2014 BRD Arad Challenger – Singles
Adrian Ungur was the defending champion, but lost in the second round to Martín Cuevas.
Singles | |
---|---|
2014 BRD Arad Challenger | |
Champion | ![]() |
Runner-up | ![]() |
Final score | 6–4, 7–6(7–3) |
Damir Džumhur won the title, defeating Pere Riba in the final, 6–4, 7–6(7–3)
Seeds
Pere Riba (Final) Victor Hănescu (Second round) Frank Dancevic (First round) Adrian Ungur (Second round) Damir Džumhur (Champion) Gerald Melzer (Second round) Guido Andreozzi (First round) Marius Copil (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
Finals
Semifinals | Final | ||||||||||||
1 | ![]() | 6 | 7 | ||||||||||
![]() | 2 | 5 | |||||||||||
1 | ![]() | 4 | 63 | ||||||||||
5 | ![]() | 6 | 77 | ||||||||||
![]() | 6 | 0 | 1 | ||||||||||
5 | ![]() | 3 | 6 | 6 | |||||||||
Top Half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
1 | ![]() | 6 | 6 | ||||||||||||||||||||||||
![]() | 2 | 1 | 1 | ![]() | 6 | 6 | |||||||||||||||||||||
PR | ![]() | 7 | 6 | PR | ![]() | 2 | 1 | ||||||||||||||||||||
![]() | 5 | 2 | 1 | ![]() | 6 | 6 | |||||||||||||||||||||
WC | ![]() | 5 | 6 | 6 | WC | ![]() | 2 | 1 | |||||||||||||||||||
Q | ![]() | 7 | 4 | 1 | WC | ![]() | 6 | 4 | 6 | ||||||||||||||||||
Q | ![]() | 3 | 6 | 77 | Q | ![]() | 0 | 6 | 4 | ||||||||||||||||||
7 | ![]() | 6 | 3 | 63 | 1 | ![]() | 6 | 7 | |||||||||||||||||||
4 | ![]() | 6 | 6 | ![]() | 2 | 5 | |||||||||||||||||||||
![]() | 4 | 2 | 4 | ![]() | 3 | 3 | |||||||||||||||||||||
WC | ![]() | 3 | 4 | Q | ![]() | 6 | 6 | ||||||||||||||||||||
Q | ![]() | 6 | 6 | Q | ![]() | 2 | 1 | ||||||||||||||||||||
![]() | 7 | 2 | 4 | ![]() | 6 | 6 | |||||||||||||||||||||
![]() | 5 | 6 | 6 | ![]() | 6 | 6 | |||||||||||||||||||||
WC | ![]() | 1 | 2 | 6 | ![]() | 1 | 4 | ||||||||||||||||||||
6 | ![]() | 6 | 6 |
Bottom Half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
8 | ![]() | 77 | 4 | 3 | |||||||||||||||||||||||
Alt | ![]() | 62 | 6 | 6 | Alt | ![]() | 3 | 77 | 3 | ||||||||||||||||||
![]() | 5 | r | ![]() | 6 | 64 | 6 | |||||||||||||||||||||
![]() | 7 | ![]() | 6 | 6 | |||||||||||||||||||||||
![]() | 2 | 0r | Q | ![]() | 3 | 3 | |||||||||||||||||||||
Q | ![]() | 6 | 1 | Q | ![]() | 79 | 1 | 7 | |||||||||||||||||||
![]() | 6 | 5 | 6 | ![]() | 67 | 6 | 5 | ||||||||||||||||||||
3 | ![]() | 2 | 7 | 3 | ![]() | 6 | 0 | 1 | |||||||||||||||||||
5 | ![]() | 7 | 3 | 6 | 5 | ![]() | 3 | 6 | 6 | ||||||||||||||||||
![]() | 5 | 6 | 3 | 5 | ![]() | 6 | 6 | ||||||||||||||||||||
![]() | 6 | 6 | ![]() | 4 | 2 | ||||||||||||||||||||||
![]() | 2 | 2 | 5 | ![]() | 6 | 6 | |||||||||||||||||||||
![]() | 4 | 4 | ![]() | 4 | 4 | ||||||||||||||||||||||
![]() | 6 | 6 | ![]() | 6 | 4 | 6 | |||||||||||||||||||||
WC | ![]() | 1 | 2 | 2 | ![]() | 1 | 6 | 4 | |||||||||||||||||||
2 | ![]() | 6 | 6 |
gollark: Except I can't really just multiply by two, because that limits me to some small amount of digits supported by floating points.
gollark: * pi digits
gollark: I mean digits unconstrained by floating point inaccuracy or whatever, like those formuale for Pi.
gollark: Is there a formula for getting digit(s) of Tau? I want potatOS's tau functionality to not be limited by the amount hardcoded in wherever I put it.
gollark: Does anyone use the new forums? At all?
References
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