Alexander Berezovics vs Jonas Rosner
11. Kesarovski-Stanchev Memorial, 2012 · Result 1–0 · Queen's Pawn Game: Zukertort Variation (D02).
Turn this game into your next win
Replay Alexander Berezovics vs Jonas Rosner with deep analysis, save the moments that matter, fold the ideas into your own opening repertoire, and drill the positions until they're second nature. CipherChess turns the games you study into the results you get — free to start.
Start Free on CipherChessMore Games By These Players
Game details
- White
- Alexander Berezovics (2225)
- Black
- Jonas Rosner (2314)
- Result
- 1–0
- Event
- 11. Kesarovski-Stanchev Memorial
- Year
- 2012
- Opening
- Queen's Pawn Game: Zukertort Variation (D02)
About this chess game
This chess game between Alexander Berezovics (2225) and Jonas Rosner (2314) was played at 11. Kesarovski-Stanchev Memorial in 2012 and finished 1–0. The opening was the Queen's Pawn Game: Zukertort Variation (D02). You can replay the full game move by move on the interactive board above, or open it on the CipherChess analysis board to study every move with the Stockfish engine.
Looking for more Alexander Berezovics games or Jonas Rosner games? This Alexander Berezovics vs Jonas Rosner encounter is one of millions of chess games indexed in the CipherChess mega database. Browse both players' full records, the openings they play most, and head-to-head results, then load any game onto the board to prepare your own lines against the Queen's Pawn Game: Zukertort Variation.
Frequently asked questions
Who won Alexander Berezovics vs Jonas Rosner?
Alexander Berezovics vs Jonas Rosner (2012) finished 1–0, a win for Alexander Berezovics.
What opening was played in Alexander Berezovics vs Jonas Rosner?
The game opened with the Queen's Pawn Game: Zukertort Variation (ECO D02).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Alexander Berezovics vs Jonas Rosner, or open it on the CipherChess analysis board to review it with the Stockfish engine.