I'm trying to calculate the sum of the prime numbers in a given number. For instance, for the number 123456, the result will be 10 because 2+3+5 = 10.
I tried to write a code that does that in Lua but I had some issues.
First, here is the code:
function isPrime(num)
if(num == 1 or (num ~= 2 and num%2 == 0)) then
return false
end
for i=3, math.sqrt(num), 2 do
if(num%i == 0) then
return false
end
end
return true
end
function sumOfPrimes(num)
local sum = 0
for digit in string.gmatch(num,"%d") do
local prime = isPrime(digit)
if(isPrime(digit)) then
sum = sum + digit
end
print(digit)
end
return sum
end
function main()
print(sumOfPrimes(123456))
end
main()
It returnes 9 instead of 10. Another thing I've noticed is it adds 1 also to sum, but 1 isn't a prime. What's the problem here?
string.gmatch returns a string, you need to convert it to number before doing calculations
Btw, you are doing the prime validation twice on your loop.
This is a fixed version (returns 10 as expected):
...
function sumOfPrimes(num)
local sum = 0
for digit in string.gmatch(num, "%d") do
digit = tonumber(digit) --needed conversion
local prime_digit = isPrime(digit)
if prime_digit then
sum = sum + digit
end
end
return sum
end
Related
I'm having problem returning spesific amount of decimal numbers from this function, i would like it to get that info from "dec" argument, but i'm stuck with this right now.
Edit: Made it work with the edited version bellow but isn't there a better way?
local function remove_decimal(t, dec)
if type(dec) == "number" then
for key, num in pairs(type(t) == "table" and t or {}) do
if type(num) == "number" then
local num_to_string = tostring(num)
local mod, d = math.modf(num)
-- find only decimal numbers
local num_dec = num_to_string:sub(#tostring(mod) + (mod == 0 and num < 0 and 3 or 2))
if dec <= #num_dec then
-- return amount of deciamls in the num by dec
local r = d < 0 and "-0." or "0."
local r2 = r .. num_dec:sub(1, dec)
t[key] = mod + tonumber(r2)
end
end
end
end
return t
end
By passing the function bellow i want a result like this:
result[1] > 0.12
result[2] > -0.12
result[3] > 123.45
result[4] > -1.23
local result = remove_decimal({0.123, -0.123, 123.456, -1.234}, 2)
print(result[1])
print(result[2])
print(result[3])
print(result[4])
I tried this but it seems to only work with one integer numbers and if number is 12.34 instead of 1.34 e.g, the decimal place will be removed and become 12.3. Using other methods
local d = dec + (num < 0 and 2 or 1)
local r = tonumber(num_to_string:sub(1, -#num_to_string - d)) or 0
A good approach is to find the position of the decimal point (the dot, .) and then extract a substring starting from the first character to the dot's position plus how many digits you want:
local function truncate(number, dec)
local strnum = tostring(number)
local i, j = string.find(strnum, '%.')
if not i then
return number
end
local strtrn = string.sub(strnum, 1, i+dec)
return tonumber(strtrn)
end
Call it like this:
print(truncate(123.456, 2))
print(truncate(1234567, 2))
123.45
1234567
To bulk-truncate a set of numbers:
local function truncate_all(t, dec)
for key, value in pairs(t) do
t[key] = truncate(t[key], dec)
end
return t
end
Usage:
local result = truncate_all({0.123, -0.123, 123.456, -1.234}, 2)
for key, value in pairs(result) do
print(key, value)
end
1 0.12
2 -0.12
3 123.45
4 -1.23
One could use the function string.format which is similar to the printf functions from C language. If one use the format "%.2f" the resulting string will contain 2 decimals, if one use "%.3f" the resulting string will be contain 3 decimals, etc. The idea is to dynamically create the format "%.XXXf" corresponding to the number of decimal needed by the function. Then call the function string.format with the newly created format string to generate the string "123.XXX". The last step would be to convert back the string to a number with the function tonumber.
Note that if one want the special character % to be preserved when string.format is called, you need to write %%.
function KeepDecimals (Number, DecimalCount)
local FloatFormat = string.format("%%.%df", DecimalCount)
local String = string.format(FloatFormat, Number)
return tonumber(String)
end
The behavior seems close to what the OP is looking for:
for Count = 1, 5 do
print(KeepDecimals(1.123456789, Count))
end
This code should print the following:
1.1
1.12
1.123
1.1235
1.12346
Regarding the initial code, it's quite straight-forward to integrate the provided solution. Note that I renamed the function to keep_decimal because in my understanding, the function will keep the requested number of decimals, and discard the rest.
function keep_decimal (Table, Count)
local NewTable = {}
local NewIndex = 1
for Index = 1, #Table do
NewTable[NewIndex] = KeepDecimal(Table[Index], Count)
NewIndex = NewIndex + 1
end
return NewTable
end
Obviously, the code could be tested easily, simply by copy and pasting into a Lua interpreter.
