Loop nest optimization
Encyclopedia
Loop nest optimization (LNO) applies a set of loop transformations for the purpose of locality optimization or parallelization or other loop overhead reduction of the loop nests. One classical usage is to reduce memory access latency or the cache bandwidth necessary due to cache reuse for some common linear algebra algorithm
Algorithm
In mathematics and computer science, an algorithm is an effective method expressed as a finite list of well-defined instructions for calculating a function. Algorithms are used for calculation, data processing, and automated reasoning...

s.

Example: Matrix multiply

Many large mathematical operations on computers end up spending much of their time doing matrix multiplication
Matrix multiplication
In mathematics, matrix multiplication is a binary operation that takes a pair of matrices, and produces another matrix. If A is an n-by-m matrix and B is an m-by-p matrix, the result AB of their multiplication is an n-by-p matrix defined only if the number of columns m of the left matrix A is the...

. The operation is:

C = A×B

where A, B, and C are N×N arrays. Subscripts, for the following
description, are in the form C[row][column].

The basic loop is:


for (i = 0; i < N; i++)
{
for (j=0; j < N; j++)
{
C[i][j] = 0;
for (k = 0; k < N; k++)
C[i][j] += A[k][j] * B[i][k];
}
}


There are three problems to solve:
  • Floating point additions take some number of cycles to complete. In order to keep an adder
    Adder (electronics)
    In electronics, an adder or summer is a digital circuit that performs addition of numbers.In many computers and other kinds of processors, adders are used not only in the arithmetic logic unit, but also in other parts of the processor, where they are used to calculate addresses, table indices, and...

     with multiple cycle latency busy, the code must update multiple accumulators in parallel.

  • Machines can typically do just one memory operation per multiply–add, so values loaded must be reused at least twice.

  • Typical PC memory systems can only sustain one 8-byte doubleword per 10–30 double-precision multiply–adds, so values loaded into the cache must be reused many times.


The original loop calculates the result for one entry in the result matrix at a time. By calculating a small block of entries simultaneously, the following loop reuses each loaded value twice, so that the inner loop has four loads and four multiply–adds, thus solving problem #2. By carrying four accumulators simultaneously, this code can keep a single floating point adder with a latency of 4 busy nearly all the time (problem #1). However, the code does not address the third problem. (Nor does it address the cleanup work necessary when N is odd. Such details will be left out of the following discussion.)


for (i = 0; i < N; i += 2)
{
for (j = 0; j < N; j += 2)
{
acc00 = acc01 = acc10 = acc11 = 0;
for (k = 0; k < N; k++)
{
acc00 += A[k][j + 0] * B[i + 0][k];
acc01 += A[k][j + 1] * B[i + 0][k];
acc10 += A[k][j + 0] * B[i + 1][k];
acc11 += A[k][j + 1] * B[i + 1][k];
}
C[i + 0][j + 0] = acc00;
C[i + 0][j + 1] = acc01;
C[i + 1][j + 0] = acc10;
C[i + 1][j + 1] = acc11;
}
}


This code has had both the i and j iterations blocked by a factor of two, and had both the resulting two-iteration inner loops completely unrolled.

This code would run quite acceptably on a Cray Y-MP (built in the early 1980s), which can sustain 0.8 multiply–adds per memory operation to main memory. A machine like a 2.8 GHz Pentium 4, built in 2003, has slightly less memory bandwidth and vastly better floating point, so that it can sustain 16.5 multiply–adds per memory operation. As a result, the code above will run slower on the 2.8 GHz Pentium 4 than on the 166 MHz Y-MP!

A machine with a longer floating-point add latency or with multiple adders would require more accumulators to run in parallel. It is easy to change the loop above to compute a 3x3 block
instead of a 2x2 block, but the resulting code is not always faster. The loop requires registers to hold both the accumulators and the loaded and reused A and B values. A 2x2 block requires 7 registers. A 3x3 block requires 13, which will not work on a machine with just 8 floating point registers in the ISA
Instruction set
An instruction set, or instruction set architecture , is the part of the computer architecture related to programming, including the native data types, instructions, registers, addressing modes, memory architecture, interrupt and exception handling, and external I/O...

. If the CPU does not have enough registers, the compiler will schedule extra loads and stores to spill the registers into stack slots, which will make the loop run slower than a smaller blocked loop.

Matrix multiplication is like many other codes in that it can be limited by memory bandwidth, and that more registers can help the compiler and programmer reduce the need for memory bandwidth. This register pressure is why vendors of RISC CPUs, who intended to build machines more parallel than the general purpose x86 and 68000 CPUs, adopted 32-entry floating-point register file
Register file
A register file is an array of processor registers in a central processing unit . Modern integrated circuit-based register files are usually implemented by way of fast static RAMs with multiple ports...

s.

The code above does not use the cache very well. During the calculation of a horizontal stripe of C results, one horizontal stripe of B is loaded and the entire matrix A is loaded. For the entire calculation, C is stored once (that's good), B is loaded into the cache once (assuming a stripe of B fits in the cache with a stripe of A), but A is loaded N/ib times, where ib is the size of the strip in the C matrix, for a total of N3/ib doubleword loads from main memory. In the code above, ib is 2.

