2005 AP Tourism Hyderabad Open – Doubles

Doubles
	2005 AP Tourism Hyderabad Open
Champions	Yan Zi ; Zheng Jie
Runners-up	Li Ting ; Sun Tiantian
Final score	6–4, 6–1

Liezel Huber and Sania Mirza were the defending champions, but only Mirza chose to participate that year. She paired up with Shikha Uberoi but lost in the quarterfinals to Yan Zi and Zheng Jie.

Yan and Zheng won the title, defeating Li Ting and Sun Tiantian 6–4, 6–1 in an all-Chinese final.

Seeds

Anna-Lena Grönefeld / Martina Navratilova (Quarterfinals, Retired)
Li Ting / Sun Tiantian (Final)
Yan Zi / Zheng Jie (Champions)
Maria Kirilenko / Tamarine Tanasugarn (Semifinals)

Draw

Key

Draw

First Round					Quarterfinals					Semifinals				Final

1	A-L Grönefeld M Navratilova	6	6
	S Hisamatsu R Tezuka	2	1		1	A-L Grönefeld M Navratilova	2^r
Q	A Amanmuradova T Panova	6	6		Q	A Amanmuradova T Panova	3
	A Bondarenko K Bondarenko	0	4							Q	A Amanmuradova T Panova	1	4
3	Z Yan J Zheng	6	6							3	Z Yan J Zheng	6	6
	M Czink L Němečková	4	1		3	Z Yan J Zheng	7	6
WC	S Mirza S Uberoi	6	7		WC	S Mirza S Uberoi	5	1
	C-w Chan N Li	4	5											3	Z Yan J Zheng	6	6
	M Santangelo A Vanc	6	1	6										2	T Li T Sun	4	1
	E Bychkova T Poutchek	2	6	3		M Santangelo A Vanc	2	6	4
	N Bratchikova E Dominikovic	6	5	5	4	M Kirilenko T Tanasugarn	6	4	6
4	M Kirilenko T Tanasugarn	3	7	7						4	M Kirilenko T Tanasugarn	1	0
	T Obziler S Pe'er	2	2							2	T Li T Sun	6	6
	A Rodionova M E Salerni	6	6			A Rodionova M E Salerni	4	3
	M Adamczak N Kriz	6	0	1	2	T Li T Sun	6	6
2	T Li T Sun	2	6	6

gollark: It's currently generating a regex for `hyperbee`.

gollark: `b(e(sä|äs)|s(eä|äe)|ä(es|se))|e(b(sä|äs)|s(bä|äb)|ä(bs|sb))|s(b(eä|äe)|e(bä|äb)|ä(be|eb))|ä(b(es|se)|e(bs|sb)|s(be|eb))`

gollark: Want another?

gollark: See, this is perfect and without flaw except that the time requirement to make the regex seems to increase exponentially due to greenery.

gollark: Here you go, `b(d(e(e(io|oi)|i(eo|oe)|o(ei|ie))|i(e(eo|oe)|oe{2})|o(e(ei|ie)|ie{2}))|e(d(e(io|oi)|i(eo|oe)|o(ei|ie))|e(d(io|oi)|i(do|od)|o(di|id))|i(d(eo|oe)|e(do|od)|o(de|ed))|o(d(ei|ie)|e(di|id)|i(de|ed)))|i(d(e(eo|oe)|oe{2})|e(d(eo|oe)|e(do|od)|o(de|ed))|o(de{2}|e(de|ed)))|o(d(e(ei|ie)|ie{2})|e(d(ei|ie)|e(di|id)|i(de|ed))|i(de{2}|e(de|ed))))|d(b(e(e(io|oi)|i(eo|oe)|o(ei|ie))|i(e(eo|oe)|oe{2})|o(e(ei|ie)|ie{2}))|e(b(e(io|oi)|i(eo|oe)|o(ei|ie))|e(b(io|oi)|i(bo|ob)|o(bi|ib))|i(b(eo|oe)|e(bo|ob)|o(be|eb))|o(b(ei|ie)|e(bi|ib)|i(be|eb)))|i(b(e(eo|oe)|oe{2})|e(b(eo|oe)|e(bo|ob)|o(be|eb))|o(be{2}|e(be|eb)))|o(b(e(ei|ie)|ie{2})|e(b(ei|ie)|e(bi|ib)|i(be|eb))|i(be{2}|e(be|eb))))|e(b(d(e(io|oi)|i(eo|oe)|o(ei|ie))|e(d(io|oi)|i(do|od)|o(di|id))|i(d(eo|oe)|e(do|od)|o(de|ed))|o(d(ei|ie)|e(di|id)|i(de|ed)))|d(b(e(io|oi)|i(eo|oe)|o(ei|ie))|e(b(io|oi)|i(bo|ob)|o(bi|ib))|i(b(eo|oe)|e(bo|ob)|o(be|eb))|o(b(ei|ie)|e(bi|ib)|i(be|eb)))|e(b(d(io|oi)|i(do|od)|o(di|id))|d(b(io|oi)|i(bo|ob)|o(bi|ib))|i(b(do|od)|d(bo|ob)|o(bd|db))|o(b(di|id)|d(bi|ib)|i(bd|db)))|i(b(d(eo|oe)|e(do|od)|o(de|ed))|d(b(eo|oe)|e(bo|ob)|o(be|eb))|e(b(do|od)|d(bo|ob)|o(bd|db))|o(b(de|ed)|d(be|eb)|e(bd|db)))|o(b(d(ei|ie)|e(di|id)|i(de|ed))|d(b(ei|ie)|e(bi|ib)|i(be|eb))|e(b(di|id)|d(bi|ib)|i(bd|db))|i(b(de|ed)|d(be|eb)|e(bd|db))))|i(b(d(e(eo|oe)|oe{2})|e(d(eo|oe)|e(do|od)|o(de|ed))|o(de{2}|e(de|ed)))|d(b(e(eo|oe)|oe{2})|e(b(eo|oe)|e(bo|ob)|o(be|eb))|o(be{2}|e(be|eb)))|e(b(d(eo|oe)|e(do|od)|o(de|ed))|d(b(eo|oe)|e(bo|ob)|o(be|eb))|e(b(do|od)|d(bo|ob)|o(bd|db))|o(b(de|ed)|d(be|eb)|e(bd|db)))|o(b(de{2}|e(de|ed))|d(be{2}|e(be|eb))|e(b(de|ed)|d(be|eb)|e(bd|db))))|o(b(d(e(ei|ie)|ie{2})|e(d(ei|ie)|e(di|id)|i(de|ed))|i(de{2}|e(de|ed)))|d(b(e(ei|ie)|ie{2})|e(b(ei|ie)|e(bi|ib)|i(be|eb))|i(be{2}|e(be|eb)))|e(b(d(ei|ie)|e(di|id)|i(de|ed))|d(b(ei|ie)|e(bi|ib)|i(be|eb))|e(b(di|id)|d(bi|ib)|i(bd|db))|i(b(de|ed)|d(be|eb)|e(bd|db)))|i(b(de{2}|e(de|ed))|d(be{2}|e(be|eb))|e(b(de|ed)|d(be|eb)|e(bd|db))))` matches all anagrams of `beeoid`.

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