Causing a major upset in the first-round tiebreaker, India’s woman International Master K Priyanka defeated higher-ranked Zsoka Gaal of Hungary in the blitz games in the first round of the FIDE World Women’s Chess Cup.
After losing the first game under Classical time control, Priyanka staged a strong comeback by forcing Gaal into the tiebreaker. She lost one, but won in the rapid tiebreak games. In the subsequent blitz games, Priyanka was unstoppable as she won two games in a row to reach the round of 64 in the knockout event, which will see top players in action from the second round onwards.
Nine Indians are in the fray in the tournament with four of them – Koneru Humpy, D Harika, R Vaishali and Divya Deshmukh – getting direct entry in the second round. Out of the remaining five, Kiran Manisha Mohanty has bowed out, while Vantika Agrawal, Padmini Rout, P V Nandhidhaa and Priyanka have all reached the last-64 stage.
Fourth ranked Koneru Humpy will be in action in the second round against Afruza Khamdamova of Uzbekistan. The Indian is expected to win and secure the last-32 spot in this knockout event.
An all-Indian contest is also on the cards with Harika taking on Nandhidhaa, starting as a huge favourite. Priyanka, after her spectacular victory, will face Kulon Claudia of Poland who was a little subdued in her first-round victory. Padmini Rout, on the other hand, will face a more formidable opponent in former women’s world champion Aleksandra Kosteniuk, who now represents Switzerland.
The biggest prize for the top-three finishers is a place in the Candidates tournament with Humpy of India a favourite to clinch one of these spots. The total prize pool of the event is US $6,91,250 with US $50,000 going to the winner. Those who bow out after the second round will get US $5,000.
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