7
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[Appeal: I'm trying to figure out a fun puzzle formula. There are many out there ("Who owns the fish", sudoku, etc). I'm interested in feedback on whether this works or is interesting for people. Possible to solve by hand. Techniques you found helpful, etc. Also a rating for this if you solve it: Easy; Tricky; Difficult; Fiendish]

After a long hiatus, you are back on the case. You arrive on the island to find a dead body and a number of possible suspects. You separate them and take their statements.

This is what you know from all those training sessions at the academy:

  1. Guilty people sometimes lie and sometimes don't.
  2. Guilty people will always say their co-conspirators are innocent (if they mention them).
  3. Innocent people tell the truth (when they know it).
  4. Except for when they make up a statement to try to impress you. But they never make up more than one statement.
  5. The smallest conspiracy that is consistent with the statements is the answer.

Suspect Names: Barbara, Donnie, Jackeline, Luisa, Mauricio, Roy, Shakia, Terence, Vanessa.

Statements of Barbara:
- I am innocent
- Roy is innocent
- Vanessa is innocent
- Shakia is innocent

Statements of Donnie:
- I am innocent
- Luisa is innocent
- Terence is guilty
- Shakia is innocent

Statements of Jackeline:
- I am innocent
- Shakia is innocent
- Luisa is innocent
- Terence is innocent
- Vanessa is guilty

Statements of Luisa:
- I am innocent
- Mauricio is innocent
- Shakia is guilty
- Donnie is guilty
- Jackeline is innocent

Statements of Mauricio:
- I am innocent
- Roy is innocent

Statements of Roy:
- I am innocent
- Mauricio is innocent
- Terence is guilty

Statements of Shakia:
- I am innocent
- Barbara is innocent
- Donnie is innocent
- Terence is guilty
- Roy is guilty

Statements of Terence:
- I am innocent
- Vanessa is guilty
- Mauricio is guilty
- Jackeline is innocent
- Donnie is innocent
- Shakia is guilty

Statements of Vanessa:
- I am innocent
- Donnie is innocent
- Mauricio is innocent
- Barbara is innocent
- Terence is innocent

Here's a table summarizing all the information:

Statements     | B | D | J | L | M | R | S | T | V | 
----------------------------------------------------
Barbara        | I |   |   |   |   | I | I |   | I |
----------------------------------------------------
Donnie         |   | I |   | I |   |   | I | G |   |
----------------------------------------------------
Jackeline      |   |   | I | I |   |   | I | I | G |
----------------------------------------------------
Luisa          |   | G | I | I | I |   | G |   |   |
----------------------------------------------------
Mauricio       |   |   |   |   | I | I |   |   |   |
----------------------------------------------------
Roy            |   |   |   |   | I | I |   | G |   |
----------------------------------------------------
Shakia         | I | I |   |   |   | G | I | G |   |
----------------------------------------------------
Terence        |   | I | I |   | G |   | G | I | G |
----------------------------------------------------
Vanessa        | I | I |   |   | I |   |   | I | I |
----------------------------------------------------
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4
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There must be 3 conspirators.

Let's start by looking at Terence. He has accused 3 people and 3 people have accused him.

If we make an initial assumption (Assumption#1) that he is innocent, then either (Assumption#2) has identified 3 guilty conspirators and one of his other statements may be a guess, or (Assumption#3) he has identified 2 guilty conspirators and incorrectly guessed at a third, in which case his other statements must be true.

Starting from Assumption#3:

Terence has declared that 2 other people are innocent, so we can mark them all as innocent.
enter image description here

We can now see that Donnie thought that Thomas was guilty. This was an incorrect guess, so Donnie's other statements must be accurate. We can now mark more people as innocent.
enter image description here

If we look at Luisa, we can see that she has incorrectly accused Donnie and Shakia. So Assumption#3 cannot be correct.
enter image description here

So let's try Assumption#2:

Terence has correctly identified 3 guilty people
enter image description here

These people would not have collared any of their co-conspirators, so we can now mark Roy as innocent.
enter image description here

But Roy has incorrectly accused Terence as guilty and incorrectly identified Mauricio as innocent, so Assumption#2 must be incorrect.
enter image description here

Now we have disproven Assumption#3 and Assumption#2, so Assumption#1 must also be incorrect.

