local times=0
function rTA(v)
times=times+1
if times % 3 <= 0 then
print(v)
end
end
or
local times=0
function rTA(v)
times=times+1
if times == 3 then
print(v)
times=0
end
end
rTA("N1")
rTA("N2")
rTA("N3")
rTA("N4")
rTA("N5")
rTA("N6")
rTA("N7")
rTA("N8")
rTA("N9")
They both return the same output (N3, N6, N9), but I can't seem to understand the difference in both of them..
As pointed out both are checking if "times" is multiple of 3, although the first version is a little more "elegant" it costs more in terms of processing. The second is a little less readable in terms of meaning (you can understand that it is trying to check for multiples of 3, but it is not a first sight thing, you have to think for a moment).
Cheers
Related
I am working on Project Euler, and my code is just taking way too long to compute. I am supposed to find the sum of all primes less than 2,000,000 , but my program would take years to complete. I would try some different ways to find primes, but the problem is that I only know one way.
Anyways, here is my code:
sum=2
flag=0
prime=3
while prime<2000000 do
for i=2,prime-1 do
if prime%i==0 then
flag=1
end
end
if flag==0 then
print(prime)
sum=sum+prime
end
prime=prime+1
flag=0
if prime==2000000 then
print(sum)
end
end
Does anyone know of any more ways to find primes, or even a way to optimize this? I always try to figure coding out myself, but this one is truly stumping me.
Anyways, thanks!
This code is based on Sieve of Eratosthenes.
Whenever a prime is found, its multiples are marked as non-prime. Remaining integers are primes.
nonprimes={}
max=2000000
sum=2
prime=3
while prime<max do
if not nonprimes[prime] then
-- found a prime
sum = sum + prime
-- marks multiples of prime
i=prime*prime
while i < max do
nonprimes[i] = true
i = i + 2*prime
end
end
-- primes cannot be even
prime = prime + 2
end
print(sum)
As an optimization, even numbers are never considered. It reduces array size and number of iterations by 2. This is also why considered multiple of a found prime are (2k+1)*prime.
Your program had some bugs and computing n^2 divisions is very expensive.
How do I generate a sequence of integer numbers based on first and last number for for to loop over?
The following pseudocode
for i in sequence(4,9) do
print(i)
end
should produce the following output
4
5
6
7
8
9
Please include a short explanation what the solution does in the background and what terminology would have allowed one to find the solution.
Search attempts lead to unsearchable huge pages of documentation.
You can use numeric for loop to do that. You will find details in Programming in Lua section I referenced or the Lua manual section on For statement.
Just for the full record, there are basically 3 ways you could do this loop, one with a slightly different syntax, and 2 with the exact syntax as your pseudocode. Links point to relevant chapters in Programming in Lua (which is a great book to read, by the way).
1) Using a simple numeric for loop - in this case you won't use sequence:
for i=4,9 do
print(i)
end
2) Implement sequence as a closure:
function sequence(from,to)
local i = from - 1
return function()
if i < to then
i = i + 1
return i
end
end
end
for i in sequence(4,9) do print(i) end
3) Implement sequence as a coroutine:
function sequence(from, to)
return coroutine.wrap(function()
for i=from,to do
coroutine.yield(i)
end
end)
end
for i in sequence(4,9) do print(i) end
This is my code:
def is_prime(i)
j = 2
while j < i do
if i % j == 0
return false
end
j += 1
end
true
end
i = (600851475143 / 2)
while i >= 0 do
if (600851475143 % i == 0) && (is_prime(i) == true)
largest_prime = i
break
end
i -= 1
end
puts largest_prime
Why is it not returning anything? Is it too large of a calculation going through all the numbers? Is there a simple way of doing it without utilizing the Ruby prime library(defeats the purpose)?
All the solutions I found online were too advanced for me, does anyone have a solution that a beginner would be able to understand?
"premature optimization is (the root of all) evil". :)
Here you go right away for the (1) biggest, (2) prime, factor. How about finding all the factors, prime or not, and then taking the last (biggest) of them that is prime. When we solve that, we can start optimizing it.
A factor a of a number n is such that there exists some b (we assume a <= b to avoid duplication) that a * b = n. But that means that for a <= b it will also be a*a <= a*b == n.
