I try to draw something in OCaml (try.ocamlpro.com).
I'm not sure how to draw a line, for example x=y using the function "plot x y". Eventually I tried something like this which clearly doesn't work:
open Graphics;;
Graphics.open_graph " 200x200";;
for i = 0 to x do
plot i (f i)
done
let g x = 2*x
read_line ();;
Any help (or examples) ?
Thank you.
There is also Graphics.lineto, which is based on a turtle-type system:
Graphics.open_graph " 200x200";;
Graphics.lineto 100 100;;
Changes the current point (defaults to 0, 0) to the supplied x y point and draws a line connecting the two.
You can set the current point with moveto : int -> int -> unit.
See the docs for more.
I can't see what is your problem. When I try your code, modified a little bit, I can draw a line as you want. First you need to initialize your window:
open Graphics;;
Graphics.open_graph " 200x200";;
Then you need to define your function f:
let f x = x + 1;;
And then just draw the line using the function plot
for i = 0 to 200 do
plot i (f i)
done;;
Voilà !
Related
I am trying to write some codes to find the global maximum of an equation, e.g. f = -x**4.
Here is what I have got at the moment.
import sympy
x = sympy.symbols('x')
f = -x**4
df = sympy.diff(f,x)
ans = sympy.solve(df,x)
Then I am stuck. How should I substitute ans back into f, and how would I know if that would be the maximum, but not the minimum or a saddle point?
If you are just looking for the global maximum and nothing else, then there is already a function for that. See the following:
from sympy import *
x = symbols('x')
f = -x**4
print(maximum(f, x)) # 0
If you want more information such as the x value that gives that max or maybe local maxima, you'll have to do more manual work. In the following, I find the critical values as you have done above and then I show the values as those critical points.
diff_f = diff(f, x)
critical_points = solve(diff_f, x)
print(critical_points) # x values
for point in critical_points:
print(f.subs(x, point)) # f(x) values
This can be extended to include the second derivative test as follows:
d_f = diff(f, x)
dd_f = diff(f, x, 2)
critical_points = solve(d_f, x)
for point in critical_points:
if dd_f.subs(x, point) < 0:
print(f"Local maximum at x={point} with f({point})={f.subs(x, point)}")
elif dd_f.subs(x, point) > 0:
print(f"Local minimum at x={point} with f({point})={f.subs(x, point)}")
else:
print(f"Inconclusive at x={point} with f({point})={f.subs(x, point)}")
To find the global max, you would need to take all your critical points and evaluate the function at those points. Then pick the max from those.
outputs = [f.subs(x, point) for point in critical_points]
optimal_x = [point for point in critical_points if f.subs(x, point) == max(outputs)]
print(f"The values x={optimal_x} all produce a global max at f(x)={max(outputs)}")
The above should work for most elementary functions. Apologies for the inconsistent naming of variables.
If you are struggling with simple things like substitution, I suggest going through the docs for an hour or two.
doc.text("msg", x, y);
writes msg below the position (x, y).
How to write it above this position?
Thank you
Why don't you just use doc.text("msg", x, y-100), or something similar to accomplish this?
You already have the x, and y coordinates. Just make adjustments to always modify this number as needed when writing text.
Thank-you Anon, but I needed a generic solution.
I have found yesterday a solution :
in the file pdfkit.js, in the "_fragment: fonction (...)", replace the line
y = this.page.height - y - ( this._font.ascender / 1000 * this._fontSize );
by
y = this.page.height - y;
This way it works all the time; even if a transformation matrix is applied to the text.
#+
Use the baseline option
doc.text("msg", x, y, { baseline:"bottom" })
For baseline, PdfKit supports sames values as Canvas textBaseLine. The default for baseline is 'top'.
I want to know how to get the x-coordinate position or the y-coordinate position of the mouse individually on pygame.
Like just the x and just the y. I think It would use
pygame.mouse.get_pos
Pygame doesn't have an API that will get you only one coordinate, you always get both. But it returns them in a 2-tuple, so you can index to get just one value if you want to:
x = pygame.mouse.get_pos()[0]
If there's any chance you might need the y coordinate as well, it might make sense to unpack as normal anyway, and just ignore the y value in the part of the code where you don't need it:
x, y = pygame.mouse.get_pos()
# do stuff with x, ignore y
if something_rare_happens():
# do stuff with y too
It might even be clearer to do the unpacking even if you'll never use y, but that's really up to you.
Could someone tell me if I've coded this correctly? This is my code for solving for the sides of a triangle given its perimeter, altitude, and angle (for the algebra see http://www.analyzemath.com/Geometry/challenge/triangle_per_alt_angle.html)
Prompt P
Prompt H
Prompt L [the angle]
(HP^2)/(2H(1+cos(L))+2Psin(L))→Y
(-P^2-2(1+cos(L))Y/(-2P)→Z
(Z+sqrt(Z^2-4Y))/2→N
[The same as above but Z-sqrt...]→R
If N>0
N→U
If R>0
R→U
Y/U→V
sqrt(U^2+V^2-2UVcos(L))→W
Disp U
Disp V
Disp W
Also, how would I fix this so that I can input angle = 90?
