Details
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Bug
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Status: Closed
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Major
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Resolution: Fixed
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None
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None
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Description
Suppose that there is a subset S1 sorted on [0, 1] and we create a subset S2 in the same set that is sorted on [0]. All of the relational expressions that are in S1 will fall into S2. Therefore if S1 has a "best" relational expression, S2 should have one too.
We were failing to do this. S2's bestCost was set correctly, but best was left null.
---------------- Imported from GitHub ----------------
Url: https://github.com/julianhyde/optiq/issues/137
Created by: julianhyde
Labels:
Created at: Fri Feb 14 03:49:11 CET 2014
State: closed