GATE2010 q1 - Revision history

Arjun Suresh at 11:39, 15 July 2014

2014-07-15T11:39:16Z

2014-07-13T23:04:04Z

2014-07-13T22:50:04Z

2014-06-29T14:35:18Z

2014-06-29T14:34:15Z

2014-06-29T14:33:48Z

2014-06-29T14:27:06Z

2014-06-29T12:16:23Z

2014-04-15T17:33:48Z

2014-04-15T07:11:34Z

← Older revision		Revision as of 11:39, 15 July 2014
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	[[Category: GATE2010]]		[[Category: GATE2010]]
−	[[Category: Graph Theory questions]]	+	[[Category: Graph Theory questions from GATE]]

← Older revision		Revision as of 23:04, 13 July 2014
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	[[Category: GATE2010]]		[[Category: GATE2010]]
	[[Category: Graph Theory questions]]		[[Category: Graph Theory questions]]
−	~~[[Category:Discrete Mathematics questions]]~~

@@ Line 24: / Line 24: @@
 [[Category: GATE2010]]
 [[Category: Graph Theory questions]]
 [[Category:Discrete Mathematics questions]]

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-Let G=(V, E) be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then
+Let $G=(V, E)$ be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then
-(A) |S| = 2|T|
+(A) $|S| = 2|T|$
-(B) |S| = |T| - 1
+(B) $|S| = |T| - 1$
-'''(C) |S| = |T| '''
+'''(C)$ |S| = |T| $'''
-(D) |S| = |T| + 1
+(D) $|S| = |T| + 1$
 ==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
@@ Line 14: / Line 14: @@
 		 <br>
 		 <ul>
-			 <li>'''Base case''' : When $\xi(S) = \xi(T) = 1$, then we have only single node in both trees and hence |S| = |T|.</li>
+			 <li>'''Base case''' : When $\xi(S) = \xi(T) = 1$, then we have only single node in both trees and hence $|S| = |T|$.</li>
-			 <li>'''Induction hypothesis''' : Assume, for $\xi(S) = \xi(T) = k$, we have |S| = |T|.</li>
+			 <li>'''Induction hypothesis''' : Assume, for $\xi(S) = \xi(T) = k$, we have $|S| = |T|$.</li>
-			 <li>'''Induction step''' : For $\xi(S) = \xi(T) = k+1$, we must add a leaf vertex in both trees, and hence |S| = |T|.</li>
+			 <li>'''Induction step''' : For $\xi(S) = \xi(T) = k+1$, we must add a leaf vertex in both trees, and hence $|S| = |T|$.</li>
 		 </ul>
-		 So |S| = |T|. So option (C) is correct.
+		 So $|S| = |T|$. So option (C) is correct.

@@ Line 1: / Line 1: @@
 Let G=(V, E) be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then
-(A) <|S| = 2|T|
+(A) |S| = 2|T|
 (B) |S| = |T| - 1

@@ Line 27: / Line 27: @@
 [[Category: GATE2010]]
 [[Category: Graph Theory questions]]
-[[Category:Mathematics Questions from GATE]]
+[[Category:Mathematics Questions]]

← Older revision		Revision as of 12:16, 29 June 2014
Line 27:		Line 27:
	[[Category: GATE2010]]		[[Category: GATE2010]]
	[[Category: Graph Theory questions]]		[[Category: Graph Theory questions]]
		+	[[Category:Mathematics Questions from GATE]]

@@ Line 26: / Line 26: @@
 [[Category: Graph Theory]]
 [[Category: GATE2010]]
-[[Category: Discrete Mathematics questions]]
+[[Category: Graph Theory questions]]