2013 Rai Open – Doubles
Dustin Brown and Jonathan Marray were the defending champions but Marray decided not to participate.
Brown played alongside Jamie Delgado, but they lost in the quarterfinals to Dominik Meffert and Philipp Oswald.
Andreas Beck and Martin Fischer defeated Martin Emmrich and Rameez Junaid 7–6(7–2), 6–0 in the final to win the title.
Doubles | |
---|---|
2013 Rai Open | |
Champions | |
Runners-up | |
Final score | 7–6(7–2), 6–0 |
Seeds
Dustin Brown / Jamie Delgado (Quarterfinals) Martin Emmrich / Rameez Junaid (Final) Tomasz Bednarek / Andreas Siljeström (First Round) Uladzimir Ignatik / Mateusz Kowalczyk (First 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
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | 4 | 1 | |||||||||||||||||||||||||
4 | 2 | 6 | 6 | ||||||||||||||||||||||||
6 | 6 | 1 | 67 | ||||||||||||||||||||||||
3 | 6 | 5 | [8] | 6 | 79 | ||||||||||||||||||||||
1 | 7 | [10] | 1 | 63 | |||||||||||||||||||||||
63 | 2 | 6 | 77 | ||||||||||||||||||||||||
77 | 6 | 77 | 6 | ||||||||||||||||||||||||
1 | 0 | 2 | 62 | 0 | |||||||||||||||||||||||
6 | 6 | 6 | 79 | ||||||||||||||||||||||||
3 | 6 | [10] | 3 | 67 | |||||||||||||||||||||||
4 | 6 | 3 | [8] | 6 | 4 | [6] | |||||||||||||||||||||
6 | 6 | 2 | 4 | 6 | [10] | ||||||||||||||||||||||
3 | 1 | 2 | 62 | ||||||||||||||||||||||||
4 | 3 | 2 | 6 | 77 | |||||||||||||||||||||||
2 | 6 | 6 |
gollark: Great, send me the code and information?!?!?!
gollark: 1.00003.
gollark: If anyone sees any flaws with this, they're wrong because there aren't any.
gollark: My "accursedsort" would work by:- initializing a counter to 0- repeatedly subtracting the smallest value from each element of the list and adding it to the counter- when a list item is 0, append it + the counter to a secondary list and remove it from the existing one- when list contains no items you are doneO(n) time*!
gollark: And that weird instruction for JS floating point conversion.
References
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