Ok, here it goes another Euler problem question.
I've started to learn Lua by solving Euler project problems and got stuck on Euler problem 12.
It looks to me very straightforward and I don't understand why is my result incorrect?
Here is my solution so far:
-- return triangular number of the specified number
function get_tri_num(num)
local n = 0
for i=1, num do
n = n + i
end
return n
end
-- return all factors of the specifeid number
function factors(num)
local factors = {}
for i=1, num/2 do
if num%i == 0 then
factors[#factors+1] = i
end
end
factors[#factors+1] = num
return factors
end
-- get the first triangle number with >500 divisors
function euler12()
local n = 0
local trinum = 1
while true do
n = n + 7
trinum = get_tri_num(n)
if #factors(trinum) > 500 then break end
end
print(trinum, n)
end
euler12()
This problem is computation intensive, well, at least the way I am solving it, so I use luajit.
time luajit euler12.lua
103672800 14399
real 3m14.971s
user 3m15.033s
sys 0m0.000s
First, I try this solution on the toy example provided in the problem description. Changing the line of euler12() to if #factors(trinum) > 5 then break end, I get:
28 7
Which corresponds to the results shown in the problem example.
Second, after I see that the toy example is working I run euler12() with >500 condition. According to my solution the answer is 103672800 and yes, if I separately check the number of divisors for this result is >500:
print(#factors(103672800))
648
But...
The problem is here:
while true do
n = n + 7
Why does n increaments 7 each time? That doesn't make sense, change it to 1, and you could get the correct answer.
However, the performance is still poor. Several places that could be improved:
Every time the function get_tri_num is called, it's calculating
from scratch, that's not necessary.
You don't need the factors of a number, you only need the number of
factors of a number, so why return a table in factors?
for i=1, num/2 do is not necessary. Iterating to the square root of
num is enough to get the number of factors.
Refer to my code for the same problem.
Related
I am trying to find a solution to the problem "Two Sum" if you recognize it , and I've run into a problem and I cannot figure it out (Lua)
Code:
num = {2,7,11,15}
target = 9
current = 0
repeat
createNum1 = tonumber(num[math.random(1,#num)])
createNum2 = tonumber(num[math.random(1,#num)])
current = createNum1 + createNum2
until current == target
print(table.find(num,createNum1), table.find(num,createNum2))
Error:
lua5.3: HelloWorld.lua:9: attempt to call a nil value (field 'find')
stack traceback:
HelloWorld.lua:9: in main chunk
[C]: in ?
Thank you!
Lua has no table.find function in its very small standard library; just take a look at the reference manual.
You could implement your own table.find function, but that would just be monkey-patching an overall broken algorithm. There is no need to use a probabilistic algorithm that probably runs in at least quadratic time if there only is one pair of numbers that adds up to the desired number. Instead, you should leverage Lua's tables - associative arrays - here. First build an index of [number] = last index:
local num = {2,7,11,15}
local target = 9
local idx = {}
for i, n in ipairs(num) do idx[n] = i end
then loop over the numbers; given a number m you just need to look for target - m in your idx lookup:
for i, n in ipairs(num) do local j = idx[target - n]; if j then print(i, j) break end end
if you want to exit early - sometimes without building the full idx table - you can fuse the two loops:
local idx = {}
for i, n in ipairs(num) do
local j = idx[target - n]
if j then
print(j, i)
break
end
idx[n] = i
end
other solutions exist (e.g. using sorting, which requires no auxiliary space), but this one is elegant in that it runs in O(n) time & O(n) space to produce a solution and leverages Lua's builtin data structures.
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.
I am working on programming a Markov chain in Lua, and one element of this requires me to uniformly generate random numbers. Here is a simplified example to illustrate my question:
example = function(x)
local r = math.random(1,10)
print(r)
return x[r]
end
exampleArray = {"a","b","c","d","e","f","g","h","i","j"}
print(example(exampleArray))
My issue is that when I re-run this program multiple times (mash F5) the exact same random number is generated resulting in the example function selecting the exact same array element. However, if I include many calls to the example function within the single program by repeating the print line at the end many times I get suitable random results.
This is not my intention as a proper Markov pseudo-random text generator should be able to run the same program with the same inputs multiple times and output different pseudo-random text every time. I have tried resetting the seed using math.randomseed(os.time()) and this makes it so the random number distribution is no longer uniform. My goal is to be able to re-run the above program and receive a randomly selected number every time.
You need to run math.randomseed() once before using math.random(), like this:
math.randomseed(os.time())
From your comment that you saw the first number is still the same. This is caused by the implementation of random generator in some platforms.
The solution is to pop some random numbers before using them for real:
math.randomseed(os.time())
math.random(); math.random(); math.random()
Note that the standard C library random() is usually not so uniformly random, a better solution is to use a better random generator if your platform provides one.
Reference: Lua Math Library
Standard C random numbers generator used in Lua isn't guananteed to be good for simulation. The words "Markov chain" suggest that you may need a better one. Here's a generator widely used for Monte-Carlo calculations:
local A1, A2 = 727595, 798405 -- 5^17=D20*A1+A2
local D20, D40 = 1048576, 1099511627776 -- 2^20, 2^40
local X1, X2 = 0, 1
function rand()
local U = X2*A2
local V = (X1*A2 + X2*A1) % D20
V = (V*D20 + U) % D40
X1 = math.floor(V/D20)
X2 = V - X1*D20
return V/D40
end
It generates a number between 0 and 1, so r = math.floor(rand()*10) + 1 would go into your example.
(That's multiplicative random number generator with period 2^38, multiplier 5^17 and modulo 2^40, original Pascal code by http://osmf.sscc.ru/~smp/)
math.randomseed(os.clock()*100000000000)
for i=1,3 do
math.random(10000, 65000)
end
Always results in new random numbers. Changing the seed value will ensure randomness. Don't follow os.time() because it is the epoch time and changes after one second but os.clock() won't have the same value at any close instance.
There's the Luaossl library solution: (https://github.com/wahern/luaossl)
local rand = require "openssl.rand"
local randominteger
if rand.ready() then -- rand has been properly seeded
-- Returns a cryptographically strong uniform random integer in the interval [0, n−1].
randominteger = rand.uniform(99) + 1 -- randomizes an integer from range 1 to 100
end
http://25thandclement.com/~william/projects/luaossl.pdf
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
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );