// Now we define methods that give our pipeline several different
// schedules.
public void ScheduleForCpu()
{
// Compute the look-up-table ahead of time.
Lut.ComputeRoot();
// Compute color channels innermost. Promise that there will
// be three of them and unroll across them.
Curved.Reorder(C, X, Y)
.Bound(C, 0, 3)
.Unroll(C);
// Look-up-tables don't vectorize well, so just parallelize
// curved in slices of 16 scanlines.
var yo = new HSVar("yo");
var yi = new HSVar("yi");
Curved.Split(Y, yo, yi, 16)
.Parallel(yo);
// Compute sharpen as needed per scanline of curved.
Sharpen.ComputeAt(Curved, yi);
// Vectorize the sharpen. It's 16-bit so we'll vectorize it 8-wide.
Sharpen.Vectorize(X, 8);
// Compute the padded input as needed per scanline of curved,
// reusing previous values computed within the same strip of
// 16 scanlines.
Padded.StoreAt(Curved, yo)
.ComputeAt(Curved, yi);
// Also vectorize the padding. It's 8-bit, so we'll vectorize
// 16-wide.
Padded.Vectorize(X, 16);
// JIT-compile the pipeline for the CPU.
Curved.CompileJit();
}
public static int Main(string[] args)
{
// First we'll declare some Vars to use below.
var x = new HSVar("x");
var y = new HSVar("y");
// Let's examine various scheduling options for a simple two stage
// pipeline. We'll start with the default schedule:
{
var producer = new HSFunc("producer_default");
var consumer = new HSFunc("consumer_default");
// The first stage will be some simple pointwise math similar
// to our familiar gradient function. The value at position x,
// y is the sin of product of x and y.
producer[x, y] = HSMath.Sin(x * y);
// Now we'll add a second stage which averages together multiple
// points in the first stage.
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// We'll turn on tracing for both functions.
consumer.TraceStores();
producer.TraceStores();
// And evaluate it over a 4x4 box.
Console.WriteLine("\nEvaluating producer-consumer pipeline with default schedule");
consumer.Realize <float>(4, 4);
// There were no messages about computing values of the
// producer. This is because the default schedule fully
// inlines 'producer' into 'consumer'. It is as if we had
// written the following code instead:
// consumer(x, y) = (sin(x * y) +
// sin(x * (y + 1)) +
// sin((x + 1) * y) +
// sin((x + 1) * (y + 1))/4);
// All calls to 'producer' have been replaced with the body of
// 'producer', with the arguments substituted in for the
// variables.
// The equivalent C code is:
var result = new float[4, 4];
for (int yy = 0; yy < 4; yy++)
{
for (int xx = 0; xx < 4; xx++)
{
result[yy, xx] = (float)((Math.Sin(xx * yy) +
Math.Sin(xx * (yy + 1)) +
Math.Sin((xx + 1) * yy) +
Math.Sin((xx + 1) * (yy + 1))) / 4);
}
}
Console.WriteLine();
// If we look at the loop nest, the producer doesn't appear
// at all. It has been inlined into the consumer.
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
}
// Next we'll examine the next simplest option - computing all
// values required in the producer before computing any of the
// consumer. We call this schedule "root".
{
// Start with the same function definitions:
var producer = new HSFunc("producer_root");
var consumer = new HSFunc("consumer_root");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// Tell Halide to evaluate all of producer before any of consumer.
producer.ComputeRoot();
// Turn on tracing.
consumer.TraceStores();
producer.TraceStores();
// Compile and run.
Console.WriteLine("\nEvaluating producer.compute_root()");
consumer.Realize <float>(4, 4);
// Reading the output we can see that:
// A) There were stores to producer.
// B) They all happened before any stores to consumer.
// See figures/lesson_08_compute_root.gif for a visualization.
// The producer is on the left and the consumer is on the
// right. Stores are marked in orange and loads are marked in
// blue.
