{j - 1} \\\end{align}$$

where $c^{\prime}$ is some positive integer such that $1 \leq c^{\prime} < M_{j}$.

Thus,

$$\begin{align}c \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} &= c^{\prime} \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{j - 1} \\\end{align}$$

$$\begin{align}c &= c^{\prime} \cdot M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1} \\\end{align}$$

To make the above equation valid, we must show

$$\begin{align}c^{\prime} \cdot \frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1} = 1\end{align}$$

However, because $M_{k} > 1$ for $k \in [0, \alpha]$, $\frac{M_{i}}{c} > 1$, and $c^{\prime} \geq 1$, it is not possible to have the above equation valid. This raises a contradiction. Therefore, $i$ is unique.

In the case where $j = \alpha$, we will also have $i < \alpha$. Then we have

$$\begin{align}d &= c \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \\\end{align}$$

where $c$ is some positive integer such that $1 \leq c < M_{i}$.

Similarly,

$$\begin{align}d &= c^{\prime} \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \\\end{align}$$

where $c^{\prime}$ is some positive integer.

Thus,

$$\begin{align}c \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} &= c^{\prime} \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \\\end{align}$$

$$\begin{align}c &= c^{\prime} \cdot M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} \\\end{align}$$

To make the above equation valid, we must show

$$\begin{align}c^{\prime} \cdot \frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} = 1\end{align}$$

However, because $M_{k} > 1$ for $k \in [0, \alpha]$, $\frac{M_{i}}{c} > 1$, and $c^{\prime} \geq 1$, it is not possible to have the above equation valid. This raises a contradiction. Therefore, $i$ is unique.

Taken together, $i$ is unique if it exists.

This concludes the proof. $\square$

If $i$ exists, we will refer to $i$ as the division index and write $\widehat{M} = d \cdot \widehat{M}^{\prime}$, where $\widehat{M}^{\prime}$ is endowed with the following induced factorization:

To see this, in the case where $0 \leq i < \alpha$, we have

$$\begin{align}\widehat{M} &= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty \\&= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \cdot M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty \\&= \frac{d}{c} \cdot M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty \\&= d \cdot \frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty \\&= d \cdot M_{0}^{\prime} \cdot M_{1}^{\prime} \cdot \ldots \cdot M_{\alpha - i - 1}^{\prime} \cdot \infty \\&= d \cdot \widehat{M}^{\prime}\end{align}$$

where $\widehat{M}^{\prime} = M_{0}^{\prime} \cdot M_{1}^{\prime} \cdot \ldots \cdot M_{\alpha - i - 1}^{\prime} \cdot \infty$ with $M_{0}^{\prime} = \frac{M_{i}}{c} > 1$ and $M_{j}^{\prime} = M_{i + j}$ for $0 < j < \alpha - i$.

In the case where $i = \alpha$, we have

$$\begin{align}\widehat{M} &= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty \\&= \frac{d}{c} \cdot \infty \\&= d \cdot \infty \\&= d \cdot \widehat{M}^{\prime}\end{align}$$

where $\widehat{M}^{\prime} = \infty$.

Furthermore, we say that $M$ is weakly left divisible by $d$ if there exists $0 \leq i \leq \alpha$ such that the conditions 1 and 2 are satisfied for left divisibility, but not necessarily the condition 3.

Notice that in the proof of the uniqueness of division index $i$, we have never used the condition 3. Therefore, we could still the necessarily unique $i$ the division index for weak left divisibility, but we no longer have the factorization of $\widehat{M}$, because the factorization assumes the condition 3 of left divisibility.

Also notice that $\widehat{M}^{\prime}$ with its induced factorization can itself be considered for left divisibility or weak left divisibility (with the step or replacing the last factor by $\infty$ now being superfluous). More specifically, because $\widehat{M}^{\prime} > 0$, $\widehat{M}_{j}^{\prime} > 1$ for $j \in [0, \alpha - i - 1]$, and $\widehat{M}^{\prime} = \widehat{M}_{0}^{\prime} \cdot \widehat{M}_{1}^{\prime} \cdot \ldots \cdot \widehat{M}_{\alpha - i - 1}^{\prime} \cdot \infty$, given another positive integer $d^{\prime} > 0$, we could completely test whether the properties of left divisibility or weak left divisibility hold for $\widehat{M}^{\prime}$ with respect to $d^{\prime}$. Replacing the last factor by $\infty$ is not necessary as it is already $\infty$.

Definition 2.12 Admission for Composition - Restricted Case

We first consider composition in the restricted case of length 1 layouts for the second layout.

Let $\mathbf{S} = (M_{0}, M_{1}, \ldots, M_{\alpha})$ be a shape tuple, let $M = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha}$, and let $B = (N):(r)$ be a layout of length 1. Then we say that the pair $\{\mathbf{S}, B\}$ is admissible for composition (or simply admissible) if:

Definition 2.13 Composition - Restricted Case

The idea of admissibility is that the composition $A \circ B$ of layouts will entail “dividing $B$ along the modes of $A$”. More preciously, we have the following:

Suppose that $\mathbf{S} = (M_{0}, M_{1}, \ldots, M_{\alpha})$ is a shape tuple, and $B = (N):(r)$ is a layout of length 1 such that $\{\mathbf{S}, B\}$ is admissible for composition. Let $\mathbf{D} = (d_{0}, d_{1}, \ldots, d_{\alpha})$ be any stride tuple and let $A = (\mathbf{S}:\mathbf{D})$ be a coalesced layout.

Note that in Jay Shah’s original paper, the layout $A$ was not specified to be coalesced. It will result in some compositions not being valid.

As in Definition 2.11, let $M = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha}$ and $\widehat{M} = r \cdot \widehat{M}^{\prime}$ with division index $0 \leq i \leq \alpha$. We separate the definition of $A \circ B$ into two cases.

First suppose that $0 \leq i < \alpha$, so that

$$\begin{align}r &= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \cdot c \\\widehat{M}^{\prime} &= \frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1} \cdot \infty\end{align}$$

Then if $N \leq \frac{M_{i}}{c}$, we let $A \circ B = (N):(cd_{i})$.

Otherwise, there exists a $j \in [i + 1, \alpha]$ such that $N = \frac{M_{i}}{c} \cdot \ldots \cdot M_{j-1} \cdot c^{\prime}$, where $1 \leq c^{\prime} < M_{j}$ if $j \neq \alpha$ (when $j = i + 1$, $N = \frac{M_{i}}{c}  \cdot c^{\prime}$).

Note that here is an important fact that $c^{\prime}$ must be an integer because of the second condition for admission for composition, that is, $\widehat{M}^{\prime}$ is weakly left divisible by $N$. We must have $\frac{M_{i}}{c} \cdot \ldots \cdot M_{j-1}$ divides $N$, resulting in $c^{\prime}$ being an integer.

We let

$$\begin{align}A \circ B =\begin{cases}\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1}, c^{\prime} \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1}, d_{j} \right) & \text{if } c^{\prime} > 1 \\\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1} \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1} \right) & \text{if } c^{\prime} = 1\end{cases}\end{align}$$

If instead $i = \alpha$, then we have $r = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot c$ as before but $\widehat{M}^{\prime} = \infty$, and we let $A \circ B = (N):(cd_{\alpha})$.

Let’s look at this definition more closely.

Essentially, we are taking the one-dimensional coordinates $k \cdot r$ along the layout $A$ where $k \in [0, N - 1]$. Because we have $r = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \cdot c$, and $c$ divides $M_{i}$.

Let first consider the case of $0 \leq i < \alpha$.

If $N \leq \frac{M_{i}}{c}$, then the first mode in the layout $A$ is sufficient for dividing $B$. Consequently, the composition layout $A \circ B = (N):(cd_{i})$.

Otherwise if $N > \frac{M_{i}}{c}$, more modes in the layout $A$ will be involved for dividing $B$, and consequently the composition layout

$$\begin{align}A \circ B =\begin{cases}\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1}, c^{\prime} \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1}, d_{j} \right) & \text{if } c^{\prime} > 1 \\\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1} \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1} \right) & \text{if } c^{\prime} = 1\end{cases}\end{align}$$

Let’s then consider the case of $i = \alpha$. We have $r = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot c$ and $\widehat{M}^{\prime} = \infty$. Here $c$ is an positive integer that can be infinitely large. So only the last mode in the layout $A$ is involved for dividing $B$, and consequently the composition layout $A \circ B = (N):(cd_{\alpha})$.

Note that by this definition, $\text{size}(A \circ B) = \text{size}(B)$. This is a critical property which we will use later.

Proposition 2.14

In the situation of Definition 2.13, we have that $f_{A \circ B} = \widehat{f}_A \circ f_B$.

Proof

This more formally proves the intuition we explained to Definition 2.13.

We carry over the notation from Definition 2.13.

Given an index $0 \leq k \leq \alpha$, let $\delta_{k} \in \mathbb{N}^{\times(\alpha + 1)}$ denote the coordinate that is zero everywhere except in the $k$-th position, where it is 1. Concretely,

$$\begin{align}\delta_{0} &= \underbrace{(1, 0, 0, \ldots, 0)}_{\alpha + 1} \\\delta_{1} &= \underbrace{(0, 1, 0, \ldots, 0)}_{\alpha + 1} \\&\vdots \\\delta_{\alpha} &= \underbrace{(0, 0, 0, \ldots, 1)}_{\alpha + 1}\end{align}$$

With respect to the the isomorphism of the extended layout $A$, we have

$$\begin{align}\widehat{\iota}: \mathbb{N} \cong [0, M_{0}) \times [0, M_{1}) \times \ldots \times [0, M_{\alpha - 1}) \times \mathbb{N}\end{align}$$

Because $B = (N):(r)$, we have

$$\begin{align}f_B(k) &= k \cdot r \\&= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1} \cdot k \cdot c \\\end{align}$$

where $k \in [0, N - 1]$.