The next step to reducing the memory traffic is to make ib as large as
possible. We want it to be larger than the "balance" number reported
by streams. In the case of one particular 2.8 GHz Pentium-4 system
used for this example, the balance number is 16.5. The
second code example above can't be extended directly, since
that would require many more accumulator registers. Instead, we block
the loop over i. (Technically, this is actually the second time we've
blocked i, as the first time was the factor of 2.)


for (ii = 0; ii < N; ii += ib)
{
for (j = 0; j < N; j += 2)
{
for (i = ii; i < ii + ib; i += 2)
{
acc00 = acc01 = acc10 = acc11 = 0;
for (k = 0; k < N; k++)
{
acc00 += A[k][j + 0] * B[i + 0][k];
acc01 += A[k][j + 1] * B[i + 0][k];
acc10 += A[k][j + 0] * B[i + 1][k];
acc11 += A[k][j + 1] * B[i + 1][k];
}
C[i + 0][j + 0] = acc00;
C[i + 0][j + 1] = acc01;
C[i + 1][j + 0] = acc10;
C[i + 1][j + 1] = acc11;
}
}
}


With this code, we can set ib to be anything we like, and the number of loads of the A matrix will be reduced by that factor. This freedom has a cost: we are now keeping a Nxib slice of the B matrix in the cache. So long as that fits, this code will not be limited by the memory system.

So what size matrix fits? Our example system, a 2.8 GHz Pentium 4, has a 16KB primary data cache. With ib=20, the slice of the B matrix in this code will be larger than the primary cache when N > 100. For problems larger than that, we'll need another trick.

That trick is reducing the size of the stripe of the B matrix by blocking
the k loop, so that the stripe is of size ib x kb. Blocking the k loop
means that the C array will be loaded and stored N/kb times, for a total
of 2*N^3/kb memory transfers. A is still transferred N/ib times, for N^3/ib
transfers. So long as
2*N/kb + N/ib < N/balance
the machine's memory system will keep up with the floating point unit and
the code will run at maximum performance. The 16KB cache of the Pentium
4 is not quite big enough: we might choose ib=24 and kb=64, thus using 12KB
of the cache -- we don't want to completely fill it, since the C and A
arrays have to have some room to flow through. These numbers comes within
20% of the peak floating-point speed of the processor.

Here is the code with loop k blocked.


for (ii = 0; ii < N; ii += ib)
{
for (kk = 0; kk < N; kk += kb)
{
for (j=0; j < N; j += 2)
{
for(i = ii; i < ii + ib; i += 2 )
{
if (kk 0)
acc00 = acc01 = acc10 = acc11 = 0;
else
{
acc00 = C[i + 0][j + 0];
acc01 = C[i + 0][j + 1];
acc10 = C[i + 1][j + 0];
acc11 = C[i + 1][j + 1];
}
for (k = kk; k < kk + kb; k++)
{
acc00 += A[k][j + 0] * B[i + 0][k];
acc01 += A[k][j + 1] * B[i + 0][k];
acc10 += A[k][j + 0] * B[i + 1][k];
acc11 += A[k][j + 1] * B[i + 1][k];
}
C[i + 0][j + 0] = acc00;
C[i + 0][j + 1] = acc01;
C[i + 1][j + 0] = acc10;
C[i + 1][j + 1] = acc11;
}
}
}
}


The above code examples do not show the details of dealing with values of N which are not multiples of the blocking factors. Compilers which do loop nest optimization emit code to clean up the edges of the computation. For example, most LNO compilers would probably split
Loop splitting
Loop splitting is a compiler optimization technique. It attempts to simplify a loop or eliminate dependencies by breaking it into multiple loops which have the same bodies but iterate over different contiguous portions of the index range.-Loop peeling:...

the kk 0 iteration off from the rest of the kk iterations, in order to remove the if statement from the i loop. This is one of the values of such a compiler: while it is straightforward to code the simple cases of this optimization, keeping all the details correct as the code is replicated and transformed is an error-prone process.

The above loop will only achieve 80% of peak flops on the example system when blocked for the 16KB L1 cache size. It will do worse on systems with even more unbalanced memory systems. Fortunately, the Pentium 4 has 256KB (or more, depending on the model) high-bandwidth level-2 cache as well as the level-1 cache. We are presented with a choice:
  • We can adjust the block sizes for the level-2 cache. This will stress the processor's ability to keep many instructions in flight simultaneously, and there is a good chance it will be unable to achieve full bandwidth from the level-2 cache.

  • We can block the loops again, again for the level-2 cache sizes. With a total of three levels of blocking (for the register file, for the L1 cache, and for the L2 cache), the code will minimize the required bandwidth at each level of the memory hierarchy. Unfortunately, the extra levels of blocking will incur still more loop overhead, which for some problem sizes on some hardware may be more time consuming than any shortcomings in the hardware's ability to stream data from the L2 cache.

External links

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