Getting started with loops

Last updated on 2024-11-18 | Edit this page

Estimated time: 90 minutes

Overview

Questions

“How do I run the same piece of code repeatedly?”

Objectives

“Learn to use for loops to run over every element of an iterand.”
“Learn the difference between using for loops and using a while statement to repeatedly execute a code block.

To compute the new temperature, i.e. each element of temp_new, we need to add all the surrounding elements in temp and divide the result by 4. And, essentially, we need to repeat this process for all the elements of temp_new, or, in other words, we need to iterate over the elements of temp_new. 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:

// single-statement version
for index in iterand do
  instruction;

// multi-statement version
for index in iterand
{instructions}

The iterand is a function or 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 expressed by the iterand.

This for loop, for example

// calculate the new temperatures (temp_new) using the past temperatures (temp)
for i in 1..rows
{
  // do this for every row
}

will allow us to iterate over the rows of temp_new. Now, for each row we also need to iterate over all the columns in order to access every single element of temp_new. This can be done with nested for loops like this:

// calculate the new temperatures (temp_new) using the past temperatures (temp)
for i in 1..rows
{
  // do this for every row
  for j in 1..cols
  {
    // and this for every column in the row i
  }
}

Now, inside the inner loop, we can use the indices i and j to perform the required computations as follows:

// calculate the new temperatures (temp_new) using the past temperatures (temp)
for i in 1..rows
{
  // do this for every row
  for j in 1..cols
  {
    // and this for every column in the row i
    temp_new[i,j] = (temp[i-1,j] + temp[i+1,j] + temp[i,j-1] + temp[i,j+1]) / 4;
  }
}
temp=temp_new;

Note that at the end of the outer for loop, when all the elements in temp_new are already calculated, we update temp with the values of temp_new; this way everything is set up for the next iteration of the main while statement.

We’re ready to execute our code, but the conditions we have initially set up will not produce interesting output, because the plate has a temperature value of 25 everywhere. We can change the boundaries to have temperature 0 so that the middle will start cooling down. To do this, we should change the declaration of temp to:

var temp: [0..rows+1, 0..cols+1] real = 0; // the whole plate starts at 0
temp[1..rows,1..cols] = 25;                // set the non-boundary coordinates to 25

Now let’s compile and execute our code again:

BASH

chpl base_solution.chpl
./base_solution

OUTPUT

The simulation will consider a matrix of 100 by 100 elements,
it will run up to 500 iterations, or until the largest difference
in temperature between iterations is less than 0.0001.
You are interested in the evolution of the temperature at the
position (50,50) of the matrix...

and here we go...
Temperature at iteration 0: 25.0
Temperature at iteration 20: 25.0
Temperature at iteration 40: 25.0
Temperature at iteration 60: 25.0
Temperature at iteration 80: 25.0
Temperature at iteration 100: 25.0
Temperature at iteration 120: 25.0
Temperature at iteration 140: 25.0
Temperature at iteration 160: 25.0
Temperature at iteration 180: 25.0
Temperature at iteration 200: 25.0
Temperature at iteration 220: 24.9999
Temperature at iteration 240: 24.9996
Temperature at iteration 260: 24.9991
Temperature at iteration 280: 24.9981
Temperature at iteration 300: 24.9963
Temperature at iteration 320: 24.9935
Temperature at iteration 340: 24.9893
Temperature at iteration 360: 24.9833
Temperature at iteration 380: 24.9752
Temperature at iteration 400: 24.9644
Temperature at iteration 420: 24.9507
Temperature at iteration 440: 24.9337
Temperature at iteration 460: 24.913
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.

Challenge 1: Can you do it?

What would be the temperature at the top right corner of the plate? In our current setup we have a layer of ghost points around the internal grid. While the temperature on the internal grid was initially set to 25.0, the temperature at the ghost points was set to 0.0. Note that during our iterations we do not compute the temperature at the ghost points – it is permanently set to 0.0. Consequently, any point close to the ghost layer will be influenced by this zero temperature, so we expect the temperature near the border of the plate to decrease faster. Modify the code to see the temperature at the top right corner.

