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Russian Superfinal-2011 (Women)

User Rating: / 0
Written by Administrator   
Wednesday, 17 August 2011

The 61st Russian Superfinal (Women) will take place in Moscow from August 18 to 28. Ten players will be taking part in the round robin. The prize fund is about $50,000 with the winner getting roughly $14,000.

Here is the line-up (the Kosintseva sisters declined their invitations due to an intensive tournament schedule): Kosteniuk (2497). Qualification: by rating. Galliamova (2492). Qualification: gold at Russian Superfinal 2010. Gunina (2487). Qualification: Top League 2011. Pogonina (2442). Qualification: silver at Russian Superfinal 2010. Bodnaruk (2431). Qualification: by rating. Zaiatz (2419). Qualification: Top League 2011. Shadrina (2373). Qualification: Top League 2011. Girya (2371). Qualification: Top League 2011. Kovanova (2354). Qualification: Top League 2011. Charochkina (2310). Qualification: Top League 2011.

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Comments (7)
1. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 10:00 17 2011 .
2. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 15:42 17 2011 .
1. Pogonina 
2. Gunina 
3. Kosteniuk
3. Written by on 19:52 17 2011 .
, .
4. Written by on 11:01 18 2011 .
-2011 ()
! !!! ! ! - .
5. Written by on 16:54 18 2011 .
. - , , !
6. Written by Seth on 23:48 18 2011 .
Good luck! Your fans are rooting you on.
7. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 20:21 27 2012 .
Squares:1 nxn square2 2 (n-1)x(n-1) seuarqs jXj (n-j+1)X(n-j+1) seuarqstotal = sum_{j=1}^n j^2 = n(n+1)(2n+1)/6 seuarqsRectangles:How many jxk rectangles are there?Put one in the top left corner, then slide it over until it hits the top right. That makes (n+1-k) along the top row.Sliding down, there are (n+1-j) down the side.The total number of rectangles issum_{j,k=1}^n (n+1-k)(n+1-j)first fix j and sum over k. you getsum_{j=1}^n (n+1-k)n(n+1)/2This is just a constant times the same sum you just did, so it comes to.n^2(n+1)^2/4.Another way to think of it is that any given rectangle can be specified by four numbers, x1, x2, y1, y2. x1 gives the starting row, x2 the ending row, y1 the starting column, y2 the ending column. Require0

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Last Updated ( Wednesday, 17 August 2011 )
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