Vidit Gujrathi will take on USA’s Levon Aronian and P. Harikrishna plays Spain’s Alexie Shrov in their campaign-openers in the first leg of FIDE Grand Prix in Berlin.
The two Indians are part of a 16-player field, divided into four groups. After three legs of the Grand Prix, the top two finishers will complete the eight-player Candidates line-up. The winner of the Candidates tournament will challenge Magnus Carlsen for the world title.
In what is the last chance for the aspirants to be a Candidate, Vidit and Hari will have to punch above their weight to top the group and qualify for the semifinals. The double round-robin league ends on February 10. The event ends on February 17, with the last day reserved for the tiebreaker of the final.
The other legs are due in Belgrade (February 28-March 14) and back in Berlin (March 21-April 4). A player has to play two of these three events to be eligible for qualification.
ALSO READ |
Vidit, drawn in Group ‘C’ will face Aronian, Daniil Dubov and rising German talent Vincent Keymer. Harikrishna is in Group ‘D’ with Wesley So, Leinier Dominguez Perez and veteran Alexie Shirov.
Ding Liren, the Chinese top seed for the event, had to pull out after applying too late for the visa. Russia’s Dmitry Andreikin tested positive for Covid and became the second player to withdraw. They have been replaced by Poland’s Radoslaw Wojtaszek and Russia’s Andrey
Esipenko.
Six spots in the Candidates stood decided after two qualifying events - World Cup and the Grand Swiss. These qualifiers are, Fabiano Caruana, Jan-Krzysztof Duda, Alireza Firouzja and Sergey Karjakin besides Ian Nepomniachtchi (runner-up to Carlsen in their 2021 World title clash) and Teimour Radjabov (awarded a spot by FIDE since he pulled out of the last Candidates due to the onset of the pandemic).
Comments
SHARE