It’s difficult to do this in gaming, though. If one character is always the star of the story, the other players will feel neglected. Even if Willow gets as much stage time as Buffy, and even if she gets significant personal story arcs and love interests, the show is still about the slayer. Certainly, when one player is the one whose character is consistently featured in flashbacks and solo missions, the other players tend to lose out. It can be unfulfilling when a player character’s whole purpose is to tell a story about another player character.

In my experience, then, and in the vast majority of what I’ve heard from others, RP groups tend to use the “ensemble cast” model. Plots are tailored to try to involve every character as equally. Any given scene might focus on one character over the others, but part of the GM’s job is to make sure that everyone gets an opportunity to shine. This is much rarer in fiction, especially done well, but we do occasionally see it happen. One example is Battlestar Galactica. It’s difficult to say whether the main character of the series is Starbuck, President Roslyn, Gaius or one of the Adamas. Each regularly gets the spotlight, and the dozen or so other major characters gets it fairly frequently too. In that sense the show really is about the whole cast.

So what makes it so difficult to do this in a game? Unless the characters are designed to have explicitly harmonious sets of skills and goals, setting up a story that plays to everyone’s motivations very often winds up being watered down: only tangentially interesting to each character, or else tremendously contrived (“Tarek, you need to collect an important spell component from the eastern forest. The forest is near the tribe of orcs that used to terrorize your home town, Mika, and Nale, you realize that this is the forest rumoured to be the home of the most revered elder druid in the kingdom.”). Otherwise, we often use generic adventure plots, so that each character is equally un-special and can find their own motivation for rescuing the princess from the trap-filled dungeon of the evil monster that has been turning the villagers into thralls who polish its pile of treasure while the monster plans to unravel the fabric of existence.

Wouldn’t it be great to be able to play with the sort of dynamic we find in books and on the screen? As a player, I love to spend time wrestling with my character’s personal demons. I’ve played with groups where this worked out really well, because the other players loved that too. But not all players are happy to sit by for half an hour while one character takes another’s confession.

One solution is to run a game with just one other player. I’ve done this, and it works really well. The GM doesn’t feel guilty for dedicating the entire session to one character, or for letting them spend it all soloing in a VR rig or dream sequence. This is suboptimal for other reasons, though, chiefly because one of the main attractions of gaming is often the group dynamic, doing something fun with a bunch of other people, and it’s nice for there to be more than just the two of you.

So I was chatting recently with the GM of the Starblazers game I’m playing in, specifically about this. We came up with a couple of ideas, and I’m hoping to implement something like this at some point.

**Penumbra Model:**PCs are supporting cast for an NPC protagonist, who is never centre-stage.Here, the PCs’ actions drive the story, but the plot revolves around the NPC. One way to run this would be for the protagonist to not really be all that competent, so that the PCs are the real heroes — although they may never get recognized for it. Or the protagonist could be constrained in some way, unable or not allowed to do the really interesting work.

I can see this working really well, or else getting tiresome. You might inadvertently wind up with a “band of operatives” dynamic, where the PCs all work for some third party and go on designated missions without necessarily being personally motivated to do so. Not that there’s anything wrong with playing a party of FBI operatives or interstellar couriers, but it doesn’t tend to produce plots characters are very personally invested in. Chalk this suggestion up to “maybe”.

**Rotating Spotlight model:**Each session, a different character is explicitly designated as the hero, and the others as supporting cast.The idea here is to let the GM develop stories around one of the characters at a time, and let the dynamic of the session build around it. I think it may well be more satisfying to everyone if there’s a goal that is really important to one person, rather than spreading the focus around too much. And unless the party is totally randomly thrown together (and perhaps even then), there will always be plenty for the supporting players to do. If this week the gunslinging Jack Hustler goes back to Montana to face his fear of ghosts and finally put his father’s spirit to rest, at first glance the other characters are out of their element. But the big-city gumshoe Duncan Dirk is the first to notice that Jack Senior was the victim of foul play, the smooth British socialite John Howard convinces Jack to stick to his guns when the chips are down, and the Shinto-Jesuit monk Friar Takanawa is pivotal in granting the restless spirit its repose.

This may work really well. In fact, BSG did something like this. Listen to the opening sequence of any episode. Usually the actor who says “previously, on Battlestar Galactica” is the one who is in the spotlight for the episode. If your group likes cheese, you could even have the spotlighted character of the week do a quick recap at the beginning of the session.

