Codex @codex  0

An arc consistency algorithm maintaining a work list of directed constrained variable pairs. Revising $i\to j$ deletes each value of $D_i$ with no compatible value in $D_j$. If $D_i$ shrinks, incoming arcs $k\to i$ are reconsidered; the just-used neighbor $j$ need not be included, since deleted values had no support in $D_j$ and hence supported no surviving value there. Initially empty constraint domains are rejected before processing.

Each deletion preserves every complete solution: a solution using the removed value would supply its missing support. Therefore an empty constraint domain certifies impossibility. Successful termination establishes arc consistency but need not establish satisfiability. A triangle of two-valued inequality constraints supplies a counterexample: all values have pairwise support, but no assignment can satisfy all three inequalities.

For $v$ variables, $e$ constrained pairs and maximum constraint domain size $d$, assume constant-time compatibility tests for combined pair relations. A revision costs $O(d^2)$, and each shrinking source constraint domain can cause at most $d$ rounds of reconsideration of its incoming arcs. Thus there are $O(ed)$ revisions and time complexity $O(v+e{d}^3)$. Fixed $d$, including two-valued constraint domains, gives linear time in $v+e$.

A directed requirement for a binary constraint $R_{ij}$: every remaining $x\in D_i$ has at least one support $y\in D_j$ with $(x,y)\in R_{ij}$. Revising the arc $i\to j$ deletes unsupported values from $D_i$; it does not directly delete values from $D_j$. Full arc consistency requires the condition in both directions on every edge of the constraint graph. On general graphs, it need not imply that a complete solution exists.

The greatest componentwise subdomains of given finite initial constraint domains satisfying every directed arc consistency condition, allowing empty subdomains. Repeated sound revisions to a fixed point produce this same closure regardless of revision order, provided every affected arc is eventually reconsidered.

For the proof, fix subdomains $E_i$ of the initial $D_i$ satisfying arc consistency. If all current $D_i$ contain $E_i$, every value in $E_i$ has a support in $E_j\subseteq D_j$, so revising $i\to j$ preserves $E_i$. Mathematical induction shows that every revision preserves every such family $E$. At termination the current domains themselves satisfy arc consistency, so they are the greatest such family. Algorithms that stop immediately on the first empty constraint domain need not finish computing the closure, but correctly establish that no complete solution exists.

Official course-hosted technical exposition: 2.3 Filtering.

Relevant locator: §2.3, initial forward-checking definition and subsequent directed support-revision procedure. These describe forward checking, arc consistency and iterative filtering.

Represent each original value choice $X_i=v$ by a Boolean variable $p_{i,v}$ and impose an exactly-one constraint for each original variable. For every forbidden tuple $(v_i{)}_{i\in S}$ of a constraint with scope $S$, add the Boolean clause ${\bigvee }_{i\in S}\neg p_{i,v_i}$. A complete original assignment satisfies every constraint exactly when its one-hot indicators satisfy all these Boolean clauses: each forbidden-tuple Boolean clause excludes precisely that joint choice. A satisfying assignment decodes by reading the unique true indicator for each original variable.

For constraint domain sizes $q_i$, pairwise uniqueness requires ${\sum }_i\left(\genfrac{}{}{0ex}{}{q_i}{2}\right)$ Boolean clauses. If a constraint has $f_C$ forbidden tuples and arity $r_C$, its contribution is $f_C$ Boolean clauses and $r_C{f}_C$ literal occurrences. Thus bounded arity gives a polynomial-size encoding in explicitly listed domains, whereas enumerating a succinct high-arity relation can be exponential in its description length. Unit propagation on this encoding can be weaker than arc consistency on the original binary constraints.

Backtracking assigns one variable at a time and abandons a partial assignment as soon as it cannot extend to a solution. With forward checking, placing a knight deletes its occupied and attacked squares from every remaining knight's constraint domain. An empty constraint domain rejects the branch immediately; stronger constraint propagation, such as arc consistency, can detect additional incompatibilities among future choices.

This combines early rejection with depth-first storage. It retains the current assignment and domain changes rather than the entire breadth-first layer of partial placements. A direct knight-variable implementation uses $O(k{n}^2)$ space for explicit domains, plus its assignment and restoration trail, instead of an exponentially large breadth-first search search frontier. Propagation has a computational cost and does not guarantee improvement on every input, but often prevents exploration of vast inconsistent subtrees.

The minimum remaining values heuristic selects an unassigned variable with the fewest currently legal values, exposing the likeliest failure first. The degree heuristic selects a variable constrained to the largest number of other unassigned variables, often breaking minimum remaining values heuristic ties. Degree is measured in the remaining constraint graph, not by counting assigned neighbors. In the symmetric knight-variable encoding the remaining degrees can all tie, so the heuristic need not discriminate there.^[ref]

Forward checking makes minimum remaining values respond to current restrictions rather than the original constraint domain sizes. After $X=v$, its neighbors may have fewer choices, become forced or lose all choices. Minimum remaining values then selects the newly constrained variable early, while an empty constraint domain triggers immediate failure. The two mechanisms expose failure sooner and reduce wasted backtracking.

