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.

Lesson 11: From C To Shining C

Welcome to the penultimate post on strongly asteroidal graphs. We've seen category A constructions, in which a neighbor is added to one of the asteroidal vertices of S₃, and category B constructions, which modify the path between two asteroidal vertices. Today, we will look at category C constructions, which do both.

In a category C construction, a vertex c₁ is added which is adjacent to a₁, b₁, b₂ and b₃, and a category B construction is added to the path X₁. In addition, edges are added between b₁ and every vertex in the category B construction. The modified category B construction still creates an a₁-light path, because there will not be consecutive vertices adjacent to c₁. Equivalently, this can be thought of as adding a regular category B construction, and then making c₁ adjacent to every vertex except a₂ and a₃. However, it is more convenient, in terms of adding later constructions, to modify the category B construction so that all of the vertices are adjacent to b₁. Each category B construction has a corresponding category C construction, so the category C constructions can intuitively be labeled C1-C4. The constructions are in the figure below.

Because c₁ is adjacent to b₂ and b₃, removing b₁ from construction Cn will make a graph isomorphic to one with construction Bn applied. Therefore, in order to create a new minimal sAT, it is necessary for b₁ to be part of another light path.

In construction C1, this means that a category B or C construction must be applied to either a₂ or a₃, but in constructions C2 and C4, b₁ is already part of at least one light path, which includes one vertex from the category B construction (it is difficult to see, but these paths are colored red and blue in the diagram).

Construction C1 does not combine with construction A3 (a parasol is created), but we do get new minimal sATs by applying construction B1 or B2 to the path X₂, and applying constructions A1, A2, or B1 to a₃ (or X₃); these are graphs 37-41. In addition, construction C1 can be combined with constructions B1 and B3 to create graph 42; construction B4 can be used instead of B3 to create graph 44, which we'll get to in a second. In the above figure, note that several of these graphs have been rotated/reflected.

Applying C1 a second time can lead to several new graphs. If the vertices c₁ and c₂ are not adjacent to each other, then the vertices v₁ and v₂ become unnecessary. If c₁ and c₂ are adjacent, but v₁ and c₂ are not, then v₂ is still unnecessary. Finally, we can allow c₁ to be adjacent to c₂ and v₂, and c₂ to be additionally adjacent to v₁.

Adding construction B1 or B2 to any of these combinations, or B3 to either of the latter two combinations, will create a new minimal strongly asteroidal graph. Graphs 43-45 are those that occur using construction B2, graphs 46-48 use B2, and graphs 49 and 50 use B3. The category A constructions and construction B4 do not lead to new minimal graphs.

Finally, there is one way to apply construction C1 to all three asteroidal vertices. The vertices c₁, c₂, and c₃ are adjacent to each other, and only one additional vertex v is necessary; it is adjacent to b₁, b₂, and b₃. This creates graph 51. Other modifications of the adjacencies between c₁, c₂, c₃, and v₁, v₂, and v₃ do not create new minimal graphs; the most common result is a 3-sun.

Only a few minimal graphs arise from the other category C constructions. Construction C2 actually is a minimal sAT by itself, as the adjacencies between b₁ and the B2 portion of the construction create a₂- and a₃-light paths. This is graph 52. Construction C3 does not lead to any new minimal graphs; because the only added vertex in construction B3 is already adjacent to b₁, the addition of vertex c₁ is unnecessary to create an a₁-light path. Finally, construction C4, like construction C2, contains an a₂-light path. Category B constructions that modify the path X₃ do not combine with this, for the same reason they do not combine with construction B4. However, constructions A1 and A2 can be applied to a₃ to make graphs 53 and 54.

These are the final minimal sATs that arise from the graph S₃. Next time, we'll finish up by looking at the infinite families of asteroidal graphs, and seeing how to modify those to create strongly asteroidal graphs.

