如何用递归函数绘制5个方块堆叠？求draw_stack函数编写思路

Hey there! Let's break this down step by step—recursion can feel a bit abstract at first, but once you nail the base case and recursive step, it'll make perfect sense.

First: Recursion Core Mindset #

Recursion works by splitting a big problem into smaller, identical subproblems until you hit a "base case" (a problem so small you can solve it immediately with no more recursion). For stacking 5 blocks:

• Big problem: Draw 5 blocks anchored at (100, 200)
• Subproblem: Draw 1 block, then stack 4 blocks on top of it
• Even smaller subproblem: Draw 1 block, then stack 3 blocks...
• Base case: Draw 0 blocks (do nothing, just return)

Second: Clarify Coordinate Logic #

First, let's assume your draw_block(screen, x, y) function draws a block with its top-left corner at (x, y). Most graphics libraries (like Pygame) use a coordinate system where y increases downward, so stacking blocks upward means decreasing the y-value by the block's height each time.

Let's define a constant for block height first (adjust this to your desired size):

BLOCK_HEIGHT = 50  # Height of each individual block

Third: Write the Recursive Function #

Here's how to implement draw_stack to handle stacking 5 blocks at your anchor point:

def draw_stack(screen, x, y, count):
    # Base case: Stop recursion if we don't need to draw any more blocks
    if count <= 0:
        return
    
    # Step 1: Draw the current (bottom-most) block at the given coordinates
    draw_block(screen, x, y)
    
    # Step 2: Recursively draw the remaining blocks stacked above this one
    # We adjust the y-coordinate upward by subtracting the block height
    draw_stack(screen, x, y - BLOCK_HEIGHT, count - 1)

To draw 5 blocks anchored at (100, 200), call it like this:

draw_stack(screen, 100, 200, 5)

Let's Walk Through the Execution #

For clarity, here's how the recursion plays out when count=5:

1. First call: Draws a block at (100, 200), then calls draw_stack(screen, 100, 150, 4)
2. Second call: Draws at (100, 150), calls draw_stack(screen, 100, 100, 3)
3. Third call: Draws at (100, 100), calls draw_stack(screen, 100, 50, 2)
4. Fourth call: Draws at (100, 50), calls draw_stack(screen, 100, 0, 1)
5. Fifth call: Draws at (100, 0), calls draw_stack(screen, 100, -50, 0)
6. Sixth call: Hits the base case (count=0), returns, and the recursion unwinds to finish.

Adjustments If Needed #

• If your height parameter was meant to be the block's height (not the count), modify the function to accept both:

  def draw_stack(screen, x, y, block_height, count=5):
      if count <= 0:
          return
      draw_block(screen, x, y)
      draw_stack(screen, x, y - block_height, block_height, count - 1)
  

  Call it with: draw_stack(screen, 100, 200, 50)
• If you want to stack downward instead of upward, change y - BLOCK_HEIGHT to y + BLOCK_HEIGHT.

内容的提问来源于stack exchange，提问作者user14050624