Conditional and Nested Conditional Statements – Programming Algorithms (Senior 4 ICT) | Ganzaa

Programming Algorithms

Lessons

Senior 4 · ICT

Programming Algorithms

5 lessons

Lesson 4 of 5

Conditional and Nested Conditional Statements

25 min

Algorithms that decide

When you send money by mobile money, the system does not simply move it. It first asks questions: is the PIN correct? Is the balance enough to cover the amount and the fee? Only if every answer is yes does the transfer go ahead; otherwise it shows a message and stops. Each question is a condition, and the part of the algorithm that chooses what to do next is a selection (conditional) statement. The algorithms in lessons 2 and 3 always did the same steps; from now on they can choose.

Definition

Condition: a Boolean expression, TRUE or FALSE, such as mark ≥ 50 or balance ≥ amount AND pinOK.

Selection: a control structure that runs different steps depending on a condition: IF…THEN (one-way), IF…THEN…ELSE (two-way), ELSE IF or CASE (multi-way).

Nested condition: an IF statement placed inside another IF, so the inner question is asked only when the outer one has a certain answer.

One-way and two-way selection

IF…THEN does something only when the condition is TRUE, and nothing otherwise. IF…THEN…ELSE always does exactly one of two things.

plaintext

// One-way
IF balance < 1000 THEN
    OUTPUT "Warning: low balance"
ENDIF

// Two-way
INPUT mark
IF mark >= 50 THEN
    OUTPUT "Pass"
ELSE
    OUTPUT "Fail"
ENDIF

Code Explanation

The lines between THEN and ELSE run when the condition is TRUE; the lines between ELSE and ENDIF run when it is FALSE. Indent them so the structure is visible.

Multi-way selection

When there are more than two outcomes, test the conditions one after another with ELSE IF. The first TRUE condition wins and the rest are skipped. A school's grade bands (illustrative) make a good example.

plaintext

INPUT mark
IF mark >= 80 THEN
    grade ← "A"
ELSE IF mark >= 70 THEN
    grade ← "B"
ELSE IF mark >= 60 THEN
    grade ← "C"
ELSE IF mark >= 50 THEN
    grade ← "D"
ELSE
    grade ← "F"
ENDIF
OUTPUT grade

Code Explanation

Because the tests go from the highest boundary down, a mark of 72 fails the first test, passes the second and gets B; the remaining tests are never asked. When the choice depends on one value matching fixed cases (such as a menu option 1, 2 or 3), many languages offer CASE choice OF 1: … 2: … OTHERWISE: … ENDCASE instead.

Worked example 1: test with boundary data

Good testing uses three kinds of data: normal (typical values), boundary (values exactly at and just beside each limit) and erroneous (values that should be rejected).

Test mark	Kind	Expected grade	Path taken
65	normal	C	fails ≥ 80 and ≥ 70, passes ≥ 60
80	boundary	A	passes ≥ 80 at once
79	boundary	B	fails ≥ 80, passes ≥ 70
50	boundary	D	passes ≥ 50
49	boundary	F	fails every test, ELSE
−5 or 130	erroneous	should be rejected	the algorithm above gives F and A: it needs a validity check

The last row exposes a bug. Fix it by checking validity first: IF mark < 0 OR mark > 100 THEN OUTPUT "Invalid mark" ELSE … the grading … ENDIF.

Nested conditions: the leap-year rule

In the Gregorian calendar a year is a leap year if it divides by 4, except that century years (dividing by 100) are leap years only if they also divide by 400. Each question only makes sense after the one before has been answered, so the IFs are nested.

plaintext

INPUT year
IF year MOD 4 = 0 THEN
    IF year MOD 100 = 0 THEN
        IF year MOD 400 = 0 THEN
            leap ← TRUE
        ELSE
            leap ← FALSE
        ENDIF
    ELSE
        leap ← TRUE
    ENDIF
ELSE
    leap ← FALSE
ENDIF
OUTPUT leap

Code Explanation

Each inner IF sits inside a branch of the outer one and has its own ENDIF. Count them: three IFs, three ENDIFs.

Worked example 2: trace the leap-year rule, then shorten it

year	MOD 4 = 0?	MOD 100 = 0?	MOD 400 = 0?	leap
2026	No (remainder 2)	not asked	not asked	FALSE
2024	Yes	No	not asked	TRUE
1900	Yes	Yes	No (remainder 300)	FALSE
2000	Yes	Yes	Yes	TRUE

The same rule fits in one compound condition, using AND and OR from Unit 2: $\text{leap} = (\text{year MOD } 4 = 0 \text{ AND year MOD } 100 \ne 0) \text{ OR year MOD } 400 = 0$ Nesting shows the reasoning step by step; the compound condition is shorter. Both give the same output for every year.

python

# The leap-year rule in Python. Press Run and type a year.
year = int(input("Year: "))
if year % 4 == 0:
    if year % 100 == 0:
        if year % 400 == 0:
            leap = True
        else:
            leap = False
    else:
        leap = True
else:
    leap = False
print(year, "is a leap year:", leap)

Code Explanation

Python writes MOD as %, comparison as ==, and marks each branch by indentation after a colon instead of ENDIF. Try 2026, 2024, 1900 and 2000 and compare with the trace table. Unit 4 teaches this syntax properly.

