3
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[This is continuing from previous puzzles. #4 was quite tricky. This one is meant to be much easier (I'm testing the boundaries here). I've been testing various versions and would love your thoughts. Playability? Enjoyment? Difficulty? Suggestions?]

There's been another murder on an isolated island. The suspects are Armandina, Connie, Donnie, Ligia, Margit, Ouida, Roy, Shizue, Terence.

This is what you know from all those training sessions at the police 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.

In the following table, each line represents the statements made by one of the characters. For example: The I in the first line (Armandina) under the letter M represents a statement by Armandina that Margit is innocent.

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   |
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   |
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I |
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   |
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   |
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   |
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   |
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I |
----------------------------------------------------
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6
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TL;DR

How many people are guilty?

3

Who is guilty?

Donnie, Ouida and Terence.

Who is innocent?

Armandina, Connie, Ligia, Margit, Roy and Shizue.

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 $Y_1$, $Y_2$ and $Y_3$ are guilty, then $Y_1$, $Y_2$ and $Y_3$ are all surely innocent. Then, we have a rule $\overline{X} \rightarrow Y_1 Y_2 Y_3$ (i.e. if $X$ is guilty, $Y_1$, $Y_2$ and $Y_3$ are innocent). Further, since sentences in the form $P \rightarrow Q$ can be expressed as $\overline{P} \lor Q$, so, $\overline{X} \rightarrow Y_1 Y_2 Y_3$ turns out into $X \lor Y_1 Y_2 Y_3$.

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 $\overline{X} \lor Y_1 Y_2 Y_3 \lor \overline{Y_1} Y_2 Y_3 \lor Y_1 \overline{Y_2} Y_3 \lor Y_1 Y_2 \overline{Y_3}$.

And of course:

If $X$ is guilty and states that $Y$ is innocent, we can't conclude anything from that. So this makes no rule at all.

Converting the board to rules:

$1.\;\overline{A} \lor M \lor \overline{M}$
$2.\;A \lor True$
$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$4.\;C \lor True$
$5.\;\overline{D} \lor \overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST \lor \overline{CRS}T \lor \overline{CR}S\overline{T}$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$9.\;\overline{M} \lor S \lor \overline{S}$
$10.\;M \lor True$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$14.\;R \lor True$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17.\;\overline{T} \lor \overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{ACDM}OR \lor \overline{AC}DMOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$
$18.\;T \lor ACM$

3. Simplifying the rules

Which rules are useless?

1, 2, 4, 9, 10 and 14. 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.

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:

$19.\;\overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST \lor \overline{CRS}T \lor \overline{CR}S\overline{T} \lor CR$ (from 5 or 6)
$20.\;\overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R} \lor D$ (from 7 or 8)
$21.\;D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS \lor LS$ (from 11 or 12)
$22.\;CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO \lor O$ (from 15 or 16)
$23.\;\overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{ACDM}OR \lor \overline{AC}DMOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R} \lor ACM$ (from 17 or 18)

Simplifying those by using...

K-maps

Results in:

$24.\;CR \lor ST \lor \overline{CR}T$ (simplification of 19)
$25.\;D \lor M \lor R$ (simplification of 20)
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$ (simplification of 21)
$27.\;C \lor M \lor O$ (simplification from 22)
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$ (simplification from 23. Ugh, 3D K-map needed because there are 6 variables)

The resulting board is:

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5.\;\overline{D} \lor \overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST \lor \overline{CRS}T \lor \overline{CR}S\overline{T}$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17.\;\overline{T} \lor \overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{ACDM}OR \lor \overline{AC}DMOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$
$18.\;T \lor ACM$
$24.\;CR \lor ST \lor \overline{CR}T$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$

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...

$17.\;\overline{T} \lor \overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{ACDM}OR \lor \overline{AC}DMOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$

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: Terence accused three people, so it is impossible that he 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 $\overline{XY}$.

Then if we:

Check out to what 17 is reduced.

We get that:

$29.\;\overline{T}$ (reducting from 17 assuming only one variable might be false).

Whoa! This seems to solve the case. But...

This is incompatible with 5, which would then reduce to false.

What happened is that:

Terence is accusing three people, which is incompatible with he being innocent and having only one guilty person. However, Donnie is also accusing two other different people and at least one of them should be guilty. Either way, there would be too much people lieing.