This means that Terence must be guilty.
enter image description here

From this starting point

we can mark all the people he has accused as being innocent (as he wouldn't have named a co-conspirator).
enter image description here

Similarly, anyone that has accused Terence must be innocent too.
enter image description here

Now Luisa has accused 2 innocent people, so she must be guilty.
enter image description here

Shakia has incorrectly guessed about Roy, so her statement about Barbara must be true, making Barbara innocent too.
enter image description here

Which leaves Jackeline who has made 2 false statements, so she must be a conspirator too.
enter image description here

So the guilty people are:

Terence, Luisa and Jackeline

Interesting puzzle. As far as I can tell, rule number 5 is unnecessary in this case. It was straightforward enough to solve by hand without running into trouble - just making assumptions and disproving them to arrive at a unique solution. So I would rate it as Tricky just because of the need to disprove other possible solutions.

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  • $\begingroup$ Thanks and great job! You have found the only possible solution with 3 co-conspirators and there are no solutions with only 1 or 2. But there are a couple of possible solutions with 4 (e.g. R or V could also be guilty with the 3 you found). There are none with 5. And a couple again with 6. $\endgroup$ – Dr Xorile Dec 10 '18 at 16:49
  • $\begingroup$ @DrXorile I don't believe there are any other solutions. I've shown that T has to be guilty, so from rule #2 R can't be (as they have accused T) and V can't be (as they have been accused by T). I'd be happy to see another solution that proves me wrong! $\endgroup$ – Gordon K Dec 10 '18 at 16:55
  • $\begingroup$ Great point. I was running my tests with old code that allowed conspirators to occasionally accuse each other. $\endgroup$ – Dr Xorile Dec 10 '18 at 17:21
  • $\begingroup$ You should accept the answer then. $\endgroup$ – Marc Z Dec 11 '18 at 10:51
2
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TL;DR

How many people are guilty?

3

Who is guilty?

Jackeline, Luisa and Terence.

Who is innocent?

Barbara, Donnie, Mauricio, Roy, Shakia and Vanessa.

1. Preliminaries

Since each person is either innocent or guilty and there are 9 people, so there is only $2^9 = 512$ possible assignments of innocent-or-guilty. Those are distributed in that way:

  • 1 possibility of everybody being innocent.
  • 9 possibilities of 1 guilty person.
  • 36 possibilities of 2 guilty people.
  • 84 possibilities of 3 guilty people.
  • 126 possibilities of 4 guilty people.
  • 126 possibilities of 5 guilty people.
  • 84 possibilities of 6 guilty people.
  • 36 possibilities of 7 guilty people.
  • 9 possibilities of 8 guilty person.
  • 1 possibility of everybody being guilty.

This is not hard to brute-force with a computer. However, the puzzle states . So, let's proceed only with pencil-and-paper logic.

2. Formalizing

Let's formalize the rules. True variables denotes innocent people, false variables denotes guilty people.

But how?

Let's observe that:

Nobody says that himself/herself is guilty, even if being an innocent telling a lie. So, we can neglect the main diagonal of the matrix to make things simpler. The main diagonal is all filled with 'I's.

Also:

If X is guilty and tells that Y is also guilty, then Y is surely innocent. Then, we have a rule $\overline{X} \rightarrow Y$ (i.e. if X is guilty, Y is innocent). Further, since sentences in the form $P \rightarrow Q$ can be expressed as $\overline{P} \lor Q$, so, $\overline{X} \rightarrow Y$ turns out into $X \lor Y$.

On the other hand...

If X is innocent and tells something about $Y_1$, $Y_2$ and $Y_3$, we have that only one of those can be a lie (or none of them). Hence, we build a rule $X \rightarrow Y_1 Y_2 Y_3 \lor \overline{Y_1} Y_2 Y_3 \lor Y_1 \overline{Y_2} Y_3 \lor Y_1 Y_2 \overline{Y_3}$. This can be simplified to $\overline{X} \lor Y_1 Y_2 Y_3 \lor \overline{Y_1} Y_2 Y_3 \lor Y_1 \overline{Y_2} Y_3 \lor Y_1 Y_2 \overline{Y_3}$.