Show me the solution

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

BASH

chpl base_solution.chpl
./base_solution

OUTPUT

The simulation will consider a matrix of 100 by 100 elements,
it will run up to 500 iterations, or until the largest difference
in temperature between iterations is less than 0.0001.
You are interested in the evolution of the temperature at the position (1,100) of the matrix...

and here we go...
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

Challenge 2: Can you do it?

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, and that the temperature on the rest of the border decreases linearly as one gets farther form the corner (see the image below). Utilise for loops to setup the described boundary conditions. Compile and run your code to see how the temperature is changing now.

Show me the solution

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:

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

BASH

chpl base_solution.chpl
./base_solution

OUTPUT

The simulation will consider a matrix of 100 by 100 elements, it will run
up to 500 iterations, or until the largest difference in temperature
between iterations is less than 0.0001. You are interested in the evolution
of the temperature at the position (1,100) of the matrix...

and here we go...
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

Challenge 3: Can you do it?

Let us increase the maximum number of iterations to niter = 10_000. The code now does 10_000 iterations:

OUTPUT

...
Temperature at iteration 9960: 0.79214
Temperature at iteration 9980: 0.792139
Temperature at iteration 10000: 0.792139

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

Show me the solution

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

// update delta, the greatest difference between temp_new and temp
delta=0;
for i in 1..rows
{
  for j in 1..cols
  {
    tmp = abs(temp_new[i,j]-temp[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.

BASH

chpl base_solution.chpl
./base_solution

OUTPUT

The simulation will consider a matrix of 100 by 100 elements,
it will run up to 10000 iterations, or until the largest difference
in temperature between iterations is less than 0.0001.
You are interested in the evolution of the temperature at the
position (1,100) of the matrix...

and here we go...
Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
Temperature at iteration 40: 1.42663
...
Temperature at iteration 7460: 0.792283
Temperature at iteration 7480: 0.792281
Temperature at iteration 7500: 0.792279

Final temperature at the desired position after 7505 iterations is: 0.792279
The difference in temperatures between the last two iterations was: 9.99834e-05

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

// print final information
writeln('\nFinal temperature at the desired position after ',c,' iterations is: ',temp[x,y]);
writeln('The difference in temperatures between the last two iterations was: ',delta,'\n');

and compile and execute our final code,

BASH

chpl base_solution.chpl
./base_solution

OUTPUT

The simulation will consider a matrix of 100 by 100 elements,
it will run up to 500 iterations, or until the largest difference
in temperature between iterations is less than 0.0001.
You are interested in the evolution of the temperature at the
position (1,100) of the matrix...

and here we go...
Temperature at iteration 0: 25.0
Temperature at iteration 20: 2.0859
Temperature at iteration 40: 1.42663
Temperature at iteration 60: 1.20229
Temperature at iteration 80: 1.09044
Temperature at iteration 100: 1.02391
Temperature at iteration 120: 0.980011
Temperature at iteration 140: 0.949004
Temperature at iteration 160: 0.926011
Temperature at iteration 180: 0.908328
Temperature at iteration 200: 0.894339
Temperature at iteration 220: 0.88302
Temperature at iteration 240: 0.873688
Temperature at iteration 260: 0.865876
Temperature at iteration 280: 0.85925
Temperature at iteration 300: 0.853567
Temperature at iteration 320: 0.848644
Temperature at iteration 340: 0.844343
Temperature at iteration 360: 0.840559
Temperature at iteration 380: 0.837205
Temperature at iteration 400: 0.834216
Temperature at iteration 420: 0.831537
Temperature at iteration 440: 0.829124
Temperature at iteration 460: 0.826941
Temperature at iteration 480: 0.824959
Temperature at iteration 500: 0.823152

Final temperature at the desired position after 500 iterations is: 0.823152
The greatest difference in temperatures between the last two iterations was: 0.0258874

Key Points

“You can organize loops with for and while statements. Use a for loop to run over every element of the iterand, e.g. for i in 1..rows { ...} will run over all integers from 1 to rows. Use a while statement to repeatedly execute a code block until the condition does not hold anymore, e.g. while (c < niter && delta >= tolerance) {...} will repeatedly execute the commands in curly braces until one of the two conditions turns false.”