// Equivalent C:
var result = new float[4, 4];
// Allocate some temporary storage for the producer.
var producer_storage = new float[5, 5];
// Compute the producer.
for (int yy = 0; yy < 5; yy++)
{
for (int xx = 0; xx < 5; xx++)
{
producer_storage[yy, xx] = (float)Math.Sin(xx * yy);
}
}
// Compute the consumer. Skip the prints this time.
for (int yy = 0; yy < 4; yy++)
{
for (int xx = 0; xx < 4; xx++)
{
result[yy, xx] = (producer_storage[yy, xx] +
producer_storage[yy + 1, xx] +
producer_storage[yy, xx + 1] +
producer_storage[yy + 1, xx + 1]) / 4;
}
}
// Note that consumer was evaluated over a 4x4 box, so Halide
// automatically inferred that producer was needed over a 5x5
// box. This is the same 'bounds inference' logic we saw in
// the previous lesson, where it was used to detect and avoid
// out-of-bounds reads from an input image.
// If we print the loop nest, we'll see something very
// similar to the C above.
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
}
// Let's compare the two approaches above from a performance
// perspective.
// Full inlining (the default schedule):
// - Temporary memory allocated: 0
// - Loads: 0
// - Stores: 16
// - Calls to sin: 64
// producer.compute_root():
// - Temporary memory allocated: 25 floats
// - Loads: 64
// - Stores: 41
// - Calls to sin: 25
// There's a trade-off here. Full inlining used minimal temporary
// memory and memory bandwidth, but did a whole bunch of redundant
// expensive math (calling sin). It evaluated most points in
// 'producer' four times. The second schedule,
// producer.compute_root(), did the mimimum number of calls to
// sin, but used more temporary memory and more memory bandwidth.
// In any given situation the correct choice can be difficult to
// make. If you're memory-bandwidth limited, or don't have much
// memory (e.g. because you're running on an old cell-phone), then
// it can make sense to do redundant math. On the other hand, sin
// is expensive, so if you're compute-limited then fewer calls to
// sin will make your program faster. Adding vectorization or
// multi-core parallelism tilts the scales in favor of doing
// redundant work, because firing up multiple cpu cores increases
// the amount of math you can do per second, but doesn't increase
// your system memory bandwidth or capacity.
// We can make choices in between full inlining and
// compute_root. Next we'll alternate between computing the
// producer and consumer on a per-scanline basis:
{
// Start with the same function definitions:
var producer = new HSFunc("producer_y");
var consumer = new HSFunc("consumer_y");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// Tell Halide to evaluate producer as needed per y coordinate
// of the consumer:
producer.ComputeAt(consumer, y);
// This places the code that computes the producer just
// *inside* the consumer's for loop over y, as in the
// equivalent C below.
// Turn on tracing.
producer.TraceStores();
consumer.TraceStores();
// Compile and run.
Console.WriteLine("\nEvaluating producer.ComputeAt(consumer, y)");
consumer.Realize <float>(4, 4);
// See figures/lesson_08_compute_y.gif for a visualization.
// Reading the log or looking at the figure you should see
// that producer and consumer alternate on a per-scanline
// basis. Let's look at the equivalent C:
var result = new float[4, 4];
// There's an outer loop over scanlines of consumer:
for (int yy = 0; yy < 4; yy++)
{
// Allocate space and compute enough of the producer to
// satisfy this single scanline of the consumer. This
// means a 5x2 box of the producer.
var producer_storage = new float[2, 5];
for (int py = yy; py < yy + 2; py++)
{
for (int px = 0; px < 5; px++)
{
producer_storage[py - yy, px] = (float)Math.Sin(px * py);
}
}
// Compute a scanline of the consumer.
for (int xx = 0; xx < 4; xx++)
{
result[yy, xx] = (producer_storage[0, xx] +
producer_storage[1, xx] +
producer_storage[0, xx + 1] +
producer_storage[1, xx + 1]) / 4;
}
}
// Again, if we print the loop nest, we'll see something very
// similar to the C above.
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
// The performance characteristics of this strategy are in
// between inlining and compute root. We still allocate some
// temporary memory, but less that compute_root, and with
// better locality (we load from it soon after writing to it,
// so for larger images, values should still be in cache). We
// still do some redundant work, but less than full inlining:
// producer.ComputeAt(consumer, y):
// - Temporary memory allocated: 10 floats
// - Loads: 64
// - Stores: 56
// - Calls to sin: 40
}
// We could also say producer.ComputeAt(consumer, x), but this
// would be very similar to full inlining (the default
// schedule). Instead let's distinguish between the loop level at
// which we allocate storage for producer, and the loop level at
// which we actually compute it. This unlocks a few optimizations.
