Past exam of the computer science course of the University of Oxford / 2023 / Artificial intelligence / 3 / f / Solution

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.