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#98414
Setup and Rule Diagram Explanation

This is a Defined-Fixed, Overloaded, Grouping Game.

This setup is still in progress. Please post any questions below!
 roppo@ualberta.ca
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#98757
When will this game be explained?

Thank you!
 Luke Haqq
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#98796
Hi roppo!

Here is a basic setup for this game:

Paintings (5 OF 8)

F, G, H, I, Q, R, S, V
Rules

G or H :arrow: F

V :arrow: Q & R

H :arrow: Q & I

G & S :arrow: I

F :arrow: S OR V (only one)
Inferences

In general, it's helpful to write out contrapositives, whether separately as inferences or perhaps in parentheses next to each rule. With the first rule, for example, the contrapositive is:

F :arrow: G and H
It can also helpful to rewrite statements using the double-not arrow. If one has a generic statement A :arrow: B, this can be rewritten as A :dblline: B. In other words, this indicates that A and B can never be chosen together. One can see something like that in the third rule, which could be rewritten as:

H :dblline: Q & I
And it is also important in general to look for randoms or constrained variables. In this game, there are no random variables, as the rules draw on all of them. It might also appear initially that there aren't especially constrained variables. However, it becomes clear that some are more constrained than others. For example, if H is included, then we know Q and I are not included, and we also know that if Q is not included, then neither is V--so if H is included, then Q, I, and V, all are not included. Consider putting all the variables in a vertical line, with columns next to them to batch them into IN and OUT bins:

_______IN_______||_______OUT_______
F
G
H
I
Q
R
S
V
We can fill this out using the rules:

F :arrow: S/V, __, __, __ || S/V, __, __
G :arrow: F, S/V, __, __ || S/V, __, __
H :arrow: F, S, G, R || Q, I, V
I :arrow: Q, R, __, __ || H, G, __
Q :arrow: __, __, __, __ || H, __, __
R :arrow: __, __, __, __ || __, __, __
S :arrow: F/V, __, __, __ || F/V, __, __
V :arrow: Q, R, __, __ || H, __, __
The last rule tells us that if F occurs, then either S or V occurs, which is why we can put one S/V in the IN bin and another S/V in the OUT bin whenever F occurs. Because of the same rule, if S is included, we can do the same with an V/F in the IN and OUT bins.

For variable H, it becomes clear that there's only only possible combination of variables if it is chosen. Variable I also proves to be somewhat restricted. To start, we known from the third rule that if I is chosen, then H will be in the OUT bin. With that, there are a few combinations that work:

I :arrow: F, Q, R, V || H, G, S
I :arrow: F, Q, R, S || H, G, V
I :arrow: Q, R, S, V || H, G, F
These all include Q and R in the IN bin, as well as H and G in the OUT bin.
 vrodriguez2@ymail.com
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#100714
Hello, can you please explain how you got to the two deductions below?

I :arrow: Q, R, __, __ || H, G, __
S :arrow: F/V, __, __, __ || F/V, __, __
 Robert Carroll
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#100723
vrodriguez2,

Those are explained in the initial post:
The last rule tells us that if F occurs, then either S or V occurs, which is why we can put one S/V in the IN bin and another S/V in the OUT bin whenever F occurs. Because of the same rule, if S is included, we can do the same with an V/F in the IN and OUT bins.

For variable H, it becomes clear that there's only only possible combination of variables if it is chosen. Variable I also proves to be somewhat restricted. To start, we known from the third rule that if I is chosen, then H will be in the OUT bin. With that, there are a few combinations that work:

I :arrow: F, Q, R, V || H, G, S
I :arrow: F, Q, R, S || H, G, V
I :arrow: Q, R, S, V || H, G, F
These all include Q and R in the IN bin, as well as H and G in the OUT bin.
Robert Carroll
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 ArielA
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#101906
Hello,

I saw the initial post, but I am still confused about the answer to Rodriguez's question. Particularly, I am having trouble understanding this part:

S :arrow: F/V, __, __, __ || F/V, __, __

the original rule was:

F :arrow: S OR V (only one)

Following that line of logic, I thought that F being "in" is what triggered S or V to be in. I can't understand why S takes on that role that F has in the explanation.
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 ArielA
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#101907
ArielA wrote: Tue May 23, 2023 6:55 pm Hello,

I saw the initial post, but I am still confused about the answer to Rodriguez's question. Particularly, I am having trouble understanding this part:

S :arrow: F/V, __, __, __ || F/V, __, __

the original rule was:

F :arrow: S OR V (only one)

Following that line of logic, I thought that F being "in" is what triggered S or V to be in. I can't understand why S takes on that role that F has in the explanation.
I guess what I am saying is that I don't understand why S being "in" makes F "in" as well.

When I did the contrapositives, this is what I got:

"If F :arrow: S or V
if not S and not V :arrow: not F

If F :arrow: not
if :arrow: not F"

Maybe I did something wrong when writing the contrapositives?
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 ArielA
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#101908
correction (The computer made an error in my last message. Here are the contrapositives I meant to write):

If F :arrow: S or V
if not S and not V :arrow: not F

If F :arrow: not (S and V)
if (S and V) :arrow: not F
 Luke Haqq
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#101937
Hi ArielA!

On why it's diagrammed that way for S, to start, think of it this way: If it's given that S is chosen, then can we have both F and V also chosen? The answer is no. This would violate the final rule: "If Funscape is selected, exactly one of Salvation or Verisimilitude is selected." So we can't have both F and V in the "in" bin.

That leaves a possibility that both F and V are "out."

S :arrow: _ _ _ _ || F, V, _
This scenario, however, wouldn't work. For example, the first rule tells us, "If either or both of Golem and Helios are selected, Funscape is selected." Based on how the above diagram starts off, there's only spot left in the out bin, so we know that at least one of G or H is chosen. But one of them being chosen would mean that F is also chosen, which doesn't work in this diagram.

If it's given that S is chosen, we can therefore eliminate the possibility that F and V are both chosen, and we can also eliminate the possibility that both are not chosen. The only remaining option is that one of F and V is chosen, while one of them is not chosen, reflected by the F/V in the in and out bins.

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