4. The Accumulator Pattern¶

In the previous example, we wrote a function that computes the square of a number. The algorithm we used in the function was simple: multiply the number by itself.

In this section we will re-implement the square function and use a different algorithm, one that relies on addition instead of multiplication.

If you want to multiply two numbers together, the most basic approach is to think of it as repeating the process of adding one number to itself. The number of repetitions of this addition process is where the second number comes into play. For example, if we wanted to multiply 3 and 5, we could think about it as adding 3 to itself five times. 3 plus 3 is 6, plus 3 is 9, plus 3 is 12, and (finally) plus 3 is 15. Generalizing from this, if we want to implement the idea of squaring a number, let’s call it n, we would add `n to itself n times.

Do this by hand first and try to isolate exactly what steps you take. You’ll find you need to keep some “running total” of the sum so far, either on a piece of paper, or in your head. Remembering things from one step to the next is precisely why we have variables in a program. This means that we will need some variable to remember the “running total”. It should be initialized with a value of zero. Then, we need to update the “running total” the correct number of times. For each repetition, we’ll want to update the running total by adding the number to it.

Let’s put it another way. To square the value of n, we will repeat the process of updating a running total n times. To update the running total, we take the old value of the “running total” and add n to it. That sum becomes the new value of the “running total”.

Here is the program in activecode. Note that the function definition is the same as it was before. All that has changed is the implementation details, i.e., how the squaring is done. This is a great example of “black box” design. We can change out the details inside of the box and still use the function exactly as we did before.

def square(x): running_total = 0 # initialize the accumulator! for counter in range(x): running_total = running_total + x return running_total num = 10 result = square(num) print("The result of", num, "squared is", result)

In the program above, notice that the variable running_total starts out with a value of 0. Then the iteration is performed x times. Inside the for loop, with each iteration, the update occurs: running_total is reassigned a new value which is the old value plus the value of x.

This pattern of iterating the updating of a variable is commonly referred to as the accumulator pattern. We refer to the variable as the accumulator. This pattern will come up over and over again. Remember that the key to making it work successfully is to be sure to initialize the variable before you start the iteration. Once inside the iteration, it is required that you update the accumulator.

Here is the same program in codelens. Step through the function and watch running_total accumulate the result.

Check your understanding

Consider the following code:

def square(x):
    running_total = 0
    for counter in range(x):
        running_total = running_total + x
    return running_total

What happens if you put the initialization of running_total (the line running_total = 0) inside the for loop as the first instruction in the loop?

• The square function will return x instead of x * x
• The variable running_total will be reset to 0 each time through the loop. However because this assignment happens as the first instruction, the next instruction in the loop will set it back to x. When the loop finishes, it will have the value x, which is what is returned.
• The square function will cause an error
• Assignment statements are perfectly legal inside loops and will not cause an error.
• The square function will work as expected and return x * x
• By putting the statement that sets running_total to 0 inside the loop, that statement gets executed every time through the loop, instead of once before the loop begins. The result is that running_total is "cleared" (reset to 0) each time through the loop.
• The square function will return 0 instead of x * x
• The line running_total = 0 is the first line in the for loop, but immediately after this line, the line running_total = running_total + x will execute, giving running_total a non-zero value (assuming x is non-zero).

        func-4-2: Rearrange the code statements so that the program will add up the first n odd numbers, starting from 1, where n is provided by the user.n = int(input('How many odd numbers would
you like to add together?'))
sum = 0
odd_number = 1
---
for counter in range(n):
---
   sum = sum + odd_number
   odd_number = odd_number + 2
---
print(sum)