Unit 2 · Lesson 5
Nested Conditionals
Sometimes one decision depends on another. A shipping cost depends on weight and then on destination. Time to put conditionals inside conditionals.
Nest
if statements inside other if blocksKnow when to nest vs. when to chain
Decompose complex problems into steps
Build decision-tree programs
Decisions Inside Decisions
So far, your if/elif/else chains have been "flat" — one level of decisions. But real problems often require multi-level thinking. Consider shipping costs:
weight = 3.5 # kg
destination = "international"
if weight <= 2:
# Light package
if destination == "domestic":
cost = 5.00
else:
cost = 15.00
else:
# Heavy package
if destination == "domestic":
cost = 10.00
else:
cost = 25.00
print(f"Shipping: ${cost:.2f}") # Shipping: $25.00
The outer if checks weight. The inner if (nested inside) then checks destination. Python uses indentation to keep track of which level you're on.
Indentation is your map: Each nested level adds 4 more spaces. The outer if is at 0 spaces, its body at 4, the inner if's body at 8. If you get lost, count the spaces!
Nest vs. Chain: When to Use Which
You could rewrite the shipping example using and in a flat chain:
# Flat version with and — also works!
if weight <= 2 and destination == "domestic":
cost = 5.00
elif weight <= 2 and destination == "international":
cost = 15.00
elif weight > 2 and destination == "domestic":
cost = 10.00
else:
cost = 25.00
Both work! Here's when to prefer each:
Use nesting when: The second decision only makes sense after the first is resolved, or when you want to group related logic together. Think: "First check A, then within that, check B."
Use chaining when: All conditions are independent and you're picking from a flat list of outcomes. If nesting goes deeper than 2–3 levels, consider refactoring to a flat chain.
Use chaining when: All conditions are independent and you're picking from a flat list of outcomes. If nesting goes deeper than 2–3 levels, consider refactoring to a flat chain.
Problem Decomposition
Complex decisions are easiest when you break them into steps before writing code:
Step 1: Identify the first/biggest decision.
Step 2: For each outcome, ask "is there another decision?"
Step 3: Write the outer if/else first, then fill in the inner logic.
Step 4: Test each path independently.
Step 2: For each outcome, ask "is there another decision?"
Step 3: Write the outer if/else first, then fill in the inner logic.
Step 4: Test each path independently.
Practice Time
Challenges
Challenge 1: Shipping Calculator
Guided
Calculate shipping cost based on weight, destination, AND membership status.
Instructions: Set cost based on these rules:
Domestic: $5 if weight ≤ 5kg, $10 if heavier
International: $15 if weight ≤ 5kg, $25 if heavier
Members get 50% off the final cost (multiply by 0.5)
Use nested if/else: first check destination, then weight inside each, then apply discount.
Domestic: $5 if weight ≤ 5kg, $10 if heavier
International: $15 if weight ≤ 5kg, $25 if heavier
Members get 50% off the final cost (multiply by 0.5)
Use nested if/else: first check destination, then weight inside each, then apply discount.
Hint 1: Outer if:
Hint 2: After the nested blocks set
Hint 3:
if destination == "domestic": then inner if checks weight. Same structure for the else (international).Hint 2: After the nested blocks set
cost, add a separate if is_member: check to halve it.Hint 3:
if is_member: then cost = cost * 0.5. This should be at the same level as the outer if — not nested inside it.
Challenge 2: Movie Recommender
Guided
Recommend a movie genre based on two preference questions.
Instructions: Given likes_action and likes_romance (both booleans), set genre:
• Likes action AND romance → "Action Romance"
• Likes action only → "Thriller"
• Likes romance only → "Rom-Com"
• Neither → "Documentary"
Use nested if/else: outer checks action, inner checks romance.
• Likes action AND romance → "Action Romance"
• Likes action only → "Thriller"
• Likes romance only → "Rom-Com"
• Neither → "Documentary"
Use nested if/else: outer checks action, inner checks romance.
Hint 1:
Hint 2: The outer
Hint 3: You could also solve this with
if likes_action: — inside, check if likes_romance: → "Action Romance", else → "Thriller".Hint 2: The outer
else: means they don't like action. Inside that, check romance: if yes → "Rom-Com", else → "Documentary".Hint 3: You could also solve this with
and/or in a flat chain — both approaches work!
Challenge 3: Progressive Tax
Solo
Calculate income tax using progressive brackets — each bracket only taxes income within that range.
Instructions: Given income, calculate tax using these brackets:
• First $10,000: taxed at 10%
• Next $30,000 ($10,001–$40,000): taxed at 20%
• Anything above $40,000: taxed at 30%
Example: Income of $50,000 → $10,000×0.10 + $30,000×0.20 + $10,000×0.30 = $1,000 + $6,000 + $3,000 = $10,000
Use if/elif/else to determine which brackets apply, then calculate accordingly.
• First $10,000: taxed at 10%
• Next $30,000 ($10,001–$40,000): taxed at 20%
• Anything above $40,000: taxed at 30%
Example: Income of $50,000 → $10,000×0.10 + $30,000×0.20 + $10,000×0.30 = $1,000 + $6,000 + $3,000 = $10,000
Use if/elif/else to determine which brackets apply, then calculate accordingly.
Hint 1: If income ≤ 10000, tax is just income × 0.10. If income ≤ 40000, tax is $1000 (first bracket) + the remaining taxed at 20%.
Hint 2: For income > 40000: tax = 10000 × 0.10 + 30000 × 0.20 + (income - 40000) × 0.30
Hint 3: Three cases:
Hint 2: For income > 40000: tax = 10000 × 0.10 + 30000 × 0.20 + (income - 40000) × 0.30
Hint 3: Three cases:
if income <= 10000: → tax = income * 0.10elif income <= 40000: → tax = 10000 * 0.10 + (income - 10000) * 0.20else: → tax = 10000 * 0.10 + 30000 * 0.20 + (income - 40000) * 0.30
Challenge 4: Animal Guesser
Stretch
Build a mini decision tree that identifies an animal using 3 yes/no questions.
Instructions: Given three boolean answers, set animal using nested conditionals:
Q1: has_legs — Does it have legs?
Q2a (if legs): can_fly — Can it fly? → "eagle" if yes, then check Q3
Q3a (if legs, no fly): is_big — Is it big? → "elephant" if yes, "cat" if no
Q2b (if no legs): can_fly — Can it fly? → "bat" if yes (it's special!)
Q3b (if no legs, no fly): is_big — Is it big? → "whale" if yes, "snake" if no
This requires 3 levels of nesting!
Q1: has_legs — Does it have legs?
Q2a (if legs): can_fly — Can it fly? → "eagle" if yes, then check Q3
Q3a (if legs, no fly): is_big — Is it big? → "elephant" if yes, "cat" if no
Q2b (if no legs): can_fly — Can it fly? → "bat" if yes (it's special!)
Q3b (if no legs, no fly): is_big — Is it big? → "whale" if yes, "snake" if no
This requires 3 levels of nesting!
Hint 1: Start with
Hint 2: The full structure has 3 levels: has_legs → can_fly → is_big (for some paths). The "eagle" case is when has_legs AND can_fly (no need to check is_big).
Hint 3:
if has_legs:, then nest if can_fly: inside that.Hint 2: The full structure has 3 levels: has_legs → can_fly → is_big (for some paths). The "eagle" case is when has_legs AND can_fly (no need to check is_big).
Hint 3:
if has_legs:if can_fly:animal = "eagle"else:if is_big: → elephant, else → catelse:if can_fly: → batelse:if is_big: → whale, else → snake