Indian prodigy R Praggnanandhaa, FIDE Circuit 2025 winner, will compete in the 2026 FIDE Candidates Tournament, aiming to challenge champion Gukesh Dommaraju.
R Praggnanandhaa (Image Credits:X)
With a dominating performance at the FIDE Circuit 2025, the 20, year, old R Praggnanandhaa has sent a clear signal that he has arrived to play at the level of the best players in the world and can win a FIDE Top Chess event, the FIDE Candidates Tournament 2026. The young Indian is one of the driving forces behind India's fast emergence as a chess powerhouse, along with players like D Gukesh and Arjun Erigaisi.
In accordance to an article in Chessbase Praggnanandhaa as a 20, year, old, has done the feat of winning FIDE Circuit 2025 with a staggering 115. 17 points. After this, he has become the only Indian player in the Open section of this year's Candidates Tournament. The winner of the event secures the right to go against the world champion for the title.
During the interview with Chess. com, Praggnanandhaa talked about the difficulty of playing against world-class players. For him, most of the top players already know each other's games and their strategies to such an extent that it is very hard to surprise them. So, according to him, the secret to always being ahead of your rivals at the top level is to continuously come up with new ideas and unconventional moves.
The journey to the 2026 World Chess Championship starts this spring with the FIDE Candidates Tournament, which will feature eight of the world's best grandmasters vying for the opportunity to defeat the current champion, Gukesh Dommaraju. This exciting tournament is set to take place from March 28 to April 16 2026, at the Cap St Georges Hotel and Resort in Pegeia, Cyprus. Competitors will demonstrate their profound tactical skills and creative thinking, and the victor will be granted the highly desired chance to fight for the World Championship, the beginning of the path to the most respected chess title in the world.
