2005 Swedish Open – Singles

Mariano Zabaleta was the defending champion, but lost in the quarterfinals this year.

Singles
2005 Swedish Open
Champion Rafael Nadal
Runner-up Tomáš Berdych
Final score2–6, 6–2, 6–4
Main article: 2005 Swedish Open

Rafael Nadal won the tournament, beating Tomáš Berdych in the final, 2–6, 6–2, 6–4.

Seeds
  1. Rafael Nadal (Champion)
  2. Joachim Johansson (First Round)
  3. Tommy Robredo (Semifinals)
  4. Carlos Moyá (Second Round)
  5. Juan Carlos Ferrero (Quarterfinals)
  6. Mikhail Youzhny (Quarterfinals)
  7. Robin Söderling (Second Round)
  8. José Acasuso (First Round)

Draw

Key

Finals
Semifinals Final
          
1 Rafael Nadal 6 6  
3 Tommy Robredo 3 3  
1 Rafael Nadal 2 6 6
  Tomáš Berdych 6 2 4
  Tomáš Berdych 7 6  
  Jiří Vaněk 5 1  

Top Half
First Round Second Round Quarterfinals Semifinals
1 R Nadal 6 6  
  J Mónaco 1 1   1 R Nadal 6 6  
Q A Calleri 1 1     A Martín 2 4  
  A Martín 6 6   1 R Nadal 6 6  
  G García-López 7 6   5 JC Ferrero 3 3  
WC N Lapentti 5 1     G García-López 3 4  
Q L Dlouhý 4 1   5 JC Ferrero 6 6  
5 JC Ferrero 6 6   1 R Nadal 6 6  
3 T Robredo 7 6   3 T Robredo 3 3  
  M Tabara 5 0   3 T Robredo 6 4  
  J Björkman 2 2     G Monfils 4 2r  
  G Monfils 6 6   3 T Robredo 3 6 6
WC A Vinciguerra 6 6   6 M Youzhny 6 4 3
WC M Ryderstedt 3 1   WC A Vinciguerra 6 64 5
  J Haehnel 1 1   6 M Youzhny 4 77 7
6 M Youzhny 6 6  

Bottom Half
First Round Second Round Quarterfinals Semifinals
8 J Acasuso 4 1  
  N Almagro 6 6     N Almagro 6 3 4
  T Berdych 6 6     T Berdych 4 6 6
  K Kim 2 3     T Berdych 6 6  
  R Sluiter 6 1 4   M Zabaleta 3 2  
  M Zabaleta 2 6 6   M Zabaleta 77 6  
  Á Calatrava 5 2   4 C Moyá 63 2  
4 C Moyá 7 6     T Berdych 7 6  
7 R Söderling 6 6     J Vaněk 5 1  
  F Mayer 2 4   7 R Söderling 6 3 2
  Ó Hernández 3 6 6   Ó Hernández 1 6 6
Q G Fraile 6 2 2   Ó Hernández 4 2  
  J Nieminen 5 1     J Vaněk 6 6  
Q F Serra 7 6   Q F Serra 4 6 3
  J Vaněk 62 77 6   J Vaněk 6 1 6
2 J Johansson 77 62 2
gollark: (original is https://github.com/TomSmeets/FractalArt/blob/master/src/Main.hs#L127)
gollark: <@!319753218592866315> fix bug || <:bees:724389994663247974>
gollark: [REDACTED]
gollark: ```haskellfoldM' :: (Monad m) => (a -> b -> m a) -> a -> [b] -> m afoldM' _ z [] = return zfoldM' f z (x:xs) = do z' <- f z x z' `seq` foldM' f z' xs```
gollark: Huh, I just realized that the Haskell program I'm porting defines its own weird monadic combinator thing and then *doesn't use it*.

References
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