Hilbert curve using turtle graphics and recursion
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0
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I'm trying to implement an L-System generated Hilbert curve ,making use of python turtle graphics and recursion. My code seems to be working for the first two levels of recursion n=1 and n=2 but beyond that , the graphics just entangled (although I´m able to observe further modules within them), and I can´t seem to grasp what might be wrong here, do I need some intermediate steps to regenerate the Hilbert modules for deeper levels of recursion? Please see my code below , its relatively simple:
import turtle
def Hilbert_curve(A,rule,t,n):
if n>=1:
if rule:
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
t.right(90)
Hilbert_curve(A, rule,t, n-1)
t.forward(A)
Hilbert_curve(A,rule,t, n-1)
t.right(90)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
else:
t.right(90)
Hilbert_curve(A,rule,t, n-1)
t.forward(A)
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
t.forward(A)
Hilbert_curve(A, rule,t, n-1)
t.right(90)
def main():
A=10
t=turtle.Turtle()
my_win=turtle.Screen()
n=2
rule=True
Hilbert_curve(A,rule,t,n)
my_win.exitonclick()
main()
Hilbert n=2
Hilbert n=3
python recursion turtle-graphics fractals hilbert-curve
edited Nov 11 at 9:07
cdlane
16.6k21042
asked Nov 10 at 22:18
JFT
12
add a comment |
up vote
0
down vote
favorite
I'm trying to implement an L-System generated Hilbert curve ,making use of python turtle graphics and recursion. My code seems to be working for the first two levels of recursion n=1 and n=2 but beyond that , the graphics just entangled (although I´m able to observe further modules within them), and I can´t seem to grasp what might be wrong here, do I need some intermediate steps to regenerate the Hilbert modules for deeper levels of recursion? Please see my code below , its relatively simple:
import turtle
def Hilbert_curve(A,rule,t,n):
if n>=1:
if rule:
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
t.right(90)
Hilbert_curve(A, rule,t, n-1)
t.forward(A)
Hilbert_curve(A,rule,t, n-1)
t.right(90)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
else:
t.right(90)
Hilbert_curve(A,rule,t, n-1)
t.forward(A)
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
t.forward(A)
Hilbert_curve(A, rule,t, n-1)
t.right(90)
def main():
A=10
t=turtle.Turtle()
my_win=turtle.Screen()
n=2
rule=True
Hilbert_curve(A,rule,t,n)
my_win.exitonclick()
main()
Hilbert n=2
Hilbert n=3
python recursion turtle-graphics fractals hilbert-curve
edited Nov 11 at 9:07
cdlane
16.6k21042
asked Nov 10 at 22:18
JFT
12
if you like R, here's the one line coden=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s")just changenin for loop, e.g.for(k in 1:5)z=...This doesn't solve your problem here, but just in case you want to look around :)
– Dejan Marić
Nov 10 at 22:34
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to implement an L-System generated Hilbert curve ,making use of python turtle graphics and recursion. My code seems to be working for the first two levels of recursion n=1 and n=2 but beyond that , the graphics just entangled (although I´m able to observe further modules within them), and I can´t seem to grasp what might be wrong here, do I need some intermediate steps to regenerate the Hilbert modules for deeper levels of recursion? Please see my code below , its relatively simple:
import turtle
def Hilbert_curve(A,rule,t,n):
if n>=1:
if rule:
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
t.right(90)
Hilbert_curve(A, rule,t, n-1)
t.forward(A)
Hilbert_curve(A,rule,t, n-1)
t.right(90)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
else:
t.right(90)
Hilbert_curve(A,rule,t, n-1)
t.forward(A)
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
t.forward(A)
Hilbert_curve(A, rule,t, n-1)
t.right(90)
def main():
A=10
t=turtle.Turtle()
my_win=turtle.Screen()
n=2
rule=True
Hilbert_curve(A,rule,t,n)
my_win.exitonclick()
main()
Hilbert n=2
Hilbert n=3
python recursion turtle-graphics fractals hilbert-curve
edited Nov 11 at 9:07
cdlane
16.6k21042
asked Nov 10 at 22:18
JFT
12