So I think I’m going to try implementing the “rotating spotlight” method. There are a few details to work out, such as:

- Do the characters rotate in order through the spotlight, is it randomly chosen, or do I choose based on what plots I have brewing? Either way, I’d need to keep track of how many times each has been in the spotlight.
- Do I tell the characters beforehand which of them will be the star?
- Is it necessary to sweeten the deal a bit for the supporting characters? For instance, in FATE I might give each supporting character an extra Fate Point or something.

If you’ve read this far, cool. Please let me know if you have any ideas on this.

]]>Anyway, most of the work is done, and my German is now a little more fluent. That is, up from “barely at all”. Most of the language is translated, leaving mostly just notation, plus a couple of technical terms like *Asymptotenpuncte* (asymptotic point) and *Ebenenbüscheln* (sheaves). You can see the original and the results here:

Steiner, J. – Über die Flächen dritten Grades [1857] (PDF)

Steiner, J. – Über die Flächen dritten Grades (text, edited OCR, German)

Steiner, J. – On surfaces of the third degree (text, edited and Google-translated OCR, English)

It’s some cool stuff, and it kinda makes me look forward to learning to do Math in German. But holy shit was this a pain in the ass.

]]>- Thursday, August 6th. A shopkeeper in Pakistan’s southern Sindh province claims that a 60-year-old muslim woman “disrespectfully flung about pages of the Koran at his shop”, according to the BBC. The woman claims “it wasn’t the Koran she flung to the ground but a register in which the shopkeeper had listed her credit”. A crowd of people gathered at the woman’s house and threw rocks at it until police broke it up.
- Two days earlier, Tuesday, August 4th. The owner of a leather factory in Punjab province takes an old calendar down from the wall. Unfortunately for him, a lot of calendars contain verses from the Koran. A factory supervisor got incensed and stirred up the workers and local residents. The owner and one other person were killed in the ensuing violence.
- Three days before that, Saturday, August 1. A week after three christian youths in Gojra, eastern Pakistan are accused of burning a copy of the Koran, clerics call for their deaths. Fanatics stream in from surrounding districts. 40 houses belonging to christians are burned down, killing at least eight people, and possibly dozens.

So what’s happening here? According to local officials, none of the allegations were credible. But the BBC reports that there is “recurring evidence” that people are using Pakistan’s blasphemy laws and, in the case of the 40 christian houses, the government’s indifference to the rights of non-muslims, to settle personal scores by falsely accusing people of blasphemy. A shopkeeper and a customer have a dispute over accounts, and so the shopkeeper intimidates the customer by having people throw rocks at her house. Factory workers have a beef about their wages, so they incite violence and kill their boss. People don’t like having 50,000 christians in their town so they send thousands of them running for their lives.

This is what happens when governments pander to religious nuts. This is what happens when a government takes a stance that says “this portion of our population is better than the rest.” The saner religious folks are fine, but by and large they condone the extremists and people die. Whether it’s Najeeb Zafar in the Pakistani leather factory, or George Tiller in a Kansas church, or any number of genocides and “ethnic cleansings” around the world, or Aqsa Parvez right here in Toronto, religion kills.

]]>Today I have a proof I wrote recently, of a relatively fundamental fact: principal congruence subgroups of the modular group contain no elliptic elements.

The proof is available as a PDF or below the fold.

**Theorem.** The congruence subgroup $${}_{m}hatGamma$$ contains elliptic elements if and only if $${}_{m}hatGamma = Gamma$$, the modular group.

*Proof.* In three parts.

- It is easy to see that $$Gamma$$ contains elliptic elements. For instance, the following matrix is elliptic, with fixed points $$pm i$$:

$$! left[begin{array}{cc} 0 & -1 \ 1 & 0 end{array}right] $$ - Assume that $$m>1$$. Let:

$$! P = left[begin{array}{cc} a & b \ c & d end{array}right] = left[begin{array}{cc} alpha m + 1 & beta m \ gamma m & delta m + 1 end{array}right] in {}_{m}hatGamma quad ; quad alpha, beta, gamma, delta in mathbb{Z} $$The characteristic polynomial of $$P$$ is $$cz^2 + (d-a)z – b$$. Thus, using the fact that $$ad-bc = 1$$, the discriminant is:

$$! (d-a)^2 + 4cb quad = quad (a + d)^2 – 4 $$To show that $$P$$ is elliptic, we aim to establish that its fixed points are two distinct complex conjugate points, ie, that its discriminant is negative. For this to be true it is necessary and sufficient for the trace $$(a+d)$$ of $$P$$ to be between -2 and 2. Since all variables here are integers, we make use of the stronger statement:

$$! (alpha + delta) m in { -3, -2, -1 } $$This implies that $$m | 1, m | 2$$ or $$m | 3$$. The only possibilities are that $$m in {2,3}$$ and $$alpha + delta = -1$$.

