DAPHNE — Python Tutor

1

Strings

40 pts ▼

Problem

Produce the output:

Today, the temperature is 23.0 degrees Fahrenheit

Variable temperature = -5 (Celsius)
Variable start = "Today, the temperature is "
Variable end = " degrees Fahrenheit"
Convert using: F = (C × 9) / 5 + 32
Assemble everything into display using concatenation

Starter Code

# Strings

temperature =        # the temperature
temp_F =             # the formula
start =              # the beginning of the string
end =                # the end of the string

display =            # assemble the string here

print(display)

variable assignment arithmetic operators str() conversion string concatenation (+)

💡 Hint 1 — What type is temp_F?

temp_F will be a float after the formula. You need str(temp_F) to concatenate it with strings.

💡 Hint 2 — Watch the math

(-5 * 9) / 5 + 32 = -9 + 32 = 23.0 — division with / always returns a float in Python.

💡 Hint 3 — Spaces matter

The spaces are baked into start and end. start ends with a space, end starts with a space. Just concatenate: start + str(temp_F) + end

▶ Show Solution

# Strings — Solution

temperature = -5
temp_F = (temperature * 9) / 5 + 32
start = "Today, the temperature is "
end = " degrees Fahrenheit"

display = start + str(temp_F) + end

print(display)
# Output: Today, the temperature is 23.0 degrees Fahrenheit

2

Lists & Tuples

40 pts ▼

Problem — Step 1 (40%)

Given:

month = ('Jan','Feb','Mar','Apr','May','June','July','Aug','Sep','Oct','Nov','Dec')
expenses = (1800,2700,1202,3100,2100,1750,1845,2234,2141,3010,1911,2950)

Calculate and display:

Your total expenses are: $26,743 Your average expenses are: $2,228.58

Problem — Step 2 (60%)

Set budgetMax = 3000
Find highest monthly expense
If any expense exceeds budget, show:

You have exceeded your maximum budget at least once. The month with the highest expenses is April with $3100.

Change budgetMax = 3200 to get:

Congratulations! You kept your budget under control all year.

No loops allowed — use built-in functions only.

sum() len() max() .index() f-string formatting :,.2f format spec if/else

💡 Hint 1 — Formatting currency

Use f"${total:,}" for comma-separated integers.
Use f"${avg:,.2f}" for two decimal places with commas.

💡 Hint 2 — Finding the month name

max_exp = max(expenses) gives the highest value.
idx = expenses.index(max_exp) gives its position.
month[idx] gives the month name. But watch out — 'Apr' vs 'April'!

💡 Hint 3 — The budget check (no loops!)

You don't need to loop through expenses. Just check: if max(expenses) > budgetMax: — if the highest exceeds it, at least one did.

▶ Show Solution

# Lists and Tuples — Solution

month = ('Jan','Feb','Mar','Apr','May','June','July','Aug','Sep','Oct','Nov','Dec')
expenses = (1800,2700,1202,3100,2100,1750,1845,2234,2141,3010,1911,2950)

# Step 1
total = sum(expenses)
avg = total / len(expenses)
print(f"Your total expenses are: ${total:,}")
print(f"Your average expenses are: ${avg:,.2f}")

# Step 2
budgetMax = 3000
max_exp = max(expenses)
idx = expenses.index(max_exp)

if max_exp > budgetMax:
    print("You have exceeded your maximum budget at least once.")
    print(f"The month with the highest expenses is {month[idx]} with ${max_exp}.")
else:
    print("Congratulations! You kept your budget under control all year.")

# Note: month[3] = 'Apr', but expected output says "April"
# Your prof may want the full name — adjust the tuple or hardcode

3

While Loops

40 pts ▼

Problem

Use a while loop to compute:

Sum from n=0 to 15 of n²

Validation: Sum from n=0 to 3 = 0 + 1 + 4 + 9 = 14

while condition accumulator pattern counter increment n ** 2

💡 Hint 1 — The pattern

You need two variables: a counter (n) starting at 0, and an accumulator (total) starting at 0. Each iteration: add n**2 to total, then increment n.

💡 Hint 2 — Loop boundary

while n <= 15: — use <= not < because we want to include n=15 itself.