I'm trying to implement an L-System generated Hilbert curve ,making use of python turtle graphics and recursion. My code seems to be working for the first two levels of recursion n=1 and n=2 but beyond that , the graphics just entangled (although I´m able to observe further modules within them), and I can´t seem to grasp what might be wrong here, do I need some intermediate steps to regenerate the Hilbert modules for deeper levels of recursion? Please see my code below , its relatively simple:
import turtle
def Hilbert_curve(A,rule,t,n):
if n>=1:
if rule:
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
t.right(90)
Hilbert_curve(A, rule,t, n-1)
t.forward(A)
Hilbert_curve(A,rule,t, n-1)
t.right(90)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
else:
t.right(90)
Hilbert_curve(A,rule,t, n-1)
t.forward(A)
t.left(90)
Hilbert_curve(A,not rule,t, n-1)
t.forward(A)
Hilbert_curve(A,not rule,t, n-1)
t.left(90)
t.forward(A)
Hilbert_curve(A, rule,t, n-1)
t.right(90)
def main():
A=10
t=turtle.Turtle()
my_win=turtle.Screen()
n=2
rule=True
Hilbert_curve(A,rule,t,n)
my_win.exitonclick()
main()
Hilbert n=2
Hilbert n=3
python recursion turtle-graphics fractals hilbert-curve
python recursion turtle-graphics fractals hilbert-curve
edited Nov 11 at 9:07
cdlane
16.6k21042
asked Nov 10 at 22:18
JFT
12
edited Nov 11 at 9:07
cdlane
16.6k21042
asked Nov 10 at 22:18
JFT
12
edited Nov 11 at 9:07
cdlane
16.6k21042
edited Nov 11 at 9:07
cdlane
16.6k21042
edited Nov 11 at 9:07
cdlane
16.6k21042
16.6k21042
asked Nov 10 at 22:18
JFT
12
asked Nov 10 at 22:18
JFT
12
asked Nov 10 at 22:18
JFT
12
12
if you like R, here's the one line coden=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s")just changenin for loop, e.g.for(k in 1:5)z=...This doesn't solve your problem here, but just in case you want to look around :)
– Dejan Marić
Nov 10 at 22:34
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25
add a comment |
if you like R, here's the one line coden=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s")just changenin for loop, e.g.for(k in 1:5)z=...This doesn't solve your problem here, but just in case you want to look around :)
– Dejan Marić
Nov 10 at 22:34
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25
if you like R, here's the one line code
n=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s") just change n in for loop, e.g. for(k in 1:5)z=... This doesn't solve your problem here, but just in case you want to look around :)– Dejan Marić
Nov 10 at 22:34
if you like R, here's the one line code
n=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s") just change n in for loop, e.g. for(k in 1:5)z=... This doesn't solve your problem here, but just in case you want to look around :)– Dejan Marić
Nov 10 at 22:34
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:
else:
t.right(90)
Hilbert_curve(A, not rule, t, n - 1)
t.forward(A)
t.left(90)
Hilbert_curve(A, rule, t, n - 1)
t.forward(A)
Hilbert_curve(A, rule, t, n - 1)
t.left(90)
t.forward(A)
Hilbert_curve(A, not rule, t, n - 1)
t.right(90)
However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:
from turtle import Screen, Turtle
def hilbert_curve(turtle, A, parity, n):
if n < 1:
return
turtle.left(parity * 90)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.forward(A)
turtle.right(parity * 90)
hilbert_curve(turtle, A, parity, n - 1)
turtle.forward(A)
hilbert_curve(turtle, A, parity, n - 1)
turtle.right(parity * 90)
turtle.forward(A)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.left(parity * 90)
screen = Screen()
yertle = Turtle()
yertle.speed('fastest') # because I have no patience
hilbert_curve(yertle, 10, 1, 4)
screen.exitonclick()
answered Nov 11 at 8:31
cdlane
16.6k21042
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:
else:
t.right(90)
Hilbert_curve(A, not rule, t, n - 1)
t.forward(A)
t.left(90)
Hilbert_curve(A, rule, t, n - 1)
t.forward(A)
Hilbert_curve(A, rule, t, n - 1)
t.left(90)
t.forward(A)
Hilbert_curve(A, not rule, t, n - 1)
t.right(90)
However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:
from turtle import Screen, Turtle
def hilbert_curve(turtle, A, parity, n):
if n < 1:
return
turtle.left(parity * 90)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.forward(A)
turtle.right(parity * 90)
hilbert_curve(turtle, A, parity, n - 1)
turtle.forward(A)
hilbert_curve(turtle, A, parity, n - 1)
turtle.right(parity * 90)
turtle.forward(A)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.left(parity * 90)