Let first consider the case of $0 \leq i < \alpha$.

If $N \leq \frac{M_{i}}{c}$, i.e. $N \cdot c \leq M_{i}$, then we must have $k \cdot c < M_{i}$ for all $k \in [0, N - 1]$. Because of the isomorphism of the extended layout $A$, we have

$$\begin{align}f_B(k) \mapsto \delta_{i} \cdot k \cdot c\end{align}$$

Then we have

$$\begin{align}\left(\widehat{f}_A \circ f_B \right)(k) &= \widehat{f}_A \left(f_B\left(k\right)\right) \\&= \widehat{f}_A \left(\delta_{i} \cdot k \cdot c\right) \\&= k \cdot c \cdot d_{i} \\\end{align}$$

According to Definition 2.13, we have

$$\begin{align}f_{A \circ B}(k) &= k \cdot c \cdot d_{i} \\\end{align}$$

Therefore, $f_{A \circ B} = \widehat{f}_A \circ f_B$.

Otherwise if $N > \frac{M_{i}}{c}$, i.e. $N = \frac{M_{i}}{c} \cdot \ldots \cdot M_{j-1} \cdot c^{\prime}$. Because of the isomorphism of the extended layout $A$, by definition, we have

$$\begin{align}f_B(k) &\mapsto \left(f_B(k) \mod M_{0}, \left\lfloor \frac{f_B(k)}{M_{0}} \right\rfloor \mod M_{1}, \left\lfloor \frac{f_B(k)}{M_{0} \cdot M_{1}} \right\rfloor \mod M_{2}, \ldots, \left\lfloor \frac{f_B(k)}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1}} \right\rfloor \mod M_{i}, \ldots, \left\lfloor \frac{f_B(k)}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 2}} \right\rfloor \mod M_{\alpha - 1}, \left\lfloor \frac{f_B(k)}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \right) \\&= \left(0, 0, \ldots, \left(k \cdot c\right) \mod M_{i}, \left\lfloor \frac{k \cdot c}{M_{i}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod M_{j}, \ldots, \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 2}} \right\rfloor \mod M_{\alpha - 1}, \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \right) \\&= \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod M_{j}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 2}} \right\rfloor \mod M_{\alpha - 1}, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \right) \\\end{align}$$

Note that here we used the property $\left(k \cdot c\right) \mod M_{i} = \left( k \mod \frac{M_{i}}{c} \right) \cdot c$, if $c$ divides $M_{i}$.

To see this, suppose $k \cdot c = p \cdot M_{i} + r$, where $0 \leq r < M_{i}$. Then we have $k = \frac{p \cdot M_{i} + r}{c} = p \cdot \frac{M_{i}}{c} + \frac{r}{c}$. Because $c$ divides $M_{i}$, $\frac{r}{c}$ ust be an integer. Thus, we have $\left(k \cdot c\right) \mod M_{i} = r$, and $\left( k \mod \frac{M_{i}}{c} \right) \cdot c = \frac{r}{c} \cdot c = r$. Therefore, $\left(k \cdot c\right) \mod M_{i} = \left( k \mod \frac{M_{i}}{c} \right) \cdot c$.

Further more, because $0 \leq k < N$, we have $0 \leq k \cdot c < N \cdot c = M_{i} \cdot \ldots \cdot M_{j - 1} \cdot c^{\prime}$, where $1 \leq c^{\prime} < M_{j}$. Thus $0 \leq k \cdot c < M_{i} \cdot \ldots \cdot M_{j - 1} \cdot M_{j}$.

When $c^{\prime} > 1$, we have

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j}} \right\rfloor&= \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{j}} \right\rfloor \\&= 0\end{align}$$

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j}} \right\rfloor \mod M_{j + 1}&= 0\end{align}$$

and of course for any $l \in [j + 1, \alpha - 1]$, we have

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{l}} \right\rfloor&= \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{l}} \right\rfloor \\&= 0\end{align}$$

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{l}} \right\rfloor \mod M_{l + 1}&= 0\end{align}$$

Thus, we have

$$\begin{align}f_B(k) &\mapsto \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod M_{j}, 0, 0, \ldots, 0 \right) \\\end{align}$$

What’s more, because $k \leq (N - 1)$ and $\frac{M_{i}}{c} \cdot \ldots \cdot M_{j - 1} = \frac{N}{c^{\prime}}$, we have

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor&= \left\lfloor \frac{k}{\frac{N}{c^{\prime}}} \right\rfloor \\&= \left\lfloor \frac{k}{N} \cdot c^{\prime} \right\rfloor \\&\leq \left\lfloor \frac{N - 1}{N} \cdot c^{\prime} \right\rfloor \\&\leq \left\lfloor c^{\prime} \right\rfloor \\&\leq c^{\prime}\end{align}$$

Because $c^{\prime} < M_{j}$, we have

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod M_{j}&= \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime} \\\end{align}$$

Thus, we have

$$\begin{align}f_B(k) &\mapsto \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime}, 0, 0, \ldots, 0 \right) \\\end{align}$$

Then we have

$$\begin{align}\left(\widehat{f}_A \circ f_B \right)(k) &= \widehat{f}_A \left(f_B\left(k\right)\right) \\&= \widehat{f}_A \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime}, 0, 0, \ldots, 0 \right) \\&= 0 \cdot d_{0} + 0 \cdot d_{1} + \ldots + \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime} \right) \cdot d_{j} + 0 \cdot d_{j + 1} + \ldots + 0 \cdot d_{\alpha} \\&= \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime} \right) \cdot d_{j} \\\end{align}$$

According to Definition 2.13, we have

$$\begin{align}f_{A \circ B}(k) &= \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod c^{\prime} \right) \cdot d_{j} \\\end{align}$$

Therefore, $f_{A \circ B} = \widehat{f}_A \circ f_B$.

When $c^{\prime} = 1$, we have $0 \leq k \cdot c < N \cdot c = M_{i} \cdot \ldots \cdot M_{j - 1} \cdot c^{\prime} = M_{i} \cdot \ldots \cdot M_{j - 1}$.

Thus, we have

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor&= \left\lfloor \frac{k \cdot c}{M_{i} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \\&= 0\end{align}$$

$$\begin{align}\left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 1}} \right\rfloor \mod M_{j}&= 0\end{align}$$

Thus, we have

$$\begin{align}f_B(k) &\mapsto \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 2}} \right\rfloor \mod M_{j - 1}, 0, 0, \ldots, 0 \right) \\\end{align}$$

Then we have

$$\begin{align}\left(\widehat{f}_A \circ f_B \right)(k) &= \widehat{f}_A \left(f_B\left(k\right)\right) \\&= \widehat{f}_A \left(0, 0, \ldots, \left( k \mod \frac{M_{i}}{c} \right) \cdot c, \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1}, \ldots, \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 2}} \right\rfloor \mod M_{j - 1}, 0, 0, \ldots, 0 \right) \\&= 0 \cdot d_{0} + 0 \cdot d_{1} + \ldots + \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 2}} \right\rfloor \mod M_{j - 1} \right) \cdot d_{j - 1} + 0 \cdot d_{j} + 0 \cdot d_{j + 1} + \ldots + 0 \cdot d_{\alpha} \\&= \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 2}} \right\rfloor \mod M_{j - 1} \right) \cdot d_{j - 1} \\\end{align}$$

According to Definition 2.13, we have

$$\begin{align}f_{A \circ B}(k) &= \left( k \mod \frac{M_{i}}{c} \right) \cdot c \cdot d_{i} + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c}} \right\rfloor \mod M_{i + 1} \right) \cdot d_{i + 1} + \ldots + \left( \left\lfloor \frac{k}{\frac{M_{i}}{c} \cdot M_{i + 1} \cdot \ldots \cdot M_{j - 2}} \right\rfloor \mod M_{j - 1} \right) \cdot d_{j - 1} \\\end{align}$$

Therefore, $f_{A \circ B} = \widehat{f}_A \circ f_B$.

Let’s then consider the case of $i = \alpha$.

$$\begin{align}f_B(k) &= k \cdot r \\&= k \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1} \cdot c \\\end{align}$$

where $k \in [0, N - 1]$.

Because of the isomorphism of the extended layout $A$, we have

$$\begin{align}f_B(k) &\mapsto \delta_{\alpha} \cdot k \cdot c \\\end{align}$$

Then we have

$$\begin{align}\left(\widehat{f}_A \circ f_B \right)(k) &= \widehat{f}_A \left(f_B\left(k\right)\right) \\&= \widehat{f}_A \left(\delta_{\alpha} \cdot k \cdot c\right) \\&= k \cdot c \cdot d_{\alpha} \\\end{align}$$

According to Definition 2.13, we have

$$\begin{align}f_{A \circ B}(k) &= k \cdot c \cdot d_{\alpha} \\\end{align}$$

Therefore, $f_{A \circ B} = \widehat{f}_A \circ f_B$.

Taken together, we have $f_{A \circ B} = \widehat{f}_A \circ f_B$ for all the cases in Definition 2.13.

This concludes the proof. $\square$

One might ask why the second condition for admission for composition is necessary. If we don’t have it, $c^{\prime}$ can be fractional and we can still define $A \circ B$ to be

$$\begin{align}A \circ B =\begin{cases}\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1}, \left\lceil c^{\prime} \right\rceil \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1}, d_{j} \right) & \text{if } c^{\prime} > 1 \\\left(\frac{M_{i}}{c}, M_{i + 1}, \ldots, M_{j - 1} \right) : \left(cd_{i}, d_{i + 1}, \ldots, d_{j - 1} \right) & \text{if } c^{\prime} = 1\end{cases}\end{align}$$

It’s not too difficult to show that we still have $f_{A \circ B} = \widehat{f}_A \circ f_B$ for the domain of $f_B$ when the length of $B$ is 1.

However, the critical property $\text{size}(A \circ B) = \text{size}(B)$ will not hold in this case. As we will see later, without having this property, $f_{A \circ B} = \widehat{f}_A \circ f_B$ cannot be true when $B$ is multi-modal, i.e., the length of $B$ is greater than 1.

Definition 2.16 Interval of Definition

In the situation of Definition 2.12, where layout $B$ is of length 1, let $f_B: [0, N) \to \mathbb{N}$ be the layout function, and let $I = [r, r(N - 1)]$ be the interval given by the convex closure of the image $f_B([1, N))$. Let $M^{\prime} = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}$ and $J = I \cap [1, M^{\prime})$ (so $J = \emptyset$ if $\alpha = 0$). Then the interval of definition for $\{\mathbf{S}, B\}$ is $J$.

Definition 2.17 Composition - General Case

Let $\mathbf{S} = (M_{0}, M_{1}, \ldots, M_{\alpha})$ be a shape tuple, and let $B = (N_{0}, N_{1}, \ldots, N_{\beta}) : (r_{0}, r_{1}, \ldots, r_{\beta})$ be a layout, let $B_{k} = (N_{k}) : (r_{k})$ for $0 \leq k \leq \beta$. Then we say that the pair $\{\mathbf{S}, B\}$ is admissible for composition if:

In this case, if $\mathbf{D} = (d_{0}, d_{1}, \ldots, d_{\alpha})$ is a stride tuple and $A = \mathbf{S} : \mathbf{D}$, then we define the composition $A \circ B$ to be the concatenated layout

$$\begin{align}A \circ B := \left(A \circ B_{0}, A \circ B_{1}, \ldots, A \circ B_{\beta}\right)\end{align}$$

where each $A \circ B_{k}$ is defined as in Definition 2.13.

Theorem 2.18 Composition - General Case

In the situation of Definition 2.17, we have that $f_{A \circ B} = \widehat{f}_A \circ f_B$.

Proof

By Definition 2.13, $\text{size}(A \circ B_{k}) = \text{size}(B_{k}) = N_{k}$ for all $0 \leq k \leq \beta$. We have the following isomorphism for both the layout $A \circ B$ or the layout $B$.

$$\begin{align}\iota: [0, N_{0} \cdot N_{1} \cdot \ldots \cdot N_{\beta}) \cong [0, N_{0}) \times [0, N_{1}) \times \ldots \times [0, N_{\beta})\end{align}$$

Given any $x \in [0, N_{0} \cdot N_{1} \cdot \ldots \cdot N_{\beta})$, because of the isomorphism $\iota$, we have

$$\begin{align}x &\mapsto \left(x_{0}, x_{1}, \ldots, x_{\beta}\right)\end{align}$$

By Lemma 2.19, we have

$$\begin{align}\widehat{f}_A \circ f_B(x) &= \widehat{f}_A \left(f_B(x)\right) \\&= \widehat{f}_A \left(f_{B_{0}}(x_{0}) + f_{B_{1}}(x_{1}) + \ldots + f_{B_{\beta}}(x_{\beta})\right) \\\end{align}$$

By Definition 2.17, Lemma 2.19, and Definition 2.13, we have

$$\begin{align}f_{A \circ B}(x) &= f_{A \circ B_{0}}(x_{0}) + f_{A \circ B_{1}}(x_{1}) + \ldots + f_{A \circ B_{\beta}}(x_{\beta}) \\&= \widehat{f}_A \circ f_{B_{0}}(x_{0}) + \widehat{f}_A \circ f_{B_{1}}(x_{1}) + \ldots + \widehat{f}_A \circ f_{B_{\beta}}(x_{\beta}) \\&= \widehat{f}_A \left(f_{B_{0}}(x_{0})\right) + \widehat{f}_A \left(f_{B_{1}}(x_{1})\right) + \ldots + \widehat{f}_A \left(f_{B_{\beta}}(x_{\beta})\right) \\\end{align}$$

Normally, we don’t have $\widehat{f}_A \left(x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta}\right) = \widehat{f}_A \left(x_{A, 0}\right) + \widehat{f}_A \left(x_{A, 1}\right) + \ldots + \widehat{f}_A \left(x_{A, \beta}\right)$, because the layout function $\widehat{f}_A$ is not linear. For example, suppose $A = (2, 3) : (1, 4)$, and we have $\widehat{f}_A(1) = 1$ and $\widehat{f}_A(3) = 5$. $\widehat{f}_A(3) = \widehat{f}_A(1 + 1 +1) \neq \widehat{f}_A(1) + \widehat{f}_A(1) + \widehat{f}_A(1)$.

However, there are some special cases where the above equation holds. For example, for simplicity, suppose $\beta = \alpha$, if we have

$$\begin{align}x_{A, 0} &\in [0, M_{0}) \\x_{A, 1} &\in \{0, 1 \cdot M_{0}, 2 \cdot M_{0}, \ldots, \infty \cdot M_{0}\} \cap [0, M_{1}) \\x_{A, 2} &\in \{0, 1 \cdot M_{0} \cdot M_{1}, 2 \cdot M_{0} \cdot M_{1}, \ldots, \infty  \cdot M_{0} \cdot M_{1}\} \cap [0, M_{2}) \\&\vdots \\x_{A, \alpha} &\in \{0, 1 \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}, 2 \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}, \ldots, \infty  \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}\} \cap [0, M_{\alpha}) \\\end{align}$$

By definition,

$$\begin{align}\widehat{f}_A \left(x\right)&= \left( x \mod M_{0} \right) \cdot d_{0} + \left( \left\lfloor \frac{x}{M_{0}} \right\rfloor \mod M_{1} \right) \cdot d_{1} + \ldots + \left( \left\lfloor \frac{x}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \mod M_{\alpha} \right) \cdot d_{\alpha} \\\end{align}$$

So in our case, we have

$$\begin{align}\widehat{f}_A \left(x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta}\right)&= \left( \left( x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta} \right) \mod M_{0} \right) \cdot d_{0} + \left( \left\lfloor \frac{x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta}}{M_{0}} \right\rfloor \mod M_{1} \right) \cdot d_{1} + \ldots + \left( \left\lfloor \frac{x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta}}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \mod M_{\alpha} \right) \cdot d_{\alpha} \\&= \left( x_{A, 0} \mod M_{0} \right) \cdot d_{0} + \left( \left\lfloor \frac{x_{A, 1}}{M_{0}} \right\rfloor \mod M_{1} \right) \cdot d_{1} + \ldots + \left( \left\lfloor \frac{x_{A, \beta}}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha - 1}} \right\rfloor \mod M_{\alpha} \right) \cdot d_{\alpha} \\&= \widehat{f}_A \left(x_{A, 0}\right) + \widehat{f}_A \left(x_{A, 1}\right) + \ldots + \widehat{f}_A \left(x_{A, \beta}\right) \\\end{align}$$

The idea of having the second condition for admission for composition, i.e., the interval of definition for the pairs $\{\mathbf{S}, B_{k}\}_{0 \leq k \leq \beta}$ are disjoint, are exactly the same.

Because $r_{k} = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c$, for $x_{k} \in [0, N_{k})$, we have

$$\begin{align}f_{B_{k}}(x_{k}) &\in [0, 1 \cdot r, 2 \cdot r, \ldots, (N_{k} - 1) \cdot r] \\&= [0, M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c, 2 \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c, \ldots, (N_{k} - 1) \cdot M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c] \\\end{align}$$

Because of the isomorphism of the layout $A$, we have

$$\begin{align}f_{B_{k}}(x_{k}) &\mapsto \left(x_{A, 0}, x_{A, 1}, \ldots, x_{A, \alpha}\right) \\\end{align}$$

where

$$\begin{align}x_{A, i} = \left\lfloor \frac{f_{B_{k}}(x_{k})}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i - 1}} \right\rfloor \mod M_{i}\end{align}$$

Then we must have some integer $p_{k}$ and $q_{k}$, $p_{k} < q_{k}$, where $x_{A, i} = 0$ for $i < p_{k}$ and $i > q_{k}$.

The second condition for admission for composition ensures that $[p_{k}, q_{k}]$ are disjoint for all $0 \leq k \leq \beta$. Therefore, we can have the equation:

$$\begin{align}\widehat{f}_A \left(x_{A, 0} + x_{A, 1} + \ldots + x_{A, \beta}\right) = \widehat{f}_A \left(x_{A, 0}\right) + \widehat{f}_A \left(x_{A, 1}\right) + \ldots + \widehat{f}_A \left(x_{A, \beta}\right)\end{align}$$

This concludes the proof. $\square$

Going back to the discussion why the second condition for admission for composition in the restricted case in Proposition 2.14 is necessary.

Because the critical property $\text{size}(A \circ B) = \text{size}(B)$ will not hold, we will have two completely different isomorphisms for the layout $A \circ B$ and the layout $B$, respectively.

Lemma 2.19 Concatenation of Layouts

Let $C = (C_{0}, C_{1}, \ldots, C_{\gamma})$ be a concatenated layout. Let

$$\begin{align}\iota: [0, \text{size}(C)) \cong [0, \text{size}(C_{0})) \times \cdots \times [0, \text{size}(C_{\gamma}))\end{align}$$

be the usual isomorphism (as in Definition 2.3). Then the following diagram commutes:

$$\begin{equation}\begin{CD}[0, \text{size}(C)) @>{\iota}>{\cong}> [0, \text{size}(C_0)) \times \dots \times [0, \text{size}(C_\gamma)) \\@V{f_C}VV @VV{(f_{C_0}, \dots, f_{C_\gamma})}V \\\mathbb{N} @<+<< \mathbb{N} \times \dots \times \mathbb{N}\end{CD}\end{equation}$$

Proof

If $C_{0}, \ldots, C_{\gamma}$ are all length 1 layouts, then this is immediate from Definition 2.3.

