One-Way Selection: the IF Statement

A conditional statement (an "if statement") runs a block of code only when a condition is true. The condition is a Boolean expression, so it evaluates to either true or false. A one-way selection has a single IF and no ELSE: when the condition is true the block runs, and when it is false nothing happens and the program simply moves on.

battery = 15
if battery < 20:
    print("low battery")
print("done")

battery <- 15
IF (battery < 20)
{
    DISPLAY("low battery")
}
DISPLAY("done")

Here battery is 15, so battery < 20 is true and the program prints low battery and then done. If battery were 80, the condition would be false, the indented block would be skipped, and only done would print. Notice that in Python the block is marked by indentation, while AP pseudocode uses curly braces. Both show exactly which statements belong to the IF.

k = 7
if k > 10:
    k = 0
print(k)

Two-Way Selection: IF / ELSE

A two-way selection adds an ELSE. Now the program always does exactly one of two things: the IF block when the condition is true, or the ELSE block when it is false. There is no way for both to run and no way for neither to run. Exactly one branch executes, every time.

code = 4321
if code == 4321:
    print("unlocked")
else:
    print("locked")

code <- 4321
IF (code = 4321)
{
    DISPLAY("unlocked")
}
ELSE
{
    DISPLAY("locked")
}

a = 3
b = 3
if a = b:
    print("equal")

Building the Condition

The power of a conditional is in its condition. A condition is any Boolean expression, and you build it from two families of operators from the exam reference sheet.

Relational operators

Relational operators compare two values and produce a Boolean. The exam reference sheet lists six: =, ≠, >, <, ≥, and ≤. In Python these become ==, !=, >, <, >=, and <=. The two you must watch are the boundary pairs: > versus ≥ and < versus ≤. A single wrong boundary is the most common conditional bug on the exam, and it fails for exactly one input value.

Logical operators: NOT, AND, OR

Logical operators combine or flip Boolean values. NOT reverses a condition. AND is true only when both sides are true. OR is true when at least one side is true. Mixing up AND and OR is the second most common conditional error, because an OR is far easier to satisfy than an AND.

age = 70
# kids under 13 OR seniors 65+
if age < 13 or age >= 65:
    print("half price")
else:
    print("full price")

age <- 70
// kids under 13 OR seniors 65+
IF (age < 13 OR age >= 65)
{
    DISPLAY("half price")
}
ELSE
{
    DISPLAY("full price")
}

MOD for divisibility (even, odd, multiples)

A condition often needs to know if a number divides evenly. The MOD operator (written % in Python) gives the remainder after division. n MOD 2 is 0 when n is even and 1 when n is odd, so n MOD 2 = 0 is the standard test for "even." More generally, n MOD d = 0 means n is a multiple of d. You will use this constantly in conditions.

t = 6
if t < 3 or t > 6:
    print("out")
else:
    print("in")

Two Separate IFs Are Not the Same as IF / ELSE

This distinction is a favorite exam trap. With IF/ELSE, the two blocks are mutually exclusive, so exactly one runs. With two separate IF statements, each condition is checked on its own, so zero, one, or both blocks can run. The difference becomes visible whenever the first block changes a variable that the second condition then tests.

n = 6
if n > 5:
    n = n - 4
if n > 5:
    n = n + 10
print(n)   # 2

n <- 6
IF (n > 5)
{
    n <- n - 4
}
ELSE
{
    n <- n + 10
}
DISPLAY(n)   // 2

In the two-IF version on the left, n starts at 6, the first IF is true so n becomes 2, and then the second IF checks the new value: 2 is not greater than 5, so it is skipped and the result is 2. In this specific case the IF/ELSE version also gives 2, because only the first condition was ever true. But change the starting value and they diverge: for a value where the first block pushes n back above 5, the two-IF version runs the second block too, while IF/ELSE never would. The lesson: with separate IFs, always re-check each condition using the most current value.

How to Trace a Conditional

Tracing selection is a repeatable procedure. Do it the same way every time and these questions stop being guesswork:

1. Write down the current value of every variable the condition uses.
2. Substitute those values into the Boolean expression and evaluate it to true or false. Evaluate NOT first, then AND, then OR.
3. If the condition is true, run the IF block. If it is false, run the ELSE block, or skip entirely when there is no ELSE.
4. Update your variable table with any changes the block made.
5. Continue to the next statement. If it is another IF, start over at step 1 with the updated values.

if c > 100:
    print("discount")
else:
    print("full price")

Key Vocabulary