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 is them simplified to:

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5b.\;\overline{D} \lor \overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11b.\;\overline{O} \lor D\overline{L}R\overline{S} \lor DLR\overline{S} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17b.\;\overline{T} \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{AC}DMOR$
$18.\;T \lor ACM$
$24.\;CR \lor ST \lor \overline{CR}T$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28b.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{AC}DOR$

I.e, only those rules changed:

11, 17 and 28.

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:

$5c.\;\overline{D}$ (from 5b)

This give us our other guilty person! Is that fine?

$11b.\;\overline{O} \lor D\overline{L}R\overline{S} \lor DLR\overline{S} \lor D\overline{L}RS$

Oops. This tell us that:

At least one third guilty person would be needed among O, L and S.

Hence:

Having two guilty people with T being one of them leads to a contradiction. Hence, if there are indeed exactly two guilty people, T is not one of them.

Applying that to our board:

By replacing every T with true.

We note that:

5 and 17b were simplified. 18 and 24 became tautologic. The board now have only 8 variables instead of 9.

Then:

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5c.\;\overline{D} \lor \overline{CR}S \lor C\overline{R}S \lor \overline{C}RS$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11b.\;\overline{O} \lor D\overline{L}R\overline{S} \lor DLR\overline{S} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17c.\;A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{AC}DMOR$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28b.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{AC}DOR$

We can simplify:

$17c.\;DOR \land (A\overline{CM} \lor \overline{A}C\overline{M} \lor \overline{AC}M)$

Hence:

If we are in the right path, D, O and R are all innocent. The board now have only 5 variables.

If we are in the right track, we are pretty close solving this:

$5d.\overline{C}S$
$7d.\;\overline{L} \lor M$
$11b.\;\overline{LS} \lor L\overline{S} \lor \overline{L}S$
$13d.\;CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15d.\;\overline{S} \lor CM$
$17d.\;A\overline{CM} \lor \overline{A}C\overline{M} \lor \overline{AC}M$
$26d.\;\overline{L} \lor \overline{LS} \lor S$ $28d.\;ACM \lor \overline{CM} \lor \overline{AM} \lor \overline{AC}$

Whoa, this says that:

5d says that C is guilty and S is yet another innocent!

But:

15d says that either S is guilty or C is innocent (together with M). Darn, this is a contradiction.

Conclusion:

2 is not the number of guilty people. Let's try 3.

7. Even harder

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

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5.\;\overline{D} \lor \overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST \lor \overline{CRS}T \lor \overline{CR}S\overline{T}$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17.\;\overline{T} \lor \overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{ACDM}OR \lor \overline{AC}DMOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$
$18.\;T \lor ACM$
$24.\;CR \lor ST \lor \overline{CR}T$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR \lor \overline{AC}D\overline{MO}R \lor \overline{AC}D\overline{M}O\overline{R}$

There are a few places which have 4 guilty people.

Let's cut those.

So:

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5.\;\overline{D} \lor \overline{CR}ST \lor C\overline{R}ST \lor \overline{C}RST \lor \overline{CRS}T \lor \overline{CR}S\overline{T}$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17e.\;\overline{T} \lor \overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{AC}DMOR$
$18.\;T \lor ACM$
$24.\;CR \lor ST \lor \overline{CR}T$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR$

We have 84 possibilities. So, let's presume that:

T is innocent.

For the same reason as mentioned above.

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5f.\;\overline{D} \lor \overline{CR}S \lor C\overline{R}S \lor \overline{C}RS \lor \overline{CRS}$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$17f.\;\overline{AC}D\overline{M}OR \lor A\overline{C}D\overline{M}OR \lor \overline{A}CD\overline{M}OR \lor \overline{AC}DMOR$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR$

We can:

Simplify 17f to:
$17f.\;DOR \land (\overline{ACM} \lor A\overline{CM} \lor \overline{A}C\overline{M} \lor \overline{AC}M)$

So:

D, O and R would also be innocent.

Hence:

$5g.\;\overline{C}S$
$7g.\;\overline{L} \lor M$
$8g.\;L$
$11g.\;\overline{LS} \lor L\overline{S}$
$12g.\;LS$
$13g.\;CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15g.\;\overline{S} \lor CM$
$17g.\;\overline{ACM} \lor A\overline{CM} \lor \overline{A}C\overline{M} \lor \overline{AC}M$
$26g.\;\overline{L} \lor \overline{LS} \lor S$
$28g.\;ACM \lor \overline{CM} \lor \overline{AM} \lor \overline{AC}$

It states that:

12g says that L and S are innocent. But 11g says that L must be guilty. Contradiction!

