Reply to Re: How to visualize network

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Posted by itschy on 01/20/08 01:13

On 19 Jan., 20:57, The Natural Philosopher <a...@b.c> wrote:
> itschy wrote:
> > On 19 Jan., 17:19, The Natural Philosopher <a...@b.c> wrote:
> >> itschy wrote:
> >>> On 18 Jan., 23:31, Jensen Somers <jensen.som...@gmail.com> wrote:
> >>>> PHP has support to create images, so my guess is - if I understand the
> >>>> problem correctly - you'll just need to find a way to draw dots and
> >>>> lines, making them thicker depending on the weight they have.
> >>> Not quite.
> >>> You missed the part, where I mentiond that I do not have any
> >>> coordinates of the nodes. Thus I'cant just draw them, cause I don't
> >>> know where. :)
> >>> What I seek is some solution, which finds a representation according
> >>> to the information I have about the edges (that is, connected or not,
> >>> and if, what weight/line length do they have).
> >>> E.g.:
> >>> I have three nodes A,B,C and 3 connections a(A,B,3), b(B,C,4), c(A,C,
> >>> 5) (each with infos about from_node, to_node and weight/length.
> >>> In this szenario, there are only two possible ways to draw that
> >>> network correctly:
> >>> 5
> >>> A-----C B
> >>> 3\ /4 3/ \4
> >>> B A-----C
> >>> 5
> >>> And their rotations.
> >>> I try to find some tool/algorithm which is able to do this (and draw
> >>> it) in php.
> >>> Tough luck I guess... :(
> >>> itschy
> >> Didn't they teach you basic geometry at school?
>
> >> three nodes with three paths, defines a unique shape, which is constant.
> >> It can be rotated and mirrored,without violating the original
> >> definition, but the shape remains constant.
>
> >> To plot it, simply assume one point (A) is at 0,0, the next one - say
> >> 'B' is at the correct distance along the X axis, so in your case its
> >> co-ordinates are 3,0.
>
> >> Solving the position of the third point C involves solving a pair of
> >> simultaneous quadratic equations. Cf Pythagoras and the dropping of
> >> perpendiculars.
>
> >> Namely:-
>
> >> Cx^2+Cy^2=(5)^2
> >> (3-Cx)^2 +Cy^2=(4)^2
>
> >> Now there are either two, or none, solutions for Cy as there always are
> >> for quadratics. One solution is the mirror of the other.
>
> >> Now go solve it.
>
> >> Its your homework, not mine.
>
> > Thanks for teaching me.
> > Now, have you read anything I wrote?
> > In short words i will summarize:
> > 0. They did not only teach me basic geometry in school, but also
> > analytical geometry, graph theory and such in university.
> > 1. I try to find an EXISTING solution not write my own.
> > 2. Jensen Somers misunderstood my initial post, so I clarified, what
> > kind of solution I seek. I described to him exactly what you did to me
> > (in a visual way).
> > 3. I know well how to do it, I wanted to know if it was done before
> > (because with all the drwaing stuff it is not worth it to implement it
> > by myself)
> > 4. Your answer is still welcome, maybe someone searching and finding
> > this thread can use it. but next time please don't assume everyone
> > else is stupid! :)
>
> I don't assume. I go on the evidence.
>
> Maths packages exist.
> Graphics packages exists.
> The mathematics of quadratics is well known.
>
> So all it needs s an unstupid unlazy person to put them all together.
>
> Php doesn't have a package to calculate the net loss in weight of a
> rutting elephant after sex, either.
>
> If you want soenmthing that obscure, write it.

So, the answer to my question would be:
"No, it doesn't exist"
All I wanted to know...

PS: I have to admit I never wondered about that elephant thing ;)

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