Working with the data from top players, professionals, is already problematic, I can only imagine how awful (and at the same time interesting) the full data is. But I think that would be too much for me.

Adding 4 fake draws against the same average opposition, and 3 fake draws agains 2300, and then adding 43 points? I am sure this makes sense in his scheme of things, but I don't think it's a great way (or the best way) to calculate TPRs. The traditional ways are imo better, namely:

(i) use elo rating (in case of infinity we have a problem, but adding a fake draw against youserlf is probably better than nothing - alexmagnus had some ideas concerning this issue, but I don't remember very well now)

(ii) add or subtract 400 points (+op. rating) for win or a loss, take the average etc.

in Linares 2001 we had one player with a +5, and five players with a -1. If we take 0.25 from Kasparov, and add +0.25 to the others, we will end with 31 points for 30 games.

If we use (1/2-%)*0.5, we add as much negative as positive points (zero sum), because the mean % is always 1/2 -- so we end with 30 points for 30 games, which I guess it's better.

Pearl Spring would look like:
Carlsen: 7.85
Topalov: 5.48
Wang Yue: 4.52
Others: 4.05

If Nanjing were 5 games long (with the same final %), we would have:

Carlsen: 3.85
Topalov: 2.73
Wang Yue: 2.27
Others: 2.05

Carlsen still loses the same amount of points in both cases, as in the original idea, but only 0.15. (5 games: 77%, 10 games: 78.5%, infinite games: 80%)

Let's start by arranging the numbers in decreasing order, x_1, x_2,...,x_n say, and form what we refer to as <the list>, call it L. L=(x_1,x_2,...,x_n).

Next, let's consider the differences between consecutive elements in the list: a_1=x_1-x_2, a_2=x_2-x_3, and so on. Thus the numbers a_1, a_2, ... , a_(n-1) are non-negative (that is, positive or zero).

Let us now fix some weights: w_1, w_2, ..., w_(n-1). We require these weights to be non-negative real numbers, but other than that we choose them as we please, unlike the initial numbers x_1,... which are <given>. Let W denote this sequence of weights, W=(w_1,...,w_(n-1)). We refer to W as <the weight>.

Next, multiply the differences a_1,... by the corresponding weights w_1,... and add them up. Denote the sum by D(W,L).

D(W,L)=w_1a_1 + w_2a_2+...+w_(n-1)a_(n-1).

We interpret D(W,L) as a measure of how much x_1 dominates the rest of the list L <with respect to the weight W>.

Example 1. Let W be given by: w_1=w_2=...=w_9=1, and let all the other weights (w_10, w_11 ,...) be equal to zero. Then:

D(W,L) = a_1+a_2+...+a_9 = x_1-x_10. So this is the difference between #1 and #10 on the list, as considered by <alexmagnus> above.

Example 2. Let W be given by w_1=...=w_99=1, and all the other weights equal to zero. Then this time D(W,L) measures the difference between #1 and #100 on the given list.

Example 3. Let W be given by w_1=1, and all the other weights zero. In this case D(W,L) measures the gap between #1 and #2 on the given list, as I've done on that puzzle a few days ago on Carlsen's page, and applied it for each country individually.

So this is an example of the type considered in my previous post. This example also shows that the weight W should not be set in stone. There was reasonable motivation for <alexmagnus> to consider the difference between #1 and #100, or #1 and #10, in that context. But in this context, those weights may not be the most appropriate.

Thus in each context we might want to first decide which weight we feel is the most appropriate.

Danailov has a clear idea in mind when arguing that Topalov deserves the chess Oscar more than Anand. He uses smaller weights on those categories which favor Anand (such as holding the world champion title), and higher weights on those categories which favor Topalov (such as ranking and rating). Thus, when he speaks about the Anand-Kramnik match, he does not refer to it as a world championship match, but as a match between #5 and #6, and so on.

Let's apply the same idea. Suppose we are given several lists L_1, L_2, ..., L_m, and we secretly favor one of them, call it L (so L is one of the above). We need to choose a weight W, and then use this same weight to compute each of D(W,L_1), D(W,L_2),..., D(W,L_m).

As an example, think of L_1 being a rating list showing Kasparov as #1, L_2 another rating list with Kasparov on top, L_3 a rating list with Topalov as #1, and L_4 a rating list with Anand as #1. Suppose I want to choose a weight W which will make Topalov look good. So in this example m=4 and L=L_3.