screen = Screen()
yertle = Turtle()
yertle.speed('fastest') # because I have no patience
hilbert_curve(yertle, 10, 1, 4)
screen.exitonclick()
answered Nov 11 at 8:31
cdlane
16.6k21042
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
add a comment |
up vote
2
down vote
The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:
else:
t.right(90)
Hilbert_curve(A, not rule, t, n - 1)
t.forward(A)
t.left(90)
Hilbert_curve(A, rule, t, n - 1)
t.forward(A)
Hilbert_curve(A, rule, t, n - 1)
t.left(90)
t.forward(A)
Hilbert_curve(A, not rule, t, n - 1)
t.right(90)
However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:
from turtle import Screen, Turtle
def hilbert_curve(turtle, A, parity, n):
if n < 1:
return
turtle.left(parity * 90)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.forward(A)
turtle.right(parity * 90)
hilbert_curve(turtle, A, parity, n - 1)
turtle.forward(A)
hilbert_curve(turtle, A, parity, n - 1)
turtle.right(parity * 90)
turtle.forward(A)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.left(parity * 90)
screen = Screen()
yertle = Turtle()
yertle.speed('fastest') # because I have no patience
hilbert_curve(yertle, 10, 1, 4)
screen.exitonclick()
answered Nov 11 at 8:31
cdlane
16.6k21042
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
add a comment |
up vote
2
down vote
up vote
2
down vote
The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:
else:
t.right(90)
Hilbert_curve(A, not rule, t, n - 1)
t.forward(A)
t.left(90)
Hilbert_curve(A, rule, t, n - 1)
t.forward(A)
Hilbert_curve(A, rule, t, n - 1)
t.left(90)
t.forward(A)
Hilbert_curve(A, not rule, t, n - 1)
t.right(90)
However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:
from turtle import Screen, Turtle
def hilbert_curve(turtle, A, parity, n):
if n < 1:
return
turtle.left(parity * 90)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.forward(A)
turtle.right(parity * 90)
hilbert_curve(turtle, A, parity, n - 1)
turtle.forward(A)
hilbert_curve(turtle, A, parity, n - 1)
turtle.right(parity * 90)
turtle.forward(A)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.left(parity * 90)
screen = Screen()
yertle = Turtle()
yertle.speed('fastest') # because I have no patience
hilbert_curve(yertle, 10, 1, 4)
screen.exitonclick()
answered Nov 11 at 8:31
cdlane
16.6k21042
The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:
else:
t.right(90)
Hilbert_curve(A, not rule, t, n - 1)
t.forward(A)
t.left(90)
Hilbert_curve(A, rule, t, n - 1)
t.forward(A)
Hilbert_curve(A, rule, t, n - 1)
t.left(90)
t.forward(A)
Hilbert_curve(A, not rule, t, n - 1)
t.right(90)
However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:
from turtle import Screen, Turtle
def hilbert_curve(turtle, A, parity, n):
if n < 1:
return
turtle.left(parity * 90)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.forward(A)
turtle.right(parity * 90)
hilbert_curve(turtle, A, parity, n - 1)
turtle.forward(A)
hilbert_curve(turtle, A, parity, n - 1)
turtle.right(parity * 90)
turtle.forward(A)
hilbert_curve(turtle, A, - parity, n - 1)
turtle.left(parity * 90)
screen = Screen()
yertle = Turtle()
yertle.speed('fastest') # because I have no patience
hilbert_curve(yertle, 10, 1, 4)
screen.exitonclick()
answered Nov 11 at 8:31
cdlane
16.6k21042
answered Nov 11 at 8:31
cdlane
16.6k21042
answered Nov 11 at 8:31
cdlane
16.6k21042
answered Nov 11 at 8:31
cdlane
16.6k21042
16.6k21042
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
add a comment |
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
Thank you , this works very well. Thank you for giving me the parity variable idea , I hadn´t considered before . My idea was to generate two cases for the two generation rules given in the L-System. Your parity solution is much more elegant
– JFT
Nov 11 at 15:38
add a comment |
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if you like R, here's the one line code
n=scan();a=1+1i;b=1-1i;z=0;for(k in 1:n)z=c((w<-1i*Conj(z))-a,z-b,z+a,b-w)/2;plot(z,t="s")just changenin for loop, e.g.for(k in 1:5)z=...This doesn't solve your problem here, but just in case you want to look around :)– Dejan Marić
Nov 10 at 22:34
Well I´m not familiar with R so I don´t really understand what´s your line of code but I´ll check it out
– JFT
Nov 11 at 15:25