Gaussian moats: why bounded-step prime walks must stop

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Gaussian moats: why bounded-step prime walks must stop

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Clase 54 de 100

Clase 54 / 100

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Suppose a walk through the Gaussian integers must avoid zero modulo some chosen Gaussian primes at every step. This class shows how to count, in bits, what that constraint costs, and why that cost proves no bounded-step walk through distinct Gaussian primes can go on forever.

The idea in one picture

Each step of a walk with step bound D is one of K_D lattice vectors, so one increment carries at most log K_D of information. This is a fixed budget per step. Zero avoidance charges a fee against it. If enough independent fees pile up, the budget runs out and the walk has to stop.

The worked case [1:35]

Modulo 2 + i, the Gaussian integers collapse to ℤ/5, because i ≡ −2 ≡ 3. The unit steps 1, i, −1, −i land on the residues 1, 3, 4, 2. Read the word (1, i, i). Its running sums are 1, 4, 2. A walk that starts at x and avoids zero needs x + sum ≢ 0 at every step, which rules out x = 4, 1, 3. Only x = 2 survives [2:36]: three increments gave away both bits of the start. Under zero avoidance, the word remembers where it began.

From one prime to a contradiction

Real batches [2:56]: take the primes p ≡ 1 (mod 4) in [T, 2T]. Combine repeated word tests, a coverage estimate (each offset hits a residue with probability about 1/p) and random choices of factors. Proposition 7.1 then gives at least c/log T of information per increment for each batch.
One wallet [3:34]: the batches are nested and the word lengths are powers of two. Block subadditivity makes the charges telescope, so their sum is at most log K_D + o(1). A walk with no sieve pays nothing. A single batch fits in the budget.
Divergence [4:25]: the batches have T = B^j and log T in [X_m, 1.05X_m], with X_m = 100^m. Each window adds about c/log B, and there are about M/2 windows. The fees grow without bound while the budget stays fixed, so no infinite walk avoids the sieve.
Uniform bound [5:16]: the sieve is periodic modulo Q. A finite cluster cannot equal its own shifted copy, so it fits inside (ℤ/Qℤ)² and has at most Q² points. Adding back the finitely many deleted primes gives B_D: finite, the same for every starting prime, and nonexplicit.

Whenever a proof claims a constraint holds at every step, ask how much information each step pays for it and what the per-step budget is [5:54]. The next thing to read is the geometric half: distinct differences and the torus-degree argument.

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