Returning to the general case, we are now ready to define the Cauchy-Danailov weights. Let L=(x_1,...,x_n) be our secretly preferred list, and denote as usual a_1=x_1-x_2, a_2=x_2-x_3,... Next, for each positive real number t, we define a Cauchy-Danailov weight W by letting:

W=(ta_1,ta_2,...,ta_(n-1)).

In other words, W and the sequence of differences (a_1,...,a_(n-1)) are proportional. This follows Danailov's idea, in the sense that those differences among a_1,...,a_(n-1) which are large receive large weights, and those difference which are small receive correspondingly small weights. Sometime, by accident, this weight W may also help other lists among L_1, L_2,...,L_m. Usually it doesn't. But it always helps our secret list L, and that's the whole point of the Cauchy-Danailov weights.

Why is Cauchy mentioned in this context? Because Cauchy's inequality accurately describes the effect of this type of weights.

For any list L=(x_1,...,x_n), and the corresponding differences a_1,...,a_(n-1), and for any weight W=(w_1,...,w_(n-1)), Cauchy's inequality states that

D(W,L)=w_1a_1+...+w_(n-1)a_(n-1) is less than or equal to

(w_1^2+...+w_(n-1)^2)^(1/2)(a_1^2+...+a_(n-1)^2)^-
(1/2).

Next, there is a notion of angle between any two vectors V and W, which is consistent with Cauchy's inequality. More precisely, if we denote this angle by theta, and denote the Euclidean norms (lengths) of V and W by ||V|| and ||W||, then the scalar product is given by:

D(W,L)=||W|| ||V|| cos(theta).

Cauchy's inequality states that the scalar product in absolute value is less than or equal to ||W|| times ||V||. This is consistent with the equality above since cos(theta) in absolute value is at most 1. Moreover, cos(theta)=1 exactly when theta=0, so exactly when the vectors W and V have the same direction (are proportional).

Now if we have several lists L_1,...,L_m and a fixed W, then

D(W,L_1)=||W|| ||V_1|| cos(theta_1), and similarly for the others.

Here the length ||W|| will appear in all of D(W,L_1),...,D(W,L_m), so it won't make a difference on the question of which of these numbers is the largest. Also, the lengths ||V_1||,...||V_m|| , which are not equal to each other, are anyway fixed, so there is nothing we can do about them, they are given to us from the beginning.

7-8.Anand 95 Jul 98
9.Topalov 84 Jul 06

Let's see if we can help Topalov move in front of Anand by changing the weight as discussed above. First, we copy the July 2006 list, from #1 Topalov 2813 to #10 Gelfand 2729:

L=(2813, 2779, 2761, 2743, 2742, 2738, 2734, 2732, 2731, 2729).

Taking differences between consecutive elements in the list, we find:

V=(34, 18, 18, 1, 4, 4, 2, 1, 2).

Let us compute the Euclidean norm of V. Its square is given by:

||V||^2=1156+324+324+1+16+16+4+1+4=1846.

Next, consider the July 1998 list. Anand is #2, and we help him a bit by considering the difference between #2 and #11, which is 105, instead of the difference between #2 and #10, which is 95 as computed by <alexmagnus>. We copy the list:

L_1=(2795, 2780, 2730, 2725, 2720, 2720, 2715, 2710, 2700, 2690).

Taking differences between consecutive elements, we obtain:

V_1=(15, 50, 5, 5, 0, 5, 5, 10, 10).

The square of its Euclidean norm is:

||V_1||^2=225+2500+25+25+0+25+25+100+100=3025.

Not surprisingly, the length of V_1 is larger than the length of V. Also, we know from <alexmagnus> dominator list that if we define U=(1,1,1,1,1,1,1,1,1), then U has a larger scalar product with V_1 than with V. More precisely, the scalar product of V with U equals

34+18+18+1+4+4+2+1+2 = 84,

while the scalar product of U with V_1 equals

15+50+5+5+0+5+5+10+10 = 105.

But, if we take a vector W proportional to V,

W=tV=(34t, 18t, 18t, t, 4t, 4t, 2t, t, 2t), where t is an arbitrary positive real number, then the scalar product of W with V_1 will be smaller than with V. In other words, D(W,L) > D(W,L_1), and hence Topalov moves in front of Anand if we use the Cauchy-Danailov weight W. Indeed,

D(W,L) = t||V||^2 = 1846 t,

while

In connection with the data obtained by computing the difference between #1 and #10 on a given set of lists, some people may consider appropriate to also add the differences #10-#11, #11-#12, and so on, with smaller and smaller weights. One alternative would be to proceed as follows.

