= Solution
A <Gibbs sampler> updates one coordinate by drawing from its <full conditional distribution>, leaving the others unchanged and using their current values. Each coordinate kernel $K_i$ preserves the target joint law $P$: integrate first over the unchanged coordinates, then over the old coordinate, and finally draw the new coordinate from the same <conditional distribution>. The resulting joint law is again $P$. In the discrete notation, if two states share the remaining coordinates,
$$
P(x)K_i(x,x')
=P_{-i}(x_{-i})P(x_i\mid x_{-i})P(x_i'\mid x_{-i}),
$$
which is symmetric in $x_i,x_i'$. Thus a coordinate update satisfies <detailed balance>. A systematic full sweep has kernel $K_1\cdots K_k$ and remains invariant because each factor preserves $P$. A fixed-probability random-scan mixture also preserves $P$; the mixture is reversible, whereas a systematic composition need not be.
In the <autologistic binary-image model>, count each neighboring pair once. Changing one pixel from zero to one raises $n_1$ by one and $n_2$ by the number $z_{ij}$ of neighboring ones. The ratio of its two conditional weights is $\exp(\alpha+\beta z_{ij})$. Therefore
$$
\boxed{p_{ij}=\frac{e^{\alpha+\beta z_{ij}}}{1+e^{\alpha+\beta z_{ij}}}
=\operatorname{logistic}(\alpha+\beta z_{ij}).}
$$
For single-site <Gibbs sampling>, select a pixel, compute its current neighbor sum and replace it by a draw from a <Bernoulli distribution> with this probability. One may use random scans or complete ordered sweeps. The <normalizing constant> is not needed. For finite $\alpha,\beta$, every <conditional probability> lies strictly between zero and one; the finite chain is irreducible and has self-transitions, so its stationary law is unique and iterating converges to it.
For <checkerboard Gibbs sampling>, no edge joins two black pixels or two white pixels. Conditional on the whites, the black-pixel law factorizes:
$$
P(B=b\mid W=w)=
\prod_{v\in B}p_v(w)^{b_v}(1-p_v(w))^{1-b_v}.
$$
Draw all black pixels independently with these probabilities, then draw all white pixels independently conditional on the newly drawn blacks. Each color block is a full conditional update, so the two-block sweep preserves the joint target. Pixels of the same color can be updated together because their <conditional distribution> factorizes.
For the noisy image, the log likelihood contributes $-\tfrac12\sum_v(y_v-x_v)^2$. Since $x_v^2=x_v$, dropping terms independent of $x$ gives
$$
P(x\mid y)\ \propto\
\exp\!\left\{\sum_v(\alpha+y_v-\tfrac12)x_v
+\beta\sum_{\{v,w\}\text{ edge}}x_vx_w\right\}.
$$
The <Gaussian-noise posterior for an autologistic image> therefore has <conditional probabilities>
$$
\boxed{P(X_{ij}=1\mid X_{-(ij)},Y=y)
=\operatorname{logistic}\!\left(\alpha+\beta z_{ij}+y_{ij}-\tfrac12\right).}
$$
Use this probability in both single-pixel and sequential color-block schemes. The independent observation factors change only the local field, so conditional independence within each color is retained. The same invariance proof then targets the <posterior distribution> rather than the prior.
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