city block (2)

Does anybody know the difference between the euclidean distance on a
triangular lattice vs. on a rectangular lattice?

It seems that the ED on a rectangular lattice is pretty much equivalent to
the city block function on a triangular lattice, with some intervals being
weighted by V2's and V3's.

Imagine a weighted triangular lattice actually drawn to scale, with rungs
of different lengths. Does anybody know what the euclidean distance would
be like on such a thing?

Carl

On Fri, 12 Mar 1999, Carl Lumma wrote:
> It seems that the ED on a rectangular lattice is pretty much equivalent to
> the city block function on a triangular lattice, with some intervals being
> weighted by V2's and V3's.

And then later:
> Gee, that's not right. I was only thinking of cutting one unit cube at a
> time. Still, Euclidean distance hasn't been mentioned in this thread.
> What does anybody think?

You can't get any city-block function to come out the same as Euclidian
distance, no matter what kind of weighting you use.

I don't really see that ED correlates to anything meaningful in this
context.

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

NOTE: dehyphenate node to remove spamblock. <*>

> Carl Lumma wrote:
>
>It seems that the ED on a rectangular lattice is pretty much equivalent to
>the city block function on a triangular lattice, with some intervals being
>weighted by V2's and V3's.

[ED is Euclidian distance]

15/8 and 6/5 still come out the same on a rectangular lattice.

>Imagine a weighted triangular lattice actually drawn to scale, with rungs
>of different lengths. Does anybody know what the euclidean distance would
>be like on such a thing?

I tried this a while back. I don't have figures to hand, but I remember the
heirarchy of intervals being pretty much the same as with the city block.
As the ED is much easier to calculate on a computer, I have no problem with
it as a pragmatic approximation. Others, it appears, disagree.

Graham Breed wrote,

>I tried this a while back. I don't have figures to hand, but I
remember the
>heirarchy of intervals being pretty much the same as with the city
block.
>As the ED is much easier to calculate on a computer, I have no problem
with
>it as a pragmatic approximation. Others, it appears, disagree.

The city block distance has the very important property that if an
interval arises most simply as the sum of two simpler intervals, the
metric of the first interval is the sum of the metrics of the other two.
Plus I've given my other reasons for the triangular city-block metric,
which is even easier to calculate on a computer.

Paul Erlich wrote:

>The city block distance has the very important property that if an
>interval arises most simply as the sum of two simpler intervals, the
>metric of the first interval is the sum of the metrics of the other two.

On a triangular lattice? What kind of sum do you mean?

>Plus I've given my other reasons for the triangular city-block metric,
>which is even easier to calculate on a computer.

You may be right, can I see the algorithm? Looks like I just happen to be
more familiar with continuous spaces.

>>The city block distance has the very important property that if an
>>interval arises most simply as the sum of two simpler intervals, the
>>metric of the first interval is the sum of the metrics of the other
two.

>On a triangular lattice? What kind of sum do you mean?

Well, 15:8 is the result of stacking a 5:4 and a 3:2, which is an
intervallic sum as would be normally understood. On a triangular _or_
rectangular lattice, the city block metric makes this the sum of the
metric of 5:4 and the metric of 3:2.

>>Plus I've given my other reasons for the triangular city-block metric,
>>which is even easier to calculate on a computer.

>You may be right, can I see the algorithm? Looks like I just happen to
be
>more familiar with continuous spaces.

Wait while Paul H, Manuel and I iron this one out.