Past exam of the computer science course of the University of Oxford / 2015 / Intelligent systems / 3 / c

Always true. An all-different constraint of arity $k$ requires $k$ distinct values. If $|D|<k$, the pigeonhole principle makes even one such constraint impossible. Since $m>0$, at least one is present, and consequently

True in some cases and false in others: the two numerical inequalities do not guarantee a solution.

For an unsatisfiable example satisfying both inequalities, take six variables, a constraint domain with five values, and $k=4$, $m=3$, with all-different constraints onAny pair of the six variables occurs together in at least one of these scopes: view the variables as the three pairs $\{1,2\},\{3,4\},\{5,6\}$, and each scope as the union of two pairs. All six values must therefore be distinct. This is impossible with five available values, even though $5>4$ and $5>3$.

For a satisfiable example with the same $k,m,|D|$, take three disjoint groups of four variables and impose one all-different constraint on each group. Assign values $1,2,3,4$ separately in every group. All three constraints hold with $D=\{1,2,3,4,5\}$.These examples concern the intended cases in which the antecedent holds; they do not rely on vacuous truth of an implication.

True in some cases and false in others. Equality of the constraint domain size and treewidth does not decide satisfiability.

For an unsatisfiable example, let $D=\{0,1\}$ and impose the three binary all-different constraints $X_1\ne X_2$, $X_2\ne X_3$, and $X_3\ne X_1$. The primal graph is a triangle, with treewidth two. It needs three different values, so no solution exists despite $|D|=\mathrm{tw}(P)=2$.

For a satisfiable example, take four variables with the same constraint domain and constraintsIts primal graph is a four-cycle of treewidth two: bags $\{1,2,3\}$ and $\{1,3,4\}$ give width two, while width one would require a forest. The assignment $(0,1,0,1)$ satisfies all constraints.

Always true. Replace every all-different constraint by the pairwise inequalities on its scope. Then solutions are precisely the proper graph colourings of the primal graph using values from $D$ as colours. Put $t=\mathrm{tw}(P)$.

A graph of treewidth $t$ has a proper graph colouring with at most $t+1$ colours. Here is a constructive proof. In a width-$t$ tree decomposition, remove any leaf bag contained in its neighbor. If more than one bag remains, a leaf bag has a vertex absent from its neighbor. The running-intersection property means that vertex occurs only in the leaf bag, so every neighbor of the vertex belongs to that bag. It has at most $t$ neighbors. If only one bag remains, it has at most $t+1$ vertices and the same degree bound holds.

Remove the chosen vertex and repeat; deleting vertices cannot increase the width of the restricted tree decomposition. Now colour in reverse removal order. At most $t$ already coloured neighbors forbid colours, so one of $t+1$ colours is always available. Since the integer inequality $|D|>t$ gives $|D|\ge t+1$, this produces a satisfying assignment:The common unrestricted constraint domain and the exclusively all-different constraints are essential: arbitrary constraint relations do not reduce to graph colouring in this way.

Always true for a nonempty finite constraint domain, as conventionally assumed. Write $q=|D|$ and identify its values with $\{0,\dots ,q-1\}$. For each satisfying assignment $(x_1,\dots ,x_n)$, simultaneously replace every value byThis permutation of the common constraint domain preserves every equality and inequality, hence every all-different constraint. The resulting $q$ assignments are distinct: equality of even one coordinate before and after a shift forces $a=0(\mathrm{mod}q)$. There is at least one variable, since $m>0$ and $k\ge 2$.

The cyclic group of shifts therefore partitions the solution set into disjoint orbits of size $q$. It follows thatThe conclusion includes an unsatisfiable instance, whose solution count is zero. Full permutation group orbits need not have size $q!$, because unused values can be permuted without changing an assignment; the cyclic-shift argument avoids that mistake.

If an empty constraint domain is permitted, there are zero assignments here, but the wording “divisible by $|D|$” then asks about divisibility by zero, which is ordinarily left undefined. The asserted divisibility uses the standard nonempty-domain convention, and the no-solution conclusion still holds for that degenerate input.