2013 McDonald's Burnie International – Men's Singles

Danai Udomchoke was the defending champion but decided not to participate.
John Millman defeated Stéphane Robert 6–2, 4–6, 6–0 in the final to win the title.

Men's Singles
2013 McDonald's Burnie International
Champion John Millman
Runner-up Stéphane Robert
Final score6–2, 4–6, 6–0
Main article: 2013 McDonald's Burnie International

Seeds
  1. Lu Yen-hsun (Withdrew, due to a tooth abscess)
  2. Peter Polansky (Quarterfinals)
  3. John Millman (Champion)
  4. Samuel Groth (Second Round)
  5. Brydan Klein (Second Round)
  6. James Duckworth (First Round)
  7. Alessandro Giannessi (First Round)
  8. John-Patrick Smith (First Round)

Draw

Key

Finals
Semifinals Final
          
  Jose Statham 3 2  
3 John Millman 6 6  
3 John Millman 6 4 6
  Stéphane Robert 2 6 0
  Stéphane Robert 77 77  
  Matthew Barton 65 65  

Top Half
First Round Second Round Quarterfinals Semifinals
LL R Agar 6 2 5
  É Chvojka 3 6 7   É Chvojka 3 4  
WC J Andrijic 3 4   WC A Whittington 6 6  
WC A Whittington 6 6   WC A Whittington 77 5 1
  D Propoggia 6 64 4   J Statham 65 7 6
  J Statham 4 77 6   J Statham 6 77  
  G Jones 4 6 6   G Jones 4 64  
7 A Giannessi 6 3 3   J Statham 3 2  
3 J Millman 4 6 6 3 J Millman 6 6  
  B Mitchell 6 2 2 3 J Millman 6 6  
  Alexander Lobkov 6 3 0 Q J Thompson 4 4  
Q J Thompson 3 6 6 3 J Millman 7 6  
Q J Lemke 6 6   Q J Lemke 5 1  
  I Klec 1 1   Q J Lemke 4 6 6
  J Ward 1 6 6   J Ward 6 3 3
6 J Duckworth 6 3 4

Bottom Half
First Round Second Round Quarterfinals Semifinals
5 B Klein 4 7 6
  L Vanni 6 6 4 5 B Klein 2 2  
  S Robert 6 6     S Robert 6 6  
Q M Venus 3 3     S Robert  3  6  6
  A Feeney 6 3 6   A Feeney  6  3  2
WC B Mousley 4 6 2   A Feeney 6 6  
  J Eysseric 3 64   4 S Groth 3 1  
4 S Groth 6 77     S Robert 77 77  
8 J-P Smith 2 2     M Barton 65 65  
  M Barton 6 6     M Barton 5 6 6
  A Bolt 77 5 3   M Reid 7 3 1
  M Reid 63 7 6   M Barton 2 77 6
Q Michael Look 6 6   2 P Polansky 6 61 3
WC Harry Bourchier 4 4   Q M Look 2 6 3
  A Arnaboldi 2 5   2 P Polansky 6 3 6
2 P Polansky 6 7  
gollark: Technically functors have `fmap`, actually.
gollark: Functor: has `map`, lets you run an `a → b` over a `f a` to get a `f b`Applicative: has `<*>`, lets you run a `f (a → b)` over a `f a` to get a `f b` and `pure`, which lets you get a `f a` from an `a`Monad: has `join`, which does `f (f a)) → f a` or alternately `bind`, which is `f a → (a → f b) → f b`.
gollark: Ah yes.
gollark: An applicative is a functor with, er, `<*>` or something.
gollark: A monad is an applicative with bind/join.

References
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