L1 norm contour plot in matlab

Question

I am trying to plot the level set of L1 norm in matlab. But it is not at all working, and I am stuck. Any help ?

x = linspace(-1,1,10);
y = linspace(-1,1,10);
[xm,ym] = meshgrid(x,y);
z = sum(abs(xm-ym));
surfc(x,y,z)

Luis Mendo · Accepted Answer · 2015-05-04T21:30:55.773

5

Change your fourth line into this, in accordance with the definition of L1 norm:

z = abs(xm)+abs(ym);

You could do it more efficiently using with bsxfun to avoid generating the matrices xm, ym:

x = linspace(-1,1,10);
y = linspace(-1,1,10);
z = bsxfun(@plus, abs(x), abs(y).');
surfc(x,y,z)

Either of the two approaches produces:

enter image description here

For a better picture you should increase sampling, and perhaps remove surface edges:

x = linspace(-1,1,100);
y = linspace(-1,1,100);
z = bsxfun(@plus, abs(x), abs(y).');
surfc(x,y,z,'edgecolor','none')

enter image description here

edited May 04 '15 at 21:30

answered May 04 '15 at 21:11

Luis Mendo

I thought it would be z = abs(xm) - abs(ym); – kaz May 04 '15 at 21:14
@kaz A norm cannot be negative :-) – Luis Mendo May 04 '15 at 21:15
but what i see from other references, the contour is rectangular. here is it not rectangular. – Shew May 04 '15 at 21:28
The contour is (and should be) _square_. Try `x = linspace(-1,1,100); y = linspace(-1,1,100)` for better resolution – Luis Mendo May 04 '15 at 21:29
why do you use abs(y).' instead of abs(y)' ?. What effect does . have ? – Shew May 04 '15 at 21:38
2

That's one of my favourite questions :-) That's to use a _transpose_, instead of a _conjugate transpose_. See a discussion [here](http://stackoverflow.com/q/25150027/2586922) – Luis Mendo May 04 '15 at 21:41
50K! Quite an acheivement huh! Been a long way up there :) – Divakar May 05 '15 at 07:28
@Divakar Thanks! You're not that far from that :-) – Luis Mendo May 05 '15 at 12:16

1 Answers1