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This course material is now made available for public usage.Special acknowledgement to School of Computing, National University of Singapore

for allowing Steven to prepare and distribute these teaching materials.

Com etitive Pro rammin

Week 05 Graph 1 (Basics)

CS3233 Competitive Programming,Steven Halim, SoC, NUS

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Mini Contest #4 + Break + Discussion

Admins CS2010/2020 Reviews (not discussed in details except red ones )

Graph: Preliminary & Motivation rap raversa gor ms

DFS/BFS: Connected Components /Flood Fill DFS only: Toposort/Cut Vertex+Bridges/ Strongly Connected Components

Minimum Spanning Tree Algorithm

Kruskals and Prims: Plus Various Applications


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CS2010 2020 Reviews (not discussed in details except red ones ) Single Source Shortest Paths

Dijkstras: SSSP on Weighted (no ve cycle) graph/Variants

AllPairs Shortest Paths

Floyd Warshalls + Variants Special Graphs Part 1

, ,


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Vertices Nodes

Edges

(Strongly) Connected Component

Un/Weighted Un/Directed

u rap Complete Graph

In/Out Degree Self Loop/Multiple Edges

recte

cyc c

rap Tree/Forest

Multigraph vs Simple

Graph

Euler Hamiltonian

Path/Cycle parse ense Path, Cycle

par e rap

so ate , eac a eCS3233 Competitive Programming,

Steven Halim, SoC, NUS

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Kaohsiun 2006Standings & Felixs Analysis

A. Print Wor s in LinesB. The Bug Sensor Problem

A. DPB. Graph, MST

. D. Lucky and Good MonthsE. Route Planning

. D. Tedious Ad HocE. Search?

Shift Cipher (N/A in Live Archive)F. A Scheduling Problem

Complete Search + STLF. Greedy?

. ec t e nesH. Perfect Service

.H. DA Graph, DP on Tree


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Sin a ore 2007Standings & Felixs Analysis

A. MODEXB. JONES

A. Mat , Mo u o Arit meticB. DP

.D. TUSKE. SKYLINE

. D. GeometryE. DS, Segment Tree

F. USHERG. RACING

F. Graph, APSP, Min Weighted Cycle++G. Graph, Maximum Spanning Tree


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Jakarta 2008Standings & Suhendrys Analysis

A. Anti Brute Force LocB. Bonus Treasure

A. Grap , MSTB. Recursion

. D. Disjoint PathsE. Expert Enough?

.D. DA Graph, DP on TreeE. Complete Search (is enough)

F. Free ParenthesesG. Greatest KPalindrome Sub

F. DP, harder than problem IG. String

. yper oI. ICPC Team Strategy

. atI. DP, medium

. . ,


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A. ASCII Diamon iB. Match Maker

A. A HocB. DP, Stable Marriage Problem?

. D. Irreducible FractionsE. Gun Fight

. D. MathE. Graph, MCBM (AlternatingPath)

F. Unlock the LockG. Ironman Race in Treeland

F. Graph, SSSP (BFS)G. DA Graph, Likely DP on Tree?

. oot ng t e onsterI. Addition Substraction Game

. omp eoI. ?

.

K. Triangle Hazard.

K. Math


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A. Sa esB. String Popping

A. Brute ForceB. Recursive Backtracking

.D. MinesE. Binary Search Tree

. D. Geometry + SCCs (Graph)E. Graph, BST, Math, Combinatoric

F. Tour BeltG. String Phone

F. Graph, MST (modified)G. ?

. nsta at onsI. Restaurant

.I. ?

. .


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Depth First Search (DFS)Breadth First Search (BFS)

Reachability

Flood FillTopological Sort

Finding Cycles

(Back

Edges)

Finding Articulation Points & Bridges

GRAPH TRAVERSAL ALGORITHMS


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How to solve these UVa problems:

469 Wetlands of Florida Similar problems: 260, 352, 572, 782, 784, 785, etc

Similar problems: 1263, 11709, etc

ou am ar y w ep rs earc

algorithm and its variants, they look hard


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How to solve these UVa problems:

336 A Node Too Far Similar problems: 383, 439, 532, 762, 10009, etc

graph traversal algorithm , they look hard


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Given a graph, we want to traverse it!

