PPA-2

Question

Consider the piece-wise function given below.

\[ f(x) = \left\{ \begin{array}{ll} x + 2 & \quad 0 < x < 10 \\ x^2 + 2 & \quad 10 \leq x \\ 0 & \quad \text{otherwise} \end{array} \right. \]

Accept the value of \(x\) as input and print the value of \(f(x)\) as output. Note that both the input and output are real numbers. Your code should reflect this aspect. That is, both \(x\) and \(f(x)\) should be float values.

Hint

Real numbers are represented as float values. So, you would have to accept the input as follows:

x = float(input())

The output in this case is the value of \(f(x)\), which is again to be treated as a float value according to the question. Here is an incomplete code snippet that you can use to complete the solution:

if 0 < x < 10:
    print(x + 2)

Notice the usage 0 < x < 10. This is called operator chaining. This actually corresponds to 0 < x and x < 10. We have chained two operations into a single one. Chaining enables us to smoothly translate the mathematical expression \(0 < x < 10\) to the Python expression 0 < x < 10.

Few more things to consider:

• If you use print statements within the conditional blocks, like the way it is shown here, how many of them would you require to solve this problem?
• Is there a way to solve this problem that requires just one print statement? Can you come up with that solution if it exists?

Solutions

You have to be careful about printing the value \(0\). The expected value is \(0.0\), the float value, and not \(0\), the integer value.

Solution-1

x = float(input())
if 0 < x < 10:
    print(x + 2)
elif 10 <= x:
    print(x ** 2 + 2)
else:
    print(0.0)

Solution-2

x = float(input())
if 0 < x < 10:
    y = x + 2
elif 10 <= x:
    y = x ** 2 + 2
else:
    y = 0.0
print(y)

Solution-3

Here is a slight variation of solution-2. Can you see what has changed?

x = float(input())
y = 0.0
if 0 < x < 10:
    y = x + 2
elif 10 <= x:
    y = x ** 2 + 2
print(y)

Solution-4

This is different from all three solutions before. Instead of using operator chaining, this uses the and operator:

x = float(input())
y = 0.0
if 0 < x and x < 10:
    y = x + 2
elif 10 <= x:
    y = x ** 2 + 2
print(y)

Solution-5

This is the last variation. Instead of having elif, we have an if in both places. This is also a valid solution for this problem. This is because the conditions in the two if-blocks are mutually exclusive. That is, if \(0 < x < 10\), then the body of the first if-block gets executed. There is no way for the second if-block also to get triggered since \(10 \leq x\) will evaluate to false.

x = float(input())
y = 0.0
if 0 < x < 10:
    y = x + 2
if 10 <= x:
    y = x ** 2 + 2
print(y)

It is better to stick to if-elif-else ladders wherever possible and avoid a sequence of if conditions. The if-elif-else ladder is more efficient. As soon as one of the conditions is satisfied, the control will exit from the ladder. An if-if-if ladder on the other hand will end up checking every one of the if conditions for every possible input.

Video Solution