tag:blogger.com,1999:blog-3802979373213475425.post1002406027507939398..comments2019-04-20T15:33:02.879+05:30Comments on Critical Thinking Puzzles: Secret agentsUnknownnoreply@blogger.comBlogger20125tag:blogger.com,1999:blog-3802979373213475425.post-22283008654601124862014-11-25T21:47:11.620+05:302014-11-25T21:47:11.620+05:30should be (n-1)!should be (n-1)!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-11931978517410066492011-02-22T14:55:49.027+05:302011-02-22T14:55:49.027+05:30This comment has been removed by the author.Affan Abdul Kadarhttps://www.blogger.com/profile/03283493782274166388noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-9163091765284730892010-06-22T16:50:40.252+05:302010-06-22T16:50:40.252+05:30"Anonymous" of April 18 is correct.
4 a..."Anonymous" of April 18 is correct.<br /><br />4 agents clearly only require 4 conversations, not 6:<br /><br /><b>1)</b> 1<>2 (1 knows 1,2 | 2 knows 1,2)<br /><b>2)</b> 3<>4 (3 knows 3,4 | 4 knows 3,4)<br /><b>3)</b> 1<>3 (1 knows 1,2,3,4 | 3 knows 1,2,3,4)<br /><b>4)</b> 2<>4 (2 knows 1,2,3,4 | 4 knows 1,2,3,4)<br /><br />Thereafter, each time you add an agent to the previous group of agents, two extra conversations are required - one at the start between the last added agent and a member of the previously established group, and one at the end between the last added agent and the group.<br /><br />Therefore 4 agents require 4 conversations, 5 req 6, 6 req 8, 7 req 10, etc.<br /><br />So, for 4 or more agents, 2n-4 conversations will suffice.Secret Squïrrelnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-89379819075432262252010-06-08T14:43:30.886+05:302010-06-08T14:43:30.886+05:30Solution:
a
1 Agent = 0 Calls
a b
a = b
2 Ag...Solution:<br /><br />a <br />1 Agent = 0 Calls<br /><br />a b <br />a = b<br />2 Agent= 1 Calls<br /><br />a b c <br />a = b c<br />b = c<br />3 Agent = 3 Calls<br /><br />a b c d <br />a = b c d<br />b = c d<br />c = d<br />4 Agent = 6 Calls<br /><br />a b c d e <br />a = b c d e<br />b = c d e<br />c = d e<br />d = e<br />5 Agent = 10 Calls<br /><br />a b c d e f <br />a = b c d e f<br />b = c d e f<br />c = d e f<br />d = e f<br />e = f<br />6 Agent = 15 Calls<br /><br /><br /><br />((n^2 + n) / 2 ) - n = No. of callsAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-71271961206086532512010-06-05T14:32:41.192+05:302010-06-05T14:32:41.192+05:302n-32n-3Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-75832894797940641882010-05-27T15:53:17.370+05:302010-05-27T15:53:17.370+05:30feddas it doesnt say that there are 5 people it sa...feddas it doesnt say that there are 5 people it says n... <br /><br />if its a,b,c,d,e that means that the amount of secret agents are given in the question but its notKevinhttps://www.blogger.com/profile/13957483128423077472noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-70591733085501408062010-05-23T07:37:12.282+05:302010-05-23T07:37:12.282+05:30i think the answer would be..
2n-3..
for minimum n...i think the answer would be..<br />2n-3..<br />for minimum nos of calls all of the agents must call to a particular agent so total nos of calls to that agent will be (n-1)..nd now this agent nd the agent calling last has full information...now the other left out agents (n-2) must again phone to him nd gather the full information....so<br />total nos of calls= (n-1)+(n-2)=2n-3harendrahttps://www.blogger.com/profile/15601985027454792675noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-33054419931480635662010-04-18T12:32:35.874+05:302010-04-18T12:32:35.874+05:30To start with, let us say how many calls does one ...To start with, let us say how many calls does one take in case of 4 folks::<br /><br />1 2 3 4<br />=== ===<br />Group1 Group2<br />Now 1 and 2 talk, 3 and 4 talk<br />As such, <br /><br />1-> knows (1 and 2) <br />2 -> 12<br /><br />3-> 34<br />4-> 34<br /><br />Now say 1 from Group1 and 4 from Group2 talk.<br /><br />Now the case is <br /><br />1> 12 +34 = 1234<br />4-> 12 + 34 = 1234<br /><br />Now let 2 and 3 talk,<br /><br />After this again<br /><br />3-> 34 + 12 = 1234<br />4-> 34 + 12 = 1234<br /><br />So in a group of 4 only 4 converstaions need to be done to let everybody know of the other 3 folks' secret.<br /><br />Now take the case of 5, <br /><br />1234 5<br />Group1 Group2<br /><br />Suppose 4 talks to 5 and gets his secret. Now let the group of 4 ( Group 1) talk among themselves.So 4 convs need to be made to let 1,2,3,4 know -> 1234 + 5'secret. <br /><br />Now 4 goes and talks to 5 . Now everybody knows everybody's secret.