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        riddles >> medium >> Drug Traffickers
        
(Message started by: ThudnBlunder on Sep 9th, 2011, 4:36pm)
Title: Drug Traffickers
Post by ThudnBlunder on Sep 9th, 2011, 4:36pm
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?  
Title: Re: Drug Traffickers
Post by aicoped on Sep 11th, 2011, 8:15pm
would you please explain exactly what each is saying. is suspect 1 saying person 5 had drugs and that he had no drugs?
Title: Re: Drug Traffickers
Post by towr on Sep 11th, 2011, 10:02pm

on 09/11/11 at 20:15:46, 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.
Title: Re: Drug Traffickers
Post by Noke Lieu on Sep 11th, 2011, 11:41pm
???

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....
Title: Re: Drug Traffickers
Post by Grimbal on Sep 12th, 2011, 2:16am
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 09/11/11 at 23:41:52, Noke Lieu wrote:
???
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.
Title: Re: Drug Traffickers
Post by ThudnBlunder on Sep 12th, 2011, 2:46am

on 09/12/11 at 02:16:39, 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.)  :P
Title: Re: Drug Traffickers
Post by Grimbal on Sep 12th, 2011, 3:14am
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.
Title: Re: Drug Traffickers
Post by Hippo on Sep 12th, 2011, 6:39am

on 09/12/11 at 03:14:27, Grimbal wrote:
So, the verdict is: #1,#4,#5,#7,#8,#10 guilty.

I agree :). 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.


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