Zoot Suit Riotous

WARNING: Minor Glee spoliers ahead. If you haven't watched this week's episode, you may want to come back tomorrow.

It's rehearsal time at McKinley, and we acknowledge the fact that we've had Sue as co-director for a week, but now that's over, and we're back on track for Sectionals. Apparently, Sectionals have been postponed, because they should have been this past weekend. Remember that, writers? When you said that Sectionals were "in two weeks," and then you packed two weeks' worth of "show" time into the next two episodes?

That doesn't matter, though, because now there's trouble brewing at football practice. Coach Tanaka has his knickers in a twist, and has declared that the team must practice this Thursday, when the glee club normally rehearses. The issue of why the team is still practicing at all, given that the football season ended three weeks ago, is not brought up. Remember that, writers? When you said the football team was 0-6, and then ran through two months without showing any more football? High school teams don't play into December. We will note that the students are beginning to wear cold-weather clothing.

Anyway, back to the stated premise of these posts. Let's assume that the original rehearsal is on the Monday following last week's episode (so, November 16). The Thursday practice happens. Let's further assume that the Rachel-Puck scene occurs at the end of that particular practice (even though that means Rachel changed outfits post-rehearsal), which will allow Finn to return to glee club on Friday. This packs a heck of a lot of subplot into one week, but as usual, we are generous here at the Preschool. The date is Friday, November 20. We've missed Halloween, and are getting ready for Thanksgiving break.

And that's what you've missed on Glee. It sounds like there will be a few weeks without new episodes, which might allow real time to catch up with show time. Sectionals, no doubt, will be postponed without explanation again. See you in a month.

Eight Days a Week?

WARNING: Minor Glee spoilers ahead. If you haven't seen this week's episode, you may want to stop reading now. Of course, if you have normal human eyeballs, it's probably already too late. Sorry.

So... was "Keep Holding On" the sectionals song?

As you'll recall, last week's episode began with sectionals two weeks away, and ended a week (plus a few days) later. So, presumably, sectionals will be the next weekend.

This week, we begin with Sue and Schu fighting like Mom and Dad on payday, and then there's a flashback. A lot happens: Sue splits up the Glee Club, there's practices for a singing competition, Will flunks all the Cheerios (except for Quinn and Santana and Brittany, apparently), we find out that Puck is one of the Chosen People (shalom!), there's a big fight in Principal Figgins' office.

We could try to figure out dates for all of these. For example, assuming the previous episode ended on Thursday, the original meeting with Figgins has to be, at the very earliest, Monday - surely, he's not trying to "take the temperature" after a single practice? Fortunately, trying to figure out how all of these events shoehorn into a few days is totally unnecessary, thanks (as usual) to the B-plot - Terri's increasingly absurd fake pregnancy. Without going into any additional commentary on the ideas, writing, and character development surrounding this donnybrook, I'll just point out that the ultrasound appointment takes place (per Will) on Friday, just after the big shouting match with Sue. Assuming everything else did take place in the course of a single week (and, since there is no mention of sectionals, we have to assume that), we have a date for these events - it's Friday, November 13.

And this is where the show (finally) begins to lose date continuity. Will comes into Sue's office, and they apologize, and talk about the set list. We still haven't seen sectionals, so this scene takes place... when? Saturday morning? The sectionals were, per previous exposition, scheduled for this particular weekend. The gang sings us out, but it's unclear if this is the competition, or a practice, or a daydream. So, Will and Sue either make up Saturday, before sectionals, or Monday after sectionals which we didn't see (given the competition, plausible), or sectionals got postponed and no one told us, or most of the McKinley students and faculty own time-turners.

And that's what you missed on Glee. Tune in next week for football three weeks after the end of football season, and perhaps Halloween decorations just in time for Thanksgiving.