So:

If we have 3 guilty people, T must be one of them.

Backtracking that:

$3.\;\overline{C} \lor LR \lor \overline{L}R \lor L\overline{R}$
$5h.\;\overline{D} \lor \overline{CR}S$
$6.\;D \lor CR$
$7.\;\overline{L} \lor \overline{D}MR \lor DMR \lor \overline{DM}R \lor \overline{D}M\overline{R}$
$8.\;L \lor D$
$11.\;\overline{O} \lor D\overline{L}R\overline{S} \lor \overline{DL}R\overline{S} \lor DLR\overline{S} \lor D\overline{LRS} \lor D\overline{L}RS$
$12.\;O \lor LS$
$13.\;\overline{R} \lor CMS \lor \overline{C}MS \lor C\overline{M}S \lor CM\overline{S}$
$15.\;\overline{S} \lor CM\overline{O} \lor \overline{C}M\overline{O} \lor C\overline{MO} \lor CMO$
$16.\;S \lor O$
$18h.\;ACM$
$24h.\;CR$
$25.\;D \lor M \lor R$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S} \lor DRS$
$27.\;C \lor M \lor O$
$28.\;ACM \lor \overline{C}D\overline{M}OR \lor \overline{A}D\overline{M}OR \lor \overline{ACM}OR \lor \overline{AC}DOR$

Now, we have only:

A, C, M and R are innocent (by 18f and 24f). So, we have only 4 variables and 6 possibilities left. And exactly 2 of those variables are false (guilty) and exactly 2 are true (innocent).

So, applying substitution, eliminating tautologies and eliminating clauses with too much innocent or too much guilty people, we get:

$5i.\;\overline{D}$
$8.\;L \lor D$
$11i.\;\overline{O} \lor D\overline{LS} \lor DL\overline{S} \lor D\overline{L}S$
$12.\;O \lor LS$
$16.\;S \lor O$
$26.\;D\overline{L} \lor \overline{D}S \lor \overline{L}R\overline{S}$

It says that:

D is our second guilty person (5i). Only 3 variables left!

Then:

$8j.\;L$
$11j.\;\overline{O}$
$12.\;O \lor LS$
$16.\;S \lor O$
$26j.\;S \lor \overline{L}R\overline{S}$

And then:

L is innocent and O is our third guilty. Only one variable left!

Now:

$12k.\;S$
$16k.\;S$
$26k.\;S$

So:

I.E. The three remaining expressions converged into saying that S is also innocent and not to some contradiction.

This means that:

The puzzle is solved!

8. Puzzle solved

How many people are guilty?

3

Who is guilty?

Donnie, Ouida and Terence.

Who is innocent?

Armandina, Connie, Ligia, Margit, Roy and Shizue.

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  • $\begingroup$ Wow. Just wow! K-maps! I love it. Nice job. $\endgroup$ – Dr Xorile Dec 11 '18 at 4:04
  • $\begingroup$ Any thoughts on how you would rate it from a playability, fun, and difficulty perspective? $\endgroup$ – Dr Xorile Dec 11 '18 at 4:06
  • 1
    $\begingroup$ @DrXorile I am also solving your previous puzzle with the same method to get a comparison. Hold on! $\endgroup$ – Victor Stafusa Dec 11 '18 at 4:59
  • $\begingroup$ @DrXorile In fact, both are of similar difficulty, but your previous puzzle seemed to be a bit harder (not sure if this is a confirmation bias, effect of I am being much more tired or if it was indeed more difficulty). They are both very tricky and laborious. However, once you develop a method to solve one of them, you may be able to mechanically think to solve them all, which is not necessarily bad, since many other logic puzzles like sudoku and Einstein challenge are exactly like that. I also think that puzzles like this can be generated through a fairly simple algorithm. $\endgroup$ – Victor Stafusa Dec 11 '18 at 7:52
  • $\begingroup$ @DrXorile Also, I recommend you to give a look on this puzzle and this other one for some inspiration about how to make it harder by requireing something more complicated than boolean algebra to solve. $\endgroup$ – Victor Stafusa Dec 11 '18 at 7:53
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@Victor Stafusa gave a very detailed answer to essentially deduce the answer. The point of the puzzle, at least the human way of solving it, is to use induction or abduction. Obviously, Victor's solution shows that deduction is possible here and there is a bias among puzzlers to do so (there was a recent puzzle on this forum demonstrating a reticence to make a guess in Sudoku).