{
var producer = new HSFunc("producer_root_y");
var consumer = new HSFunc("consumer_root_y");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// Tell Halide to make a buffer to store all of producer at
// the outermost level:
producer.StoreRoot();
// ... but compute it as needed per y coordinate of the
// consumer.
producer.ComputeAt(consumer, y);
producer.TraceStores();
consumer.TraceStores();
Console.WriteLine("\nEvaluating producer.store_root().ComputeAt(consumer, y)");
consumer.Realize <float>(4, 4);
// See figures/lesson_08_store_root_compute_y.gif for a
// visualization.
// Reading the log or looking at the figure you should see
// that producer and consumer again alternate on a
// per-scanline basis. It computes a 5x2 box of the producer
// to satisfy the first scanline of the consumer, but after
// that it only computes a 5x1 box of the output for each new
// scanline of the consumer!
//
// Halide has detected that for all scanlines except for the
// first, it can reuse the values already sitting in the
// buffer we've allocated for producer. Let's look at the
// equivalent C:
var result = new float[4, 4];
{
// producer.store_root() implies that storage goes here:
var producer_storage = new float[5, 5];
// There's an outer loop over scanlines of consumer:
for (int yy = 0; yy < 4; yy++)
{
// Compute enough of the producer to satisfy this scanline
// of the consumer.
for (int py = yy; py < yy + 2; py++)
{
// Skip over rows of producer that we've already
// computed in a previous iteration.
if (yy > 0 && py == yy)
{
continue;
}
for (int px = 0; px < 5; px++)
{
producer_storage[py, px] = (float)Math.Sin(px * py);
}
}
// Compute a scanline of the consumer.
for (int xx = 0; xx < 4; xx++)
{
result[yy, xx] = (producer_storage[yy, xx] +
producer_storage[yy + 1, xx] +
producer_storage[yy, xx + 1] +
producer_storage[yy + 1, xx + 1]) / 4;
}
}
}
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
// The performance characteristics of this strategy are pretty
// good! The numbers are similar compute_root, except locality
// is better. We're doing the minimum number of sin calls,
// and we load values soon after they are stored, so we're
// probably making good use of the cache:
// producer.store_root().ComputeAt(consumer, y):
// - Temporary memory allocated: 10 floats
// - Loads: 64
// - Stores: 39
// - Calls to sin: 25
// Note that my claimed amount of memory allocated doesn't
// match the reference C code. Halide is performing one more
// optimization under the hood. It folds the storage for the
// producer down into a circular buffer of two
// scanlines. Equivalent C would actually look like this:
{
// Actually store 2 scanlines instead of 5
var producer_storage = new float[2, 5];
for (int yy = 0; yy < 4; yy++)
{
for (int py = yy; py < yy + 2; py++)
{
if (yy > 0 && py == yy)
{
continue;
}
for (int px = 0; px < 5; px++)
{
// Stores to producer_storage have their y coordinate bit-masked.
producer_storage[py & 1, px] = (float)Math.Sin(px * py);
}
}
// Compute a scanline of the consumer.
for (int xx = 0; xx < 4; xx++)
{
// Loads from producer_storage have their y coordinate bit-masked.
result[yy, xx] = (producer_storage[yy & 1, xx] +
producer_storage[(yy + 1) & 1, xx] +
producer_storage[yy & 1, xx + 1] +
producer_storage[(yy + 1) & 1, xx + 1]) / 4;
}
}
}
}
// We can do even better, by leaving the storage outermost, but
// moving the computation into the innermost loop:
{
var producer = new HSFunc("producer_root_x");
var consumer = new HSFunc("consumer_root_x");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// Store outermost, compute innermost.
producer.StoreRoot().ComputeAt(consumer, x);
producer.TraceStores();
consumer.TraceStores();
Console.WriteLine("\nEvaluating producer.store_root().ComputeAt(consumer, x)");
consumer.Realize <float>(4, 4);
// See figures/lesson_08_store_root_compute_x.gif for a
// visualization.
