A. Programming Concepts and Skills · Programming Basics: Data, Decisions and LoopsNested and Compound Selection
A2.2 A2.3 A1.4 A4.4About 70 minutes
Learning goals- I can write and trace an
if statement nested inside another if. - I can decide when to nest and when to combine conditions with
and or or. - I can use selection to protect a calculation from bad input, such as dividing by zero.
- I can draw a decision tree and turn it into code.
Words to know
nested selection
An if statement placed inside a block of another if, elif or else.
compound condition
A condition made from smaller ones with and, or and not.
decision tree
A branching diagram of questions and outcomes, used to plan nested selection.
guard
A condition that checks a value is safe before it is used, such as if den != 0: before dividing.
run-time error
An error that happens while the program runs, such as ZeroDivisionError. Lines before it have already run.
Questions inside questions
A movie theatre's prices: children under 13 pay $8, seniors (65 or over) pay $9, and everyone else pays $14, except on Tuesdays, when they pay $10. To find a price, you first ask about age, and only for one age group do you ask about the day. A question asked only after another question is answered is nested selection.
Expectation A2.3 asks for algorithms with nested structures. In this lesson you nest selection inside selection; in Lesson 13 you will nest selection inside loops.
Writing a nested if
An if inside a block is indented one more level (4 more spaces). Its whole if/elif/else belongs to the outer block.
- The outer decision runs first.
- The inner decision runs only when Python enters that outer block.
Indentation shows exactly which else belongs to which if, so line it up carefully.
if goes.Worked exampleExample 1: Movie prices
Write code that sets price from age and a Boolean is_tuesday, using the theatre's rules.
- Draw the decision tree (above). The first question is age, with three branches.
- Only the middle branch (13 to 64) has a second question, so put the nested
if there. - Code:
-
if age < 13:
- price = 8
- elif age >= 65:
- price = 9
- else:
- if is_tuesday:
- price = 10
- else:
- price = 14 - Trace: age 30, Tuesday: not < 13, not ≥ 65, so the
else; inside, is_tuesday is True → $10. - Trace: age 70, Tuesday:
age >= 65 → $9. The inner question is never asked.
Children $8, seniors $9, others $10 on Tuesday and $14 otherwise.
Nest, or combine with and?
The nested else in Example 1 could also be written flat:
elif is_tuesday: → price = 10else: → price = 14
Because the chain only reaches that elif when the age is 13 to 64, it means the same thing. In general:
if A: with if B: nested directly inside (and nothing else) is the same as if A and B:. Prefer the flat version: less indentation, easier to read.- Nest when the outer branch does something of its own, or when the inner question only makes sense, or is only safe, after the outer one is True.
Guards: protecting your program
Some operations crash with certain values. Dividing by zero raises a ZeroDivisionError, a run-time error: the program stops in the middle. An if can guard the dangerous step so it runs only with safe values. This is the most important use of nested selection.
num = int(input("Numerator: "))
den = int(input("Denominator: "))
if den != 0:
if num % den == 0:
print(f"{num} / {den} divides evenly: {num // den}")
else:
print(f"{num} / {den} has a remainder of {num % den}")
else:
print("Cannot divide by 0.")
With a denominator of 0, the guard sends the program to the outer else and it prints Cannot divide by 0. instead of crashing. Output, line by line: Numerator: 17 | Denominator: 5 | 17 / 5 has a remainder of 2 (the typed input is shown after each prompt)
Worked exampleExample 2: Trace a nested structure
Trace this code for x = 4, y = -2 and for x = -1, y = 5.
if x > 0: if y > 0: q = "I" else: q = "IV"else: if y > 0: q = "II" else: q = "III"print(q)
- It reports the quadrant of the point (x, y) on a coordinate grid (points on an axis go to the
else branches; a real program would check them first). - x = 4, y = −2: outer
x > 0 is True, so enter the first block. Inner y > 0 is False, so the inner else: q = "IV". - x = −1, y = 5: outer is False, so enter the outer
else. Inner y > 0 is True: q = "II".
IV for (4, −2) and II for (−1, 5).
Just for funIf it is snowing, and if it is a school day, and if the buses are cancelled, and if your homework is done, then you get to sleep in.
Programmers call that a nested if. Everyone else calls it "very unlikely".
Hands-on activityDecision tree for a board game or sport
You need: paper, a pencil, and Python. Choose a rule set you know: who goes first in a board game, which hockey age division a player is in, or how a curling end is scored.
- Draw a decision tree with at least two levels of questions. Label every branch.
- Decide which questions are independent (use
and/or or separate ifs) and which depend on an earlier answer (nest them). - Write the program with
input() for each piece of information. Use a guard for any input that could break a calculation. - Make a test table: one row for every leaf (end point) of your tree, with the inputs and the expected output.
- Run every test. Ask a family member to try to "break" your program with odd inputs.
What to look for: the number of test rows equals the number of leaves in the tree; every else lines up under its own if; there is no nested if that could be replaced by a simple and.
