I am trying to write a script that determines stardate using the formula
a = LastLeapYearShort (if year is leap year make 4 years ago)
b = 366 + (365 * ((CurrentYearShort - 1) - LastLeapYearShort)
c = DayOfYear - DayOfMonth
d = DayOfMonth
e = (SecondOfMinute + (MinuteOfHour * 60))/1440
f = 36525
st = ((a + b + c + d + e)/f)*100000
separate x.y into x and y
if the century is greater than 19 add 1- to the beginning of x and get the first to digits of y
date = x.y
however I can't seem to determine a way to get DayOfYear. The Current Script I have is
function isLeapYear(year)
return year%4==0 and (year%100~=0 or year%400==0)
end
function lastLeapYear(year)
if(isLeapYear(year))
result = strsub(year,2,4) - 4
else
year = year - 1
if(isLeapYear(year))
result = strsub(year,2,4)
else
year = year - 1
if(isLeapYear(year))
result = strsub(year,2,4)
else
year = year - 1
if(isLeapYear(year))
result = strsub(year,2,4)
else
result = "Invalid"
end
end
end
end
return result
end
function stardate()
yearf = os.date("%Y")
yearh = os.date("%y")
a = lastLeapYear(yearf)
b = (366 + (365 * (yearh - a)))
c = (!!DayOfYear!! - os.date("%d"))
d = os.date("%d")
e = (os.date("%S") + (os.date("%M") * 60))/1440
f = 36525
st = ((a + b + c + d + e)/f)*100000
!!Separate st into x and y!!
if(strsub(yearf,0,2) > 19)
diff = strsub(yearf,0,2) - 19
lead = diff "-" lead
end
return lead.dec
end
if there are any other errors in my code please point them out as I have very little Lua experience.
The day of year is the value of os.date("*t").yday or os.date("%j").
The first expression gives you a number; the second one gives you a string (which can be converted explicitly to a number with tonumber or implicitly when used in an arithmetic operation).
For any one who can't find the proper format string a more advanced list than that on http://www.lua.org/pil/22.1.html (which is where I was looking) can be found on http://docs.rainmeter.net/manual/measures/time
Related
Quick explanation, I have recently started using codewars to further improve my programming skills and my first challenge was to make a roman numeral decoder, I went through many versions because I wasnt satisfied with what I had, So I am asking if there is an easier way of handling all the patterns that roman numerals have, for example I is 1 but if I is next to another number it takes it away for example V = 5 but IV = 4.
here is my CODE:
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local number = 0
local i = 1
while i < #roman + 1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
if roman:sub(i,i) == "I" and roman:sub(i + 1,i + 1) ~= "I" and roman:sub(i + 1,i + 1) ~= "" then -- Checks for the I pattern when I exists and next isnt I
number = number + (Dict[roman:sub(i +1,i + 1)] - Dict[roman:sub(i,i)]) -- Taking one away from the next number
i = i + 2 -- Increase the counter
else
number = number + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
i = i + 1
end
end
return number
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII
at the moment I am trying to get 1049 (MXLIX) to work but I am getting 1069, obviously I am not following a rule and I feel like its more wrong then it should be because usually if its not correct its 1 or 2 numbers wrong.
The algorithm is slightly different: you need to consider subtraction when the previous character has less weight than the next one.
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local num = 0
local i = 1
for i=1, #roman-1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
local letter_p = roman:sub(i+1,i+1)
if (Dict[letter] < Dict[letter_p]) then
num = num - Dict[letter] -- Taking one away from the next number
print("-",Dict[letter],num)
else
num = num + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
print("+",Dict[letter],num)
end
end
num = num + Dict[roman:sub(-1)];
print("+",Dict[roman:sub(-1)], num)
return num
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII
this problem seems very simple but I cannot find a solution for it, actually I don't even know what is wrong!!!
So basically I have this Lua code:
io.write("\nPlease provide the message to be decyphered: ")
message = io.read()
seq = #message
ffib = {}
a = 0
b = 1
c = a + b
fib = 0
while c < (seq - 10) do
fib = fib + 1
ffib[fib] = c
a = b
b = c
c = a + b
end
decyphered = ""
for i = 1,seq do
decyphered = table.concat{decyphered, message:sub(ffib[i],ffib[i])}
end
io.write("\nDecyphered message: ", decyphered, "\n\n")
and trying to access ffib[fib] returns nil. So trying to message:sub(ffib[i]... later throws an error.
When I try accessing ffib's values manually, ffib[1] for example, it works alright, it's only when trying to access it with an iterator that it screws up.
