Coding BigInt Division [closed] - division

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 11 years ago.
Question: A good and efficient algorithm for BigInt division?
Attempted: Polynomial long division, division with ints (overflow remainder), binary long division (any good? Not sure, that's what I have in the post below), quotient guessing division (too many subtractions with large quotients).
I have been trying to code BigInt division for a while. My latest algorithm uses binary division, but I don't think that this is the best method.
So I am looking for some ideas as to what algorithms may be out there ; ).
The language I'm working in doesn't support passing items such as arrays around or various data types. I'm stuck with integers and booleans and global arrays as well as local arrays declared at the top of the function.
I'm working with a word size of 32768 for increased speed, which happens to be 2^15. Because of this, I can quickly and efficiently convert to base 2 and back, which is why I decided to try a binary division algorithm approach.
My first approach caused the remainder to overflow in some situations, but it was extremely fast. My next approach was an idea of polynomial long division. I also tried the quotient idea, although it would fail on extremely big numbers as there would be way too many subtractions involved. Overall, I think that the crappy binary division algorithm may be the best bet ; |.
Numbers get smaller towards the end of the divisor and remainder arrays. They are smaller towards the beginning of the dividend array.
The final answer is stored in binaryDividendBuffer with size binaryDividendBufferSize (quotient) and remainder with size remainderSize (remainder). This thing works with 0 bugs, but I have a feeling that it is just really bad :o.
globals
private static integer array binaryDividendBuffer //division binary buffer #1 (to be divided)
private static integer binaryDividendBufferSize //division binary count #1
private static integer array binaryBufferDivisor //division binary buffer #2 (to divide)
private static integer binaryBufferDivisorSize //division binary count #2
endglobals
Code:
local integer currentDividendDigit = binaryDividendBufferSize //to be divided int digit count
local integer tempDigit2 //temp digit 2
local integer tempDigit3 //temp digit 3
local integer array remainder //remainder
local integer remainderSize = 0 //remainder count
local boolean remainderLessThanDividend //is the remainder < divisor?
local integer binaryBufferDividendDigitOffset //subtract -1 or 0 (shift the divisor by 1 bit for extra digit)
local boolean gatheredDigits //were bits gatheredDigits?
loop
//gather bits equal to the length of the divisor only if the current remainder isn't equal to length of divisor and there are bits remaining
set gatheredDigits = false
set gatheredDigits = remainderSize != binaryBufferDivisorSize and 0 != currentDividendDigit
if (gatheredDigits) then
loop
exitwhen remainderSize == binaryBufferDivisorSize or 0 == currentDividendDigit
set currentDividendDigit = currentDividendDigit - 1
set remainder[remainderSize] = binaryDividendBuffer[currentDividendDigit]
set remainderSize = remainderSize + 1
set binaryDividendBuffer[currentDividendDigit] = 0
endloop
endif
//if the remainder is smaller than the divisor and there are no bits left to gather, exit
if (remainderSize < binaryBufferDivisorSize and 0 == currentDividendDigit) then
set binaryDividendBuffer[currentDividendDigit] = 0
exitwhen true
endif
//compare the remainder and the divisor to see which one is greater
set tempDigit2 = 0
set remainderLessThanDividend = false
loop
set remainderLessThanDividend = remainder[tempDigit2] < binaryBufferDivisor[tempDigit2]
set tempDigit2 = tempDigit2 + 1
exitwhen tempDigit2 == binaryBufferDivisorSize or remainderLessThanDividend or remainder[tempDigit2] > binaryBufferDivisor[tempDigit2]
endloop
//if remainderLessThanDividend and there are bits remaining, add an additional bit
//set the dividend's current bit to 0 IF bits were gatheredDigits (division taking place)
//if bits weren't gatheredDigits, then setting it to 0 will set an already divided bit
if (remainderLessThanDividend) then
exitwhen 0 == currentDividendDigit
if (gatheredDigits) then
set binaryDividendBuffer[currentDividendDigit] = 0
endif
set currentDividendDigit = currentDividendDigit - 1
set remainder[remainderSize] = binaryDividendBuffer[currentDividendDigit]
set remainderSize = remainderSize + 1
set binaryBufferDividendDigitOffset = -1 //shift divisor's bits by 1 to account for extra digit in remainder
else
set binaryBufferDividendDigitOffset = 0 //don't shift as there is no extra digit in remainder
endif
//subtract
set binaryDividendBuffer[currentDividendDigit] = 1
set tempDigit2 = remainderSize
loop
set tempDigit2 = tempDigit2 - 1
//if only subtract if the divisor actually has a bit to do subtracting (remainder might have 1 more bit than divisor)
if (tempDigit2 + binaryBufferDividendDigitOffset > -1) then
//if the remainder's current bit is remainderLessThanDividend than the divisor's bit, borrow
if (remainder[tempDigit2] < binaryBufferDivisor[tempDigit2 + binaryBufferDividendDigitOffset]) then
set remainder[tempDigit2 - 1] = remainder[tempDigit2 - 1] - 1
set remainder[tempDigit2] = remainder[tempDigit2] + 2
endif
//subtract them
set remainder[tempDigit2] = remainder[tempDigit2] - binaryBufferDivisor[tempDigit2 + binaryBufferDividendDigitOffset]
endif
exitwhen 0 == tempDigit2
endloop
//cut out all of the 0s in front of the remainder and shift it over
//000033 -> 33
//this first loop goes through all of the 0s
loop
exitwhen 0 != remainder[tempDigit2] or tempDigit2 == remainderSize
set tempDigit2 = tempDigit2 + 1
endloop
//this loop removes the 0s by shifting over
if (0 < tempDigit2) then
if (tempDigit2 == remainderSize) then
set remainderSize = 0
set remainder[0] = 0
else
set tempDigit3 = 0
set remainderSize = remainderSize-tempDigit2
loop
set remainder[tempDigit3] = remainder[tempDigit3+tempDigit2]
set remainder[tempDigit3+tempDigit2] = 0
set tempDigit3 = tempDigit3 + 1
exitwhen tempDigit3 == remainderSize
endloop
endif
endif
exitwhen 0 == currentDividendDigit
endloop
//cut out all of the 0s in front of dividend
loop
exitwhen 0 != binaryDividendBuffer[binaryDividendBufferSize]
set binaryDividendBufferSize = binaryDividendBufferSize - 1
endloop