Concretely, suppose $C_{k} = (M_{k}) : (d_{k})$ for all $0 \leq k \leq \gamma$. The concatenated layout becomes

$$\begin{align}C &= (C_{0}, C_{1}, \ldots, C_{\gamma}) \\&= (M_{0}:d_{0}, M_{1}:d_{1}, \ldots, M_{\gamma}:d_{\gamma}) \\&= (M_{0}, M_{1}, \ldots, M_{\gamma}) : (d_{0}, d_{1}, \ldots, d_{\gamma})\end{align}$$

Because of the isomorphism of the layout $C$, we have

$$\begin{align}x &\mapsto \left(x_{0}, x_{1}, \ldots, x_{\gamma}\right) \\\end{align}$$

Then by definition, the concatenated layout function is

$$\begin{align}f_C(x) &= x_{0} \cdot d_{0} + x_{1} \cdot d_{1} + \ldots + x_{\gamma} \cdot d_{\gamma} \\\end{align}$$

For each of the length 1 layouts $C_{k}$, by definition, the layout function is

$$\begin{align}f_{C_{k}}(x_{k}) &= x_{k} \cdot d_{k} \\\end{align}$$

Therefore, we have

$$\begin{align}f_C(x) &= f_{C_{0}}(x_{0}) + f_{C_{1}}(x_{1}) + \ldots + f_{C_{\gamma}}(x_{\gamma}) \\\end{align}$$

In the case where some of the layouts $C_{k}$ are not length 1, we can apply the same argument to each of the sublayouts $C_{k}$, and the result follows by induction.

Concretely, suppose $C_{k}$ are not length 1 and $C_{k} = (C_{k, 0}, C_{k, 1}, \ldots, C_{k, \gamma_{k}})$, where $C_{k, 0}, \ldots, C_{k, \gamma_{k}}$ are length 1 layouts. Based on what we have proved above, we have

$$\begin{align}f_{C_{k}}(x_{k}) &= f_{C_{k, 0}}(x_{k, 0}) + f_{C_{k, 1}}(x_{k, 1}) + \ldots + f_{C_{k, \gamma_{k}}}(x_{k, \gamma_{k}}) \\\end{align}$$

where

$$\begin{align}x_{k} &\mapsto \left(x_{k, 0}, x_{k, 1}, \ldots, x_{k, \gamma_{k}}\right) \\\end{align}$$

Suppose the layout $C$ can be maximally decomposed into layouts of length 1.

$$\begin{align}C &= (C_{0}, C_{1}, \ldots, C_{\gamma}) \\&= (C_{0, 0}, C_{0, 1}, \ldots, C_{0, \gamma_{0}}, C_{1, 0}, C_{1, 1}, \ldots, C_{1, \gamma_{1}}, \ldots, C_{\gamma, 0}, C_{\gamma, 1}, \ldots, C_{\gamma, \gamma_{\gamma}}) \\\end{align}$$

Then we have

$$\begin{align}f_C(x) &= f_{C_{0,0}}(x_{0,0}) + f_{C_{0,1}}(x_{0,1}) + \ldots + f_{C_{0,\gamma_{0}}}(x_{0,\gamma_{0}}) + f_{C_{1,0}}(x_{1,0}) + f_{C_{1,1}}(x_{1,1}) + \ldots + f_{C_{1,\gamma_{1}}}(x_{1,\gamma_{1}}) + \ldots + f_{C_{\gamma,0}}(x_{\gamma,0}) + f_{C_{\gamma,1}}(x_{\gamma,1}) + \ldots + f_{C_{\gamma,\gamma_{\gamma}}}(x_{\gamma,\gamma_{\gamma}}) \\&= f_{C_{0}}(x_{0}) + f_{C_{1}}(x_{1}) + \ldots + f_{C_{\gamma}}(x_{\gamma}) \\\end{align}$$

where

$$\begin{align}x &\mapsto \left(x_{0}, x_{1}, \ldots, x_{\gamma}\right) \\\end{align}$$

This concludes the proof. $\square$

Definition 2.21 CUTLASS Admission for Composition - Restricted Case

The CUTLASS admission for composition in the restricted case is more restrictive.

Let $\mathbf{S} = (M_{0}, M_{1}, \ldots, M_{\alpha})$ be a shape tuple, let $M = M_{0} \cdot M_{1} \cdot \ldots \cdot M_{\alpha}$, and let $B = (N):(r)$ be a layout of length 1. Then we say that the pair $\{\mathbf{S}, B\}$ is admissible for composition (or simply admissible) if:

Note that the second condition is the left divisibility, instead of the weak left divisibility in Definition 2.12.

For example, suppose $A = (8, 6, 8) : (1, 16, 108)$ and $B = (8) : (4)$. According to Definition 2.12, $A \circ B = (2, 4) : (4, 16)$. However, if we run composition for $A$ and $B$ in CUTLASS, we will encounter an error because CUTLASS requires left divisibility for the second condition.

More specifically, in the CUTLASS composition layout algebra implementation, we have

12345678910

void shape_div(int* shapeA, int N, int& strideB) {   for (int i = 0; i < N; ++i) {      assert(shapeA[i] %   strideB == 0 or               strideB % shapeA[i] == 0);      int new_shape  = ceil_div(shapeA[i], strideB);      int new_stride = ceil_div(strideB, shapeA[i]);      shapeA[i] = new_shape;      strideB   = new_stride;   }}

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void shape_mod(int* shapeA, int N, int& shapeB) {   for (int i = 0; i < N; ++i) {      assert(shapeA[i] %    shapeB == 0 or                shapeB % shapeA[i] == 0);      int new_shapeA =      min(shapeA[i], shapeB);      int new_shapeB = ceil_div(shapeB, shapeA[i]);      shapeA[i] = new_shapeA;      shapeB    = new_shapeB;   }}

The reason why CUTLASS enforces this is because of the logical division operation. Without this restriction, the logical division operation will not be defined in some cases.

Logical Division

Definition 2.22 Logical Division

Let $A = \mathbf{S} : \mathbf{D}$ and $B$ be layouts, and let $M$ be the size of $A$. Suppose that the pairs $\{B, M\}$ and $\{\mathbf{S}, B\}$ are admissible (for complementation and composition, respectively). Then we define the logical division $A / B$ to be the layout

$$\begin{align}A / B := A \circ \left(B, \text{complement}(B, M)\right)\end{align}$$

Note that here the conditions of admission for composition follows Definition 2.21 rather than Definition 2.12.

Implicitly Lemma 2.23 is used in Definition 2.22.

Lemma 2.23 Logical Division Implication

Suppose $A = \mathbf{S} : \mathbf{D}$, $M = \text{size}(A)$, and $B$ are as in Definition 2.22. Then $\{\mathbf{S}, \left(B, \text{complement}(B, M)\right)\}$ is admissible for composition.

Proof

We denote $A = \mathbf{S} : \mathbf{D} = (M_{0}, M_{1}, \ldots, M_{\alpha}) : (d_{0}, d_{1}, \ldots, d_{\alpha})$, and $B = (N_{0}, N_{1}, \ldots, N_{\beta}) : (r_{0}, r_{1}, \ldots, r_{\beta})$. Let

$$\begin{align}\varphi: [0, \beta] \xrightarrow{\cong} [0, \beta]\end{align}$$

be the automorphism such that $B^{\varphi} := (N_{\varphi(0)}, N_{\varphi(1)}, \ldots, N_{\varphi(\beta)}) : (r_{\varphi(0)}, r_{\varphi(1)}, \ldots, r_{\varphi(\beta)})$ is sorted.

Then by Definition 2.6, we have

$$\begin{align}B^{\prime}&= \text{complement}(B, M) \\&= \left(r_{\varphi(0)}, \frac{r_{\varphi(1)}}{N_{\varphi(0)}r_{\varphi(0)}}, \frac{r_{\varphi(2)}}{N_{\varphi(1)}r_{\varphi(1)}}, \ldots, \frac{r_{\varphi(\beta)}}{N_{\varphi(\beta - 1)}r_{\varphi(\beta - 1)}}, \frac{M}{N_{\varphi(\beta)}r_{\varphi(\beta)}}\right) : (1, N_{\varphi(0)}r_{\varphi(0)}, N_{\varphi(1)}r_{\varphi(1)}, \ldots, N_{\varphi(\beta - 1)}r_{\varphi(\beta - 1)}, N_{\varphi(\beta)}r_{\varphi(\beta)})\end{align}$$

Now we denote each mode of $B^{\prime}$ as

$$\begin{align}B^{\prime}_{k}&=\begin{cases}\left(r_{\varphi(0)}\right) : (1) & \text{if } k = 0 \\\left(\frac{r_{\varphi(k)}}{N_{\varphi(k - 1)}r_{\varphi(k - 1)}}\right) : (N_{\varphi(k - 1)}r_{\varphi(k - 1)}) & \text{if } 1 \leq k \leq \beta \\\left(\frac{M}{N_{\varphi(\beta)}r_{\varphi(\beta)}}\right) : (N_{\varphi(\beta)}r_{\varphi(\beta)}) & \text{if } k = \beta + 1\end{cases}\end{align}$$

for $k \in [0, \beta + 1]$.

Because the pair $\{\mathbf{S}, B\}$ is admissible for composition, for each mode in $B$, $B_{k} = (N_{k}) : (r_{k})$ for $k \in [0, \beta]$, by Definition 2.17, the pair $\{\mathbf{S}, B_{k}\}$ is admissible for composition. Therefore, by Definition 2.12, $M$ is left divisible by $r_{k}$ and the quotient $\frac{M}{r_{k}}$ is left divisible (not weakly left divisible) by $N_{k}$ for all $k \in [0, \beta]$.

It is trivial to see $M$ is left divisible by $1$. Let’s see if $M$ is also left divisible by $N_{\varphi(k - 1)}r_{\varphi(k - 1)}$ for all $k \in [1, \beta + 1]$.

