View
223
Download
0
Embed Size (px)
Citation preview
CPSC 311, Fall 2009 1
CPSC 311Analysis of Algorithms
More Graph AlgorithmsProf. Jennifer Welch
Fall 2009
CPSC 311, Fall 2009 2
Minimum Spanning Tree
7
1645
6 8
11
15
14
17
10
13
3
12
29
18
find subset of edges that span all the nodes,create no cycle, and minimize sum of weights
CPSC 311, Fall 2009 3
Facts About MSTs There can be many spanning trees of
a graph In fact, there can be many minimum
spanning trees of a graph But if every edge has a unique
weight, then there is a unique MST
CPSC 311, Fall 2009 4
Uniqueness of MST Suppose in contradiction there are 2
MSTs, M1 and M2. Let e be edge with minimum weight that
is one but not the other (say it is in M1).
If e is added to M2, a cycle is formed. Let e' be an edge in the cycle that is not
in M1
CPSC 311, Fall 2009 5
Uniqueness of MST
e: in M1 but not M2
e': in M2 but not M1; wt is less than wt of e
M2:
Replacing e with e' creates a new MST M3 whose weight is less than that of M2
CPSC 311, Fall 2009 6
Generic MST Algorithm input: weighted undirected graph
G = (V,E,w) T := empty set while T is not yet a spanning tree of G
find an edge e in E s.t. T U {e} is a subgraph of some MST of G
add e to T return T (as MST of G)
CPSC 311, Fall 2009 7
Kruskal's MST algorithm
7
1645
6 8
11
15
14
17
10
13
3
12
29
18
consider the edges in increasing order of weight,add in an edge iff it does not cause a cycle
CPSC 311, Fall 2009 8
Kruskal's Algorithm as a Special Case of Generic Alg. Consider edges in increasing order of
weight Add the next edge iff it doesn't cause
a cycle At any point, T is a forest (set of
trees); eventually T is a single tree
CPSC 311, Fall 2009 9
Kruskal's Algorithm input: G = (V,E,w) // weighted graph T := Ø // subset of E sort E by increasing weights for each (u,v) in E in sorted order do
if T U {u,v} has no cycle then T := T U {(u,v)}
return (V,T)
CPSC 311, Fall 2009 10
Why is Kruskal's Greedy? Algorithm manages a set of edges s.t.
these edges are a subset of some MST At each iteration:
choose an edge so that the MST-subset property remains true
subproblem left is to do the same with the remaining edges
Always try to add cheapest available edge that will not violate the tree property locally optimal choice
CPSC 311, Fall 2009 11
Correctness of Kruskal's Alg. Let e1, e2, …, en-1 be sequence of
edges chosen Clearly they form a spanning tree Suppose it is not minimum weight Let ei be the edge where the
algorithm goes wrong {e1,…,ei-1} is part of some MST M but {e1,…,ei} is not part of any MST
CPSC 311, Fall 2009 12
Correctness of Kruskal's Alg.
white edges are part of MST M, which contains e1 to ei-1,but not ei
M: ei, forms a cycle in M
e* :min wt. edge in cycle not ine1 to ei-1
replacing e* w/ ei formsa spanning tree withsmaller weight than M,contradiction!
wt(e*) > wt(ei)
CPSC 311, Fall 2009 13
Note on Correctness Proof Argument on previous slide works for
case when every edge has a unique weight
Algorithm also works when edge weights are not necessarily correct
Modify proof on previous slide: contradiction is reached to assumption that ei is not part of any MST
CPSC 311, Fall 2009 14
Implementing Kruskal's Alg. Sort edges by weight
efficient algorithms known How to test quickly if adding in the
next edge would cause a cycle? use Disjoint Sets data structure…
CPSC 311, Fall 2009 15
Disjoint Sets in Kruskal's Alg. Use Disjoint Sets data structure to test whether
adding an edge to a set of edges would cause a cycle
Each set represents nodes that are connected using edges chosen so far
Initially put each node in a set by itself using Make-Set
When testing edge (u,v), call Find-Set to check if u and v are in the same set if so, then a cycle would result if (u,v) is chosen
When edge (u,v) is chosen, call Union(u,v)
CPSC 311, Fall 2009 16
Kruskal's Algorithm #2 input: G = (V,E,w) // weighted graph T := Ø // subset of E sort E by increasing weights for each v in V do MakeSet(v) for each (u,v) in E in sorted order do
a := FindSet(u) b := FindSet(v) if a ≠ b then
T := T U {(u,v)} Union(a,b)
return (V,T)
CPSC 311, Fall 2009 17
Running Time of Kruskal's MST Algorithm Sorting the edges takes O(E log E) time The rest of the time is proportional to the total time taken by all
the calls to MakeSet, FindSet, and Union. Number of calls to MakeSet is |V| If we unravel the for loop, we can see:
total number of calls to FindSet is O(E) total number of calls to Union is O(V)
So number of operations is O(V+E), with |V| MakeSet ops. If the Disjoint Sets data structure is implemented using union by
rank and path compression, the time for all the Disjoint Sets operations is O((V+E) log*V)
Total time is O(E log E), since graph is connected, so V ≤ E + 1, and this equals O(E log V), since E is O(V2) and log V2 = 2log V
CPSC 311, Fall 2009 18
Another Greedy MST Alg. Kruskal's algorithm maintains a forest
that grows until it forms a spanning tree Alternative idea is keep just one tree and
grow it until it spans all the nodes Prim's algorithm
At each iteration, choose the minimum weight outgoing edge to add greedy!
