Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Partially Defined Grouping game.

This Game was considered to be fairly difficult by test takers, in part because they struggled to appropriately show the “on” and “off” designations, and also because the last rule was seen by many as confusing. To represent “on” and “off,” simply use a slash through the number when that light is off.

O05_Game_#2_setup_diagram 1.png

When combined with the first two rules, the last rule can be fairly restrictive. Consider the hypothetical where five switches are on. That means that switch number 5 must be on. If 5 is on, then 1 and 4 are off, so the five switches that are on must be 2, 3, 5, 6, and 7.

Note too that you must always have at least two switches off, since 1 cannot be on with either 3 or 5, and 4 cannot be on with either 2 or 5. If 1 and 4 are off, then the other five can be on, and if either 1 or 4 are on then the maximum number of switches that can be on is four. For instance, if 1 is on, then 3 and 5 are off, and either 2 or 4 must also be off since they cannot be on together (that is three switches off: 3, 5, and 2/4).

[miruke]

Hello,

I'm finding this switch game hard.I don't get how question 7 is switches 2 and 7.I picked E.I am rereading the rationale and don't get the explanation.I thought 1 and 3 are double not arrowed? I'm having a hard time understanding the switch rule which is probably why I missed question 10 as well.Also,since the rules are conditional,initially i was going to put a slash on the off channels (right side) of the arrows.Are conditional statements such as this automatically create double not arrows?


Thanks!

[Robert Carroll]

Miruke,

Let's start with diagramming the rules. For example:

If switch 1 is on, then switch 3 and switch 5 are off.

We can split this into:

If switch 1 is on, then switch 3 is off.

If switch 1 is on, then switch 5 is off.

Taking the first rule, it can be diagrammed as follows:

1 3

Now take the contrapositive:

3 1

In other words, if 1 is on, 3 is off, and if 3 is on, 1 is off. What this means is that we can't have both 1 and 3 on, which we represent by a double not-arrow:

1 3

The double not-arrow follows from the original conditional and its contrapositive. The reason we use the double not-arrow is that it contains those two pieces of information (the conditional and its contrapositive) in one rule. It's also easier to understand what it means and what its implications are for grouping games: the double not-arrow here means "1 and 3 can't both be on". The double not-arrow is an inference from the conditional - whenever you have something like "if 1, then not 3", where the sufficient condition leads to the negation of the necessary condition, you will indeed have a double not-arrow. Your instincts were right in that regard! It's not as if the conditional way of representing it is wrong, it just doesn't display all the information in an easy way like the double not-arrow does.

Moving on to question 7, 1 and 3 are related by a double not-arrow, so they can't both be on. That's a global rule. Question 7 adds a local condition that tells us neither is on, so instead of possibly having 1 on or possibly having 3 on (but never both, as that violates the double not-arrow), for question 7, we have a local rule stating both are definitely off. Since 1 and 3 are off, and there are only 7 switches, the maximum number of switches on would seem to be 5. But remember that 2 and 4 are related by a double not-arrow (global rule), so they can't both be on. So we're down to a maximum of 4 switches on. If 4 switches were on, because of the rule "The switch whose number corresponds to the circuit load of the panel is itself on", switch 4 would have to be on. What else could be on? 1 can't, 2 can't (double not-arrow with 4!), 3 can't, 5 can't (double not-arrow with 4), 6 could, 7 could; so we can't get 4 switches on with 4. Therefore, our maximum of 4 was wrong. How about 3? Again, the rule "The switch whose number corresponds to the circuit load of the panel is itself on" means that if 3 total switches were on, switch number 3 itself would need to be on, but our local rule tells us this isn't true. The same works for 1, so we know a total of 2 switches must be on, and switch number 2 must be one of them.