Converting the board to rules:

$1.\;\overline{B} \lor RSV \lor \overline{R}SV \lor R\overline{S}V \lor RS\overline{V}$
$2.\;V \lor True$
$3.\;\overline{D} \lor LS\overline{T} \lor \overline{L}S\overline{T} \lor L\overline{ST} \lor LST$
$4.\;D \lor T$
$5.\;\overline{J} \lor LST\overline{V} \lor \overline{L}ST\overline{V} \lor L\overline{S}T\overline{V} \lor LS\overline{TV} \lor LSTV$
$6.\;J \lor V$
$7.\;\overline{L} \lor \overline{D}JM\overline{S} \lor DJM\overline{S} \lor \overline{DJ}M\overline{S} \lor \overline{D}J\overline{MS} \lor \overline{D}JMS$
$8.\;L \lor S$
$9.\;\overline{M} \lor R \lor \overline{R}$
$10.\;M \lor True$
$11.\;\overline{R} \lor M\overline{T} \lor \overline{MT} \lor MT$
$12.\;R \lor T$
$13.\;\overline{S} \lor BD\overline{RT} \lor \overline{B}D\overline{RT} \lor B\overline{DRT} \lor BDR\overline{T} \lor BD\overline{R}T$
$14.\;S \lor RT$
$15.\;\overline{T} \lor DJ\overline{MSV} \lor \overline{D}J\overline{MSV} \lor D\overline{JMSV} \lor DJM\overline{SV} \lor DJ\overline{M}S\overline{V} \lor DJ\overline{MS}V$
$16.\;T \lor MSV$
$17.\;\overline{V} \lor BDMT \lor \overline{B}DMT \lor B\overline{D}MT \lor BD\overline{M}T \lor BDM\overline{T}$
$18.\;V \lor True$

3. Simplifying the rules

Which rules are useless?

2, 9, 10 and 18. They don't tell us anything at all because they are tautologic. I.E.: They reduce to $True$ without giving us any value to any variable; Every possible combination of values for their variables produces $True$ regardless of anything.

Simplifying those by using...

K-maps

Results in:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3b.\;\overline{D} \lor LS \lor L\overline{T} \lor S\overline{T}$
$4.\;D \lor T$
$5b.\;\overline{J} \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11b.\;\overline{R} \lor M \lor \overline{T}$
$12.\;R \lor T$
$13b.\;\overline{S} \lor D\overline{RT} \lor B\overline{RT} \lor BD\overline{T} \lor BD\overline{R}$
$14.\;S \lor RT$
$15b.\;\overline{T} \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$ (Ugh, 3D K-map needed because there are 6 variables)
$16.\;T \lor MSV$
$17b.\;\overline{V} \lor BDM \lor BDT \lor BMT \lor DTM$

How can we start to combine rules?

Let's start by the rules of $X \lor Y$ and $\overline{X} \lor Z$. We can combine them into $Y \lor Z$.

So:

$19.\;T \lor LS \lor L\overline{T} \lor S\overline{T}$ (from 3b or 4)
$20.\;V \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$ (from 5 or 6)
$21.\;S \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$ (from 7 or 8)
$22.\;T \lor M \lor \overline{T}$ (from 11 or 12)
$23.\;RT \lor D\overline{RT} \lor B\overline{RT} \lor BD\overline{T} \lor BD\overline{R}$ (from 13 or 14)
$24.\;MSV \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$ (from 15 and 16)

We can further simplify that. Results in:

$19b.\;L \lor S \lor T$
$20b.\;LS \lor LT \lor ST \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$22b. True$ (tautology)
$23b.\;RT \lor BD \lor B\overline{RT} \lor D\overline{RT}$

The resulting board is:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3b.\;\overline{D} \lor LS \lor L\overline{T} \lor S\overline{T}$
$4.\;D \lor T$
$5b.\;\overline{J} \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11b.\;\overline{R} \lor M \lor \overline{T}$
$12.\;R \lor T$
$13b.\;\overline{S} \lor D\overline{RT} \lor B\overline{RT} \lor BD\overline{T} \lor BD\overline{R}$
$14.\;S \lor RT$
$15b.\;\overline{T} \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$
$16.\;T \lor MSV$
$17b.\;\overline{V} \lor BDM \lor BDT \lor BMT \lor DTM$
$19b.\;L \lor S \lor T$
$20b.\;LS \lor LT \lor ST \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23b.\;RT \lor BD \lor B\overline{RT} \lor D\overline{RT}$
$24.\;MSV \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$

Improving from the previous board is hard because so far we ignored something:

The minimum number of guilty people is the correct.

So let's try with...

4. The trivial case

Zero!

This leads to:

All the variables being truth. We can assign false to any clause with an $\overline{X}$.

But...

$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$

I.e:

No clause of that is free from a $\overline{X}$. This would be false if nobody is guilty. So zero isn't our number. The reason is simple: Luisa accused two people, so it is impossible that she is innocent and didn't told two or more lies.

