# A Matrix for the New HPL-AI Benchmark

My colleagues are looking for a matrix to be used in a new benchmark. They've come to the right place.

### Contents

A couple of weeks ago I got this email from Jack Dongarra at the University of Tennessee.

`For this new HPL-AI benchmark, I'm looking for a matrix that is not symmetric, is easily generated (roughly O(n^2) ops to generate), is dense, doesn’t require pivoting to control growth, and has a smallish condition number (<O(10^5)).`

`There are two steps for the benchmark:`

`1) do LU on the matrix in short precision and compute an approximation solution x. (On Nvidia this is 6 times faster in half precision then in DP.)`

`2) do GMRES on with L and U as preconditions with x as the starting point (this usually takes 3-4 iterations to converge.)`

`The benchmark details are here: http://bit.ly/hpl-ai.`

`Any ideas for a matrix?`

#### Symmetric Hilbert Matrix

One of the first matrices I ever studied was the Hilbert matrix. With the singleton expansion features now in MATLAB, the traditional symmetric Hilbert matrix can be generated by

```
format rat
n = 5;
i = 1:n
j = i'
H = 1./(i+j-1)
```

i = 1 2 3 4 5 j = 1 2 3 4 5 H = 1 1/2 1/3 1/4 1/5 1/2 1/3 1/4 1/5 1/6 1/3 1/4 1/5 1/6 1/7 1/4 1/5 1/6 1/7 1/8 1/5 1/6 1/7 1/8 1/9

When you add the row vector `i` to the column vector `j`, the resulting `i+j` is a matrix. Compare this with the *inner product*, `i*j`, which produces a scalar, and the *outer product*, `j*i`, which produces another matrix. The expression `i+j` can be thought of as an "outer sum".

Two more scalar expansions complete the computation; `i+j-1` subtracts 1 from each element of the sum and `1./(i+j-1)` divides 1 by each element in the denominator. Both of these scalar expansions have been in MATLAB from its earliest days.

The elements of Hilbert matrices depend only on the sum `i+j`, so they are constant on the antidiagonals. Such matrices are known as Cauchy matrices. There are fast algorithms for solving system of equations involving Cauchy matrices in $O(n^2)$ time.

#### Nonsymmetric Hilbert Matrix

Some years ago, I wanted a nonsymmetric test matrix, so I changed the plus sign in the denominator of the Hilbert matrix to a minus sign. But this created a division by zero along the subdiagonal, so I also changed the -1 to +1/2. I call this the "nonsymmetric Hilbert matrix."

H = 1./(i-j+1/2)

H = 2 2/3 2/5 2/7 2/9 -2 2 2/3 2/5 2/7 -2/3 -2 2 2/3 2/5 -2/5 -2/3 -2 2 2/3 -2/7 -2/5 -2/3 -2 2

The elements depend only on `i-j` and so are constant on the diagonals. The matrices are Toeplitz and, again, such systems can be solved with $O(n^2)$ operations.

I was astonished when I first computed the singular values of nonsymmetric Hilbert matrices -- most of them were nearly equal to $\pi$. Already with only the 5-by-5 we see five decimal places.

```
format long
sigma = svd(H)
```

sigma = 3.141589238341321 3.141272220456508 3.129520593545258 2.921860834913688 1.461864493324889

This was a big surprise. The eventual explanation, provided by Seymour Parter, involves a theorem of Szego from the 1920's relating the eigenvalues of Toeplitz matrices to Fourier series and square waves.

#### Benchmark Matrix

I have been working with Jack's colleague Piotr Luszczek. This is our suggestion for the HPL-AI Benchmark Matrix. It depends upon two parameters, `n` and `mu`. As `mu` goes from 1 to -1, the core of this matrix goes from the symmetric to the nonsymmetric Hilbert matrix. Adding `n` in the denominator and 1's on the diagonal produces diagonal dominance.

A = 1./(i + mu*j + n) + double(i==j);

clear A type A

function Aout = A(n,mu) % A(n,mu). Matrix for the HPL-AI Benchmark. mu usually in [-1, 1]. i = 1:n; j = i'; Aout = 1./(i+mu*j+n) + eye(n,n); end

If `mu` is not equal to +1, this matrix is not Cauchy. And, if `mu` is not equal to -1, it is not Toeplitz. Furthermore, if `mu` is greater than -1, it is diagonally dominant and, as we shall see, very well-conditioned.