Related

Looking for a fast way, to represent the value of an integer, by a number in the range 0 to 3 (without branching?)

Given that val is some random integer,
and number the possible outcome:
if the value is less then &H100 ; the number is 0
if the value is less then &H10000 ; the number is 1
if the value is less then &H1000000 ; the number is 2
else ; the number is 3
I've got this:
If (val And &HFFFF0000) = 0 Then
If (val And &HFF00) = 0 Then
num = 0
Else
numb = 1
End If
ElseIf (**val** And &HFF000000) = 0 Then
numb = 2
Else
numb = 3
End If
I believe to remember that I could achieve this with a simple calculation, but I can not
wrap my head around it...
cheers..
Jhonny
edit:--- after reaction of video.baba ---
Here is half a solution:
The result is a number from 0 to 7, of witch the bits represent a non-zero byte.
A lookuptable could be used to translate it to a 2-bit value.
val >>= 8 ' move to the right, so the first byte can hold identification-bits
val += &H3FF0000 'set a bit in the first byte, if the second one is not zero
val = val And &H400FFFF
val += &H1FFFF00 'set a bit in the first byte, if the third one is not zero
val = val And &H60000FF
val += &HFFFFFF'set a bit in the first byte, if the fourth one is not zero
val >>= 24 'put the result in the last byte
number=lookuptable(val)
have not tested it for speed yet, but it feels over-complicated?
Do you mean something like:
Select Case Value
Case < &H100
Number = 0
Case < &H10000
Number = 1
Case < &H1000000
Number = 2
Case Else
Number = 3
End Select

