Optimal Substructure of Prim's Algorithm

Proof:
There is an undirected graph $G=(\mathbf{V},E)$ , whereV = { v 1 , v 2 , … , vn } \mathbf V=\{ v_1,v_2,\dots,v_n\}V={ v1​,v2​,…,vn​}，E = { e 1 , e 2 , … , em } \mathbf E=\{ e_1,e_2,\dots,e_m\}E={ e1​,e2​,…,em​} .The spanning tree is$G^{{}^{\prime }}=(V,E^{{}^{\prime }})$,if$E^{{}^{\prime }}=\{e_{t_1},e_{t_2},\dots ,e_{t_n-1}\}$ .
(optimal substructure):
without loss of generality, let the initial node specified by the prim algorithm be$v_1$, Figure $G$ has some minimum spanning tree$T$ , where with$v_1$The connected node with the smallest edge weight is $v_a$, the corresponding side is $e_{1a}=(v_1,v_a)$ . Will$G$ andTTv 1 v_1in$T$$v_1$sum $v_a$merged into a new vertex $v_c$, delete edge $e_{1a}$, and for both and $v_1$sum $v_a$Connected nodes, keep the edge with the smallest weight and $v_c$Get G 1 G_1 after connecting$G_1$and $T_1$. Now need to prove $T_1$for $G_1$The minimum spanning tree of .
Let $T_1$Not $G_1$The minimum spanning tree of $T_2$for $G_1$The minimum spanning tree of $W(T_2)<W\left(T_1\right)$ . T$T_2$and $T_1$vc v_c in$v_c$Node reverts to $v_1$sum $v_a$节点，则有$W\left(T_2\right)+W\left({ie}_{1a}\right)<W\left(T_1\right)+W\left({ie}_{1a}\right)=W\left(T\right)$ and$T$ is the minimum spanning tree contradiction. So the optimal substructure holds.