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Re: Plot Joined Intelligently

  • To: mathgroup at
  • Subject: [mg30517] Re: [mg30501] Plot Joined Intelligently
  • From: Daniel Lichtblau <danl at>
  • Date: Fri, 24 Aug 2001 20:58:11 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

Hugh Goyder wrote:
> Dear Mathgroup,
> A list of {x, y} values supplied to ListPlot, with PlotJoined -> True,
> sensibly joins the points in the order of the list. Sometimes I produce
> points that lie on a set of smooth curves but do not produce the points in
> any particular order. How can I join the points to form a line
> intelligently, for example, joining each point to its nearest neighbour?
> An example. I solve for the roots of a polynomial with a parameter g that I
> vary. I convert the complex values to {x,y} pairs. I then plot using
> ListPlot which clearly shows that the roots lie along definite curves. How
> can I join the points up to make the curves lines?
> data = {Re[#], Im[#]} & /@ Flatten[Table[s /. NSolve[s^10 + g^10 (1 + s) ==
> 0, s], {g, 0, 2, 0.05}]];
> ListPlot[ data];
> Thanks
> Hugh Goyder

If the points come from, say, a multivalued algebraic function, it may
be better to work directly with the multiple branches. One approach in
that setting is to try to match up the (algebraic) functions at crossing
points. Some ideas for how one might try to find crossing points are in
a MathGroup post from a year ago, at the URL below.

Between crossings you can simply connect points that come from a
particular piece of the solution set. At crossings in effect you connect
pieces by matching up derivatives.

If all you have is a set of points partitioned by common x values, you
can simulate the above ideas as follows.

(i) Assume we are at an x value, x[0] not "near" to a crossing. Then
finding successor points to each {x[0],y[k,0]} pair is just a matter of
finding, for each point at the next x-value (x[1]), the closest of the
candidate y[1] values; it becomes y[k,1]. This assumes no crossing
occurred in between, which in turn is an assumption that derivatives are
not excessive.

(ii) When you get to an x value for which some y values are "close", the
appropriate connections are now obtained via derivative approximations.
You want the least bend, so (y[k,n+1]-y[k,n])/(x[n+1]-x[n]) should most
closely approximate (y[k,n]-y[k,n-1])/(x[n]-x[n-1]) in order to
correctly extend the kth curve.

Note that branches that "go complex" will cause trouble, as they join
together at a vertical tangent at the crossings. Also an tangential
crossing will be dificult to properly match. In fact, the method I show
at the above URL will not even find such crossings.

Daniel Lichtblau
Wolfram Research

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