2010 Nielsen Pro Tennis Championship – Singles

Alex Kuznetsov was the defender of championship title, but he lost to Michael Russell in the first round.
Brian Dabul defeated Tim Smyczek 6–1, 1–6, 6–1 in the final.

Singles
2010 Nielsen Pro Tennis Championship
Champion Brian Dabul
Runner-up Tim Smyczek
Final score6–1, 1–6, 6–1
Main article: 2010 Nielsen Pro Tennis Championship

Seeds
  1. Björn Phau (Quarterfinal)
  2. Michael Russell (Semifinal)
  3. Somdev Devvarman (First Round)
  4. Brian Dabul (Champion)
  5. Donald Young (Semifinal)
  6. Édouard Roger-Vasselin (First Round)
  7. Go Soeda (Second Round)
  8. Kevin Kim (Second Round)

Draw

Key

Finals
Semifinals Final
          
  Tim Smyczek 6 6  
5 Donald Young 4 2  
  Tim Smyczek 1 6 1
4 Brian Dabul 6 1 6
4 Brian Dabul 6 6  
2 Michael Russell 1 4  

Top Half
First Round Second Round Quarterfinals Semifinals
1 B Phau 6 7  
  B Reynolds 4 64   1 B Phau 4 6 6
Q P-L Duclos 7 0 6 Q P-L Duclos 6 1 3
  V Estrella 64 6 4 1 B Phau 5 6 0
  J Witten 1 4     T Smyczek 7 2 6
  T Smyczek 6 6     T Smyczek 6 6  
Q A Sitak 1 2   8 K Kim 4 1  
8 K Kim 6 6     T Smyczek 6 6  
3 S Devvarman 6 3 4 5 D Young 4 2  
  L Cook 4 6 6   L Cook 6 6  
Q F Wolmarans 7 6   Q F Wolmarans 2 3  
  I Merwe 68 4     L Cook 63 1  
WC A Krajicek 3 2   5 D Young 7 6  
WC S Stadler 6 6   WC S Stadler 3 5  
WC R Harrison 3 2   5 D Young 6 7  
5 D Young 6 6  

Bottom Half
First Round Second Round Quarterfinals Semifinals
6 É Roger-Vasselin 7 3 3
  P Polansky 5 6 6   P Polansky 6 2 3
  T Ito 1 2   Q M Raonic 3 6 6
Q M Raonic 6 6   Q M Raonic 7 3 4
  A Begemann 6 6   4 B Dabul 5 6 6
  M Yani 3 4     A Begemann 2 0  
WC JP Fruttero 4 3   4 B Dabul 6 6  
4 B Dabul 6 6   4 B Dabul 6 6  
7 G Soeda 7 6   2 M Russell 1 4  
  R DeHeart 67 1   7 G Soeda 4 4  
  R de Voest 6 6     R de Voest 6 6  
  A Peya 2 1     R de Voest 6 4 3
ALT N Okun 64 3   2 M Russell 4 6 6
  A Brugués-Davi 7 6     A Brugués-Davi 2 0  
  A Kuznetsov 4 2   2 M Russell 6 6  
2 M Russell 6 6  
gollark: I see.
gollark: I actually just copypastaized that from a thing I'm reading on Y combinators.
gollark: Of course it is.
gollark: Now, here's the puzzle: what if you were asked to define the factorial function in Scheme, but were told that you could not use recursive function calls in the definition (for instance, in the factorial function given above you cannot use the word factorial anywhere in the body of the function). However, you are allowed to use first-class functions and higher-order functions any way you see fit. With this knowledge, can you define the factorial function?
gollark: 2013, after Incident 2971.

References
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