Quickest way to determine if a number is a prime number or not VB

I have been working on this for quite a bit now, I have a task of creating a program which calculates if a number entered by a user is prime number or not, the program calculates the time taken and displays this to the user, however I have found two method, one takes more time than the other but it produce accurate numbers, the other one calculates very quickly however it is wrong, I am hoping if someone can help me and tell me the quickest way of calculating this, here are my two codes
Code1:
Dim ch As String
ch = "y"
While ch = "y"
If (num Mod 2 = 0) Then
Console.WriteLine("Is not a prime number!")
Else
Console.WriteLine("Is a prime number!")
End If
Code2:
check = 1 'initilizing a check point to use it in the program to determine prime number
Dim Value As Long
Console.Write(vbLf & "Enter a number To check Whater it is Prime or Not :")
Value = Long.Parse(Console.ReadLine())
start_time = Now
Dim ch As ULong
ch = 0
Dim i As ULong
i = 2
While (i <= Value / 2)
If (Value Mod i = 0) Then
ch = 1
Exit While
End If
i = i + 1
End While
If (ch = 0) Then
Console.WriteLine("Prime Number")
Else
Console.WriteLine("Not Prime Number")
End If
There are a great many prime testers out there, many of them on this site. For checking a single number I use a faster variant of your Code2 with a little extra checking. Here is the pseudocode:
boolean function isPrime(num)
//1, 0 and negatives cannot be prime.
if (num < 2) then
return false
endif
// 2 is the only even prime.
if (num MOD 2 = 0) then
return (num = 2)
endif
// Check for odd factors.
limit <- sqrt(num)
for (factor <- 3; factor <= limit; factor <- factor + 2) do
if (num MOD factor = 0) then
return false
endif
endfor
// If we reach this point then the number is prime.
return true
endfunction
As #user448810 said, you should use the square root of your target number as the limit of your testing loop. You can basically halve the number of tests you do by treating even numbers separately. Once you have taken out the even numbers, then you only have to test odd factors: 3, 5, 7, ...

Generalizing increasing number of nested loop algorithm

Sorry for the terrible title, but I have no clue on how to generalize (or simplify) my loop case here.
I have a program that iterates to a sequence of integer, for example dimension=1 to 5.
In each iteration, there will be a main loop, and inside the main loop, there will be a nested loop. The number of the nested loop will be [dimension].
For example, in dimension=1, there is a For loop. In dimension=2, there is a For loop inside a For loop. And so on.
Is there any possible way to simplify the algorithm? currently I'm manually write totally different code for each value of [dimension]. Imagine if dimension=1 to 100? I'll be dead.
Here's my piece of program (written in VB.NET)
for dimension=2
Dim result(2) As Integer
For i = 0 To 1
For j = 0 To 1
result(0)=i
result(1)=j
Next
Next
For dimension=3
Dim result(3) As Integer
For i = 0 To 1
For j = 0 To 1
For k = 0 To 1
result(0)=i
result(1)=j
result(2)=k
Next
Next
Next
For dimension=4
Dim result(4) As Integer
For i = 0 To 1
For j = 0 To 1
For k = 0 To 1
For l = 0 To 1
result(0)=i
result(1)=j
result(2)=k
result(3)=l
Next
Next
Next
Next
And so on..
Any suggestion?
Thanks!
There are plenty of solutions:
Recursion
Idk, if vb.net supports methods, but if it does, this would probably be the simplest:
void nestedLoop(int lower , int upper , int remaining_loops , int[] values)
if(remaining_loops == 0)
//process values list
else
for int i in [lower , upper)
values[remaining_loops] = i
nestedLoop(lower , upper , remaining_loops - 1)
Integer Transformation
In theory, a number can be represented by any radix:
d_i * radix ^ i + d_i-1 * radix ^ (i - 1) ... + d_0 * radix ^ 0
Consider each digit the value of one of the nested loops:
for int i in [0 , max)
for int j in [0 , max)
for int k in [0 , max)
...
Could be represented by a 3-digit number with radix max, where d_0 = i, d_1 = j, etc.. Basically how each digit is mapped to one of the values can be arbitrary and will only affect the order of the output.
void nestedLoops(int upper , int dimension)
for int i in [0 , pow(upper , dimension))
int[] values
int digit_sub = 1
int tmp = i
for int j in [0 , dimension)
values[j] = tmp % dimension
tmp /= dimension
//all values of the loops are now in values
//process them here
There would be a few other options aswell, but these are the most common.
Please do note that when you do
Dim result(2) As Integer
You are actually declaring an array of 3 elements see this question for why. It's a subtle difference in VB.NET
That being said, I'll assume that you meant to declare an array of only 2 elements. If this is the case then you could build and call a recursive function like this
LoopOver(result)
Sub LoopOver(ByRef array() As Integer, ByVal Optional level As Integer = 0)
If array.Length = level Then
Return
Else
array(level) = 1
LoopOver(array, level + 1)
End If
End Sub
This recursive function will call itself (i.e., it will loop) for as many times as the array's size.

