We do this in two steps, first we make the graph regular (and then we convert it into an expander). For each vertex v in graph, we generate an expander (of degree d₀ and expansion at least h₀) on d(v) vertices and replace it with v (each old edge of v will become adjacent to exactly one of the new vertices which have replaced v). All the old edges will have their previous constraint and all the new edges (edges of expanders) will be assigned an equality constraint (i.e. C(e)={(x,x) : x ∈ Σ }). Let's call this new graph G', trivially G' is regular (of degree d₀+1), is of size at most a constant times size of G and generating it, is possible in polytime. Moreover since every assignment to vertices of G can be naturally transformed (assign the value of vertex v to all the vertices of expander which is replaced v) to an assignment of vertices of G', then UNSAT(G') ≤ UNSAT(G).
Now consider a "best" assignment σ' to vertices of G', so exactly UNSAT(G') edges of G' will remain unsatisfied under this assignment. Build assignment σ to vertices of G by taking plurality (majority), i.e. for every vertex v in G, look at values (under σ') of all vertices of the expander which replaced v and set σ(v) to the most frequently visited value among them. Assume β fraction of edges of G remain unsatisfied under σ, then we claim that (for some constant c) at least cβ fraction of edges of G' is unsatisfied under σ' (Note that this will complete what we expect from this black box to perform, except the expansion property). Let F be the set of unsatisfied edges of G, F' be the set of unsatisfied edges of G' and S be the set of vertices of G' that their assigned value is different from majority vote among vertices of the expander they belong to. (It is trivial to see |F'|+|S| ≥ |F| ). Now if |F'|≥ 0.5 |F|, the claim will be trivial (note that by the construction |E'|=(d₀+1).|E| ). On the other hand if |F'| < 0.5|F|, then by previous inequality we must have |S| > 0.5|F|. Now group vertices in S that have been assigned the same value under σ' and belong to the same expander. By definition of S (majority), each of these groups has a size less than half of the vertices of that expander, and so expands with a constant factor. But this means that the whole S also must expand with a constant factor, which immediately completes the proof.
Now we have a regular graph, to convert it into an expander we just consider a (d,h₀)-expander on the same vertex set of the graph and take the union of edges. It is trivial to see every thing that we want from this black box is now established.

Let's define powering first. For input GSP G and integer t, G^t is the constraint graph on the same vertex set as G. Each uv walk of length t in G is converted into a uv edge in G^t. Each vertex v of G^t takes values from alphabet Σ^d1+t/2, we veiw these as "the opinions of v " about vertices of distance at most t/2 to v. Finally we create a constraint on edge e=uv of G^t such that e is satisfied iff opinions of u and v together, satisfy all the edges in G which can be evaluated just using opinions of u and v.

To analyse this construction we need two lemmas:

Lemma1: For any non-negative random varible X we have:

$\[ \frac{E^2\left(X\right)}{E\left(X^2\right)}\le \frac{E^2\left(X|X>0\right)P^2\left(X>0\right)}{E\left(X^2|X>0\right)P\left(X>0\right)}\le P\left(X>0\right) \]$ where last inequality is true because: $$0\le \text{Var}\left(X|X>0\right)=E\left(X^2|X>0\right)-E^2\left(X|X>0\right)$$

Lemma2: Given an expander G=(V,E) with degree d and 2nd eigenvalue λ and a set of "bad edges" F, odds that a random walk starting at a random edge in F has its i+1-st edge in F, is at most:

$$\frac{\left|F\right|}{\left|E\right|}+{\left(\frac{\lambda }{d}\right)}^i$$ We will prove this using similar techniques we saw in class for analysing random walks and then continue with the description of black box.
Let K be the distribution imposed on vertices of G by first selecting an edge in F and then selecting a vertex adjacent to that edge. Let B be the vertices adjacent to at least one edge of F and A the normalized adjacency matrix of G. Let x be the vector such that x_v=Pr_K[v] and y_v be the probability that a random step from v falls in F, we have: x_v≤ d / (2|F|) and y=2|F| x / d. The lemma seeks the following probability: $$\sum _{v\in B}{y}_v{\left(A^{i}x\right)}_v=⟨y,A^{i}x⟩$$ Write $x=x^{\parallel }+x^{\perp }$ where $x^{\parallel }=\frac{1}{n}1$. Using properties of eigenvalues and Cauchy-Schwartz inequality (respectively) we have: $$\parallel A^i{x}^{\perp }\parallel \le |\frac{\lambda }{d}{|}^i\parallel x^{\perp }\parallel \le |\frac{\lambda }{d}{|}^i\parallel x\parallel$$ $$⟨y,A^i{x}^{\perp }⟩\le \parallel y\parallel .\parallel A^i{x}^{\perp }\parallel \le 2\frac{\left|F\right|}{d}\parallel x\parallel .{\left|\frac{\lambda }{d}\right|}^i\left|\right|x\parallel \le {\left|\frac{\lambda }{d}\right|}^i$$ Now by combining everything above: $$⟨y,A^{i}x⟩=⟨y,A^i{x}^{\parallel }⟩+⟨y,A^i{x}^{\perp }⟩\le 2\frac{\left|F\right|}{dn}+|\frac{\lambda }{d}{|}^i=\frac{\left|F\right|}{\left|E\right|}+{\left(\frac{\left|\lambda \right|}{d}\right)}^i$$