// You should see that producer and consumer now alternate on
// a per-pixel basis. Here's the equivalent C:
var result = new float[4, 4];
// producer.store_root() implies that storage goes here, but
// we can fold it down into a circular buffer of two
// scanlines:
var producer_storage = new float[2, 5];
// For every pixel of the consumer:
for (int yy = 0; yy < 4; yy++)
{
for (int xx = 0; xx < 4; xx++)
{
// Compute enough of the producer to satisfy this
// pixel of the consumer, but skip values that we've
// already computed:
if (yy == 0 && xx == 0)
{
producer_storage[yy & 1, xx] = (float)Math.Sin(xx * yy);
}
if (yy == 0)
{
producer_storage[yy & 1, xx + 1] = (float)Math.Sin((xx + 1) * yy);
}
if (xx == 0)
{
producer_storage[(yy + 1) & 1, xx] = (float)Math.Sin(xx * (yy + 1));
}
producer_storage[(yy + 1) & 1, xx + 1] = (float)Math.Sin((xx + 1) * (yy + 1));
result[yy, xx] = (producer_storage[yy & 1, xx] +
producer_storage[(yy + 1) & 1, xx] +
producer_storage[yy & 1, xx + 1] +
producer_storage[(yy + 1) & 1, xx + 1]) / 4;
}
}
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
// The performance characteristics of this strategy are the
// best so far. One of the four values of the producer we need
// is probably still sitting in a register, so I won't count
// it as a load:
// producer.store_root().ComputeAt(consumer, x):
// - Temporary memory allocated: 10 floats
// - Loads: 48
// - Stores: 56
// - Calls to sin: 40
}
// So what's the catch? Why not always do
// producer.store_root().ComputeAt(consumer, x) for this type of
// code?
//
// The answer is parallelism. In both of the previous two
// strategies we've assumed that values computed on previous
// iterations are lying around for us to reuse. This assumes that
// previous values of x or y happened earlier in time and have
// finished. This is not true if you parallelize or vectorize
// either loop. Darn. If you parallelize, Halide won't inject the
// optimizations that skip work already done if there's a parallel
// loop in between the store_at level and the ComputeAt level,
// and won't fold the storage down into a circular buffer either,
// which makes our store_root pointless.
// We're running out of options. We can make new ones by
// splitting. We can store_at or ComputeAt at the natural
// variables of the consumer (x and y), or we can split x or y
// into new inner and outer sub-variables and then schedule with
// respect to those. We'll use this to express fusion in tiles:
{
var producer = new HSFunc("producer_tile");
var consumer = new HSFunc("consumer_tile");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// We'll compute 8x8 of the consumer, in 4x4 tiles.
var x_outer = new HSVar("x_outer");
var y_outer = new HSVar("y_outer");
var x_inner = new HSVar("x_inner");
var y_inner = new HSVar("y_inner");
consumer.Tile(x, y, x_outer, y_outer, x_inner, y_inner, 4, 4);
// Compute the producer per tile of the consumer
producer.ComputeAt(consumer, x_outer);
// Notice that I wrote my schedule starting from the end of
// the pipeline (the consumer). This is because the schedule
// for the producer refers to x_outer, which we introduced
// when we tiled the consumer. You can write it in the other
// order, but it tends to be harder to read.
// Turn on tracing.
producer.TraceStores();
consumer.TraceStores();
Console.WriteLine("\nEvaluating:");
Console.WriteLine("consumer.tile(x, y, x_outer, y_outer, x_inner, y_inner, 4, 4);");
Console.WriteLine("producer.ComputeAt(consumer, x_outer);");
consumer.Realize <float>(8, 8);
// See figures/lesson_08_tile.gif for a visualization.
