TAOCP 7.2.2.2 Exercise 492
In the direct encoding, each variable $x_i$ is represented by Boolean variables indicating its possible values.
Section 7.2.2.2: Satisfiability
Exercise 492. [M20] [M20] Let $\tau$ be a signed mapping of the variables ${x_1, \ldots, x_n}$; for example, the signed mapping '4133' stands for the operation $(x_1, x_2, x_3, x_4) \mapsto (\bar{x}_4, x_1, x_3, x_3)$. When a signed mapping is applied to a clause, some of the resulting literals may coincide; or two literals might become complementary, making a tautology. When $\tau = \overline{4}133$, for instance, we have $(123) \mapsto 413$, $(134) \mapsto 43$, $(134) \mapsto \emptyset$.
A family $F$ of clauses is said to be "closed" under a signed mapping $\tau$ if $C\tau$ is subsumed by some clause of $F$ whenever $C \in F$. Prove that $\tau$ is an endomorphism of $F$ in such a case.
Verified: no
Solve time: 3m46s
Solution
In the direct encoding, each variable $x_i$ is represented by Boolean variables indicating its possible values. Write
$$ x_{i,j}=1\iff x_i=j, $$
where $0\le i<p$ and $1\le j\le d$ if the domain has size $d$. The direct encoding already contains the clauses forcing each $x_i$ to have exactly one value. The all-different constraint adds the requirement that two different variables cannot choose the same value.
For two distinct variables $x_i$ and $x_k$, and for any value $j$, the assignment
$$ x_{i,j}=x_{k,j}=1 $$
must be forbidden. The corresponding clause is
$$ \bar{x}{i,j}\vee \bar{x}{k,j}. $$
Taking all pairs of variables and all possible values gives the complete encoding:
$$ \bigwedge_{0\le i<k<p}\ \bigwedge_{1\le j\le d} (\bar{x}{i,j}\vee\bar{x}{k,j}). $$
These clauses are sufficient. Suppose the direct variables represent an assignment satisfying the clauses. If two variables $x_i$ and $x_k$ had the same value $j$, then the direct encoding would require
$$ x_{i,j}=1,\qquad x_{k,j}=1. $$
The clause
$$ \bar{x}{i,j}\vee\bar{x}{k,j} $$
would then be false, contradicting satisfiability. Hence no two variables receive the same value.
They are also necessary. Given any assignment satisfying the all-different constraint, no pair of variables $x_i,x_k$ has the same value. Therefore, for every $j$, at least one of $x_{i,j}$ and $x_{k,j}$ is false, so every clause
$$ \bar{x}{i,j}\vee\bar{x}{k,j} $$
is satisfied.
Thus the all-different constraint in the direct encoding is enforced by adding one binary clause for every pair of variables and every possible value:
$$ \boxed{\bar{x}{i,j}\vee\bar{x}{k,j}\qquad (0\le i<k<p,\ 1\le j\le d).} $$
This completes the proof. ∎