Dag and minimum spanning trees

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T l i l S tTopological SortMinimum Spanning Treesp g

M. Shoaib Farooq 1

Directed Acyclic Graphsy p

● A directed acyclic graph or DAG is a directedA directed acyclic graph or DAG is a directed graph with no directed cycles:

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DFS and DAGs

● Argue that a directed graph G is acyclic if aArgue that a directed graph G is acyclic if a DFS of G yields no back edges:■ Forward: if G is acyclic will be no back edges■ Forward: if G is acyclic, will be no back edges

○ Trivial: a back edge implies a cycle

■ Backward: if no back edges, G is acyclic■ Backward: if no back edges, G is acyclic○ Argue contrapositive: G has a cycle ⇒ ∃ a back edge

Let v be the vertex on the cycle first discovered, and u be the d f h lpredecessor of v on the cycle

When v discovered, whole cycle is whiteMust visit everything reachable from v before returning from

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DFS-Visit()So path from u→v is yellow→yellow, thus (u, v) is a back edge

Topological Sortp g

● Topological sort of a DAG:Topological sort of a DAG:■ Linear ordering of all vertices in graph G such that

vertex u comes before vertex v if edge (u, v) ∈ Gg ( , )■ TS can be viewed as an ordering of its vertices

along a horizontal line so that all directed edges go g g gfrom left to right.

● Example Given a set of jobs, courses, etc., with i it t i t t t th j b i d th tprerequisite constraints, output the jobs in an order that

does not violate any of the prerequisites.l ld l i d d

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● Real-world example: getting dressed

Existence of Topological Sortp gLemma G can be topologically sorted iff it has no cycle, that

i iff it i d (di t d li h)is, iff it is a dag (directed acyclic graph).

Proof ⇒ If G has a cycle, then it cannot be topologically sorted.y p g y

⇐ If G has no cycle, then it can be topologically sorted.

Constructive proof: An algorithm that sorts any dag.

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Topological Sort of Digraphp g g p

Ordering < over V(G) such that u < v whenever (u, v) ∈ E(G).

a b Some topological sorts:

c d1. a, c, e, b, d, g, f2. a, b, c, d, g, f, ec

g f3. b, d, g, a, c, f, e

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Getting Dressedg

UndeShorts 11/16 Socks 17/18UndeShorts 11/16

Shoes 13/14Pants 12/15

Watch 9/10

Belt 6/7

Shirt 1/8

Tie 2/5

Jacket 3/4Jacket 3/4

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Getting Dressed

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Topological Sort AlgorithmTopological Sort Algorithm

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Topological Sort Algorithmp g g

Initialize a global queue L ← ⟨ ⟩ within DFS(G)

Add a line to DFS-visit

DFS(G)for each vertex u ∈ V[G]

color[u] = WHITE;Π[u]= NIl

time = 0;for each vertex u ∈ V[G]

if (color[u] == WHITE)if (color[u] == WHITE)DFS_Visit(u);

L ← insert(v, L) // insert v in the front of L

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Topological Sort Algorithmp g g

● Depth-first search takes O(V+E) time and itDepth first search takes O(V E) time and it takes O(1) time to insert each of the |V| vertices onto the front of the link list.ve t ces o to t e o t o t e st.

Ti O(V+E)● Time: O(V+E)

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Minimum Spanning Tree (MST)p g ( )Problem Select edges in a connected and undirected graph to

form a tree that connects all the vertices (spanning tree).minimize the total edge weight of the spanning tree.

1f b g7

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c e4Total weight of tree edges: 14

Applications of MSTpp

Telephone wiring (use as little wire as possible)p g ( p )

Electronic circuit board design

Cancer imaging (MSTs describe arrangements of nucleiin skin cells)

Biomedical image analysis (detect actin fibers in cell images)

Astronomy (find clusters of quasars and Seyfert galaxies)

Finding road networks in satellite and aerial imagery

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Finding road networks in satellite and aerial imagery

Minimum Spanning Tree

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Minimum Spanning Treep g

● Problem: given a connected, undirected,Problem: given a connected, undirected, weighted graph, find a spanning tree using edges that minimize the total weightedges t at e t e tota we g t

6 45 9

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● The phrase “ minimum spanning tree” is a shortened p p gform of the phrase “minimum-weight spanning tree”.

● All spanning trees have exactly |v| -1 edges.● MSTs satisfy the optimal substructure property: an

optimal tree is composed of optimal subtrees

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Free TreeMi i S i T (T )Minimum Spanning Tree (Terms)

● A free tree is a tree with no vertex designated as the groot vertex.

● A free tree with n vertices has exactly n- 1 edges. ● There exists a unique path between any two vertices

of a free tree.● Adding any edge to a free tree creates a unique cycle.

● Breaking any edge on this cycle restores the free tree.

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Viable SetMi i S i T (T )Minimum Spanning Tree (Terms)

● A subset A subset of E is viable if 1. A is a subset of edges of some MST 2. A cannot contains a cycle . ca ot co ta s a cyc e

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Safe EdgeMi i S i T (T )Minimum Spanning Tree (Terms)

● A subset A subset of E is viable if ● A is a subset of edges of some MST. ● An edge (u, v) subset of { E- A} is safe if edge (u, v) subset o { } s safe● A [ {(u, v)} is viable. ● In other words the choice (u v) is a safe choice to● In other words, the choice (u, v) is a safe choice to

add so that A can still be extended to form a MST.OROR

● A edge is safe if it will not create a cycle in MST

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CUTMi i S i T (T )Minimum Spanning Tree (Terms)

● A cut (S, V- S) is just a partition of vertices into two ( , ) j pdisjoint subsets.

● An edge (u, v) crosses the cut if one endpoint is in S and the other is in V-S.

