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For a connected undirected weighted graph, a spanning tree connects all vertices without a graph cycle. A minimum spanning tree minimizes the sum of its edge weights. Disconnected input has no spanning tree, although the same methods produce a minimum spanning forest. Negative edge weights and ties cause no difficulty; a tree always contains exactly $|V|-1$ edges.

The central exchange argument is the minimum spanning tree cut property. Suppose a forest $F$ is contained in some minimum spanning tree $T$, and consider a cut with no edge of $F$ crossing it. Let $e$ be a lightest crossing edge. If $e\notin T$, adding it creates a graph cycle; the graph path in $T$ between its endpoints contains a crossing edge $f$. That $f$ is not in $F$, and $w(e)\le w(f)$. Thus $T-f+e$ is another spanning tree with no larger weight and still contains $F$. It is also minimum and contains $F\cup \{e\}$. This proves that the chosen edge is safe, rather than merely asserting a greedy choice.

Kruskal's algorithm sorts all edges by increasing weight, starts with the empty forest, and adds an edge precisely when its endpoints are in different components. Each added edge is lightest across a component cut respected by the current forest. The exchange proof inductively keeps the forest inside a minimum spanning tree; connectedness makes the final forest a spanning tree, hence an optimum. Sorting takes $O(|E|\log |E|)$ comparisons, and even a simple component-merging implementation takes only polynomial additional work. Efficient component tracking improves that overhead further.

Prim's algorithm instead grows a single tree, repeatedly adding a lightest edge from its vertex set to the outside. The same cut argument proves correctness. A dense implementation using an array runs in $O(|V{|}^2)$ comparisons. The dual minimum spanning tree cycle property says a uniquely heaviest edge of a graph cycle belongs to no minimum spanning tree: exchanging it for a lighter crossing edge strictly improves any tree containing it. Ties must be handled carefully; a merely nonunique heaviest edge can occur in an optimum.

Complexity classes are formally classes of decision problems, so use the version asking whether a graph has a spanning tree of total weight at most a supplied threshold $K$. Assume integer or rational weights and threshold are encoded in binary. The input length includes all these bits, not only the numbers of vertices and edges. P consists of decision problems decidable by a deterministic algorithm in time polynomial in that input length. NP consists of those with polynomial-size complexity certificates verified in deterministic polynomial time, equivalently decidable in polynomial time by a nondeterministic machine.

Kruskal's algorithm constructs the optimum and compares its weight with $K$. Binary integer comparisons and sums are polynomial-time operations. Rational comparisons can be made by cross multiplication, and their summed numerator and denominator have polynomially bounded bit length: a product of input denominators has bit length no greater than the sum of their bit lengths. The combinatorial algorithm makes polynomially many such operations. Hence the minimum-spanning-tree decision problem is in P; the optimization version likewise has a polynomial-time algorithm.

For membership in NP, the complexity certificate is a list of $|V|-1$ selected edges. Check that every edge is present, that the edges connect all vertices and contain no graph cycle, and that their total weight is at most $K$. A graph search and binary arithmetic do this in polynomial time. A successful complexity certificate need not establish that its tree is minimum: it proves the threshold question's yes-answer. Conversely, a yes-instance has such a tree as a complexity certificate. ThereforeMembership in NP does not arithmetic mean NP-hardness; here the explicit greedy algorithm provides the stronger tractability result.