Getting started with 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 “Basic.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 “Basic.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 “Basic.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 “Basic.3” we should have a working program to simulate our heat transfer equation. Let’s 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