2012 Città di Caltanissetta – Singles

Andreas Haider-Maurer was the defending champion but decided not to participate.
Tommy Robredo won the title by defeating Gastão Elias 6–3, 6–2 in the final.

Singles
2012 Città di Caltanissetta
Champion Tommy Robredo
Runner-up Gastão Elias
Final score6–3, 6–2
Main article: 2012 Città di Caltanissetta

Seeds
  1. Alejandro Falla (First Round)
  2. Paolo Lorenzi (First Round)
  3. Frederico Gil (Second Round)
  4. Roberto Bautista-Agut (Second Round)
  5. Antonio Veić (Second Round)
  6. Rogério Dutra da Silva (Second Round)
  7. Daniel Gimeno-Traver (First Round)
  8. Grega Žemlja (Second Round)

Draw

Key

Finals
Semifinals Final
          
  Iñigo Cervantes Huegun 4 2  
WC Tommy Robredo 6 6  
WC Tommy Robredo 6 6  
  Gastão Elias 3 2  
Q Alejandro González 7 4 3
  Gastão Elias 5 6 6

Top Half
First Round Second Round Quarterfinals Semifinals
1/WC A Falla 3 3  
  T Schoorel 6 6     T Schoorel 6 6  
Q B Pašanski 5 4     S Vagnozzi 4 2  
  S Vagnozzi 7 6     T Schoorel 3 1  
  Y Mertens 6 1 2   I Cervantes Huegun 6 6  
  I Cervantes Huegun 4 6 6   I Cervantes Huegun 6 6  
  M González 6 6     M González 4 2  
7 D Gimeno-Traver 3 4     I Cervantes Huegun 4 2  
4 R Bautista-Agut 6 6   WC T Robredo 6 6  
LL Omar Giacalone 2 0   4 R Bautista-Agut 6 63 3
WC T Robredo 6 7   WC T Robredo 4 77 6
  D Lajović 1 5   WC T Robredo 6 2 6
  M Viola 6 6     M Viola 4 6 4
WC Marco Cecchinato 3 2     M Viola 64 6 6
  M Alund 2 0   6 R Dutra da Silva 77 2 2
6 R Dutra da Silva 6 6  

Bottom Half
First Round Second Round Quarterfinals Semifinals
8 G Žemlja 7 7  
  G Pella 5 66   8 G Žemlja 0 3  
Q A González 4 6 6 Q A González 6 6  
  I Navarro 6 0 2 Q A González 77 77  
  T Alves 3 2   Q D Kosakowski 63 61  
Q D Kosakowski 6 6   Q D Kosakowski 6 6  
  M Authom 6 3 3 3 F Gil 3 4  
3 F Gil 3 6 6 Q A González 7 4 3
5 A Veić 6 6     G Elias 5 6 6
WC G Naso 3 4   5 A Veić 7 64 5
  J Silva 1 5   Q D Meffert 5 77 7
Q D Meffert 6 7   Q D Meffert 4 69  
  J Martí 6 0 2   G Elias 6 711  
  D Junqueira 1 6 6   D Junqueira 63 6 1
  G Elias 78 6     G Elias 77 3 6
2 P Lorenzi 66 3  
gollark: More keywords → more complexity in the language/parsing/whatever, more stuff programmers have to know.
gollark: For all (values of) f there exists a (value) g such that f (x, y) = (g x) y. In other words, you can convert any function which takes two values as a tuple or something to a curried one. I think.
gollark: I knew it would eventually be useful setting that as my status!
gollark: What part of ∀f ∃g (f (x,y) = (g x) y) did you not understand?
gollark: Also written as ∀, I think.

References
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