This is my last post about processors and performance, I swear! Plus, my wrists are starting to hurt from this bloodpact thing (as I'm diagnosed with RSI), so I think this will be a light one.

As I've discussed previously, modern desktop processors work really hard to exploit the inherent parallelism in your programs. This is called instruction-level parallelism, and is one of the techniques processors use to get more performance out of slower clock rates (along with data-level parallelism (SIMD) or multiple cores (MIMD)*). Previously, I waved my hands a bit and said "The processor makes independent operations run in parallel." Now I'm going to teach you how to count cycles in the presence of latency and parallelism.

Traditionally, when analyzing the cost of an algorithm, you would simply count the operations involved, sum their costs in cycles, and call it a day. These days, it's not that easy. Instructions have two costs: dependency chain latency and reciprocal throughput.

Reciprocal throughput is simply the reciprocal of the maximum throughput of a particular instruction. Throughput is measured in instructions/cycle, so reciprocal throughput is cycles/instruction.

OK, that sounds like the way we've always measured performance. So what's dependency chain latency? When the results of a previous calculation are needed for another calculation, you have a dependency chain. In a dependency chain, you measure the cost of an instruction by its latency, not its reciprocal throughput. Remember that our processors are working really hard to exploit parallelism in our code. When there is no instruction-level parallelism available, we get penalized.

Let's go back to our sum 10000 numbers example, but unroll it a bit: 

float array[10000];
float sum = 0.0f;
for (int i = 0; i < 10000; i += 8) {
    sum += array[i+0];
    sum += array[i+1];
    sum += array[i+2];
    sum += array[i+3];
    sum += array[i+4];
    sum += array[i+5];
    sum += array[i+6];
    sum += array[i+7];
}
return sum;
In x86: 
xor ecx, ecx     ; ecx  = i   = 0
mov esi, array
xorps xmm0, xmm0 ; xmm0 = sum = 0.0

begin:
addss xmm0, [esi+0]
addss xmm0, [esi+4]
addss xmm0, [esi+8]
addss xmm0, [esi+12]
addss xmm0, [esi+16]
addss xmm0, [esi+20]
addss xmm0, [esi+24]
addss xmm0, [esi+28]

add esi, 32
add ecx, 1
cmp ecx, 10000
jl begin ; if ecx < 10000, goto begin

; xmm0 = total sum

Since each addition to sum in the loop depends on the previous addition, these instructions are a dependency chain. On a modern processor, let's say the reciprocal throughput of addss is 1 cycle. However, the minimum latency is 4 cycles. Since every instruction depends on the previous, each addition costs 4 cycles.

As before, let's try summing with four temporary sums: 

xor ecx, ecx     ; ecx  = i    = 0
mov esi, array
xorps xmm0, xmm0 ; xmm0 = sum1 = 0.0
xorps xmm1, xmm1 ; xmm1 = sum2 = 0.0
xorps xmm2, xmm2 ; xmm2 = sum3 = 0.0
xorps xmm3, xmm3 ; xmm3 = sum4 = 0.0

; top = sum0

begin:
addss xmm0, [esi+0]
addss xmm1, [esi+4]
addss xmm2, [esi+8]
addss xmm3, [esi+12]
addss xmm0, [esi+16]
addss xmm1, [esi+20]
addss xmm2, [esi+24]
addss xmm3, [esi+28]

add esi, 32
add ecx, 1
cmp ecx, 10000
jl begin ; if ecx < 10000, goto begin

; accumulate sums
addss xmm0, xmm1
addss xmm2, xmm3 ; this instruction happens in parallel with the one above
addss xmm0, xmm2

Here, the additions in the loop that depend on each other are 4 cycles apart, meaning the minimum latency is no longer a problem. This lets us hit the maximum addition rate of one per cycle.

Removing dependency chains is a critical part of optimizing on today's processors. The Core 2 processor has six execution units, three of which are fully 128-bit SIMD ALUs. If you can restructure your algorithm so calculations happen independently, you can take advantage of all of them. (And if you can pull off making full use of the Core 2's ALU capacity, you win.)

* BTW, it's sort of unrelated, but I couldn't help but link this article. Greg Pfister has an interesting comparison and history of SIMD vs. MIMD here. Ignore the terminology blathering and focus on the history of and influences on SIMD and MIMD over time.