Chapel

• is a modern, open-source programming language developed at Cray Inc.
• offers simplicity and readability of scripting languages such as Python or Matlab
• is a compiled language
• features speed and performance of Fortran and C
• supports high-level abstractions for data distribution/parallelism, and for task parallelism
  • allow users to express parallel computations in a natural, almost intuitive, manner
  • can achieve anything you can do with MPI and OpenMP
• was designed around a multi-resolution philosophy: users can incrementally add more detail to their original code, to bring it as close to the machine as required
• has its source code stored in text files with the extension .chpl

Running Chapel codes on Cedar / Graham / Beluga

On Compute Canada clusters Cedar and Graham we have two versions of Chapel, one is a single-locale (single-node) Chapel, and the other is a multi-locale (multi-node) Chapel. For now, we will start with single-locale Chapel. If you are logged into Cedar or Graham, you’ll need to load the single-locale Chapel module:

$ module spider chapel     # list all Chapel modules
$ module load gcc chapel-single/1.15.0

Running Chapel codes inside a Docker container

If you are familiar with Docker and have it installed, you can run multi-locale Chapel inside a Docker container (e.g., on your laptop, or inside an Ubuntu VM on Arbutus):

docker pull chapel/chapel-gasnet   # will emulate a cluster with 4 cores/node
mkdir -p ~/tmp
docker run -v /home/ubuntu/tmp:/mnt -it -h chapel chapel/chapel-gasnet  # map host's ~/tmp to container's /mnt
cd /mnt
apt-get update
apt-get install nano    # install nano inside the Docker container
nano test.chpl   # file is /mnt/test.chpl inside the container and ~ubuntu/tmp/test.chpl on the host VM
chpl test.chpl -o test
./test -nl 8

Running Chapel codes on the training cluster cassiopeia.c3.ca

If you are working on cassiopeia.c3.ca training cluster, please load Chapel from the shared project directory:

$ source /project/60302/shared/startSingleLocale.sh

Let’s write a simple Chapel code, compile and run it:

$ cd ~/tmp
$ nano test.chpl
    writeln('If you can see this, everything works!');
$ chpl test.chpl -o test
$ ./test

You can optionally pass the flag --fast to the compiler to optimize the binary to run as fast as possible in the given architecture.

Depending on the code, it might utilize one / several / all cores on the current node. The command above implies that you are allowed to utilize all cores. This might not be the case on an HPC cluster, where a login node is shared by many people at the same time, and where it might not be a good idea to occupy all cores on a login node with CPU-intensive tasks. Therefore, we’ll be running test Chapel codes inside submitted jobs on compute nodes.

Let’s write the job script serial.sh:

#!/bin/bash
#SBATCH --time=00:05:00   # walltime in d-hh:mm or hh:mm:ss format
#SBATCH --mem-per-cpu=1000   # in MB
./test

and then submit it:

$ chpl test.chpl -o test
$ sbatch serial.sh
$ squeue -u $USER
$ cat slurm-jobID.out

Case study: solving the Heat transfer problem

• have a square metallic plate with some initial temperature distribution (initial conditions)
• its border is in contact with a different temperature distribution (boundary conditions)
• want to simulate the evolution of the temperature across the plate

To solve the 2nd-order heat diffusion equation, we need to discretize it, i.e., to consider the plate as a grid of points, and to evaluate the temperature on each point at each iteration, according to the following finite difference equation:

Tnew[i,j] = 0.25 * (T[i-1,j] + T[i+1,j] + T[i,j-1] + T[i,j+1])

• Tnew = new temperature computed at the current iteration
• T = temperature calculated at the past iteration (or the initial conditions at the first iteration)
• the indices (i,j) indicate the grid point located at the i-th row and the j-th column

So, our objective is to:

1. Write a code to implement the difference equation above.The code should have the following requirements: (a) it should work for any given number of rows and columns in the grid, (b) it should run for a given number of iterations, or until the difference between Tnew and T is smaller than a given tolerance value, and (c) it should output the temperature at a desired position on the grid every given number of iterations.
2. Use task parallelism to improve the performance of the code and run it on a single cluster node.
3. Use data parallelism to improve the performance of the code and run it on multiple cluster nodes using hybrid parallelism.

Variables

A variable has three elements: a name, a type, and a value. When we store a value in a variable for the first time, we say that we initialized it. Further changes to the value of a variable are called assignments, in general, x=a means that we assign the value a to the variable x.

Variables in Chapel are declared with the var or const keywords. When a variable declared as const is initialized, its value cannot be modified anymore during the execution of the program.