CPSC 311, Fall 2009 19
Idea of Prim's Algorithm Instead of growing the MST as possibly
multiple trees that eventually all merge, grow the MST from a single node, so that there is only one tree at any point.
Also a special case of the generic algorithm: at each step, add the minimum weight edge that goes out from the tree constructed so far.
CPSC 311, Fall 2009 20
Prim's Algorithm input: weighted undirected graph G = (V,e,w) T := empty set S := {any node in V} while |T| < |V| - 1 do
let (u,v) be a min wt. outgoing edge (u in S, v not in S) add (u,v) to T add v to S
return (S,T) (as MST of G)
CPSC 311, Fall 2009 22
Correctness of Prim's Algorithm Let Ti be the tree represented by (S,T)
at the end of iteration i. Show by induction on i that Ti is a
subtree of some MST of G. Basis: i = 0 (before first iteration). T0
contains just a single node, and thus is a subtree of every MST of G.
CPSC 311, Fall 2009 23
Correctness of Prim's Algorithm Induction: Assume Ti is a subtree of
some MST M. We must show Ti+1 is a subtree of some MST.
Let (u,v) be the edge added in iteration i+1.
u vTi
Ti+1
Case 1: (u,v) is in M.Then Ti+1 is also asubtree of M.
CPSC 311, Fall 2009 24
Correctness of Prim's AlgorithmCase 2: (u,v) is not in M. There is a path P in M from u to v, since M
spans G. Let (x,y) be the first edge in P with one
endpoint in Ti and the other not in Ti.
u v
Ti
x
yP
CPSC 311, Fall 2009 25
Correctness of Prim's Algorithm Let M' = M - {(x,y)} U {(u,v)} M' is also a spanning tree of G. w(M') = w(M) - w(x,y) + w(u,v) ≤ w(M)
since (u,v) is min wt outgoing edge So M' is also an MST and
Ti+1 is subtree
M' u v
Ti
x
y
Ti+1
CPSC 311, Fall 2009 26
Implementing Prim's Algorithm How do we find minimum weight
outgoing edge? First cut: scan all adjacency lists at
each iteration. Results in O(VE) time. Try to do better.
CPSC 311, Fall 2009 27
Implementing Prim's Algorithm Idea: have each node not yet in the
tree keep track of its best (cheapest) edge to the tree constructed so far.
To find min wt. outgoing edge, find minimum among these values use a priority queue to store the best edge
info (insert and extract-min operations)
CPSC 311, Fall 2009 28
Implementing Prim's Algorithm When a node v is added to T, some
other nodes might have their best edges affected, but only neighbors of v add decrease-key operation to the priority
queueu
vTi
Ti+1 w
w's best edge to Ti
check if this edge is cheaper for w
x
x's best edge to Ti
v's best edge to Ti
CPSC 311, Fall 2009 29
Details on Prim's AlgorithmAssociate with each node v two fields: best-wt[v] : if v is not yet in the tree, then
it holds the min. wt. of all edges from v to a node in the tree. Initially infinity.
best-node[v] : if v is not yet in the tree, then it holds the name of the node u in the tree s.t. w(v,u) is v's best-wt. Initially nil.