I also make a better door than a window

A while back, our regular trivia host took sick, and was missing for two weeks. One of the regulars got in touch with him, picked up the clues, and read them in his stead. In an effort to help out, I offered to write and read a round of trivia, and was allowed to do so the second week. I came up with ten questions that I thought were great. Unfortunately, it turns out I'm a better trivia-taker than trivia writer; the winning group in my round got one (1) question right, and they were the only ones to get that right. To be sure, I did a poor job. Hard questions are fine, but they're for separating the wheat from the chaff; you have to throw a couple of softballs out to keep people playing. Surely, though, you can do better than one.

1. Name the Pep Boys.
2. In Calvin and Hobbes, what is Calvin's favorite bedtime story?
3. Who was the 23rd President of the United States, serving between Grover Cleveland's non-consecutive terms?
4. What dog won the first three Frisbee World Championships, which are now named after him?
5. In the Velvet Underground song, "Waiting for the Man," how much money is in the singer's hand?
6. How many Canadian provinces border at least one Great Lake?
7. Who was the first African-American woman awarded the Nobel Prize in Literature?
8. On which island did Napoleon die, on May 5, 1821?
9. What is the name of the theme music for The Daily Show?
10. What Dutch driver holds the Indy 500 records for fastest practice lap, qualifying lap, four-lap qualifying time, and 200-lap race?

Lesson 10: All This Has Happened B4

It has been two months since I posted lesson 9. There are a few reasons for the delay, but one reason is that the final category B construction has been a thorn in my side for months. I originally thought that it gave rise to no new minimal graphs. Then, I found one new sAT, then two, then three, then went back down to two, and so on. Every time I thought that I had thoroughly examined the construction, another combination occurred to me. Over the last month, I have examined each combination of constructions that included B4, one by one, and have officially convinced myself whether each combination does or does not produce a new minimal sAT.

Construction B4: The vertex v₁ is adjacent to a₂. In addition, a vertex u₁ is added, which is adjacent to v₁ and b₃. This creates the a₁-light path a₂-v₁-u₁-b₃-a₃.

The great difficulty in construction B4 lies in the fact that it lacks symmetry. With the previous category B constructions, adding a construction to a₂ was no different than adding it to a₃. For example, two combinations produce graph 16. With B4, this is no longer the case; when we apply other constructions, it is important whether they are applied to a₂ or a₃. In addition, if we apply B4 a second time, it matters which of the asteroidal vertices is adjacent to the vertex v₂.

One important thing to note is that applying this construction, and then removing the vertex b₂, is equivalent to just applying construction B1. To create a minimal sAT, the other two constructions must be applied in such a way that either b₂ is part of every a₃-light path, or that every a₂-light path would be heavy if b₂ were removed. This means that the construction applied to a₂ cannot be a category A construction, or construction B3. In addition, applying construction B1 or B2 to the path X₃ creates either a graph in which b₂ can be removed, or one with a chordless cycle. Applying construction A3 to either a₂ or a₃ creates a copy of the parasol.

If we apply a category B construction to the path X₂, the added vertices must be adjacent to v₁, in order to avoid suns and parasols. Construction B1 will create new minimal graphs when constructions A1, A2, or B3 are applied to a3; these are graphs 33-35 below. Construction B2 results in a graph in which u₁ is removable; in fact, it creates a category C construction, which we'll get to next time.

Attempting to apply B4 a second time creates problems. If it is applied to the path X₂, we must decide whether the vertex v₂ is adajcent to a₁ or a₃. If it is adjacent to a₁, then we do not get a minimal sAT, as (depending on the adjacencies) the resulting graph will contain suns, chordless cycles, or a category C construction. Suppose, then, that v₂ is adjacent to a₃. Then applying a category A construction to a₃, or construction B3 to the path X₃ is no longer possible (for the same reasons they could not be applied to a₂ or X₂ earlier). However, allowing v₁ to be adjacent to u₂ and v₂, and v₂ to additionally be adjacent to u₁, creates the a₃-light path a₁-b₁-u₂-v₁-a₂, while also preventing the removal of b₂ or b₃. This is, therefore, a minimal sAT; this is graph 36 at left.