// The producer and consumer now alternate on a per-tile
// basis. Here's the equivalent C:
var result = new float[8, 8];
// For every tile of the consumer:
for (int yy_outer = 0; yy_outer < 2; yy_outer++)
{
for (int xx_outer = 0; xx_outer < 2; xx_outer++)
{
// Compute the x and y coords of the start of this tile.
int x_base = xx_outer * 4;
int y_base = yy_outer * 4;
// Compute enough of producer to satisfy this tile. A
// 4x4 tile of the consumer requires a 5x5 tile of the
// producer.
var producer_storage = new float[5, 5];
for (int py = y_base; py < y_base + 5; py++)
{
for (int px = x_base; px < x_base + 5; px++)
{
producer_storage[py - y_base, px - x_base] = (float)Math.Sin(px * py);
}
}
// Compute this tile of the consumer
for (int yy_inner = 0; yy_inner < 4; yy_inner++)
{
for (int xx_inner = 0; xx_inner < 4; xx_inner++)
{
int xx = x_base + xx_inner;
int yy = y_base + yy_inner;
result[yy, xx] =
(producer_storage[yy - y_base, xx - x_base] +
producer_storage[yy - y_base + 1, xx - x_base] +
producer_storage[yy - y_base, xx - x_base + 1] +
producer_storage[yy - y_base + 1, xx - x_base + 1]) / 4;
}
}
}
}
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
// Tiling can make sense for problems like this one with
// stencils that reach outwards in x and y. Each tile can be
// computed independently in parallel, and the redundant work
// done by each tile isn't so bad once the tiles get large
// enough.
}
// Let's try a mixed strategy that combines what we have done with
// splitting, parallelizing, and vectorizing. This is one that
// often works well in practice for large images. If you
// understand this schedule, then you understand 95% of scheduling
// in Halide.
{
var producer = new HSFunc("producer_mixed");
var consumer = new HSFunc("consumer_mixed");
producer[x, y] = HSMath.Sin(x * y);
consumer[x, y] = (producer[x, y] +
producer[x, y + 1] +
producer[x + 1, y] +
producer[x + 1, y + 1]) / 4;
// Split the y coordinate of the consumer into strips of 16 scanlines:
var yo = new HSVar("yo");
var yi = new HSVar("yi");
consumer.Split(y, yo, yi, 16);
// Compute the strips using a thread pool and a task queue.
consumer.Parallel(yo);
// Vectorize across x by a factor of four.
consumer.Vectorize(x, 4);
// Now store the producer per-strip. This will be 17 scanlines
// of the producer (16+1), but hopefully it will fold down
// into a circular buffer of two scanlines:
producer.StoreAt(consumer, yo);
// Within each strip, compute the producer per scanline of the
// consumer, skipping work done on previous scanlines.
producer.ComputeAt(consumer, yi);
// Also vectorize the producer (because sin is vectorizable on x86 using SSE).
producer.Vectorize(x, 4);
// Let's leave tracing off this time, because we're going to
// evaluate over a larger image.
// consumer.TraceStores();
// producer.TraceStores();
var halide_result = consumer.Realize <float>(160, 160);
// See figures/lesson_08_mixed.mp4 for a visualization.
// Here's the equivalent (serial) C:
var c_result = new float[160, 160];
// For every strip of 16 scanlines (this loop is parallel in
// the Halide version)
for (int yyo = 0; yyo < 160 / 16 + 1; yyo++)
{
// 16 doesn't divide 160, so push the last slice upwards
// to fit within [0, 159] (see lesson 05).
int y_base = yyo * 16;
if (y_base > 160 - 16)
{
y_base = 160 - 16;
}
// Allocate a two-scanline circular buffer for the producer
var producer_storage = new float[2, 161];
// For every scanline in the strip of 16:
for (int yyi = 0; yyi < 16; yyi++)
{
int yy = y_base + yyi;
for (int py = yy; py < yy + 2; py++)
{
// Skip scanlines already computed *within this task*
if (yyi > 0 && py == yy)
{
continue;
}
// Compute this scanline of the producer in 4-wide vectors
for (int x_vec = 0; x_vec < 160 / 4 + 1; x_vec++)
{
int x_base = x_vec * 4;
// 4 doesn't divide 161, so push the last vector left
// (see lesson 05).
