1. Page 620 Single-Source Shortest Paths
Problem Definition
Shortest paths and Relaxation
Dijkstra’s algorithm (can be viewed as a greedy algorithm)
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2. Find a shortest path from station A to station B.
-need serious thinking to get a correct algorithm.
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3. Adjacency-list representation
Let G=(V, E) be a graph.
V– set of nodes (vertices)
E– set of edges.
For each uV, the adjacency list Adj[u] contains all nodes in V that are adjacent to u. 2 1 5 4 3 /
(a)
(b)
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4. Adjacency-matrix representation
Assume that the nodes are numbered 1, 2, …, n.
The adjacency-matrix consists of a |V||V| matrix A=(aij) such that
aij= 1 if (i,j) E, otherwise aij= 0. 2 1 5 4 3
(a)
(c)
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5. Problem Definition:
Real problem: A motorist wishes to find the shortest possible route from Chicago to Boston.Given a road map of the United States on which the distance between each pair of adjacent intersections is marked, how can we determine this shortest route?
Formal definition: Given a graph G=(V, E, W), where each edge has a weight, find a shortest path from s to v for some interesting vertices s and v.
s—source
v—destination.
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6. The cost of the shortest path from s to v is denoted as (s, v).
The weight of path p=<v0,v1,…,vk > is the sum of the weights of its constituent edges:
The cost of the shortest path from s to v is denoted as (s, v).
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7. Negative-Weight edges:
Edge weight may be negative.
negative-weight cycles– the total weight in the cycle (circuit) is negative.
If no negative-weight cycles reachable from the source s, then for all v V, the shortest-path weight remains well defined,even if it has a negative value.
If there is a negative-weight cycle on some path from s to v, we define =
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8. a b -4 h i 3 -1 2 4 3 c d 6 8 5 -8 3 5 11 g s -3 e 3 f 2 7 j -6
Figure1 Negative edge weights in a directed graph.Shown within each vertex is its
shortest-path weight from source s.Because vertices e and f form a negative-weight
cycle reachable from s,they have shortest-path weights of Because vertex g is
reachable from a vertex whose shortest path is ,it,too,has a shortest-path weight
of Vertices such as h, i ,and j are not reachable from s,and so their shortest-path
weights are , even though they lie on a negative-weight cycle.
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9. Representing shortest paths:
we maintain for each vertex vV , a predecessor
[ v] that is the vertex in the shortest path
right before v.
With the values of , a backtracking process can give the shortest path. (We will discuss that after the algorithm is given)
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10. Observation: (basic)
Suppose that a shortest path p from a source s to a vertex v can be decomposed into
s u v
for some vertex u and path p’. Then, the weight of a shortest path from s to v is
We do not know what is u for v, but we know u is in V and we can try all nodes in V in O(n) time.
Also, if u does not exist, the edge (s, v) is the shortest.
Question: how to find (s, u), the first shortest from s to some node?
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11. Relaxation:
The process of relaxing an edge (u,v) consists of testing whether we can improve the shortest path to v found so far by going through u and,if so,updating d[v] and [v].
RELAX(u,v,w)
if d[v]>d[u]+w(u,v)
then d[v] d[u]+w(u,v) (based on observation)
[v] u
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12. u v u v 2 2 5 9 5 6 RELAX(u,v) RELAX(u,v) u v u v 2 2 5 7 5 6 (a) (b)
Figure2 Relaxation of an edge (u,v).The shortest-path estimate of each vertex
is shown within the vertex. (a)Because d[v]>d[u]+w(u,v) prior to relaxation,
the value of d[v] decreases. (b)Here, d[v] d[u]+w(u,v) before the relaxation
step,so d[v] is unchanged by relaxation.
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13. Initialization:
For each vertex v  V, d[v] denotes an upper bound on the weight of a shortest path from source s to v.
d[v]– will be (s, v) after the execution of the algorithm.
initialize d[v] and [v] as follows: .
INITIALIZE-SINGLE-SOURCE(G,s)
for each vertex v  V[G]
do d[v]
[v] NIL
d[s]
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14. Dijkstra’s Algorithm:
Dijkstra’s algorithm assumes that w(e)0 for each e in the graph.
maintain a set S of vertices such that
Every vertex v S, d[v]=(s, v), i.e., the shortest-path from s to v has been found. (Intial values: S=empty, d[s]=0 and d[v]=)
(a) select the vertex uV-S such that
d[u]=min {d[x]|x V-S}. Set S=S{u}
D[u]= (s, u) at this moment! Why?
(b) for each node v adjacent to u do RELAX(u, v, w).
Repeat step (a) and (b) until S=V.