- Now we suppose that $$m in {2,3}$$, and return to:

$$! begin{align} ad – bc & = 1 \ (alpha m + 1)((- alpha – 1 ) m + 1) – beta gamma m^2 & = 1 \ (-alpha^2 – alpha – beta gamma) m^2 – m & = 0 \ (-alpha^2 – alpha – beta gamma) m & = 1 end{align} $$This implies that $$m | 1$$, a contradiction.

Therefore, the only principal congruence subgroup that contains elliptic elements is the modular group itself. **QED**

The implications of this fact have yet to be fully realized. The question is still open as to whether it is possible to generate all the congruence subgroups of a given level, but the notion of decomposing subgroup generators into constituent generators, then manipulating those one by one is an interesting prospect.

]]>Briefly, I’m exploring properties of subsets of the modular group, defined as the set of 2×2 matrices with determinant 1. I work with an eye toward the problem of generating congruence subgroups, which is not adequately solved. It is possible to check whether a given group is a subgroup, but it has shown to be difficult to work in the opposite direction.

This week I’m delving into Topology. Any modular subgroup can be described as a polygon on the hyperbolic plane, which in turn can be turned into a Riemann surface by identifying corresponding edges (see for example Wohlfahrt). Of particular interest is the nature of the covering of that surface. Read beneath the fold for diffeomorphisms.

The following is adapted from Schlichenmaier, pp.13-14:

First, suppose we are given two differentiable manifolds and an isomorphism between them:

$$!(M,T), (M’,T’)$$

$$!f : M rightarrow M’$$

Now we take a point in the first manifold, and its image in the second. We take a neighbourhood around each of these two points and define the function that maps each neighbourhood to coordinates in the reals:

$$! a in U subset M quad ; quad b = f(a) in V subset M’ $$

$$! phi : U rightarrow mathbb{R}^n quad ; quad psi : V rightarrow mathbb{R}^n $$

Here $$(U,phi)$$ and $$(V,psi)$$ are known as “coordinate patches”.

Now, we can define a function that takes values in $$mathbb{R}^n$$, maps them into $$U$$, then $$V$$, then back to $$mathbb{R}^n$$, like so:

$$! f_{UV} : mathbb{R}^n rightarrow mathbb{R}^n $$

$$! f_{UV} = psi circ f circ phi^{-1} $$

This of course is defined on all the coordinates of the pre-image of $$V$$ in $$U$$ by $$f$$. It gives us $$n$$-dimensional real values that are coordinates of the image of $$U$$ in $$V$$.

Now, if the mapping $$ f_{UV} $$ is differentiable for all possible choices of $$a$$ and $$b$$, $$f$$ is called a differentiable map. If this is true of the inverse function as well, then $$f$$ is a **diffeomorphism**.

Diffeomorphisms help us classify manifolds. It’s easy to see that we can define an equivalence relation based on diffeomorphisms, and therefore divide manifolds into equivalence classes. The special case occurs when we’re talking about compact two-dimensional orientable manifolds (which includes all 1-dimensional complex manifolds). For these, there is exactly one diffeomorphic class for each genus. This is why we can classify such surfaces by genus, and why we get bad jokes like the following:

An employee at a coffee shop filled a topologist’s order, a coffee and donut. The topologist looked up, distraught, and asked, “But which is which?”

In higher dimensions, this is not true. Schlichenmeier gives the example of the 7-sphere, which has 28 inequivalent structures. Fortunately, I am only dealing with Riemann surfaces (1-dimensional complex) here

Finally, when dealing with complex manifolds, if all of our $$ f_{UV} : mathbb{C}^n rightarrow mathbb{C}^n $$ are holomorphic (analytic), and $$f$$ is bijective and holomorphic in both directions, then we call the map $$ f : M rightarrow M’ $$ an **analytic isomorphism**.