Mr Long Education: Selection Programming, Nested If Statements. Watch how an IF placed inside another IF handles several scenarios, and how the indentation shows which ELSE belongs to which IF.

Warning

The order of ELSE IF tests matters. If the grade algorithm tested mark >= 50 first, every mark from 50 to 100 would get D, because the first TRUE condition wins. Test from the most demanding condition down.

Match every IF with its ENDIF. In nested IFs a missing ENDIF makes it unclear which ELSE belongs to which IF. Indent each level.

Watch the boundary symbol. mark > 50 and mark >= 50 differ for exactly one mark, 50, and that is the one that gets tested.

Tip

Draw the number line first. For any banded decision (grades, fares, tax rates), sketch the ranges on a line, mark which side owns each boundary, and choose test data on both sides of every boundary.

✦

Practice

Explain the difference between IF…THEN and IF…THEN…ELSE, with an example of each.
Write pseudocode that inputs a temperature in °C and outputs "Hot" if it is above 30, otherwise "Not hot".
Trace the grade algorithm for the marks 70, 69, 100 and 0.
Rewrite the grade algorithm so that marks below 0 or above 100 output "Invalid mark".
Use the leap-year rule to decide whether 2100 and 2400 are leap years.
Write pseudocode with a nested IF: input age and whether a student has a ticket (TRUE/FALSE); output "Enter" only if the student has a ticket and is at least 12; if there is no ticket output "Buy a ticket"; if there is a ticket but the student is under 12 output "Come with an adult".
Choose six test values for a condition age >= 18 where valid ages are 0 to 120, and label each normal, boundary or erroneous.
A student orders the grade tests as >= 50, >= 60, >= 70, >= 80. Evaluate the algorithm by tracing the marks 45 and 85.

Show answersHide answers

IF…THEN runs its steps only when the condition is TRUE and does nothing otherwise (IF balance < 1000 THEN OUTPUT "Low"); IF…THEN…ELSE always runs exactly one of two branches (IF mark ≥ 50 THEN "Pass" ELSE "Fail").
INPUT temp; IF temp > 30 THEN OUTPUT "Hot" ELSE OUTPUT "Not hot" ENDIF.
70 → B; 69 → C; 100 → A; 0 → F.
INPUT mark; IF mark < 0 OR mark > 100 THEN OUTPUT "Invalid mark" ELSE (the grade IFs) OUTPUT grade ENDIF.
2100: divides by 4 and by 100 but not by 400 (remainder 100): not a leap year. 2400: divides by 400: leap year.
INPUT age, hasTicket; IF hasTicket THEN IF age ≥ 12 THEN OUTPUT "Enter" ELSE OUTPUT "Come with an adult" ENDIF ELSE OUTPUT "Buy a ticket" ENDIF.
For example: 30 and 5 (normal); 17 and 18 (boundary); −1 and 150 (erroneous).
Wrong: 45 correctly gets F, but 85 passes the first test (≥ 50) and gets D, and the other tests are skipped. The tests must start from the highest boundary.

Quick check

With the tests ordered ≥ 80, ≥ 70, ≥ 60, ≥ 50, ELSE, which grade does a mark of 70 get?

Key points

A condition is TRUE or FALSE; IF…THEN is one-way, IF…THEN…ELSE two-way, ELSE IF and CASE multi-way.
In an ELSE IF chain the first TRUE condition wins, so order the tests from the most demanding down.
Nested IFs ask an inner question only after an outer answer; each IF has its own ENDIF. A compound condition with AND/OR can often replace nesting.
Test with normal, boundary and erroneous data; validate inputs before using them.

Sources and credits

Conditional (computer programming), Leap year and Boundary-value analysis, Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0. Adapted by Ganzaa.
Video: Selection Programming Part 3, Nested If Statements, Mr Long Education, embedded from YouTube.
The scenario, grade bands (illustrative), number line, flowchart, pseudocode, code, worked examples and practice items were written by Ganzaa.

This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Algorithms that decide

2One-way and two-way selection

3Multi-way selection

4Nested conditions: the leap-year rule

Algorithms that decide

One-way and two-way selection

Multi-way selection

Nested conditions: the leap-year rule