region_plot with "or" expressions

DrNick
21 ●1 ●1 ●5

I'm trying to do a region_plot of a bunch of things that involve an "or" and getting behavior I don't understand.

Here's an example (not so relevant to mine) of the kind of thing I'm encountering:

region_plot([x < y or x < -y], (x,-1,1), (y, -1,1))

If you run this, clearly the output is messed up. It also changes when you switch the order of the two conditions, so it seems to pay attention to just one of them.

I tried carious things like making a function of (x,y) that returns the relevant boolean that works correctly when you just feed it numbers, but exhibits the same odd behavior when you put it in region_plot.

Any idea what's going on here?

Thanks.

Comments

By the way, the problem here was that using the `or` meant you were asking Sage/Python to evaluate `x<y` and `x<-y` - try `sage: x<y or="" x<-y`="" and="" see="" what="" you="" get="" :)="" since="" sage="" can't="" guarantee="" that="" the="" first="" one="" is="" true="" or="" false="" (`bool(x<y)`="" is="" `false`,="" as="" is="" `bool(x<-y)`),="" the="" outcome="" of="" `or`="" will="" automatically="" be="" the="" second="" thing.<="" p="">

kcrisman ( 12 years ago )

Sorry, I couldn't get the less than signs to work out right in the comment...

kcrisman ( 12 years ago )

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answered 12 years ago

vdelecroix
7640 ●20 ●85 ●168 http://www.labri.fr/pe...

Hello,

It is documented within the function region_plot (and you can access directly the documentation of a function through question mark). To plot a region with an 'or' statement you must use the following

region_plot([lambda x,y: x < y or x < -y], (x,-1,1), (y, -1,1))

For an 'and' statement you have the choice between

region_plot([lambda x,y: x < y and x < -y], (x,-1,1), (y, -1,1))

and

region_plot([x < y,x < -y], (x,-1,1), (y, -1,1))

Vincent

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Comments

I noticed that in the documentation, and I got the former to work for one function. I have a large family of "or"'s indexed, by say p in P and somehow this broke it. Should region_plot([lambda x,y: A(x,y,p) or B(x,y,p) for p in P], (x,0,1), (y,0,1)] work?

DrNick ( 12 years ago )

It depends on what you want : a list of graphics ? the intersection for p in P ? What you wrote will plot (A(x,y,p1) or B(x,y,p1)) and (A(x,y,p2) or B(x,y,p2)) and ... The lambda operator is very lazy, it takes the arguments and replace them in the expression you gave it.

vdelecroix ( 12 years ago )

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