Behind the scenes of the proof that level 5 is impossible.

Level five of what? Geometry Dash, of course.

JK, Back on Track is harder.

I’m talking about Conway’s Soldiers.

And today, just for comfort, instead of calling the soldiers soldiers, I will be calling them pegs.

Conway’s Soldiers is simple. There are an infinite number of pegs behind a line that divides the land between the pegs and the enemy. The pegs may jump another peg vertically or horizontally, but never diagonally.

A number, Φ, exists so that:

{\displaystyle \varphi ={\frac {{\sqrt {5}}-1}{2}}\approx 0.61803\,39887\ldots }
If you couldn’t tell, Φ is the reciprocal of the golden ratio.

Select the target square, which, in our case is any square on level 5, and mark it with Φ^0, or just 1 for simplicity. All the other squares on the enemy territory will be marked with Φ^n, where n is the Manhattan distance to Φ^0. Also, for simplicity, Φ^0’s square will be called T, for target.

We can define a “score” for the configuration of the pegs. If two pegs are lined up to reach T after one jump, the score is Φ^1 + Φ^2. The score can change in three ways:

  • If a jump moves a peg towards T, the change in score is Φ^(n – 2) – Φ^(n – 1) – Φ^n. That is because The original pegs are at Φ^(n – 2) and Φ^(n – 1), and when the peg jumps, the jumped peg, Φ^(n – 1) is removed, hence subtracting the score, and Φ^(n – 2) moves to Φ^n, hence subtracting Φ^(n – 2) and adding Φ^n, which leads to 0.
  • If a jump doesn’t impact the value of the jumping peg, the score is changed by -Φ^(n – 1).
  • If a jump moves a peg away from T, the score is changed by -2Φ^(n + 1).

Therefore, the score can never increase.

Now, let’s think about it at a different approach. If there was a row of pegs right behind T, the score would be:

{\displaystyle \varphi +2\varphi ^{2}+2\varphi ^{3}+\ldots }
and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on…

The score of a line of pegs n spaces away would be {Φ^(n – 1)} * (Φ + 2Φ^2 + 2Φ^3 + 2Φ^4 + …). Two things:

\sum _{{n=2}}^{{\infty }}\varphi ^{n}=1
and also,
{\displaystyle \varphi ^{2}=1-\varphi }

That means that if S(1) was the starting score for a board with T at level one, S(1) = 5 + 3Φ. Since the difference of two squares directly next to each other is logarithmically Φ, S(n + 1) = S(n) * Φ.

So:

  • S(2) = 3 + 2Φ
  • S(3) = 2 + Φ
  • S(4) = 1 + Φ
  • S(5) = 1

Now, let there be an end configuration score E, where E = 1 + ϵ, where 1 is the contribution of the peg on T, and ϵ the contribution of the peg behind the line. ϵ is positive, but can be as small as possible.

Now, for S(5), which is one, since ϵ > 0, E > S(5). But the score never increases, so therefore that must mean that S(5) ≥ E. We have reached a contradiction, since if a > b, then b ≥ a cannot be true.

Q.E.D.

Credits to Wikipedia for the information. For detail, go check out the page.

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