CPSC 311, Fall 2009 30
Details on Prim's Algorithm input: G = (V,E,w)// initialization initialize priority queue Q to contain all
nodes, using best-wt values as keys let v0 be any node in V decrease-key(Q,v0,0)
// last line means change best-wt[v0] to 0 and adjust Q accordingly
CPSC 311, Fall 2009 31
Details on Prim's Algorithm while Q is not empty do
u := extract-min(Q) // node w/ smallest best-wt
if u is not v0 then add (u,best-node[u]) to T for each neighbor v of u do
if v is in Q and w(u,v) < best-wt[v] then best-node[v] := u decrease-key(Q,v,w(u,v))
return (V,T) // as MST of G
CPSC 311, Fall 2009 32
Running Time of Prim's AlgorithmDepends on priority queue implementation. Let Tins be time for insert Tdec be time for decrease-key Tex be time for extract-minThen we have |V| inserts and one decrease-key in the
initialization: O(VTins+Tdec) |V| iterations of while
one extract-min per iteration: O(VTex) total
CPSC 311, Fall 2009 33
Running Time of Prim's Algorithm Each iteration of while includes a for
loop. Number of iterations of for loop varies,
depending on how many neighbors the current node has
Total number of iterations of for loop is O(E).
Each iteration of for loop: one decrease key, so O(ETdec) total
CPSC 311, Fall 2009 34
Running Time of Prim's Algorithm O(V(Tins + Tex) + ETdec) If priority queue is implemented with
a binary heap, then Tins = Tex = Tdec = O(log V) total time is O(E log V)
(Think about how to implement decrease-key in O(log V) time.)
CPSC 311, Fall 2009 35
Shortest Paths in a Graph We already saw the Floyd-Warshall
algorithm to compute all pairs of shortest paths
Let's review two important single-source shortest path algorithms: Dijkstra's algorithm Bellman-Ford algorithm
CPSC 311, Fall 2009 36
Single Source Shortest Path Problem Given: directed or undirected graph
G = (V,E,w) and source node s in V Find: For each t in V, a path in G from
s to t with minimum weight Warning! Negative weights are a
problem:
s t4
5
CPSC 311, Fall 2009 37
Shortest Path Tree Result of a SSSP algorithm can be
viewed as a tree rooted at the source Why not use breadth-first search? Works fine if all weights are the same:
weight of each path is (a multiple of) the number of edges in the path
Doesn't work when weights are different
CPSC 311, Fall 2009 38
Dijkstra's SSSP Algorithm Assumes all edge weights are nonnegative Similar to Prim's MST algorithm Start with source node s and iteratively
construct a tree rooted at s Each node keeps track of tree node that
provides cheapest path from s (not just cheapest path from any tree node)
At each iteration, include the node whose cheapest path from s is the overall cheapest
CPSC 311, Fall 2009 40
Implementing Dijkstra's Alg. How can each node u keep track of its best
path from s? Keep an estimate, d[u], of shortest path
distance from s to u Use d as a key in a priority queue When u is added to the tree, check each of
u's neighbors v to see if u provides v with a cheaper path from s: compare d[v] to d[u] + w(u,v)
CPSC 311, Fall 2009 41
Dijkstra's Algorithm input: G = (V,E,w) and source node s// initialization d[s] := 0 d[v] := infinity for all other nodes v initialize priority queue Q to contain
all nodes using d values as keys
CPSC 311, Fall 2009 42
Dijkstra's Algorithm while Q is not empty do
u := extract-min(Q) for each neighbor v of u do
if d[u] + w(u,v) < d[v] thend[v] := d[u] + w(u,v)decrease-key(Q,v,d[v])parent(v) := u
CPSC 311, Fall 2009 43
Dijkstra's Algorithm Example
a b
d e
c
2
84 9
2
4
12
10 6 3
source is node a
0 1 2 3 4 5Q abcd
ebcde cde de d Ø
d[a]
0 0 0 0 0 0
d[b]
∞ 2 2 2 2 2
d[c] ∞ 12 10 10 10 10
d[d]
∞ ∞ ∞ 16 13 13
d[e]
∞ ∞ 11 11 11 11
iteration
CPSC 311, Fall 2009 44
Correctness of Dijkstra's Alg. Let Ti be the tree constructed after i-th
iteration of while loop: nodes not in Q edges indicated by parent variables
Show by induction on i that the path in Ti
from s to u is a shortest path and has distance d[u], for all u in Ti.
Basis: i = 1. s is the only node in T1 and d[s] = 0.
CPSC 311, Fall 2009 45
Correctness of Dijkstra's Alg. Induction: Assume Ti-1 is a correct shortest path
tree and show for Ti. Let u be the node added in iteration i. Let x = parent(u).
s x
Ti-1
u
Ti
Need to show path in Ti from s to u is a shortest path, and has distance d[u]
CPSC 311, Fall 2009 46
Correctness of Dijkstra's Alg
s
x
Ti-1
u
Ti
P, path in Ti
from s to u
(a,b) is first edge in P' thatleaves Ti-1 (i.e., a is in Ti-1 but b is not)
a
b
P', anotherpath from s to u
CPSC 311, Fall 2009 47
Let P1 be part of P' before (a,b).
