Fix the two piecewise-linear regression calculation #66
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shallawa
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WebKit#66 The current implementation of the regression calculation has these flaws: When processing (x[0], y[0]), L1 must be any line through (x[0], y[0]) which meets L2 at a point (x’, y’) where x[0] < x' < x[1]. L1 has no error. When processing (x[n - 2], y[n - 2]), L2 must be any line through (x[n - 1], y[n - 1]) which meets L1 at a point (x’, y’) where x[n - 2] < x' < x[n - 1]. L2 has no error. The lambda calculation is incorrect. It includes a term called H which is equal to C - I. Looking at the algorithm of Kundu/Ubhaya, this should be just C. lambda should to be used with calculating L1 and (1 - lambda) should to be used with calculating L2. Currently (1 - lambda) is used in calculating L1 and L2. The current calculation has this condition if (t1 != t2) continue; This condition is almost always true even if t1 and t2 are essentiallyEqual.
shallawa
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Aug 30, 2024
WebKit#66 The current implementation of the regression calculation has these flaws: When processing (x[0], y[0]), L1 must be any line through (x[0], y[0]) which meets L2 at a point (x’, y’) where x[0] < x' < x[1]. L1 has no error. When processing (x[n - 2], y[n - 2]), L2 must be any line through (x[n - 1], y[n - 1]) which meets L1 at a point (x’, y’) where x[n - 2] < x' < x[n - 1]. L2 has no error. The lambda calculation is incorrect. It includes a term called H which is equal to C - I. Looking at the algorithm of Kundu/Ubhaya, this should be just C. lambda should to be used with calculating L1 and (1 - lambda) should to be used with calculating L2. Currently (1 - lambda) is used in calculating L1 and L2. The current calculation has this condition if (t1 != t2) continue; This condition is almost always true even if t1 and t2 are essentiallyEqual.
shallawa
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Aug 30, 2024
WebKit#66 The current implementation of the regression calculation has these flaws: When processing (x[0], y[0]), L1 must be any line through (x[0], y[0]) which meets L2 at a point (x’, y’) where x[0] < x' < x[1]. L1 has no error. When processing (x[n - 2], y[n - 2]), L2 must be any line through (x[n - 1], y[n - 1]) which meets L1 at a point (x’, y’) where x[n - 2] < x' < x[n - 1]. L2 has no error. The lambda calculation is incorrect. It includes a term called H which is equal to C - I. Looking at the algorithm of Kundu/Ubhaya, this should be just C. lambda should to be used with calculating L1 and (1 - lambda) should to be used with calculating L2. Currently (1 - lambda) is used in calculating L1 and L2. The current calculation has this condition if (t1 != t2) continue; This condition is almost always true even if t1 and t2 are essentiallyEqual.
shallawa
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Sep 3, 2024
WebKit#66 The current implementation of the regression calculation has these flaws: 1. When processing (x[0], y[0]), L1 must be any line through (x0, y0) which meets L2 at a point (x’, y’) where x[0] < x' < x[1]. L1 has no error. 2. When processing (x[n - 2], y[n - 2]), L2 must be any line through (x[n - 1], y[n - 1]) which meets L1 at a point (x’, y’) where x[n - 2] < x' < x[n - 1]. L2 has no error. 3. The lambda calculation is incorrect. It includes a term called H which is equal to C - I. Looking at the algorithm of Kundu/Ubhaya, this should be just C. 4. lambda should to be used with calculating L1 and (1 - lambda) should to be used with calculating L2. Currently (1 - lambda) is used in calculating L1 and L2. 5. The current calculation has this condition if (t1 != t2) continue; This condition is almost always true even if t1 and t2 are essentiallyEqual.
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The current implementation of the regression calculation has these flaws:
Hwhich is equal toC - I. Looking at the algorithm of Kundu/Ubhaya, this should be justC.if (t1 != t2) continue;This condition is almost always true even if t1 and t2 areessentiallyEqual.The text was updated successfully, but these errors were encountered: