FAQ

On Tue, May 31, 2011 at 2:54 PM, Tom Lane wrote:
Prevent problems by clamping negative penalty values to
zero.  (Just to be really sure, I also made it force NaNs to zero.)
Do gistchoose et al expect the triangle function to obey the triangle
inequality? If so isn't it possible treating NaNs as zero would fail
that? I'm not sure there's any safe assumption for NaN

--
greg

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  • Tom Lane at May 31, 2011 at 11:12 pm

    Greg Stark writes:
    On Tue, May 31, 2011 at 2:54 PM, Tom Lane wrote:
    Prevent problems by clamping negative penalty values to
    zero.  (Just to be really sure, I also made it force NaNs to zero.)
    Do gistchoose et al expect the triangle function to obey the triangle
    inequality?
    Don't think so.
    If so isn't it possible treating NaNs as zero would fail
    that? I'm not sure there's any safe assumption for NaN
    Well, leaving it as NaN is almost certain to not work desirably.

    regards, tom lane
  • Greg Stark at May 31, 2011 at 11:58 pm

    On Tue, May 31, 2011 at 4:11 PM, Tom Lane wrote:
    Do gistchoose et al expect the triangle function to obey the triangle
    inequality?
    Don't think so.
    I guess it was obvious but that was "expect the *penalty* function to
    obey the triangle inequality"

    --
    greg
  • Alexander Korotkov at Jun 1, 2011 at 7:20 am

    On Wed, Jun 1, 2011 at 3:57 AM, Greg Stark wrote:

    I guess it was obvious but that was "expect the *penalty* function to
    obey the triangle inequality"
    Actually, penalty function for boxes is even not commutative. Fox example:
    A = ((0,0)-(1,1))
    B = ((0,0)-(2,2))

    penalty(A,B) = 3
    penalty(B,A) = 0

    ------
    With best regards,
    Alexander Korotkov.

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