Result = keep_decimal({0.123, -0.123, 123.456, -1.234}, 2)
for Index = 1, #Result do
print(Result[Index])
end
This should print the following:
0.12
-0.12
123.46
-1.23
Edit due to the clarification of the need of truncate:
function Truncate (Number, Digits)
local Divider = Digits * 10
local TruncatedValue = math.floor(Number * Divider) / Divider
return TruncatedValue
end
On my computer, the code is working as expected:
> Truncate(123.456, 2)
123.45
I have what I believe is a proper implementation of the miller-rabin algorithm using Lua, and I am trying to get a consistent return for prime numbers. It seems my implementation only works half of the time. Although if I try implementing similar code within python, that code works 100% of the time. Could someone point me in the right direction?
--decompose n-1 as (2^s)*d
local function decompose(negOne)
exponent, remainder = 0, negOne
while (remainder%2) == 0 do
exponent = exponent+1
remainder = remainder/2
end
assert((2^exponent)*remainder == negOne and ((remainder%2) == 1), "Error setting up s and d value")
return exponent, remainder
end
local function isNotWitness(n, possibleWitness, exponent, remainder)
witness = (possibleWitness^remainder)%n
if (witness == 1) or (witness == n-1) then
return false
end
for _=0, exponent do
witness = (witness^2)%n
if witness == (n-1) then
return false
end
end
return true
end
--using miller-rabin primality testing
--n the integer to be tested, k the accuracy of the test
local function isProbablyPrime(n, accuracy)
if n <= 3 then
return n == 2 or n == 3
end
if (n%2) == 0 then
return false
end
exponent, remainder = decompose(n-1)
--checks if it is composite
for i=0, accuracy do
math.randomseed(os.time())
witness = math.random(2, n - 2)
if isNotWitness(n, witness, exponent, remainder) then
return false
end
end
--probably prime
return true
end
if isProbablyPrime(31, 30) then
print("prime")
else
print("nope")
end
Python has arbitrary length integers. Lua doesn't.
The problem is in witness = (possibleWitness^remainder)%n.
Lua is unable to calculate exact result of 29^15 % 31 directly.
There is a workaround working for numbers n < sqrt(2^53):
witness = mulmod(possibleWitness, remainder, n)
where
local function mulmod(a, e, m)
local result = 1
while e > 0 do
if e % 2 == 1 then
result = result * a % m
e = e - 1
end
e = e / 2
a = a * a % m
end
return result
end
I need Help adding a if clause to my Change Maker, so that if say I have denominations of coins that can't equal the input coin value. For Example I have Coins worth 2,4,6 and I have a Value of 1. I Want it to return Change Not Possible I tried adding a clause to it below but when I test it I get 1.#INF
I also am curious how I can find the optimal coin solution, So on top of saying the minimum number of coins it returns the optimal coin setup if there is one.
function ChangeMaking(D,n)
--[[
//Applies dynamic programming to find the minimum number of coins
//of denominations d1< d2 < . . . < dm where d1 = 1 that add up to a
//given amount n
//Input: Positive integer n and array D[1..m] of increasing positive
// integers indicating the coin denominations where D[1]= 1
//Output: The minimum number of coins that add up to n
]]
F = {} -- F is List Array of Coins
m = tablelength(D)
F[0] = 0
for i =1,n do
temp = math.huge
j = 1
while j <= m and i >= D[j] do
temp = math.min(F[ i - D[j] ], temp)
j = j + 1
end
F[i] = temp + 1
end
--I wanted to Catch the failed Solution here but I return 1.#INF instead
--if F[n] <= 0 and F[n] == 1.#INF then print("No Change Possible") return end
return F[n]
end
function main()
--[[
//Prints a Greeting, Asks for Denominations separated by spaces.
// Iterates through the input and assigns values to table
// Table is then input into ChangeMaker, and a while loop takes an n value for user input.
// User Enters 0 to end the Loop
]]
io.write("Hello Welcome the to Change Maker - LUA Edition\nEnter a series of change denominations, separated by spaces: ")
input = io.read()
deno = {}
for num in input:gmatch("%d+") do table.insert(deno,tonumber(num)) end
local i = 1
while i ~= 0 do
io.write("Please Enter Total for Change Maker, When Finished Enter 0 to Exit: ")
input2 = io.read("*n")
if input2 ~= 0 then io.write("\nMinimum # of Coins: " .. ChangeMaking(deno,input2).."\n") end
if input2 == 0 then i=0 print("0 Entered, Exiting Change Maker") end
end
end
function tablelength(T)
--[[
//Function for grabbing the total length of a table.
