is an edge that connects a vertex in one set with a vertex in the other. The following properties lead to a number of MST algorithms.
Let $e$ be the edge we modify to get $G'$, and let $T'$ be the tree computed according to the algorithm.
We know that weight of $T'$ is less than or equal to the weight of $T$.
Prim's algorithm works by attaching a new edge to a single growing tree at each step: Start with any vertex as a single-vertex tree; then add V-1 edges to it, always taking next (coloring black) the minimum-weight edge that connects a vertex on the tree to a vertex not yet on the tree (a crossing edge for the cut defined by tree vertices).
We use a priority queue to hold the crossing edges and find one of minimal weight.
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In general, only the information that you provide, or the choices you make while visiting a web site, can be stored in a cookie.There is also a path from one endpoint of $e'$ to the other endpoint via $e''$ (around the cycle), all within $T'$.Then we can construct a path from $u$ to $v$ via $e''$ in $T'$ by merging these three paths and removing the overlap (although a walk is sufficient for connectivity).But we must do more: any edge connecting the vertex just added to a tree vertex that is already on the priority queue now becomes (it is no longer a crossing edge because it connects two tree vertices).The lazy implementation leaves such edges on the priority queue, deferring the ineligibility test to when we remove them.To implement Kruskal's algorithm, we use a priority queue to consider the edges in order by weight, a union-find data structure to identify those that cause cycles, and a queue to collect the MST edges.Program Kruskal implements Kruskal's algorithm along these lines.For example, the site cannot determine your email name unless you choose to type it.Allowing a website to create a cookie does not give that or any other site access to the rest of your computer, and only the site that created the cookie can read it.Suppose that in $T$ (which is definitely a spanning tree), the path from $u$ to $v$ did not involve $e'$, then the same path exists in $T'$.Alternatively, suppose that it did use $e'$, then there is a path (without loss of generality) from $u$ to one endpoint of $e'$ and from the other endpoint of $e'$ to $v$.