Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is an Advanced Linear: Balanced game.

In this game seven suspects are each questioned once, on consecutive days with one suspect questioned each day. This configuration creates a Basic Linear game that is Balanced. However, the end of the game scenario introduces another variable set—the confessions—that connects to each suspect, and turns this into an Advanced Linear game. The confessions variable set is easy to miss if you read too quickly. There are different ways to represent the confessions, but we will use “C” for confess and “NC” for not confess, leading to the following basic setup:

PT53-Dec2007_LGE-G3_srd1.png

With this basic setup in mind, we can analyze each of the rules.

Rule #1. This rule is easy to represent by placing T on the third day:

PT53-Dec2007_LGE-G3_srd2.png

Of course, with T removed from consideration, there are now only six suspects to consider.

Rule #2. This rule is also easy to represent, with a NC on the fourth day:

PT53-Dec2007_LGE-G3_srd3.png

Rule #3. This rule creates a W S sequence, which then yields two Not Laws:

PT53-Dec2007_LGE-G3_srd4.png

Rule #4. This rule creates another sequence, which can be represented as:

PT53-Dec2007_LGE-G3_srd5.png

This sequence yields Not Laws showing X and V cannot be questioned first, and Z cannot be questioned on the last two days:

PT53-Dec2007_LGE-G3_srd6.png

Rule #5. This rule establishes that any suspect questioned after W does not confess. By itself, this rule can be diagrammed as:

PT53-Dec2007_LGE-G3_srd7.png

There are other ways to diagram this rule as well, but none is preferable to another.

By combining this rule with the third rule, we can deduce that S does not confess:

PT53-Dec2007_LGE-G3_srd8.png

By itself, this rule is somewhat powerful. However, this rule will become considerably more useful once we examine the sixth rule.

Rule #6. This rule establishes that exactly two suspects confessed after T was questioned. From the first rule we know that T was questioned on day three. Thus, on exactly two days out of days four through seven the suspects confessed, and on the other two days out of days four through seven the suspects did not confess. Rule #2 already established that on day four the suspect did not confess, which means that out of days five, six, and seven, suspects confessed on two days and did not confess on the other day.

The information above is powerful, but do not stop linking the rules just yet. From the combination of the third and fifth rules we know that S does not confess and that S is questioned after W. Because no suspects confess after W is questioned, and two suspects must confess within the grouping of days five, six, and seven, we can infer that S must be questioned last, and that W must be questioned on day six. Of course, because S is questioned last, S does not confess. With S not confessing, we can infer that the suspects questioned on days five and six do confess.

This powerful information leads to the following setup:

PT53-Dec2007_LGE-G3_srd9.png

While the information above is significant, the placement of W and S leads to new inferences when the fourth rule is revisited. Specifically, only four variables remain unplaced: Y, V, X, and Z. Y is random, and thus can be questioned on any of the four remaining open days. V, X, and Z are all involved in the fourth rule, and from that rule we can deduce that Z cannot be questioned on day four or five, the last two remaining open days (otherwise there would be no room to question X and V). Thus, Z must be questioned on day one or two. Further, since X and V cannot be questioned on day one, only Y or Z can be questioned on day one, leading to a Y/Z dual-option on day one.

When the above information is added to the prior setup, we arrive at the final setup for this game:

PT53-Dec2007_LGE-G3_srd10.png

While arriving at this setup takes time and work, now that we have this setup and all of the information it contains, the questions should not be difficult. To work with this setup most efficiently, focus on the four unplaced variables— Y, V, X, and Z—and on the confession status of the suspects questioned on the first three days.

[crharke42]

I am needing help setting up this game.

So far I have...

T is questioned on day 3 (obviously)

W > S, so S cannot be questioned on Day 1 and W cannot be questioned on Day 7.

Z>X and V so X and V cannot be questioned on Day 1 and Z cannot be questioned on Day 6 or 7.

Then I have (C,C,N) for whether suspects 5-7 confessed.

Help please!!!

Thanks

[Dave Killoran]

This is one of those interesting setups where the information keeps adding up, and in the end the result is a setup with a lot more variables placed than you'd think initially.

From what you've said, you've got some of the pieces already down. For example, in the Confessions row, it appears you already have the following correct information:


N .....C C N
1 2 3 4 5 6 7

The Suspects row seems to have given you more trouble. Here's some inferences for that row, to help get you started:

• Z is quite restricted, and can only be first or second. And, only Z or Y can be first.
  
  W must be sixth, and S must be seventh (this results form combining the 5th, 3rd, and 6th rules.

Please check those out, and let me know if that gets you rolling on this game. Thanks!

[CEF]

Can you please explain the set up on a game like this? I missed the entire game and I think it falls back on the fact that I didn't full understand the set up and how the variables worked together.

Thank you!

[Nikki Siclunov]

The key to this Advanced Linear game is to recognize that there are two variable sets to keep track of: the order in which the suspects were questioned, but also whether each suspect confessed or did not confess. The best way to do that would be to create two separate "stacks": the bottom one for the order of suspects being questioned, the top - whether they confessed (C) or did not (N).

We know that T was questioned on day 3, and also that exactly two suspects confessed after that. So, there are two confessions and two non-confessions after day 3 (C, C, N, N). One of the non-confessions is on day 4 (second rule):

C/N: _ _ _ N (C,C,N)

Susp: _ _ T_ _ _ _

Well, the second to last rule states that no suspects confessed after W was questioned, so W was questioned either 6th or 7th. But, we know that S was questioned after W, so W cannot have been questioned last. Clearly, then W must have been questioned 6. Thus, our final set-up would look like this:

C/N: _ _ _ N C C N

Susp: _ _ T_ _ W S

Hope this gets you started

[CEF]

I had the set up correct on this but I was struggling to fully understand the W rule.

In my set up I missed why W couldn't be 5th, so I had a split between the 5th and 6th spots. I looked over the fact that if W was 5th, it would conflict with the rule that two suspects had to confess after T and that would be impossible if W was 5th because no suspects confessed after W!

Also, in the set up, couldn't you also put a split for Z on the 1st and 2nd spots since Z has to be before X and V?

Thanks for the clarification on this!

[Dave Killoran]

Hi CEF,

Yes, you got it right as far as W not being able to on day 5, which then forces W to be on day 6 (and S on day 7).

I think that what Nikki was trying to do was give you the starting elements of the setup, not a complete setup (the graphical limitations of the forum sometimes make complete setups a challenge). Z is indeed split between day 1 and 2, and day 1 is actually a dual-option between Y and Z.

Thanks!

[CEF]

Great. Thanks for all the help!

[SLF]

I understand the possibilities and the constraints of the suspects...but I am confused about the rules of the confession part.

Are slots 1,2,3 open for confession or non-confession depending upon the local rules...or is the game fixed at slots 4 and 7 being non-confessions and slots 5 and 6 being confessions?

Question 13: "If Z was the second suspect to confess..." Does this mean that Z is in slot 2 and a confession happened in slot 1 and 2?

[Dave Killoran]

Hi SLF,

Yes, the first three slots are undetermined as to whether it is a confession or non-confession. It can change depending on the local conditions.

As far as #13, yes. Here's why: The key to this problem is to read the question stem closely, because it states that “Z is the second suspect to confess.” As Z must be questioned first or second (which is inferred during the setup of the game), from the question stem we can infer that Z is questioned second and confesses, and that the suspect questioned first also confesses. As also determined in the setup, only Y or Z can be questioned first, so if Z is questioned second then Y must be questioned first. Thus, Y is questioned first and must confess. Accordingly, answer choice (E) cannot be true and is correct.

Does that help? Please let me know. Thanks!