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Engineering LibreTexts

20.3: The while Statement

  • Page ID
    15392
  • Computers are often used to automate repetitive tasks. Repeating identical or similar tasks without making errors is something that computers do well and people do poorly.

    We have seen two programs, countdown and print_n, that use recursion to perform repetition, which is also called iteration. Because iteration is so common, Python provides several language features to make it easier. One is the for statement we saw in Section 4.2. We’ll get back to that later.

    Another is the while statement. Here is a version of countdown that uses a while statement:

    def countdown(n):
        while n > 0:
            print n
            n = n-1
        print 'Blastoff!'
    

    You can almost read the while statement as if it were English. It means, “While n is greater than 0, display the value of n and then reduce the value of n by 1. When you get to 0, display the word Blastoff!

    More formally, here is the flow of execution for a while statement:

    1. Evaluate the condition, yielding True or False.
    2. If the condition is false, exit the while statement and continue execution at the next statement.
    3. If the condition is true, execute the body and then go back to step 1.

    This type of flow is called a loop because the third step loops back around to the top.

    The body of the loop should change the value of one or more variables so that eventually the condition becomes false and the loop terminates. Otherwise the loop will repeat forever, which is called an infinite loop. An endless source of amusement for computer scientists is the observation that the directions on shampoo, “Lather, rinse, repeat,” are an infinite loop.

    In the case of countdown, we can prove that the loop terminates because we know that the value of n is finite, and we can see that the value of n gets smaller each time through the loop, so eventually we have to get to 0. In other cases, it is not so easy to tell:

    def sequence(n):
        while n != 1:
            print n,
            if n%2 == 0:        # n is even
                n = n/2
            else:               # n is odd
                n = n*3+1
    

    The condition for this loop is n != 1, so the loop will continue until n is 1, which makes the condition false.

    Each time through the loop, the program outputs the value of n and then checks whether it is even or odd. If it is even, n is divided by 2. If it is odd, the value of n is replaced with n*3+1. For example, if the argument passed to sequence is 3, the resulting sequence is 3, 10, 5, 16, 8, 4, 2, 1.

    Since n sometimes increases and sometimes decreases, there is no obvious proof that n will ever reach 1, or that the program terminates. For some particular values of n, we can prove termination. For example, if the starting value is a power of two, then the value of n will be even each time through the loop until it reaches 1. The previous example ends with such a sequence, starting with 16.

    The hard question is whether we can prove that this program terminates for all positive values of n. So far, no one has been able to prove it or disprove it! (See http://en.Wikipedia.org/wiki/Collatz_conjecture.)

    Exercise \(\PageIndex{1}\)

    Rewrite the function print_n from Section 5.8 using iteration instead of recursion.

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