📅  最后修改于: 2020-11-25 04:48:04             🧑  作者: Mango
S中的一组点的凸包是最小凸区域的边界,该最小凸区域包含S内部或边界上的所有S点。
要么
令$ S \ subseteq \ mathbb {R} ^ n $ S的凸包,表示为$ Co \ left(S \ right)$是S的所有凸组合的集合,即$ x \ in Co \ left (S \ right)$当且仅当$ x \ in \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_ix_i $,其中$ \ displaystyle \ sum \ limits_ {1} ^ n \ lambda_i = 1 $并且$ \ lambda_i \ geq 0 \ forall x_i \ in S $
备注-平面中S中的一组点的凸包定义了一个凸多边形,多边形边界上的S点定义了多边形的顶点。
定理$ Co \ left(S \ right)= \ left \ {x:x = \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_ix_i,x_i \ in S,\ displaystyle \ sum \ limits_ {i = 1 } ^ n \ lambda_i = 1,\ lambda_i \ geq 0 \ right \} $显示凸包是凸集。
设$ x_1,x_2 \ in Co \ left(S \ right)$,然后$ x_1 = \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_ix_i $和$ x_2 = \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_ \ gamma x_i $其中$ \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_i = 1,\ lambda_i \ geq 0 $和$ \ displaystyle \ sum \ limits_ {i = 1} ^ n \ gamma_i = 1,\ gamma_i \ geq0 $
对于$ \ theta \ in \ left(0,1 \ right),\ theta x_1 + \ left(1- \ theta \ right)x_2 = \ theta \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_ix_i + \ left (1- \ theta \ right)\ displaystyle \ sum \ limits_ {i = 1} ^ n \ gamma_ix_i $
$ \ theta x_1 + \ left(1- \ theta \ right)x_2 = \ displaystyle \ sum \ limits_ {i = 1} ^ n \ lambda_i \ theta x_i + \ displaystyle \ sum \ limits_ {i = 1} ^ n \ gamma_i \左(1- \ theta \ right)x_i $
$ \ theta x_1 + \ left(1- \ theta \ right)x_2 = \ displaystyle \ sum \ limits_ {i = 1} ^ n \ left [\ lambda_i \ theta + \ gamma_i \ left(1- \ theta \ right)\右] x_i $
考虑这些系数
$ \ displaystyle \ sum \ limits_ {i = 1} ^ n \ left [\ lambda_i \ theta + \ gamma_i \ left(1- \ theta \ right)\ right] = \ theta \ displaystyle \ sum \ limits_ {i = 1 } ^ n \ lambda_i + \ left(1- \ theta \ right)\ displaystyle \ sum \ limits_ {i = 1} ^ n \ gamma_i = \ theta + \ left(1- \ theta \ right)= 1 $
因此,$ \ theta x_1 + \ left(1- \ theta \ right)x_2 \ in Co \ left(S \ right)$
因此,凸包是凸集。