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.
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