Kruskal's algorithm finds a specific kind of tree, a tree being a connected network that has no loops and a designated root node, so every point will be connected to every other, and there won't be any loops.
@@CatoSierraWasTaken trees don't need a designated root node. you can take any tree and change which node is the root and it'll still be a tree (though it might not have the same properties)
@@asdfghyter The algorithm starts by treating every node as a separate tree. Every time it looks for a new edge, it only accepts ones that connect two different trees, then updates the list of trees to replace the two old trees with the single tree formed by connecting them.
@@atreidesson Kruskal's algorithm only creates a stitch if the stitch would connect two "parts" that are not already connected, which guarantees the no-loop condition
What prevents this from creating loops? How does it ensure that there is a solid path from start to end?
Kruskal's algorithm finds a specific kind of tree, a tree being a connected network that has no loops and a designated root node, so every point will be connected to every other, and there won't be any loops.
@@CatoSierraWasTaken trees don't need a designated root node. you can take any tree and change which node is the root and it'll still be a tree (though it might not have the same properties)
@@CatoSierraWasTaken how does it do that?
@@asdfghyter The algorithm starts by treating every node as a separate tree. Every time it looks for a new edge, it only accepts ones that connect two different trees, then updates the list of trees to replace the two old trees with the single tree formed by connecting them.
fascinating; this appears to be easy to code too
Why don't they circle?
because lines
@@chrishunter1109 I meant, not "circles" but "circle"
having no loops is a guarantee of the algorithm used (Kruskal's algorithm)
@@CatoSierraWasTaken Oh, that looked like almost random stitches
@@atreidesson Kruskal's algorithm only creates a stitch if the stitch would connect two "parts" that are not already connected, which guarantees the no-loop condition
is this loss?
Blocked