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The only non-trivial claim that we had about this box is that it amplifies the amount of unsatisfaction (where UNSAT(G) is small of course). Let G^t=G' and let σ' be a "best" assignment for vertices of G'. Build the assignment σ for vertices of G this way: for every vertex v in G start a random walk of length t/2 in G and set σ(v) such that the probability that the opinion of the other end of random walk about v becomes σ(v), gets maximum. Now let F be the subset of unsatisfied edges of G under σ (but if |F| is more that 1/t fraction of the edges of G then throw out some of the members of F such that |F|/|E| ≤ 1/t ). From now on we refer to an edge e in G' and its corresponding t-walk in G by the same notation e. Let's name vertices of e by v₀,v₁,…,v_t-1,v_t. We say an edge e is hit in its i-th edge iff v_iv_i+1 ∈ F, opinion of v₀ about v_i is σ(v_i) and opinion of v_t about v_i+1 is σ(v_i+1). A hit is in the middle if $\frac{t}{2}-\sqrt{t}\le i\le \frac{t}{2}+\sqrt{t}$ . Let N(e) show the number of times an edge e is hit in the middle. Obviously every edge which is hit is unsatisfied in G', so P_e(N(e)>0) ≤ UNSAT(G') (over e taken uniformly). Now we will prove the following two lemmas, note that these two combined with Lamma1 and the previous ineqaulity will finish the proof.

Lemma3: $E\left(N\left(e\right)>0\right)\ge \frac{\left|F\right|}{\left|E\right|}\Omega \left(\sqrt{t}\right)$
Proof of this doesn't use any property of expanders, and in fact it uses standard analysis methods (defining indicator random variables and …). It relies on the fact that two poisson experiments with the same probability of success and "almost" the same number of experiments have the same distribution around their mean.

Lemma4: $E\left(N{\left(e\right)}^2>0\right)\le \frac{\left|F\right|}{\left|E\right|}O\left(\sqrt{t}\right)$
Define Z(e) to be the number of "middle" edges of e which are in F, obviously N(e) < Z(e), so we will upper-bound E(Z(e)²). Let's define an indicator variable Z_i which indicates that edge e is hit in its i-th edge. We will have: $$E\left(Z{\left(e\right)}^2\right)=\sum E\left({Z_i}^2\right)+2\sum E\left(Z_i{Z}_j\right)=4\sqrt{t}\frac{|F{|}^2}{|E{|}^2}+2\sum E\left(Z_i{Z}_j\right)$$ Since |F|/|E| ≤ 1/t, the first term of the last inequality approaches zero. The second term is upperbounded by Lemma2.

Here we accept existence of an algorithm which is called assignment tester. For our purpose this definition of an assignment tester suffices:

Assignment Tester: An A.T. is an algorithm A with a fixed alphabet Σ₀ and constant ε such that given any Boolean formula φ over a set of Boolean variables X, generates a GSP over the same set of Boolean variables X and some auxiliary variables Y which take values from Σ₀ (so these new variable are not boolean) such that

• Every assignment to X that satisfies φ, has an extension to Y that satisfies the GSP.
• Every extension b of every assignment a to X that doesn't satisfy φ, satisfies this property UNSAT(GSP) ≥ ε d(a,,SAT(φ)) where SAT(φ) indicates the set of all satisfying assignments of φ and d(.,.) is the relative hamming distance between two strings.

So suppose we have such an assignment tester T (explicit constructions of T exist, but for our purpose as far as we have a proof of existence of A.T.'s, we can use brute force search to find one). Since we will always run T on inputs of constant size, we don't need T to be efficient.
Now we build our black box. The idea is to apply T on each edge of G (input GSP) and take the union of all generated GSPs. But each edge of G is a constraint over a finite alphabet Σ and T just works on "Boolean functions with Boolean input", so we first encode each vertex of G using binary variables. Using an efficient error correcting code with relative distance ρ (which means any two valid code words are in relative distance at least ρ), we will need O(log |Σ|) new variables per vertex of G. Now we apply T to each edge of G and take the union of outputs, this union is our output GSP G'.
So far this black box does almost everything that we asked for (trivial to see), except the condition on UNSAT value. We will prove for some constant c, c.UNSAT(G) ≤ UNSAT(G') (proving UNSAT(G') ≤ UNSAT(G) will be similar). First see that w.l.o.g. we can assume that all the GSP's that we took their unions had the same size (just add extra edges when necessary). Take a "best" assignment σ' in G', we will build an assignment σ in G such that every unsatisfied edge in G generates at least c unsatisfied edges under application of T to that (for a constant c). Note that this will complete the proof. To build σ, take a vertex v in G and look at all O(log |Σ|) variables that encode v, these variable under σ' have assigned binary values (though they might not be a correct code word). Assign to σ(v) the simple in Σ with the nearest valid code word to those variables. Now look at an unsatisfied edge e=uv in G. Since e is unsatisfied, and since we have an error correcting code, at least half of bits in u or v must change to make e satisfied, which means d (σ(u,v),SAT(e)) ≥ ρ/4. (σ(u,v) indicates the assignment under σ to all variables that encode u and v). But this using the properties of T shows that at least ερ/4 fraction of edges remain unsatisfied when we apply T to e.