Bonus Food 🌟

🎯 Learning Objectives

• Use a for loop to iterate over a list of bonus food positions
• Understand range() for generating integer sequences
• Apply the filter pattern (list comprehension) to remove items from a list safely
• Use time.time() to track elapsed game time

💡 The Big Idea — Annotated Code

import time

bonus_foods = []   # list of [x, y, spawn_time]

BONUS_INTERVAL = 5   # spawn a new bonus food every 5 seconds
last_bonus_time = time.time()

# --- Each frame ---
now = time.time()

# Spawn new bonus food on interval
if now - last_bonus_time >= BONUS_INTERVAL:
    bx = random.randint(0, GRID_COLS - 1)
    by = random.randint(0, GRID_ROWS - 1)
    bonus_foods.append([bx, by, now])   # position + spawn time
    last_bonus_time = now

# Draw all bonus foods
for bf in bonus_foods:
    bx, by, _ = bf   # _ = "don't need spawn_time here"
    pygame.draw.rect(screen, GOLD,
        (bx * CELL_SIZE, by * CELL_SIZE, CELL_SIZE, CELL_SIZE))

# Check if snake ate any bonus food
for bf in bonus_foods:
    if snake[0] == (bf[0], bf[1]):
        score += 3

# Filter pattern — remove eaten AND expired bonus foods
bonus_foods = [
    bf for bf in bonus_foods
    if snake[0] != (bf[0], bf[1])   # not eaten
    and now - bf[2] < 8             # not expired (< 8 seconds old)
]

for bf in bonus_foods:

Iterates over every item in the list; bf is each bonus food.

time.time()

Returns current time in seconds (float). now - bf[2] = seconds since spawned.

[x, y, spawn_time]

Each bonus food carries its own metadata.

Filter pattern

[item for item in list if condition] creates a NEW list with only matching items. This safely removes items without modifying the list mid-iteration.

Why not list.remove() inside the loop?

Modifying a list during iteration skips elements. The filter pattern is safe because it creates a brand-new list.

🔢 Activity Answers (step-5-activity.html)

Tier 1 — Check Understanding

Tier 2 — Fix the Bug

Effect: Skips items — some bonus foods are never checked or removed correctly.

Fix: Use the filter pattern AFTER the loop instead of removing inside it.

Tier 3 — Extend

Make bonus food worth 5 points and grow the snake by 1 extra cell.

# When bonus food is eaten:
score += 5
tail = snake[-1]   # duplicate the tail
snake.append(tail)

Tier 4 — New Task

Add a countdown effect — bonus foods flash (draw only on even frames) when less than 2 seconds remain.

frame_num = 0

for bf in bonus_foods:
    time_left = 8 - (now - bf[2])
    if time_left > 2 or frame_num % 2 == 0:
        pygame.draw.rect(screen, GOLD,
            (bf[0] * CELL_SIZE, bf[1] * CELL_SIZE, CELL_SIZE, CELL_SIZE))

frame_num += 1

Reflective Journal — Suggested Strong Answers

⚠️ Mistakes Students Actually Make

🗣️ Teaching & Facilitation Notes

Opening hook

“How does a game spawn enemies over time and clean up dead ones? Both use the pattern you’re implementing now.”

Key pedagogical moment — The filter pattern

Spend 5 minutes with a concrete example on the board. Write numbers = [1, 2, 3, 4, 5], then show [n for n in numbers if n % 2 == 0] → [2, 4]. Ask students: “What happened to the odd numbers?” — they were never included, not deleted.

CT connection — Abstraction

Each bonus food is a self-contained data bundle (position + time). Ask: “What other properties might a game object carry?” This plants the seed for Step 9’s OOP refactor.

Differentiation

Fast finishers can add a second type of bonus food (e.g., a “slow” power-up) stored in a separate list with a different colour — this reinforces the pattern.