2013 Open Féminin de Marseille – Doubles
Séverine Beltrame and Laura Thorpe were the defending champions, having won the event in 2012, but both players chose not to defend their title.
Doubles | |
---|---|
2013 Open Féminin de Marseille | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 1–6, 6–4, [10–5] |
Sandra Klemenschits and Andreja Klepač won the title, defeating Asia Muhammad and Allie Will in the final, 1–6, 6–4, [10–5].
Seeds
Julia Cohen / Tatjana Maria (first round) Irina Buryachok / Julia Glushko (first round) Sandra Klemenschits / Andreja Klepač Lyudmyla Kichenok / Nadiya Kichenok (quarterfinals)
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
First round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 3 | 3 | ||||||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 4 | 0 | ||||||||||||||||||||||
![]() ![]() | 5 | 6 | [10] | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
WC | ![]() ![]() | 7 | 1 | [5] | ![]() ![]() | 6 | 6 | ||||||||||||||||||||
4 | ![]() ![]() | w/o | ![]() ![]() | 4 | 2 | ||||||||||||||||||||||
![]() ![]() | 4 | ![]() ![]() | 5 | 6 | [7] | ||||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 7 | 3 | [10] | |||||||||||||||||||||
WC | ![]() ![]() | 4 | 2 | ![]() ![]() | 6 | 4 | [5] | ||||||||||||||||||||
![]() ![]() | 6 | 6 | 3 | ![]() ![]() | 1 | 6 | [10] | ||||||||||||||||||||
![]() ![]() | 3 | 3 | ![]() ![]() | 4 | 2 | ||||||||||||||||||||||
WC | ![]() ![]() | 0 | 4 | 3 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||
3 | ![]() ![]() | 6 | 6 | 3 | ![]() ![]() | 6 | 7 | ||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 2 | 5 | ||||||||||||||||||||||
![]() ![]() | 2 | 2 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
![]() ![]() | 6 | 1 | [10] | ![]() ![]() | 1 | 2 | |||||||||||||||||||||
2 | ![]() ![]() | 4 | 6 | [6] |
gollark: 3.1 Using 2x2 matricesThe argument of iterateabove is a linear transformation, so we can represent it as matrix and compute the nth power of this matrix with O(log n) multiplications and additions.For example, using the simple matrix implementation in Prelude extensions,fib n = head (apply (Matrix [[0,1], [1,1]] ^ n) [0,1])
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gollark: d o n ' t c h a n g e p e r
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References
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