Applying construction B4 to path X3 does not create new strongly asteroidal graphs; in any strongly chordal graph where this is applied, either a1 or b2 will be removable.

Next time, I'll start on the category C constructions. It's all downhill from here, kids. Two more posts max, and then we can move to a new topic.

What Time Is It, Mr. Fox?

If you're like me, you spend your Wednesdays from 9-10 p.m. watching Glee (if you're not like me, you still watch it, but you have a DVR, so you watch it later and without commercials). I love the singing, love the one-liners, and I'd watch Jane Lynch read "Sinners in the Hands of an Angry God," because I think she'd read it in a sardonic tone that would just crack me up the whole time. The relationships (all of them) are AWK-ward, but I prefer to view that as intentional, like it's a whole school of Michael Scotts. One thing has been bothering me, though. A couple weeks ago, I made the following comment over at ALOTT5MA:

I'm having trouble with the tiimeline for this show. If there's enough time passing that it's noticeable that Mr. Shuster only shows up for rehearsal once a week, where are we in the semester?

(This was referring to the "Acafellas" episode. And yes, I've misspelled "Schuester.")

Fortunately, the next week, we got an answer to exactly where we were: towards the beginning of the episode, Ken Tanaka made reference to the football team being 0-6. Assuming (generously) that it took one week for the following:

1. Kurt joins the football team as its kicker (Monday)
2. Kurt and Finn convince the entire football team to take dance lessons (Tuesday)
3. The team choreographs and rehearses a dance routine to "Single Ladies" to the point of presentability (Wednesday, Thursday)
4. The football team, with the help of their newfound dancing skills, wins their game on Friday.

Then the team has run its record to 1-6. High school football in the North generally starts the last Friday in August, so, assuming there were no bye weeks, the game took place on October 9. Finally, we have pinned down a date!

(NOTE: At ALOTT5MA, I said this game would have taken place on October 16, based on the fact that football starts the week before Labor Day. However, with the late Labor Day this year, that was not the case. This jibes with my own high school's football schedule.)

Having figured out what day it is for a specific episode, we can keep track of the chronology by estimating how long each new episode takes. Again, I want to be generous with these estimates - a lot more happens in a typical high schooler's day than most of us remember. Anyway, "The Rhodes Not Taken" begins with Rachel having quit the Glee Club to join the school play (this happened at the end of the previous episode). Let's assume that this happened on Saturday, at a weekend rehearsal. Then the episode opens at Monday's reharsal. In the course of this episode:

1. Will talks to Emma about having found April's transcripts, then gets in touch with April.
2. April returns to high school and meets the gang. This, obviously, couldn't happen until Tuesday.
3. The Glee Club rehearses two songs. Along the way, April wins over the teenagers by contributing to their delinquency. This, plausibly, takes place over the course of the rest of the week. Given the number of outfit changes, this can't just be Tuesday and Wednesday. This is important, because:
4. They perform two songs at their invitational. This, presumably, takes place on Saturday - given the previous assumption, it can't only be Thursday, and the football team (and Cheerios) would have a game on Friday. April leaves after the first song. Fortunately, Rachel has gotten fed up with Sandy Ryerson (after, what, two rehearsals? Three?), and re-joins Glee for the second song.

