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Suppose backtracking assigns each of finitely many variables once along each branch, branching only on distinct values in its initial nonempty constraint domain $D_i$. Then the number of leaves is at most ${\prod }_i|D_i|$, even if inconsistent partial assignments terminate early. Ignore constraints and associate a leaf with every full domain assignment extending its partial assignment. Each leaf has at least one such extension. Different leaves have disjoint extension families: at their first diverging choice they assigned different values to the same selected variable. These nonempty disjoint families lie in a Cartesian product of cardinality ${\prod }_i|D_i|$, proving the bound. The selected next variable can depend on the preceding assignments, provided each full assignment follows a unique branch.

The bound is exact for an unpruned enumeration tree, but need not be attained when actual constraints cause search pruning. Counting complete satisfying assignments alone also does not count failed partial leaves.

Assign values from specified domains to variables so that all stated constraints hold simultaneously. A partial assignment can be rejected when it violates a constraint or leaves an unassigned variable with no permissible value. Backtracking applies this test during a depth-first search.

Dependency resolution chooses compatible software package versions and their transitive requirements for a requested installation. It is a constraint satisfaction problem: choosing a version can introduce additional requirements, which may invalidate choices made earlier. Pip uses backtracking to reconsider such choices.^[ref]

If $A$ requires library $L$ at version at least 2 but below 4, and $B$ requires $L$ at least 3 but below 5, the shared allowed interval is versions at least 3 but below 4. If $B$ instead requires at least 4, no single version satisfies both. Downloading more candidates cannot remove that contradiction. Recording the selected versions preserves a solution to those constraints; it does not by itself record all execution conditions.

Repeatedly apply a constraint propagation operation that only deletes values from finite constraint domains, preserves their order, and preserves every valid assignment. The sum of domain sizes is a nonnegative loop variant at every changing step, so repetition reaches a fixed point. For $N$ initial candidate values there are at most $N$ changing steps. An empty domain proves failure; a non-singleton fixed point may still require backtracking or explicit enumeration. Preservation of solutions, rather than mere deletion, is essential.

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]

For a finite constraint satisfaction problem, remove values violating unary restrictions before starting. The following backtracking algorithm makes independent copies of constraint domains at each branch, so failed forward checking cannot contaminate a sibling branch.

SOLVE(P):
    D := domains of P after unary filtering
    if some D[X] is empty: return failure
    return SEARCH({}, D)

SEARCH(assignment, D):
    if every variable is assigned: return assignment
    X := SELECT-UNASSIGNED-VARIABLE(assignment, D, P)
    for v in ORDER-DOMAIN-VALUES(X, assignment, D, P):
        if X=v is compatible with the current assignment:
            next_assignment := assignment extended by X=v
            next_domains := a deep copy of D
            next_domains[X] := {v}
            if INFERENCES(X, v, next_assignment, next_domains, P) succeeds:
                result := SEARCH(next_assignment, next_domains)
                if result is not failure: return result
    return failure

SELECT-UNASSIGNED-VARIABLE implements minimum remaining values: choose an unassigned variable with the smallest current constraint domain. Break ties by the degree heuristic, choosing the variable constrained to the greatest number of other unassigned variables; a fixed final tie rule makes execution reproducible. ORDER-DOMAIN-VALUES may use any complete ordering; ascending values suffice here.

INFERENCES implements forward checking: for each unassigned neighbor $Y$ of $X$, remove every value incompatible with $X=v$, and fail if any constraint domain becomes empty. For higher-arity constraints, reject any completely instantiated violation and use a sound chosen filtering rule; the stated neighbor test is the usual binary case. Forward checking never removes a value used by a possible extension of the current assignment. Exhausting every surviving value therefore makes this finite backtracking procedure both sound and complete.^[ref][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.

One solution is $(a,b,c,d,e)=(1,2,3,2,1)$. Use ascending value order; after minimum remaining values and the degree heuristic, break any remaining variable tie alphabetically. Apply the unary restrictions first:

Minimum remaining values selects $a$, even though $c$ has greater initial vertex degree. Set $a=1$. Forward checking removes $1$ from the three neighbors $b,c,d$:Now $b$ is the unique minimum remaining values choice. Set $b=2$. Forward checking removes $2$ from $c,e$, givingSelect $c$ and set $c=3$. Its remaining neighbors are $d,e$, so forward checking givesThe two remaining variables tie on constraint domain size and each has one unassigned neighbor, so the degree heuristic also ties. Alphabetical order selects $d$; take $d=2$. Since $2$ is already absent from $D_e$, forward checking makes no further deletion. Finally choose $e=1$. No empty constraint domain or violation occurs, so this branch needs no backtracking.

All eight binary constraints are checked explicitly by the unequal value pairsThe unary restrictions also hold. Equalities $a=e$ and $b=d$ are permitted because those pairs have no edges in the constraint graph.