if (x_base > 161 - 4)
{
x_base = 161 - 4;
}
// If you're on x86, Halide generates SSE code for this part:
int[] xx = { x_base, x_base + 1, x_base + 2, x_base + 3 };
float[] vec = { (float)Math.Sin(xx[0] * py), (float)Math.Sin(xx[1] * py),
(float)Math.Sin(xx[2] * py), (float)Math.Sin(xx[3] * py) };
producer_storage[py & 1, xx[0]] = vec[0];
producer_storage[py & 1, xx[1]] = vec[1];
producer_storage[py & 1, xx[2]] = vec[2];
producer_storage[py & 1, xx[3]] = vec[3];
}
}
// Now compute consumer for this scanline:
for (int x_vec = 0; x_vec < 160 / 4; x_vec++)
{
int x_base = x_vec * 4;
// Again, Halide's equivalent here uses SSE.
int[] xx = { x_base, x_base + 1, x_base + 2, x_base + 3 };
float[] vec =
{
(producer_storage[yy & 1, xx[0]] +
producer_storage[(yy + 1) & 1, xx[0]] +
producer_storage[yy & 1, xx[0] + 1] +
producer_storage[(yy + 1) & 1, xx[0] + 1]) / 4,
(producer_storage[yy & 1, xx[1]] +
producer_storage[(yy + 1) & 1, xx[1]] +
producer_storage[yy & 1, xx[1] + 1] +
producer_storage[(yy + 1) & 1, xx[1] + 1]) / 4,
(producer_storage[yy & 1, xx[2]] +
producer_storage[(yy + 1) & 1, xx[2]] +
producer_storage[yy & 1, xx[2] + 1] +
producer_storage[(yy + 1) & 1, xx[2] + 1]) / 4,
(producer_storage[yy & 1, xx[3]] +
producer_storage[(yy + 1) & 1, xx[3]] +
producer_storage[yy & 1, xx[3] + 1] +
producer_storage[(yy + 1) & 1, xx[3] + 1]) / 4
};
c_result[yy, xx[0]] = vec[0];
c_result[yy, xx[1]] = vec[1];
c_result[yy, xx[2]] = vec[2];
c_result[yy, xx[3]] = vec[3];
}
}
}
Console.WriteLine("Pseudo-code for the schedule:");
consumer.PrintLoopNest();
Console.WriteLine();
// Look on my code, ye mighty, and despair!
// Let's check the C result against the Halide result. Doing
// this I found several bugs in my C implementation, which
// should tell you something.
for (int yy = 0; yy < 160; yy++)
{
for (int xx = 0; xx < 160; xx++)
{
float error = halide_result[xx, yy] - c_result[yy, xx];
// It's floating-point math, so we'll allow some slop:
if (error < -0.001f || error > 0.001f)
{
Console.WriteLine("halide_result(%d, %d) = %f instead of %f",
xx, yy, halide_result[xx, yy], c_result[yy, xx]);
return(-1);
}
}
}
}
// This stuff is hard. We ended up in a three-way trade-off
// between memory bandwidth, redundant work, and
// parallelism. Halide can't make the correct choice for you
// automatically (sorry). Instead it tries to make it easier for
// you to explore various options, without messing up your
// program. In fact, Halide promises that scheduling calls like
// compute_root won't change the meaning of your algorithm -- you
// should get the same bits back no matter how you schedule
// things.
// So be empirical! Experiment with various schedules and keep a
// log of performance. Form hypotheses and then try to prove
// yourself wrong. Don't assume that you just need to vectorize
// your code by a factor of four and run it on eight cores and
// you'll get 32x faster. This almost never works. Modern systems
// are complex enough that you can't predict performance reliably
// without running your code.
// We suggest you start by scheduling all of your non-trivial
// stages compute_root, and then work from the end of the pipeline
// upwards, inlining, parallelizing, and vectorizing each stage in
// turn until you reach the top.
// Halide is not just about vectorizing and parallelizing your
// code. That's not enough to get you very far. Halide is about
// giving you tools that help you quickly explore different
// trade-offs between locality, redundant work, and parallelism,
// without messing up the actual result you're trying to compute.
Console.WriteLine("Success!");
return(0);
}