<br /><br />Total convos = 1 + 4 + 1 = 6<br /><br />This can be best visualised like this ::<br /><br /> ------<br />| 1234 | --> 5 <br /> <--<br /> ------<br />4 convos + 2 convos<br /><br />For 6 <br /><br /> ------<br />| 1234 | --> 5 --> 6 <br /> <-- <--<br /> ------<br /><br />4 + 2 + 2 <br /><br />In case of 6, let 6 talk to 5, then 5 talks to 4 , now let the group of 4 talk to each other. After these 6 calls, 1234 know - > 123456 . After that 4 ( from 1234) talks to 5 ( here 5 knows -> 123456 ). <br />And then 5 talks to 6 (or 4 to 6) (after this 6 too knows 123456 )<br /><br /><br />Hence it is like<br /><br />4 - > 4 == ( 2*4 - 4) <br />5 -> 6 = 2*5 -4 <br />6-> 8 = 2*6 -4<br /><br />The magic is 4 . For 3 , it takes 3 convos ( not 2 ) for each to know of the others' secret. ( 2*3 - 3) <br />For 2 -> 1 ( 2*2 - 3 )<br /><br />so for 2 and 3, it is (2n -3)<br /><br />And for any no >=4 it is (2n -4) <br /><br />The grouping of 4 is the key.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-62902946002068595172010-04-06T11:52:00.799+05:302010-04-06T11:52:00.799+05:30the answer is (n!)the answer is (n!)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-12701624071726943682010-04-06T08:07:37.926+05:302010-04-06T08:07:37.926+05:30Trinadh, what would the calls look like for n = 5?...Trinadh, what would the calls look like for n = 5? Can you find a way where than make no more than n-1 calls?<br /><br />I need at least 7.<br />(a,b) (c,d) (a,e) (c,e) (a,d) (b,c) (e,anyone)feddashttps://www.blogger.com/profile/17863547444259520885noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-77438278621683460342010-03-21T11:10:20.017+05:302010-03-21T11:10:20.017+05:30The worst case of the calls to connect each other ...The worst case of the calls to connect each other is n-1.<br /><br />All of the persons except lastone should recieve calls twice to know all secrets.<br /><br />So Total calls = 2(n-1) - 1 = 2n-3.<br /><br />Also the sharing of the secret can be minimised for the first 4 persons.<br /><br />To share a secret for 4 persons (say a,b,c,d), only 4 calls are sufficient.<br />(a,b) (c,d) (a,c) (b,d).<br /><br />2(4) - 3 = 5.<br />(actually 4 calls are sufficent for first 4 persons)<br /><br />One call can be optimized there.<br /><br />So total calls = (2n-3) - 1 = 2n-4.Trinadhnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-25288319983755364722010-03-18T16:12:22.173+05:302010-03-18T16:12:22.173+05:30to Audrey,
it should be 2n-1, because the last per...to Audrey,<br />it should be 2n-1, because the last person who called the first person will already share everything.. hence the first person need not call the last person again..Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-31958787416803901882010-03-18T12:55:31.742+05:302010-03-18T12:55:31.742+05:30the answer is n.the answer is n.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-80279037606122365692010-03-17T18:44:16.727+05:302010-03-17T18:44:16.727+05:30i think 2n-2 because everyone calls one person and...i think 2n-2 because everyone calls one person and shares their info (n-1) phone calls, then that person calls everyone back with all the info (n-1) phone calls, so (n-1)+(n-1)= 2n-2. I don't know if there's a more efficient way, but I'm not sure how everyone can have all the info with just n-1 phone callsAudreyhttps://www.blogger.com/profile/07540585455840869844noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-57091780879372265622010-03-17T12:29:54.633+05:302010-03-17T12:29:54.633+05:30n-1 is the answern-1 is the answerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-1036061967824631492010-03-16T19:58:09.392+05:302010-03-16T19:58:09.392+05:30My answer is a series...
mod(x,y) is the remainder...My answer is a series...<br />mod(x,y) is the remainder of the division x/y<br /><br />at the end of the equation if s(1) equals 0 and s(2) equals 1...André Meneseshttps://www.blogger.com/profile/03406954406655867824noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-68279504596842595152010-03-16T15:06:38.465+05:302010-03-16T15:06:38.465+05:30i dun really understand what André Meneses's e...i dun really understand what André Meneses's equation is..<br />but i think answer is 2n-1...?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-18779143930411106042010-03-15T01:31:49.985+05:302010-03-15T01:31:49.985+05:30It will be N-1 as every person has to be contacted...It will be N-1 as every person has to be contacted atleast once...Rahulhttps://www.blogger.com/profile/01614454450922936436noreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-21353802874170866682010-03-10T03:45:11.152+05:302010-03-10T03:45:11.152+05:30Well... that's quiet interessting but frankly ...Well... that's quiet interessting but frankly i have a hard time figuring it... wonder how others think about this..Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3802979373213475425.post-21699312913116166482010-03-09T18:48:29.186+05:302010-03-09T18:48:29.186+05:30S(1) = 0
S(2) = 1
Solution:
(2*S((N-mod(N,2))/2)...S(1) = 0<br />S(2) = 1<br /><br />Solution:<br /><br />(2*S((N-mod(N,2))/2))+((N-mod(N,2))/2)+((N-mod(N,2))*2)André Meneseshttps://www.blogger.com/profile/03406954406655867824noreply@blogger.com