So, for each b = n/2, n/2-1, ... the potential corresponding factor is known automatically as a = n / b, there's no need to test a for divisibility at all ... and perhaps you can figure out which of as don't have to be tested for primality as well.
Lastly, if p is the smallest prime factor of n, then the prime factors of n are p and all the prime factors of n / p. Right?
Now you can complete the task.
update: you can find more discussion and a pseudocode of sorts here. Also, search for "600851475143" here on Stack Overflow.
I'll address not so much the answer, but how YOU can pursue the answer.
The most elegant troubleshooting approach is to use a debugger to get insight as to what is actually happening: How do I debug Ruby scripts?
That said, I rarely use a debugger -- I just stick in puts here and there to see what's going on.
Start with adding puts "testing #{i}" as the first line inside the loop. While the screen I/O will be a million times slower than a silent calculation, it will at least give you confidence that it's doing what you think it's doing, and perhaps some insight into how long the whole problem will take. Or it may reveal an error, such as the counter not changing, incrementing in the wrong direction, overshooting the break conditional, etc. Basic sanity check stuff.
If that doesn't set off a lightbulb, go deeper and puts inside the if statement. No revelations yet? Next puts inside is_prime(), then inside is_prime()'s loop. You get the idea.
Also, there's no reason in the world to start with 600851475143 during development! 17, 51, 100 and 1024 will work just as well. (And don't forget edge cases like 0, 1, 2, -1 and such, just for fun.) These will all complete before your finger is off the enter key -- or demonstrate that your algorithm truly never returns and send you back to the drawing board.
Use these two approaches and I'm sure you'll find your answers in a minute or two. Good luck!
Do you know you can solve this with one line of code in Ruby?
Prime.prime_division(600851475143).flatten.max
=> 6857
Im a amatuer at coding. So, mind me if i face palmed some things.
Anyways, im making a alpha phase for a OS im making right? I'm making my installer. Two questions. Can i get a code off of pastebin then have my lua script download it? Two. I put the "print" part of the code in cmd. I get "Illegal characters". I dont know what went wrong. Here's my code.
--Variables
Yes = True
No = False
--Loading Screen
print ("1")
sleep(0.5)
print("2")
sleep(0.5)
print("Dowloading OS")
sleep(2)
print("Done!")
sleep(0.2)
print("Would you like to open the OS?")
end
I see a few issues with your code.
First of all, True and False are both meaningless names - which, unless you have assigned something to them earlier, are both equal to nil. Therefore, your Yes and No variables are both set to nil as well. This isn't because true and false don't exist in lua - they're just in lowercase: true and false. Creating Yes and No variables is redundant and hard to read - just use true and false directly.
Second of all, if you're using standard lua downloaded from their website, sleep is not a valid function (although it is in the Roblox version of Lua, or so I've heard). Like uppercase True and False, sleep is nil by default, so calling it won't work. Depending on what you're running this on, you'll want to use either os.execute("sleep " .. number_of_seconds) if you're on a mac, or os.execute("timeout /t " .. number_of_seconds) if you're on a PC. You might want to wrap these up into a function
function my_sleep_mac(number_of_seconds)
os.execute("sleep " .. number_of_seconds)
end
function my_sleep_PC(number_of_seconds)
os.execute("timeout /t " .. number_of_seconds)
end
As for the specific error you're experiencing, I think it's due to your end statement as the end of your program. end in lua doesn't do exactly what you think it does - it doesn't specify the end of the program. Lua can figure out where the program ends just by looking to see if there's any text left in the file. What it can't figure out without you saying it is where various sub-blocks of code end, IE the branches of if statements, functions, etc. For example, suppose you write the code
print("checking x...")
if x == 2 then
print("x is 2")
print("Isn't it awesome that x is 2?")
print("x was checked")
lua has no way of knowing whether or not that last statement, printing the x was checked, is supposed to only happen if x is 2 or always. Consequently, you need to explicitly say when various sections of code end, for which you use end. For a file, though, it's unnecessary and actually causes an error. Here's the if statement with an end introduced
print("checking x...")
if x == 2 then
print("x is 2")
print("isn't it awesome that x is 2?")
end
print("x was checked")
although lua doesn't care, it's a very good idea to indent these sections of code so that you can tell at a glance where it starts and ends:
print("checking x...")
if x == 2 then
print("x is 2")
print("isn't it awesome that x is 2?")
end
print("x was checked")
with regards to your "pastebin" problem, you're going to have to be more specific.
