>> r8btree_bracket_test R8BTREE_NODE_ADD_RANDOM_TEST Given a BTREE of data and a value X, we seek to find "bracket values" so that X(L) <= X <= X(R). J I Left Right Parent 1 1 2 3 -1 2 2 -1 4 1 3 3 5 -1 1 4 4 6 7 2 5 5 8 9 3 6 6 10 11 4 7 7 12 13 4 8 8 14 15 5 9 9 16 17 5 10 10 18 19 6 11 11 20 21 6 12 12 22 23 7 13 13 24 25 7 14 14 26 27 8 15 15 28 29 8 16 16 30 31 9 17 17 32 33 9 18 18 -1 -1 10 19 19 -1 -1 10 20 20 -1 -1 11 21 21 -1 -1 11 22 22 -1 -1 12 23 23 -1 -1 12 24 24 -1 -1 13 25 25 -1 -1 13 26 26 -1 -1 14 27 27 -1 -1 14 28 28 -1 -1 15 29 29 -1 -1 15 30 30 -1 -1 16 31 31 -1 -1 16 32 32 -1 -1 17 33 33 -1 -1 17 J Tree data 1 0.5000 0.2500 2 0.0000 0.0000 3 1.0000 1.0000 4 0.2500 0.0625 5 0.7500 0.5625 6 0.1250 0.0156 7 0.3750 0.1406 8 0.6250 0.3906 9 0.8750 0.7656 10 0.0625 0.0039 11 0.1875 0.0352 12 0.3125 0.0977 13 0.4375 0.1914 14 0.5625 0.3164 15 0.6875 0.4727 16 0.8125 0.6602 17 0.9375 0.8789 18 0.0312 0.0010 19 0.0938 0.0088 20 0.1562 0.0244 21 0.2188 0.0479 22 0.2812 0.0791 23 0.3438 0.1182 24 0.4062 0.1650 25 0.4688 0.2197 26 0.5312 0.2822 27 0.5938 0.3525 28 0.6562 0.4307 29 0.7188 0.5166 30 0.7812 0.6104 31 0.8438 0.7119 32 0.9062 0.8213 33 0.9688 0.9385 X(33) = 0.968800 F(X) = 0.938500 X = 0.972086 X(3) = 1.000000 F(X) = 1.000000 X(3) = 1.000000 F(X) = 1.000000 X = 1.108688 No right bracket. No left bracket. X = -0.059520 X(2) = 0.000000 F(X) = 0.000000 X(3) = 1.000000 F(X) = 1.000000 X = 1.120064 No right bracket. X(15) = 0.687500 F(X) = 0.472700 X = 0.698539 X(29) = 0.718800 F(X) = 0.516600 No left bracket. X = -0.103689 X(2) = 0.000000 F(X) = 0.000000 X(20) = 0.156200 F(X) = 0.024400 X = 0.167747 X(11) = 0.187500 F(X) = 0.035200 X(14) = 0.562500 F(X) = 0.316400 X = 0.570322 X(27) = 0.593800 F(X) = 0.352500 X(3) = 1.000000 F(X) = 1.000000 X = 1.186260 No right bracket. X(3) = 1.000000 F(X) = 1.000000 X = 1.197333 No right bracket. No left bracket. X = -0.013580 X(2) = 0.000000 F(X) = 0.000000 X(3) = 1.000000 F(X) = 1.000000 X = 1.205889 No right bracket. X(3) = 1.000000 F(X) = 1.000000 X = 1.185750 No right bracket. X(25) = 0.468800 F(X) = 0.219700 X = 0.478063 X(1) = 0.500000 F(X) = 0.250000 X(17) = 0.937500 F(X) = 0.878900 X = 0.950421 X(33) = 0.968800 F(X) = 0.938500 R8BTREE_BRACKET_TEST Normal end of execution. >>