In Chapel, to declare a variable we must specify the type of the variable, or initialize it in place with some value. The common variable types in Chapel are:

• integer int,
• floating point number real,
• boolean bool, or
• string string

If a variable is declared without a type, Chapel will infer it from the given initial value, for example (let’s store this in file baseSolver.chpl)

const rows = 100, cols = 100;      // number of rows and columns in a matrix
const niter = 500;     // number of iterations
const iout = 50, jout = 50;     // some row and column numbers

All these constant variables will be created as integers, and no other values can be assigned to these variables during the execution of the program.

On the other hand, if a variable is declared without an initial value, Chapel will initialize it with a default value depending on the declared type. The following variables will be created as real floating point numbers equal to 0.0.

var delta: real;    // the greatest temperature difference from one iteration to next
var tmp: real;      // for temporary results when computing the temperatures

Of course, we can use both, the initial value and the type, when declaring a varible as follows:

const tolerance = 0.0001: real;   // temperature difference tolerance
var count = 0: int;                // the iteration counter
const nout = 20: int;             // the temperature at (iout,jout) will be printed every nout interations

Lets print out our configuration after we set all parameters:

writeln('Working with a matrix ', rows, 'x', cols, ' to ', niter, ' iterations or dT below ', tolerance);

Ranges and Arrays

A series of integers (1,2,3,4,5, for example), is called a range in Chapel. Ranges are generated with the .. operator, and are useful, among other things, to declare arrays of variables. For example, the following variables

var T: [0..rows+1,0..cols+1] real;      // current temperatures
var Tnew: [0..rows+1,0..cols+1] real;   // newly computed temperatures

are 2D arrays (matrices) with (rows + 2) rows and (cols + 2) columns of real numbers, all initialized as 0.0. The ranges 0..rows+1 and 0..cols+1 used here, not only define the size and shape of the array, they stand for the indices with which we could access particular elements of the array using the [ , ] notation. For example, T[0,0] is the real variable located at the first row and first column of the array T, while T[3,7] is the one at the 4th row and 8th column; T[2,3..15] access columns 4th to 16th of the 3th row of T, and T[0..3,4] corresponds to the first 4 rows on the 5th column of T. Similarly, with

T[1..rows,1..cols] = 25;     // set the initial temperature

we assign an initial temperature of 25 degrees across all points of our metal plate.

We must now be ready to start coding our simulations … here is what we are going to do:

• this simulation will consider a matrix of rows by cols elements
• it will run up to niter iterations, or until the largest difference in temperature between iterations is less than tolerance
• at each iteration print out the temperature at the position (iout,jout)

Conditional statements

Chapel, as most high level programming languages, has different staments to control the flow of the program or code. The conditional statements are: the if statement, and the while statement.

The general syntax of a while statement is:

while condition do 
{instructions}

The code flows as follows: first, the condition is evaluated, and then, if it is satisfied, all the instructions within the curly brackets are executed one by one. This will be repeated over and over again until the condition does not hold anymore.

The main loop in our simulation can be programmed using a while statement like this

delta = tolerance;   // safe initial bet; could also be a large number
while (count < niter && delta >= tolerance) do {
  // specify boundary conditions for T
  count += 1;      // increase the iteration counter by one
  Tnew = T;    // will be replaced: calculate Tnew from T
  // update delta, the greatest difference between Tnew and T
  T = Tnew;    // update T once all elements of Tnew are calculated
  // print the temperature at [iout,jout] if the iteration is multiple of nout
}

Let’s focus, first, on printing the temperature every 20 interations. To achieve this, we only need to check whether count is a multiple of 20, and in that case, to print the temperature at the desired position. This is the type of control that an if statement give us. The general syntax is:

if condition then 
  {instructions A} 
else 
  {instructions B}

In our case we print the temperature at the desired position if the iteration is multiple of niter:

if count%nout == 0 then writeln('Temperature at iteration ', count, ': ', T[iout,jout]);

To print 0th iteration, we can insert just before the time loop

writeln('Temperature at iteration ', count, ': ', T[iout,jout]);

Note that when only one instruction will be executed, there is no need to use the curly brackets. % returns the remainder after the division (i.e. it returns zero when count is multiple of 20).

Let’s compile and execute our code to see what we get until now, using the job script serial.sh:

#!/bin/bash
#SBATCH --time=00:05:00   # walltime in d-hh:mm or hh:mm:ss format
#SBATCH --mem-per-cpu=1200   # in MB
#SBATCH --output=solution.out
./baseSolver

and then submitting it:

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
Temperature at iteration 20: 25.0
Temperature at iteration 40: 25.0
...
Temperature at iteration 500: 25.0

Of course the temperature is always 25.0, as we haven’t done any computation yet.