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15. Continue: DIJKSTRA(G,w,s): INITIALIZE-SINGLE-SOURCE(G,s) S Q V[G]
while Q
do u EXTRACT -MIN(Q)
S S {u}
for each vertex v  Adj[u]
do RELAX(u,v,w)
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16. Implementation:
a adaptable priority queue Q stores vertices in V-S, keyed by their d[] values.
the graph G is represented by adjacency lists so that it takes O(1) time to find an edge (u, v) in (b).
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17. u v 10 5 2 1 3 4 6 9 7 8 s
Single Source Shortest Path Problem x y
(a)
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18. u v 1 10 8 10 9 s 2 3 4 6 7 5 5 8 2 x y
(b)
(s,x) is the shortest path using one edge. It is also the shortest path from s to x.
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19. Assume EXTRACT -MIN(Q)=x. (s,x) is the shortest path using one edge.
Why? Since (s, x) is the shortest among all edges starting from s.
It is also the shortest path from s to x.
Proof: (1) Suppose that path P: s->u…->x is the shortest path. Then w (s,u) w(s, x).
(2) Since edges have non-negative weight, the total weight of path P is at least w(s,u) w(s, x).
(3) So, the edge (s, x) is the shortest path from s to x.
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20. u v 1 8 14 10 9 s 2 3 4 6 7 5 5 7 2 x y
(c)
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21. Statement: Suppose S={s, x} and d[y]=min d(v). ……(1) vV-S
Then d[y] is the cost of the shortest, i.e., either s->x->y or (s->y) is the shortest path from s to y.
Why? If (s, y) is the shortest path, d[y] is correct.
Consider that case: s->x->y is the shortest path for y.
Proof by contradiction.
Assume that s->x->y is not the shortest path and that
P1: s->yy->…->y is a shortest path and yyS.
(At this moment, we already tried the case for yy=s and yy=x in the alg.)
Thus, w(P1)< w(s->x->y). (from assumption: s->x-Y is not the shortest)
Since w(e)0 for any e, w(s->yy)<w(P1)<w(s->x->y).
Therefore, d[yy]<d[y] and (1) is not true.
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22. u v 1 8 13 10 9 s 2 3 4 6 7 5 5 7 2 x y
(d)
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23. 7 9 5 8 10 2 1 3 4 6 s u v x y
(e)
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24. u v 1 8 9 10 9 s 2 3 4 6 7 5 5 7 2 x y
(f)
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25. Theorem: Let S be the set in algorithm and d[y]=min d(v). ……(1) vV-S
Then d[y] is the cost of the shortest. (hard part)
Proof: Assume that (1) for any v in S, y[v] is the cost of the shortest path. We want to show that d[y] is also the cost of the shortest path after execution of Step (a)
If the shortest path from s to v contains vertices in S ONLY, then d[v] is the length of the shortest.
Assume that (2) d[y] is NOT the cost of the shortest path from s to y and that P1: s…->yy->…->y is a shortest path and yyS is the 1st node in P1 not in S. Thus, w(P1)<d[y].
So, w(s…->yy)<w(P1). (weight of edge >=0)
Thus, w(s…->yy)<w(P1)<d[y].
From (1) and (2), after execution of step (a) d[yy]w(s…->yy).
Therefore, d[yy]<d[y] and (1) is not true.
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26. Time complexity of Dijkstra’s Algorithm:
Time complexity depends on implementation of the adaptable priority.
Method 1: Use an array to story the Queue
EXTRACT -MIN(Q) --takes O(|V|) time.
Totally, there are |V| EXTRACT -MIN(Q)’s.
time for |V| EXTRACT -MIN(Q)’s is O(|V|2).
RELAX(u,v,w) --takes O(1) time.
Totally |E| RELAX(u, v, w)’s are required.
time for |E| RELAX(u,v,w)’s is O(|E|).
Total time required is O(|V|2+|E|)=O(|V|2)
Backtracking with [] gives the shortest path in inverse order.
Method 2: The adaptable priority queue is implemented as a heap. It takes O(log |V|) time to do EXTRACT-MIN(Q) and O(log |V|) time for each RELAX(u, v, w)’s. The total running time is O((|V|+|E|)log |V|)=O(|E|log |V|) assuming the graph is connected.
When |E| is O(|V|) the second implementation is better. If |E|=O(|V|2), then the first implementation is better.
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27. Method 3: The priority queue is implemented as a Fibonacci heap
Method 3: The priority queue is implemented as a Fibonacci heap. It takes O(log |V|) time to do EXTRACT-MIN(Q) and O(1) time to decrease the key value of an entry. The total running time is O(|V|log |V|+|E|). (not required)
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