A solution is $(A,B,C,D,E,F,G)=(1,2,1,3,4,2,3)$. Here is a backtracking search with forward checking, listing the chosen value order rather than assuming ascending values. No backtrack is needed on this branch.

Start with the constraint domains in (a). Assign $A=1$; remove $1$ from $B,G,E$. Choose $E=4$; remove $4$ from each remaining neighbour. Now $B,G$ both have domain $\{2,3\}$ and $C,D,F$ each have $\{1,2,3\}$. Choose $B=2$. Its inequality removes $2$ from $G$, forcing $G=3$, and also removes $2$ from $D$, leaving $D\in \{1,3\}$. Assigning $G=3$ removes $3$ from $F$, leaving $F\in \{1,2\}$.

Choose $D=3$. The edge $DC$ reduces $C$ to $\{1,2\}$; $DF$ makes no further deletion because $3$ is already absent from $F$. Choose $F=2$ and then $C=1$. All variables are assigned, and every constrained pair has different values.

In time order, the schedule is

The Head of Department attends slots $1,2,3$ for $A,B,G$; first-year attendees see $C,D$ in slots $1,3$; second-year attendees see $F,G$ in $2,3$; the third-year Keble group sees $B,D$ in $2,3$; Physics attendees see $D,F$ in $3,2$. Slot $4$ is free of every competing talk, so all assumed society overlaps are respected.

For a finite constraint satisfaction problem with $v$ variables, all complete assignments have depth $v$ when each step assigns one new variable. Thus the shallowest-solution property of breadth-first search offers no advantage: every complete solution is already at that depth.

Depth-first search with backtracking stores a path and pending choices, using $O(vd)$ auxiliary space with ordinary explicit choice bookkeeping, or $O(v)$ when each level stores only a current value index. Breadth-first search may store an exponentially large frontier, up to $O(d^v)$ partial assignments. Depth-first search can also combine naturally with immediate constraint propagation and undoing changes. The finite depth avoids the infinite-branch obstruction of unrestricted depth-first tree search. Neither traversal removes the possible exponential time complexity.

The conventional worst-case estimate is the backtracking leaf bound given by the product of the initial constraint domain sizes. Before exploiting any attacking-queen constraints, the three models give respectivelyModel 1 has $N^2$ binary choices. Model 2 has $N$ labeled queens with $N^2$ possible cells each. Model 3 has $N$ row variables with $N$ possible columns each. The rule of product gives the displayed bounds; assigning variables in one prescribed order avoids multiplying again by the number of variable orderings.

With $k$ mutually nonattacking prescribed queens incorporated as singleton constraint domains, the corresponding bounds improve toIn Model 1 only the $k$ prescribed occupied cells have been fixed in this estimate; immediate deletion of attacked cells tightens it further. In Model 3 consistent clues occupy $k$ distinct rows. If the clues are inconsistent, preprocessing can reject the instance immediately.

These are upper bounds on leaves, rather than exact attained counts for a fully specified pruning implementation. Ordinary backtracking rejects inconsistent partial assignments and often creates far fewer leaves; a stopped branch is itself a leaf even if not all variables were assigned. Such early leaves still obey the backtracking leaf bound, since distinct leaves describe disjoint nonempty families of full domain assignments when constraints are ignored. Counting only possible complete nonattacking assignments is a different question. For example, row/column restrictions alone reduce full assignments to at most $N!$ in Model 3, but that number does not directly count failed partial leaves. Exact counts depend on variable ordering and constraint propagation, neither of which is specified.

Use Model 3 and keep every prescribed row at its prescribed column. Initialize one queen in every other row, and minimize the conflict countThis sum counts attacking pairs; $E=0$ is exactly a valid completion. A neighbor changes the column of one unfixed row. Evaluate such neighbors and move to one with lower $E$, breaking equal choices arbitrarily. Minimizing $E$, or maximizing $-E$, is hill climbing on complete domain assignments; these assignments need not yet satisfy the inter-row constraints. Fixed rows are never moved.

The min-conflicts algorithm chooses a conflicted unfixed queen and places it in a column giving the smallest conflict count against the other queens. Recomputing only the affected pair contributions makes a move cheaper than reevaluating every pair. Equal-cost moves can help cross plateaus; bounded sideways moves or repeated random starts of only the unfixed queens can escape some traps. These modifications still preserve the prescribed positions.

Return a board only after checking $E=0$ and every prescribed position. A local minimum with positive $E$ is not proof of impossibility. A queens completion problem may be unsatisfiable even when its prescribed queens do not attack each other; unrestricted restart policies give no finite worst-case success bound. If every row is fixed, simply validate the one board. For a guaranteed decision procedure use finite backtracking, possibly with constraint propagation, instead of relying on hill climbing.

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.