Also, in this code does it matter if the altitude is the one between b and c (refer to the website again)?
Thanks in advance
The code already works with L=90°.
Yes, the altitude must be the distance from point A to the base a between points B and C, forming a right-angle with that base. The derivation made that assumption, specifically with respect to the way it used h and a in the second area formula 1/2 h a. That exact formula would not apply if h was drawn differently.
The reason your second set of inputs resulted in a non-real answer is that sometimes a set of mathematical parameters can be inconsistent with each other and describe an impossible construct, and your P, h, and L values do exactly that. Specifically, they describe an impossible triangle.
Given an altitude h and angle L, the smallest perimeter P that can be achieved is an isosceles triangle split down the middle by h. With L=30, this would have perimeter P = a + b + c = 2h tan15 + h/cos15 + h/cos15, which, plugging in your h=3, results in P=7.819. You instead tried to use P=3+sqrt(3)=4.732. Try using various numbers less than 7.819 (plus a little; I've rounded here) and you'll see they all result in imaginary results. That's math telling you you're calculating something that cannot exist in reality.
If you fill in the missing close parenthesis between the Y and the / in line 5, then your code works perfectly.
I wrote the code slightly differently from you, here's what I did:
Prompt P
Prompt H
Prompt L
HP²/(2H(1+cos(L))+2Psin(L))→Y
(HP-Ysin(L))/H→Z
Z²-4Y→D
If D<0:Then
Disp "IMAGINARY"
Stop
End
(Z+√(D))/2→C
Y/C→B
P-(B+C)→A
Disp A
Disp B
Disp C
Edit: #Gabriel, there's nothing special (with respect to this question) about the angles 30-60-90; there is an infinite number of sets of P, h, and L inputs that describe such triangles. However, if you actually want to arrive at such triangles in the answer, you've actually changed the question; instead of just knowing one angle L plus P and h, you now know three angles (30-60-90) plus P and h. You've now over-specified the triangle, so that it is pretty well certain that a randomly generated set of inputs will describe an impossible triangle. As a contrived example, if you specified h as 0.0001 and P as 99999, then that's clearly impossible, because a triangle with a tiny altitude and fairly unextreme angles (which 30-60-90 are) cannot possibly achieve a perimeter many times its altitude.
If you want to start with just one of P or h, then you can derive equations to calculate all parameters of the triangle from the known P or h plus the knowledge of the 30-60-90 angles.
To give one example of this, if we assume that side a forms the base of the triangle between the 90° and 60° angles, then we have L=30 and (labelling the 60° angle as B) we have h=b, and you can get simple equations for all parameters:
P = a + h + c
sin60 = h/c
cos60 = a/c
=> P = c cos60 + c sin60 + c
P = c(cos60 + sin60 + 1)
c = P/(cos60 + sin60 + 1)
b = h = c sin60
a = c cos60
Plugging in P=100 we have
c = 100/(cos60 + sin60 + 1) = 42.265
b = h = 36.603
a = 21.132
If you plug in P=100, h=36.603, and L=30 into the code, you'll see you get these exact results.
Always optimize for speed, then size.
Further optimizing bgoldst's code:
Prompt P,H,L
HP²/(2H(1+cos(L))+2Psin(L
.5(Z+√((HP-sin(L)Ans)/H)²-4Ans
{Y/C→B,P-B-Ans,Ans
In my attempts to practice Julia, I've made a program which draws a bifurcation diagram. My code is as follows:
function bifur(x0,y0,a=1.3,b=0.4,n=1000,m=10000)
i,x,y=1,x0,y0
while i < n && abs(x) < m
x,y = a - x^2 + y, b * x
i += 1
end
if abs(x) < m
return x
else
return 1000
end
end
la = Float64[];
lx = Float64[];
for a=0:400
for j = 1:1000
x0 = rand()
y0 = rand()
x = bifur(x0,y0,a/100)
if x != 1000
push!(la,a/100)
push!(lx,x)
end
end
end
using Gadfly
myplot = Gadfly.plot( x=la, y=lx , Scale.x_discrete, Scale.y_continuous, Geom.point)
draw(PNG("myplot.png",10inch,8inch),myplot)
The output I get is this image:
In order to make my plot look more like this:
I need to be able to set point sizes to as small as one pixel. Then by increasing the iteration length I should be able to get a better bifurcation diagram. Does anyone know how to set the point sizes in Gadfly diagrams in Julia?
[Just to encapsulate the comments as an answer...]
Gadfly's Theme defaults can be changed. In particular, point_size is probably what you are looking for.
For changing the automatic plot scale/range settings, have a look at Gadfly's Scale params.