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Among six suspects (C, D, G, H, J, T), two worked together to commit a murder.

  • H said C and G did it.

  • J said D and T did it.

  • D said T and C did it.

  • G said H and C did it.

  • C said D and J did it.

  • T refused to say anything.

Four of the five suspects that said anything named one person correctly and one person incorrectly. The fifth suspect that said something answered both people incorrectly.

Who commited the murder?


I encountered this puzzle a while ago through a friend of a friend but I'm not entirely sure how to approach this problem. I tried drawing a graph with six vertices in which (u, v) is an edge provided that person u accused person v. This was a hint to the puzzle. However, I can't really proceed from here.

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    $\begingroup$ Welcome to Puzzling! Congratulations on your first post: looks like a nice logic puzzle. :) $\endgroup$
    – boboquack
    Oct 8, 2020 at 9:32
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    $\begingroup$ @Wan This is a great little puzzle, and you've seen a variety of approaches :) One tip ... I assume that the letters are derived from actual names in the original? Reducing to letters is a good choice, but once you've done so I would reduce further to letters A-F $\endgroup$
    – Brondahl
    Oct 9, 2020 at 5:57

6 Answers 6

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It seems like

the murderers are C and J.

That way, H, D, G and C are the four people who named one person correctly and J was the one who got both wrong.

One simple method to approach this is through trial and error. You know that one person got both of his statements wrong. So, you can assign one of the people to be this person and then see if the rest got 1 of their statements correct.

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    $\begingroup$ "The best way to approach this is through trial and error." Why do you think other ways are necessarily worse than trial and error? $\endgroup$
    – Stef
    Oct 9, 2020 at 13:48
  • $\begingroup$ I changed it to "one simple method" instead to avoid any comparisons. $\endgroup$
    – Alaiko
    Oct 9, 2020 at 15:42
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The consensus of answers is correct.

The diagram above is based on the hint received by the puzzle’s poser but with a different interpretation than that in the puzzle statement.

I tried drawing a graph with six vertices in which (u,v) is an edge provided that person u accused person v. This was a hint to the puzzle.

Here comes perhaps the intended interpretation of that hint: a six-vertex graph of links based purely on which pairs of suspects are accused in each sentence. Speakers’ identities add no information and may be disregarded.

This amounts to the following links, along with two possible situations of the guilty vertices. The two guilty vertices are endpoints of a combined total of four links, all to innocent vertices. At this stage of deduction it is uncertain whether or not the guilty vertices have any linked neighbors in common.

Of the two ways to link four innocent vertices to two guilty vertices, only the second way can be overlaid onto the original graph. That overlay is readily obtained.

   

There it is, the structure underlying the solution at the top of this answer.

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I think the answer is

C and J.

My steps to approach the problem were as follows:

First observe that the five suspects mentioned 10 people in total, out of which only four are right and the other six are wrong. Now count how many times each suspect was mentioned: C was mentioned three times, D and T were mentioned twice each, and G, H, and J were mentioned only once each.

Given that two suspects committed a murder, and four mentions were correct, there are only a few possible combinations of suspects: CG, CH, CJ, DT. Now check for each combination if anyone mentioned the two suspects at once (which makes that combination wrong, given that no one got both right). CG was mentioned by H, CH was mentioned by G, DT was mentioned by J. Therefore the only possible combination is CJ.

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  • $\begingroup$ This is exactly how I figured it out quite quickly (in less than a minute). +1 $\endgroup$
    – Paul Evans
    Oct 9, 2020 at 15:09
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Even though all the other answers correctly specify

C and J

I think my reasoning is a bit more elegant and intuitive.

1. First, notice it doesn't matter at all who said what. The only important information is the five pairs. I've reordered them to note the pattern that three include C, the remaining two include D:

CG, CT, CH, DJ, DT

We know that at least one person is wrong in each of these pairs. Since three pairs include C, and the other person is different in each case, C must be one of the robbers. (C cannot be the incorrect person in all three cases or there would be three robbers, G, T & H). There is one other case, but we will come back to that.

D seems like the obvious choice for the other robber, but we know one pair is double wrong, so it can't be D. That means it's either J or T, but it can't be T, because T is in a pair with C, and no pair is double right.

So this gives us an answer. But what about the case that one of C's pairings is double wrong? We didn't consider that. But that can't be, because if C is wrong, two of his potential partners must be the robbers, which means that at least one of D's pairings would also be double wrong, which we know is not the case.

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A possible answer is :

C and J where J is the fifth suspect.

How to formally solve it:

So you have 4 correct mentions of the culprits. One culprit would have been mentioned at least twice otherwise if both are mentioned less than 2 times, they can't add up to 4. Then we just do a frequency analysis.

C - 3 , T - 2, D - 2. If C is a culprit, TD can't be one. So, one pair is CJ. If T is a culprit, there are no possibilities. If D is a culprit, then also there are no possibilities.

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The only possible answer is :

C and J committed the murder.

Reasoning:

Since only 1 person was entirely wrong that means that in any pair of accusations, at least 1 of the accused is guilty. Thus any pair of people that is entirely absent from any pair of accusations is safe.

Accusation pair | guilty pair(s) ruled out:
HJ|HJ
HD|(all HJD combos)
HG|(all JDT combos)
HC|HT
JD|(all HJG combos)
JG|JG
JC|(all HCG combos)
DG|(all JDG combos)
DC|GH
GC|GT

After ruling all of those out, we're left with only 3 options:
CD or CJ, CT

CT is explicitly accused, which proves that it cannot be correct.

CD would require ALL answers to give 1 correct name and is thus disallowed.

Thus CJ is the answer.

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  • $\begingroup$ Nicer ways to format inside spoiler tags are welcome! Ideally, I'd have new lines, and tables :( $\endgroup$
    – Brondahl
    Oct 8, 2020 at 16:08

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