2017 China Open – Women's Doubles

Bethanie Mattek-Sands and Lucie Šafářová were the defending champions, but neither player could participate this year due to injury.

Women's Doubles
2017 China Open
Champions Chan Yung-jan
Martina Hingis
Runners-up Tímea Babos
Andrea Hlaváčková
Final score6–1, 6–4

Chan Yung-jan and Martina Hingis won the title, defeating Tímea Babos and Andrea Hlaváčková in the final, 6–1, 6–4.

Seeds

The top four seeds received a bye into the second round.

  1. Chan Yung-jan / Martina Hingis (Champions)
  2. Ekaterina Makarova / Elena Vesnina (Semifinals)
  3. Sania Mirza / Peng Shuai (Semifinals)
  4. Tímea Babos / Andrea Hlaváčková (Final)
  5. Kateřina Siniaková / Barbora Strýcová (Quarterfinals)
  6. Ashleigh Barty / Casey Dellacqua (Second round)
  7. Gabriela Dabrowski / Xu Yifan (Quarterfinals)
  8. Anna-Lena Grönefeld / Květa Peschke (Second round)

Draw

Key

Finals

Semifinals Final
          
1 Chan Yung-jan
Martina Hingis
2 6 [10]
3 Sania Mirza
Peng Shuai
6 1 [5]
1 Chan Yung-jan
Martina Hingis
6 6
4 Tímea Babos
Andrea Hlaváčková
1 4
4 Tímea Babos
Andrea Hlaváčková
7 64 [10]
2 Ekaterina Makarova
Elena Vesnina
5 77 [8]

Top Half

First Round Second Round Quarterfinals Semifinals
1 Y-j Chan
M Hingis
3 6 [12]
S Cîrstea
J Ostapenko
4 3 S-w Hsieh
M Niculescu
6 4 [10]
S-w Hsieh
M Niculescu
6 6 1 Y-j Chan
M Hingis
6 6
Alt L Davis
A Riske
7 2 [6] 7 G Dabrowski
Y Xu
3 2
L Kichenok
R Olaru
5 6 [10] L Kichenok
R Olaru
5 5
N Kichenok
An Rodionova
6 2 [8] 7 G Dabrowski
Y Xu
7 7
7 G Dabrowski
Y Xu
1 6 [10] 1 Y-j Chan
M Hingis
2 6 [10]
3 S Mirza
S Peng
6 1 [5]
3 S Mirza
S Peng
7 6
A Spears
C Vandeweghe
77 63 [10] E Mertens
D Schuurs
5 2
E Mertens
D Schuurs
65 77 [12] 3 S Mirza
S Peng
4 6 [10]
M Ninomiya
R Voráčová
6 67 [9] 5 K Siniaková
B Strýcová
6 2 [7]
H-c Chan
A Cornet
1 79 [11] H-c Chan
A Cornet
6 4 [7]
A Kudryavtseva
A Rosolska
2 2 5 K Siniaková
B Strýcová
4 6 [10]
5 K Siniaková
B Strýcová
6 6

Bottom Half

First Round Second Round Quarterfinals Semifinals
8 A-L Grönefeld
K Peschke
3 6 [12]
N Melichar
A Smith
6 2 [10] 8 A-L Grönefeld
K Peschke
6 65 [5]
WC Q Wang
Y Wang
5 3 K Bertens
J Görges
2 77 [10]
K Bertens
J Görges
7 6 K Bertens
J Görges
6 3 [9]
A Sevastova
D Vekić
6 4 [7] 4 T Babos
A Hlaváčková
3 6 [11]
S Stephens
H Watson
4 6 [10] S Stephens
H Watson
3 4
4 T Babos
A Hlaváčková
6 6
4 T Babos
A Hlaváčková
7 64 [10]
6 A Barty
C Dellacqua
6 6 2 E Makarova
E Vesnina
5 77 [8]
O Savchuk
K Srebotnik
1 2 6 A Barty
C Dellacqua
5 4
S Aoyama
Z Yang
4 3 A Klepač
MJ Martínez Sánchez
7 6
A Klepač
MJ Martínez Sánchez
6 6 A Klepač
MJ Martínez Sánchez
6 0 [7]
WC X Han
C Liang
715 6 2 E Makarova
E Vesnina
4 6 [10]
R Atawo
D Jurak
613 3 WC X Han
C Liang
1 2
2 E Makarova
E Vesnina
6 6
gollark: Basically, you wanted three levels or something, so store those directly and multiply when actually doing IO.
gollark: (and replace `total -= 254./3.;` accordingly, obviously)
gollark: Make `total` into an int. Replace `total += 254./3.;` with `total = min(2, max(0, total + 1))` or something, if the arduinos' weird language has that. Do `analogWrite(LED, total * 85)`. QED.
gollark: Make the total an integer from 0 to 2 or something and enforce this, then multiply by 85 in the analogWrite bit.
gollark: The main issue is that data is just *data*, and can't corrupt itself in some way if you do stuff wrong or enforce timeouts, only the programs operating on it can (and generally do).

References

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