Darius Ruzele vs Axel Horstmann
4. Open, 1997 · Result 1–0 · Old Indian Defense: Normal Variation (A55).
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Game details
- White
- Darius Ruzele (2525)
- Black
- Axel Horstmann (1806)
- Result
- 1–0
- Event
- 4. Open
- Year
- 1997
- Opening
- Old Indian Defense: Normal Variation (A55)
About this chess game
This chess game between Darius Ruzele (2525) and Axel Horstmann (1806) was played at 4. Open in 1997 and finished 1–0. The opening was the Old Indian Defense: Normal Variation (A55). 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 Darius Ruzele games or Axel Horstmann games? This Darius Ruzele vs Axel Horstmann 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 Old Indian Defense: Normal Variation.
Frequently asked questions
Who won Darius Ruzele vs Axel Horstmann?
Darius Ruzele vs Axel Horstmann (1997) finished 1–0, a win for Darius Ruzele.
What opening was played in Darius Ruzele vs Axel Horstmann?
The game opened with the Old Indian Defense: Normal Variation (ECO A55).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Darius Ruzele vs Axel Horstmann, or open it on the CipherChess analysis board to review it with the Stockfish engine.