2017 Pekao Szczecin Open – Singles
Alessandro Giannessi was the defending champion but lost in the first round to Artem Smirnov.
Singles | |
---|---|
2017 Pekao Szczecin Open | |
Champion | ![]() |
Runner-up | ![]() |
Final score | 7–6(7–3), 7–6(7–4) |
Richard Gasquet won the title after defeating Florian Mayer 7–6(7–3), 7–6(7–4) in the final.
Seeds
Richard Gasquet (Champion) Florian Mayer (Final) Alessandro Giannessi (First round) Carlos Berlocq (First round) Marco Cecchinato (First round) Jerzy Janowicz (Quarterfinals) Casper Ruud (First round) Renzo Olivo (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/WC | ![]() | 6 | 6 | ||||||||||
![]() | 4 | 4 | |||||||||||
1/WC | ![]() | 77 | 77 | ||||||||||
2 | ![]() | 63 | 64 | ||||||||||
SE | ![]() | 7 | 3 | 4 | |||||||||
2 | ![]() | 5 | 6 | 6 | |||||||||
Top half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
1/WC | ![]() | 6 | 6 | ||||||||||||||||||||||||
WC | ![]() | 3 | 0 | 1/WC | ![]() | 6 | 6 | ||||||||||||||||||||
![]() | 6 | 6 | ![]() | 3 | 2 | ||||||||||||||||||||||
Q | ![]() | 2 | 3 | 1/WC | ![]() | 4 | 77 | 6 | |||||||||||||||||||
![]() | 77 | 6 | ![]() | 6 | 62 | 4 | |||||||||||||||||||||
WC | ![]() | 63 | 2 | ![]() | 6 | 6 | |||||||||||||||||||||
![]() | 3 | 6 | 6 | ![]() | 3 | 3 | |||||||||||||||||||||
5 | ![]() | 6 | 4 | 4 | 1/WC | ![]() | 6 | 6 | |||||||||||||||||||
4 | ![]() | 4 | 62 | ![]() | 4 | 4 | |||||||||||||||||||||
![]() | 6 | 77 | ![]() | 3 | 6 | 6 | |||||||||||||||||||||
SE | ![]() | 4 | 6 | 6 | SE | ![]() | 6 | 2 | 3 | ||||||||||||||||||
Q | ![]() | 6 | 3 | 2 | ![]() | 3 | 2 | ||||||||||||||||||||
![]() | 6 | 6 | ![]() | 6 | 6 | ||||||||||||||||||||||
WC | ![]() | 2 | 4 | ![]() | 7 | 7 | |||||||||||||||||||||
![]() | 6 | 4 | 3 | 8 | ![]() | 5 | 5 | ||||||||||||||||||||
8 | ![]() | 3 | 6 | 6 |
Bottom half
First Round | Second Round | Quarterfinals | Semifinals | ||||||||||||||||||||||||
7 | ![]() | 5 | 77 | 2 | |||||||||||||||||||||||
SE | ![]() | 7 | 64 | 6 | SE | ![]() | 7 | 6 | |||||||||||||||||||
LL | ![]() | 4 | 1 | ![]() | 5 | 4 | |||||||||||||||||||||
![]() | 6 | 6 | SE | ![]() | 6 | 4 | 6 | ||||||||||||||||||||
LL | ![]() | 3 | 4 | ![]() | 2 | 6 | 4 | ||||||||||||||||||||
![]() | 6 | 6 | ![]() | 77 | 4 | 77 | |||||||||||||||||||||
Q | ![]() | 77 | 6 | Q | ![]() | 64 | 6 | 65 | |||||||||||||||||||
3 | ![]() | 64 | 1 | SE | ![]() | 7 | 3 | 4 | |||||||||||||||||||
6 | ![]() | 6 | 6 | 2 | ![]() | 5 | 6 | 6 | |||||||||||||||||||
Q | ![]() | 4 | 1 | 6 | ![]() | 6 | 77 | ||||||||||||||||||||
![]() | 7 | 6 | ![]() | 1 | 62 | ||||||||||||||||||||||
![]() | 5 | 2 | 6 | ![]() | 2 | 63 | |||||||||||||||||||||
![]() | 3 | 6 | 6 | 2 | ![]() | 6 | 77 | ||||||||||||||||||||
LL | ![]() | 6 | 2 | 3 | ![]() | 4 | 4 | ||||||||||||||||||||
![]() | 7 | 1 | 2 | 2 | ![]() | 6 | 6 | ||||||||||||||||||||
2 | ![]() | 5 | 6 | 6 |
gollark: If you multiply the `(x-1)` by `(ax^3+bx^2+cx+d)` it should expand out into having an x^4 term.
gollark: I'm probably explaining this badly, hmmm.
gollark: Then set the x^4/x^3/x^2/x^1 coefficients and constant terms on each side to be equal and work out a/b/c/d.
gollark: Set it equal to `(x-1)(ax^3+bx^2+cx+d)` (the thing you know it's divisible by times the generalized cubic thingy), and expand that out/simplify.
gollark: It would be annoying and inconsistent if it was 0. It's 1.
References
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