#### Example

Here is the 5x5 with rational output. When `mu` is +1, we get a section of a symmetric Hilbert matrix plus 1's on the diagonal.

```
format rat
mu = 1;
A(n,mu)
```

ans = 8/7 1/8 1/9 1/10 1/11 1/8 10/9 1/10 1/11 1/12 1/9 1/10 12/11 1/12 1/13 1/10 1/11 1/12 14/13 1/14 1/11 1/12 1/13 1/14 16/15

And when `mu` is -1, the matrix is far from symmetric. The element in the far southwest corner almost dominates the diagonal.

mu = -1; A(n,mu)

ans = 6/5 1/6 1/7 1/8 1/9 1/4 6/5 1/6 1/7 1/8 1/3 1/4 6/5 1/6 1/7 1/2 1/3 1/4 6/5 1/6 1 1/2 1/3 1/4 6/5

#### Animation

Here is an animation of the matrix without the addition of `eye(n,n)`. As `mu` goes to -1 to 1, `A(n,mu)-eye(n,n)` goes from nonsymmetric to symmetric.

#### Gaussian elimination

If mu > -1, the largest element is on the diagonal and `lu(A(n,mu))` does no pivoting.

```
format short
mu = -1
[L,U] = lu(A(n,mu))
mu = 1
[L,U] = lu(A(n,mu))
```

mu = -1 L = 1.0000 0 0 0 0 0.2083 1.0000 0 0 0 0.2778 0.1748 1.0000 0 0 0.4167 0.2265 0.1403 1.0000 0 0.8333 0.3099 0.1512 0.0839 1.0000 U = 1.2000 0.1667 0.1429 0.1250 0.1111 0 1.1653 0.1369 0.1168 0.1019 0 0 1.1364 0.1115 0.0942 0 0 0 1.1058 0.0841 0 0 0 0 1.0545 mu = 1 L = 1.0000 0 0 0 0 0.1094 1.0000 0 0 0 0.0972 0.0800 1.0000 0 0 0.0875 0.0729 0.0626 1.0000 0 0.0795 0.0669 0.0580 0.0513 1.0000 U = 1.1429 0.1250 0.1111 0.1000 0.0909 0 1.0974 0.0878 0.0800 0.0734 0 0 1.0731 0.0672 0.0622 0 0 0 1.0581 0.0542 0 0 0 0 1.0481

#### Condition

Let's see how the 2-condition number, $\sigma_1/\sigma_n$, varies as the parameters vary. Computing the singular values for all the matrices in the following plots takes a little over 10 minutes on my laptop.

#### sigma_1

load HPLAI.mat s1 sn n mu surf(mu,n,s1) view(25,12) xlabel('mu') ylabel('n') set(gca,'xlim',[-0.9 1.0], ... 'xtick',[-0.9 -0.5 0 0.5 1.0], ... 'ylim',[0 2500]) title('\sigma_1')

The colors vary as `mu` varies from -0.9 up to 1.0. I am staying away from `mu` = -1.0 where the matrix looses diagonal dominance. The other horizontal axis is the matrix order `n`. I have limited `n` to 2500 in these experiments.

The point to be made is that if `mu` > -0.9, then the largest singular value is bounded, in fact `sigma_1` < 2.3.

#### sigma_n

surf(mu,n,sn) view(25,12) xlabel('mu') ylabel('n') set(gca,'xlim',[-0.9 1.0], ... 'xtick',[-0.9 -0.5 0 0.5 1.0], ... 'ylim',[0 2500]) title('\sigma_n')

If `mu` > -0.9, then the smallest singular value is bounded away from zero, in fact `sigma_n` > .75.

#### sigma_1/sigma_n

surf(mu,n,s1./sn) view(25,12) xlabel('mu') ylabel('n') set(gca,'xlim',[-0.9 1.0], ... 'xtick',[-0.9 -0.5 0 0.5 1.0], ... 'ylim',[0 2500]) title('\sigma_1/\sigma_n')

If `mu` > -0.9, then the condition number in the 2-norm is bounded, `sigma_1/sigma_n` < 2.3/.75 $\approx$ 3.0 .

#### Overflow

The HPL-AI benchmark would begin by computing the LU decomposition of this matrix using 16-bit half-precision floating point arithmetic, FP16. So, there is a signigant problem if the matrix order `n` is larger than 65504. This value is `realmax`, the largest number that can be represented in FP16. Any larger value of `n` overflows to `Inf`.

Some kind of scaling is going to have to be done for `n` > 65504. Right now, I don't see what this might be.

The alternative half-precision format to FP16 is Bfloat16, which has three more bits in the exponent to give `realmax` = 3.39e38. There should be no overflow problems while generating this matrix with Bfloat16.

#### Thanks

Thanks to Piotr Luszczek for working with me on this. We have more work to do.

## 评论

要发表评论，请点击 此处 登录到您的 MathWorks 帐户或创建一个新帐户。