Depth First Search (DFS)

sua y mp emen e us ng recurs on

More natural ost requent y use to traverse a grap

Breadth First Search (BFS) Usually implemented using queue (+ map), use STL Can solve special case* of shortest paths problem!


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O(V + E) if using Adjacency List

O(V2) if using Adjacency Matrixt ypedef pai r i i ; t ypedef vect or vi ;

voi d df s ( i nt u) { / / DFS f or nor mal usagepr i nt f ( " %d" , u) ; / / t hi s ver t ex i s vi si t eddf s_num[ u] = DFS_BLACK; / / mar k as vi s i t edf or ( i nt j = 0; j < ( i nt ) Adj Li s t [ u] . s i ze; j ++) {

i i v = Adj Li s t [ u] [ j ] ; / / t r y al l nei ghbor s v of ver t ex ui f ( df s_ num[ v. f i r st ] == DFS_WHI TE) / / avoi d cycl e

df s( v. f i r s t ) ; / / v i s a ( nei ghbor , wei ght ) pai r}

}


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Complexity: also O(V + E) using Adjacency List

map n , n s ; s sour ce = ;queue q; q. push( sour ce) ; / / st ar t f r om sour ce

w e q. emp yi nt u = q. f r ont ( ) ; q. pop( ) ; / / queue: l ayer by l ayer !f or ( i nt j = 0; j < ( i nt ) Adj Li st [ u] . si ze( ) ; j ++) {

v = s u ; or eac ne g our s o ui f ( ! di st . count ( v. f i r s t ) ) {

di st [ v. f i r st ] = di st [ u] + 1; / / unvi si t ed + r eachabl eq. pus v. r s ; enqueue v. r s or nex s eps

}}


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st

DFS (and BFS) can find connected components

A call of dfs u visits onl vertices connected to u

i nt numComp = 0;

df s_num . assi gn ( V, DFS_WHI TE) ;REP ( i , 0, V - 1) / / f or each ver t ex i i n [ 0. . V- 1]

== _ _ pr i nt f ( " Component %d, vi si t " , ++numComp) ;

df s( i ) ; / / one component f oundpr i nt f ( " \ n" ) ;

}r i nt f ( " Ther e ar e %d connect ed com onent s\ n" , numCom ) ;


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n ng opo og ca or see ex ooFinding Articulation Points and Bridges (see text book)Findin Stron l Connected Com onent

TARJANS DFS ALGORITHMS


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Tarjans SCC 0 DFSpann ng

TreeInput: A Directed Graphdfs_num: visitation counter

dfs_num(0) = 0 dfs_num(1) = 1 dfs_num(3) = 2 dfs_num(4) = 4 dfs_num(5) = 5

s_ ow: owes s_num reac a e rom a ver exusing the current DFS spanning tree

dfs_low(0) = 0 dfs_low(1) = 1 dfs_low(3) = 1 dfs_low(4) = 4 dfs_low(5) = 4

1 30 4 5

dfs_num(2) = 3dfs_low(2) = 1



1320

4576

DAG aftercontracting


6SCCs

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Code: Tarjans SCC (not in IOI syllabus)vi dfs_num, dfs_low, S, visited; / / gl obal var i abl es

voi d t ar j anSCC( i nt u) {dfs_low[u] = dfs_num[u] = dfsNumberCounter++; / / df s_ l ow[ u]

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DFS

BFS

Slightly easier to Can solve SSSP on

Use less memory

(discussed later)

Cons:

Cons: Slightly longer to

unweighted graphs code


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Prims algorithm Read textbook on your own or revise CS2010 CS2020 material

KRUSKALS ALGORITHM FOR MINIMUM SPANNING TREE


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Given this graph, select some edges s.t

the ra h is connectedbut with minimal total weight!225

MST!1 35

58

458

5


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Given a connected undirected graph G,

select E G such that a tree is formed and this tree spans (covers) all V G!

