Four Indians in the Paris leg
There will be four Indians in the Paris leg of Freestyle Grand Slam Chess Tour 2025. It will take place from 7th to 14th April 2025. They are - GM D Gukesh, GM Arjun Erigaisi, GM R Praggnanandhaa and GM Vidit Gujrathi.
Vidit Gujrathi defeated Richard Rapport (HUN) 1.5-0.5 in the Finals to qualify to the Freestyle Grand Slam Tour 2025 Paris | Photo: Anmol Bhargav
Interview with Vidit Gujrathi | Video: ChessBase India
Brackets | Graphic: chess.com
Vidit's performance | Graphic: chess.com
Play-in prize details | Graphic: chess.com
Freestyle Grand Slam Tour Paris leg participants | Graphic: chess.com
Replay Knockout stage live commentary by GM David Howell (ENG) and IM David Pruess (USA) | Video: Chess.com
Arjun Erigaisi wins Freestyle Friday three in-a-row, Mitrabha Guha stuns Magnus Carlsen
Arjun Erigaisi has won Freestyle Friday online tournament for the third consecutive time. He scored 10/11, finishing a full point ahead of the rest. The inaugural and reigning Commonwealth Rapid Gold medalist, Mitrabha Guha defeated Magnus Carlsen (NOR) to finish sixth scoring 8.5/11. Mitrabha finished ahead of D Gukesh (10th), S L Narayanan (13th), B Adhiban (17th) and R Praggnanandhaa (18th).
Arjun Erigaisi won Freestyle Friday on 14th March | Graphic: chess.com
GM Mitrabha Guha finished sixth in Freestyle Friday, ahead of D Gukesh, S L Narayanan, B Adhiban and R Praggnanandhaa | Photo: Shahid Ahmed
GM Mitrabha Guha analyzes his win over the World no.1 - GM Magnus Carlsen | Video: GMMitrabha
GM Mitrabha Guha defeated the World no.1, GM Magnus Carlsen (NOR) | Source: chess.com
Final standings