Structured iterations with for-loops

To compute the new temperature Tnew at any point, we need to add temperatures T at all the surronding points, and divide the result by 4. And, esentially, we need to repeat this process for all the elements of Tnew, or, in other words, we need to iterate over the elements of Tnew. When it comes to iterating over a given number of elements, the for-loop is what we want to use. The for-loop has the following general syntax:

for index in iterand do
  {instructions}

The iterand is a statement that expresses an iteration; it could be the range 1..15, for example. index is a variable that exists only in the context of the for-loop, and that will be taking the different values yielded by the iterand. The code flows as follows: index takes the first value yielded by the iterand, and keeps it until all the instructions inside the curly brackets are executed one by one; then, index takes the second value yielded by the iterand, and keeps it until all the instructions are executed again. This pattern is repeated until index takes all the different values exressed by the iterand.

We need to iterate both over all rows and all columns in order to access every single element of Tnew. This can be done with nested for loops like this

for i in 1..rows do {   // process row i
  for j in 1..cols do {   // process column j, row i
    Tnew[i,j] = (T[i-1,j] + T[i+1,j] + T[i,j-1] + T[i,j+1])/4;
  }
}

Now let’s compile and execute our code again:

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
...
Temperature at iteration 200: 25.0
Temperature at iteration 220: 24.9999
Temperature at iteration 240: 24.9996
...
Temperature at iteration 480: 24.8883
Temperature at iteration 500: 24.8595

As we can see, the temperature in the middle of the plate (position 50,50) is slowly decreasing as the plate is cooling down.

Exercise 1

What would be the temperature at the top right corner (row 1, column cols) of the plate? The border of the plate is in contact with the boundary conditions, which are set to zero (default boundary values for T), so we expect the temperature at these points to decrease faster. Modify the code to see the temperature at the top right corner.

Exercise 2

Now let’s have some more interesting boundary conditions. Suppose that the plate is heated by a source of 80 degrees located at the bottom right corner (row rows, column cols), and that the temperature on the rest of the border on adjacent sides (to this bottom right corner) decreases linearly to zero as one gets farther from that corner. Utilize for loops to setup the described boundary conditions. Compile and run your code to see how the temperature is changing now.

Exercise 3

So far, delta has been always equal to tolerance, which means that our main while loop will always run the 500 iterations. So let’s update delta after each iteration. Use what we have studied so far to write the required piece of code.

Now, after Exercise 3 we should have a working program to simulate our heat transfer equation. Let’s just print some additional useful information:

writeln('Final temperature at the desired position [', iout, ',', jout, '] after ', count, ' iterations is: ', T[iout,jout]);
writeln('The largest temperature difference between the last two iterations was: ', delta);

and compile and execute our final code

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
...
Temperature at iteration 500: 0.823152
Final temperature at the desired position [1,100] after 500 iterations is: 0.823152
The largest temperature difference between the last two iterations was: 0.0258874

Using command line arguments

From the last run of our code, we can see that 500 iterations is not enough to get to a steady state (a state where the difference in temperature does not vary too much, i.e. delta<tolerance). Now, if we want to change the number of iterations we would need to modify niter in the code, and compile it again. What if we want to change the number of rows and columns in our grid to have more precision, or if we want to see the evolution of the temperature at a different point (iout,jout)? The answer would be the same, modify the code and compile it again!

No need to say that this would be very tedious and inefficient. A better scenario would be if we can pass the desired configuration values to our binary when it is called at the command line. The Chapel mechanism for this is the use of config variables. When a variable is declared with the config keyword, in addition to var or const, like this:

config const niter = 500;    //number of iterations

$ chpl baseSolver.chpl -o baseSolver       # using the default value 500
$ ./baseSolver --niter=3000                # passing another value from the command line

Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
...
Temperature at iteration 2980: 0.793969
Temperature at iteration 3000: 0.793947
Final temperature at the desired position after 3000 iterations is: 0.793947
The greatest difference in temperatures between the last two iterations was: 0.00142546

Exercise 4

Make rows, cols, nout, iout, jout, tolerance configurable variables, and test the code simulating different configurations. What can you conclude about the performance of the code.

Timing the execution of code in Chapel

The code generated after Exercise 4 is the basic implementation of our simulation. We will be using it as a benchmark, to see how much we can improve the performance when introducing the parallel programming features of the language in the following lessons.

But first, we need a quantitative way to measure the performance of our code. Maybe the easiest way to do it, is to see how much it takes to finish a simulation. The UNIX command time could be used to this effect

$ time ./baseSolver --rows=650 --cols=650 --iout=200 --jout=300 --niter=10000 --tolerance=0.002 --nout=1000

Working with a matrix 650x650 to 10000 iterations or dT below 0.002
Temperature at iteration 0: 25.0
Temperature at iteration 1000: 25.0
Temperature at iteration 2000: 25.0
Temperature at iteration 3000: 25.0
Temperature at iteration 4000: 24.9998
Temperature at iteration 5000: 24.9984
Temperature at iteration 6000: 24.9935
Temperature at iteration 7000: 24.9819
Final temperature at the desired position after 7750 iterations is: 24.9671
The greatest difference in temperatures between the last two iterations was: 0.00199985

real   0m9.206s
user   0m9.122s
sys    0m0.040s

The real time is what interest us. Our code is taking around 9.2 seconds from the moment it is called at the command line until it returns. Sometimes, however, it could be useful to take the execution time of specific parts of the code. This can be achieved by modifying the code to output the information that we need. This process is called instrumentation of the code.

