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Trees and graphs

The document discusses various concepts related to trees and graphs including: - Definitions of trees, rooted trees, subtrees, and tree traversal methods like preorder, inorder and postorder searches. - Minimum spanning trees, algorithms to find them like Prim's and Kruskal's, and definitions of spanning trees. - Definitions of graphs, types of graphs, Euler's formula, planar graphs, Hamiltonian graphs, and graph isomorphism.

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TREES AND GRAPHS
TREES: LET A RELATION T ON SET A={V0,V1,V2,…VN} IS SAID TO BE A TREE , IF THERE IS AN UNIQUE PATH FROM V0 TO VK V K=1,2…N BUT NO PATH FROM V0 TO VO. THE VERTEX V0 IS CALLED ROOT OF THE TREE AND DUE TO THIS REASON SOMETIMES THE TREE IS CALLED ROOTED TREE. HERE THE VERTEX V0 IS DIFFERENT FROM THE OTHER VERTICES. A TREE IS IRREFLEXIVE ,NOT SYMMETRIC AND NOT TRANSITIVE.
THEOREM 1: LET (T,V0) BE A ROOTED TREE. THEN (A)THERE ARE NO CYCLES IN T. (B)V0 IS THE ONLY ROOT OF T. (C)EACH VERTEX IN T, OTHER THAN V0,HAS AN IN-DEGREE ZERO.
PATH LENGTH: THE NUMBER OF EDGES BETWEEN THE VERTEX AND THE ROOT IS CALLED PATH LENGTH BETWEEN THE VERTEX AND ROOT. SUBTREE: THE SUBSET OF TREE T WHICH IS ITSELF A TREE IS CALLED SUBTREE OF T. THE FOLLOWING IS AN EXAMPLE OF THESE CONCEPTS.
Tree of algebraic expressions
TO REPRESENT A BINARY TREE WE REQUIRED THE EXTENDED VERSION OF THIS LINKED LIST CALLED DOUBLY LINKED LIST. FOR EXAMPLE: Q1. CONSTRUCT A DIGRAPH OF THIS TREE WITH EACH VERTEX LABELLED AS INDICATE. DOUBLY LINKED LIST
SOLUTION:INDEX LEFT DATA RIGHT 1 9 - 0 2 10 M 7 3 0 Q 0 4 8 T 0 5 3 V 4 6 0 X 2 7 0 K 0 8 0 D 0 9 6 G 5 10 0 C 0 DOUBLY LINKED LIST
Q2. CONSTRUCT THE TREE OF THE FOLLOWING ALGEBRAIC EXPRESSIONS. (𝑋 + 𝑌 − 𝑋 + 𝑌 ) × ((3 ÷ 2 × 7 ) × 4) SOLUTION: DOUBLY LINKED LIST
THREE DIFFERENT METHODS OF SEARCHING A TREE: 1) PREORDER SEARCH OR VLR: 2) POSTORDER SEARCH OR LRV: 3) INORDER SEARCH OR LVR: FOR EXAMPLE: Q1. PERFORM PREORDER,POSTORDER AND INORDER SEARCH FOR THE FOLLOWING TREES. TREE SEARCHING (TRAVERSING):
SOLUTION :- PREORDER(VLR): 1 2 3 4 8 9 10 5 11 12 6 7 POSTORDER(LRV): 4 3 9 10 8 2 12 11 7 6 5 1 INORDER(LVR): 4 3 2 9 8 10 1 11 12 5 6 7 TREE SEARCHING (TRAVERSING):
A SUBGRAPH T OF A CONNECTED GRAPH OF RELATION R IS CALLED A SPANNING TREE OF THE GRAPH IF T IS A TREE AND T INCLUDES ALL VERTICES OF THE GRAPH. A MINIMUM SPANNING TREE OF A WEIGHTED GRAPH IS A SPANNING TREE WHOSE TOTAL WEIGHT IS AS SMALL AS POSSIBLE. SPANNING TREE: MINIMUMSPANNING TREE:
ALGORITHMS TO FIND MINIMAL SPANNING TREES:  PRIM’S ALGORITHM: STEP1 : CHOOSE ANY ONE VERTEX OF THE GRAPH. STEP2:CHOOSE NEAREST MINIMUM WEIGHT . THEN CHOOSE NEAREST EDGES HAVING MINIMUM WEIGHT. STEP3: REPEAT STEP2 UNTIL N-1 EDGES. THEN THIS MINIMAL SPANNING TREE CONTAINS N VERTICES AND N-1 EDGES. STEP4: END. Q1. FIND MINIMAL SPANNING TREE OF THE WEIGHTED GRAPH. USING PRIM’S ALGORITHM. SOLUTION: MINIMUMSPANNING TREE:
