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Topic: Unreliable Witnesses (Read 1724 times) |
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ThudnBlunder
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Unreliable Witnesses
« on: Oct 16th, 2004, 11:21pm » |
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The Drugs Squad are prosecuting 10 suspected drugs dealers and have 5 witnesses whom they don't completely trust.
For each suspect, the 5 witnesses are asked whether or not they saw the suspect sell drugs.
SUSPECT SOLD DIDN'T SELL
DRUGS 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
A) What is the fewest number of lies that can possibly have been told?
B) Given only that the total number of lies told by all of the witnesses is exactly 9, and that most of the lies claim 'sold no drugs' when the truth is 'sold drugs', which suspects actually sold drugs?
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towr
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Re: Unreliable Witnesses
« Reply #1 on: Oct 17th, 2004, 8:16am » |
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The answer to A is 0. Not every witness needs to have seen everything (which is what was asked).
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ThudnBlunder
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Re: Unreliable Witnesses
« Reply #2 on: Oct 17th, 2004, 9:40am » |
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on Oct 17th, 2004, 8:16am, towr wrote:| The answer to A is 0. Not every witness needs to have seen everything (which is what was asked). |
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A) asks the for the minimum possible number of lies that were told in total about all of the suspects by all of the witnesses.
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TenaliRaman
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Re: Unreliable Witnesses
« Reply #3 on: Oct 17th, 2004, 11:38am » |
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A> 6
B> Suspects 1,5,7,8,10 sold drugs
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ThudnBlunder
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Re: Unreliable Witnesses
« Reply #4 on: Oct 17th, 2004, 11:57am » |
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Quote:
:Nope, A [smiley=ngtr.gif] 6 
Quote:| B> Suspects 1,5,7,8,10 sold drugs :: |
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:So they were all lying about Suspect 4?
Method?
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| « Last Edit: Oct 17th, 2004, 1:19pm by ThudnBlunder » |
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TenaliRaman
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Re: Unreliable Witnesses
« Reply #5 on: Oct 17th, 2004, 1:42pm » |
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Oops yes i walked right over 4 ....
::
As for the method,
i made 2 assumptions
1> every witness knows everything abt every suspect
2> at any instant all the witnesses don't lie
the implications of these two assumptions are direct ...
any row that has (sold,not sold) = (5,0) or (0,5) is out of suspicion.
All other rows were under suspicion
for a, i just added the mins of (sold,not sold)
for b, i first looked at the not sold column
and added all those values which were <5
that adds to 10
but we are looking for 9
a bit of trial and error as to what combo of sold/not sold gives 9 leads to the solution ....
::
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ThudnBlunder
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Re: Unreliable Witnesses
« Reply #6 on: Oct 17th, 2004, 6:24pm » |
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Quote:| a bit of trial and error as to what combo of sold/not sold gives 9 leads to the solution .... :: |
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Trial and error? But are you sure that is an appropriate method to use in a criminal case like this?
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| « Last Edit: Oct 18th, 2004, 5:22am by ThudnBlunder » |
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towr
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Re: Unreliable Witnesses
« Reply #7 on: Oct 18th, 2004, 12:24am » |
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on Oct 17th, 2004, 9:40am, THUDandBLUNDER wrote:
A) asks the for the minimum possible number of lies that were told in total about all of the suspects by all of the witnesses. |
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I don't know what you call a lie, but if someone didn't see something, and he says he didn't see it, I wouldn't call that a lie..
If you want the minimum possible number of mistaken inferences about who sold drugs, that's 6, the sum of the minimum(pro, con) of for each suspect. But none of those accounts need to be lies.
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| « Last Edit: Oct 18th, 2004, 12:38am by towr » |
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ThudnBlunder
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Re: Unreliable Witnesses
« Reply #8 on: Oct 18th, 2004, 4:33am » |
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Quote:| don't know what you call a lie, but if someone didn't see something, and he says he didn't see it, I wouldn't call that a lie.. |
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I suppose you mean that, just as seeing is believing, not seeing is not believing. 
But, if one assumes your interpretation of the problem,
A) is trivial
and
B) is insoluble
Next puzzle, please!
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| « Last Edit: Oct 18th, 2004, 6:32am by ThudnBlunder » |
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Three Hands
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Re: Unreliable Witnesses
« Reply #9 on: Oct 18th, 2004, 5:54am » |
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Well, for A, either Towr is correct in his statement, or (assuming that all witnesses have seen all the suspects a sufficient number of times to have equivalent to full information) the fewest number of lies would be 6.
For B, all I could say is that, given that most of the lies told were saying "Didn't Sell" when it should be "Sold", there is at least one witness for Suspects 1, 4 and 8 - presumably sufficient evidence for the Drugs Squad. For all the others, they are possible sellers, but I can't be bothered to work out all the different possibilities now.
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Keith H
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Well, nobody has provided any explanation for this old problem, so I shall. The answer is already in the clear on this thread so I won't hide it.
Suspects 3,5,7, and 10 are under consideration. Suspect 3 contributes either 2 or 3 lies, which I'll rewrite as contributing {2,3} lies (order is insignificant in this notation). Suspect 5 contributes {1,4} lies. Suspect 7 contributes {2,3} lies, and suspect 10 contributes {1,4} lies. Thus, if we just look at suspects 3 and 5 (call this group A), they contribute a total of {3,4,6,7} lies ({2+1,3+1,2+4,3+4}). Similarly suspects 7 and 10 (group B) contribute {3,4,6,7} lies.
We know that all four suspects together contribute 9 lies, and that only happens in two situations: group A contributes 6 lies and group B contributes 3, or A contributes 3 and B contributes 6.
Examining both cases (trace back to where the 6's and 3's came from), we see that in the first case, the breakdown of lies is 7 false positives (regarding suspects 3, 5, and 10) and 2 false negatives (regarding suspect 7). That contradicts the assumption that false negatives were more common. A quick symmetric check shows that in the second case, 7 false negatives and 2 false positives occur.
Thus the only solution is when group A contributes 3 lies (3 is not guilty, 5 is guilty) and both members of group B are guilty (for 6 group B lies). In other words, of the four suspects under consideration, 5, 7, and 10 are guilty.
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Deedlit
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Re: Unreliable Witnesses
« Reply #11 on: Apr 6th, 2005, 4:46pm » |
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Well, we aren't given that the remaining suspects aren't under consideration. So you're about one line from a complete proof. 
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Keith H
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Re: Unreliable Witnesses
« Reply #12 on: Apr 12th, 2005, 2:13pm » |
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on Apr 6th, 2005, 4:46pm, Deedlit wrote:| Well, we aren't given that the remaining suspects aren't under consideration. So you're about one line from a complete proof. |
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I was just filling in the "trial and error" gap from an earlier proof, so I left out the obvious fact that there are too many contradicting testimonies to allow one of the (0,5) or (5,0) suspects to be incorrectly identified.
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Deedlit
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Re: Unreliable Witnesses
« Reply #13 on: Apr 13th, 2005, 8:31pm » |
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Fair enough.
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