This, was the most probable outcome anyway, so let's move on...

5. Not so trivial

So let's try exactly one guilty person.

Which means:

Assign false to any clause with some $\overline{XY}$.

So, we get that:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3b.\;\overline{D} \lor LS \lor L\overline{T} \lor S\overline{T}$
$4.\;D \lor T$
$5b.\;\overline{J} \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$
$6.\;J \lor V$
$7c.\;\overline{L} \lor JM\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11b.\;\overline{R} \lor M \lor \overline{T}$
$12.\;R \lor T$
$13c.\;\overline{S} \lor BD\overline{T} \lor BD\overline{R}$
$14.\;S \lor RT$
$15c.\;\overline{T}$
$16.\;T \lor MSV$
$17b.\;\overline{V} \lor BDM \lor BDT \lor BMT \lor DTM$
$19b.\;L \lor S \lor T$
$20b.\;LS \lor LT \lor ST \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23b.\;RT \lor BD$
$24c.\;MSV$

This tells us that:

M, S and V are innocent (24c). T is guilty (15c). Everyone else should be innocent.

However...

$7c.\;\overline{L} \lor JM\overline{S} \lor \overline{D}JM$

So what?

T is guilty by 15c. But 7c tells that at least one among D, L and S is guilty. This contradicts the supposition that there should be only one guilty.

That is unfortunate, so move on.

6. Getting hard

We try as the number of guilty people:

2

So...

Discard everything with a $\overline{XYZ}$.

The resulting board (from the end of part 3) is them simplified to:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3b.\;\overline{D} \lor LS \lor L\overline{T} \lor S\overline{T}$
$4.\;D \lor T$
$5b.\;\overline{J} \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11b.\;\overline{R} \lor M \lor \overline{T}$
$12.\;R \lor T$
$13b.\;\overline{S} \lor D\overline{RT} \lor B\overline{RT} \lor BD\overline{T} \lor BD\overline{R}$
$14.\;S \lor RT$
$15d.\;\overline{T} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$
$16.\;T \lor MSV$
$17b.\;\overline{V} \lor BDM \lor BDT \lor BMT \lor DTM$
$19b.\;L \lor S \lor T$
$20b.\;LS \lor LT \lor ST \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23b.\;RT \lor BD \lor B\overline{RT} \lor D\overline{RT}$
$24d.\;MSV \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$

I.e, only those rules changed:

15 and 24.

We have 36 possibilities for that (as stated in the preliminaries), how to sort them out?

Let's suppose that T is guilty and eliminate all the clauses with two guilty people that are not T or with T being innocent.

Why?

He already busted the previous guess. Also, he is the person who most both accuses and absolves other people, so he is a good heuristic.

Then:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3e.\;\overline{D} \lor L \lor S$
$4e.\;D$
$5e.\;\overline{J} \lor LS\overline{V}$
$6.\;J \lor V$
$7e.\;\overline{L} \lor JM\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$12e.\;R$
$13e.\;\overline{S} \lor D\overline{R} \lor B\overline{R} \lor BD$
$14e.\;S$
$16e.\;MSV$
$17e.\;\overline{V} \lor BDM$
$19e.\;L \lor S$
$20e.\;LS \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23e.\;BD$
$24e.\;MSV$

Then:

B, D, M, R, S and V are all innocent.

Hence:

$5f.\;\overline{J}$
$7f.\;\overline{L}$

Which means:

J and L are guilty. However, we got three guilty people, but we assumed that there was only two! So we have a contradiction and T being guilty is incompatible with 2 is not the correct number of guilty people.

Ok, so...

T must be innocent.

Then:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3g.\;\overline{D} \lor LS$
$5g.\;\overline{J} \lor LS \lor L\overline{V} \lor S\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11g.\;\overline{R} \lor M$
$13g.\;\overline{S} \lor BD\overline{R}$
$14g.\;S \lor R$
$15g.\;DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$
$17g.\;\overline{V} \lor BD \lor BM \lor DM$
$20g.\;L \lor S \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23g.\;R \lor BD$
$24d.\;MSV \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$

And also:

Rule 20g is redundant due to 8, so we can discard it.

We can...

Simplify rule 15g.

Like this:

$15h.\;DJ \land (\overline{SV} \lor \overline{MV} \lor \overline{MS})$

So:

D and J are innocent.

