By IANS
Leon (Spain) : World No. 1 Viswanathan Anand has made it to the final of the 20th Ciudad de León chess tournament with a 3-1 win over Ruslan Ponomariov of Ukraine in the first semi-final.
Anand, a six-time winner here, will meet the winner of the other semi-final between Veselin Topalov of Bulgaria and Rustam Kasimdzhanov of Uzbekistan.
On the first day, Anand outplayed Ponomariov in the first two games and then drew the remaining encounters to make it a clear 3-1 verdict.
The first game was a French Opening in which Anand had white pieces and he used the space advantage very well in the end. Though Ponomariov played a surprise move as early as the sixth, Anand nevertheless built an advantage early by the 13th move.
In the middle game, Anand tried to gain control of the black squares and then amid some very fast play, the Indian star also managed to create a passed pawn.
Ponomariov tried to stop that pawn with his rook, but it was a losing battle and he gave up after 46 moves to give Anand the lead.
In the second game, which was Ruy Lopez, Anand with black played the Marshall and Ponomariov avoided the mainline theory. The Ukrainian sacrificed a pawn to open up Anand's king, but the Indian fought back well.
Anand traded rooks and also returned the pawn. Over the next few moves, he created a decisive advantage and by the 31st he had a passed pawn.
Anand completed a fine win in 50 moves after a great opening variation, excellent middle game and smooth ending.
The next two games were drawn and Anand was through 3-1.
The four invited players have all held the title of FIDE World Champions. They play four games in the semi-finals and the winners clash in final.
All games are to be played at a rate of 20 minutes for the whole game, with an increment of 10 seconds after each move. In case of a final draw, there will be a tiebreak in the form of blitz games – five minutes for each player.