There's no beating Judy's ham!

There's not much to say about this week's Glee. It's hard to stay with a timeline, as I can't find any point in this week's episode that specifically references an occurrence in the previous week's episode. There is a temporal clue, though, as we know that Finn is invited to the Fabrays' house for Sunday dinner. This does, of course, lead to more time-continuity issues, as Rachel appears to run into Susie Pepper in a restroom at school, immediately after dinner. That could be Monday morning, but then Rachel is wearing a different outfit before rehearsing with Mr. Schuester. Is that Tuesday?

Let's be generous, and assume that some of this week's episode happened in parallel with the previous episode's events. This seems unlikely, given that the gang would be spending a lot of time in wheelchairs, but if we don't start crediting the show a week here and there, we're going to hit Holiday Break in a hurry, and Sectionals (which are now "in a few weeks," by the way) will be happening around Valentine's Day. (This, as I alluded to last week, would actually be more in line with the schedule of show choir events in Ohio, but that's a whole other kettle of home-cured meat.) Sunday dinner is still going to happen after last week, but I'll assume that the final Schu-Rachel scene does take place on Monday (perhaps Rachel was slushied after lunch and had to change). That gives us a current date of December 7. Last day of school is the 18th, people, so sectionals better get here fast.

Back then, of course, if the fight lasted less than fifty rounds, we demanded our nickel back!

It's been a while since I put up trivia, and I've been stocking up on misses. Unfortunately, I've misplaced most of that stock, but here are a couple of the tougher ones from the last few weeks. Get these, and you too could be taking your local pub for a bottle of wine and a gift certificate every week.

1. According to Mattel, what is Barbie's last name?
2. Who is the only Major League Baseball player to hit a home run for both leagues in the All-Star Game? (Not in the same game, obviously)
3. What African dog breed is also known as "The Barkless Dog?"
4. What comedian introduced Johnny Carson as the new host of The Tonight Show on October 1, 1962?
5. What fruit forms the base for the liqueur creme de cassis?
6. Who was the only boxer to defeat John L. Sullivan by knockout?

It's beginning to look a lot like...

WARNING: Minor Glee spoliers ahead. If you haven't watched this week's episode, you may want to come back tomorrow. Also, there's a spot where I link to a blog post that links back here, so you could get caught in a Recursive Internet Loop if you're not careful.

Previously on Glee: The football team won its first game with three weeks left in the season, and then continued to practice for six more weeks. Meanwhile, Mr. Schuester announced that we were two weeks away from Sectionals, then led his merry band of singing, dancing teens through three weeks of mash-ups and throwdowns, all in preparation for the aforementioned Sectionals, which still haven't happened as of November 19.

They won't happen this week, either, as we find out that the team can't afford to rent the short bus to travel to Sectionals. There's cupcakes, and fighting, and more Artie than we're used to, which is a good thing, and more Brittany than we're used to, which is probably also a good thing. The plots all advance a little, but the main point of this episode is to showcase the Rachel-Kurt "Defying Gravity" performance, which is quite spectacular. Everyone feels good, everyone's happy, and there's no mention of fake pregnancy or any of the uncomfortable adult relationships or why anyone ever mentioned that Sectionals were in two weeks when most Ohio show choir competitions won't take place before next semester anyway.

So, what's the date? As usual, the show gives us one temporal clue, as Schu tells everyone to "be ready on Thursday" for the diva-off. We can assume, as usual, that all of the subplots and outfit changes take place over as short a time frame as possible, given the previous episode. The Thursday after November 19 would be the 26th, but that happens to be Thanksgiving. Thus, it has to be the following Thursday, December 3. We will note that the Cheerios have begun to practice indoors, so there is some attempt to keep the show in a universe in which Ohio actually gets cold in autumn and winter.

This week, it occurred to me that maybe the show wasn't totally serialized, and thus it could be earlier. After all, most of the subplots, particularly the various teen interactions, could easily have run in parallel. Kurt could join the football team and come out to his dad, and the whole Kristen Chenoweth thing could have been the week before, and that could still have been in a later episode, because they don't really have anything to do with each other. But every single episode has a moment, often just a few lines, that mark it as happening after the previous episode. This week, it's Quinn being kicked out of Cheerios. I get the need to tell the story as one continuous arc — it's easier for the viewer to keep track of everyone's relationships if we don't have to worry about whether today's events happened before or after the stuff we saw last week — but you have to account for the fact that time always moves forward. You can't just have everyone be a junior again for the second season.

Obviously, I'm more into the date-continuity issues of this show than most folks, but what really bothers me about them is how easy they would have been to avoid. If you don't say that the football team is 0-6, you can still claim it's early September. "Sectionals are in two weeks" was part of a throwaway line meant to show us that the kids weren't focusing, but details like that are exactly the sort of thing writers need to pay attention to. If you say "two weeks," you've committed yourself to a date, and that's a problem four episodes down the road, when you've had a month's worth of subplot, and still four more episodes until the actual competition.

