Posted by The Natural Philosopher on 01/19/08 16:19

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.

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