Verify Gamefield VB.NET

So I'm developing a minesweeper game and im assigning the mines, but I've got to check where are the mines now, in order to generate the numbers. The problem is that when I'm verifying the columns and lines I need the program not to get out of the game field.
Here's how my code looks like now:
Public Sub avisinhos(ByVal line, ByVal column)
If mat(line, column) = 0 Then
mat(line, column) = -1
numbandeiras = numbandeiras + 1
End If
For auxlinha = -1 To 1
For auxcolumn = -1 To 1
Next
Next
End Sub
How do I create a IF function to verify that I don't get out of the game field?
Best regards, joao.
pseudo code
int linestart = -1;
int lineend = 1;
int colstart = -1;
int colend = 1;
Assuming a 10 x 10 grid (zero based)
if line < 2 linestart = 0
if line > 8 lineend = 0
if column < 2 colstart = 0
if column > 8 colend = 0
For auxlinha = linestart To lineend
For auxcolumn = colstart To colend
// check
Next
Next
Personally though I wouldn't bother with the loops, they add very little to nothing
HasMineAbove = (line > 1) and (gamefield[line -1,column] = MinePresentValue
would be my approach, do it all in one.
Not to mention the huge potential confusion when auxlinha and auxcolumn are both zero...
I'm not sure exactly what your code is saying. It's a bit cryptic since you're using abbreviations and all lowercase names. You might want to try camelCasing and spelling out the words more completely, intellisense is your friend. =)
But coding style aside, if you are trying to loop through a limited range of values, you can keep your values bounded by using the modulus operator (%). For example, if you need to keep you values between 0-7 and you end up with a value of 12, just take the modulus of 8 to loop back to within range with a value of 4:
12 % 8 = 4
9 % 8 = 1
15 % 8 = 7
24 % 8 = 0
I realize this doesn't answer your specific question, but it's a handy technique might find useful.

How can I convert a decimal to a fraction?