There can be several spanning trees in G The one where total cost is minimum

is called the Minimum Spanning Tree (MST) UVa: 908 (Reconnecting Computer Sites)


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pannng reeCost: 4+4+6+6 = 20

nCost: 4+6+6+2 = 18

e r g na rap

124

124

124

0 24

68

0 24

68

0 24

68

369

369

369

4 4 4

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Prims (Greedy Algorithm)

At ever iteration choose an ed e withminimum cost that does not form a cycle

rows an MST from a root

Kruskals (also Greedy Algorithm) Repeate y in s e ges wit minimum costs

that does not form a cycle forms an MST by connecting forests

Which one is easier to code?CS3233 Competitive Programming,

Steven Halim, SoC, NUS

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In my opinion, Kruskals algorithm is simplersor t edges by i ncr easi ng wei ght O( E l og E)while t her e ar e unpr ocessed edges l ef t O( E)

pi ck an edge e wi t h mi ni mum costif addi n e t o MST does not f or m a c cl e

add e t o MST

(EdgeList) and sort them, or use Priority Queue

es or cyc es us ng s o n e s n on n


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onnec anAs this edge is smallest

onnec anNo cycle is formed

e or g na grap ,no edge is selected

124

124

124

0 24

68

0 24

68

0 24

68

369

369

369

4 4 4

the MST formed. Observe that we can also choose toconnect vertex 2 and 0 also with weight 4!

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onnec anMST is formed

onnec an The next smallest edge

anno connec anAs it will form a cycle

124

124

124

0 24

68

0 24

68

0 24

68

369

369

369

444,

determines how the MST formed; Connecting0 and 4 is also a valid next move

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s s e naMST with cost 18

owever, w nomodify anything else

u s an ar rus a salgorithm will still continue

124

124

124

0 260 24

68

0 24

68

36369

369

444

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sor t e y e ge costvect or < pai r > EdgeLi st ;

/ / i nser t edges i n f or mat ( wei ght , ( u, v) ) t o EdgeLi stsor t ( EdgeLi s t . begi n( ) , Edgel i s t . End( ) ;

_

f or ( i nt I = 0; I < E; i ++) { / / whi l e mor e edgespai r f r ont = EdgeLi s t [ i ] ;i f ( ! i sSameSet ( f r ont . second. f i r s t , f r ont . second. second) ) {

/ / i f addi ng e t o MST does not f or m a cycl emst cost += f r ont . f i r s t add t he wei ht of e t o MST _ uni onSet ( f r ont . second. f i r st , f r ont . second. second) ;

}


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You have not teach us Union Find DS in CS3233??

It is

also

only

covered

briefly

in

CS2010/CS2020

Yeah, we choose to skip that DS in CS3233

learn Union Find DS on your own (Sec 2.3.2) ,

mini contests problems A + B


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BFS (unweighted)s ra s non ve cyc e

Bellman Fords (may have ve cycle), not discussedFlo d Warshalls all airs

SHORTEST PATHS


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SSSP is a classical problem in Graph theory:

Find shortest aths from one source to the rest^ Special case: UVa 336 (A Node Too Far)

Problem Description:

Given an un weighted & un directed Graph,a starting vertex v , and an integer TTL

or has distance > TTL from v

. . _ ,


E l (1)0 1 2 3

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Example (1)0 1 2 3

4 765

8 Q = {5}D[5] = 0

9 10 11 12

5

0 1 2 3 E l (2)

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0 1 2 3 Example (2)4 765

8 Q = {5}Q = {1, 6, 10} D[5] = 0

9 10 11 12= =

D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1

1

5 6

10

0 1 2 3 E l (3)