]]
local count = 0
for _ in pairs(T) do count = count + 1 end
return count
end
main()
The following script finds prime numbers in a range from 1 to 13.
When I explicitly iterate over the table that contains the results I can see that the script works as expected. However, if I use unpack() function on the table only the first 3 numbers get printed out.
From docs: unpack is "a special function with multiple returns. It receives an array and returns as results all elements from the array, starting from index 1".
Why is it not working in the script below?
t = {}
for i=1, 13 do t[i] = i end
primes = {}
for idx, n in ipairs(t) do
local isprime = true
for i=2, n-1 do
if n%i == 0 then
isprime = false
break
end
end
if isprime then
primes[idx] = n
end
end
print('loop printing:')
for i in pairs(primes) do
print(i)
end
print('unpack:')
print(unpack(primes))
Running
$ lua5.3 primes.lua
loop printing:
1
2
3
5
7
13
11
unpack:
1 2 3
Change
primes[idx] = n
to
primes[#primes+1] = n
The reason is that idx is not sequential as not every number is a prime.
Here is my Lua code for taking user input, and checking if the number entered is prime. My issue is that the program thinks that any even number is not prime, and any odd number is.
print("Enter a number.")
local number = io.read("*n")
function prime(n)
for i = 2, n^(1/2) do
if (n % i) == 0 then
return false
end
return true
end
end
if prime(number) == true then
print("Your number is prime!")
end
if prime(number) == false then
print("Your number is not prime!")
end
Move return true out of the loop.
Hence:
function prime(n)
for i = 2, n^(1/2) do
if (n % i) == 0 then
return false
end
end
return true
end
You return true too soon. You return true as soon as any i meets the condition. You must place the return after the loop.
I know it's an old post but since it's near the top on google I figured it can't hurt to post up my prime finder. It basically does a few simple checks of the obvious stuff and then loops through whats left in a similar fashion to the first example in Jon Ericson' post. Haven't benchmarked it but it seems to cope well enough.
--returns true if prime
function isPrime(n)
local n = tonumber(n)
--catch nil, 0, 1, negative and non int numbers
if not n or n<2 or (n % 1 ~=0) then
return false
--catch even number above 2
elseif n>2 and (n % 2 == 0) then
return false
--primes over 5 end in 1,3,7 or 9
--catch numbers that end in 5 or 0 (multiples of 5)
elseif n>5 and (n % 5 ==0) then
return false
--now check for prime
else
--only do the odds
for i = 3, math.sqrt(n), 2 do
--did it divide evenly
if (n % i == 0) then
return false
end
end
--can defeat optimus
return true
end
end
If you are going to be checking primality, you might as well pick an efficient algorithm. As one answer (cryptically) pointed out, all even numbers greater than 2 are not prime. Therefore, you can short-circuit the check for half the numbers, which doubles the speed to check any particular number:
function check_prime (x)
-- Negative numbers, 0 and 1 are not prime.
if x < 2 then
return false
end
-- Primality for even numbers is easy.
if x == 2 then
return 2
end
if x%2 == 0 then
return false
end
-- Since we have already considered the even numbers,
-- see if the odd numbers are factors.
for i = 3, math.sqrt(x), 2 do
if x%i == 0 then
return false
end
end
return x
end
There are all sorts of optimizations we could apply, but let's take a shot at doing this in a more Lua manner:
function sieve (x)
if x < 2 then
return false
end
-- Assume all numbers are prime until proven not-prime.
local prime = {}
prime[1] = false
for i = 2, x do
prime[i] = true
end
-- For each prime we find, mark all multiples as not-prime.
for i = 2, math.sqrt(x) do
if prime[i] then
for j = i*i, x, i do
prime[j] = false
end
end
end
return prime
end
To use the sieve function:
prime = sieve(number)
if prime[number] then
print("Your number is prime!")
else
print("Your number is not prime!")
end
In my tests, the sieve version is about 6 times faster than the previous algorithm for generating all the primes less than 1 million. (Your mileage may vary.) You can easily check the primality of all numbers less than number at no extra cost. On the other hand, it uses more memory and if you really want check the primality of just one number, it's less efficient.
I would check for primes by dividing the number with 2 and checking if the floor of the division is equal to the division. It looks like this.
if (input/2 == math.floor(input/2)) then
print("is prime")
else
print("is not prime")
end