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1
Take a single person A out of the 10 people. The remaining 9 people has to be either friends or enemy with A. Therefore acc to pigeon hole principle there atleast⎾9/2⏋(i.e., 5) friends or enemy of A. Lets assume that A has five friends.Now, if at ... are four mutual enemies). similarly we can prove that there are either three mutual enemies or four mutual friends by assuming A has 5 enemies.
posted Apr 24 in Study Materials Joyoshish Saha 1,518 views
3
B is False. If it were a preemptive scheduling then there would have been a transition from Running state to Ready state.
posted Jan 28, 2019 in Databases Ashish Goyal 517 views
5
Yes u are correct it should be D damm my java concepts are broken :(( here is a code class g{ void he(){ System.out.println("g"); } } class h extends g{ void he(){ System.out.println("h"); } } public class Nitin{ public static void main(String args[]) { h obj = new h(); ((g)obj).he(); ((g)new h()).he(); } OUTPUT: h h
posted Jan 2, 2019 in Compiler Design Arjun 1,504 views
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