After this episode, I estimate the date as being October 17. We're now caught up to this week's episode, which begins with Mr. Schuester saying that sectionals are in two weeks, and that the kids have been coasting since last week, when they got the notice that their only competition would be a school for the deaf and an all-girls alternative school. I had originally given the benefit of the doubt and decided this announcement took place before the invitational, despite April's absence, but Rachel's presence (and outfit - she did re-join Glee for a day in the last episode, but quit again partway through the rehearsal) indicates that this is post-invitational, and thus we begin on October 26 at the earliest. Will then pits the boys and girls against each other in a singing competition, to be started, and I quote, "one week from today." The boys sing on Tuesday (so it was actually the 27th at the beginning of the episode), the girls sing on Wednesday, a drug scandal hits the program on Thursday. Sue Sylvester is now co-coach of the Glee Club as of, at the earliest, November 5. Two things to note from this:

1. It is still sunny and warm in Lima, Ohio - as evidenced by none of the Cheerios wearing tights or long-sleeved shirts under their uniforms.
2. The Ohio high school football playoffs begin on November 6. Even if the football team has won out to finish their season, they are, at best, 4-6, and not making the playoffs. Football season is over.

And that's what you've missed on Glee. Stay tuned in the following weeks, as I keep everyone up to date on the date!

Back to Bedlam

We're back, after a two-month hiatus. I hope that I have a little more energy to keep this thing going this time.

Anyway, the last couple of graph theory lessons are ready (oh joy!), I've got a stockpile of trivia questions, and there's one new piece that needs to be somewhere other than ALOTT5MA comment threads. Stick around.

Sports and Spice (but not Sporty Spice)

While we're waiting for the next graph theory lesson, here's some sports and music trivia to tide you over. We've been rocking the Monday trivia lately, so this ought to catch me up. The last few weeks have been in our wheelhouse, but the writer at this particular establishment is pretty in tune to that, so I expect things to get difficult in short order.

1. Who is the only player to hit a grand slam in the Major League Baseball All-Star Game?

2. In the forty-three Super Bowls to date, only once has either team failed to score a touchdown. Which team was it? (Bonus: which year?)

3. Which team lost Super Bowl II?

4. Which NHL team plays in an arena known as "The Igloo?"

5. Which island nation produces 90% of the world's cinnamon (as of 2006)?

6. What was the name of John Lennon and Paul McCartney's first band?

7. John Lennon was shot in front of what New York hotel?

Be Afraid. Be Very Afraid.

Lesson 10 is coming. Really. Re-checking this construction is a bear.

Sarek Knows Best

This week's trivia misses themes: science fiction and Oscars trivia!

 

1. According to starwars.com, how old was Jabba the Hutt?

2. What year did the original Star Trek first premiere on television?

3. What actress played Spock's mother on both the Star Trek television series and in the fourth film?

4. What science fiction novel, now one of the top-selling sci-fi novels of all time, won the 1985 Nebula Award for best novel?

5. What song, from The Poseidon Adventure, won the 1972 Academy Award for Best Original Song?

6. What Best Picture-winning comedy also earned Billy Wilder Academy Awards for directing and writing?

Lesson 9: Construction B3

We're still looking for minimal strongly asteroidal graphs, and it's time to move on to the next category B construction. Recall (from last time) that this construction will involve adding a vertex which is adjacent to all three interior vertices; we will call this vertex v₁.

Construction B3: v₁ is adjacent to both a₂ and a₃, thus creating the path a₂-v₁-a₃, which is a₁-light. Combining this construction with previous constructions is tricky. Adding constructions A1 or A2 to both a₂ and a₃ creates a copy of the bad aster. However, if we modify these constructions by making the added vertices adjacent to v₁ as well, this creates a minimal graph (if a₂ uses the modified construction, but a₃ does not, then b₂ could be removed to make a smaller strongly asteroidal graph). Similarly, adding construction B1 to both X₂ and X₃ creates a sun, unless we modify the construction by making the added vertices adjacent to v₁. This modification does not work for construction A3, however, as it creates a copy of the parasol.

If we combine constructions of type A with construction B1, we must still use the modified construction of B1; otherwise, we will create a graph in which a₁ can be removed to make a smaller strongly asteroidal graph. We may, however, use modified or unmodified type A constructions. Graphs 19-21 use only type A constructions with construction B3, and graphs 22-26 use construction B1 at least once.

