I have this code in Pine script for tradingview:
if close >= base_bar_low and close <= base_bar_high
//Check for one or more neighboring bars within 50-50 range of Stochastic or Williams %R
i := 0
for i in range(2, stoch_length + 1):
if (stoch_D[i] <= 50) and (stoch_D[i] >= 50) or (willR[i] >= -50) and (willR[i] <= 50):
//Check for last bar close within 30% of base bar's high
last_bar_high = high[1] + (atr * 0.3)
if close <= last_bar_high:
plot("W1-Long")
end
and I have this error: 'i' is not a valid type keyword in variable declaration - how can I avoid it?
Related
I am trying to make a program in turtle that creates a Lyapunov fractal. However, as using timeit shows, this should take around 3 hours to complete, 1.5 if I compromise resolution (N).
import turtle as t; from math import log; from timeit import default_timer as dt
t.setup(2000,1000,0); swid=t.window_width(); shei=t.window_height(); t.up(); t.ht(); t.tracer(False); t.colormode(255); t.bgcolor('pink')
def lyapunov(seq,xmin,xmax,ymin,ymax,N,tico):
truseq=str(bin(seq))[2:]
for x in range(-swid//2+2,swid//2-2):
tx=(x*(xmax-xmin))/swid+(xmax+xmin)/2
if x==-swid//2+2:
startt=dt()
for y in range(-shei//2+11,shei//2-1):
t.goto(x,y); ty=(y*(ymax-ymin))/shei+(ymax+ymin)/2; lex=0; xn=prevxn=0.5
for n in range(1,N+1):
if truseq[n%len(truseq)]=='0': rn=tx
else: rn=ty
xn=rn*prevxn*(1-prevxn)
prevxn=xn
if xn!=1: lex+=(1/N)*log(abs(rn*(1-2*xn)))
if lex>0: t.pencolor(0,0,min(int(lex*tico),255))
else: t.pencolor(max(255+int(lex*tico),0),max(255+int(lex*tico),0),0)
t.dot(size=1); t.update()
if x==-swid//2+2:
endt=dt()
print(f'Estimated total time: {(endt-startt)*(swid-5)} secs')
#Example: lyapunov(2,2.0,4.0,2.0,4.0,10000,100)
I attempted to use yield but I couldn't figure out where it should go.
On my slower machine, I was only able to test with a tiny N (e.g. 10) but I was able to speed up the code about 350 times. (Though this will be clearly lower as N increases.) There are two problems with your use of update(). The first is you call it too often -- you should outdent it from the y loop to the x loop so it's only called once on each vertical pass. Second, the dot() operator forces an automatic update() so you get no advantage from using tracer(). Replace dot() with some other method of drawing a pixel and you'll get back the advantage of using tracer() and update(). (As long as you move update() out of innermost loop as I noted.)
My rework of your code where I tried out these, and other, changes:
from turtle import Screen, Turtle
from math import log
from timeit import default_timer
def lyapunov(seq, xmin, xmax, ymin, ymax, N, tico):
xdif = xmax - xmin
ydif = ymax - ymin
truseq = str(bin(seq))[2:]
for x in range(2 - swid_2, swid_2 - 2):
if x == 2 - swid_2:
startt = default_timer()
tx = x * xdif / swid + xdif/2
for y in range(11 - shei_2, shei_2 - 1):
ty = y * ydif / shei + ydif/2
lex = 0
xn = prevxn = 0.5
for n in range(1, N+1):
rn = tx if truseq[n % len(truseq)] == '0' else ty
xn = rn * prevxn * (1 - prevxn)
prevxn = xn
if xn != 1:
lex += 1/N * log(abs(rn * (1 - xn*2)))
if lex > 0:
turtle.pencolor(0, 0, min(int(lex * tico), 255))
else:
lex_tico = max(int(lex * tico) + 255, 0)
turtle.pencolor(lex_tico, lex_tico, 0)
turtle.goto(x, y)
turtle.pendown()
turtle.penup()
screen.update()
if x == 2 - swid_2:
endt = default_timer()
print(f'Estimated total time: {(endt - startt) * (swid - 5)} secs')
screen = Screen()
screen.setup(2000, 1000, startx=0)
screen.bgcolor('pink')
screen.colormode(255)
screen.tracer(False)
swid = screen.window_width()
shei = screen.window_height()
swid_2 = swid//2
shei_2 = shei//2
turtle = Turtle()
turtle.hideturtle()
turtle.penup()
turtle.setheading(90)
lyapunov(2, 2.0, 4.0, 2.0, 4.0, 10, 100)
screen.exitonclick()
I try to use z3 do some easy math, condition is if without else.