Here are my badges — so far:

]]>Show that $$n^2 – n$$ is even, for any given integer

n.

And here’s the proof: we can easily see that $$n^2 – n = n(n-1)$$. Since either $$n$$ or $$n-1$$ must be even, and the product of an even number with any other number is even, their product must also be even.

This sort of problem has a property similar to many geometry problems, which is that the original statement is easily comprehensible to people without a mathematical background. In this case, the solution also is widely accessible. There are some more complex proofs, though, and I’ve put one below the fold if you’re interested.

It is impossible to pick two integers a and b greater than two such that $$2^b – 1$$ divides evenly into $$2^a + 1$$.

This is what I mean: I think this is a statement that is easily understood by anyone who remembers any 8th-grade math. Finding a proof of this, though, is trickier. We use an iterative approach, manipulating the expression until we get results.

First of all, for the sake of brevity, I’m going to use a bit of shorthand. When I say $$x|y$$ what I mean is “x divides evenly into y”. This is the same as saying $$y = px$$ for some integer value of p. The opposite of this is $$xnot|y$$. Now the proof itself.

*Proof.* Let a and b be any (not necessarily different) integers greater than 2.

Suppose for the time being that $$2^b – 1|2^a + 1$$. We are going to accept this assumption and show that it leads to a contradiction. This will prove that the assumption is false.

Now,

$$!begin{align}2^a+1 & = (2^{a-b} times 2^{b}) + 1 \ &= (2^{a-b} times 2^{b} – 2^{a-b} + 2^{a-b}) + 1 \ &= 2^{a-b}(2^{b} – 1 + 1) + 1 \ &= 2^{a-b}(2^{b} – 1) + (2^{a-b} + 1)end{align}$$

Since we’ve assumed that $$2^b – 1$$ divides evenly into $$2^a + 1$$, we must be able to subtract any multiple of the first from the second, and the result will still be divisible by the first (if this is not obvious, try it with some integers and you’ll see why it works). So we can subtract $$2^{a-b}(2^{b} – 1)$$ from the above expression, and it should still be divisible by $$2^b – 1$$.

Here, we’ve shown that if $$2^b – 1|2^a + 1$$ then $$2^b – 1|2^{a-b} + 1$$.

Now, we have three possible cases:

- $$a-b=b$$

In this case, we have $$2^b – 1|2^{a-b} + 1 = 2^b + 1$$. We can do the same thing as above, where we subtract a multiple of the first number, to get:

$$!2^b – 1|(2^b + 1) – (2^b – 1)$$

$$!2^b – 1|2$$

This means that $$2^b – 1$$ is a factor of 2, namely either 1 or 2. Thus $$2^b = 2$$ or $$2^b = 3$$. 3 is impossible if $$b$$ is an integer, and 2 is only possible if $$b=1$$, but we know that $$ b ge 3 $$ so this is impossible. - $$a-b < b$$

Since $$2^b - 1|2^{a-b} + 1$$, the second term must be equal to or greater than the first. We can't have $$x|y$$ if $$y < x$$. So this gives us:

$$!2^b - 1 le 2^{a-b} + 1$$

$$!2^b - 2^{a-b} le 2$$

Now since $$2^b ge 8$$, this means $$2^{a-b} ge 6$$ and therefore $$a-b > 2$$. But we divide the last line by two to get:

$$!2^{b-1} - 2^{a-b-1} le 1$$

Note that this difference cannot be zero, since the exponents are not equal. If both $$b-1$$ and $$a-b-1$$ are greater than 0 (and not equal), then their difference will be a multiple of 2. But it clearly is not. So one of the two must be less than or equal to 0. Which means either $$b$$ or $$a-b$$ is less than or equal to 1. It can't be $$b$$, because we know it's greater than 2. And it can't be $$a-b$$ because we already determined that it's greater than 2. So we have another impossibility. - $$a-b > b$$

This is the only possibility left, therefore it must be true (if our initial assumption is true).

Now we have $$2^b – 1|2^{a-b} + 1$$, and we know that $$a-b > 2$$, since it’s greater than $$b$$. Thus, we can plug $$a-b$$ into this whole process, in place of $$a$$.

