KARGER'S MINCUT ALGORITHM
NORTHERN LIGHTS
YUN LIU DONGFENG YU TIANXIANG GAO ZHIYUAN LI XIAOYANG ZHENG
ALGORITHM NORTHERN LIGHTS YUN LIU DONGFENG YU TIANXIANG GAO - - PowerPoint PPT Presentation
KARGER'S MINCUT ALGORITHM NORTHERN LIGHTS YUN LIU DONGFENG YU TIANXIANG GAO ZHIYUAN LI XIAOYANG ZHENG SOME SOLUTIONS For all pairs of vertexes u and v , run maximum flow. toooo slow Stoer Wagner Algorithm nice
NORTHERN LIGHTS
YUN LIU DONGFENG YU TIANXIANG GAO ZHIYUAN LI XIAOYANG ZHENG
G=(V,E) is a partition of the vertices V into two non-empty, disjoint sets S∪T=V.
weight of all edges between S and T.
weight among all of the cuts.
A CUT WITH VALUE 3 A MINCUT WITH VALUE 2
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
into a single node.
Repeat until just two nodes remain:
mincut(G) = mincut(G/e)
𝑜.
average vertex degree is 2 𝑛
𝑜. Therefore, there exists at least
𝑜.
𝑛 𝑜.
containing a single node u, and the other containing the remaining n − 1 nodes. The size of this cut is degree(u). Since we already prove there is a vertex with degree not greater than 2
𝑛 𝑜 , there exists a cut with value at most
2
𝑛 𝑜. So, that the size of the mincut is at most 2 𝑛 𝑜.
𝑜−2 𝑜
𝑛 𝑜, so P(e is in the
2𝑛 𝑜 × 1 𝑛 = 2 𝑜.
2 𝑜 = 𝑜−2 𝑜
𝑜2
mincut) × Pr(second selected edge is not in mincut) × · · · = 𝑜−2
𝑜 × 𝑜−3 𝑜−1 × 𝑜−4 𝑜−2 … × 2 4 × 1 3 = 2 𝑜(𝑜−1) ≥ 2 𝑜2
𝑜
𝑜2 2 𝑜2 2 ×ln 𝑜 ≥
1 − 1 − 1 𝑓ln 𝑜 = 1 − 1 𝑜
2 𝑜 is almost 1 when n is
large, but decreases to
1 3 when n=3.
𝑃(𝑜4𝑚𝑝𝑜)
𝑃(𝑜2𝑚𝑝3𝑜)
𝑃(𝑜2𝑚𝑝3𝑜)
[1] D. R. Karger and C. Stein. A new approach to the minimum cut problem. J. ACM, 43(4):601–640, 1996.