Suppose $\varphi(k - 1) = h$ and Because $M$ is left divisible by $r_{h}$, we have

$$\begin{align}r_{\varphi(k - 1)} &= r_{h} \\&= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k - 1} - 1} \cdot c_{k - 1}\end{align}$$

where $c_{k}$ divides $M_{i_{k}}$.

$$\begin{align}N_{\varphi(k - 1)} &= N_{h} \\&= \frac{M_{i_{k}}}{c_{k - 1}} \cdot M_{i_{k - 1} + 1} \cdot \ldots \cdot M_{j_{k - 1} - 1} \cdot c_{k - 1}^{\prime}\end{align}$$

where $c_{k - 1}^{\prime}$ divides $M_{j_{k - 1}}$.

Thus, $M$ is also left divisible by $N_{\varphi(k - 1)}r_{\varphi(k - 1)}$, because

$$\begin{align}N_{\varphi(k - 1)}r_{\varphi(k - 1)} &= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k - 1} - 1} \cdot c_{k - 1} \cdot \frac{M_{i_{k}}}{c_{k - 1}} \cdot M_{i_{k} + 1} \cdot \ldots \cdot M_{j_{k - 1} - 1} \cdot c_{k - 1}^{\prime} \\&= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{j_{k - 1} - 1} \cdot c_{k - 1}^{\prime}\end{align}$$

where $c_{k - 1}^{\prime}$ divides $M_{j_{k - 1}}$.

Next, we will have to show $M$ is left divisible by $\frac{r_{\varphi(k)}}{N_{\varphi(k - 1)}r_{\varphi(k - 1)}}$ for all $k \in [1, \beta]$.

$$\begin{align}r_{\varphi(k)} &= M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c_{k} \\\end{align}$$

Because $r_{\varphi(k)} \geq N_{\varphi(k - 1)}r_{\varphi(k - 1)}$, we must have $i_{k} \geq j_{k - 1}$. Thus,

$$\begin{align}\frac{r_{\varphi(k)}}{N_{\varphi(k - 1)}r_{\varphi(k - 1)}}&= \frac{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c_{k}}{M_{0} \cdot M_{1} \cdot \ldots \cdot M_{j_{k - 1} - 1} \cdot c_{k - 1}^{\prime}} \\&= \frac{M_{j_{k - 1}} \cdot M_{j_{k - 1} + 1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c_{k}}{c_{k - 1}^{\prime}} \\&= \frac{M_{j_{k - 1}}}{c_{k - 1}^{\prime}} \cdot M_{j_{k - 1} + 1} \cdot \ldots \cdot M_{i_{k} - 1} \cdot c_{k} \\\end{align}$$

Thus, $M$ is left divisible by $\frac{r_{\varphi(k)}}{N_{\varphi(k - 1)}r_{\varphi(k - 1)}}$ for all $k \in [1, \beta]$.

It is trivial to see $M$ is left divisible by $\frac{M}{N_{\varphi(\beta)}r_{\varphi(\beta)}}$.

Therefore, the pair $\{\mathbf{S}, B^{\prime}_{k}\}$ is admissible for composition for all $k \in [0, \beta + 1]$.

By Definition 2.17, in order to show $\{\mathbf{S}, (B, \text{complement}(B, M))\}$ is admissible for composition, we also have to show the interval of definition for the pairs $\{\mathbf{S}, B_{k}\}_{0 \leq k \leq \beta}$ and $\{\mathbf{S}, B^{\prime}_{k}\}_{0 \leq k \leq \beta + 1}$ are disjoint.

By Proposition 2.7, the concatenated layout $(B, \text{complement}(B, M))$ is automatically satisfied with the disjoint argument.

Therefore, $\{\mathbf{S}, (B, \text{complement}(B, M))\}$ is admissible for composition.

This concludes the proof. $\square$

Note that in Definition 2.22, if the conditions of admission for composition follows Definition 2.12, our proof above will not be valid. That’s why CUTLASS enforces the conditions of admission for composition follows Definition 2.21.

Permutation Expressible As Layout Functions

This section explains how to retrieve all permutations that are expressible as layout functions in a structured way. The basic language of category theory is used to describe the process.

Definition 3.1 Ordered Factorization

We define the set $\text{ob}(\textbf{Fact})$ of ordered factorizations to consists of all expressions $[p_1 \ldots p_k]$ where $k \geq 0$ and the $p_{i}$ are primes (not necessarily distinct). The case $k = 0$ corresponds to the empty factorization, which we denote as $[\ ]$.

For example, the set $\text{ob}(\textbf{Fact})$ includes expressions such as $[\ ]$, $[2]$, $[3]$, $[22]$, $[23]$, $[32]$, $[232]$, etc.

Notation 3.3

Let $\underline{k}$ denote the set $\{1, 2, \ldots, k\}$ consisting of $k$ elements. (If $k = 0$, then $\underline{0} = \emptyset$ is the empty set.)

Definition 3.4 Category of Ordered Factorizations

We define the category $\textbf{Fact}$ of ordered factorizations as follows:

$$\begin{align}E^{\alpha} = [p_{\alpha(1)} \ldots p_{\alpha(n)}] \xrightarrow{\alpha_{E}} E = [p_1 \ldots p_k]\end{align}$$

in $\textbf{Fact}$. This defines the set of all morphisms with codomain $E$, and ranging over all $E$ thus defines the set of all morphisms in $\textbf{Fact}$.

$$\begin{align}E^{\alpha} = [p_{\alpha(1)} \ldots p_{\alpha(n)}] = [q_1 \ldots q_n]\end{align}$$

Let $\gamma = \alpha \circ \beta: \underline{m} \to \underline{k}$. Then the composition of morphisms

$$\begin{align}\alpha_{E}: E^{\alpha} = [p_{\alpha(1)} \ldots p_{\alpha(n)}] \xrightarrow{} E = [p_1 \ldots p_k]\end{align}$$

$$\begin{align}\beta_{E^{\alpha}}: E^{\beta} = [q_{\beta(1)} \ldots q_{\beta(m)}] \xrightarrow{} E^{\alpha} = [q_1 \ldots q_n]\end{align}$$

is given by $\gamma_{E}: E^{\gamma} \xrightarrow{} E$, where we used that $[q_{\beta(1)} \ldots q_{\beta(m)}] = [p_{\gamma(1)} \ldots p_{\gamma(m)}]$.

It’s easy to check that the composition of morphisms in $\textbf{Fact}$ is associative and has identities, which are the two axioms that composition in a category must satisfy, so Definition 3.4 really does define a category.

To see why the composition of morphisms is associative, suppose we have morphisms of finite sets $\alpha: \underline{n} \to \underline{k}$, $\beta: \underline{m} \to \underline{n}$, and $\gamma: \underline{l} \to \underline{m}$. Then we have

$$\begin{align}\alpha \circ (\beta \circ \gamma) = (\alpha \circ \beta) \circ \gamma\end{align}$$

To see why the composition of morphisms has identities, suppose for every $n$ we have a morphism of finite sets $\text{id}_{\underline{n}}: \underline{n} \to \underline{n}$, such that $\text{id}_{\underline{n}}(i) = i$ for all $i \in \underline{n}$. Then we have

$$\begin{align}E^{\text{id}_{\underline{n}}} = [p_{\text{id}_{\underline{n}}(1)} \ldots p_{\text{id}_{\underline{n}}(n)}] \xrightarrow{} E = [p_{1} \ldots p_{n}]\end{align}$$

For every morphism $\alpha: \underline{n} \to \underline{k}$, we have

$$\begin{align}\alpha \circ \text{id}_{\underline{n}} = \text{id}_{\underline{k}} \circ \alpha = \alpha\end{align}$$

Therefore, $\text{id}_{\underline{n}}$ is the identity morphism for $\underline{n}$.

Notation 3.5

Let $\Sigma_{k}$ denote the symmetric group on $k$ letters. Given an element $\varphi \in \Sigma_{k}$, we also denote the associated automorphism of $\underline{k}$ by $\varphi$.

In mathematics, a group is a set with an operation that associates an element of the set to every pair of elements of the set (as does every binary operation) and satisfies the following constraints: the operation is associative, it has an identity element, and every element of the set has an inverse element.

In this sense, the symmetric group is a set of all permutations of a set of $k$ elements with an operation of composition of permutations (applying one permutation after another).

Suppose $E = [222]$. Then every permutation $\varphi \in \Sigma_{3}$ defines an automorphism $E^{\varphi} = E \xrightarrow{} E$ in $\textbf{Fact}$.

Suppose $E = [232]$. Then the transposition $\sigma = (13) \in \Sigma_{3}$ defines an automorphism $E^{\sigma} = E \xrightarrow{} E$ in $\textbf{Fact}$. On the other hand, the transposition $\tau = (12) \in \Sigma_{3}$ defines a morphism $E^{\tau} = [322] \xrightarrow{} E = [232]$ in $\textbf{Fact}$.

Remark 3.7

Let $\textbf{FinSet}$ denote the category of finite sets (or rather a skeleton, with objects given by the sets $\underline{n}$ for $n \geq 0$). Given an object $\underline{k} \in \textbf{FinSet}$, let $\textbf{FinSet}^{/\underline{k}}$ denote the overcategory, whose objects are morphisms $[\alpha: \underline{n} \to \underline{k}]$ and whose morphisms are commuting triangles. Recall that this category has a final object given by the identity morphism $[\text{id}_{\underline{k}}]$.

Then for every expression $E = [p_1 \ldots p_k]$ of length $k$, we have a functor

$$\begin{align}F_{E}: \textbf{FinSet}^{/\underline{k}} \to \textbf{Fact}\end{align}$$

that sends the object $[\alpha: \underline{n} \to \underline{k}]$ to the expression $E^{\alpha}$ and the unique morphism $[\alpha] \xrightarrow{} [\text{id}_{\underline{k}}]$ to $\alpha_{E}: E^{\alpha} \xrightarrow{} E$. This functor has every morphism in $\textbf{Fact}$ with codomain $E$ in its image.

