我想绘制以下不等式: y < p 2 (1 - p 1 )和 x < p 1 (1 - (y /(1 - p 1 )))。
鉴于第一个是满意的,我想绘制满足两者的区域 p 1 和 p 2 可以在[0,1]内变化。
我将不胜感激任何帮助!
答案 0 :(得分:17)
试试这个:红色区域是两个不等式都满足的地方。
[X,Y]=meshgrid(0:0.01:1,0:0.01:1); % Make a grid of points between 0 and 1
p1=0.1; p2=0.2; % Choose some parameters
ineq1 = Y<p2*(1-p1);
ineq2 = X<p1*(1-(Y./(1-p1)));
colors = zeros(size(X))+ineq1+ineq2;
scatter(X(:),Y(:),3,colors(:),'filled')
答案 1 :(得分:8)
另一种解决方案(与Sam Robert相似)将使用contourf
:
[X, Y] = meshgrid((0:999) / 1000, (0:999) / 1000);
p = rand(2, 1); %# In this example p = [0.1, 0.2]
ineq1 = Y < p(2) * (1 - p(1)); %# First inequation
ineq2 = X < p(1) * (1 - (Y / (1 - p(1)))); %# Second inequation
both = ineq1 & ineq2; %# Intersection of both inequations
figure, hold on
c = 1:3; %# Contour levels
contourf(c(1) * ineq1, [c(1), c(1)], 'b') %# Fill area for first inequation
contourf(c(2) * ineq2, [c(2), c(2)], 'g') %# Fill area for second inequation
contourf(c(3) * both, [c(3), c(3)], 'r') %# Fill area for both inequations
legend('First', 'Second', 'Both')
set(gca, ... %# Fixing axes ticks
'XTickLabel', {t(get(gca, 'XTick'))}, 'YTickLabel', {t(get(gca, 'YTick'))})
这就是结果:
红色区域(如图例中所述)表示满足两个不等式的位置。
请注意,第二次和第三次contourf
调用只是为了说明,以显示只有一个不等式被满足的地方。
答案 2 :(得分:2)
我认为这种方法很容易理解。制作曲面图并将其旋转到顶视图。
v = -5:0.1:5;
p1 = 0.1;
p2 = 0.2;
[x,y] = meshgrid(v);
ineq1 = y<p2*(1-p1);
ineq2 = x<p1*(1-(y./(1-p1)));
ineq = double(ineq1 & ineq2); % intersection of the inequalities
surf(x,y,ineq);
view(0,90) % rotate surface plot to top view