Lua in ComputerCraft 1.5
This seems to work but the recursive loops is breaking after 4 or 5 times running.
Cannot seem to see why.
Am i doing something incredibly wrong here?
Full Code
Snippet for the loop:
x = 1
function loop()
if x > 0 then
getTarg()
derp1()
sleep(2.9)
monInit()
loop()
end
end
loop()
It looks as though you are not even using the X var... so why don't you try this... this is a more effective way to keep doing the same thing over and over
while true do
getTarg()
derp1()
sleep(2.9)
monInit()
end
Related
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
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
I'm currently working on a game using Roblox (which uses Lua). It is a basically made up of several minigames. At the beginning of each round, all the players in game are put in a table and teleported to an area. That is where the coroutine comes into play. As the round is in progress, I want a coroutine to start. Every second that coroutine checks if the player's health is below zero, and removes them from the currentPlayer table if it is.
Sorry if I am not describing the problem correctly, but the coroutine will not yield. I haven't used coroutines before, so I am probably trying to yield it the wrong way. I know most of you will not be familiar with Roblox, but the Lua syntax is the same.
Can someone please give me an example of how I would end a looping coroutine?
currentPlayers = {}
roundTime = 60
local lookForWinners = coroutine.create(function()
while coroutine.running do
wait(1)
for i, v in pairs(currentPlayers) do
if v.Character.Humanoid.Health <= 0 then
table.remove(currentPlayers, v)
end
end
end
end)
while wait() do
repeat display("Two or more players need to be in the game.", 1) until #_G.plrs > 1 --Ignore, just checks if two+ players are in game.
display("Picking a map...", 3) pickMap()
teleport(0, 500, 0)
coroutine.resume(lookForWinners)
wait(roundTime)
print("Round over")
coroutine.yield(lookForWinners)
end
Lua is a single-threaded language. Coroutines do not cause functions to execute in parallel.
Coroutines are effectively just a way to make a function that can pause its own execution (using coroutine.yield), that can be resumed from outside (using coroutine.resume). There is no "coroutine.running": there's only one line "running" at any given time.
If Roblox were meant for you to use wait() to jump out of the Lua thread, you would write this as a series of loops that check their condition and then call wait():
local currentPlayers={}
local roundTime = 60
while #_G.plrs > 1 do
display("Two or more players need to be in the game.", 1)
wait()
end
display("Picking a map...", 3) pickMap()
teleport(0, 500, 0)
for i=0, roundTime do
for i, v in pairs(currentPlayers) do
if v.Character.Humanoid.Health <= 0 then
table.remove(currentPlayers, v)
end
end
wait(1)
end
print("Round over")
However, this is bad code. (Whenever you write code, let loops with a "wait" function in them serve to indicate that something is being done incorrectly.) You should be using Roblox's Events to handle your game's logic.
Check to see if the game should start only when the number of players changes.
"Look For Winners" only when a Humanoid's health changes (the HealthChanged event).
Run the timer on some kind of timer or interval (don't forget that you'll probably want to end your game early once somebody has won).
Events have many, many advantages over a busy loop; the most visible one will be that your checks occur when the thing they're checking for happens, and not later.
I suggest you follow Stuart's advice to use events; this is mostly to provide additional information on what coroutines are to help you use them correctly.
Think of coroutines as functions that may return values, but with a twist: while a "normal" function completes when it executes return, when you yield from a coroutine, it saves its state, so that resume can then continue from the point where you yielded as if nothing happened. Note that you only yield from a coroutine and only to the point where the resume of that coroutine was done (this is no different from calling a function and returning from it; the control returns to the point where you called the function).
In addition to that, resume and yield calls allow you to pass values to the coroutine and return (intermediate) values from the coroutine. See this SO answer for an example of how this can be used.
One can still return from a coroutine and it's no different from returning from a function, which completes its execution. If you check the status of the coroutine at that time (coroutine.status), it should be "dead".
So, to answer your question how you can end a looping coroutine: you can return from it, you can yield() from it (and never resume it again), or you can call error(), which you can then catch and check for in the result of the resume call. Having said that, I agree with Stuart that it may be the wrong way to solve your problem.
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 use very simple Lua scripting in an online game called ROBLOX. My problem is that values in my scripts aren't changing! Example:
num = 0
while true do
num = num + 1
print(num)
wait(1)
end
That should count up starting on 0, but the number won't change. Could this be from the ROBLOX website? I can't figure out what else it might be.
What happens with
local num = 0
while true do
num = num + 1
print(num)
wait(1)
end
?
Maybe some other part of the system is changing the global num.
I just put your code in the Lua demo and it works fine if you remove the wait() function call. I'm assuming you defined this function somewhere?
There is nothing wrong with the code. You must be running it wrong. Also, wait is a function defined in the Roblox API. It is legit.
There is no error in your code. If you are using ROBLOX, then I'm not sure how you're running it wrong as it is a fairly simple interface. I'll try it in ROBLOX and see if it errors for me.
To the people who were wondering about wait(): it's a ROBLOX-specific global function, that pauses the current task the amount of seconds in the parentheses.
Try this:
local num = 0
while true do
num = num + 1
print(num)
print(type(num))
wait(1)
end