wu :: forums
« wu :: forums - Drug Traffickers »

Welcome, Guest. Please Login or Register.
Jun 2nd, 2024, 3:14am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   medium
(Moderators: Eigenray, SMQ, Grimbal, ThudnBlunder, towr, william wu, Icarus)
   Drug Traffickers
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Drug Traffickers  (Read 853 times)
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Drug Traffickers  
« on: Sep 9th, 2011, 4:36pm »
Quote Quote Modify Modify

The police have 10 suspected drug traffickers and 5 untrustworthy witnesses. For each suspect the witnesses stated whether or not they saw them with drugs. The statements can be summarized as follows:
 
SUSPECT    HAD DRUGS   NO DRUGS
     1                 5                  0  
     2                 0                  5                                                                                  
     3                 2                  3
     4                 5                  0
     5                 4                  1
     6                 0                  5
     7                 3                  2
     8                 5                  0
     9                 0                  5
    10                1                  4
 
It is known that the total number of lies is either 8 or 9 and that most of these claim 'no drugs' when the truth is 'had drugs'. Which suspects are guilty of trafficking?  
 
« Last Edit: Sep 10th, 2011, 3:17am by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
aicoped
Junior Member
**





   


Gender: male
Posts: 57
Re: Drug Traffickers  
« Reply #1 on: Sep 11th, 2011, 8:15pm »
Quote Quote Modify Modify

would you please explain exactly what each is saying. is suspect 1 saying person 5 had drugs and that he had no drugs?
IP Logged
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
Posts: 13730
Re: Drug Traffickers  
« Reply #2 on: Sep 11th, 2011, 10:02pm »
Quote Quote Modify Modify

on Sep 11th, 2011, 8:15pm, aicoped wrote:
would you please explain exactly what each is saying. is suspect 1 saying person 5 had drugs and that he had no drugs?
No, the five witnesses are all saying suspect 1 had drugs. For each suspect the witnesses make a claim about whether (s)he did or didn't have drugs on them, which is why the last two columns always add to 5.
IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Noke Lieu
Uberpuzzler
*****



pen... paper... let's go! (and bit of plastic)

   
WWW

Gender: male
Posts: 1884
Re: Drug Traffickers  
« Reply #3 on: Sep 11th, 2011, 11:41pm »
Quote Quote Modify Modify

Huh
 
The symmetry keeps pushing me back to square one.
 
If- for example all 5  lied for 6,  the 3 lied for 3 and the 1 lied for 10- that fits the information. That's 9 lies.
 
But equally, it could have been liars for 3,5,6, totalling 8 lies... or...
 
 
Ah, so then it goes into a string of statements such as IF suspect 2 DOES have drugs, then 10 does... Okay....
« Last Edit: Sep 11th, 2011, 11:45pm by Noke Lieu » IP Logged

a shade of wit and the art of farce.
Grimbal
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 7527
Re: Drug Traffickers  
« Reply #4 on: Sep 12th, 2011, 2:16am »
Quote Quote Modify Modify

Hmm.  Does it count as a lie if a witness said "didn't see him with drugs", which is true, even though the suspect was dealing with drugs?
 
If it doesn't count, they could all be guilty, except that most witnesses didn't see them.
 
on Sep 11th, 2011, 11:41pm, Noke Lieu wrote:
Huh
If- for example all 5  lied for 6,  the 3 lied for 3 and the 1 lied for 10- that fits the information. That's 9 lies.

There must have been lies for 5 and 7 also.
 
 
IP Logged
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: Drug Traffickers  
« Reply #5 on: Sep 12th, 2011, 2:46am »
Quote Quote Modify Modify

on Sep 12th, 2011, 2:16am, Grimbal wrote:
Hmm.  Does it count as a lie if a witness said "didn't see him with drugs", which is true, even though the suspect was dealing with drugs?

If a suspect was not seen with drugs we will assume (s)he was not dealing in drugs.
 
(However, I know of a kangaroo court that is looking for a good programmer.)  Tongue
 
IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
Grimbal
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 7527
Re: Drug Traffickers  
« Reply #6 on: Sep 12th, 2011, 3:14am »
Quote Quote Modify Modify

So there is at least 1 lie about #5 and #10 each and 2 lies about #3 and #7.
 
If the verdict follows the majority, we have 6 lies.  We need to adjust it to add 2 or 3 more lies.
 
For suspects with 0 or 5 "votes", changing the verdict adds 5 extra lies, that is too many.
 
Changing the verdict for #5 or #10 adds 3 lies.  Changing #7 or #3 adds 1 lie each.
To reach the "lie budget", we need to change #5, #10, or #7 and #3.
Changing #5 adds 3 lies of the inculpatory type.  That makes more inculpatory than disculpatory lies.  That cannot be.
Changing #10 adds 3 lies for a total of 9.  7 disculpatory and 2 inculpatory.  That works.
Changing #7 and #3 adds 2 lies, but this results in 4 inculpatory and 4 disculpatory lies.  Not possible.
 
So, the verdict is: #1,#4,#5,#7,#8,#10 guilty.
« Last Edit: Sep 12th, 2011, 4:39am by Grimbal » IP Logged
Hippo
Uberpuzzler
*****





   


Gender: male
Posts: 919
Re: Drug Traffickers  
« Reply #7 on: Sep 12th, 2011, 6:39am »
Quote Quote Modify Modify

on Sep 12th, 2011, 3:14am, Grimbal wrote:

So, the verdict is: #1,#4,#5,#7,#8,#10 guilty.

I agree Smiley. 6 forced lies so need for 2 or 3 more. So 5/0 resp 0/5 should remain unchanged. We have 1/4,2/3,3/2 and 4/1 cases to make 2 ro 3 difference by  swapping. 2/3 and 3/2 swap adds 1, 1/4 and 4/1 swap adds 3. So one option is both 2/3 and 3/2 swaps, but it leads  to 4+ and 4- contradiction.
 
The other option is to swap either 1/4 or 4/1. Just one is compatible with the "most" condition.
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board