Somewhere else in my code I have this:
io.write("\nPlease provide the message to be cyphered: ")
message = io.read()
cyphered = ""
seq = #message
ffib = {}
a = 0
b = 1
c = a + b
for fib = 1,seq do
ffib[fib] = c
a = b
b = c
c = a + b
end
which is basically the same thing but instead of using a while loop, it uses a for loop, and it works just fine!
Please help me solve this I am going insane.
Alright, I figured it out!
io.write("\nPlease provide the message to be decyphered: ")
message = io.read()
seq = #message
ffib = {}
a = 0
b = 1
c = a + b
fib = 0
while c < (seq - 10) do
fib = fib + 1
ffib[fib] = c
a = b
b = c
c = a + b
end
decyphered = ""
for i = 1,seq do <--------------
decyphered = table.concat{decyphered, message:sub(ffib[i],ffib[i])}
end
io.write("\nDecyphered message: ", decyphered, "\n\n")
I was using the wrong variable in the for loop, so it was looping through the entire message length instead of the fibonacci array length, the "nil" values were indexes out of bounds!
To correct this, I simply changed seq for #ffib in that For Loop, marked by an arrow.
Thanks everyone who tried to help me anyway!
this part doesn't make much sense I think
while c < (seq - 10) do
Why the minus 10? ffib will have less entries than seq while in the loop after that you expect a value in ffib from 1 to seq
And even if you change it to
while c < seq do
Then there still won't be enough for messages larger than length 2.
If anything, you might want to do
while c < (seq + 10) do
But even there you will run into an issue when the message is a certain length.
I'm also not familiar with that algorithm, but it looks pretty weird to me and I wonder what it actually establishes
i am working on a calculator running in pure lua but i need help with making the out put decimals in to fractions
This solution uses continued fraction to exactly restore fractions with denominator up to 107
local function to_frac(num)
local W = math.floor(num)
local F = num - W
local pn, n, N = 0, 1
local pd, d, D = 1, 0
local x, err, q, Q
repeat
x = x and 1 / (x - q) or F
q, Q = math.floor(x), math.floor(x + 0.5)
pn, n, N = n, q*n + pn, Q*n + pn
pd, d, D = d, q*d + pd, Q*d + pd
err = F - N/D
until math.abs(err) < 1e-15
return N + D*W, D, err
end
local function print_frac(numer,denom)
print(string.format("%.14g/%d = %d/%d + %g", numer, denom, to_frac(numer/denom)))
end
print_frac(1, 4) --> 1/4 = 1/4 + 0
print_frac(12, 8) --> 12/8 = 3/2 + 0
print_frac(4, 2) --> 4/2 = 2/1 + 0
print_frac(16, 11) --> 16/11 = 16/11 + 5.55112e-17
print_frac(1, 13) --> 1/13 = 1/13 + 0
print_frac(math.sqrt(3), 1) --> 1.7320508075689/1 = 50843527/29354524 + -4.44089e-16
print_frac(math.pi, 1) --> 3.1415926535898/1 = 80143857/25510582 + 4.44089e-16
print_frac(0, 3) --> 0/3 = 0/1 + 0
print_frac(-10, 3) --> -10/3 = -10/3 + -1.11022e-16
This is not possible. You need a class which stores fractions for that.
You can achieve an approximate solution. It works nicely for things that can be expressed as fraction and blows up for everything else
local function gcd(a, b)
while a ~= 0 do
a, b = b%a, a;
end
return b;
end
local function round(a)
return math.floor(a+.5)
end
function to_frac(num)
local integer = math.floor(num)
local decimal = num - integer
if decimal == 0 then
return num, 1.0, 0.0
end
local prec = 1000000000
local gcd_ = gcd(round(decimal*prec), prec)
local numer = math.floor((integer*prec + round(decimal*prec))/gcd_)
local denom = math.floor(prec/gcd_)
local err = numer/denom - num
return numer, denom, err
end
function print_frac(numer,denom)
print(string.format("%d/%d = %d/%d + %g", numer, denom, to_frac(numer/denom)))
end
print_frac(1,4)
print_frac(12,8)
print_frac(4,2)
print_frac(16,11)
print_frac(1,13)
Output:
1/4 = 1/4 + 0
12/8 = 3/2 + 0
4/2 = 2/1 + 0
16/11 = 290909091/200000000 + 4.54546e-10
1/13 = 76923077/1000000000 + 7.69231e-11
If X = weekday that the month started on (for example, this month would be 4, or Wednesday)
Y = some other day of the month (for example, 21)
Find Z, which is the weekday (1-7) of Y
I thought this would work:
Z = (Y-X) % 7
In the example above Z = (21-4) % 7 = 3, which is correct (Oct. 21st is a Tuesday)
But it fails for November 8th: Z = (8-7) % 7 = 1, incorrect because Nov. 8th is a Saturday (weekday=7).
