Answer / guest

(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.

Answer / harsh

N-1

Answer / m.n.prakash

N-1
EX:
take N=4 then.....
N1 call to N2,N3,N4.so N1 can know information from those
three.then N2 call N3,N4 and know info from both.finally
N3 call to N4 and know the info from N4.so totally need the
only 4 calls to know each others.
so i can that only they need 4 phone calls.

Answer / max

The minimum number of calls required are: (n-1)+(n-2) = 2n-3

First agent needs to call n-1 agents to get all the
information. At the end of his last call i.e. (n-1)'th call,
the first and the n'th agent know all the information.
Now, if the first agent calls the remaining (n-2) agents,
all the information is shared between all agents.
It would not matter if the number of agents were even or odd.

Answer / manish kumar verma

a--->b = a calls to b
let say n peoples are {n1,n2,n3....nN}

step 1:n1--->n2 , n3-->n4 , n5-->n6....
step 2:n1-->n3, n5-->n7, n9-->n11 ..
step 3:n1-->n5, n9-->n13...
.
.
.
so on.
So in total minimum n-1 calls in both n even or odd..

Answer / fayaz

ANS:2(N-1)

Answer / abhinay

if we suppose n=2
then minimum 1 call is needed to share the message to each
other.
if n=3 then 2+1 call
if n=4 then 3+2+1 call
similarly for N=n then (n-1)+(n-2)+(n-3)+.........+1 call
needed to share the message to each other

Answer / guest

N(N-1)

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