An American in Paris

Paul Charles Morphy – Adolf Anderssen
Paris, december 1858
King’s Gambit C39

1. e4 e5 2. f4 exf4 3. Nf3 g5 4. h4 g4 5. Ne5 Nf6 6. Nxg4 Nxe4 7. d3 Ng3 8. Bxf4 Nxh1 9. Qe2+. On 9. Bg5 Be7 10. Qe2 h5 11. Qe5 f6! 12. Nxf6+ Kf7 Black wins (Steinitz’s analysis). 9. ... Qe7 10. Nf6+ Kd8 11. Bxc7+ Kxc7 12. Nd5+ Kd8 13. Nxe7 Bxe7. “[A]nd Black should win”, Robert James Fischer wrote in his best-selling book “My 60 Memorable Games”. 14. Qg4 d6 15. Qf4 Rg8 16. Qxf7. This looks stronger than 16. Nc3 Be6 17. Be2 Nc6 18. 0-0-0 Ng3 19. Bf3 Nf5 20. d4 Nxh4 21. Kb1 Nxf3 22. gxf3 d5 23. Nb5 Rc8 24. c4 Kd7 25. cxd5 Bxd5 26. Nc3 Bd6 27. Qf5+ Be6 28. Qb5 Kc7 29. Ka1 Bd7 30. Nd5+ Kb8 31. Nf6 Rgd8 32. d5 Ne5 33. Qe2 Bf5 34. Ne4 Bxe4 35. fxe4 Nc4 36. Rc1 b5 37. e5 Nxe5 38. Qxb5+ Ka8 39. Rd1 Rc2 40. Qa6 f5 41. Kb1 Rc7 42. Rf1 Nc4 43. Rf2 Rb8 44. b3 Ba3 0 : 1 Baucher – Morphy, Paris 1858. 16. ... Bxh4+. Géza Maróczy also considers 16. ... Rf8 17. Qxh7 Ng3 followed by ... Bc8-f5. 17. Kd2 Re8 18. Na3

18. ... Na6!? This is most likely a questionable move by Anderssen and, according to Maróczy – a little exaggeratedly – the decisive mistake. He recommended 18. ... Ng3 19. Nb5 Be7 and now we’d interpolate 20. Qf4 (20. Nc7? loses to 20. ... Rf8) 20. ... Nf5 with Black still maintaining more than sufficient compensation for the Queen. 19. Qh5 Bf6 20. Qxh1 Bxb2? Indeed, this does seem to be the decisive mistake! After 20. ... Bg5+ 21. Kc3 Be6 22. b3 Rc8+ followed by ... Bg5-f6+ Black should be able to hold his own. 21. Qh4+ Kd7. Or 21. ... Kc7 22. Nb5+ Kb8 23. Rb1 and wins. 22. Rb1 Bxa3 23. Qa4+ 1 : 0.