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Re: [CSSWG] Minutes and Resolutions 2009-02-04: box-shadow and border-image

From: Tab Atkins Jr. <jackalmage@gmail.com>
Date: Tue, 17 Feb 2009 19:58:02 -0600
Message-ID: <dd0fbad0902171758s41ce9bdr208ecb797db18f67@mail.gmail.com>
To: Aryeh Gregor <Simetrical+w3c@gmail.com>
Cc: fantasai <fantasai.lists@inkedblade.net>, "www-style@w3.org" <www-style@w3.org>
On Tue, Feb 17, 2009 at 5:17 PM, Aryeh Gregor <Simetrical+w3c@gmail.com> wrote:
> On Tue, Feb 17, 2009 at 5:13 PM, Tab Atkins Jr. <jackalmage@gmail.com> wrote:
>> On Tue, Feb 17, 2009 at 2:51 PM, fantasai <fantasai.lists@inkedblade.net> wrote:
>>>                     ______
>>>                    /      \
>>>                   /        \
>>>                  /    /\    \
>>>                 /    /  \    \
>> Huh.  Yeah, you're right.  Don't know why I didn't see that
>> immediately.  And since applying a manhattan distance metric really
>> *is* just like tracing it with a square brush, this is exactly what
>> would result.
> Are you certain about that?  It seems to me that you'd have to trace
> with a diamond shape (centered around the edge of the figure you're
> tracing) to get Manhattan distance.  The curve formed by remaining
> within a fixed Manhattan distance of a given point (a Manhattan
> circle) is certainly a diamond shape, not a square.  In the diagram
> given, the pointy thing would remain pointy.

Gah, also right.  Man, I'm zero for two.  I was thinking of the metric
where diagonals are distance 1, rather than distance 2.  I forget what
the name of that metric is.  In any case, it's probably a simpler
metric to apply than a true manhattan distance metric (distance is
max(x2-x1,y2-y1) rather than (x2-x1)+(y2-y1), and it corresponds very
easily to a simple loop).

> However, I believe the corners of a square would get flattened, by the
> same logic.  So even if I'm right, Manhattan distance isn't going to
> preserve sharp edges.


Received on Wednesday, 18 February 2009 01:58:44 UTC

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