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0 1 2 3 Example (3)4 765

8 Q = {5}Q = {1, 6, 10} D[5] = 0

9 10 11 12= , , ,

Q = {10, 0, 2, 11 }Q = {0, 2, 11, 9}

= =D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1

1 200 2

D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2

= =

5 6

10 119

0 1 2 3 Example (4)

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0 1 2 3 Example (4)4 765

8 Q = {5}Q = {1, 6, 10} D[5] = 0

9 10 11 12= , , ,

Q = {10, 0, 2, 11 }Q = {0, 2, 11, 9}

= =D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1

1 20

, , ,

Q = {11, 9, 4, 3}Q = {9, 4, 3, 12 }=

0 2 3

D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2= =

5 6

4

D[4] = D[0] + 1 = 3D[3] = D[2] + 1 = 3D 12 =D 11 +1 =3

8

D[8] = D[9] + 1 = 3

10 119 12 77

0 1 2 3 Example (5)

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0 1 2 3 Example (5)4 765

8 Q = {5}Q = {1, 6, 10} D[5] = 0

9 10 11 12= , , ,

Q = {10, 0, 2, 11 }Q = {0, 2, 11, 9}

= =D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1

1 20

, , ,

Q = {11, 9, 4, 3}Q = {9, 4, 3, 12 }=

0 2 3

D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2= =

5 6

Q = {3, 12, 8}Q = {12, 8, 7}

= 8 74 7

D[4] = D[0] + 1 = 3D[3] = D[2] + 1 = 3D 12 =D 11 +1 =3

Q = {7}Q = {}8

D[8] = D[9] + 1 = 3D[7] = D[3] + 1 = 4

10 119 12 This is the BFS = SSSP spanningtree when BFS is started from vertex 5

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For SSSP on Wei hted Gra h but without Ne ative Wei ht C cle

DIJKSTRAs


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If the graph is un weighted , we can use BFS

But what if the ra h is wei hted ? UVa 341 (Non Stop Travel)

Solution: Dijkstra

O((V+E)

log

V) A Greedy Algorithm

Use Priority Queue


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3pq = { (0, 2) }

221

7We store this pair of information to thepriority queue: (D[vertex], vertex),sorted by increasing D[vertex], and

06 6then if ties, by vertex number

See that our priority queue is clean at

4

algorithm, it only contains (0, thesource s)

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3pq = {(0, 2)}

21

727

2pq = , , , , ,

We greedily take the vertex in the front of the ueue here it is vertex 2 the source

066

6

and then successfully relax all its neighbors(vertex 0, 1, 3).

4

r or y ueue w or er ese ver ces as1, 0, 3, with shortest path estimate of 2, 6, 7, respectively.

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Vertex 3 appears twice in thepriority queue, but this does

pq = {(0, 2)}3

not matter, as we will takeonly the first (smaller) one

pq = , , , , ,pq = { (5, 3) , (6, 0), (7, 3), (8, 4) }2

172

52

066

6

the queue (now, it is vertex 1), thensuccessfully relax all its neighbors (vertex 3and 4).

4Priority Queue will order the items as

3, 0, 3, 4 with shortest path estimate of , , , , .

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pq = {(0, 2)}3 pq = , , , , ,

pq = {(5, 3), (6, 0), (7, 3), (8, 4)}pq = {(6, 0), (7, 3), (8, 4)}2

172

52

066

6We greedily take the vertex in the front of the queue (now, it is vertex 3), then try torelax all its nei hbors onl vertex 4 .

4

However D[4] is already 8. Since D[3] +w(3, 4) = 5 + 5 is worse than 8, we do not

do anything.

Priority Queue will now have these items0, 3, 4 with shortest path estimate of 6 7 8 res ectivel .

