CSE 421 Algorithms

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# CSE 421 Algorithms - PowerPoint PPT Presentation

CSE 421 Algorithms. Richard Anderson Lecture 21 Shortest Path Network Flow Introduction. Announcements. Friday, 11/18, Class will meet in CSE 305 Reading 7.1-7.3, 7.5-7.6 Section 7.4 will not be covered. Find the shortest paths from v with exactly k edges. 7. y. v. 5. 1. -2. -2.

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### CSE 421Algorithms

Richard Anderson

Lecture 21

Shortest Path

Network Flow Introduction

Announcements
• Friday, 11/18, Class will meet in CSE 305
• Section 7.4 will not be covered
Express as a recurrence
• Optk(w) = minx [Optk-1(x) + cxw]
• Opt0(w) = 0 if v=w and infinity otherwise
Algorithm, Version 1

foreach w

M[0, w] = infinity;

M[0, v] = 0;

for i = 1 to n-1

foreach w

M[i, w] = minx(M[i-1,x] + cost[x,w]);

Algorithm, Version 2

foreach w

M[0, w] = infinity;

M[0, v] = 0;

for i = 1 to n-1

foreach w

M[i, w] = min(M[i-1, w], minx(M[i-1,x] + cost[x,w]))

Algorithm, Version 3

foreach w

M[w] = infinity;

M[v] = 0;

for i = 1 to n-1

foreach w

M[w] = min(M[w], minx(M[x] + cost[x,w]))

Correctness Proof for Algorithm 3
• Key lemma – at the end of iteration i, for all w, M[w] <= M[i, w];
• Reconstructing the path:
• Set P[w] = x, whenever M[w] is updated from vertex x

7

y

v

5

1

-2

-2

3

1

x

3

z

If the pointer graph has a cycle, then the graph has a negative cost cycle
• If P[w] = x then M[w] >= M[x] + cost(x,w)
• Equal after update, then M[x] could be reduced
• Let v1, v2,…vk be a cycle in the pointer graph with (vk,v1) the last edge added
• Just before the update
• M[vj] >= M[vj+1] + cost(vj+1, vj) for j < k
• M[vk] > M[v1] + cost(v1, vk)
• 0 > cost(v1,v2) + cost(v2,v3) + … + cost(vk, v1)

v1

v4

v2

v3

Negative Cycles
• If the pointer graph has a cycle, then the graph has a negative cycle
• Therefore: if the graph has no negative cycles, then the pointer graph has no negative cycles
Finding negative cost cycles
• What if you want to find negative cost cycles?
Network Flow Definitions
• Capacity
• Source, Sink
• Capacity Condition
• Conservation Condition
• Value of a flow
Flow Example

u

20

10

30

s

t

10

20

v

Residual Graph

u

u

15/20

0/10

5

10

15

15/30

s

t

15

15

s

t

5

5/10

20/20

5

20

v

v