Question of the Day | Logical Reasoning #20
Question of the Day
Logical Reasoning #20
There are N secret agents each know a different piece of secret information. They can telephone each other and exchange all the information they know. After the telephone call, they both know anything that either of them knew before the call.
What are the minimum number of telephone calls needed so that all of the them know everything?
(2N – 3) telephone calls, for N = 2,3
(2N – 4) telephone calls, for N > 3
Divide the N secret agents into two groups. If N is odd, one group will contain one extra agent. Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and so on.
Similarly in second group, agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. After (N – 2) calls, two agents in each the group will know anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and Z1 & Z2 from group 2. Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4 agents will know everything.
Now (N – 4) telephone calls are reqiured for remaining (N – 4) secret agents.
Total telephone calls require are = (N – 2) + 2 + (N – 4)
= 2N – 4
Let\’s take an example. Say there are 4 secret agents W, X, Y & Z. Divide them into two groups of 2 each i.e. (W, X) and (Y, Z). Here, 4 telephone calls are required.
1. W will call up X.
2. Y will call up Z.
3. W, who knows WX will call up Y, who knows YZ.
4. X, who knows WX will call up Z, who knows YZ.
Take an another example. Say there are 5 secret agents J, K, L, M & N. Divide them into two
groups i.e. (J, K) and (L, M, N). Here, 6 telephone calls are required.
1. J will call up K.
2. L will call up M.
3. M will call up N. Now M and N know LMN.
4. J, who knows JK will call up M, who knows LMN.
5. K, who knows JK will call up N, who knows LMN.
6. L will call up to anyone of four.