1. Single-Source Shortest Paths Revisited
   1.1 The Single-Source Shortest Path Problem.
   Input: Directed graph G=(V,E) G = ( V , E ) , edge lengths ce c e for each e∈E e ∈ E , source vertex s∈V s ∈ V . [Can assume no parallel edges].
   1.2 Goal: For every destination v∈V v ∈ V , compute the length(sum of edge costs) pf a shortest s-v path.
2. On Dijkstra’s Algorithm
   2.1 Good news: O(mlogn) O ( m l o g n ) running time using heaps.
   Bad news:
   (1) Not always correct with negative edge lengths.
   (2)Not very distributed (relevant for Internet routing)
   Solution: The Bellman-Ford algorithm.
   On Negative Cycles.
   Question:How to define shortest path when G G has a negative cycle?
   Soution #1: Compute the shortest s-v path, which cycles allowed.
   Problem: Undefined or −∞ − ∞ .
   Solution #2: Computes shortest cycle-free s-v path.
   Problem: NP-hard (no polynomial algorithm, unless P=NP)
   Solution #3: (For now) Assume input graph has no negative cycles.
   Single-Source Shortest Path Problem, Revisited.
   Input: Directed graph G=(V,E) G = ( V , E ) , edge lengths ce c e ,
   source vertex s∈V s ∈ V .
   Goal: either
   (A) For all destinations v∈V v ∈ V ,compute the length of a shortest s−v s − v path. OR
   (B) Output a negative cycle.
   Optimal Substructure(Informal)
   Intuition: Exploit sequential nature of paths. Subpath of a shortest path should itself be shortest.
   Issue: Not clear how to define “smaller” & “larger” subproblems.
   Key idea: Artificially restrict the number of edges in a path.
   Subproblem size ⇔ ⇔ Number of permitted edges.
   Lemma: Let G=(V,E) G = ( V , E ) be a directed graph with edge lengths ce c e and source vertex s.
   [G might or might not have a negative cycle]
   For every v∈V,i∈{1,2…} v ∈ V , i ∈ { 1 , 2… } , let P= P = shortest s−v s − v path with at most i i edges. (Cycles are permitted)
   Case 1: If P has ≤(i−1) ≤ ( i − 1 ) edges, it is a shortest s−v s − v path with ≤(i−1) ≤ ( i − 1 ) edges.
   Case 2: If P has i i edges with last hop( w w , v v ), then P′ P ′ is a shortest s−w s − w path with ≤(i−1) ≤ ( i − 1 ) edges.
   The Basic Algorithm
   The Recurrence
   Notation: Let Li,v= L i , v = minimum length of a s-v path with ≤i ≤ i edges.
   − − With cycles allowed
   − − Defined as +∞ + ∞ if no s−v s − v paths with ≤i ≤ i edges.
   Recurrence: For every v∈V,i∈{1,2,...} v ∈ V , i ∈ { 1 , 2 , . . . }

Li,v=min{L(i−1),vmin(u,v)∈E{L(i−1),w+cwv},(1) (1) L i , v = m i n { L ( i − 1 ) , v m i n ( u , v ) ∈ E { L ( i − 1 ) , w + c w v } ,

If No Negative Cycles

Now: Suppose input graph

G G has negative cycles.

⇒ ⇒ Shortest paths do not have cycles.

[removing a cycle only decrease length]

⇒ ⇒ Have

≤(n−1) ≤ ( n − 1 ) egdes.

Point: If

G G has negative cycle, only need to solve subproblems up to

i=n−1 i = n − 1 .

Point: If

G G has no negative cycle, only need to solve subproblems up to

i=n−1 i = n − 1 .

Subproblems:

Compute

Li,v L i , v for all

i∈{0,1,...,n−1} i ∈ { 0 , 1 , . . . , n − 1 } and all

v∈V v ∈ V .

The Bellman-Ford Algorithm

(1) Let

A=2−D A = 2 − D array(indexed by

i i and

v v )

Base case:

A[0,s]=0;A[0,v]=+∞ A [ 0 , s ] = 0 ; A [ 0 , v ] = + ∞ for all

v≠s v ≠ s /

For

i=1,2,...,n−1 i = 1 , 2 , . . . , n − 1

For each

v∈V v ∈ V

Li,v=min{L(i−1),vmin(u,v)∈E{L(i−1),w+cwv},(2) (2) L i , v = m i n { L ( i − 1 ) , v m i n ( u , v ) ∈ E { L ( i − 1 ) , w + c w v } ,

As discussed: IF G has no negative cycle, then algorithm is correct.

[with final answer =

A[n−1,v]′s A [ n − 1 , v ] ′ s ]

Whatis the running time of the Bellman-Ford algorithm?
Reason: Total work is O(n∑v∈Vin.deg(v)) O ( n ∑ v ∈ V i n . d e g ( v ) ) = O(mn) O ( m n ) 入度。
Checking for a Negative Cycle
Question: What if the input graph G has a negative cycle.
Claim:
G has a no negative-cost cycle ⇔ ⇔ In the extended Bellman-Ford algorithm,
A[n−1,v]=A[n,v] A [ n − 1 , v ] = A [ n , v ] for all v∈V v ∈ V .
Consequence: Can check for a negative cycle just by running Bellman-Ford for one extra iteration(running time still O(mn) O ( m n ) ).
Proof of Claim
( ⇒ ⇒ ) Already proved in correctness of Bellman-Ford.
( ⇐ ⇐ ) Assume A[n−1,v]=A[n,v] A [ n − 1 , v ] = A [ n , v ] for all
v∈V v ∈ V
Consequence: Can check for a negative cycle just by jsut running Bellman-Ford for one extra iteration.
(running time still O(mn) O ( m n ) ) .
Let d(v) d ( v ) denote the common value of A[n−1,v] A [ n − 1 , v ] and A[n,v] A [ n , v ] .
Recall algorithm:

A[n,v]=min{A[i−1,v]min(w,v)∈E{A[n−1,w]+cwv},(3) (3) A [ n , v ] = m i n { A [ i − 1 , v ] m i n ( w , v ) ∈ E { A [ n − 1 , w ] + c w v } ,

其中

A[n,v]→d(v),A[n−1,w]→d(w) A [ n , v ] → d ( v ) , A [ n − 1 , w ] → d ( w )

Thus:

d(v)≤d(w)+cwv d ( v ) ≤ d ( w ) + c w v for all edges

(w,v)∈E ( w , v ) ∈ E

Equivalently :

d(v)−d(w)≤cwv d ( v ) − d ( w ) ≤ c w v