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Use sprinkling of a binomial random graph to write $G(n,p)=G_1\cup G_2$, where the rounds are independent,and $C$ is chosen large enough that $1-p=(1-p_1)(1-p_2)$. By the given theorem, $G_1$ has a Hamilton cycle $v_1{v}_2\cdots v_n{v}_1$ with high probability.

Condition on such a cycle. For each $3\le \ell \le n-1$, every chord $v_i{v}_{i+\ell -1}$ closes one of the two paths around the Hamilton cycle into a cycle of length $\ell$. There are at least $n/2$ distinct candidate chords, so the probability that $G_2$ supplies none is at mostA union bound over the fewer than $n$ lengths shows that all these cycles occur simultaneously with high probability when $c>4$. The Hamilton cycle itself supplies length $n$, so $G$ is pancyclic.

The sharp Hamilton cycle threshold for the Erdős-Rényi model says thatonly above the window $np=\log n+\log \log n+o(1)$. The hypothesis therefore places $p(n)$ above that window. The Hamiltonicity-to-pancyclicity sprinkling principle then says that three independent $G(n,p(n))$ rounds contain every cycle graph $C_{\ell }$, $3\le \ell \le n$, with high probability: one round supplies a Hamilton cycle, while the other two supply the chords and short-cycle edges used to obtain all intermediate lengths.

The union of the three rounds has individual edge probabilityBy the standard monotone coupling, it is a subgraph of $G(n,3p(n))$. Since being pancyclic is an increasing graph property, the required probability tends to one. If $3p(n)>1$, interpret the latter parameter as $\min \{3p(n),1\}$, in which case the conclusion is immediate.

A weighted Hamilton cycle is a Hamilton cycle in a weighted graph, with cost equal to the sum of its traversed edge weights. Here weights represent symmetric nonnegative costs, rather than conductances. Asking whether a Hamilton cycle exists is a feasibility question; asking for the cheapest one is an instance of the travelling salesman problem. A complete graph supplies every possible connection between distinct vertices, making feasibility automatic for $n\ge 3$, while choosing the cheapest visiting order remains a separate mathematical optimization problem.

Assign the complete graph $K_4$ the symmetric edge costs $w_{AB}=2$, $w_{BC}=3$, $w_{CD}=4$, $w_{AD}=5$, $w_{AC}=6$ and $w_{BD}=7$, in arbitrary units. Fix the start at $A$ and identify each Hamilton cycle with its reverse. The three possibilities areThese exhaust the possibilities: the permutations of the other three vertices give $3!=6$ orders, paired into three reversals. Hence the perimeter tour really is optimal for this toy instance. The diagram's geometric lengths are not costs, and no triangle inequality is assumed.

The usefulness of this model is to distinguish visiting required sites from covering required connections. If tasks are assigned to vertices and the route must return after one visit to each, a weighted Hamilton cycle captures the ordering requirement. If the job instead requires each edge once, the relevant object is an Euler circuit. The shortest path problem asks only for a cheapest route between specified endpoints. A 1954 primary research paper gave a concrete motivation for the travelling salesman problem: certify that a particular tour through 49 cities has minimum road-distance cost.^[ref]

These are abstract route models. If a cost stands for a whole graph path through an underlying network, travelling between consecutive task vertices can pass through intermediate network vertices more than once. One visit per task therefore does not imply one visit per physical junction. One-way connections require a directed graph; changing costs or extra timing constraints require a richer model than the fixed symmetric toy instance.