Unit 2 · Lesson 4

Logical Operators

Real decisions are rarely simple. "Can you ride?" depends on height and age. "Is it a holiday?" means Saturday or Sunday. Time to combine conditions.

Use and to require multiple conditions

Use or to accept alternatives

Use not to flip a condition

Build compound boolean expressions

The and Operator

and requires both conditions to be True. If either one is False, the whole expression is False:

age = 25 has_license = True if age >= 16 and has_license: print("You can drive!") else: print("You cannot drive.")

Think of and as a strict bouncer — both conditions must pass:

and — both must be True

A		B	Result
True	and	True	True
True	and	False	False
False	and	True	False
False	and	False	False

or — at least one must be True

A		B	Result
True	or	True	True
True	or	False	True
False	or	True	True
False	or	False	False

The or Operator

or only needs one condition to be True. It's only False when both are False:

day = "Saturday" if day == "Saturday" or day == "Sunday": print("It's the weekend!") else: print("It's a weekday.")

⚠️ Common mistake: Writing day == "Saturday" or "Sunday" does NOT work as expected! Python reads that as (day == "Saturday") or ("Sunday"), and since the string "Sunday" is truthy, it's always True. You must write the full comparison on each side of or.

The not Operator

not flips a boolean — True becomes False, False becomes True:

is_raining = False if not is_raining: print("No umbrella needed!") # This runs

not is great for readability. Compare: if not is_raining reads like English, while if is_raining == False is clunky.

Combining All Three

You can mix and, or, and not in the same expression. Use parentheses to make the logic clear:

age = 20 is_student = True is_member = False # Discount if: (student AND under 25) OR member if (is_student and age < 25) or is_member: print("You get a discount!") # True — student and under 25

Operator precedence: Python evaluates not first, then and, then or. But don't memorize that — just use parentheses to make your intent crystal clear!

Real-World Example: Leap Years

The leap year algorithm is a classic example of compound conditions. A year is a leap year if:

Rule: Divisible by 4, except years divisible by 100, unless they're also divisible by 400.

So 2024 is a leap year (÷4 ✓), 1900 is NOT (÷100 but not ÷400), and 2000 IS (÷400 ✓).

You'll implement this yourself in Challenge 2!

Practice Time

Challenges

🎢

Challenge 1: Ride Check

Guided

A roller coaster requires riders to be at least 48 inches tall AND at least 8 years old.

Instructions: Using and, write an if/else that sets can_ride:
• True if height >= 48 AND age >= 8
• False otherwise

Then set message to "Enjoy the ride!" or "Sorry, you can't ride yet."

Hint 1: You need both conditions to be true: height >= 48 and age >= 8
Hint 2: You can set can_ride first, then use it in the if:
can_ride = height >= 48 and age >= 8
if can_ride:
Hint 3: Or do it all in one if: if height >= 48 and age >= 8:

📅

Challenge 2: Leap Year

Guided

Implement the leap year algorithm using logical operators.

Instructions: Set is_leap to True or False based on the rules:
• Divisible by 4 → leap year
• EXCEPT if divisible by 100 → not a leap year
• UNLESS also divisible by 400 → leap year again

You can do this with a single boolean expression:
(divisible by 4 and not divisible by 100) or (divisible by 400)

Hint 1: "Divisible by 4" means year % 4 == 0. Same pattern for 100 and 400.
Hint 2: The full expression: (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
Hint 3: Test in your head: 2024 → ÷4 yes, ÷100 no → leap! 1900 → ÷4 yes, ÷100 yes, ÷400 no → not leap. 2000 → ÷400 yes → leap!

🔑

Challenge 3: Login System

Solo

Build a login checker with multiple conditions and helpful error messages.

Instructions: Check the username and password against the stored credentials. Set message to:
• "Welcome back, admin!" — if both username AND password are correct
• "Incorrect password." — if username is correct but password is wrong
• "User not found." — if username is wrong (regardless of password)

Hint 1: Think about the order of checks. Check the most specific case first (both correct), then username correct but password wrong, then everything else.
Hint 2: if username == correct_user and password == correct_pass: → welcome. elif username == correct_user: → wrong password. else: → user not found.
Hint 3: The elif only runs if the if was False, so you already know both aren't correct. If the username matches, the password must be wrong!

📐

Challenge 4: Triangle Validator

Stretch

Use the Triangle Inequality Theorem to check if three sides can form a valid triangle, and if so, what type.

Instructions: Given three side lengths, determine:
1. Set is_valid — True if all three triangle inequality conditions hold: every side must be less than the sum of the other two (a+b>c AND a+c>b AND b+c>a)
2. If valid, set triangle_type:
  • "equilateral" — all 3 sides equal
  • "isosceles" — exactly 2 sides equal
  • "scalene" — no sides equal
3. If not valid, set triangle_type to "invalid"

Hint 1: The validity check uses and three times: a + b > c and a + c > b and b + c > a
Hint 2: For equilateral: a == b and b == c. For isosceles: a == b or b == c or a == c (but not equilateral!).
Hint 3: Check validity first with if. Inside the valid branch, use if/elif/else for equilateral → isosceles → scalene. Put the invalid case in the outer else.