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Chen, Yan Xi (1149) - Altenbuchner, Johannes (888) - AUT-ch U10 (1), St Veit, 2023
B77

1.e4 c5 2.Nf3 Nc6 3.d4 cxd4 4.Nxd4 Nf6 5.Nc3 g6 6.Be3 Bg7 7.Bc4 O-O 8.f3 d6 9.Qd2 Bd7 10.Bh6 Bxh6 11.Qxh6 Nxd4 12.O-O-O Nc6 13.h4 Ne5 14.Bb3 Rc8 15.Nd5 e6 16.Nxf6+ Qxf6 17.h5 Qg7 18.Qxg7+ Kxg7 19.Rxd6 Bb5 20.hxg6 Nxg6 21.Rd2 Nf4 22.Kb1 Rfd8 23.Rf2 Rd7 24.g3 Ng6 25.Rfh2 Nf8 26.f4 Rcd8 27.Kc1 a5 28.Rh5 Be2 29.Rg5+ Kh8 30.c3 Bf3 31.Re1 Rd2 32.e5 Rh2 33.Bd1 Rxd1+ 34.Rxd1 Bxd1 35.Kxd1 Rxb2 36.a4 Ra2 37.Kc1 Rxa4 38.Kb2 Rc4 39.Kb3 b5 40.Kc2 b4 41.Kb3 Rxc3+ 42.Ka4 Rc1 43.Kb3 Nd7 44.Kb2 Rc3 45.g4 a4 46.Rh5 a3+ 47.Ka2 Rc2+ 48.Kb3 a2 49.Rh1 Rg2 50.Ra1 Rxg4 51.Kxb4 Rxf4+ 52.Kb3 Nxe5 53.Rxa2 h5 54.Rh2 Rf5 55.Kc3 Ng6 56.Kd4 h4 57.Ke4 Kh7 58.Ke3 Kh6 59.Ke4 Kh5 60.Ke3 Kg4 61.Rg2+ Kh3 62.Rg1 Kh2 63.Re1 h3 64.Rb1 Kg3 65.Rg1+ Kh4 66.Rh1 Ne5 67.Ke2 Ng4 68.Ke1 h2 69.Ke2 Kh3 70.Ke1 Kg2 71.Rxh2+ Kxh2 72.Kd1 Re5 73.Kd2 f5 74.Kd3 f4 75.Kd2 f3 76.Kd1 f2 77.Kd2 f1=Q 78.Kc3 Rd5 79.Kb4 Qb5+ 80.Kc3 Qc6+ 0-1

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