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 | |
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2017 China Open | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 6–1, 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.
Chan Yung-jan / Martina Hingis (Champions) Ekaterina Makarova / Elena Vesnina (Semifinals) Sania Mirza / Peng Shuai (Semifinals) Tímea Babos / Andrea Hlaváčková (Final) Kateřina Siniaková / Barbora Strýcová (Quarterfinals) Ashleigh Barty / Casey Dellacqua (Second round) Gabriela Dabrowski / Xu Yifan (Quarterfinals) Anna-Lena Grönefeld / Květa Peschke (Second round)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
Finals
Semifinals | Final | ||||||||||||
1 | ![]() ![]() | 2 | 6 | [10] | |||||||||
3 | ![]() ![]() | 6 | 1 | [5] | |||||||||
1 | ![]() ![]() | 6 | 6 | ||||||||||
4 | ![]() ![]() | 1 | 4 | ||||||||||
4 | ![]() ![]() | 7 | 64 | [10] | |||||||||
2 | ![]() ![]() | 5 | 77 | [8] | |||||||||
Top Half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
1 | ![]() ![]() | 3 | 6 | [12] | |||||||||||||||||||||||
![]() ![]() | 4 | 3 | ![]() ![]() | 6 | 4 | [10] | |||||||||||||||||||||
![]() ![]() | 6 | 6 | 1 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
Alt | ![]() ![]() | 7 | 2 | [6] | 7 | ![]() ![]() | 3 | 2 | |||||||||||||||||||
![]() ![]() | 5 | 6 | [10] | ![]() ![]() | 5 | 5 | |||||||||||||||||||||
![]() ![]() | 6 | 2 | [8] | 7 | ![]() ![]() | 7 | 7 | ||||||||||||||||||||
7 | ![]() ![]() | 1 | 6 | [10] | 1 | ![]() ![]() | 2 | 6 | [10] | ||||||||||||||||||
3 | ![]() ![]() | 6 | 1 | [5] | |||||||||||||||||||||||
3 | ![]() ![]() | 7 | 6 | ||||||||||||||||||||||||
![]() ![]() | 77 | 63 | [10] | ![]() ![]() | 5 | 2 | |||||||||||||||||||||
![]() ![]() | 65 | 77 | [12] | 3 | ![]() ![]() | 4 | 6 | [10] | |||||||||||||||||||
![]() ![]() | 6 | 67 | [9] | 5 | ![]() ![]() | 6 | 2 | [7] | |||||||||||||||||||
![]() ![]() | 1 | 79 | [11] | ![]() ![]() | 6 | 4 | [7] | ||||||||||||||||||||
![]() ![]() | 2 | 2 | 5 | ![]() ![]() | 4 | 6 | [10] | ||||||||||||||||||||
5 | ![]() ![]() | 6 | 6 |
Bottom Half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
8 | ![]() ![]() | 3 | 6 | [12] | |||||||||||||||||||||||
![]() ![]() | 6 | 2 | [10] | 8 | ![]() ![]() | 6 | 65 | [5] | |||||||||||||||||||
WC | ![]() ![]() | 5 | 3 | ![]() ![]() | 2 | 77 | [10] | ||||||||||||||||||||
![]() ![]() | 7 | 6 | ![]() ![]() | 6 | 3 | [9] | |||||||||||||||||||||
![]() ![]() | 6 | 4 | [7] | 4 | ![]() ![]() | 3 | 6 | [11] | |||||||||||||||||||
![]() ![]() | 4 | 6 | [10] | ![]() ![]() | 3 | 4 | |||||||||||||||||||||
4 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||||
4 | ![]() ![]() | 7 | 64 | [10] | |||||||||||||||||||||||
6 | ![]() ![]() | 6 | 6 | 2 | ![]() ![]() | 5 | 77 | [8] | |||||||||||||||||||
![]() ![]() | 1 | 2 | 6 | ![]() ![]() | 5 | 4 | |||||||||||||||||||||
![]() ![]() | 4 | 3 | ![]() ![]() | 7 | 6 | ||||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 0 | [7] | |||||||||||||||||||||
WC | ![]() ![]() | 715 | 6 | 2 | ![]() ![]() | 4 | 6 | [10] | |||||||||||||||||||
![]() ![]() | 613 | 3 | WC | ![]() ![]() | 1 | 2 | |||||||||||||||||||||
2 | ![]() ![]() | 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).
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