Construction B3 can also be combined with construction B2. If vertices u₂, V₂, and W₂ are added to create the path a₁-b₁-u₂-v₂-w₂-b₃-a₃, then there is a parasol, unless v₂ and/or w₂ is adjacent to v₁. If only w₂ is adjacent to v₁, a chordless cycle is created, and if only v₂ is adjacent to v₁, a₁ and b₁ can be removed to make a smaller strongly asteroidal graph, regardless of the construction added to a₃ (which must be a type A construction, because B2 does not combine with B1, and applying B2 twice will produce a copy of graph 18). If both are adjacent, then removing a₁ and b₁ leaves a graph equivalent to combining B3 with B1 (with u₂ and v₁ standing in for a₁ and b₁). So, u₂, v₂, and w₂ must all be adjacent to v₁. Then, either a modified or unmodified type A construction can be added to a₃, producing graphs 27-30.

If construction B3 is applied twice, with vertex v₂ adjacent to a₁ and a₃ (as well as each vertex b_i), then v₁ and v₂ must be adjacent to avoid chordless cycles. Adding construction A1 or A2 to a₃ will produce a copy of graph 9, so we must add a type B construction to X₃ to obtain a minimal graph. Applying B3 a third time will result in a 3-sun, but both B1 and B2 will work, as long as all of the added vertices are adjacent to both v₁ and v₂. Construction B1 produces graph 31, and construction B2 produces graph 32.

Next time, we face the horror that is construction B4. For those who couldn't give a rip about all of this graph theory, there will be trivia posted tonight or tomorrow.

Lesson 8: B2, or not-B2

Welcome to the latest installment in the continuing struggle to find minimal strongly asteroidal graphs. We are creating these graphs by adding vertices to the minimal asteroidal graphs, as discussed in lesson 5. Specifically, we're adding these vertices to the graph S₃; see lesson 7 for the terminology I use in referring to the various vertices in that graph. Last time, we looked at category A constructions, which add a neighbor to the vertex a_i, in order to prevent it from being a middle vertex. In today's lesson, we'll start to look at the category B constructions, which involve adding vertices to the path X_i to make it a_i-light. The number of vertices we can add is limited somewhat; I'll spare the details, but an a_i-light path in a minimal sAT will have a maximum of three a_i-heavy vertices. This is a small result, but it is important, because it a) means that there are only a small number of constructions in this category, and b) it places an upper bound on the length of the a_i-light paths, which in turn places an upper bound on the order of the minimal graphs we can get by adding vertices to S₃, which means the number of sporadic cases is finite. Before we begin, recall that, when a construction of category B (or C) is applied, the added vertices can be adjacent to the added vertices from other constructions from category B or C. This adds a level of complexity not found in the category A constructions.

Construction B1: a single vertex u₁ is added, adjacent to b₂ and b₃. This creates an a₁-light path, and thus prevents a₁ from being a middle vertex. If this construction is applied twice, say by adding a vertex u₂ to prevent a₂ from being the middle vertex, then the new vertices cannot be adjacent: this would create a chordless cycle u₁−u₂−b₁−b₂−u₁, and adding a chord between b_i and u_i (where i is 1 or 2) would make u_i a_i-heavy, defeating the purpose of the construction. Then B1 cannot be applied to all three paths, as this will create a 3-sun. It may be applied up to twice, though, in combination with constructions A1 or A2, to produce graphs 10-14. Applying constructions A3 and B1 together produces a copy of the parasol, so this is not minimal.

Starting with construction B2, the remaining constructions all involve adding one or more vertices which are adjacent to all three interior vertices, and to other vertices (possibly one or more of the asteroidal vertices) to create a_i-light paths. The other constructions of type B involve adding only one such vertex; we will call this vertex v₁.

