CS267: Lecture 2 (part 1), Jan 18 1996

Designing fast linear algebra kernels in the presence of memory hierarchies

The interfaces of these kernels have been standardized as the Basic Linear Algebra Subroutines or BLAS. The motivation for software and hardware vendors agreeing to a standard definition of these routines is their popularity: The BLAS appear in many applications (as we will see), so if vendors agree on a common interface, and supply an optimized BLAS library for each computer, then users can write programs calling the BLAS, and move their programs from one machine to another and still expect high performance.

For historical reasons, these BLAS as classified into three categories:

The Level 1 BLAS or BLAS1 operate mostly on vectors (1D arrays), or pairs of vectors. If the vectors have length n, these routines perform O(n) operations, and return either a vector or a scalar. Examples are

The Level 2 BLAS or BLAS2 operate mostly on a matrix (2D array) and a vector (or vectors), returning a matrix or a vector. If the array is n-by-n, O(n^2) operations are performed. Here are some examples.

The Level 3 BLAS or BLAS3 operate on pairs or triples of matrices, returning a matrix. Here are some examples.

Mathematically, many these operations are essentially equivalent. For example, the dot product can be thought of as the product of a 1-by-n vector and an n-by-1 vector, and each entry of A*B is the dot product of a row of A and a column of B. So essentially any linear algebra algorithm can be expressed in terms of any of these constructs. So why distinguish among them? The answer is performance, as indicated on the following plot of BLAS performance on the IBM RS 6000/590.

The slide plots the performance in megaflops of the BLAS on the RS 6000/590, versus matrix or vector dimension. These BLAS are specially optimized by IBM to take advantage of all features of the RS 6000/590 architecture (we will discuss this in more detail later). The top curve is for matrix-matrix multiply (BLAS3), the second highest curve (for large dimension) is matrix-vector multiply (BLAS2), and the lowest curve is for saxpy (BLAS1) on the RS6000. (These plots assume the data is not initially in the cache, or fast memory.) As you can see there is potentially a big speed advantage if an algorithm can be expressed in terms of the BLAS3 instead of BLAS2 or BLAS1. The top speed of the BLAS3, about 250 Mflops, is very close to the peak machine speed of 266 Mflops. Later in the course, we will explain how to reorganize algorithms, like Gaussian elimination, so that they use BLAS3 rather than BLAS1.

Historically, the BLAS1 were invented and standardized first, because they were adequate to reach the top performance of the machines at that time (early 1970s). When they no longer reached peak performance on later machines, the BLAS2 were introduced in the mid 1980s, and the BLAS3 in 1990. As we will now explain, the reason for their performance differences is how they depend on the memory hierarchy.

To quantify our compararisons, we will use a simple model of the way the memory hierarchy works: Assume we have two levels of memory hierarchy, fast and slow, and that all data initially resides in slow memory. For each of saxpy, sgemv and sgemm, applied to vectors and matrices of dimension n, we will count

       m = number of memory references to slow memory needed just to read
           the input data from slow memory, and write the output data back
       f = number of floating point operations
       q = f/m = average number of flops per slow memory reference


              m         justification for m                          f        q
   saxpy     3*n        read each x(i), y(i) once, write y(i) once  2*n      2/3
   sgemv     n^2+O(n)   read each A(i,j) once                      2*n^2      2
   sgemm   4*n^2        read each A(i,j),B(i,j),C(i,j) once,       2*n^3     n/2
                        write each C(i,j) once

There is a clear difference in q for each level of BLAS, with BLAS3 the highest. The significance of q is this: for each word read from slow memory (the expensive operation), one can hope to do at most q operations on it (on average) while it resides in fast memory. The higher q is, the more the algorithm appears to operate at the top of the memory hierarchy, where it is most efficient. Another way to describe q is in terms of the miss ratio, the fraction of all memory references (to fast or slow memory) that ``miss'' the fast memory, and need data from the slow memory; a good problem has a low miss ratio. Assuming all input data initially resides in the slow memory, that the final result will also reside in the slow memory, and that each floating point operation involves 2 references to either fast or slow memory, we may bound

                 number of references to slow memory       m       1
   miss ratio = ------------------------------------- >= ----- = ----- .
                 total number of memory references        2*f     2*q