Your turn
1What is printed for a = 5, b = 3? For a = 5, b = 8? For a = -1, b = 8?
if a > 0: if b > a: print("X") else: print("Y")else: print("Z")
2Rewrite without nesting:
if temp < 0: if wind > 30: print("Stay inside")
if temp < 0 and wind > 30: with print("Stay inside") indented under it.
3Write a guard so that average = total / count only runs when count is not 0; otherwise print No data.
Sample:
if count != 0: average = total / count print(average)else: print("No data")
4Using Example 1's rules, what is the price for (a) age 12 on a Tuesday, (b) age 13 on a Friday, (c) age 64 on a Tuesday?
(a) $8 (under 13; the day is never checked), (b) $14, (c) $10.
5Why does if count != 0 and total / count > 50: not crash when count is 0?
Short-circuit evaluation: count != 0 is False, so Python never evaluates the division.
6Write nested selection that reads a mark from 0 to 100. If it is outside that range print Invalid; otherwise print Pass (50 or more) or Not yet.
Sample:
mark = int(input("Mark: "))if 0 <= mark <= 100: if mark >= 50: print("Pass") else: print("Not yet")else: print("Invalid")
More practice: warm-up
1What is nested selection?
An if statement inside a block of another if, elif or else.
2How do you show that an if is inside another if?
Indent it one more level (4 more spaces).
3What error does 5 / 0 cause?
4Name the condition that checks a value before a risky operation.
5if A: with if B: nested directly inside and nothing else: what single condition is the same?
More practice: on your own
1Find the bug: the program should print Freezing rain only when it is precipitating and the temperature is from −3 to 0 (inclusive).
if precip: if -3 <= temp <= 0: print("Freezing rain")else: print("Snow")
The else lines up with if precip:, so Snow prints when there is no precipitation, which is wrong. The else probably belongs to the inner if (and even then, "snow" should depend on the temperature being below −3). Indentation decides which if an else belongs to.
2A curling club: members pay $25 per game; non-members pay $35, or $30 if they are students. Write the selection with variables member and student (both Booleans).
Sample:
if member: fee = 25else: if student: fee = 30 else: fee = 35
(Flat version: if member: / elif student: / else:.)
3Trace for n = 12:
if n % 2 == 0: if n % 3 == 0: print("six") else: print("two")else: print("odd")
12 is even, so enter the block; 12 % 3 is 0, so it prints six.
4How many test cases are needed to make every print in the previous question run at least once? Give one value for each.
Three: a multiple of 6 such as 12 (six), an even number not divisible by 3 such as 4 (two), and an odd number such as 7 (odd).
5Write a program that reads two numbers and an operator (+, -, * or /) and prints the result, guarding against division by zero and unknown operators.
Sample:
a = float(input("a: "))op = input("operator: ")b = float(input("b: "))if op == "+": print(a + b)elif op == "-": print(a - b)elif op == "*": print(a * b)elif op == "/": if b != 0: print(a / b) else: print("Cannot divide by 0")else: print("Unknown operator")
The division branch has a nested guard.
Challenge
1Write a program that reads the coefficients of a linear equation ax + b = 0 and prints the solution, handling a = 0 correctly (no solution, or every x is a solution).
Sample:
a = float(input("a: "))b = float(input("b: "))if a != 0: print("x =", -b / a)else: if b == 0: print("Every x is a solution") else: print("No solution")
If a = 0 the equation is just b = 0: always true when b = 0, never true otherwise.
2Rock, paper, scissors: write a condition (one expression with and/or) that is True when player 1 wins, using strings p1 and p2.
(p1 == "rock" and p2 == "scissors") or (p1 == "scissors" and p2 == "paper") or (p1 == "paper" and p2 == "rock"). A full program checks p1 == p2 first (a tie), then this condition, else player 2 wins.
3Explain why this structure can never print C, and suggest what the author probably meant:
if x > 10: print("A")elif x > 20: print("B") if x < 5: print("C")
The elif x > 20 branch is unreachable, because any x > 20 is also > 10 and is caught by the first branch. And even inside it, x < 5 can never be true when x > 20. The author probably meant to check x > 20 first (then x > 10), and to put the x < 5 test in a separate elif or else branch.
Quiz
1. Which is the same as if a: with if b: nested directly inside it (and nothing else)?
2. For x = 3: if x > 0: → (if x > 5: → print("big"), else: → print("small")), else: → print("negative"). What is printed?
4. What decides which if an else belongs to in Python?
5. A decision tree has 5 leaves (end points). What is the minimum number of tests to run every branch?
6. Which is the best reason to nest instead of using and?
Big idea: expectation A2.2 (selection) extended to nested structures (A2.3), using compound Boolean conditions (A1.4) and tracing (A4.4).
How to help: ask your teen to explain a decision tree for something familiar (a sports league's age divisions, a bus fare chart) and count its leaves. Then ask how many tests are needed.
Common difficulties:
- An
else at the wrong indentation, so it belongs to the wrong if. - Deep nesting where a simple
and would do. - Forgetting a guard, so the program crashes on 0 or on an empty value.
- Branches that can never be reached because an earlier condition already caught those values.