Moving on to question 10, if 5 and 6 are on, 1 can't be on and 4 can't be on (double not-arrows in both cases). So the maximum load is 5. If only 5 and 6 were on, the load would be 2, so 2 would have to be on, so we know the load can't be 2. So it's 3, 4, or 5; but if it were 4, switch 4 would have to be on, so the load is 3 or 5. If it's 3, switch 3 itself is on, and if it's 5, switches 2, 3, 5, 6, and 7 are on. In either case, switch 3 is on, so that's the answer.

The rule about the circuit load just means that the total number of "on" switches has to be the same number as one of the "on" switches. It's an additional rule to consider besides the double not-arrows you get from the two other rules.

Robert

[lexigibbs]

On Page 3-22 Game #2 from October 2005 questions 2-4 I am confused on. I feel i set my standard graph up correctly but those 3 questions I am confused with. For 2 I marked B or E and the correct answer is A, for 3 I marked E or D, and number 4 I was completely at a loss for.

[Adam Tyson]

Before we offer any advice on this one, it would be very helpful if you would describe for us your diagram. Rather than walking you through the whole setup, we could then focus on finding where you may have gone wrong - perhaps a misunderstood rule, a missed inference, etc. So, tell us more, try to show us visually as much as you can using the tools available here, and we will follow up from there!

[adlindsey]

I would like to see a set up for this game too. I was only able to diagram the conditional rules.

[Adam Tyson]

The conditional rules are most of what you need here, adlindsey. They key is to create conditional chains out of them. Link variables together, showing the "domino effect" of one switch being on or off. For example, two rules require that switch 5 be off, so try a hypothetical solution where 5 is on.

The thing that takes this game to another level is the stuff in the scenario and rules about the circuit load. Diagramming that is tough, and I am not sure there is any one clean, clear pictogram that will capture it. First, the circuit load is the number of switches that are on. If 3 switches are on, the circuit load is 3; if 5 switches are on, the circuit load is 5, etc. Then they throw in that the switch "whose number corresponds to the circuit load of the panel is itself on". That means if the circuit load is 3, switch 3 must be on; if the circuit load is 5, switch 5 must be on.

I might diagram that with words rather than pictures, honestly, if I were to diagram it at all (and I didn't bother the first time I did the game - I just memorized it and internalized it, which is risky and could allow you to forget it). How about:

If 2 switches, 2 is on
If 3 switches, 3 is on
etc

Go back and try out those conditional chains. and come up with a few hypotheticals. What if just 1 switch is on? What if 2 switches are on? Then, when you have a good sense of how the rules work together, move on to the questions. If you like, come back and share your diagram here and we'll see what else can be done with it.

A complete explanation, including the (very bare bones) diagram, can be found in the homework section for Lesson 6 in our full length course books.

Thanks for asking! You are not alone in having issues with this one, so this discussion will surely be useful to others.

[Tamirra]

Hi,

I've gone back to try to makes sense of the final rule and nothing is working. Can someone please rephrase?

Thanks,
Tamirra

[KelseyWoods]

Hi Tamirra!

Let's breakdown that last rule:

"The switch whose number corresponds to the circuit load of the panel is itself on."

What's the "circuit load"? The stimulus defines it for us:

"The circuit load of the panel is the total number of switches that are on."

So the rule is telling us that whatever switch that has the same number as the total number of switches that are on, also has to be on.

So if the circuit load (or number of switches ON) = n, then switch #n must be one of those ON switches. Meaning:

If there is exactly 1 switch ON, that switch has to be #1.
If there are exactly 2 switches ON, switch #2 must be ON.
If there are exactly 3 switches ON, switch #3 must be ON.
If there are exactly 4 switches ON, switch #4 must be ON.
If there are exactly 5 switches ON, switch #5 must be ON.
If there are exactly 6 switches ON, switch #6 must be ON.
If there are exactly 7 switches ON, well, clearly switch #7 would have to be one of those ON switches.
(Note: due to the other rules, not all of these scenarios are actually possible.)

Hope this helps!

Best,
Kelsey