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pq = {(0, 2)}3 pq = , , , , ,

pq = {(5, 3), (6, 0), (7, 3), (8, 4)}pq = {(6, 0), (7, 3), (8, 4)}2

172

52

pq = {(7, 3), (7, 4) , (8, 4)}

06 6 6 We greedily take the vertex in the front of

4

, ,successfully relax all its neighbors (onlyvertex 4).

Priority Queue will now have these items3, 4, 4 with shortest path estimate of 7, 7, 8, respectively.

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Remember that vertex 3appeared twice in the priority

pq = {(0, 2)}3

,algorithm will only consider thefirst (shorter) one

pq = , , , , ,pq = {(5, 3), (6, 0), (7, 3), (8, 4)}pq = {(6, 0), (7, 3), (8, 4)}2

172

52

pq = {(7, 3), (7, 4), (8, 4)}pq = {(7, 4), (8, 4)}

= (8, 4)06 6 6

pq = {}

4Similarly for vertex 4. The one

with shortest path estimate 7will be processed first and theone with shortest pathestimate 8 will be ignored,

anymore

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v st V, I NF ; st s = ; I NF = Bpr i or i t y_queue< i i , vect or , gr eat er > pq;

pq. push( i i ( 0, s) ) ; / / sor t based on i ncr easi ng di st ancewhi l e ( ! pq. empt y( ) ) { / / mai n l oopi i t op = pq. t op( ) ; pq. pop( ) ; / / gr eedyi nt d = t o . f i r st , u = t o . second;

i f ( d == di st [ u] ) {f or ( i nt j = 0; j < ( i nt ) Adj Li st [ u] . si ze( ) ; j ++) {

i f ( di st [ u] + v. second < di st [ v. f i r st ] ) {

di s t [ v. f i r s t ] = di s t [ u] + v. second; / / r el axpq. push( ( d st [ v. f r st ] , v. f r st ) ) ;} / / enqueue t hi s nei ghbor r egar dl ess i t i s

} / / al r eady i n pq or not

}} CS3233 Competitive Programming,Steven Halim, SoC, NUS

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For AllPairs Shortest Paths

FLOYD WARSHALLs


UVa 11463 Commandos 1

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UVa 11463 Commandos 1AlKhawarizmi, Malaysia National Contest 2008

Given: A table that stores the amount of minutes to travel

between buildings (there are at most 100 buildings) 2 special buildings: startB and endB K soldiers to bomb all the K buildings in this mission Each of them start at the same time from startB,

c oose one ui ing B t at as not een om e y ot er soldier (bombing time negligible),

.

What is the minimum time to complete the mission?



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How long do you need to solve this problem?

Solution: The answer is determined by sp fromstarting building, detonate furthest building ,and sp from that furthest building to end building

max(dist[start][i] + dist[i][end]) for all i V

How to compute sp for many pairs of vertices?



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This problem is called: AllPairs Shortest Paths

Two options

to

solve

this: Call SSSP algorithms multiple times

Dijkstra O(V * (V+E) * log V), if E = V2 O(V3 log V) Bellman Ford O(V * V * E), if E = V2 O(V4) Slow to code

Use F oy Wars a , a c ever DP a gorit m O(V3) algorithm

In this problem, V is

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O(V3) since we have three nested loops!

Use adjacency

matrix:

G[ MAX_V] [ MAX_V] ; So that weight of edge(i, j) can be accessed in O(1)

f or ( i nt i = 0; i < V; i ++)f or ( i nt j = 0; j < V; j ++)

G[ i ] [ j ] = mi n( G[ i ] [ j ] , G[ i ] [ k] + G[ k] [ j ] ) ;

ee more exp ana on o s ree ner algorithm in CP


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ree, u er rap , rec e cyc c rap as csDAG is also re visited (next week, Week 06)Bi artite Gra h Week 08

SPECIAL GRAPHS (Part 1)


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4 special graphs frequently appear in contest: Tree, keywords: connected, E = V1, unique path!

, Actually also rare now

Directed Acyclic Graph, keywords: no cycle

Bipartite, keywords: 2 sets, no edges within set!