● Given a subset of edges A, a cut respects A if no edge in A crosses the cut.

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Growing a MSTg

GENERIC-MST(G, w)1. A ← Φ1. A Φ2. while A doesnot form a spanning tree3. do find an edge(u, v) that is safe for A4. A ← A U {(u, v)}5. return A

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

S V - S

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

S V - S

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Growing MSTg

b c d8 7

ei 144

h g f 10

8 7h g f

Total Weight = 37

Two Algorithms are used to solve minimum spanning tree problem

Kruskal’s algorithm

Two Algorithms are used to solve minimum spanning tree problemLet A be a subset of edges included in some MST

A is a forest.The safe edge added to A has least weight among all

The safe edge added to A has least weight among alledges connecting its two components.

Prim’s algorithmPrim’s algorithmA is a tree. The safe edge is a light edge connecting a vertex in A

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to one not in A.

Kruskal’s Algorithm (Operations)g

● Make-set(u): Create a set containing a singleMake set(u): Create a set containing a single item u.

● Find-set(u): Find the set that contains u● Find-set(u): Find the set that contains u

● Union(u,v): merge the set containing u and set containing v into a common set.

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Kruskal’s Algorithmg

Kruskal()2 19

Run the algorithm:

T = ∅;

for each v ∈ VMakeSet(v);

sort E by increasing edge weight w

sort E by increasing edge weight wfor each (u,v) ∈ E (in sorted order)

if FindSet(u) ≠ FindSet(v)if FindSet(u) ≠ FindSet(v)T = T U {{u,v}};Union(u, v);

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2? 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19?

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Kruskal()2 19

Run the algorithm:

T = ∅;

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Return T

Prim’s Algorithm – an Exampleg p

2choosen

f b g11 34

c e 74

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edge candidates

f b g11 34

c e 74

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f b g11 34

c e 74

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f b g11 34

c e 74

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f b g11 34

c e 74

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Prim’s Algorithm – an ExampleTotal weight of the MST: 14

f b g11 34

c e 74

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Prim’s Algorithmg

MST-Prim(G, w, r)Q = V[G];for each u ∈ Q

key[u] = ∞;ykey[r] = 0;p[r] = NULL;while (Q not empty)while (Q not empty)

u = ExtractMin(Q);for each v ∈ Adj[u]

if (v ∈ Q and w(u,v) < key[v])p[v] = u;key[v] = w(u,v);

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Prim’s Algorithmg

MST-Prim(G, w, r) 6 49Q = V[G];

for each u ∈ Qkey[u] = ∞; 14 10

ykey[r] = 0;p[r] = NULL;while (Q not empty)

8while (Q not empty)u = ExtractMin(Q);for each v ∈ Adj[u]

Run on example graph

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Prim’s Algorithmg

MST-Prim(G, w, r) ∞6 49Q = V[G];

for each u ∈ Qkey[u] = ∞;

∞ ∞ ∞

Run on example graph

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Prim’s Algorithmg

∞ ∞ ∞

0 ∞ ∞

while (Q not empty)u = ExtractMin(Q);for each v ∈ Adj[u]

Pick a start vertex r

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Prim’s Algorithmg

∞ ∞ ∞

0 ∞ ∞

Red vertices have been removed from Q

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Prim’s Algorithmg

∞ ∞ ∞

0 ∞ ∞

Red arrows indicate parent pointers

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Prim’s Algorithmg

14 ∞ ∞

0 ∞ ∞

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Prim’s Algorithmg

14 ∞ ∞

0 ∞ ∞

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Prim’s Algorithmg

14 ∞ ∞

0 8 ∞

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Prim’s Algorithmg

10 ∞ ∞

0 8 ∞

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Prim’s Algorithmg

10 ∞ ∞

0 8 ∞

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Prim’s Algorithmg

10 2 ∞

0 8 ∞

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Prim’s Algorithmg

10 2 ∞

0 8 15

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Prim’s Algorithmg

MST-Prim(G, w, r) ∞6 49

Q = V[G];for each u ∈ Q

key[u] = ∞;

10 2 ∞

0 8 15

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Prim’s Algorithmg

MST-Prim(G, w, r) ∞6 49

key[u] = ∞;

10 2 9

0 8 15

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Prim’s Algorithmg

MST-Prim(G, w, r) 46 49

key[u] = ∞;

10 2 9

0 8 15

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Prim’s Algorithmg

key[u] = ∞;

0 8 15

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Prim’s Algorithmg

key[u] = ∞;

0 8 15

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Prim’s Algorithmg

key[u] = ∞;

0 8 15

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Prim’s Algorithmg

key[u] = ∞;

0 8 15

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Prim’s Algorithmg

MST-Prim(G, w, r) 46 49Q = V[G];

uykey[r] = 0;p[r] = NULL;while (Q not empty)

0 8 15

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Review: Prim’s Algorithmg

key[u] = ∞;ykey[r] = 0;p[r] = NULL;while (Q not empty)

What is the hidden cost in this code?while (Q not empty)

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key[u] = ∞;ykey[r] = 0;p[r] = NULL;while (Q not empty)while (Q not empty)

if (v ∈ Q and w(u,v) < key[v])p[v] = u;DecreaseKey(v, w(u,v));

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key[u] = ∞; How often is ExtractMin() called?ykey[r] = 0;p[r] = NULL;while (Q not empty)

How often is ExtractMin() called?How often is DecreaseKey() called?

if (v ∈ Q and w(u,v) < key[v])p[v] = u;DecreaseKey(v, w(u,v));

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key[u] = ∞;What will be the running time?A: Depends on queuey

key[r] = 0;p[r] = NULL;while (Q not empty)

A: Depends on queuebinary heap: O(E lg V)Fibonacci heap: O(V lg V + E)

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Dag and minimum spanning trees

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