Let us use a very simple problem to illustrate why a large q means we can hope to run faster. Suppose one can do operations at 1 Mflop on data in the fast memory, and fetching data from slow memory takes 10 times longer per word. Suppose our simple algorithm is

   s = 0
   for i = 1, n
      s = s + f(x(i))
   end for

Here are some more details of our simple memory hierarchy model. We will use them to analyze several different algorithms for matrix multiplication (sgemm) in more detail, and in fact design one that uses the memory hierarchy ``optimally''.

1. There are just two levels in the hierarchy, fast and slow.
2. The small, fast memory has size M words, where M << n^2, so we can only fit a small part of an entire n-by-n matrix, but M >= 4*n, so we can fit several whole rows or columns.
3. Each word is read from slow memory individually (in practice, larger groups of words are read, such as cache lines or memory pages, but this doesn't change the basic analysis).
4. We have complete control over which words are transferred between the two levels. This is a best-case assumption, since often the hardware (cache or virtual memory system) makes this decision for us. In parallel computing, however, when the two levels are local processor memory and remote processor memory, we often have explicit control, whether we want it or not.

We consider 3 implementations of matrix multiply and compute q=f/m for each.

Algorithm 1: Unblocked matrix multiply

       for i=1 to n
          {Read row i of A into fast memory}
          for j=1 to n
             {Read C(i,j) into fast memory}
             {Read column j of B into fast memory}
             for k=1 to n
                 C(i,j)=C(i,j) + A(i,k)*B(k,j)
             end for
             {Write C(i,j) back to slow memory}
          end for
       end for

Also, one can describe two innermost loops (on j and k) as doing a vector-matrix multiplication of the ith row of A times the matrix B, to get the i-th row of C. This is a hint that we will not perform any better than these operations, since they are within the inner loop. Here is the detailed accounting of memory references:

     m = # slow memory refs =   n^3     read each column of B n times
                              + n^2     read each row of A once for each i,
                                          and keep it in fast memory during
                                          the execution of the two inner loops
                              + 2*n^2   read/write each entry of C once
                            = n^3 + 3*n^2

Thus q = f/m = (2*n^3)/(n^3 + 3*n^2) ~ 2. This is the same q as for sgemv, which is no surprise as mentioned above. It is far below the upper bound of n/2. We leave it to the reader to show that reading B into fast memory differently cannot significantly improve m and q.

Algorithm 2: Column blocked matrix multiply

       for j=1 to N
           {Read Bj into fast memory}
           {Read Cj into fast memory}
           for k=1 to n
               {Read column k of A into fast memory}
               Cj = Cj + A( :,k ) * Bj( k,: )  ...  rank-1 update of Cj
           end for
           {Write Cj back to slow memory}
       end for

The inner loop does a rank-one update of Cj, multiplying column k of A times row k of Bj. We assume that the fast memory is large enough to let us keep Bj, Cj and one column of A in it at a time; this means M >= 2*n^2/N + n, or N >= 2*n^2/(M-n) ~ 2*n^2/M, so that N is bounded below. Here is the detailed accounting of memory references:

     m = # memory refs =   n^2     read each Bj once
                         + N*n^2   read each column of A N times
                         + 2*n^2   read/write each Cj once
                       = (N+3)*n^2

Algorithm 3: Square blocked matrix multiply (also called 2D blocked, or panel blocked)

        for i=1 to N
           for j=1 to N
              {Read Cij into fast memory}
              for k=1 to N
                 {Read Aik into fast memory}
                 {Read Bkj into fast memory}
                 Cij = Cij + Aik * Bkj
              end for
              {Write Cij back to slow memory}
           end for
        end for