An easy way to instrument our code with Chapel is by using the module Time. Modules in Chapel are libraries of useful functions and methods that can be used in our code once the module is loaded. To load a module we use the keyword use followed by the name of the module. Once the Time module is loaded we can create a variable of the type Timer, and use the methods start, stopand elapsed to instrument our code.

use Time;
var watch: Timer;
watch.start();
while (count < niter && delta >= tolerance) do {
  ...
}
watch.stop();
writeln('The simulation took ', watch.elapsed(), ' seconds');

$ chpl --fast baseSolver.chpl -o baseSolver
$ ./baseSolver --rows=650 --cols=650 --iout=200 --jout=300 --niter=10000 --tolerance=0.002 --nout=1000

Working with a matrix 650x650 to 10000 iterations or dT below 0.002
Temperature at iteration 0: 25.0
Temperature at iteration 1000: 25.0
Temperature at iteration 2000: 25.0
Temperature at iteration 3000: 25.0
Temperature at iteration 4000: 24.9998
Temperature at iteration 5000: 24.9984
Temperature at iteration 6000: 24.9935
Temperature at iteration 7000: 24.9819
The simulation took 8.03206 seconds
Final temperature at the desired position after 7750 iterations is: 24.9671
The greatest difference in temperatures between the last two iterations was: 0.00199985

Exercise 5

Try recompiling without --fast and see how it affects the execution time. If it becomes too slow, try reducing the problem size. What is the speedup factor with --fast?

Solutions

Solution to Exercise 1

To see the evolution of the temperature at the top right corner of the plate, we just need to modify iout and jout. This corner correspond to the first row (iout=1) and the last column (jout=cols) of the plate.

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
Temperature at iteration 20: 1.48171
Temperature at iteration 40: 0.767179
...
Temperature at iteration 460: 0.068973
Temperature at iteration 480: 0.0661081
Temperature at iteration 500: 0.0634717

Solution to Exercise 2

To get the linear distribution, the 80 degrees must be divided by the number of rows or columns in our plate. So, the following couple of for loops at the start of time iteration will give us what we want:

// boundary conditions
for i in 1..rows do
  T[i,cols+1] = i*80.0/rows;   // right side
for j in 1..cols do
  T[rows+1,j] = j*80.0/cols;   // bottom side

Note that 80 degrees is written as a real number 80.0. The division of integers in Chapel returns an integer, then, as rows and cols are integers, we must have 80 as real so that the result is not truncated.

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
Temperature at iteration 40: 1.42663
...
Temperature at iteration 460: 0.826941
Temperature at iteration 480: 0.824959
Temperature at iteration 500: 0.823152

Solution to Exercise 3

The idea is simple, after each iteration of the while loop, we must compare all elements of Tnew and T, find the greatest difference, and update delta with that value. The next nested for loops do the job:

// update delta, the greatest difference between Tnew and T
delta = 0;
for i in 1..rows do {
  for j in 1..cols do {
    tmp = abs(Tnew[i,j] - T[i,j]);
    if tmp > delta then delta = tmp;
  }
}

Clearly there is no need to keep the difference at every single position in the array, we just need to update delta if we find a greater one.

$ chpl baseSolver.chpl -o baseSolver
$ sbatch serial.sh
$ tail -f solution.out

Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
Temperature at iteration 40: 1.42663
...
Temperature at iteration 460: 0.826941
Temperature at iteration 480: 0.824959
Temperature at iteration 500: 0.823152

Solution to Exercise 4

For example, lets use a 650 x 650 grid and observe the evolution of the temperature at the position (200,300) for 10000 iterations or until the difference of temperature between iterations is less than 0.002; also, let’s print the temperature every 1000 iterations.

$ chpl --fast baseSolver.chpl -o baseSolver
$ ./baseSolver --rows=650 --cols=650 --iout=200 --jout=300 --niter=10000 --tolerance=0.002 --nout=1000

Temperature at iteration 0: 25.0
Temperature at iteration 1000: 25.0
Temperature at iteration 2000: 25.0
Temperature at iteration 3000: 25.0
Temperature at iteration 4000: 24.9998
Temperature at iteration 5000: 24.9984
Temperature at iteration 6000: 24.9935
Temperature at iteration 7000: 24.9819
Final temperature at the desired position after 7750 iterations is: 24.9671
The greatest difference in temperatures between the last two iterations was: 0.00199985

Solution to Exercise 5

Without --fast the calculation will become slower by ~95X.