This leads to:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3i.\;LS$
$5g.\;\overline{J} \lor LS \lor L\overline{V} \lor S\overline{V}$
$6.\;V$
$7i.\;\overline{L} \lor M\overline{S}$
$8.\;L \lor S$
$11g.\;\overline{R} \lor M$
$13i.\;\overline{S} \lor B\overline{R}$
$14g.\;S \lor R$
$15i.\;\overline{SV} \lor \overline{MV} \lor \overline{MS}$
$17i.\;\overline{V} \lor B \lor M$
$21i.\;S \lor M$
$23i.\;R \lor B$
$24i.\;MSV \lor \overline{SV} \lor \overline{MV} \lor \overline{MS}$

And...

24i became redundant to 15i. 5g and 8 are redundant to 3i. L, S and V are also innocent.

But let's look closer:

$3i.\;LS$
$7i.\;\overline{L} \lor M\overline{S}$

Oops. This is:

A contradiction! 3i says that L and S are both innocent. 7i says that at least one of them is guilty.

What this means?

Assuming that 2 people are guilty fatally leads to a contradiction. So this can't be right

Now what?

Let's try 3 guilty people.

7. Even harder

Let's grab the board from the end of part 3 again:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3b.\;\overline{D} \lor LS \lor L\overline{T} \lor S\overline{T}$
$4.\;D \lor T$
$5b.\;\overline{J} \lor LST \lor LS\overline{V} \lor LT\overline{V} \lor ST\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$11b.\;\overline{R} \lor M \lor \overline{T}$
$12.\;R \lor T$
$13b.\;\overline{S} \lor D\overline{RT} \lor B\overline{RT} \lor BD\overline{T} \lor BD\overline{R}$
$14.\;S \lor RT$
$15b.\;\overline{T} \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$
$16.\;T \lor MSV$
$17b.\;\overline{V} \lor BDM \lor BDT \lor BMT \lor DTM$
$19b.\;L \lor S \lor T$
$20b.\;LS \lor LT \lor ST \lor V$
$21b.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23b.\;RT \lor BD \lor B\overline{RT} \lor D\overline{RT}$
$24.\;MSV \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$

We have no clause with 4 or more guilty people and 84 possibilities. So, let's assume that:

T is guilty.

For the same reason as mentioned earlier. This is the result:

$1b.\;\overline{B} \lor RS \lor SV \lor RV$
$3g.\;\overline{D} \lor L \lor S$
$4g.\;D$
$5g.\;\overline{J} \lor LS\overline{V}$
$6.\;J \lor V$
$7b.\;\overline{L} \lor JM\overline{S} \lor \overline{D}J\overline{S} \lor \overline{D}M\overline{S} \lor \overline{D}JM$
$8.\;L \lor S$
$12g.\;R$
$13g.\;\overline{S} \lor D\overline{R} \lor B\overline{R} \lor BD$
$14g.\;S$
$16g.\;MSV$
$17g.\;\overline{V} \lor BDM$
$19g.\;L \lor S$
$20g.\;LS \lor V$
$21g.\;S \lor \overline{D}J \lor JM \lor \overline{D}M$
$23g.\;BD \lor B\overline{R} \lor D\overline{R}$
$24.\;MSV \lor J\overline{MSV} \lor D\overline{MSV} \lor DJ\overline{SV} \lor DJ\overline{MV} \lor DJ\overline{MS}$

This means that:

D, M, R, S, V are innocent. Also, we get rid of 6 variables, there are only 3 left.

So:

$5h.\;\overline{J}$
$7h.\;\overline{L}$
$13h.\;B$
$17h.\;B$
$23h.\;B$

This means that:

I.E. The five remaining expressions converged into saying that J and L are the other two guilty and B is innocent. Very good that it didn't resulted in some contradiction.

This means that:

The puzzle is solved!

8. Puzzle solved

How many people are guilty?

3

Who is guilty?

Jackeline, Luisa and Terence.

Who is innocent?

Barbara, Donnie, Mauricio, Roy, Shakia and Vanessa.

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  • $\begingroup$ Interesting alternative approach to the problem. I think you've got an 'I' where you should have an 'R' in each instance of clue 1. I also think you've made a supposition about T in stage 6 and stage 7 that you would additionally need to disprove to be thorough. +1 for effort. $\endgroup$ – Gordon K Dec 11 '18 at 11:05
  • $\begingroup$ @GordonK Fixed everything but the disprove on stage 7. I am too tired now and need to take a break, will try to revisit stage 7 later. $\endgroup$ – Victor Stafusa Dec 11 '18 at 14:14

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