New Delhi: Star shuttler Saurabh Verma of India made his place in the semi-finals of Vietnam Open with a brilliant performance. He reached the men's singles semi-finals of the BWF Tour Super 100 Tournament. Second seed Saurabh defeated local player Tien Minh Nguyen in straight sets. The Indian won the match 21–13, 21–18 for 43 minutes. The national champions will take on either the sixth seed Tanorongsak Senseombonsuke of Japan or Minoru Koga of Thailand in the final. Saurabh dominated from the beginning of the first game.

He took a 4–1 lead without losing time and increased it to 11–6 by the break. Even then, he did not give any chance to the opposing player in the first game. However, in the second game, Nguyen gave a tough challenge to the Indian player and took an 11–10 lead till the break. After the break, Saurabh regained the rhythm and extended his lead to 17-12. Nguyen, however, did not lose and came back once again, taking the score to 19-18. The twenty-six-year-old Indian took advantage of the experience to win the game by scoring two points and winning the game 21-18. Saurabh lost in the final 16 at the Chinese Taipei Open last month.

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