Suppose we have an object of morphism $[\alpha: \underline{n} \to \underline{k}]$ and another object of morphism $[\beta: \underline{m} \to \underline{k}]$ in $\textbf{FinSet}^{/\underline{k}}$. Then the morphism of the overcategory is a commuting triangle, whose remaining morphism is $[\gamma: \underline{m} \to \underline{n}]$, that maps $[\alpha: \underline{n} \to \underline{k}]$ to $[\beta: \underline{m} \to \underline{k}]$.

The identity morphism $[\text{id}_{\underline{k}}]$ is the final object of $\textbf{FinSet}^{/\underline{k}}$ because every other object of morphism in $\textbf{FinSet}^{/\underline{k}}$ has a unique morphism to $[\text{id}_{\underline{k}}]$.

In this case, given an object of morphism $[\alpha: \underline{n} \to \underline{k}]$, the commuting triangle has a remaining morphism $[\gamma: \underline{k} \to \underline{n}]$, and we will have to show that this remaining morphism in the commuting triangle is unique.

Given $\underline{k} = \{1, 2, \ldots, k\}$, the identity morphism maps $i \in \underline{k}$ to $i \in \underline{k}$. Given a morphism $[\alpha: \underline{n} \to \underline{k}]$, in which without loss of generality we assume $\alpha(i) = i$ for all $i \in [1, k]$, the remaining morphism $[\gamma: \underline{k} \to \underline{n}]$ must have $i = \alpha(i)$ for all $i \in [1, k]$, so that we have a commuting triangle that $i \in \underline{k} \xrightarrow{\gamma} \xrightarrow{\alpha} i \in \underline{k}$. Otherwise, for the remaining morphism $[\gamma: \underline{k} \to \underline{n}]$, if there exists an $i \in \underline{k}$ such that $i \neq \alpha(i)$, then the commuting triangle will not be valid because $i \in \underline{k} \xrightarrow{\gamma} \xrightarrow{\alpha} j \in \underline{k}$ and $i \neq j$. Therefore, the morphism of commuting triangle is unique.

By definition, let $C$ and $D$ be categories. A functor $F$ from $C$ to $D$ is a mapping that

In the category $\textbf{FinSet}^{/\underline{k}}$, we have $X = [\alpha: \underline{n} \to \underline{k}]$ and $Y = [\gamma: \underline{m} \to \underline{k}]$, $Z = [\text{id}_{\underline{k}}]$, and the morphisms are commuting triangles $f = [\alpha] \xrightarrow{} [\gamma]$, $g = [\gamma] \xrightarrow{} [\text{id}_{\underline{k}}]$, and $g \circ f = [\alpha] \xrightarrow{} [\text{id}_{\underline{k}}]$.

In the category $\textbf{Fact}$, by the functor $F_{E}$, we have $F_{E}(X) = E^{\alpha} = [p_{\alpha(1)} \ldots p_{\alpha(n)}]$, $F_{E}(Y) = E^{\gamma} = [p_{\gamma(1)} \ldots p_{\gamma(m)}]$, $F_{E}(Z) = E = [p_1 \ldots p_k]$, and the morphisms are $F_{E}(f) = \alpha_{E}: E^{\alpha} \xrightarrow{} E^{\gamma}$, $F_{E}(g) = \gamma_{E}: E^{\gamma} \xrightarrow{} E$, and $F_{E}(g) \circ F_{E}(f) = \alpha_{E}: E^{\alpha} \xrightarrow{} E$.

Remark 3.8

In fact, we can identify $\textbf{Fact}$ itself as a certain overcategory (or rather, a full subcategory thereof). Namely, let $\mathcal{P}$ denote the the infinite set of primes $\{2, 3, 5 \ldots\}$, let $\textbf{Set}$ be the category of sets, and let $\textbf{FinSet}^{/\mathcal{P}}$ be the full subcategory of $\textbf{Set}^{/\mathcal{P}}$ on those morphisms $X \xrightarrow{} \mathcal{P}$ where $X$ is a finite set. Then we have an equivalence of categories

$$\begin{align}\textbf{Fact} \simeq \textbf{FinSet}^{/\mathcal{P}}\end{align}$$

that sends an expression $E = [p_1 \ldots p_k]$ to the morphism $E_{\bullet}: k \xrightarrow{} \mathcal{P}$ given by $i \mapsto p_i$.

Under this equivalence, the functor $F_{E}$ of Remark 3.7 identifies with the functor

$$\begin{align}\textbf{FinSet}^{/k} \simeq \left(\textbf{FinSet}^{/\mathcal{P}}\right)^{/E_{\bullet}} \xrightarrow{} \textbf{FinSet}^{/\mathcal{P}}\end{align}$$

that forgets the map to $E_{\bullet}$.

To understand this, let’s consider the following example.

Suppose we have an object $E = [232]$ from $\textbf{Fact}$. Then we have a morphism $E_{\bullet}: \underline{3} \xrightarrow{} \mathcal{P}$ given by

$$\begin{align}i = 1 &\mapsto p_{1} = 2 \\i = 2 &\mapsto p_{2} = 3 \\i = 3 &\mapsto p_{3} = 2\end{align}$$

Every object of morphisms in $\textbf{FinSet}^{/\mathcal{P}}$ can be mapped from an object from $\textbf{Fact}$ and thus we have an equivalence of categories $\textbf{Fact} \simeq \textbf{FinSet}^{/\mathcal{P}}$.

Because of the functor $F_{E}$ of Remark 3.7, with the equivalence above, we have

$$\begin{align}\textbf{FinSet}^{/\underline{k}} \to \textbf{FinSet}^{/\mathcal{P}}\end{align}$$

Definition 3.9

Suppose $E = [p_1 \ldots p_k]$ and $\alpha: \underline{n} \to \underline{k}$. We define a layout $L_{(E, \alpha)}$ as follows:

We also let $f_{(E, \alpha)}$ denote the associated layout function.

Suppose $E = [23]$ and $\varphi = (12) \in \Sigma_{2}$ is the nontrivial transposition. Then $L_{(E, \varphi)} = (3, 2) : (2, 1)$.

Suppose $E = [23]$, $\varphi(1) = 2$, $\varphi(2) = 1$, $\varphi(3) = 2$. Then $L_{(E, \varphi)} = (3, 2, 3) : (2, 1, 2)$. This layout seems strange because its layout function is not an injection mapping. However, it is still a valid layout by Definition 2.2.

Suppose $E = [222]$ and $\varphi = (231) \in \Sigma_{3}$, so $\varphi$ is a cycle of order $3$ with $\varphi(1) = 2$, $\varphi(2) = 3$, and $\varphi(3) = 1$. Then $L_{(E, \varphi)} = (2, 2, 2) : (2, 4, 1)$.

We could now see why $p_{i}$ is prime number. It’s used for constructing the stride tuple of any kind for the layout, because any natural number can be uniquely factored into a product of prime numbers.

Remark 3.11

Let $E = [p_1 \ldots p_k]$ and $\alpha: \underline{n} \to \underline{k}$. Let $N = p_{1} \cdot p_{2} \cdot \ldots \cdot p_{k}$ and $N^{\alpha} = p_{\alpha(1)} \cdot p_{\alpha(2)} \cdot \ldots \cdot p_{\alpha(n)}$. In what follows, consider the canonical isomorphisms

$$\begin{align}[0, N) &\cong [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k) \\[0, N^{\alpha}) &\cong [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)})\end{align}$$

Then the associated layout function $f_{(E, \alpha)}: [0, N^{\alpha}) \to [0, N) \subset \mathbb{N}$ can be described as the multilinear function

$$\begin{align}[0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)}) \xrightarrow{} [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)\end{align}$$

that sends the basis vector $\delta_{i}$ of one vector space to the basis vector $\beta_{\alpha(i)}$ of the other for $i \in [1, n]$.

In particular, if $\alpha$ is itself a bijection, then $f_{(E, \alpha)}$ restricts to an automorphism of $[0, N)$.

Proof

We noticed that $f_{(E, \alpha)}: [0, N^{\alpha}) \to [0, N) \subset \mathbb{N}$. The domain of $f_{(E, \alpha)}$ is $[0, N^{\alpha})$ and it is obvious. The codomain of $f_{(E, \alpha)}$ is $[0, N) \subset \mathbb{N}$, however, is less obvious.

$$\begin{align}f_{(E, \alpha)} = x_{\alpha_{(1)}} d_{1} + x_{\alpha_{(2)}} d_{2} + \ldots + x_{\alpha_{(n)}} d_{n}\end{align}$$

where $x_{\alpha_{(i)}} \in [0, p_{\alpha(i)})$ and $d_{i} = \prod_{j < \alpha(i)} p_{j}$.

So we have

$$\begin{align}\max \left( f_{(E, \alpha)} \right) &= (p_{\alpha(1)} - 1) d_{1} + (p_{\alpha(2)} - 1) d_{2} + \ldots + (p_{\alpha(n)} - 1) d_{n} \\&= (p_{\alpha(1)} - 1) \prod_{j < \alpha(1)} p_{j} + (p_{\alpha(2)} - 1) \prod_{j < \alpha(2)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\\end{align}$$

Without losing generality, we assume $p_{\alpha(1)} \leq p_{\alpha(2)} \leq \ldots \leq p_{\alpha(n)}$. Then we have

$$\begin{align}\max \left( f_{(E, \alpha)} \right) &= (p_{\alpha(1)} - 1) \prod_{j < \alpha(1)} p_{j} + (p_{\alpha(2)} - 1) \prod_{j < \alpha(2)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&\leq p_{\alpha(1)} \prod_{j < \alpha(1)} p_{j} + (p_{\alpha(2)} - 1) \prod_{j < \alpha(2)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&= \prod_{j < \alpha(2)} p_{j} + (p_{\alpha(2)} - 1) \prod_{j < \alpha(2)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&= p_{\alpha(2)} \prod_{j < \alpha(2)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&\leq \prod_{j < \alpha(3)} p_{j} + \ldots + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&\leq \ldots \\&\leq p_{\alpha(n)} \prod_{j < \alpha(n - 1)} p_{j} + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&= \prod_{j < \alpha(n)} p_{j} + (p_{\alpha(n)} - 1) \prod_{j < \alpha(n)} p_{j} \\&= p_{\alpha(n)} \prod_{j < \alpha(n)} p_{j} \\&= \prod_{j \leq \alpha(n)} p_{j} \\&\leq \prod_{j \leq k} p_{j} \\&= N\end{align}$$

Thus, we have $f_{(E, \alpha)}: [0, N^{\alpha}) \to [0, N) \subset \mathbb{N}$.

