Posted by The Natural Philosopher on 01/20/08 02:12

itschy wrote:
> 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"

The astounding question I have, is whatever made you think it would?

I can't think of a single application for such a program.


> All I wanted to know...
>
> PS: I have to admit I never wondered about that elephant thing ;)

Well I never wondered about drawing a shape with three co-ordinates
defined in the way yours are.

[Back to original message]