# Diagonal-move Conway soldiers: row-9 obstruction

In the diagonal variant a soldier may jump in any of the eight compass directions:
orthogonally or diagonally over an adjacent occupied cell into the empty cell two
steps away.  For a fixed target square `t`, use the Chebyshev distance

```text
d∞(c,t) = max(|cx - tx|, |cy - ty|)
```

and assign weight `α^d∞(c,t)`.

## The constant

The local obstruction is the same golden-ratio conjugate as in the ordinary
orthogonal game:

```text
φ = (√5 - 1) / 2,      φ² + φ = 1.
```

For a jump straight toward the target, the two removed soldiers can have weights
`φ^d` and `φ^(d-1)`, while the landing square has weight `φ^(d-2)`.  Potential
non-increase therefore requires

```text
φ^(d-2) ≤ φ^(d-1) + φ^d,
```

or equivalently `1 ≤ φ + φ²`.  Equality at `φ² + φ = 1` is the sharp case.  A
larger base would make the half-plane sum larger, so the sharp base is the
useful one.

The earlier worry that the diagonal game needs a different algebraic number was
misplaced: the move invariant uses the same `φ`.  What changes is the geometry
of the half-plane sum, because Chebyshev spheres have linear-width rows rather
than the Manhattan product used by the orthogonal proof.

## Half-plane sum for a row-N target

Place the target at `(0,N)` and sum over the starting half-plane `y ≤ 0`.  For a
row at vertical gap `k = N + n`, the horizontal contribution is

```text
H(k) = Σ_{x∈ℤ} φ^max(|x|, k)
     = (2k + 1) φ^k + 2 Σ_{m=k+1..∞} φ^m
     = (2k + 1) φ^k + 2 φ^(k+1)/(1-φ).
```

Using `1 - φ = φ²`, this is `(2k + 1)φ^k + 2φ^(k-1)`.

The whole half-plane below a row-`N` target is therefore

```text
S_N = Σ_{k=N..∞} ((2k + 1)φ^k + 2φ^(k-1))
    = (2N + 1) φ^N/(1-φ) + 4φ^(N+1)/(1-φ)²
    = (2N + 1)φ^(N-2) + 4φ^(N-3).
```

For `N = 9`,

```text
S_9 = 19φ^7 + 4φ^6 ≈ 0.8773075812 < 1.
```

So any finite army starting on row `0` or below has potential strictly less than
`1` against a row-9 target.  A soldier on the target contributes weight exactly
`1`, and legal diagonal/orthogonal jumps never increase the potential, so row 9
is unreachable.

For comparison, `S_8 ≈ 1.308057305 > 1`, matching the classical threshold shape:
the potential proof blocks row 9 while leaving row 8 possible.