Construction B2: In addition to v₁, vertices u₁and w₁ are added, such that u₁ is adjacent only to b₂ and v₁, and w₁ is adjacent only to b₃ and v₁. Then a₂−b₂−u₁−v₁−w₁−b₃−a₃ is a₁-light. This construction can be applied once, in combination with constructions A1 and A2, to produce graphs 15-17. Combining B2 with A3, by adding a neighbor to a₂, does not produce a minimal graph: this combination will have a copy of the parasol, unless the vertex added to a₂ is adjacent to v₁. But, in this case, a₁, a₂, and w₁ are in a strongly asteroidal triple, so a₃ can be removed to make a smaller strongly asteroidal graph (it is, in fact, graph 35, which we will see later). Combining B2 and B1 (by adding u₂ to the path X₂) also fails to make a minimal graph; adding any of the previous constructions to a₃ or X₃ will produce a graph in which a₂ (at least) can be removed to make a smaller strongly asteroidal graph. Applying B2 a second time creates an irregular construction. If vertices u₂, v₂, and w₂ are added to create the path a₁−b₁−u₂−v₂−w₂−b₃−a₃, then there is a 3-sun unless v₁ and v₂ are adjacent. Adding construction B1 to X₃ will produce a sun. Adding a construction of type A to a₃ creates a copy of either graph 13 or 14 (in which u₁, u₂, and a₃ are a strongly asteroidal triple), unless u₂ is adjacent to v₁ and/or u₁ is adjacent to v₂; we may assume the former. In this case, though, v₂ and w₂ (and the construction added to a₃) can be removed to make a smaller strongly asteroidal graph. Specifically, this is graph 18, in which, for 1 ≤ i ≤ 3, there is a vertex u_i adjacent only to b_i and v₁. Each u_i is a_j-light for j ≠ i, and so for each i, there is an a_i-light path between the other two asteroidal vertices. Applying B2 a third time creates a graph which contains either a sun, or a copy of graph 18, so this does not give us an additional minimal graph.

That wasn't so bad, was it? Things get tougher, though, with construction B3, and I may need to spend the weekend re-checking my notes before I'm ready to show you B4.

Lesson 7: A is for Adjacent

In the previous lesson, I broke down the categories of construction which we can use to change an asteroidal triple into a strongly asteroidal triple. Today, we'll examine the category A constructions. Right now, we're only looking at S₃; let's label the vertices of the asteroidal triple as a₁, a₂, and a₃, and let b₁, b₂, and b₃ be their respective neighbors. Then the path X₁, for example, would be a₂-b₂-b₃-a₃. Each path X_i is a_i-heavy; our goal is to add vertices in such a way as to make either make this path a_i-light, or to create a new path that is a_i-light.

The category A constructions involve adding a single vertex adjacent to the vertex a_i, and possibly to additional vertices, specifically to prevent a_i from being a middle vertex. There are three constructions in this category, each of which involves adding only one vertex, which we will call v.

Before we go further: when describing each construction, in this lesson and in future lessons, I will assume that vertices are being added for the explicit purpose of preventing a₁ from being a middle vertex unless otherwise noted. Some constructions may affect other vertices or paths, and some constructions have special considerations when they are being used more than once; I will discuss these considerations when they come up.

Construction A1: v is a pendant vertex; in other words, the v is adjacent only to a₁. This means that a₁ is no longer simplicial, and thus cannot be a middle vertex. This construction can be applied to all three vertices to produce a minimal sAT; this is graph 3, shown below.

 

Construction A2: v is adjacent to both a₁ and b₁. While a₁ is still simplicial, v is not adjacent to any vertex on the path X₁, which makes this path a₁-light. This construction can be applied to all three vertices, to produce graph 4; or, it can be applied in combination with construction A1 to produce graphs 5 and 6, shown below.