Fix a real number t between 0 and 1, and consider the infinite sequence W=(1,t,t^2,t^3,t^4,...). Given a list L=(x_1,...,x_n), we define as usual a_1=x_1-x_2, ... , a_(n-1)=x_(n-1)-x_n. Then we extend this finite sequence to an infinite sequence by letting a_n, a_(n+1), ... be all equal to zero. Then define D(W,L) by

D(W,L)=a_1+a_2t+a_3t^2+a_4t^3+...

Note that if t=0, D(W,L) reduces to the difference between #1 and #2 on the list L. If t=1, D(W,L) equals the difference between #1 and the last element of the list L. For 0<t<1, D(W,L) takes values between the above two extreme cases.

A nice feature of these weights W, which other weights do not possess, is that if we remove x_1 from the list L, and denote the resulting new list by L*, then:

D(W,L) = a_1 + t D(W,L*).

Recall our interpretation of D(W,L) as a measure of how much x_1 dominates the rest of the list L, with respect to the weight W. Then the above simple formula establishes a connection between how much x_1 dominates the rest of the list L, and how much x_2 dominates the remaining elements, with respect to the <same> weight W. Also, a_1 is just the rating difference x_1-x_2.

For example, if we look again at the July 1998 list, the above formula provides a relationship between how much #1 Kasparov dominates the other players and how much #2 Anand dominates the rest of the list after removing Kasparov, as a function of t and the rating difference between Kasparov and Anand.

As a side comment, there are many sequences of interest in number theory, and it would be interesting to study some of them from this dominator perspective introduced by <alexmagnus>.

As a simple example, how much would we say the number zero dominates the negative numbers? With respect to the above weights, the answer is nice and simple:

L=(0,-1,-2,-3,...), a_1=1, a_2=1,...

D(W,L) = 1+t+t^2+... = 1/(1-t).

As a variation of the above question, let's say every third number starting with -1 loses a rating point. So -1, -4, -7,... become -2,-5, -8, ... Then the domination of 0 will be slightly stronger than before. Again we can compute D(W,L) in closed form, as a rational function of t.

L=(0,-2,-2,-3,-5,-5,-6,...). The sequence of differences is (2,0,1,2,0,1,...), and

The above weights W=(1,t,t^2,...) have w_m=t^(m-1), which for fixed t decays exponentially fast as m increases. If instead we want w_m to go to zero in a more moderate way, like a power of m say, then we can fix a positive real number s and let w_m=1/m^s. Then

W=(1, 1/2^s, 1/3^s, 1/4^s, 1/5^s,...),

and for a given list L=(x_1,...) producing the sequence of differences a_1, a_2,... we have

D(W,L)= a_1+a_2/2^s+a_3/3^s+...

If L is a finite list, the case s=0 corresponds to the difference between x_1 and the last element of L, while if we let s go to infinity then D(W,L) approaches the difference x_1-x_2. These were also the two extreme cases in the previous post.

The difference is that in the previous post D(W,L) was a power series in t (respectively a polynomial if the list L was finite), and now it is a Dirichlet series.

Power series and Dirichet series behave quite differently. For example, in the problem of how much zero dominates the negative numbers, working with these new weights gives:

L=(0,-1,-2,...). Differences: 1,1,1,...

D(W,L) = 1+1/2^s+1/3^s+1/4^s+... = zeta(s),

the Riemann zeta function computed at s. This holds for s > 1. Zeta(s) has analytic continuation to the entire complex plane, except at s = 1 where it has a pole, and satisfies a functional equation relating zeta(s) to zeta(1-s).

To wrap up this entire discussion, let's also have a look at that variation where each of the numbers -1, -4, -7, ... looses a rating point.

L=(0,-2,-2,-3,-5,-5,-6,...). Differences: 2,0,1,2,0,1,...

Now the idea is to write this sequence of differences as a sum of two sequences: 1,1,1,1,1,1,... and 1,-1,0,1,-1,0,...and realize that this last sequence corresponds to the unique non-principal Dirichlet character modulo 3, call it chi. Therefore:

D(W,L) = zeta(s) + L(s,chi).

Zs. Szajbély - P. Baki (2161)

1. e4 e5 2. Nf3 d6 3. d4 Nd7 4. Bc4 c6 5. dxe5 dxe5 6. Ng5 Nh6 7. O-O Be7 8. Ne6 fxe6 9. Bxh6 Nb6 10. Qh5+ Kf8 11. f4 Qd4+ 12. Kh1 Bf6 13. fxe5 gxh6 14. Rxf6+ Ke7 15. Rf7+ 1-0

http://web.aanet.com.au/memorysport...