Let P2 be part of P' after (a,b).
w(P') = w(P1) + w(a,b) + w(P2)
≥ w(P1) + w(a,b) (nonneg wts)
≥ (wt of path in Ti-1 from s to a) + w(a,b) (inductive hypothesis)
≥ (wt of path in Ti-1 from s to x) + w(x,u) (alg chose u in
iteration i and d-values are accurate, by inductive hyp.)
= w(P).
So P is a shortest path, and d[u] is accurate after iteration i.
Correctness of Dijkstra's Alg
s
x
Ti-1
u
Ti
a
bP'
P
CPSC 311, Fall 2009 48
Running Time of Dijstra's Alg. initialization: insert each node once
O(V Tins)
O(V) iterations of while loop one extract-min per iteration => O(V Tex) for loop inside while loop has variable number of
iterations…
For loop has O(E) iterations total one decrease-key per iteration => O(E Tdec)
Total is O(V (Tins + Tex) + E Tdec)
CPSC 311, Fall 2009 49
Using Fancier Heap Implementations O(V(Tins + Tex) + E Tdec) If priority queue is implemented with a
binary heap, then Tins = Tex = Tdec = O(log V) total time is O(E log V)
There are fancier implementations of the priority queue, such as Fibonacci heap: Tins = O(1), Tex = O(log V), Tdec = O(1) (amortized) total time is O(V log V + E)
CPSC 311, Fall 2009 50
Using Simpler Heap Implementations O(V(Tins + Tex) + E Tdec) If graph is dense, so that |E| = (V2), then it
doesn't help to make Tins and Tex to be at most O(V).
Instead, focus on making Tdec be small, say constant.
Implement priority queue with an unsorted array: Tins = O(1), Tex = O(V), Tdec = O(1) total is O(V2)
CPSC 311, Fall 2009 51
What About Negative Edge Weights? Dijkstra's SSSP algorithm requires all
edge weights to be nonnegative even more restrictive than outlawing
negative weight cycles Bellman-Ford SSSP algorithm can
handle negative edge weights even "handles" negative weight cycles by
reporting they exist
CPSC 311, Fall 2009 52
Bellman-Ford Idea Consder each edge (u,v) and see if u
offers v a cheaper path from s compare d[v] to d[u] + w(u,v)
Repeat this process |V| - 1 times to ensure that accurate information propagates from s, no matter what order the edges are considered in
CPSC 311, Fall 2009 53
Bellman-Ford SSSP Algorithm input: directed or undirected graph G = (V,E,w)//initialization initialize d[v] to infinity and parent[v] to nil for all v in V other
than the source initialize d[s] to 0 and parent[s] to s// main body for i := 1 to |V| - 1 do
for each (u,v) in E do // consider in arbitrary order if d[u] + w(u,v) < d[v] then
d[v] := d[u] + w(u,v) parent[v] := u
// continued on next slide
CPSC 311, Fall 2009 54
Bellman-Ford SSSP Algorithm// check for negative weight cycles for each (u,v) in E do
if d[u] + w(u,v) < d[v] then output "negative weight cycle exists"
CPSC 311, Fall 2009 55
Running Time of Bellman-Ford O(V) iterations of outer for loop O(E) iterations of inner for loop O(VE) time total
CPSC 311, Fall 2009 56
Bellman-Ford Example
s
c
a
b3
—44
2
1
process edges in order(c,b)(a,b)(c,a)(s,a)(s,c)
<board work>
CPSC 311, Fall 2009 57
Correctness of Bellman-FordAssume no negative-weight cycles.Lemma: d[v] is never an underestimate of the
actual shortest path distance from s to v.Lemma: If there is a shortest s-to-v path
containing at most i edges, then after iteration i of the outer for loop, d[v] is at most the actual shortest path distance from s to v.
Theorem: Bellman-Ford is correct.Proof: Follows from these 2 lemmas and fact
that every shortest path has at most |V| - 1 edges.
CPSC 311, Fall 2009 58
Correctness of Bellman-Ford Suppose there is a negative weight
cycle. Then the distance will decrease even
after iteration |V| - 1 shortest path distance is negative infinity
This is what the last part of the code checks for.
CPSC 311, Fall 2009 59
Speeding Up Bellman-Ford The previous example would have
converged faster if we had considered the edges in a different order in the for loop move outward from s
If the graph is a DAG (no cycles), we can fully exploit this idea to speed up the running time
CPSC 311, Fall 2009 60
DAG Shortest Path Algorithm input: directed graph G = (V,E,w) and
source node s in V topologically sort G d[v] := infinity for all v in V d[s] := 0 for each u in V in topological sort order do
for each neighbor v of u do d[v] := min{d[v],d[u] + w(u,v)}