an optimal minimum spanning tree algorithm

An Optimal Minimum

Spanning Tree Algorithm

Seth Pettie and Vijaya Ramachadran Journal of the ACM Vol. 49, No. 1

Outline

– Cut and Cycle – DJP Algorithm

– MST in Dense Case

• Overview of the Algorithm

• Key Lemma and the Partition Algorithm • Decision Tree

Overview of the Algorithm

• Precompute the optimal decision treesdecision trees for all graphs with ≤ log(3) n vertices

• _{PartitionPartition the original graph into small subgraphs}

– soft heap

• Find the MST of each subgraph

– decision tree

• Use the small MSTs to construct the MST of the original graph

C₁ C₂ C₃ C₄ C₅ C₆ C₇

GM

C₁ C₂ C₃ C₄ C₅ C₆ C₇

Partition(G,r,



; M,C)

GM

以 DenseCase 求 MS F

GM

Partition(G,r,



; M,C)

•Boruvka2:

•將 vertex 數降為 1/4

•G_c: m 降為 1/4 •OptimalMSF(G_c)

Key Lemma

• Lemma 3.2 Let M be a set of edges in a graph G. If C is a subgraph of G that is

DJP-contractible w.r.t. GM, then MSF(G) is a subset of MSF(C)  MSF(G \ C  MC)  MC DJP-contractible w.r.t GM 即：在 G 上，使用 SoftHeap ， 以 Prim algorithm 長出來的子圖

M_c : corrupted edges with one endpoint in C

• 證法： – 要證明在 (1) 裡的每條 edge ， 在 (2) 都會存在  證明在 (2) 中不存在的 edge ， 也必不存在 (1) 把 (2) 再拆成兩部分來看 a) In C, if an edge eMSF(C), then eMSF(G) b) In G\C, if an edge eMSF(G\CMC)MC, then eMSF(G) C C M M C G MSF C MSF G MSF( ) ( )  ( \  )  (1 ) (2 ) • 欲證明： G C

• 證明：

– _eMSF(C)

 e must be the heaviest edge on some cycle in C. (cycle property)

Such a cycle exists in G as well. So eMSF(G)

a) 欲證明：

In C, if an edge eMSF(C), then eMSF(G)

G

• 證明：

– Let H=G\CM_C, show that

• no edge in HMSF(H) is in MSF(G)

– Let eHMSF(H)

 e is the heaviest edge on some cycle  in

H.

1. 若  不包含 C contract 形成的那個點 2. 若  包含 C contract 形成的那個點

( 見下頁 )

目標：證明 e is also the heaviest edge on

a cycle in G  eMSF(G) b) 欲證明： In G\C, if an edge eMSF(G\CMC)MC, then eMSF(G) G C  

• 證明 _{( 續上頁 ) ：}

2. 若  包含 C contract 形成的那個點

 x, y, w, z: nodes

 (x,w) and (y,z): end edges of path P

 Since H includes no corrupted edges with on point in C, so:

G-weight of these edges = (GM)-weight

 T: the spanning tree of CM

 Q: the path in T connecting x and y  _{g: the heaviest edge in} Q.

 w(e) > max(w(x,w), w(y,z)) (def. of e)

> (GM)-weight of g (Lemma 2.1)

 G-weights of all edges in Q.

So, w.r.t. G-weights, e is the heaviest edge on a cycle PQ and cann ot be in MSF(G). C x y w z P Q e

目標：證明 e is also the heaviest edge on a cycle in G  eMSF(G)

• 選定一個點為起點， 用 DJPDJP 演算法長 component C_i  長到 1. 大於 maxsize, 或 2. 連到別的 component C_i 就停下來

Partition & Key Lemma

• By applying Key Lemma repeatedly, we se e that after Ci is built, the MSF of G is a su

Decision Trees

• 一個 Decision Tree 為一 rooted tree ，在 每個 _{internal node 上都含有某些條件判斷} 式，不同的分支代表判斷式不同的結果。 所有 _{leaves 各自代表了不同的決策，而自} root 到一 leaf 之間所經過的 path 即代表選 擇此決策所需要的條件。

MSF Decision Trees

• 每個 internal node of MSF Decision Trees 上的判斷式是一個 _{edge-weight compariso} n （對於給定的 graph G ）。

• 每個 internal node of MSF Decision Trees 必恰好有兩個 _{children ，分別代表該} edge-weight comparison 為 true 或 false 。

• MSF Decision Trees 的 leaves 列出所有符 合 _{root 到每一 leaf 之路徑上判斷式及其結} 果的一些 _{spanning tree 。}

MSF Decision Trees (cont.)

• Correct MSF Decision Trees 的條件為每 一條從 _{root 到一個 leaf 的 path 可決定唯一} 的一個 _{spanning tree 為 G 的 MSF 。}

• Optimal MSF Decision Tree 的條件為沒有 比之深度更小的 _{Correct MSF Decision Tre} e of G 存在。

• 當我們有 graph G 的 Optimal MSF Decisio n Trees ，我們就可以將 G 的 edge weight 值代入其判斷式找出其 _{MSF 。}

Find Optimal Decision Trees

• 簡單的說，是用 brute force search (BFS) 去暴力搜 尋所有可能性並驗證。

• 分析當我們要找出所有具 r 個 vertices 的 graph 之 Op timal Decision Trees 的時間複雜度：

• 因此若取 則整個處理為 o(n) ti me 。

Complexity

MSF of the : edges and vertices graph with any on MSF the determine to needed s comparison of number optimal the : ) , ( OptimalMSF of time running the : ) , ( * xity ree comple decision-t T m n n m T n m T * m c ) n , m T( ) (C T T(m,n) i i * _{  }   ₂ 4 2 •Observations:

•If T*(m,n) = O(m), then T(m,n)=O(m) •T(m,n)=O(T*(m,n)) for many natural functions for T* (including m(m,n)) •Prove that: T(m,n)=O(T*(m,n)) holds, no matter what describing O(T*(m,n))

•Observations:

•If T*(m,n) = O(m), then T(m,n)=O(m) •T(m,n)=O(T*(m,n)) for many natural functions for T* (including m(m,n)) •Prove that: T(m,n)=O(T*(m,n)) holds, no matter what describing O(T*(m,n))