Five-time World Champion Viswanathan Anand set to return to FIDE World Cup after 15 years with an aim to seal a berth at the 2018 Candidates tournament, PTI reported.

Anand will lead a seven-member Indian team at the 128-player Knock Out tournament, where the finalists get to play the Candidates tournament.

The Candidates tournament determines the challenger for the World Chess Championship against the incumbent World Champion.

The 47-year-old Indian has won the World Cup twice at China (2000) and Hyderabad (2002) respectively. He never needed to play at the prestigious tournament ever since winning the World Championships in 2007.

Anand will open his campaign against Yeoh Li Tian (rated 2480) in the first round and if he can surpass the young Malaysian, Anand might face American Grand Master Varuzhan Akobian in the second round.

If Anand is at his best, his first major test will be against England No 1 Michael Adams in the third round, followed by Hikaru Nakamura in the next round.

Anand had an impressive outing at the Sinquefield Cup where he finished joint second early last month. However, he finished ninth in the St. Louis Rapid and Blitz Chess tournament.

Indian Grand master Pentala Harikrishna will compete against Yuri Gonzalez Vidal of Cuba in his first round game in the tournament, which will be held from September 2 to 27 at Tbilisi.

Vidit Santosh Gujrathi, Baskaran Adhiban, Deep Sengupta, Sethuraman and Karthikeyan Murali will be the other Indians in the fray.

Among others, World Champion and No1 Magnus Carlsen, defending champion Sergey Karjakin, Ruslan Ponomariov, Vladimir Kramnik will also compete in the tournament, making it the strongest World Cup in history.