Thanks. I added them up and get 735 combinations.

I think it's an interesting problem due to having so many players and trying to give them all equal playing time. Some people (read: parents) don't realize just how difficult it can be.

Thanks for the help. I'll give that problem a go when I get a bit of time.

It turns out that if you just minimize SAM-SAC you will get the same ratio between r/h which is what we were originally looking for.

We've got problems here. C-H is 6 players not 7. Wyman's calculation assumes 15 players. Also, are you saying that B is always in the game, or can you have A and no B? Also, do you want 1PG, 1G, and 3P? Wyman is assuming 1PG, 2G, and 2P.

Ah, yeah that's fine, then change all my 7's to 6's because C-H = CDEFGH = 6 players.

I didn't check his stuff for internal consistency, nor did I sanity check.

I read the problem as only A or B could play PG, so one or both of them are in at a time. And you always have to take 2 of IJKLMN, and you fill the rest (1 or 2) with CDEFGH.

It's 1 PG, 2G, 2P.

B sometimes plays G when A is PG, but B isn't always in the game for the entire duration of the game; however, A or B are ALWAYS in the game either 1 of them at a time or both at the same time.

BUT, when B is PG, A isn't in the game.

You can have:

(first letter is PG, second letter is G)
A and B
A and no B
B and no A

In other words, if A is in the game he will be playing PG and B can play G, if A is PG B is sitting on the bench, or if B is PG then A is not in the game.

540 combinations?

Quite probable my brain is just fried from too much work and this is really obvious, but...

We just got an assignment to model some markov chains in matlab. One of them was to model an election using the results from year x and comparing them to the results from year x+1. Say year x results were [0.25 0.25 05] and year x+1 were [0.33 0.33 0.34]. So the assignment is to figure out the transition matrix given two vectors. How do I do that? :S

OK, and there are 14 players or 15? C-H is 6 not 7, so only 6 can play G other than B?

14 players.

1 primary PG, 7 G, 6 P.

1 G can also play PG.

But 6 G only play G (C-H).

I see, you included B in the 7G. So there's 540 combinations.

A and no B: 1*C(6,2)*C(6,2)

B and no A: 1*C(6,2)*C(6,2)

A and B: 1*1*6*C(6,2)
===========================
540


Just for my own knowledge, I thought basketball had 2 guards, 2 forwards, and 1 center. You seem to have a point guard plus 2 other guards, 1 forward and a center. Is that more common?

Awesome, thanks!! Also, thanks to Wyman!

We just break down 1 PG at the top, 2 guards on the elbows, and 2 posts down low. It's easier to break it down into a more simpler structure.

Basically, we have 2 guys who can handle the ball really well, a few guys who are big and play big down low, and then a few smaller players who handle the ball well and can move quickly, but aren't quite big enough to be banged around down low. 5th graders...

You should probably start by writing out the definition of a transition matrix.

I am writing an algorithm to perform automatic matchmaking, and I feel like someone should have solved this problem before and wrote a good algorithm somewhere or at least talked about some central ideas behind it. I have my own ideas but I want to "compare notes" first. However, I can't find any information on this on the internet, does anyone know where I could?

To be clear, what I mean by "automatic matchmaking" is where you tell a central server you are looking for a game, and then that server pairs opponents together.

It's solved. Search for the Microsoft Tru-Skill paper. Some of it is propitiatory, but the description is pretty telling.

Basically, each person is assigned a central estimate of their skill and a confidence/error that the system has. As you play more games, the confidence/error shrinks because it thinks of the peak as being more accurate.

That's not what I meant by matchmaking.

What I meant is more of the following, at any given instant, you have a set of players searching for a game [of which their elo or trueskill or glicko or whatever rating information is known], and you may pair these players to go play the game. Your objectives in this goal are varied; for example to pair people quickly, to pair people so that they have a match with someone close to their skill level, to avoid if possible repeats in pairing people, etc.

Please tell me if I understand this concept correctly:

If I want to know the displacement of an object and I have the velocity function ,I simply integrate the velocity function and evaluate at F(t_f)-F(t_i).

If I want to find the total distance an object has traveled and I have the velocity function, I find where v(t)=0, then, as before, evaluate the integral at F(t_f)-F(t_i) over each interval created by the values where v(t)=0, then I add the absolute value of each of the evaluated intervals?

Right, for the second thing you can also simply integrate the absolute value of the velocity and avoid all the search for 0s and reversals of direction etc.

Can someone tell me how the sum of moments about A are found (F_1)?

3W - 2F1 = 0

F1 = 3/2 * W

Law of the Lever

I think it will depend on a lot of context specific things. Look at games like Heroes of Newerth or LoL. They use an elo system at the heart of their match making system for sure but then they added adjustments.

HoN specifically added time as a variable (not really context specific but probably useful in a game match making service). They added in that after x time the quality of the game would degrade (less likely to be matched fairly) but the result is less of a queue time.

As always, thank you Bruce

From Shifrin "Multivariable Mathematics", page 6.

I approached this in the following way. I called the points, in order, P, Q, and R.

I found the vectors PQ and PR. I then added this vector to P to get the fourth vertex.

I then repeated this to find the other two. QR + QP + Q and RP + RQ + R.

I got answers (4, 3, 7), (2, -1, 3) and (0, 5, -1).

Am I approaching this correctly? Can anyone verify this?

EDIT: Nvm, just realized that the selected answers are somewhat random rather than just having the odds, and this one happens to be in there. It was right

How many integrals can a function have? Intuitively it seems as though they can have an infinite # of integrals--the basic integral + an infinite many # of constants that can be substituted for C, but what is a terse way to say, technically speaking, what it is all the integrals will have in common?

Is saying that the integrals will all have the exact same terms provided those terms are not a constant correct?

The antiderivatives are all the same to within an additive constant.

In the following, assume that everything that I write down is continuous.

Suppose that F(x) and G(x) are both antiderivatives of f(x).

Then consider F(x) - G(x).

d/dx (F(x) - G(x)) = F'(x) - G'(x) = f(x) - f(x) = 0

So the derivative of (F-G) is 0. That makes F-G a constant.

Similarly, if F(x) is an antiderivative of f(x), then consider G(x) = F(x) + c

G'(x) = F'(x) = f(x), so G is an antiderivative of f. This shows that you can add a constant to any antiderivative of f and get back an antiderivative of f.

So basically this completes a categorization of antiderivatives: Let F be an antiderivative of f. Then G is an antiderivative of f if and only if there exists a constant c for which G(x) = F(x) + c.

Suppose U, V c Rn are subspaces and U c V. Prove that V-perp c U-perp.

This is clearly true but I don't know how to prove it :/ help