Here is a pictorial representation of the algorithm, for N=4:

Thus, the inner loop is an n/N-by-n/N matrix multiply. We assume that the fast memory is large enough so that the 3 subblocks Cij, Aik and Bkj fit in the fast memory; this means that M >= 3*(n/N)^2, or N >= sqrt(3/M)*n, so that N is bounded below. Here is the detailed accounting of memory references

     m = # memory refs =   N*n^2   read each Bkj N^3 times
                         + N*n^2   read each Aik N^3 times
                         + 2*n^2   read/write each Cij once
                       = (2*N+2)*n^2

Thus q = f/m = (2*n^3)/((2*N+2)*n^2) ~ n/N. As before, we can increase q, and so accelerate the algorithm, by making N as small as possible. But N is bounded below by the memory constraint N >= sqrt(3/M)*n. If we choose N as small as possible, so that N ~ sqrt(3/M)*n, then q ~ sqrt(M/3). Now q is independent of n, and in fact essentially optimal:

Theorem: ("I/O Complexity: the red blue pebble game", X. Hong and H. T. Kung, 13th Symposium on the Theory of Computing, ACM, 1981). Any blocked version of this matrix multiplication algorithm has q = O( sqrt(M) ), i.e. growing at most as fast as a constant multiple of sqrt(M).

There is a lot more to matrix multiplication, both theoretically and practically.

Fast serial matrix multiplication algorithms

     Let M = [ m11 m12 ] = [ a11 a12 ] * [ b11 b12 ]
             [ m21 m22 ]   [ a21 a22 ]   [ b21 b22 ]
  
     Let 
         m1 = ( a12 - a22 ) * ( b21 + b22 )
         m2 = ( a11 + a22 ) * ( b11 + b22 )
         m3 = ( a11 - a21 ) * ( b11 + b12 )
         m4 = ( a11 + a12 ) * b22
         m5 = a11 * ( b12 - b22 )
         m6 = a22 * ( b21 - b11 )
         m7 = ( a21 + a22 ) * b11

     Then
         m11 = m1 + m2 - m4 + m6
         m12 = m4 + m5
         m21 = m6 + m7
         m22 = m2 - m3 + m5 - m7

     M = Strassen_Matrix_Multiply(A,B,n)
        /* Return M=A*B, where A and B are n-by-n; 
           Assume n is a power of 2 */
        if n=1,
            return M=A*B   /* scalar multiplication */
        else
            Partition A = [ A11  A12 ] , B = [ B11  B12 ]
                          [ A21  A22 ]       [ B21  B22 ]
                      where the subblocks Aij and Bij are n/2-by-n/2

            M1 = Strassen_Matrix_Multiply( A12 - A22, B21 + B22, n/2 )
            M2 = Strassen_Matrix_Multiply( A11 + A22, B11 + B22, n/2 )
            M3 = Strassen_Matrix_Multiply( A11 - A21, B11 + B12, n/2 )
            M4 = Strassen_Matrix_Multiply( A11 + A12, B22, n/2 )
            M5 = Strassen_Matrix_Multiply( A11, B12 - B22, n/2 )
            M6 = Strassen_Matrix_Multiply( A22, B21 - B11, n/2 )
            M7 = Strassen_Matrix_Multiply( A21 + A22, B11, n/2 )

            M11 = M1 + M2 - M4 + M6
            M12 = M4 + M5
            M21 = M6 + M7
            M22 = M2 - M3 + M5 - M7

            return M = [ M11  M12 ]
                       [ M21  M22 ]
         end

       T(n) =    7*T(n/2)   ... the cost of the 7 recursive calls 
              + 18*(n/2)^2  ... the cost of the 18 n/2-by-n/2 matrix additions

       T'(m) =    7*T'(m-1) + 18*2^(2*m-2)

We will discuss matrix multiplication again later, for distributed memory parallel machines, where part of the algorithm design problem is to decide which processor memories store which parts of A, B and C, and which processors are responsble for computing which parts of the product A*B.