Theodore Slade vs Brandon Jiang
51. Annual World Open, 2023 · Result 1–0 · Dutch Defense: Leningrad Variation, Warsaw Variation (A88).
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Game details
- White
- Theodore Slade (2012)
- Black
- Brandon Jiang (2187)
- Result
- 1–0
- Event
- 51. Annual World Open
- Year
- 2023
- Opening
- Dutch Defense: Leningrad Variation, Warsaw Variation (A88)
About this chess game
This chess game between Theodore Slade (2012) and Brandon Jiang (2187) was played at 51. Annual World Open in 2023 and finished 1–0. The opening was the Dutch Defense: Leningrad Variation, Warsaw Variation (A88). 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 Theodore Slade games or Brandon Jiang games? This Theodore Slade vs Brandon Jiang 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 Dutch Defense: Leningrad Variation, Warsaw Variation.
Frequently asked questions
Who won Theodore Slade vs Brandon Jiang?
Theodore Slade vs Brandon Jiang (2023) finished 1–0, a win for Theodore Slade.
What opening was played in Theodore Slade vs Brandon Jiang?
The game opened with the Dutch Defense: Leningrad Variation, Warsaw Variation (ECO A88).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Theodore Slade vs Brandon Jiang, or open it on the CipherChess analysis board to review it with the Stockfish engine.