All%20Pair%20Shortest%20Path

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All Pair Shortest Path

• IOI/ACM ICPC Training
• June 2004

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All Pair Shortest Path

• Note Dijkstras Algorithm takes O((VE)logV)
  time
• All Pair Shortest Path Problem can be solved by
  executing Dijkstras Algorithm V times
• Running Time O(V(VE)log V)
• Floyd-Warshall Algorithm O(V3)

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Idea

• Label the vertices with integers 1..n
• Restrict the shortest paths from i to j to
  consist of vertices 1..k only (except i and j)
• Iteratively relax k from 1 to n.

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Definition

• Find shortest distance from i to j using vertices
  1 .. k only

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Example
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i 4, j 5, k 0
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i 4, j 5, k 1
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i 4, j 5, k 2
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i 4, j 5, k 3
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Idea
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The tables
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The code

• for i 1 to V
• for j 1 to V
• aij0 cost(i,j)
• for k 1 to V
• for i 1 to V
• for j 1 to V
• aijk min( aijk-1,
• aikk-1 akjk-1)

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Topological sort

• IOI/ACM ICPC Training
• June 2004

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Topological order

• Consider the prerequisite structure for courses
• Each node x represents a course x
• (x, y) represents that course x is a prerequisite
  to course y
• Note that this graph should be a directed graph
  without cycles.
• A linear order to take all 5 courses while
  satisfying all prerequisites is called a
  topological order.
• E.g.
• a, c, b, e, d
• c, a, b, e, d

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Topological sort

• Arranging all nodes in the graph in a topological
  order
• Applications
• Schedule tasks associated with a project

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Topological sort algorithm

• Algorithm topSort1
• n V
• Let R0..n-1 be the result array
• for i 1 to n
• select a node v that has no successor
• Rn-i v
• delete node v and its edges from the graph
• return R

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Example
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Time analysis

• Finding a node with no successor takes O(VE)
  time.
• We need to repeat this process V times.
• Total time O(V2 V E).
• We can implement the above process using DFS. The
  time can be improved to O(V E).

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Algorithm based on DFS

• Algorithm topSort2
• s.createStack()
• for (all nodes v in the graph)
• if (v has no predecessors)
• s.push(v)
• mark v as visited
• while (s is not empty)
• let v be the node on the top of the stack s
• if (no unvisited nodes are children to v) //
  i.e. v has no unvisited successor
• aList.add(1, v)
• s.pop() // blacktrack
• else
• select an unvisited child u of v
• s.push(u)
• mark u as visited

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Bipartite Matching

• IOI/ACM ICPC Training
• June 2004

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Unweighted Bipartite Matching
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Definitions
Matching
Free Vertex
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Definitions

• Maximum Matching matching with the largest
  number of edges

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Definition

• Note that maximum matching is not unique.

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Intuition

• Let the top set of vertices be men
• Let the bottom set of vertices be women
• Suppose each edge represents a pair of man and
  woman who like each other
• Maximum matching tries to maximize the number of
  couples!

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Applications

• Matching has many applications. For examples,
• Comparing Evolutionary Trees
• Finding RNA structure
• This lecture lets you know how to find maximum
  matching.

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Alternating Path

• Alternating between matching and non-matching
  edges.

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d-h-e alternating path a-f-b-h-d-i alternating
path starts and ends with free vertices f-b-h-e
not alternating path e-j alternating path starts
and ends with free vertices
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Idea

• Flip augmenting path to get better matching
• Note After flipping, the number of matched edges
  will increase by 1!

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Idea

• Theorem (Berge 1975)
• A matching M in G is maximum iffThere is no
  augmenting path
• Proof
• (?) If there is an augmenting path, clearly not
  maximum. (Flip matching and non-matching edges in
  that path to get a better matching!)

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Proof for the other direction

• (?) Suppose M is not maximum. Let M be a maximum
  matching such that MgtM.
• Consider H M?M (M?M)-(M?M)i.e. a set of
  edges in M or M but not both
• H has two properties
• Within H, number of edges belong to M gt number
  of edges belong to M.
• H can be decomposed into a set of paths. All
  paths should be alternating between edges in M
  and M.
• There should exist a path with more edges from
  M. Also, it is alternating.