How do I convert a indefinite decimal (i.e. .333333333...) to a string fraction representation (i.e. "1/3"). I am using VBA and the following is the code I used (i get an overflow error at the line "b = a Mod b":
Function GetFraction(ByVal Num As Double) As String
If Num = 0# Then
GetFraction = "None"
Else
Dim WholeNumber As Integer
Dim DecimalNumber As Double
Dim Numerator As Double
Dim Denomenator As Double
Dim a, b, t As Double
WholeNumber = Fix(Num)
DecimalNumber = Num - Fix(Num)
Numerator = DecimalNumber * 10 ^ (Len(CStr(DecimalNumber)) - 2)
Denomenator = 10 ^ (Len(CStr(DecimalNumber)) - 2)
If Numerator = 0 Then
GetFraction = WholeNumber
Else
a = Numerator
b = Denomenator
t = 0
While b <> 0
t = b
b = a Mod b
a = t
Wend
If WholeNumber = 0 Then
GetFraction = CStr(Numerator / a) & "/" & CStr(Denomenator / a)
Else
GetFraction = CStr(WholeNumber) & " " & CStr(Numerator / a) & "/" & CStr(Denomenator / a)
End If
End If
End If
End Function
As .333333333 is not 1/3 you will never get 1/3 but instead 333333333/1000000000 if you do not add some clever "un-rounding" logic.
Here is a solution for handling numbers with periodic decimal representation I remember from school.
A number 0.abcdabcd... equals abcd/9999. So 0.23572357... equals 2357/9999 exactly. Just take that many 9s as your pattern is long. 0.11111... equals 1/9, 0.121212... equals 12/99, and so on. So try just searching a pattern and setting the denominator to the corresponding number. Of course you have to stop after some digits because you will never know if the pattern is repeated for ever or just many times. And you will hit the rounding error in the last digit, so you still need some clever logic.
This only works in Excel-VBA but since you had it tagged "VBA" I will suggest it. Excel has a custom "fraction" format that you can access via "Format Cells" (or ctrl-1 if you prefer). This particular number format is Excel-Specific and so does not work with the VBA.Format function. It does however work with the Excel Formula TEXT(). (Which is the Excel equivalent of VBA.Format. This can be accessed like So:
Sub Example()
MsgBox Excel.WorksheetFunction.Text(.3333,"# ?/?")
End Sub
To show more than one digit (Example 5/12) just up the number of question marks.
Google for "decimal to fraction" and you'll get about a gazillion results.
I really like this one, because it's simple, has source code (in RPL, similar to Forth, ~25 lines), and is pretty fast (it's written to run on a 4-bit, 4MHz CPU). The docs say:
In a book called Textbook of Algebra by G. Chrystal, 1st
edition in 1889, in Part II, Chapter 32, this improved continued fraction
algorithm is presented and proven. Odd to tell, Chrystal speaks of it as if it
were ancient knowledge.
This site seem to have a really nice implementation of this in JavaScript.
I would multiply by 10000000(or whatever you want depending on the precision), then simplify the resulting fraction (ie n*10000000/10000000)
You can approximate it. Essentially cycle through all numerators and denominators until you reach a fraction that is close to what you want.
int num = 1;
int den = 1;
double limit == 0.1;
double fraction = num / den;
while(den < 1000000 ) // some arbitrary large denominator
{
den = den + 1;
for(num = 0; num <= den; num++)
{
fraction = num / den;
if(fraction < n + limit && fraction > n - limit)
return (num + "/" + den);
}
}
This is slow and a brute force algorithm, but you should get the general idea.
In general, it'll be easier if you find the repeating part of the rational number. If you can't find that, you'll have a tough time. Let's say the number if 8.45735735735...
The answer is 8 + 45/100 + 735/999/100 = 8 1523/3330.
The whole number is 8.
Add 45/100 - which is .45, the part before the repeating part.
The repeating part is 735/999. In general, take the repeating part. Make it the numerator. The denominator is 10^(number of repeating digits) - 1.
Take the repeating part and shift it the appropriate number of digits. In this case, two, which means divide by 100, so 735/999/100.
Once you figure those parts out, you just need some code that adds and reduces fractions using greatest common fractions ...
Similar to CookieOfFortune's, but it's in VB and doesn't use as much brute force.
Dim tolerance As Double = 0.1 'Fraction has to be at least this close'
Dim decimalValue As Double = 0.125 'Original value to convert'
Dim highestDenominator = 100 'Highest denominator you`re willing to accept'
For denominator As Integer = 2 To highestDenominator - 1
'Find the closest numerator'
Dim numerator As Integer = Math.Round(denominator * decimalValue)
'Check if the fraction`s close enough'
If Abs(numerator / denominator - decimalValue) <= tolerance Then
Return numerator & "/" & denominator
End If
Next
'Didn't find one. Use the highest possible denominator'
Return Math.Round(denominator * decimalValue) & "/" & highestDenominator
...Let me know if it needs to account for values greater than 1, and I can adjust it.
EDIT: Sorry for the goofed up syntax highlighting. I can't figure out why it's all wrong. If someone knows how I can make it better, please let me know.
Python has a nice routine in its fractions module. Here is the working portion that converts a n/d into the closest approximation N/D where D <= some maximum value. e.g. if you want to find the closest fraction to 0.347, let n=347,d=1000 and max_denominator be 100 and you will obtain (17, 49) which is as close as you can get for denominators less than or equal to 100. The '//' operator is integer division so that 2//3 gives 0, i.e. a//b = int(a/b).