........................................
for (i = 0; i <= 15; ++i)
{
if (s1[i] > 64 && s1[i] <= 90)
s1[i] = (s1[i] - 51) % 26 + 65;
if (s1[i] > 96 && s1[i] <= 122)
s1[i] = (s1[i] - 79) % 26 + 97;
}
for (i = 0; i <= 15; ++i)
{
result = key[i];
if (s1[i] != result)
return result;
}
from z3 import *
key = list(b"Qsw3sj_lz4_Ujw#l")
s1 = [BitVec('s1_%d' % i, 8) for i in range(len(key))]
s = Solver()
for i, n in enumerate(key):
con1 = If(Or(64 < s1[i], s1[i] <= 90), (s1[i] - 51) % 26 + 65 == key[i])
con2 = If(Or(96 < s1[i], s1[i] <= 122), (s1[i] - 79) % 26 + 97 == key[i])
s.add(con1)
s.add(con2)
print(s.check())
print(s.model())
Got error. TypeError: If() missing 1 required positional argument: 'c'
If takes three arguments; (1) condition, (2) then result, and (3) else result. You've only provided two. You can substitute True if you don't care about the else case constraint:
con1 = If(Or(64 < s1[i], s1[i] <= 90), (s1[i] - 51) % 26 + 65 == key[i], True)
and similarly for con2.
Note that there's no such thing as an If without else in z3.
While this will fix your "syntax" error, it won't fix the mistake you're making in your model; if any, of course. (As stated, your conditions are unsat, which I presume wasn't what you were expecting.) You should really consider what how the program changes (or doesn't change) when the condition is false, and instead of constraints like this, you should model how the variables themselves change after putting them into SSA (single-static assignment form.)
I'm checking for line-line intersection and need to figure out if the intersection point (x,y) is within the bounding box of a line segment l2 (consisting of points p1 and p2)
The following printout illustrates my problem:
the intersection point is (100,300)
print("x",x,">=",math.min(l2.p1.x,l2.p2.x),x >= math.min(l2.p1.x,l2.p2.x))
print("x",x,"<=",math.max(l2.p1.x,l2.p2.x),x <= math.max(l2.p1.x,l2.p2.x))
print("y",y,">=",math.min(l2.p1.y,l2.p2.y),y >= math.min(l2.p1.y,l2.p2.y))
print("y",y,"<=",math.max(l2.p1.y,l2.p2.y),y <= math.max(l2.p1.y,l2.p2.y))
which yeld:
x 100 >= 100 true
x 100 <= 100 false
y 300 >= 140 true
y 300 <= 300 false
What is going on and how can it be fixet?
(Lua version 5.2.3)
Say hello to float point arithmetic: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
For example
> string.format("%.32f", 1/3)
0.33333333333333331482961625624739
> 1/3 == 0.3333333
false
So it depends on how your calculations of X and lp* looks like.
You should use a tolerance while comparing float point numbers.
> math.abs(1/3 - 0.33) == 0
false
> math.abs(1/3 - 0.33333) < 1/10^6
false
> math.abs(1/3 - 0.33333) < 1/10^5
true
My a* path finding algorithm only works for certain cases but I don't understand why. Every node in my grid is walkable so in theory every path should work. I believe the error is in this line:
PathFindingNode *neighbor = NULL;
if ((y > 0 && x > 0) && (y < gridY - 1 && x < gridX - 1))
neighbor = [[grid objectAtIndex:x + dx] objectAtIndex:y +dy];
In function -(void)addNeighbors:, the line
if ((y > 0 && x > 0) && (y < gridY - 1 && x < gridX - 1))
neighbor = [[grid objectAtIndex:x + dx] objectAtIndex:y +dy];
has bug because if curNode is on boundary, it does not add neighbors to the queue. So that the algorithm will never reach endNode in the four corners (i.e. [0,0], [gridX-1,0], [0,gridY-1], [gridX-1,gridY-1]).
I'm creating a game and currently have to deal with some math.randomness.
As I'm not that strong in Lua, how do you think
Can you make an algorithm that uses math.random with a given percentage?
I mean a function like this:
function randomChance( chance )
-- Magic happens here
-- Return either 0 or 1 based on the results of math.random
end
randomChance( 50 ) -- Like a 50-50 chance of "winning", should result in something like math.random( 1, 2 ) == 1 (?)
randomChance(20) -- 20% chance to result in a 1
randomChance(0) -- Result always is 0
However I have no clue how to go on, and I completely suck at algorithms
I hope you understood my bad explanation of what I'm trying to accomplish
With no arguments, the math.random function returns a number in the range [0,1).
Lua 5.1.4 Copyright (C) 1994-2008 Lua.org, PUC-Rio
> =math.random()
0.13153778814317
> =math.random()
0.75560532219503
So simply convert your "chance" to a number between 0 and 1: i.e.,
> function maybe(x) if math.random() < x then print("yes") else print("no") end end
> maybe(0.5)
yes
> maybe(0.5)
no
Or multiply the result of random by 100, to compare against an int in the range 0-100:
> function maybe(x) if 100 * math.random() < x then print(1) else print(0) end end
> maybe(50)
0
> maybe(10)
0
> maybe(99)
1
Yet another alternative is to pass the upper and lower limits to math.random:
> function maybe(x) if math.random(0,100) < x then print(1) else print(0) end end
> maybe(0)
0
> maybe(100)
1
I wouldn't mess around with floating-point numbers here; I'd use math.random with an integer argument and integer results. If you pick 100 numbers in the range 1 to 100 you should get the percentages you want:
function randomChange (percent) -- returns true a given percentage of calls
assert(percent >= 0 and percent <= 100) -- sanity check
return percent >= math.random(1, 100) -- 1 succeeds 1%, 50 succeeds 50%,
-- 100 always succeeds, 0 always fails
end