Because $f_{(E, \alpha)}$ is a multilinear function, and because of the canonical isomorphisms, $f_{(E, \alpha)}$ can be described as the multilinear function

$$\begin{align}[0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)}) \xrightarrow{} [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)\end{align}$$

We denote a vector space $V = [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)})$ and a vector space $W = [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)$. Then the layout function $f_{(E, \alpha)}$ is a linear map $V \xrightarrow{} W$.

Suppose $v_{1}, v_{2}, av_{1}, bv_{2}, av_{1} + bv_{2} \in V$, $f_{(E, \alpha)}(v_{1}) = w_{1}$, and $f_{(E, \alpha)}(v_{2}) = w_{2}$. Then we have

$$\begin{align}f_{(E, \alpha)}(v_{1}) &= v_{1, 1} d_{1} + v_{1, 2} d_{2} + \ldots + v_{1, n} d_{n} \\f_{(E, \alpha)}(v_{2}) &= v_{2, 1} d_{1} + v_{2, 2} d_{2} + \ldots + v_{2, n} d_{n} \\f_{(E, \alpha)}(av_{1}) &= av_{1, 1} d_{1} + av_{1, 2} d_{2} + \ldots + av_{1, n} d_{n} \\&= af_{(E, \alpha)}(v_{1}) \\f_{(E, \alpha)}(bv_{2}) &= bv_{2, 1} d_{1} + bv_{2, 2} d_{2} + \ldots + bv_{2, n} d_{n} \\&= bf_{(E, \alpha)}(v_{2}) \\f_{(E, \alpha)}(av_{1} + bv_{2}) &= (av_{1} + bv_{2})_{1} d_{1} + (av_{1} + bv_{2})_{2} d_{2} + \ldots + (av_{1} + bv_{2})_{n} d_{n} \\&= af_{(E, \alpha)}(v_{1}) + bf_{(E, \alpha)}(v_{2})\end{align}$$

So $f_{(E, \alpha)}: V \xrightarrow{} W$ is indeed a linear (multilinear) map.

Given an index $1 \leq i \leq \alpha$, let $\delta_{i} \in \mathbb{N}^{\times \alpha}$ denote the coordinate that is zero everywhere except in the $k$-th position, where it is 1. Note here the indexing is 1-based, instead of the similar one used in Proposition 2.14 which is 0-based. $\delta_{i}$ is the basis vector of the vector space $V$ for $1 \leq i \leq \alpha$.

We send $\delta_{i}$ to $f_{(E, \alpha)}$ for $1 \leq i \leq \alpha$. Then we have

$$\begin{align}f_{(E, \alpha)}(\delta_{i}) &= d_{i} \\&= \prod_{j < \alpha(i)} p_{j}\end{align}$$

Given the canonical isomorphism $[0, N) \cong [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)$, we have the multilinear function $g: W \xrightarrow{} \mathbb{N}$

$$\begin{align}g(w) = w_{1} + w_{2} p_{1} + w_{3} p_{1} p_{2} + \ldots + w_{k} \prod_{j < k} p_{j}\end{align}$$

Given an index $1 \leq i \leq k$, let $\beta_{i} \in \mathbb{N}^{\times k}$ denote the coordinate that is zero everywhere except in the $k$-th position, where it is 1. $\beta_{i}$ is the basis vector of the vector space $W$ for $1 \leq i \leq k$.

Thus, we have

$$\begin{align}f_{(E, \alpha)}(\delta_{i})&= \prod_{j < \alpha(i)} p_{j} \\&= g(\beta_{\alpha(i)})\end{align}$$

This means the basis vector $\delta_{i}$ in the vector space $V$ is sent to the basis vector $\beta_{\alpha(i)}$ in the vector space $W$ by the multilinear function.

Suppose $v = c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha} \in V$. Then we have

$$\begin{align}f_{(E, \alpha)}(v) &= f_{(E, \alpha)}(c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha}) \\&= (c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha})_{1} d_{1} + (c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha})_{2} d_{2} + \ldots + (c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha})_{\alpha} d_{\alpha} \\&= c_{1} d_{1} + c_{2} d_{2} + \ldots + c_{\alpha} d_{\alpha} \\&= c_{1} f_{(E, \alpha)}(\delta_{1}) + c_{2} f_{(E, \alpha)}(\delta_{2}) + \ldots + c_{\alpha} f_{(E, \alpha)}(\delta_{\alpha}) \\&= c_{1} g(\beta_{\alpha(1)}) + c_{2} g(\beta_{\alpha(2)}) + \ldots + c_{\alpha} g(\beta_{\alpha(\alpha)}) \\&= g(c_{1} \beta_{\alpha(1)} + c_{2} \beta_{\alpha(2)} + \ldots + c_{\alpha} \beta_{\alpha(\alpha)}) \\\end{align}$$

Therefore, we have set up the basis vector mapping for the multilinear function $f_{(E, \alpha)}: V \xrightarrow{} W$. Given $v = c_{1} \delta_{1} + c_{2} \delta_{2} + \ldots + c_{\alpha} \delta_{\alpha} \in V$, it maps to $w = c_{1} \beta_{\alpha(1)} + c_{2} \beta_{\alpha(2)} + \ldots + c_{\alpha} \beta_{\alpha(\alpha)} \in W$.

This concludes the proof. $\square$

Lemma 3.12

Elaborating on Remark 3.11, we have the following lemma, which indicates that composition in the category $\textbf{Fact}$ is compatible with the composition of layout functions.

Suppose we have morphisms of finite sets $\alpha: \underline{n} \to \underline{k}$, $\beta: \underline{m} \to \underline{n}$, and an expression $E = [p_1 p_2 \ldots p_k]$. Write $\gamma = \alpha \circ \beta$. Consider the composition

$$\begin{align}\gamma_{E}: E^{\gamma} = (E^{\alpha})^{\beta} \xrightarrow{\beta_{E^{\alpha}}} E^{\alpha} \xrightarrow{\alpha_{E}} E\end{align}$$

in $\textbf{Fact}$. Then the associated layout functions satisfy the composition equality

$$\begin{align}f_{(E, \gamma)} = f_{(E, \alpha)} \circ f_{(E^{\alpha}, \beta)}\end{align}$$

Proof

Let $N = p_{1} \cdot p_{2} \cdot \ldots \cdot p_{k}$, $N^{\alpha} = p_{\alpha(1)} \cdot p_{\alpha(2)} \cdot \ldots \cdot p_{\alpha(n)}$, and $N^{\gamma} = p_{\gamma(1)} \cdot p_{\gamma(2)} \cdot \ldots \cdot p_{\gamma(m)}$. We use the canonical isomorphisms

$$\begin{align}[0, N) &\cong [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k) \\[0, N^{\alpha}) &\cong [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)}) \\[0, N^{\gamma}) &\cong [0, p_{\gamma(1)}) \times [0, p_{\gamma(2)}) \times \ldots \times [0, p_{\gamma(m)})\end{align}$$

to write the domains and codomains of the layout functions in question.

More specifically, we have

$$\begin{align}f_{(E, \alpha)}: [0, N^{\alpha}) &\to [0, N) \\f_{(E^{\alpha}, \beta)}: [0, N^{\gamma}) &\to [0, N^{\alpha}) \\f_{(E, \gamma)}: [0, N^{\gamma}) &\to [0, N)\end{align}$$

We are trying to equate the multilinear function

$$\begin{align}f_{(E, \gamma)}: [0, p_{\gamma(1)}) \times [0, p_{\gamma(2)}) \times \ldots \times [0, p_{\gamma(m)}) \xrightarrow{} [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)\end{align}$$

with the composition of the two multilinear functions

$$\begin{align}f_{(E, \alpha)}&: [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)}) \xrightarrow{} [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k) \\f_{(E^{\alpha}, \beta)}&: [0, p_{\gamma(1)}) \times [0, p_{\gamma(2)}) \times \ldots \times [0, p_{\gamma(m)}) \xrightarrow{} [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)})\end{align}$$

We denote a vector space $V = [0, p_{\alpha(1)}) \times [0, p_{\alpha(2)}) \times \ldots \times [0, p_{\alpha(n)})$, a vector space $W = [0, p_1) \times [0, p_2) \times \ldots \times [0, p_k)$, and a vector space $U = [0, p_{\gamma(1)}) \times [0, p_{\gamma(2)}) \times \ldots \times [0, p_{\gamma(m)})$. The basis vectors of $V$, $W$, and $U$ are $\delta_{i}$, $\sigma_{j}$, and $\tau_{l}$ for $1 \leq i \leq n$, $1 \leq j \leq k$, and $1 \leq l \leq m$.

Based on the basis vector mapping by Remark 3.11, given $u = c_{1} \tau_{1} + c_{2} \tau_{2} + \ldots + c_{m} \tau_{m} \in U$, by $f_{(E^{\alpha}, \beta)}$, it maps to $v = c_{1} \delta_{\beta(1)} + c_{2} \delta_{\beta(2)} + \ldots + c_{m} \delta_{\beta(m)} \in V$. Then by $f_{(E, \alpha)}$, it maps to $w = c_{1} \sigma_{\alpha(\beta(1))} + c_{2} \sigma_{\alpha(\beta(2))} + \ldots + c_{m} \sigma_{\alpha(\beta(m))} \in W$.