Note that $$a-b$$ is smaller than $$a$$ since $$b$$ is positive. So when we wind up at this same step, with a result of $$(a-b) – b = a-2b > b$$, we’re dealing with an even smaller number. We keep going, and eventually we get a number $$a-nb$$ that is smaller than or equal to $$b$$ (I’m not going to prove that this is inevitable, but you should be able to convince yourself of this). This leads us to possibility 1 or 2, above, which is a contradiction.

Now the contradiction cascades back through the proof. Since a contradiction is generated in one of those three options, the conclusion that $$a-nb le b$$ must be false, therefore it cannot be the case that $$2^b – 1 | 2^{a-nb} + 1$$. Thus it cannot be true that $$2^b – 1 | 2^{a-(n-1)b} + 1$$ and so on, back through the iterations of the proof. Thus it cannot be true that $$2^b – 1 | 2^{a-b} + 1$$, therefore it cannot be true that $$2^b – 1 | 2^a + 1$$. This was our initial assumption, therefore we have proved the opposite.

Now, the final step is to generalize the whole thing. Sure, this may work for some choice of $$a$$ and $$b$$, but is this always true? Well, the proof does not depend on the values of the variables, except for the restriction that they both be greater than 2. So regardless of what values you put in, the original statement (in blockquote) holds. $$blacksquare$$

One final thing: why do we require that $$a$$ and $$b$$ be greater than 2? Because you can find counterexamples otherwise. For instance, if $$ a = 3, b = 2$$, then the statement is false. And if $$b = 1$$ then the statement is always false.

Anyway, I hope I haven’t bored you too much. I find this sort of work very interesting, and this is just the basic groundwork of the field of Number Theory. Some of the more complicated theorems to do with, for instance, prime numbers, are very deep. Hopefully I’ll succeed in getting to work on this as a research project over the summer, and I can write more about it.

]]>In the latest issue of The Underground, UTSC’s campus paper, Erica Rodrigues writes “When freedom of speech goes too far”, a spittle-flecked protest of Peezy Myers‘ now-infamous host wafer desecration, or “Crackergate“. The article isn’t on the Underground website, so I’ve scanned it, and you can find it linked here.

Below the fold is my response.

To: editor@the-underground.ca

Subject: Re: When freedom of speech goes too far

Dear Underground,

Erica Rodrigues’ article “When freedom of speech goes too far” (Underground, Nov. 27, 2008) is a perfect example of the sort of idiocy people like Paul Myers and “FSM Dude” were mocking in their acts of so-called desecration.

No, Ms. Rodrigues, a host wafer is not an embodiment of Christ. It’s not *actually* a piece of Jesus’ flesh. It is a metaphor, a symbol. The object itself is not special, no matter what your priest tells you. What’s special about a religious symbol is the way religious people treat the object: the ritual surrounding it, not the object itself.

Likewise, the Qur’an itself is not special. What a non-Muslim does with it is none of anyone’s business. Whether I throw my Qur’an in the garbage, set it on fire or wipe my behind with it, that’s nobody’s business but my own.

Here’s the thing that people like Ms. Rodrigues miss as they’re crying about the imagined violation of their religious freedom: freedom of speech and freedom of religion doesn’t just mean that religious people get to teach whatever nonsense and superstition they want. It also means that people get to call it nonsense and superstition. People get to make fun of Catholics by desecrating host wafers. People get to tear pages out of the Qur’an, put a yarmulke on the ground or make cartoons about Mormon underwear. People get to do these things in part because they don’t infringe on the rights of religious people to practice their religions.

Where that freedom stops is at inciting violence or other harm. Paul Myers has never, to my knowledge, advocated harming anyone. He has never suggested that religious people have anything but freedom to believe what they want and practice their faith in non-harmful ways. However, and this brings me to an interesting omission in Ms. Rodrigues’ article, PZ Myers has received tens of thousands of letters since “Crackergate” from irate Christians, demanding that he recant, proclaiming that he should be stripped of his professorship and insisting in various ways (including death threats) that he be denied the right to speak his mind and express his views on religion.

I ask that Ms. Rodrigues and other people who think that “freedom of religion” exists solely to protect the existence of religion, and to shield it from criticism and ridicule, take their heads out of their behinds and rejoin the real world.

Thank you,

David Leaman

UTSC Undergrad

… noticed this picture from the U of T front page (you may have to refresh a few times to see this image). Something about it piqued my notice.

What is that, anyway? Some kind of printing press? That would be my guess.

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