So here's how I went about solving it (But wait, didn't I already know the answer? Not really, because I had forgotten and I'm working through quite a few of these at the moment!)

I draw a bunch of columns on the right of the table and use it to keep track of my hypotheses and deductions (because you use deduction to do induction).

Here's a photo of my working:

Working

Here's a step-by-step (I'm changing the order a bit for clarity but the principle is the same):

  1. I guess that Terence might be innocent (I actually guessed guilty but it's often good to do a reductio ad absurdam), so I mark "(I)" in brackets to track that it's a guess.
  2. I then guess that Donnie is guilty, so I mark "((G))" next to his name, to indicate that it's the second layer of guess (or hypothesis testing)
  3. Now that means that Terence already has one wrong and so everything else he said must be right. So I mark those in also.

Here's the working so far:

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | G
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   | G
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | ((G))
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | G
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | I
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   |
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (I)
----------------------------------------------------

Now, I have 4 guilty parties. But one of them (Donnie) is saying that another one of them (Connie) is guilty. Impossible. So my guess of Donnie being guilty was wrong. I, therefore, can mark him as innocent (in practice starting a new column, but for clarity I'll erase it here).

I also then guess that Connie is innocent, which Terence got wrong. Since Terence is assumed innocent, everything else Terence said must be true. So I mark that in too:

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | G
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   |((I)) 
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | I
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | G
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | I
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   |
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (I)
----------------------------------------------------

But then Roy is innocent and has got Margit wrong, so he must be right about Shizue.

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | G
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   |((I)) 
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | I
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | G
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | I
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   | I
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (I)
---------------------------------------------------- 

And then Shizue has two wrong statements, even though he's innocent. So this assumption doesn't work, and Connie must be guilty. Similar logic continues until I find that Armandina, Connie, Ligia, Margit, Roy, and Shizue are all guilty (if I assume that Terence in innocent).

On the other hand, if I assume that Terence is guilty, then I can immediately mark the people he's accused as innocent:

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | I
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   | I
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | 
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | I
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | 
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | 
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   | 
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (G)
---------------------------------------------------- 

Now it Donnie is innocent, then since he's wrong about Terence he must be right about Connie being guilty, so Donnie must be guilty. That means Roy is innocent

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | I
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   | I
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | [G]
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   |
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | I
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | 
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   | 
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (G)
---------------------------------------------------- 

Now, if Ouida is innocent, then he's wrong about Donnie, so he must be right about everything else:

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | I
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   | I
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | [G]
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   | G
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | I
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | ((I))
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   | G
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (G)
---------------------------------------------------- 

But then Ligia is guilty and is accusing Donnie of being guilty which is a contradiction. So Ouida must be guilty and therefore Ligia and Shizue are innocent:

Statements     | A | C | D | L | M | O | R | S | T |
----------------------------------------------------
Armandina      | I |   |   |   | I |   |   |   |   | I
----------------------------------------------------
Connie         |   | I |   | I |   |   | I |   |   | I
----------------------------------------------------
Donnie         |   | G | I |   |   |   | G | I | I | [G]
----------------------------------------------------
Ligia          |   |   | G | I | I |   | I |   |   | I
----------------------------------------------------
Margit         |   |   |   |   | I |   |   | I |   | I
----------------------------------------------------
Ouida          |   |   | I | G |   | I | I | G |   | [G]
----------------------------------------------------
Roy            |   | I |   |   | I |   | I | I |   | I
----------------------------------------------------
Shizue         |   | I |   |   | I | G |   | I |   | I
----------------------------------------------------
Terence        | G | G | I |   | G | I | I |   | I | (G)
---------------------------------------------------- 

So now we have a simpler solution: Donnie, Ouida, and Terence are guilty. By rule 5 this is the answer.

This seems long winded when written out step-by-step, but with pencil and paper seems to work.

As mentioned before, I would love any feedback on this or other methods of solving these puzzles and whether they are fun and might be fun on a broader scale.

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