Given $u = c_{1} \tau_{1} + c_{2} \tau_{2} + \ldots + c_{m} \tau_{m} \in U$, by $f_{(E, \gamma)}$, because $\gamma = \alpha \circ \beta$, $\gamma(i) = \alpha(\beta(i))$, it maps to $w^{\prime} = c_{1} \sigma_{\gamma(1)} + c_{2} \sigma_{\gamma(2)} + \ldots + c_{m} \sigma_{\gamma(m)} = c_{1} \sigma_{\alpha(\beta(1))} + c_{2} \sigma_{\alpha(\beta(2))} + \ldots + c_{m} \sigma_{\alpha(\beta(m))} \in W$.

Because $w = w^{\prime}$, we have $f_{(E, \gamma)} = f_{(E, \alpha)} \circ f_{(E^{\alpha}, \beta)}$.

This concludes the proof. $\square$

In Lemma 3.12, the per-mode condition of admissibility for composition (Definition 2.12) is satisfied. To see this, we have

$$\begin{align}E &= [p_1 p_2 \ldots p_k] \\E^{\alpha} &= [p_{\alpha(1)} p_{\alpha(2)} \ldots p_{\alpha(n)}] \\E^{\gamma} &= [p_{\gamma(1)} p_{\gamma(2)} \ldots p_{\gamma(m)}]\end{align}$$

$$\begin{align}L_{(E, \alpha)} &= \left(p_{\alpha(1)}, p_{\alpha(2)}, \ldots, p_{\alpha(n)}\right) : \left(d_{1}, d_{2}, \ldots, d_{n}\right) \\\end{align}$$

where $d_{i} = \prod_{j < \alpha(i)} p_{j}$.

$$\begin{align}L_{(E^{\alpha}, \beta)} &= \left(p_{\alpha(\beta(1))}, p_{\alpha(\beta(2))}, \ldots, p_{\alpha(\beta(m))}\right) : \left(d^{\prime}_{1}, d^{\prime}_{2}, \ldots, d^{\prime}_{m}\right) \\\end{align}$$

where $d^{\prime}_{i} = \prod_{j < \beta(i)} p_{\alpha(j)}$.

Because

$$\begin{align}M &= p_{\alpha(1)} \cdot p_{\alpha(2)} \cdot \ldots \cdot p_{\alpha(n)} \\&= \left( \prod_{j < \beta(i)} p_{\alpha(j)} \right) \cdot p_{\alpha(\beta(i))} \cdot p_{\alpha(\beta(i) + 1)} \cdot \ldots \cdot p_{\alpha(n)} \\&= \left( \prod_{j < \beta(i)} p_{\alpha(j)} \right) \cdot M^{\prime}\end{align}$$

Thus, $M$ is left divisible by $d^{\prime}_{i}$, $M^{\prime}$ is weakly left divisible and also left divisible by $p_{\alpha(\beta(i))}$, and the per-mode condition of admissibility for composition is satisfied.

The disjointness condition in Definition 2.17 is satisfied when $\beta: \underline{m} \to \underline{n}$ is an injective function and may be violated when it is not.

When $\beta: \underline{m} \to \underline{n}$ is injective, we have $m \leq n$, and $\beta{(i)} \neq \beta{(j)}$ for $i \neq j$. By Definition 2.16, for each mode $i \in [1, m]$, we have $N_{i} = p_{\alpha(\beta(i))}$, $I_{i} = [d^{\prime}_{i}, d^{\prime}_{i} (N_{i} - 1)]$. $M^{\prime} = p_{\alpha(1)} \cdot p_{\alpha(2)} \cdot \ldots \cdot p_{\alpha(n - 1)}$. So the interval of definition is $J_{i} = I_{i} \cap [1, M^{\prime})$. Because $d^{\prime}_{i} = \prod_{j < \beta(i)} p_{\alpha(j)} \geq 1$, $d^{\prime}_{i}(N_{i} - 1) = \prod_{j < \beta(i)} p_{\alpha(j)} \cdot (p_{\alpha(\beta(i))} - 1) = \prod_{j \leq \beta(i)} p_{\alpha(j)} - \prod_{j < \beta(i)} p_{\alpha(j)} < M^{\prime}$. Thus, $J_{i} = I_{i} = [\prod_{j < \beta(i)} p_{\alpha(j)}, \prod_{j \leq \beta(i)} p_{\alpha(j)} - \prod_{j < \beta(i)} p_{\alpha(j)}]$. Suppose we have a different mode $k$, $k \neq i$. Then $J_{k} = I_{k} = [\prod_{j < \beta(k)} p_{\alpha(j)}, \prod_{j \leq \beta(k)} p_{\alpha(j)} - \prod_{j < \beta(k)} p_{\alpha(j)}]$. Because $\beta(i) \neq \beta(k)$, without losing generality, we assume $\beta(i) < \beta(k)$. Then we have

$$\begin{align}\prod_{j < \beta(k)} p_{\alpha(j)} - \left( \prod_{j \leq \beta(i)} p_{\alpha(j)} - \prod_{j < \beta(i)} p_{\alpha(j)} \right)&= \prod_{j < \beta(k)} p_{\alpha(j)} - \prod_{j \leq \beta(i)} p_{\alpha(j)} + \prod_{j < \beta(i)} p_{\alpha(j)} \\&> 0\end{align}$$

Thus, $J_{i} \cap J_{k} = \emptyset$ for any $i \neq k$. The disjointness condition is satisfied.

When $\beta: \underline{m} \to \underline{n}$ is not injective, we don’t have $\beta{(i)} \neq \beta{(j)}$ for $i \neq j$. The disjointness condition may be violated.

So Lemma 3.12 actually proves Theorem 2.18 for layouts that has any arbitrary strides (not yet any arbitrary shapes) of the second layout that satisfies Definition 2.17.

Definition 3.14

We now define a “realization” functor from the category $\textbf{Fact}$ to the category $\textbf{FinSet}$ that sends morphisms of ordered factorizations to their associated layout functions.

Let $R: \textbf{Fact} \to \textbf{FinSet}$ be the functor defined as follows:

By Lemma 3.12, $R: \textbf{Fact} \to \textbf{FinSet}$ does indeed define a functor since it respects the composition of morphisms and identities as well.

We note that, as mentioned previously, $R$ does not contain every possible function expressible as a layout function in its image. However, it does contain every automorphism of $[0, N) \xrightarrow{\cong} [0, N)$ expressible as a layout function in its image.

Proposition 3.15

Let $N > 0$ be a positive integer and let $f: [0, N) \to [0, N)$ be an automorphism such that there exists a layout $L$ of size $N$ with $f = f_{L}$. Then $f_{L}$ is in the image of the realization functor $R$.

Proof

Without loss of generality, we may suppose that the shape tuple of $L$ is given by $(p_{1}, p_{2}, \ldots, p_{k})$ where the $p_{i}$ are all prime numbers and $N = p_{1} \cdot p_{2} \cdot \ldots \cdot p_{k}$.

In order for $f_{L}$ to be an automorphism of $[0, N)$, the sorted $L$, $L^{\varphi}$, must be of the form

$$\begin{align}L^{\varphi} = \left(p_{\varphi(1)}, p_{\varphi(2)}, \ldots, p_{\varphi(k)}\right) : \left(1, p_{\varphi(1)}, p_{\varphi(1)} p_{\varphi(2)}, \ldots, \prod_{1 \leq i < k} p_{\varphi(i)}\right)\end{align}$$

for some permutation $\varphi \in \Sigma_{k}$. This means that if we let $\psi = \varphi^{-1}$ be the inverse permutation, then

$$\begin{align}\psi_{E}: E^{\psi} = [p_{1} p_{2} \ldots p_{k}] = [p_{\psi(\varphi(1))} p_{\psi(\varphi(2))} \ldots p_{\psi(\varphi(k))}] \xrightarrow{} E = [p_{\varphi(1)} p_{\varphi(2)} \ldots p_{\varphi(k)}]\end{align}$$

is a morphism in $\textbf{Fact}$ that $R(\psi_{E}) = f_{L}$.

This concludes the proof. $\square$

Remark 3.16

One way to interpret Proposition 3.15 is that if we take the maximal subgroupoid $\textbf{Fact}^{\simeq}$ inside $\textbf{Fact}$, i.e., the subcategory of all invertible morphisms, then

$$\begin{align}R: \textbf{Fact}^{\simeq} \to \textbf{FinSet}\end{align}$$

carves out exactly those permutations expressible as layouts. Our motivation for this description is that for a fixed integer $N > 0$, the subset $\Sigma^{L}_{N}$ of $\Sigma_{N}$ on those automorphisms expressible as layout functions is typically not a subgroup (being not generally closed under the group multiplication, i.e., composition).

Instead, if we let

$$\begin{align}\textbf{Fact}^{\simeq}_{N} \subset \textbf{Fact}^{\simeq}\end{align}$$

be the full subgroupoid of those objects $[p_{1} p_{2} \ldots p_{k}]$ with $N = p_{1} \cdot p_{2} \cdot \ldots \cdot p_{k}$, then the $\Sigma^{L}_{N}$ consists of those morphisms in the image of $R$ on $\textbf{Fact}^{\simeq}_{N}$. However, we see that $\Sigma^{L}_{N}$ is closed under the operation of taking the group inverse (the objects taking permutations to their inverses are also in $\textbf{Fact}^{\simeq}_{N}$). Moreover, in the special case that $N$ is a prime power $p^{k}$, then $\Sigma^{L}_{N}$ is in fact a subgroup and is isomorphic to the symmetric group $\Sigma_{k}$. This corresponds to $\textbf{Fact}^{\simeq}_{p^{k}}$ being a groupoid with a single object $[pp \ldots p]$, i.e., a group.

References

CuTe Layout Algebra

https://leimao.github.io/article/CuTe-Layout-Algebra/

Author

Lei Mao

Posted on

10-20-2024

Updated on

10-20-2024

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