Let item $j$ have weight $a_j$ and profit $p_j$. Define $D(j,b)$ as the largest profit obtainable from the first $j$ items with capacity at most the nonnegative integer $b$. Set $D(0,b)=0$. The last item is either absent or present once, giving the dynamic programming recursionThe two alternatives exhaust the feasible binary decisions. In the second alternative, deleting item $j$ leaves exactly the smaller subproblem with capacity $b-a_j$, which proves the recurrence.

Fill rows $j=1,\dots ,4$, and within each row capacities $b=0,\dots ,7$. Store which alternative attains each maximum. Backtrack from $(4,7)$: when exclusion was chosen, set $x_j=0$ and move to $(j-1,b)$; when inclusion was chosen, set $x_j=1$ and move to $(j-1,b-a_j)$. Ties may be resolved either way. Thus the stored values yield an actual optimal selection, not just its objective value.

For $n$ items and capacity $B$, the algorithm uses $O(nB)$ arithmetic operations and $O(nB)$ space with the full backtracking table. It is a pseudo-polynomial algorithm, since $B$ is a numerical capacity rather than its bit length. A value-only version can use one row, but must update capacities in decreasing order to avoid reusing the same item.

Use trie search with a letter multiset. Counts avoid generating the same result repeatedly when the input contains equal letters. The backtracking explores a letter value once, irrespective of how many indistinguishable copies remain.At entry to visit, t is reached by prefix, and left(k) is the input multiplicity of letter $k$ minus its multiplicity in prefix. All these counts are nonnegative. Each recursive call consumes one available letter, and restoration makes sibling calls see the same counts as before that child. The recursion therefore terminates at depth at most $|ls|$, even when the dictionary contains longer words.

A completed prefix is emitted once, and the traversal continues so that longer words with the same prefix are also found. Every emitted word is in the dictionary and uses at most the available multiplicities. Conversely, every dictionary word meeting those bounds has an existing trie path whose next letter has positive remaining count at every step; the traversal reaches its terminal node. A trie has a unique root path for each word, and each child is visited at most once per reached node. ThereforeThe boxed expression specifies the set of entries in the returned list, rather than its ordering. If the dictionary includes the empty string, the method includes it once. Counting costs $O(|ls|+26)$; visiting $R$ reachable nodes costs $O(26R)$ apart from string copying and output construction. The presented immutable-prefix construction additionally costs the sum of the prefix lengths it constructs. The recursion depth is at most $|ls|$, and the letter-count array has fixed size.

Prolog expresses programs through facts and rules and answers queries by solving goals. A rule such as ancestor(X,Y) :- parent(X,Y). provides one way to establish a relationship. A recursive rule can search through intermediate parents. Unification matches a goal with a rule head and propagates bindings; backtracking explores alternative matches when a branch fails or another answer is requested.^[ref]

The logical core is associated with Horn clauses and SLD resolution, but clause order and goal order also determine a conventional Prolog program's search behavior. A logically valid recursive specification can loop before reaching an available answer. Predicates with input/output effects and control operations add further operational behavior. Arithmetic evaluation is also distinct from term matching: X = 1+2 builds a term, whereas X is 1+2 evaluates arithmetic in the usual Prolog interface.

Complete a prescribed subset of occupied cells to a solution of the N-queens problem while retaining every prescribed queen. In a row-variable constraint satisfaction problem, a clue $(r,c)$ gives the singleton constraint domain $D_r=\{c\}$. Conflicting clues prove impossibility immediately, but nonattacking clues need not extend to a full solution. Backtracking decides the finite problem; hill climbing can search large candidate boards without an exhaustive guarantee.

A rooted tree whose nodes are partial search states and whose children are possible next choices. Leaves represent completed candidates or terminated branches. Backtracking traverses this tree while storing the current path, and search pruning removes branches that cannot yield a useful completion.

SLD resolution is a goal-directed inference rule for definite first-order Horn clauses. It selects an atom from the current goal, matches it with a suitably renamed clause head using a most general unifier, and replaces it by that clause's body, applying the same first-order substitution throughout the goal. Reaching the empty goal supplies a successful derivation.^[ref]

For example, the rule ancestor(X,Y) :- parent(X,Y) reduces the query ancestor(ada,Y) to parent(ada,Y). Matching the fact parent(ada,ben) then gives Y=ben. Alternative matching clauses create a search tree. Prolog normally selects the leftmost goal and searches clauses in order with backtracking; a recursive branch can therefore run indefinitely before another branch's valid answer is reached. The logical inference rule and the strategy traversing its alternatives are separate choices.

A software dependency is another component, tool, library or service required for some aspect of a program's construction or operation. Dependencies form a dependency graph; direct requirements can introduce additional transitive requirements. Dependency resolution selects versions satisfying the combined constraints.

For example, if one component requires library versions at least 2 but below 3, while another requires at least 3, no single library version satisfies both in one shared installation. A resolver must report the conflict or use an isolation mechanism that genuinely permits separate versions. Pip documents backtracking through candidate package versions to solve such constraints.^[ref] Recording selected versions improves reproducibility, but the operating system, input data and execution settings can still affect the result.