You can implement sleep in OS-independent (but CPU-intensive) way:
local function sleep(seconds)
local t0 = os.clock()
repeat
until os.clock() - t0 >= seconds
end
I would like to evaluate a math string in my corona app. Right now I'm focusing on the trig functions, so let's let the example be the most difficult we're likely to face:
local expr = "2sin(4pi+2)+7"
My goal is for this to somehow be (either) evaluated as is with maybe a pi --> math.pi switch, or to even break it up. The breaking up would be much more difficult, however, since it COULD be as complicated a above, but could also just be sin(1).
So I would prefer to stay as close to the python eval(expr) function as possible, but if that can't happen, I am flexible.
The simplest way would be to replace sin with math.sin (pi with math.pi and so on), add missing multiplications signs, and run it through loadstring, but loadstring is not available in Corona environment.
This means you will need to write your own parser for these expressions. I found a discussion on Corona forums that may help you as a starting point: here, with some details and a demo here
This should do the trick, it is able to use the lua math functions without putting 'math.function' so just sqrt(100) works fine. I threw this together because I have seen this question asked way too many times. Hopes this helps :)
If you have any questions feel free to contact me at rayaman99#gmail.com
function evaluate(cmd,v) -- this uses recursion to solve math equations
--[[ We break it into pieces and solve tiny pieces at a time then put them back together
Example of whats going on
Lets say we have "5+5+5+5+5"
First we get this:
5+5+5+5 + 5
5+5+5 + 5
5+5 + 5
5 + 5
Take all the single 5's and do their opperation which is addition in this case and get 25 as our answer
if you want to visually see this with a custom expression, uncomment the code below that says '--print(l,o,r)'
]]
v=v or 0
local count=0
local function helper(o,v,r)-- a local helper function to speed things up and keep the code smaller
if type(v)=="string" then
if v:find("%D") then
v=tonumber(math[v]) or tonumber(_G[v]) -- This section allows global variables and variables from math to be used feel free to add your own enviroments
end
end
if type(r)=="string" then
if r:find("%D") then
r=tonumber(math[r]) or tonumber(_G[r]) -- A mirror from above but this affects the other side of the equation
-- Think about it as '5+a' and 'a+5' This mirror allows me to tackle both sides of the expression
end
end
local r=tonumber(r) or 0
if o=="+" then -- where we handle different math opperators
return r+v
elseif o=="-" then
return r-v
elseif o=="/" then
return r/v
elseif o=="*" then
return r*v
elseif o=="^" then
return r^v
end
end
for i,v in pairs(math) do
cmd=cmd:gsub(i.."(%b())",function(a)
a=a:sub(2,-2)
if a:sub(1,1)=="-" then
a="0"..a
end
return v(evaluate(a))
end)
end
cmd=cmd:gsub("%b()",function(a)
return evaluate(a:sub(2,-2))
end)
for l,o,r in cmd:gmatch("(.*)([%+%^%-%*/])(.*)") do -- iteration this breaks the expression into managable parts, when adding pieces into
--print(":",l,o,r) -- uncomment this to see how it does its thing
count=count+1 -- keep track for certain conditions
if l:find("[%+%^%-%*/]") then -- if I find that the lefthand side of the expression contains lets keep breaking it apart
v=helper(o,r,evaluate(l,v))-- evaluate again and do the helper function
else
if count==1 then
v=helper(o,r,l) -- Case where an expression contains one mathematical opperator
end
end
end
if count==0 then return (tonumber(cmd) or tonumber(math[cmd]) or tonumber(_G[cmd])) end
-- you can add your own enviroments as well... I use math and _G
return v
end
a=5
print(evaluate("2+2+2*2")) -- This still has work when it comes to pemdas; however, the use parentheses can order things!
print(evaluate("2+2+(2*2)"))-- <-- As seen here
print(evaluate("sqrt(100)"))
print(evaluate("sqrt(100)+abs(-100)"))
print(evaluate("sqrt(100+44)"))
print(evaluate("sqrt(100+44)/2"))
print(evaluate("5^2"))
print(evaluate("a")) -- that we stored above
print(evaluate("pi")) -- math.pi
print(evaluate("pi*2")) -- math.pi