June 2007, Game 2, Question 8

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Question 8 is a local question. So we would go to this question after we finished up the solution question, which was question 6. Skipping over 7 just for a moment, because we're trying to provide ourselves with some examples of how the game could work in practice. We have the master sketch, but each local question gives us the opportunity to write a couple of more sketches that we can then use for other questions.

Each local question gives you a condition that's only going to be in play just for that question. So here the condition is if limelight is never shown again during the festival once grid is shown. Now this is also a min/max question. The if tells us that it is a local question, the maximum number tells us that it is a min/max question.

Specifically the max half of min/max. If limelight is never shown again after grid is shown, they wanna know the maximum number of film showings that you could have at the festival. So over question 8 we need a new sketch and in that sketch we're going to try to maximize the number of films. So that would mean delaying G as long as possible, put down your spaces.

Put G as the last film on Saturday, put L as many times as you can in front of that. So in front of H on Thursday as the last film on Friday and in front of G on Saturday. So now we're up to 1, 2, 3, 4, 5, 5 films. So already we know the answer isn't A or B, cuz we've got at least 5 here. Our question is, is there anywhere else that we could put some extra films.

We can't put in the extra G's because if you put G's down you wouldn't be allowed to put those L's according to the if condition. We can't use any more L's because we've used L's on all three days. We've got an H on Thursday, not a lot to put an H on Saturday because the rule that says, you can't have both G and H on Saturday. And since G is there, H won't be there, but there is room to put an extra H on Friday.

Use your pencil to clean up your sketch a little bit and we have LH, HL, LG, six films, and there's no way to put an extra film on. We've run through all the options. So we know it's not answer choice C though, because we put six films down. And we do have to consider another scenario here, because the way that we got six films was by maximizing the number of L's that we used.

But we could also have tried to maximize the number of G's that we use. That is, get L out of the way on the first day and then put as many G's as you could. So right now next to your sketch that you just drew, draw another scenario. Put L on Tuesday in front of H and then fit G's on to all of the days. Well, because you're not allowed to show L after you show G that would put G in between L and H on Thursday.

Then place a G as the last film on Friday and as the last film on Saturday. So we've used as many G's as we possibly can, we've used as many L's as we possibly can. The question again is, is it possible to put any extra H's in? And as before, if we've placed a G on Saturday, we're not gonna be able to put a H there.

But Friday is still open, we could put another film on Friday, just as long as we put it in front of G. So we have LG, HH, GG. Three films on Thursday, two films on Friday, one film on Saturday for a total of six films. Since we've exhausted all our options for films, it looks like there's no way even if you maximize your G's, there's no way to get any more films than six.

That means that answer choice D is our answer.

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