- By
**issac** - Follow User

- 212 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Shortest Paths' - issac

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Shortest Paths

- Definitions
- Single Source Algorithms
- Bellman Ford
- DAG shortest path algorithm
- Dijkstra
- All Pairs Algorithms
- Using Single Source Algorithms
- Matrix multiplication
- Floyd-Warshall
- Both of above use adjacency matrix representation and dynamic programming
- Johnson’s algorithm
- Uses adjacency list representation

Single Source Definition

- Input
- Weighted, connected directed graph G=(V,E)
- Weight (length) function w on each edge e in E
- Source node s in V
- Task
- Compute a shortest path from s to all nodes in V

All Pairs Definition

- Input
- Weighted, connected directed graph G=(V,E)
- Weight (length) function w on each edge e in E
- We will typically assume w is represented as a matrix
- Task
- Compute a shortest path from all nodes in V to all nodes in V

Comments

- If edges are not weighted, then BFS works for single source problem
- Optimal substructure
- A shortest path between s and t contains other shortest paths within it
- No known algorithm is better at finding a shortest path from s to a specific destination node t in G than finding the shortest path from s to all nodes in V

Negative weight edges

- Negative weight edges can be allowed as long as there are no negative weight cycles
- If there are negative weight cycles, then there cannot be a shortest path from s to any node t (why?)
- If we disallow negative weight cycles, then there always is a shortest path that contains no cycles

Relaxation technique

- For each vertex v, we maintain an upper bound d[v] on the length of shortest path from s to v
- d[v] initialized to infinity
- Relaxing an edge (u,v)
- Can we shorten the path to v by going through u?
- If d[v] > d[u] + w(u,v), d[v] = d[u] + w(u,v)
- This can be done in O(1) time

Bellman-Ford (G, w, s)

Initialize-Single-Source(G,s)

for (i=1 to V-1)

for each edge (u,v) in E

relax(u,v);

for each edge (u,v) in E

if d[v] > d[u] + w(u,v)

return NEGATIVE WEIGHT CYCLE

-5

-5

3

3

5

5

-1

7

s

s

1

1

3

3

2

2

Bellman-Ford AlgorithmG1

G2

Running Time

- for (i=1 to V-1)
- for each edge (u,v) in E
- relax(u,v);
- The above takes (V-1)O(E) = O(VE) time
- for each edge (u,v) in E
- if d[v] > d[u] + w(u,v)
- return NEGATIVE WEIGHT CYCLE
- The above takes O(E) time

Proof of Correctness

- Theorem: If there is a shortest path from s to any node v, then d[v] will have this weight at end of Bellman-Ford algorithm
- Theorem restated:
- Define links[v] to be the minimum number of edges on a shortest path from s to v
- After i iterations of the Bellman-Ford for loop, nodes v with links[v] ≤ i will have their d[v] value set correctly
- Prove this by induction on links[v]
- Base case: links[v] = 0 which means v is s
- d[s] is initialized to 0 which is correct unless there is a negative weight cycle from s to s in which case there is no shortest path from s to s.
- Induction hypothesis: After k iterations for k ≥ 0, d[v] must be correct if links[v] ≤ k
- Inductive step: Show after k+1 iterations, d[v] must be correct if links[v] ≤ k+1
- If links[v] ≤ k, then by IH, d[v] is correctly set after kth iteration
- If links[v] = k+1, let p = (e1, e2, …, ek+1) = (v0, v1, v2, …, vk+1) be a shortest path from s to v
- s = v0, v = vk+1, ei = (vi-1, vi)
- In order for links[v] = k+1, then links[vk] = k.
- By the inductive hypothesis, d[vk] will be correctly set after the kth iteration.
- During the k+1st iteration, we relax all edges including edge ek+1 = (vk,v)
- Thus, at end of the k+1st iteration, d[v] will be correct

Negative weight cycle

- for each edge (u,v) in E
- if d[v] > d[u] + w(u,v)
- return NEGATIVE WEIGHT CYCLE
- If no neg weight cycle, d[v] ≤d[u] + w(u,v) for all (u,v)
- If there is a negative weight cycle C, for some edge (u,v) on C, it must be the case that d[v] > d[u] + w(u,v).
- Suppose this is not true for some neg. weight cycle C
- sum these (d[v] ≤ d[u] + w(u,v)) all the way around C
- We end up with Σv in C d[v] ≤ (Σu in C d[u]) + weight(C)
- This is impossible unless weight(C) = 0
- But weight(C) is negative, so this cannot happen
- Thus for some (u,v) on C, d[v] > d[u] + w(u,v)

DAG-SP (G, w, s)

Initialize-Single-Source(G,s)

Topologically sort vertices in G

for each vertex u, taken in sorted order

for each edge (u,v) in E

relax(u,v);

DAG shortest path algorithm5

3

4

-4

s

5

1

3

8

Running time Improvement

- O(V+E) for the topological sorting
- We only do 1 relaxation for each edge: O(E) time
- for each vertex u, taken in sorted order
- for each edge (u,v) in E
- relax(u,v);
- Overall running time: O(V+E)

Proof of Correctness

- If there is a shortest path from s to any node v, then d[v] will have this weight at end
- Let p = (e1, e2, …, ek) = (v1, v2, v3, …, vk+1) be a shortest path from s to v
- s = v1, v = vk+1, ei = (vi, vi+1)
- Since we sort edges in topological order, we will process node vi (and edge ei) before processing later edges in the path.

