【离散数学】平面格点跳棋放置与合法移动规则技术问询

Alright, let's break down this grid peg puzzle clearly — it's a neat twist on classic peg solitaire, restricted to the lower half of the Cartesian plane:

Core Setup & Rules #

First, let's lay out the basics so we're all on the same page:

• Initial Placement: You start with unlimited pegs, which can be placed on any lattice point (integer-coordinate spot) where the y-coordinate is ≤ 0. That means every point on the x-axis, and all points directly below it, are fair game for your starting pegs.
• Valid Moves: The only allowed moves are horizontal or vertical jumps, and they have three non-negotiable rules:
  1. A peg must jump over an adjacent peg (so directly up, down, left, or right — no diagonal jumps, and the peg being jumped over has to share a side with the jumping peg).
  2. The jump has to land exactly 2 units away from the starting position. For example: if you jump from (x, y) over (x, y+1), you land at (x, y+2); jump left from (x, y) over (x-1, y), you land at (x-2, y), and so on.
  3. The target landing spot must be empty before you make the jump.
  4. Once the jump is done, the peg that was jumped over gets removed from the grid.

Common Key Observations (For Anyone Exploring This Puzzle) #

If you're looking to figure out what's possible here (like reaching points above the x-axis, or clearing certain areas), here are a couple of go-to analysis tools people use for these kinds of peg solitaire problems:

• Checkerboard Coloring: Color the grid like a chessboard, where (x,y) is black if x+y is even, white if odd. Each valid move swaps the count of pegs on black and white squares by +1/-1 (since you remove one peg from one color and move another to the opposite color). This invariant can help prove certain positions are unreachable.
• Weighted Sum Invariant: Assign each point (x,y) a weight of (1/2)^(|x| + |y|). The total sum of weights of all occupied pegs never increases with a valid move (since you remove one peg's weight and move another peg from a spot with weight w to a spot with weight w/2, so total sum decreases by w/2). Since the initial total sum (if you fill all y≤0 points) is finite, you can only reach a finite number of points above the x-axis, and specific high-up points might be impossible to reach because their weight would require the total sum to exceed the initial value.

内容的提问来源于stack exchange，提问作者user517784