1998 ASB Classic – Singles

Marion Maruska was the defending champion but lost in the first round to Meike Babel.

Singles
1998 ASB Classic
1997 Champion Marion Maruska
Champion Dominique Van Roost
Runner-up Silvia Farina
Final score4–6, 7–6, 7–5

Dominique Van Roost won in the final 4–6, 7–6, 7–5 against Silvia Farina.

Seeds

A champion seed is indicated in bold text while text in italics indicates the round in which that seed was eliminated.

  1. Sandrine Testud (Semifinals)
  2. Lisa Raymond (Quarterfinals)
  3. Dominique Van Roost (Champion)
  4. Barbara Schett (First Round)
  5. Sandra Kleinová (First Round)
  6. Silvia Farina (Final)
  7. Tamarine Tanasugarn (Quarterfinals)
  8. Marion Maruska (First Round)

Draw

Key
First Round Second Round Quarterfinals Semifinals Final
1 S Testud 7 2 6
  M Tu 5 6 1 1 S Testud 6 6  
Q G Helgeson-Nielsen 6 7 4   J Husárová 2 0  
  J Husárová 7 5 6 1 S Testud 6 6  
  S Cacic 4 6 6   S Cacic 2 2  
LL A Gerši 6 3 3   S Cacic w/o    
  M Babel 6 6     M Babel      
8 M Maruska 1 3   1 S Testud 2 6  
4 B Schett 0 5   6 S Farina 6 7  
WC J Halard-Decugis 6 7   WC J Halard-Decugis 6 7  
  C Cristea 7 4 6   C Cristea 2 5  
Q S Reeves 6 6 3 WC J Halard-Decugis 4 3  
Q B Schwartz 6 6   6 S Farina 6 6  
  K Brandi 2 0   Q B Schwartz 6 3 1
  A D-Balleret 6 2 2 6 S Farina 3 6 6
6 S Farina 1 6 6 6 Silvia Farina 6 6 5
7 T Tanasugarn 4 6 6 3 D Van Roost 4 7 7
  F Li 6 4 2 7 T Tanasugarn 7 7  
  Y Yoshida 7 6     Y Yoshida 5 6  
  K Boogert 6 1   7 T Tanasugarn 0 3  
  L Golarsa 1 6 1 3 D Van Roost 6 6  
  L Němečková 6 2 6   L Němečková 0 4  
  F Lubiani 3 3   3 D Van Roost 6 6  
3 D Van Roost 6 6   3 D Van Roost 6 6  
5 S Kleinová 4 4   Q A Miller 2 3  
  C Torrens Valero 6 6     C Torrens Valero 5 5  
  M Saeki 4 4   Q A Miller 7 7  
Q A Miller 6 6   Q A Miller 7 1  
WC R Hudson 6 4 6 2 L Raymond 5 0 r
  R Hiraki 3 6 3 WC R Hudson 4 4  
  B Rittner 4 6 4 2 L Raymond 6 6  
2 L Raymond 6 3 6
gollark: ???
gollark: That looks like the quadratic formula.
gollark: Are you just substituting into the quadratic formula?
gollark: I'm pretty sure it is just modulo some prime for this. The Wikipedia article explains that this is done to avoid providing some information about the other points via the shape of the graph or something.
gollark: It's a polynomial in a finite field, for purposes.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.