Currently not in IOI syllabus

Some classical hard problems may have faster solution on these special graphs

Eliminates those who are not aware of the faster solution as solution for

general graph is slower (TLE)or ar er o co e s ower o ge


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TREE


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Tree is a special Graph. It: Connected Has V vertices and exactly E = V 1 edges Has no cycle

Has one

unique

path

between

two

vertices Sometimes, it has one special vertex called root

(rooted tree): Root has no parent

A vertex

in

n

ary

tree

has

either

{0,

1,,n}

children n = 2 is called binary tree (most popular one)

Has many practical applications: Organization Tree, Directories in Operating System, etc


SSSP and APSP Problems

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on Weighted Tree In general weighted graph

SSSP roblem: O V+E lo V Di kstrasor O(VE) Bellman Fords 3

In weighted tree SSSP problem: O(V+E = V+V = V) DFS or BFS

There is only 1 unique path between 2 vertices in tree APSP problem: simple V calls of DFS or BFS: O(V2)

But can be made even faster usin LCA not covered


Diameter

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of a Tree In general weighted graph

We have

to

run

O V

3 Flo d

Warshalls

and

ick

the maximum over all dist[i][j] that is not INF

Do DFS/BFS twice! From any vertex s, find furthest vertex x with DFS/BFS

Then from vertex x, find furthest vertex y with DFS/BFS

Answer is the path length of xy O(V+E = V+V = V) only two calls of DFS/BFS


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Euler Graph


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An Euler path is defined as a path in a graph which visits each edge exactly once

An Euler tour/cycle is an Euler path which starts and ends on the same vertex

A graph which has either Euler path or Euler tour is called Eulerian ra h


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To check whether an undirected graph has an Euler tour is simple Check if all its vertices have even degrees.

Similarly for the Euler path

An undirected graph has an Euler path if all except two vertices have even degrees and at most two vertices have odd degrees. This Euler path will start from one of these odd degree vertices and end in the other

Such degree check can be done in O(V + E), usually done simultaneously when reading the input graph


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Knigsberg UVa 291 Non EulerianNon Eulerian Eulerian

1 1

2 3 2 3 2 3

4 4 5 4 5 6


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DIRECTED ACYCLIC GRAPH (DAG)


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Some algorithms become simpler when used on DAGs instead of general graphs, based on the principle of topological ordering

, paths from a given starting vertex in DAGs in linear time by processing the vertices in a topological order , and calculating the path length for each vertex to be the minimum or maximum length obtained via any of its incoming edges

, slower algorithms such as Dijkstra's algorithm or the Bellman Ford

algorithm , and longest paths in arbitrary graphs are NPhard to find


Sin leSource Shortest Paths

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in DAG

In general weighted graph

A ain this

is

O V+E

lo

V

usin

Di kstrasor O(VE) using Bellman Fords

The fact that there is no cycle simplifies this

pro em su stant a y Simply relax vertices according to topological order!

s ensure s ortest pat s are compute correct y One Topological sort can be found in O(V+E)


Sin leSource Lon est Paths

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in DAG

In general weighted graph

Lon est

sim le

aths is

an

NP

com lete

roblem

In DAG T e so ution is t e same as s ortest pat s in DAG,

just that we have tweak the relax operatoror a ternat ve y, negate a e ge we g t n


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Today, we have quickly gone through various

wellknown

ra h

roblems

&

al orithms Depth First Search and Breadth First Search

Kruskals for MST (briefly) ortest at s pro ems

BFS (unweighted), Dijkstras (standard), Floyd Warshalls a pa rs, t ree ners

Special Graph: Tree, Eulerian, DAG


7/28/2019 Week05 Graph 1

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Note that just knowing these algorithm will not be too useful in contest settin

You have to practice using them t east co e eac o t e a gor t ms scusse

today on a contest problem!


Documents

Week05 Graph 1