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For the three-colourability problem, represent a candidate by a complete colour assignment $c:V\to \{r,g,b\}$, initially chosen at random. Give it the conflict objective functionA neighbor changes the colour of one vertex to either other colour. For hill climbing, maximize $-H$, or equivalently repeatedly choose a recolouring that reduces $H$. One can inspect all $2|V|$ neighbors, or focus on conflicted vertices and evaluate their alternative colours using only incident edges. Stop successfully as soon as $H=0$, which is exactly a proper graph colouring with three colours.

If no improving move exists at positive $H$, the method is stuck in a local minimum. Random restarts, limited sideways moves or other exploratory local search choices can reduce this failure mode, but do not turn a finite unsuccessful run into a proof of impossibility. On a graph with no proper graph colouring using three colours, every complete colouring has $H>0$, so this procedure cannot return a valid colouring. Conversely, a positive-conflict stopping point also occurs on some graphs admitting a proper graph colouring with three colours. Failure of hill climbing alone does not certify non-colourability. For a concrete trap, use a bipartite graph with three vertices on each side and every cross-edge, and give each side one vertex of each colour. There are three conflicts. Recolouring any single vertex leaves it with exactly one same-colour neighbor, so no single change improves $H$, even though colouring each whole side differently solves the instance.

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.

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.