The New Dress

Paul Charles Morphy – Samuel Standidge Boden
London, 1858
Scotch Gambit C44

1. e4 e5 2. Nf3 Nc6 3. d4 exd4 4. Bc4 Bc5 5. 0-0. A nuance slip. Correct is the immediate 5. c3! (5. ... Nf6 6. cxd4 Bb4+). 5. ... d6! 6. c3 Nf6. Best is 6. ... Bg4! and if 7. Qb3 then 7. ... Bxf3! 8. Bxf7+ Kf8 9. Bxg8? Rxg8 10. gxf3 g5!-+ 11. Qe6 (else 11. Qd1 Qd7 12. b4 Bb6 13. Bb2 d3! 14. Qxd3 Ne5 15. Qe2 Qh3 16. Nd2 g4! 0 : 1 Kolisch – Anderssen, Paris 1860) 11. ... Ne5 12. Qf5+ Kg7 13. Kh1 Kh8 14. Rg1 g4! 15. f4 Nf3! 16. Rxg4 Qh4!! 17. Rg2 Qxh2+! 18. Rxh2 Rg1 mate, Reiner – Steinitz, Vienna 1860. 7. cxd4 Bb6 8. Nc3 0-0 9. d5!? “The only way to forestall the counterblow 9. ... Nxe4!”, writes Grandmaster Valeri Beim in his book “Paul Morphy Una Prospettiva Moderna”, Roma, Prisma Editori, 2008, p. 95, but Géza Maróczy’s suggestion 9. Bb3 is worth considering. 9. ... Na5? Very bad judgment. After 9. ... Ne5 10. Nxe5 dxe5 11. Qd3 White stands slightly better. 10. Bd3. Threatening b2-b4. 10. ... c5 11. Bg5! h6 12. Bh4 Bg4. Not 12. ... g5? 13. Nxg5! hxg5 14. Bxg5 c4 15. e5! dxe5 16. Ne4 and wins (Maróczy’s analysis). 13. h3 Bh5. Not with happiness, but 13. ... Bxf3 14. Qxf3 c4 (Beim) is met by 15. e5! dxe5 16. Ne4 with devastating effect. 14. g4! Bg6 15. Qd2 Re8 16. Rae1! Bc7. Beim recommends 16. ... a6, but then 17. e5 looks quite unpleasant for Black. 17. Nb5! Kh7. “After 17. ... Bb8 18. b4! cxb4 19. Qxb4 a6 20. Nbd4 Ba7 21. Nf5 Bc5 22. Qb2 Black wouldn’t be happy either”, writes Beim (op. cit., p. 96). 18. Bxf6 gxf6 19. Nxc7! “Morphy exchanges first one terribly limited Black Bishop, then the other, he also repairs his opponent’s Pawn structure. Why does he do these things and why do I approve of his decision? Because, as a result of this operation, all of Black’s minor pieces disappear except for the Knight at a5, which can do nothing to hinder White’s expansion on the weakened Kingside”, writes Beim (op. cit., p. 96). 19. ... Qxc7 20. Qc3. Threatening both b2-b4 and Qc3xf6. 20. ... Qd8 21. Nh4 b6 22. f4 Kg7

23. Nxg6! A lot of people would have been contented of entombing the enemy Bishop by 23. f5 Bh7 24. Ng2, but not Morphy. 23. ... fxg6 24. e5! Rc8. Not 24. ... dxe5 25. fxe5 Qxd5 on account of 26. exf6+ Kf7 (or 26. ... Kf8 27. f7 Rxe1 28. Qh8+ Ke7 29. Rxe1+ soon mating) 27. Bxg6+! Kxg6 28. Qc2+ and White mates in short order. 25. Bb1 Kf7 26. e6+ Kg7 27. Qd3 f5. What else? 28. gxf5 Qf6. If 28. ... gxf5 then 29. Qxf5 Qf6 30. Qh7+ (Maróczy) 30. ... Kf8 31. e7+ Rxe7 32. Rxe7 Qxe7 33. Qh8+ finis. 29. fxg6 Qxb2. “29. ... Re7!? offered sterner resistance, but Black could not have held off all the threats as White would have played 30. f5 Rf8 31. b3 Nb7 32. a3 Nd8 33. b4 opening a second front on the Queenside”, writes Beim (op. cit., p. 97). 30. f5 Qf6. 30. ... c4 31. Qg3! Qf6 32. e7! is equally catastrophic. 31. e7! c4 32. Qg3 c3 33. Re6 Qd4+ 34. Qf2 Qxd5 35. f6+ 1 : 0. Because of 35. ... Kh8 36. g7+ Kg8 37. f7+ Kxg7 38. Qf6 mate.

From left: Howard Staunton, Samuel Standidge Boden, and Johann Jacob Löwenthal. Image courtesy of edochess.ca.