## June 2007, Game 2, Question 8

### Transcript

Question A is a local question that provides us with a rule. And it's also a numbers question, because it's asking us about the maximum number of films that you could show under this new rule. Like with most local questions, we're gonna begin with a sketch, starting with what we know from before. Now a lot of people have trouble with this specific if condition or rule that they've given us cuz it says, if L is never shown again once G is shown.

So what that means is if I show G here on Thursday, well then I'd have to show L in front of it. Because we already know that H is always last on Thursday, so if I showed you there on Thursday, I would no longer be allowed to show L. But if I wait to show G on Saturday, then I could throw some extra L's in front of it.

Heck, I could even throw an extra L in front of it on Saturday, so that's what the rule means, once you show G you can't show L again. Now the next challenge here is going to be we're trying to get the maximum number of films. So there's really two ways we could do that, and we do need to investigate both if we're gonna get the right answer.

One way to do it would be to try to delay G until as long as possible, and then use as many L's as we can while holding off on that G. The other option would be to try to make as many G's as possible, so throw that L in really early and then try to use as many G's as possible. Those are mutually exclusive goals because using one is gonna stop us from using the other, so we're just gonna need two sketches.

In both sketches H is gonna be the last one on Thursday. So in this top sketch let's maximize our L's, we're gonna put G as late as possible which would mean putting him there on Saturday. So with G on Saturday, we're free to use an L on Thursday in front of H, give an extra space. We're free to use an L on Friday, and we're even free to use an L on Saturday.

So right now we're up to five films, we'll see if there's any other films we could throw in here. We've used up our G's, we've put an L on every day, the only thing left is H. Now because G is on Saturday, we can't put another H there, but we could put an H in front of L on Friday. And doing that, we're up to six films, so we can get rid of A, B, C, and we've got D and E left.

Now, we have to consider the other scenario, what happens if I maximize my G's? So to do that, I'm gonna have to put the L in front of the G on Thursday. Now we can no longer use L's, but we can use as many G's as we want. So we can put a G on Friday, we can put a G on Saturday, and once again, we're out of L's and we're out of G's.

We check to see if there's anywhere else we could put an H and once again, we could put an H on to Friday in front of that G. H will be blocked from Saturday because the rule says no G H on Saturday. So if we do that it looks like one, two, three, four, five, six, we're at six again, so either way we do it, the maximum ends up being six and not seven.

Read full transcriptSo what that means is if I show G here on Thursday, well then I'd have to show L in front of it. Because we already know that H is always last on Thursday, so if I showed you there on Thursday, I would no longer be allowed to show L. But if I wait to show G on Saturday, then I could throw some extra L's in front of it.

Heck, I could even throw an extra L in front of it on Saturday, so that's what the rule means, once you show G you can't show L again. Now the next challenge here is going to be we're trying to get the maximum number of films. So there's really two ways we could do that, and we do need to investigate both if we're gonna get the right answer.

One way to do it would be to try to delay G until as long as possible, and then use as many L's as we can while holding off on that G. The other option would be to try to make as many G's as possible, so throw that L in really early and then try to use as many G's as possible. Those are mutually exclusive goals because using one is gonna stop us from using the other, so we're just gonna need two sketches.

In both sketches H is gonna be the last one on Thursday. So in this top sketch let's maximize our L's, we're gonna put G as late as possible which would mean putting him there on Saturday. So with G on Saturday, we're free to use an L on Thursday in front of H, give an extra space. We're free to use an L on Friday, and we're even free to use an L on Saturday.

So right now we're up to five films, we'll see if there's any other films we could throw in here. We've used up our G's, we've put an L on every day, the only thing left is H. Now because G is on Saturday, we can't put another H there, but we could put an H in front of L on Friday. And doing that, we're up to six films, so we can get rid of A, B, C, and we've got D and E left.

Now, we have to consider the other scenario, what happens if I maximize my G's? So to do that, I'm gonna have to put the L in front of the G on Thursday. Now we can no longer use L's, but we can use as many G's as we want. So we can put a G on Friday, we can put a G on Saturday, and once again, we're out of L's and we're out of G's.

We check to see if there's anywhere else we could put an H and once again, we could put an H on to Friday in front of that G. H will be blocked from Saturday because the rule says no G H on Saturday. So if we do that it looks like one, two, three, four, five, six, we're at six again, so either way we do it, the maximum ends up being six and not seven.