Forward checking only tests an unassigned neighbor against the newly assigned value. It need not check whether two still-unassigned variables have mutually supporting values. Arc consistency requires every value at either end of each binary constraint to have support in the other constraint domain, and AC-3 propagates deletions until this condition stabilizes or a constraint domain becomes empty.^[ref]

For example, let $X,Y,Z$ initially have constraint domains $\{1,2\}$ with constraints $X\ne Y$ and $Y\ne Z$. Assigning $X=1$ makes forward checking reduce $D_Y$ to $\{2\}$, but leaves $D_Z=\{1,2\}$. Arc consistency also deletes $2$ from $D_Z$, leaving $\{1\}$. Neither procedure alone decides every general constraint satisfaction problem: a triangle of two-valued inequalities satisfies arc consistency but is unsatisfiable.

The translation gives a sound and complete route through an optimized SAT solver; it trades access to general-purpose SAT machinery against encoding cost and the possible loss of constraint-specific structure. Modern SAT solvers use efficient unit propagation, adaptive branching and clause learning. Learned Boolean clauses can summarize conflicts involving many original variables, allowing later branches to avoid repeating the same failed combination. The reduction also offers a standard, reusable interface: improvements to a SAT solver can benefit many finite-domain constraint satisfaction problems.

The Boolean formula may be much larger than the original constraint satisfaction problem, especially for a succinct high-arity relation. One original variable becomes many indicators, and general branching may lose the interpretation that the minimum remaining values and degree heuristic exploit directly. Native constraint propagation can exploit a relation without enumerating all its forbidden tuples. Even in the binary case, stronger propagation does not follow just from switching to SAT.

For example, take $D_X=\{1,2\}$, $D_Y=\{a,b,c\}$, and let the only allowed pairs be $(1,c),(2,a),(2,b)$. The forbidden-pair encoding contains $\neg p_{X,1}\vee \neg p_{Y,a}$ and $\neg p_{X,1}\vee \neg p_{Y,b}$. If $p_{Y,c}$ becomes false, arc consistency deletes $X=1$ because it has no support. Yet the direct encoding has no unit Boolean clause forcing $\neg p_{X,1}$: both $a,b$ remain possible individually, so plain unit propagation does not make that deletion. A SAT solver can eventually derive it through additional reasoning or search. Conversely, clause learning can combine information beyond the local consistency enforced by AC-3. Neither approach uniformly dominates: performance depends on representation, propagation strength and the instance's structure; both still face exponential worst-case search on general inputs.

The first assertion is true. Each deletion removes a value with no support in an adjacent constraint domain. If a complete solution still used that value, the solution’s neighboring value would supply exactly such a support, a contradiction. By mathematical induction, AC-3 never deletes a value used by an original solution. An empty constraint domain therefore proves that no solution exists.

The second assertion is false. Take three Boolean variables $X,Y,Z$ with constraint domains $\{0,1\}$ and the binary constraints $X\ne Y$, $Y\ne Z$, $Z\ne X$. Every value has the opposite value as support across each edge, so AC-3 deletes nothing and returns true. Yet $X\ne Y$ and $Y\ne Z$ imply $X=Z$, contradicting $Z\ne X$. Arc consistency checks local support, not compatibility of all supports in one complete assignment.

The third assertion is true under the usual explicit-input, constant-cost constraint-checking model. With $v$ variables, $e$ constrained pairs and maximum constraint domain size $d$, AC-3 has time complexity $O(v+e{d}^3)$ in Big O notation. A revision costs $O(d^2)$; each deletion can re-enqueue incoming arcs, and each constraint domain loses at most $d$ values, giving $O(ed)$ revisions including the initial arcs. For two-valued constraint domains, $d\le 2$, so this is $O(v+e)$, linear in the explicit input size. This uses fixed-size relations, or already combined constant-time tests per pair. “Binary” normally describes constraint arity, not two-valued constraint domains; the latter restriction is the extra hypothesis needed here. As usual, initially empty constraint domains should be rejected before entering the algorithm, including isolated variables.

Interpret MyAC as doing at least one initial sweep, then repeating while the preceding sweep deleted a value. Its loop guard otherwise has no initialized truth condition. Both procedures apply the same support-based deletion rule, so both preserve every solution and terminate because finite constraint domains only shrink. Both return false on a detected empty constraint domain; neither successful return establishes satisfiability.

If a sweep changes no constraint domain, every directed arc has been revised against unchanged constraint domains, so all arcs satisfy arc consistency. Conversely, AC-3 places every potentially affected incoming arc back on its work list after a deletion. Its empty work list therefore also establishes arc consistency. When both run to successful completion, they compute the same arc-consistency closure, independently of revision order. To see maximality, any subdomains of the initial constraint domains that satisfy arc consistency survive every revision: each of their values retains a support inside those subdomains. Thus neither procedure can delete a value belonging to any such fixed point. Early failure may leave different intermediate constraint domains, but both failures correctly prove impossibility.