Construction A3: v is adjacent to both a₁ and b₁, and to one of the other interior vertices; in the example at left, v is adjacent to b₃. Like construction A2, this makes the path X₁ a₁-light. However, the path a₁-v-b₃-a₃ is a₂-light, so this construction also prevents a₂ from being a middle vertex! Thus, only one more construction is necessary to create a minimal sAT. If we attempt to combine this with construction A1, then we create a graph which contains the bad aster. However, combining it with construction A2 creates graph 7, below. Construction A3 can be applied twice, but the second added vertex must be adjacent to the same interior vertices (in the example above, b₁ and b₃) as v  to create a minimal sAT. The second added vertex may be adjacent to v or not. If it is not, we create graph 8; if it is adjacent to v, we create graph 9.

Other adjacencies for v do not create new constructions in this category. If v is adjacent to both b₂ and b₃, then we create no new light paths, so adding v would not prevent any of the asteroidal vertices from being the middle vertex. If v is adjacent to any of the other asteroidal vertices, then this would create a new light path, but the other adjacencies required (to keep our graph strongly chordal) will actually fall under category B or C. In this vein, construction A3 might be considered a category B construction; I have placed this construction in category A because it fits the definition of adding exactly one vertex, which is adjacent to exactly one asteroidal vertex, and which prevents that asteroidal vertex from being a middle vertex.

Next time, we'll get started on the category B constructions. Because each new construction can be combined (in most cases) with previous constructions, I'll probably break this category up into two or more parts. This also means fewer breaks if I want to get through this, so maybe it will have the added benefit of speeding up the lessons.

He only dreamed of places now and of the lions on the beach.

As I continue to catch up on our trivia misses, this week's theme is Famous Characters in Literature (and other media). The Lone Ranger's nephew's horse is still not a question.

1. Which family did Hazel work for?

2. Who is the superhero alter ego of Steve Rogers?

3. What is the name of Andy Capp's wife?

4. What is the name of Oscar Madison's wife?

5. What well-known novel features a character named Aarfy Aardvark?

6. Who is Tom Sawyer's girlfriend?

7. In The Old Man and the Sea, what is the old man's name?

Lesson 6: There is NO lesson 6

We now know a few of the minimal forbidden subgraphs for the co-TT class of graphs:

1. Chordless cycles (co-TT graphs are chordal)
2. k-suns (co-TT graphs are, in fact, strongly chordal)
3. The bad aster
4. The parasol

Last time, we learned that we can find the minimal strongly asteroidal graphs by modifying the minimal asteroidal graphs to create light paths. Let a₁, a₂, and a₃ be our asteroidal triple, and X₁, X₂, and X₃ the paths connecting them, as described previously.

The only graphs we need concern ourselves with are graphs from the III_n and IV_n families, where n > 1 (graph IV₁, recall, is the 3-sun). We will begin with graph III₂, which is S₃. This graph is unique, because of its rotational symmetry; any of the three asteroidal vertices could be a middle vertex (the rest of the graphs which we have to consider have only one middle vertex, and only one heavy path). This means we must create light paths for each of these vertices. There are several constructions that we can use to prevent an asteroidal vertex a_i from being a middle vertex. I have grouped these constructions into three categories:

• Category A constructions involve adding a vertex adjacent to a_i. This vertex may or may not be adjacent to any neighbors of a_i. In fact, adding a pendant vertex to a_i is the easiest way to prevent a_i from being a middle vertex, because it is no longer simplicial.
• Category B constructions modify the a_i-heavy path X_i by adding a_i-light vertices. When category B constructions are added to more than one of the asteroidal vertices, we also gain the option of adding edges between the newly added vertices. This has to be done carefully; many graphs produced by these constructions are not strongly chordal.
• Category C constructions combine the other two constructions, adding a new neighbor to a_i which is adjacent to the a_i-heavy vertices on X_i, and then adding vertices to X_i which are not (all) adjacent to this new neighbor. These are the most difficult constructions to find, but in at least one case, applying this construction to one asteroidal vertex prevents the entire triple from having a middle vertex.

We'll start with the category A constructions next time. They are fairly simple, so we should be able to get through all of them in one go. This will not be the case for the other two categories.