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Idea of Algorithm

• Start with an arbitrary matching
• While we still can find an augmenting path
• Find the augmenting path P
• Flip the edges in P

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Labelling Algorithm

• Start with arbitrary matching

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Labelling Algorithm

• Pick a free vertex in the bottom

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Labelling Algorithm

• Run BFS

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Labelling Algorithm

• Alternate unmatched/matched edges

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Labelling Algorithm

• Until a augmenting path is found

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Augmenting Tree
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Flip!
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Repeat

• Pick another free vertex in the bottom

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Repeat

• Run BFS

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Repeat

• Flip

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Answer

• Since we cannot find any augmenting path, stop!

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Overall algorithm

• Start with an arbitrary matching (e.g., empty
  matching)
• Repeat forever
• For all free vertices in the bottom,
• do bfs to find augmenting paths
• If found, then flip the edges
• If fail to find, stop and report the maximum
  matching.

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Time analysis

• We can find at most V augmenting paths (why?)
• To find an augmenting path, we use bfs! Time
  required O( V E )
• Total time O(V2 V E)

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Improvement

• We can try to find augmenting paths in parallel
  for all free nodes in every iteration.
• Using such approach, the time complexity is
  improved to O(V0.5 E)

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Weighted Bipartite Graph
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Weighted Matching
Score 63110
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Maximum Weighted Matching
Score 6111413
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Augmenting Path (change of definition)

• Any alternating path such that total score of
  unmatched edges gt that of matched edges
• The score of the augmenting path is
• Score of unmatched edges that of matched edges

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Note augmenting path need not start and end at
free vertices!
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Idea for finding maximum weight matching

• Theorem Let M be a matching of maximum weight
  among matchings of size M.
• If P is an augmenting path for M of maximum
  weight,
• Then, the matching formed by augmenting M by P is
  a matching of maximum weight among matchings of
  size M1.

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Overall Algorithm

• Start with an empty matching
• Repeat forever
• Find an augmenting path P with maximum score
• If the score gt 0, then flip the edges
• Otherwise, stop and report the maximum weight
  matching.

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Time analysis

• The same!
• Time required O(V2 V E)

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Stable Marriage Problem

• IOI/ACM ICPC Training
• June 2004

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Stable Marriage Problem

• Given N men and N women, each person list in
  order of preference all the people of the
  opposite sex who would like to marry.
• Problem
• Engage all the women to all the men in such a way
  as to respect all their preferences as much as
  possible.

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Stable?

• A set of marriages is unstable if
• two people who are not married both prefer each
  other than their spouses
• E.g. Suppose we have A1 B3 C2 D4 E5. This is
  unstable since
• A prefer 2 more than 1
• 2 prefer A more than C

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Naïve solution

• Starting from a feasible solution.
• Check if it is stable.
• If yes, done!
• If not, remove an unstable couple.
• Is this work?

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Naïve solution (2)

• Does not work!
• E.g.
• A1 B3 C2 D4 E5
• A2 B3 C1 D4 E5
• A3 B2 C1 D4 E5
• A3 B1 C2 D4 E5

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Solution

• Let X be the first man.
• X proposes to the best woman in the remaining on
  his list. (Initially, the first woman on his
  list!)
• If a is not engaged
• Pair up (X, a). Then, set Xnext man and goto 1.
• If a prefers X more than her fiancee Y,
• Pair up (X, a). Then, set XY and goto 1.
• Goto 1

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Example
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Time analysis

• If there are N men and N women,
• O(N2) time

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Algorithm

• prefermsw means the woman w is on the s-th
  position in the preference list of the man m
• Let nextm be the current best woman in his
  remaining list. (Initially, nextm0)
• fianceewm means the man m engaged to woman w.
  (Initially, fianceew0)
• Let rankwm is the ranking of the man m in the
  preference list of the woman w.
• For(m1mltNm) For(sms!0