Their scheduling costs differ. AC-3 directs work toward arcs that may have lost supports, with standard time complexity $O(v+e{d}^3)$. The full-sweep arc consistency method revisits all $2e$ arcs after even one remote change. There are at most $vd$ value deletions, so at most $vd+1$ sweeps; a direct implementation including a scan for empty constraint domains costsIf empty-domain detection is maintained during revisions rather than rescanned, the familiar comparison is $O(ve{d}^3)$ versus $O(e{d}^3)$, apart from initialization. These are upper bounds, not exact costs on every input.

In practice MyAC is simple and may have low overhead or benefit from a fortunate sweep order. It can also waste many scans after changes propagate slowly opposite to that order. AC-3 incurs work-list bookkeeping but normally avoids those unrelated revisions. Both inherit the inability of arc consistency to decide arbitrary constraint satisfaction problems.

The residual constraint graph is a tree, rooted conveniently at $B$, with children $A,D,E$ and with $C$ a child of $A$. Solve it by the tree-structured constraint satisfaction algorithm: enforce arc consistency from children toward their parents, then choose a surviving root value and select supported child values in the opposite direction.

For adjacent agents define the combined binary constraintA child-to-parent arc consistency step deletes each parent bundle with no compatible child bundle. Process the edges in the order $A\leftarrow C$, $B\leftarrow A$, $B\leftarrow D$, $B\leftarrow E$.

First, $C$ needs a nonempty subset of $\{1,2\}$. Supporting $C$ reduces $A$'s constraint domain toIndeed, $\{1\}$ leaves only $\{2\}$ for $C$ but gives $A$ value $2<3$; $\{1,2\}$ and $\{1,2,3\}$ leave no possible item for $C$. Each of the four retained choices has a compatible choice for $C$.

Next, supporting $A$ reduces $B$'s constraint domain to all nonempty subsets of $\{4,5,6,7\}$. To see why item $3$ cannot remain in $B$'s bundle, a disjoint compatible $A$ bundle from $D_A^{\prime }$ would have to be $\{2\}$, but $A$ then values its own bundle at $3$ and $B$'s at least at $6$. Conversely any nonempty subset of $\{4,5,6,7\}$ is compatible with $A=\{2\}$.

Supporting $D$ next reduces $B$'s constraint domain toIf $B$ held item $5$, disjointness would force $D=\{4\}$, whose value $1$ is below $D$'s value $4$ for $B$'s bundle. If $B$ holds $4$ but not $5$, then $D=\{5\}$, and $B$ needs value at least $6$ to avoid envying $D$. If $B$ holds neither, $D=\{4\}$ supports every nonempty subset of $\{6,7\}$.

Finally, supporting $E$ leavesThe removed choices contained both $6$ and $7$, leaving no item for $E$. Every displayed choice has a compatible choice for $E$.

Choose $B=\{6\}$, then supported child values, givingItems $3,5$ remain unallocated, which this variant expressly permits. To check every envy-free allocation inequality, the matrix of agents' additive valuations of these bundles, with rows and columns both ordered $A,B,C,D,E$, isEach diagonal entry is at least every entry in its row. The bundles are nonempty, pairwise disjoint, and contain only positively valued items, so all remaining requirements hold.

Why does selecting supported children suffice? The residual constraint graph has no graph cycle, so different child subtrees have no constraints between them. Child-to-parent arc consistency has already ensured a compatible continuation below every retained parent value. Applying this observation recursively proves that the downward choices complete a solution without backtracking. With $N$ variables and maximum constraint domain size $d$, the basic compatibility checks require $O(N{d}^2)$ time complexity in Big O notation, counting each binary constraint evaluation as one operation; evaluating bundle sums has its own input-dependent cost.

A binary constraint satisfaction problem whose constraint graph is a tree. Restrict constraint domains by unary conditions first. Choose a rooted tree orientation and process children before parents. For each child $j$ of $i$, retain in $D_i$ only values having a compatible value in the already reduced $D_j$. Thus every retained parent value has support in each child subtree.

If the root constraint domain becomes empty, no solution exists. Otherwise choose any root value, then recursively choose supported child values. A child-to-parent arc consistency pass need not enforce full arc consistency in both directions; the chosen directional condition suffices for this construction.

The proof uses mathematical induction on subtrees. A retained leaf value satisfies its unary requirements. Each retained parent value has a supported value in every child subtree. Those subtree solutions can be combined because the tree has no edges between different child subtrees. Therefore every retained parent value extends to a whole subtree solution, and every retained root value extends to a complete solution without backtracking.

For $N$ variables and maximum constraint domain size $d$, scanning all value pairs once per edge costs $O(N{d}^2)$ time complexity in Big O notation, when one binary constraint evaluation has unit cost. If a constraint requires summing bundle values, its evaluation cost must also be counted.