I am making a program that requires a person to guess a randomly generated number. But this is impossible as the computer is only generating extremely long decimals. Is there a way for math.random(1, 100) to only generate whole numbers?
function round(num, numDecimalPlaces)
local mult = 10^(numDecimalPlaces or 0)
return math.floor(num * mult + 0.5) / mult
end
Result:
round(1023.4345, -2) = 1000
round(1023.4345, 2) = 1023.43
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()
So I am fairly new to lua and I am learning how to use it by making a checkers app for android. When I found out that I would have to make the circles myself without a function i decided to look for other ways and it looks like the drawPoint() function is the best answer. When I made an equation to try and test it I got a error with the draw point function. It says:
attempt to call global 'drawPoint'(a nil value)
stack traceback:
main.lua54: in main chuck
am I missing something like do I have to import a library to use it? Thank you in advance and just for more background I was giving it the parameter of 2 ints. Here is my whole code below
--All the essential values
local widget = require ("widget")
local redCount = 12
local blackCount = 12
local length = 40.05
local x = length / 2
local y = 80
local startX = x
local startY = y
local allowMoves = true
--Title Display
local title = display.newText("Checkers", display.contentCenterX, 10, native.systemFontBold, 40)
--Functions
local function checkWinner()
if(redCount == 0) then
display.newText("Black Team Wins!!!", display.contentCenterX, display.contentHeight - 60, nativeSystemFontBold, 37)
elseif(blackCount == 0) then
display.newText("Red Team Wins!!!", display.contentCenterX, display.contentHeight - 60, nativeSystemFontBold, 40)
end
end
--Making the board
for i = 1, 8, 1
do
for k = 1, 4, 1
do
if i % 2 == 1 then
display.newRect( x, y, length, length, 30)
else
display.newRect( x + length, y, length, length, 30)
end
x = x + (startX * 4)
end
y = y + length
x = startX
end
--Making the outline
display.newRect(startX, startY - (length / 2), length * 15, 3, 30)
display.newRect(startX, y - (length / 2), length * 15, 3, 30)
display.newRect(startX - (length / 2), startY, 3, length * 15, 30)
display.newRect(320.4, startY, 3, length * 15, 30)
--Making the actual peices
drawPoint(2,5)
print(math.floor(12.5928))
The code above prints 12. What do you do to make it print 12.5?
print(math.floor(12.5928 * 10) / 10)
To generalize the previous answer, this function rounds a number to any number of decimals:
function round(number, decimals)
local power = 10^decimals
return math.floor(number * power) / power
end
Then, round(12.5928, 1) gives 12.5.
print(string.format("%.2f", 129.65686))
for two digits round of
print(math.floor((12.5928) * 100) / 100)
I'm looking for a rails function that could return the number to the nearest power of ten(10,100,1000), and also need to support number between 0 and 1 (0.1, 0.01, 0.001):
round(9) = 10
round(19) = 10
round(79) = 100
round(812.12) = 1000
round(0.0321) = 0.01
round(0.0921) = 0.1
I've looking on : Round number down to nearest power of ten
the accepted answer using the length of the string, that can't applied to number between 0 and 1.
updated
Round up to nearest power of 10 this one seems great. But I still can't make it work in rails.
I'm not sure about any function which automatically rounds the number to the nearest power of ten. You can achieve it by running the following code:
def rounded_to_nearest_power_of_ten(value)
abs_value = value.abs
power_of_ten = Math.log10(abs_value)
upper_limit = power_of_ten.ceil
lower_limit = power_of_ten.floor
nearest_value = (10**upper_limit - abs_value).abs > (10**lower_limit - abs_value).abs ? 10**lower_limit : 10**upper_limit
value > 0 ? nearest_value : -1*nearest_value
end
Hope this helps.
Let's simplify your problem to the following form - let the input numbers be in the range [0.1, 1), how would rounding of such numbers look like then?
The answer would be simple - for numbers smaller than 0.5 we would return the number 0.1, for larger numbers it would be 1.0.
All we have to do is to make sure that our number will be in that range. We will "move" decimal separator and remember how many moves we made in second variable. This operation is called normalization.
def normalize(fraction)
exponent = 0
while fraction < (1.0/10.0)
fraction *= 10.0
exponent -= 1
end
while fraction >= 1.0
fraction /= 10.0
exponent += 1
end
[fraction, exponent]
end
Using above code you can represent any floating number as a pair of normalized fraction and exponent in base 10. To recreate original number we will move decimal point in opposite direction using formula
original = normalized * base^{exponent}
With data property normalized we can use it in our simple rounding method like that:
def round(number)
fraction, exponent = normalize(number)
if fraction < 0.5
0.1 * 10 ** exponent
else
1.0 * 10 ** exponent
end
end
if the number is >= 1.0, this should work.
10 ** (num.floor.to_s.size - ( num.floor.to_s[0].to_i > 4 ? 0 : 1))
Try this:
def round_tenth(a)
if a.to_f >= 1
return 10 ** (a.floor.to_s.size - ( a.floor.to_s[0].to_i > 4 ? 0 : 1))
end
#a = 0.0392
c = a.to_s[2..a.to_s.length]
b = 0
c.split('').each_with_index do |s, i|
if s.to_i != 0
b = i + 1
break
end
end
arr = Array.new(100, 0)
if c[b-1].to_i > 4
b -= 1
if b == 0
return 1
end
end
arr[b-1] = 1
return ("0." + arr.join()).to_f
end
class Numeric
def name
def helper x, y, z
num = self.abs
r = 1
while true
result = nil
if num.between?(x, y)
if num >= y/2.0
result = y.round(r+1)
else
result = x.round(r)
end
return self.negative? ? -result : result
end
x *= z; y *= z; r += 1
end
end
if self.abs < 1
helper 0.1, 1, 0.1
else
helper 1, 10, 10
end
end
end
Example
-0.049.name # => -0.01
12.name # => 10
and so on, you are welcome!
For example I need number with minimum 3 digit
"512" --> 512
"24" --> 24.0
"5" --> 5.00
One option is write small function. Using answers here for my case it will be something like this
function f(value, w)
local p = math.ceil(math.log10(value))
local prec = value <= 1 and w - 1 or p > w and 0 or w - p
return string.format('%.' .. prec .. 'f', value)
end
print(f(12, 3))
But may be it is possible just using string.format() or any other simple way?
Ok, it seems this case beyond the string.format power. Thanks to #Schollii, this is my current variant
function f(value, w)
local p = math.ceil(math.log10(value))
local prec = value <= 1 and w - 1 or p > w and 0 or w - p
return string.format('%.' .. prec .. 'f', value)
end
print(f(12, 3))
There is no format code specifically for this since string.format uses printf minus a few codes (like * which would hace simplified the solution I give below). So you have to implement yourself, for example:
function f(num, w)
-- get number of digits before decimal
local intWidth = math.ceil(math.log10(num))
-- if intWidth > w then ... end -- may need this
local fmt='%'..w..'.' .. (w-intWidth) .. 'f'
return string.format(fmt, num)
end
print(f(12, 4))
print(f(12, 3))
print(f(12, 2))
print(f(512, 3))
print(f(24, 3))
print(f(5, 3))
You should probably handle case where integer part doesn't fit in field width given (return ceil or floor?).
You can't. Maximum you can reach - specify floating point precision or digit number, but you can't force output to be like your example. Lua uses C like printf with few limitations reference. Look here for full specifiers list link. Remember unsupported ones.
Writing a function would be the best and only solution, especially as your task looks strange, as it doesn't count decimal dot.