Expander Graphs and Constraint Graphs: Examples and Counterexamples

The normalized complete matrix sends every vector to its average constant vector, so it vanishes on the mean-zero space. There is one slot for every ordered pair, giving $e(S,T)=\mathrm{\mid }S\mathrm{\mid }\mathrm{\mid }T\mathrm{\mid }$ even for overlapping sets. This is exact equality in mixing, including empty or full sets and $n=1$.

For $K_{r,r}$, $M$ averages across the opposite side. Its constant vector has eigenvalue one; the vector $+1$ on one side and $-1$ on the other has eigenvalue minus one. The $2r-2$ dimensional space with zero sum on each side has eigenvalue zero. These subspaces span, so ${\mu }_2=0$ for $r\ge 2$, whereas the absolute nontrivial norm is one. A walk alternates sides, explaining why a positive algebraic gap alone does not give the absolute contraction used in mixing and walk estimates.

The Margulis bound gives $1-\alpha \ge 7\mathrm{/}80$. In the avoidance estimate, $1-\delta \le 3\mathrm{/}4$ and $1-(1-\alpha )\delta \le 1-(7\mathrm{/}80)(1\mathrm{/}4)=313\mathrm{/}320$. All factors are nonnegative, so multiplication yields the displayed estimate, including $t=0$.

For a concrete start issue take modulus $m=2$ and let $B$ be one of the four vertices. A deterministic start outside $B$ misses at time zero with probability one, whereas the stationary formula gives $3\mathrm{/}4$. Thus it does not hold unchanged for arbitrary starts. On the singleton graph a set of density at least $1\mathrm{/}4$ is the full set and avoidance is zero.

From $H_N$ form $C_N^0=A(H_N)+128I$ and $C_N^1=A(H_N)+128P_N$, where $P_N$ swaps vertices one and two for $N\ge 2$. Their entry $(1,2)$ differs by $128$, and both normalized mean-zero norms are at most $(1+{\rho }_0)\mathrm{/}2<1$. A centered indicator then bounds every normalized cut ratio below by $(1-{\rho }_0)\mathrm{/}4$ on sets of size at most half. At size one take only loops.

Assign the $i$-th polynomially clocked candidate generator size $i+2$. Choose the second matrix if its parsed output equals the first, and choose the first otherwise. The resulting family meets the same gap bound at all sizes but differs from every candidate on its assigned input. This is a counterexample to automatic constructibility of an arbitrary chosen family; the explicitly constructible family $H_N$ continues to exist.

Each original vertex has a singleton cloud. Its degree-$128$ internal graph consists of $64$ ordinary equality loops. There are therefore $128$ internal loops and one external empty-relation edge, totaling $129$. Every loop is satisfied and the external edge always fails, so every labeling has violation fraction $1\mathrm{/}129$. This matches the cloud construction's two-port count.

The degree-$128$ overlay on two vertices adds $128$ ordinary tautological edges, and the $65$ additional loops per port add $130$ more. Thus the total is $129+128+130=387$. Only the empty-relation edge fails for every labeling, yielding $1\mathrm{/}387$, exactly the overlay factor. This shows the quantitative UNSAT statement does not mean exact preservation of value. A one-symbol alphabet suffices for the example.