And that's what you missed on Glee. Join us next week, as we'll find out whether women throw themselves at Will in spite of his v-neck cardigan/skinny tie ensembles, or because of them.

Why does it happen? Because it happens.

When I lived in Michigan, a friend of mine hosted Monday Night Football parties. I expect he still hosts them, I just don't live there anymore. Anyway, a bunch of guys would show up, and we'd spend the evening drinking beer and sort-of watching the game. Since we usually only had a passing interest in the game, we would turn quickly to other distractions, usually games. It was at these parties that I was introduced to a dice game my friends called "Scat." Scat is a variation of poker dice. The rules are as follows:

• The goal for each player is to make the best "hand" (actually, it's to avoid making the worst hand, but I'll get to that in a second). A hand is the best possible set of any one rank. Sets are ranked first by cardinality (the size of the set) and then by ordinality (the rank of the dice in the set); for example, three sixes is better than three fours, but four twos is better than three sixes. Only the set itself counts, the "kickers" are irrelevant, e.g. four twos and a six is the same as four twos and a three). Ones are wild, so a one and two sixes is really three sixes. However, a hand must have a one to be valid; a hand with no ones is "scat," and is worth nothing.
• Each player has three rolls to make a hand. After each of the first two rolls, the player may choose to stop rolling any or all of the dice. However, the player must roll a one before setting aside any dice, i.e., if the player has scat, he must re-roll all five dice. If a die is set aside after the first roll, it may not be re-rolled after the second roll.
• Play passes clockwise around the table. After each player has rolled, the lowest hand is eliminated, and the remaining players begin a new round. The honor of starting each round passes counter-clockwise: the last player to roll in the previous round rolls first in this round, the first player in the previous round now rolls second, and so on. When only two players are left, the game becomes best-of-three (the first player to have the low roll twice loses).
• If, at any time, a player's dice "stack" (i.e., one die lands on top of another), that player is immediately eliminated, and the round ends.
• If, at any time, a player rolls five ones, that player wins the game, regardless of how many players are remaining.

Obviously, this is largely a game of chance, so its main purpose was to shift small sums of money around a table. However, I noticed the occasional opportunity for skillful play. For example, a player who rolled two ones, two threes, and a six would usually keep the threes, thus keeping four threes instead of three sixes. However, if the low roll was four fours, that player would need to roll a one or three on the fifth die to avoid elimination. A player who kept the six in the same situation would still need to roll a one or six, but would have two dice, thus improving his odds.

This round of math lessons will focus on this game. I'll first point out some basic probabilities related to the game, and then delve a little into strategies in specific situations. Finally, I'll look at the game theory aspects, in an attempt to come up with a basic strategy.

Lesson 12: Why Bigger Isn't Better

In the final post on strongly asteroidal graphs, we look at asteroidal graphs other than S₃; namely, graphs III_n (for n > 2) and IV_n (for n > 1). Fortunately, changing these graphs into minimal strongly asteroidal graphs is not nearly as complicated as it was for S₃, for two reasons. The first, and most important, is that there is only one potential middle vertex (a₁) in the asteroidal triple, so only one construction need be applied. The second reason is that applying the category B constructions (and, by extension, category C constructions) fails to produce a light path. Here's why:

Suppose that X₁ has three consecutive a₁-heavy vertices x-y-z, with a₁-light vertices u between x and y, and v between y and z. If x and z are not adjacent, then this creates a copy of graph II (with u, v, and a₁ as the strongly asteroidal triple). If, on the other hand, we attempt to place a single a₁-light vertex adjacent to x, y, and z, then this creates a chordless cycle with a₁. Therefore, category B constructions are minimal only if X₁ contains just two consecutive a1-heavy vertices, so we need not apply them when n > 2. Further, applying any category B construction to graph IV₂ will create a k-sun, where k is 3, 4, or 5, depending on the adjacencies between the a₁-light vertices and the neighbors of a₁.

This means we only need to focus on the category A constructions. Applying construction A1 to any of these graphs creates a copy of graph I, except in the case of graph IV₂, in which case we get graph 9. Applying construction A3 to any of these graphs results in a copy of graph II. Only construction A2 creates new minimal graphs. For the graphs in family IV_n, the added vertex must be adjacent to both neighbors of a₁; if it is only adjacent to one of the neighbors, a copy of graph II is created. This gives us two families of minimal strongly asteroidal graphs; these are 55_n and 56_n below.

There is one final construction to consider for graph IV_n; adding a pendant vertex v somewhere along the path X₁ might create a new sAT, in which a1 becomes part of a v-light path between a₂ and a₃. This construction can only work for n > 2, because v cannot be adjacent to a neighbor of a₂ or a₃. However, for n > 3, this construction will create a copy of graph II. The construction does work for graph IV₃, but the graph it creates is graph 43. We are out of ways to modify the minimal asteroidal graphs, so we conclude that we have, at long last, found all of the minimal strongly asteroidal graphs.

We have also, at long last, concluded this topic. The math talk will continue, though. Next up is some probability theory, as I examine a dice game.