💡 Hint 3 — Expected answer

The answer is 1240. (0+1+4+9+16+25+36+49+64+81+100+121+144+169+196+225)

▶ Show Solution

# While Loop — Solution

n = 0
total = 0

while n <= 15:
    total += n ** 2
    n += 1

print(total)
# Output: 1240

4

Flow Control

40 pts ▼

Problem

Given random selections from:

colors = ('green', 'red', 'white', 'yellow')
vehicletype = ('priority', 'non-priority')

Build a message string that produces the correct output:

The light is green, you may proceed. The light is red, you must stop. The light is white, you must stop. (non-priority) The light is white, you may proceed as a priority vehicle. (priority) The light is yellow, you must prepare to stop.

Use random.choice() and if/elif/else to select the right action.
Graded on efficient use of if/elif/else — minimize nesting.

random.choice() if / elif / else nested conditions f-string assembly

💡 Hint 1 — Structure

First, pick a random color and vehicle type. Then use if/elif/else on the color. White is the only color that needs a nested check on vehicle type.

💡 Hint 2 — The action variables

action_1 = "you may proceed"
action_2 = "you must stop"
action_3 = "you must prepare to stop"
action_4 = "you may proceed as a priority vehicle"
Assign the right one to an action variable based on conditions.

💡 Hint 3 — Efficient branching

Only 4 branches needed:
if color == 'green': action = action_1
elif color == 'yellow': action = action_3
elif color == 'white' and vehicle == 'priority': action = action_4
else: action = action_2 # covers red AND white+non-priority

▶ Show Solution

# Flow Control — Solution
import random as r

colors = ('green', 'red', 'white', 'yellow')
vehicletype = ('priority', 'non-priority')

action_1 = 'you may proceed'
action_2 = 'you must stop'
action_3 = 'you must prepare to stop'
action_4 = 'you may proceed as a priority vehicle'

color = r.choice(colors)
vehicle = r.choice(vehicletype)

if color == 'green':
    action = action_1
elif color == 'yellow':
    action = action_3
elif color == 'white' and vehicle == 'priority':
    action = action_4
else:
    action = action_2

message = f"The light is {color}, {action}."
print(message)

5

Boolean Tests

40 pts · paper ▼

Problem

Evaluate each expression manually (no code). This section is pen-and-paper on the real exam.

Key rule: and has higher precedence than or. Evaluate and first, then or.

Expression	Your Answer	Correct
`True or False`	Think first...	True
`False and True`	Think first...	False
`1 != 2 and ('a' == 'a' or 2 == 3)`	Think first...	True
`1 != 2 or 'a' == 'a' and 2 == 3`	Think first...	True
`(1 < 2 or 3 == 3) and (1 == 0 or 1 != 1)`	Think first...	False
`(1 < 2 and 3 == 3) or (1 == 0 and 1 != 1)`	Think first...	True

Walkthrough

💡 r1: True or False

or returns True if either side is True. Left is True → True (short-circuits, never checks right side).

💡 r3: False and True

and returns False if either side is False. Left is False → False (short-circuits).

💡 r5: 1 != 2 and ('a' == 'a' or 2 == 3)

Step 1: 1 != 2 → True
Step 2: Inside parens: 'a' == 'a' → True, so (True or False) → True
Step 3: True and True → True

💡 r7: 1 != 2 or 'a' == 'a' and 2 == 3 (tricky!)

No parentheses! and binds tighter than or.
Step 1: Evaluate the and first: 'a' == 'a' and 2 == 3 → True and False → False
Step 2: Now: 1 != 2 or False → True or False → True

This is the sneaky one — precedence makes it (1!=2) or (('a'=='a') and (2==3))

💡 r9: (1 < 2 or 3 == 3) and (1 == 0 or 1 != 1)

Step 1: Left group: 1 < 2 → True, so (True or True) → True
Step 2: Right group: 1 == 0 → False, 1 != 1 → False, so (False or False) → False
Step 3: True and False → False

💡 r11: (1 < 2 and 3 == 3) or (1 == 0 and 1 != 1)

Step 1: Left group: True and True → True
Step 2: Right group: False and False → False
Step 3: True or False → True

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