Dijkstra (G, w, s)

/* Assumption: all edge weights non-negative */

Initialize-Single-Source(G,s)

Completed = {};

ToBeCompleted = V;

While ToBeCompleted is not empty

u =EXTRACT-MIN(ToBeCompleted);

Completed += {u};

for each edge (u,v) relax(u,v);

4

15

5

7

s

1

3

2

Dijkstra’s AlgorithmRunning Time Analysis

- While ToBeCompleted is not empty
- u =EXTRACT-MIN(ToBeCompleted);
- Completed += {u};
- for each edge (u,v) relax(u,v);
- Each edge relaxed at most once: O(E)
- Need to decrease-key potentially once per edge
- Need to extract-min once per node
- The node’s d[v] is then complete

Running Time Analysis cont’d

- Priority Queue operations
- O(E) decrease key operations
- O(V) extract-min operations
- Three implementations of priority queues
- Array: O(V2) time
- decrease-key is O(1) and extract-min is O(V)
- Binary heap: O(E log V) time assuming E ≥ V
- decrease-key and extract-min are O(log V)
- Fibonacci heap: O(V log V + E) time
- decrease-key is O(1) amortized and extract-min is O(log V)

Compare Dijkstra’s algorithm to Prim’s algorithm for MST

- Dijsktra
- Priority Queue operations
- O(E) decrease key operations
- O(V) extract-min operations
- Prim
- Priority Queue operations
- O(E) decrease-key operations
- O(V) extract-min operations
- Is this a coincidence or is there something more here?

u

v

Only nodes in S

Proof of Correctness- Assume that Dijkstra’s algorithm fails to compute length of all shortest paths from s
- Let v be the first node whose shortest path length is computed incorrectly
- Let S be the set of nodes whose shortest paths were computed correctly by Dijkstra prior to adding v to the processed set of nodes.
- Dijkstra’s algorithm has used the shortest path from s to v using only nodes in S when it added v to S.
- The shortest path to v must include one node not in S
- Let u be the first such node.

u

v

Only nodes in S

Proof of Correctness- The length of the shortest path to u must be at least that of the length of the path computed to v.
- Why?
- The length of the path from u to v must be < 0.
- Why?
- No path can have negative length since all edge weights are non-negative, and thus we have a contradiction.

Computing paths (not just distance)

- Maintain for each node v a predecessor node p(v)
- p(v) is initialized to be null
- Whenever an edge (u,v) is relaxed such that d(v) improves, then p(v) can be set to be u
- Paths can be generated from this data structure

All pairs algorithms using single source algorithms

- Call a single source algorithm from each vertex s in V
- O(V X) where X is the running time of the given algorithm
- Dijkstra linear array: O(V3)
- Dijkstra binary heap: O(VE log V)
- Dijkstra Fibonacci heap: O(V2 log V + VE)
- Bellman-Ford: O(V2 E) (negative weight edges)

Two adjacency matrix based algorithms

- Matrix-multiplication based algorithm
- Let Lm(i,j) denote the length of the shortest path from node i to node j using at most m edges
- What is our desired result in terms of Lm(i,j)?
- What is a recurrence relation for Lm(i,j)?
- Floyd-Warshall algorithm
- Let Lk(i,j) denote the length of the shortest path from node i to node j using only nodes within {1, …, k} as internal nodes.
- What is our desired result in terms of Lk(i,j)?
- What is a recurrence relation for Lk(i,j)?

Shortest path using at most 2m edges

Try all possible nodes k

i

k

j

Shortest path using at most m edges

Shortest path using at most m edges

Shortest path using nodes 1 through k

i

j

Shortest path using nodes 1 through k-1

Shortest path using nodes 1 through k-1

Conceptual picturesShortest path using nodes 1 through k-1

k

OR

Running Times

- Matrix-multiplication based algorithm
- O(V3 log V)
- log V executions of “matrix-matrix” multiplication
- Not quite matrix-matrix multiplication but same running time
- Floyd-Warshall algorithm
- O(V3)
- V iterations of an O(V2) update loop
- The constant is very small, so this is a “fast” O(V3)

Johnson’s Algorithm

- Key ideas
- Reweight edge weights to eliminate negative weight edges AND preserve shortest paths
- Use Bellman-Ford and Dijkstra’s algorithms as subroutines
- Running time: O(V2 log V + VE)
- Better than earlier algorithms for sparse graphs

Reweighting

- Original edge weight is w(u,v)
- New edge weight:
- w’(u,v) = w(u,v) + h(u) – h(v)
- h(v) is a function mapping vertices to real numbers
- Key observation:
- Let p be any path from node u to node v
- w’(p) = w(p) + h(u) – h(v)

3

4

7

1

3

-7

Computing vertex weights h(v)- Create a new graph G’ = (V’, E’) by
- adding a new vertex s to V
- adding edges (s,v) for all v in V with w(s,v) = 0
- Set h(v) to be the length of the shortest path from this new node s to node v
- This is well-defined if G’ does not contain negative weight cycles
- Note that h(v) ≤ h(u) + w(u,v) for all (u,v) in E’
- Thus, w’(u,v) = w(u,v) + h(u) – h(v) ≥ 0

Algorithm implementation

- Run Bellman-Ford on G’ from new node s
- If no negative weight cycle, then use h(v) values from Bellman-Ford
- Now compute w’(u,v) for each edge (u,v) in E
- Now run Dijkstra’s algorithm using w’
- Use each node as source node
- Modify d[u,v] at end by adding h(v) and subtracting h(u) to get true path weight
- Running time:
- O(VE) [from one run of Bellman-Ford] +
- O(V2 log V + VE) [from V runs of Dijkstra]

Download Presentation

Connecting to Server..