So what would be a robust formula for this?
Note - I know there are NSDate utilities for finding the weekday of a date, but in this case all I know is X,Y as given above.
It is hard to use modular arithmetic if you don't start counting from zero.
So let's define some new variables:
W = X - 1 = the weekday number, where W = 0 means Sunday
D = Y - 1 = the day of the month, starting with 0
Then W + D is the weekday number (Sunday = 0) of day D, if W + D < 7.
So take (W + D) mod 7 to get the weekday number of day D. Add 1 to convert back to Sunday = 1, so ((W + D) mod 7) + 1.
Substitute the definitions of W and D.
Weekday number of day X (where Sunday = 1) = ((X - 1 + Y - 1) mod 7) + 1 = ((X + Y - 2) mod 7) + 1.
I am working on a client's site, and I'm writing an amortization schedule calculator in in ruby on rails. For longer loan term calculations, it doesn't seem to be breaking when the balance reaches 0
Here is my code:
def calculate_amortization_results
p = params[:price].to_i
i = params[:rate].to_d
l = params[:term].to_i
j = i/(12*100)
n = l * 12
m = p * (j / (1 - (1 + j) ** (-1 * n)))
#loanAmount = p
#rateAmount = i
#monthlyAmount = m
#amort = []
#interestAmount = 0
while p > 0
line = Hash.new
h = p*j
c = m-h
p = p-c
line["interest"] = h
line["principal"] = c
if p <= 0
line["balance"] = 0
else
line["balance"] = p
end
line["payment"] = h+c
#amort.push(line)
#interestAmount += h
end
end
And here is the view:
- #amort.each_with_index do |a, i|
%li
.m
= i+1
.i
= number_to_currency(a["interest"], :unit => "$")
.p
= number_to_currency(a["principal"], :unit => "$")
.pp
= number_to_currency(a["payment"], :unit => "$")
.b
= number_to_currency(a["balance"], :unit => "$")
What I am seeing is, in place of $0.00 in the final payment balance, it shows "-$-inf", iterates one more loop, then displays $0.00, but shows "-$-inf" for interest. It should loop until p gets to 0, then stop and set the balance as 0, but it isn't. Any idea what I've done wrong?
The calculator is here. It seems to work fine for shorter terms, like 5 years, but longer terms cause the above error.
Edit:
Changing the while loop to n.times do
and then changing the balance view to
= number_to_currency(a["balance"], :unit => "$", :negative_format => "$0.00")
Is a workaround, but i'd like to know why the while loop wouldn't work correctly
in Ruby the default for numerical values is Fixnum ... e.g.:
> 15 / 4
=> 3
You will see weird rounding errors if you try to use Fixnum values and divide them.
To make sure that you use Floats, at least one of the numbers in the calculation needs to be a Float
> 15.0 / 4
=> 3.75
> 15 / 4.0
=> 3.75
You do two comparisons against 0 , which should be OK if you make sure that p is a Float.
As the other answer suggests, you should use "decimal" type in your database to represent currency.
Please try if this will work:
def calculate_amortization_results
p = params[:price].to_f # instead of to_i
i = params[:rate].to_f # <-- what is to_d ? use to_f
l = params[:term].to_i
j = i/(12*100.0) # instead of 100
n = l * 12
m = p * (j / (1 - (1 + j) ** (-1 * n))) # division by zero if i==0 ==> j==0
#loanAmount = p
#rateAmount = i
#monthlyAmount = m
#amort = []
#interestAmount = 0.0 # instead of 0
while p > 0
line = Hash.new
h = p*j
c = m-h
p = p-c
line["interest"] = h
line["principal"] = c
if p <= 0
line["balance"] = 0
else
line["balance"] = p
end
line["payment"] = h+c
#amort.push(line)
#interestAmount += h
end
end
If you see "inf" in your output, you are doing a division by zero somewhere.. better check the logic of your calculation, and guard against division by zero.
according to Wikipedia the formula is:
http://en.wikipedia.org/wiki/Amortization_calculator
to improve rounding errors, it's probably better to re-structure the formula like this:
m = (p * j) / (1 - (1 + j) ** (-1 * n) # these are two divisions! x**-1 == 1/x
which is equal to:
m = (p * j) + (p * j) / ((1 + j) ** n) - 1.0)
which is equal to: (use this one)
q = p * j # this is much larger than 1 , so fewer rounding errors when dividing it by something
m = q + q / ((1 + j) ** n) - 1.0) # only one division
I think it has something to do with the floating point operations precision. It has already been discussed here: Ruby number precision with simple arithmetic and it would be better to use decimal format for financial purposes.
The answer could be computing the numbers in the loop, but with precomputed number of iterations and from the scratch.