 KRUSKAL’S ALGORITHM: STEP1 : CHOOSE AN EDGE WITH LEAST WEIGHT. STEP2: SELECT THE NEXT EDGE WITH MINIMUM WEIGHT. ADD THE EDGE WHICH DOES NOT FROM A CYCLE. STEP3: REPEAT STEP2 UNTIL N-1 EDGES. THEN THIS MINIMAL SPANNING TREE CONTAINS N VERTICES AND N-1 EDGES. STEP4: END. Q1. FIND MINIMAL SPANNING TREE OF THE WEIGHTED GRAPH. USING KRUSKAL’S ALGORITHM SOLUTION: MINIMUMSPANNING TREE:
GRAPH A GRAPH IS A REPRESENTATION OF A SET OF OBJECTS WHERE SOME PAIRS OF OBJECTS ARE CONNECTED BY LINKS. A GRAPH G = (V, E) COMPRISING A SET V OF VERTICES OR NODES TOGETHER WITH A SET E OF EDGES OR LINES, WHICH ARE 2-ELEMENT SUBSETS OF V
CONSIDER THE FOLLOWING GRAPH G=(V,E) & ANSWER THE QUESTION 1- FIND DEGREE OF EACH VERTEX. 2-VERIFY THE SUM OF DEGREES & NUMBER OF EDGES IN G. SOL:- 1- DEG(A)=2; DEG(B)=2; DEG(C)=2; DEG(D)=1; DEG(E)=1 2- SUM OF DEGREES = DEG(A)+DEG(B)+DEG(C)+DEG(D)+DEG(E) = 2+2+2+1+1 = 8 TOTAL EDGES = ½ (SUM OF DEGREES) = ½ (8) = 4 EXAMPLE 1.1
TYPES OF GRAPHS CONNECTED GRAPH DISCONNECTED GRAPH DISCRETE GRAPH COMPLETE GRAPH REGULAR GRAPH
EULAR’S FORMULA |V|+|R|-|E| = 2 |V|: NO. OF VERTICE = 7 |E|: NO. OF EDGES = 11 |R|: SET OF REGIONS = 6
EULAR GRAPH EXAMPLE 2.1 CONSIDER THE FOLLOWING GRAPH & DETERMINE WHETHER IT HAS AN EULAR CIRCUIT, AN EULAR PATH BUT NOT EULER CIRCUIT OR NEITHER. SOL:- EACH VERTEX OF THE GIVEN GRAPH HAVE EVEN DEGREE. THEREFORE, EULER CIRCUIT EXISTS, WHICH ARE P,R,S,T,R,Q,P OR T,R,Q,P,R,S,T p q r st
PLANER GRAPH DEFINATION:- A PLANER GRAPH IS A GRAPH WHOSE VERTICES AND EDGES CAN BE DRAWN IN A PLANE SUCH THAT NO TWO OF THE EDGES INTERSECT (I.E., EMBEDDED IN A PLANE).
Example 3 Q- IN A CONNECTED PLANER GRAPH THERE ARE 20 VERTICES, EACH OF DEGREE 3. FIND THE NO. OF REGIONS IN THE GRAPH. SOL:- HERE |V|=20 |E|= ½ (SUM OF DEGREES OF ALL VERTICES) = ½ (3 * 20) = 30 BY EULER’S FORMAULA |V|+|E|-|R|= 2 ∴ 20 +|R|- 30 = 0 ∴ |R| = 12 THERE ARE 12 REGIONS IN THE GRAPH
HAMILTONIAN GRAPH HAMILTONIAN GRAPH: A GRAPH IS SAID TO BE HAMILTON IF IT’S VERTEX IS VISITED ONLY ONCE EXCEPT FOR THE FIRST VERTEX. N ≥ 𝟑 & 𝒏 𝟐 ≤ D E G ( V )
ISOMORPHISM OF GRAPHS DEFINATION :- IN GRAPH THEORY ISOMORPHISM OF GRAPHS G & H IS BIJECTION BETWEEN THE VERTEX SETS OF G & H. f : V(G) V(H) Graph G Graph H An isomorphism between G and H f(a) = 1 f(b) = 6 f(c) = 8 f(d) = 3 f(g) = 5 f(h) = 2 f(i) = 4 f(j) = 7
EXAMPLE 4 Q – SHOW THAT THE GRAPHS G1 & G2 ARE ISOMORPHIC SOL :- HERE V1 = {A,B,C,D} & V2 = {1,2,3,4} E1 = {{A,B},{B,C},{C,D}} & E2 = {{1,2},{2,3},{3,4}} DEFINE A FUNCTION AS F :V1 V2 AS F(A) = 1, F(B) = 2, F(C) = 3 & F(D) = 4 F IS 1-1 & ONTO. HENCE, F ‘ :V’1 V’2 IS ISOMORPHISM. ∴ G1 & G2 GRAPHS ARE ISOMORPHIC. G1 G2