Popcorn

A. Popcorn Hack 1

Predictions in comments

Code Runner Challenge

UESL 3.2 Homework - grow the level collection

View IPYNB Source 
level_stars = [10, 20, 15]
# the following line will output 10. I know this because Python indexing starts at 0, so index 0 is the 1st element of the level_stars list.
print(level_stars[0])
# the following line will output 20. I know this because index 1 is the 2nd element of the level_stars list.
print(level_stars[1])
# the following line will output 3. This is because level_stars has 3 objects and thus is of length 3.
print(len(level_stars))
Lines: 1 Characters: 0
Output
Click "Run" in code control panel to see output ...

Actual output: 10, 20, 3

B. Popcorn Hack 2

Prediction: [6, 12, 12, 5] and a level count of 4. AP index 2 selects the same second level as Python index 1.

Run and Change: After changing the append value from 5 to 8, the new list will be [6, 12, 12, 8], with the new object 8 being added and the object at index 1 being changed from 9 to 12. Python (edited version, appending 8):

Code Runner Challenge

UESL 3.2 Test - empty project

View IPYNB Source 
level_stars = [6, 9, 12]
level_stars[1] = level_stars[1] + 3
level_stars.append(8)
print(level_stars)
print(len(level_stars))
Lines: 1 Characters: 0
Output
Click "Run" in code control panel to see output ...

Pseudocode (edited version):

level_stars ← [6, 9, 12]
level_stars[2] ← level_stars[2] + 3
APPEND(level_stars, 8)
DISPLAY(level_stars)
DISPLAY(LENGTH(level_stars))

Actual output: [6, 12, 12, 8], then 4. Original version (appending 5) gave [6, 12, 12, 5], then 4.

AP index: 2. Updating level_stars[1] just replaces a value that’s already there so the length stays 3, but append adds a new element so it goes up to 4.

MCQ

MCQ 3.2: 4/4 answers: B,A,C,B
  1. Which AP index selects the first level? A 0 · B 1 · C the list length. B because AP starts at one.
  2. Replacing one count changes the list length by: A 0 · B 1 · C −1. A because replacing the index doesn’t change the list length.
  3. What does APPEND(level_stars, 5) do? A Change level five · B Add five levels · C Add one element with value 5. C because append adds an element of value 5.
  4. Why use a list plus traversal for 100 levels? A It always runs faster · B It avoids a separate variable and sum edit for every level · C It prevents all errors. B - lists compress multiple different variable values into one list.

Homework

Refactored the 3.1 prototype into one list. Each element is the stars collected on one level, in level order (level 1 = 12, level 2 = 18, level 3 = 10). Then I add 5 to the second level, append a new level with 7 stars, and total it with Example C’s loop.

Prediction: [12, 23, 10, 7], 4, 52

# CODE_RUNNER: UESL 3.2 Homework - grow the level collection
level_stars = [12, 18, 10]

if len(level_stars) >= 2:
    level_stars[1] = level_stars[1] + 5
level_stars.append(7)

total_stars = 0
for stars in level_stars:
    total_stars = total_stars + stars

print(level_stars)
print(len(level_stars))
print(total_stars)

Actual output: [12, 23, 10, 7], 4, 52. Matches the prediction.

The if len(level_stars) >= 2 check skips the second-level update when there is no second level, otherwise level_stars[1] would crash on a short list.

Why a list is better than 100 separate variables: with separate variables I’d need level_1_stars through level_100_stars (100 lines) and a total line that adds all 100 by name. Adding level 101 would mean writing a new variable AND editing the total line, and it’s easy to forget one. With a list, all the stars are one line of data, a new level is one append, and the for loop doesn’t change at all because it goes through however many levels there are.

Also like in 3.1, changing the list doesn’t update a total that was already calculated. The loop has to run again (with total_stars reset to 0) to get the new total.

Tests

Language: Python. Each test starts with its own list.

Test 1: Main is the homework cell above: [12, 18, 10] → [12, 23, 10, 7], 4, 52.

Test 2: Empty project, total-only code with []

Prediction: 0, 0

# CODE_RUNNER: UESL 3.2 Test - empty project
level_stars = []

total_stars = 0
for stars in level_stars:
    total_stars = total_stars + stars

print(len(level_stars))
print(total_stars)

Actual output: 0, 0

Test List Expected Actual Pass
Main [12, 18, 10] [12, 23, 10, 7], 4, 52 [12, 23, 10, 7], 4, 52 Yes
Empty [] 0, 0 0, 0 Yes

The same loop works for 4 levels and 0 levels without changing any code. With the empty list the loop just runs 0 times so the total stays at 0.

Design Thinking

Phase What I Did
Need A game maker wants to add or reorder levels without losing track of each level’s stars
Goal Keep every level’s stars in one collection, in level order, that’s easy to read, update and add to
Representation considered Separate variables for each level (like 3.1) vs. one list + a for loop. Went with the list
Prototype level_stars = [12, 18, 10] with an update, an append and Example C’s loop for the total
Actual test Ran the main list (got 4 levels, 52 stars) and the empty list (got 0, 0)
Revision Running the update on an empty list would crash with an IndexError since there’s no level_stars[1], so I added if len(level_stars) >= 2 before it

Suggestion for Better UX

Players and makers shouldn’t have to think in indices. Add level names next to the star counts:

level_names = ["Forest", "Castle", "Mountain", "Sky"]
level_stars = [12, 23, 10, 7]

for i in range(len(level_names)):
    print(f"{level_names[i]}: {level_stars[i]} stars")