def approxFrac(n,d,max_denominator):
#give a representation of n/d as N/D where D<=max_denominator
#from python 2.6 fractions.py
#
# reduce by gcd and only run algorithm if d>maxdenominator
g, b = n, d
while b:
g, b = b, g%b
n, d = n/g, d/g
if d <= max_denominator:
return (n,d)
nn, dd = n, d
p0, q0, p1, q1 = 0, 1, 1, 0
while True:
a = nn//dd
q2 = q0+a*q1
if q2 > max_denominator:
break
p0, q0, p1, q1 = p1, q1, p0+a*p1, q2
nn, dd = dd, nn-a*dd
k = (max_denominator-q0)//q1
bound1 = (p0+k*p1, q0+k*q1)
bound2 = (p1, q1)
if abs(bound2[0]*d - bound2[1]*n) <= abs(bound1[0]*d - bound1[1]*n):
return bound2
else:
return bound1
1/ .3333333333 = 3 because 1/3 = .3333333333333, so whatever number you get do this,
double x = 1 / yourDecimal;
int y = Math.Ceil(x);
and now Display "1/" + y
It is not allways resoluble, since not all decimals are fractions (for example PI or e).
Also, you have to round up to some length your decimal before converting.
I know this is an old thread, but I came across this problem in Word VBA. There are so many limitations due to the 8 bit (16 digit) rounding, as well as Word VBA making decimals into scientific notation etc.. but after working around all these problems, I have a nice function I'd like to share that offers a few extra features you may find helpful.
The strategy is along the lines of what Daniel Buckner wrote. Basically:
1st) decide if it's a terminating decimal or not
2nd) If yes, just set the decimal tail / 10^n and reduce the fraction.
3rd) If it doesn't terminate, try to find a repeating pattern including cases where the repetition doesn't start right away
Before I post the function, here are a few of my observations of the risks and limitations, as well as some notes that may help you understand my approach.
Risks, limitations, explanations:
-> Optional parameter "denom" allows you to specify the denominator of the fraction, if you'd like it rounded. i.e. for inches you may want 16ths used. The fractions will still be reduced, however, so 3.746 --> 3 12/16 --> 3 3/4
-> Optional parameter "buildup" set to True will build up the fraction using the equation editor, typing the text right into the active document. If you prefer to have the function simply return a flat string representation of the fraction so you can store it programmatically etc. set this to False.
-> A decimal could terminate after a bunch of repetitions... this function would assume an infinite repetition.
-> Variable type Double trades off whole number digit for decimal digits, only allowing 16 digits total (from my observations anyway!). This function assumes that if a number is using all 16 of the available digits then it must be a repeating decimal. A large number such as 123456789876.25 would be mistaken for a repeating decimal, then returned as decimal number upon failing to find a pattern.
-> To express really large terminating decimal out of 10^n, VB can only handle 10^8 is seems. I round the origninal number to 8 decimal places, losing some accuracy perhaps.
-> For the math behind converting repeating patterns to fractions check this link
-> Use Euclidean Algorithm to reduce the fraction
Ok, here it is, written as a Word Macro:
Function as_fraction(number_, Optional denom As Integer = -1, Optional buildup As Boolean = True) As String
'Selection.TypeText Text:="Received: " & CStr(number_) & vbCrLf
Dim number As Double
Dim repeat_digits As Integer, delay_digits As Integer, E_position As Integer, exponent As Integer
Dim tail_string_test As String, tail_string_original As String, num_removed As String, tail_string_removed As String, removed As String, num As String, output As String
output = "" 'string variable to build into the fraction answer
number = CDbl(number_)
'Get rid of scientific notation since this makes the string longer, fooling the function length = digits
If InStr(CStr(number_), "E+") > 0 Then 'no gigantic numbers! Return that scientific notation junk
output = CStr(number_)
GoTo all_done
End If
E_position = InStr(CStr(number), "E") 'E- since postives were handled
If E_position > 0 Then
exponent = Abs(CInt(Mid(CStr(number), E_position + 1)))
num = Mid(CStr(number_), 1, E_position) 'axe the exponent
decimalposition = InStr(num, ".") 'note the decimal position
For i_move = 1 To exponent
'move the decimal over, and insert a zero if the start of the number is reached
If InStr(num, "-") > 0 And decimalposition = 3 Then 'negative sign in front
num = "-0." & Mid(num, InStr(num, ".") - 1, 1) & Mid(num, InStr(num, ".") + 1) 'insert a zero after the negative
ElseIf decimalposition = 2 Then
num = "0." & Mid(num, InStr(num, ".") - 1, 1) & Mid(num, InStr(num, ".") + 1) 'insert in front
Else 'move the decimal over, there are digits left
num = Mid(num, 1, decimalposition - 2) & "." & Mid(num, decimalposition - 1, 1) & Mid(num, decimalposition + 1)
decimalposition = decimalposition - 1
End If
Next
Else
num = CStr(number_)
End If
'trim the digits to 15, since VB rounds the last digit which ruins the pattern. i.e. 0.5555555555555556 etc.
If Len(num) >= 16 Then
num = Mid(num, 1, 15)
End If
number = CDbl(num) 'num is a string representation of the decimal number, just to avoid cstr() everywhere
'Selection.TypeText Text:="number = " & CStr(number) & vbCrLf
'is it a whole number?
If Fix(number) = number Then 'whole number
output = CStr(number)
GoTo all_done
End If
decimalposition = InStr(CStr(num), ".")
'Selection.TypeText Text:="Attempting to find a fraction equivalent for " & num & vbCrLf
'is it a repeating decimal? It will have 16 digits
If denom = -1 And Len(num) >= 15 Then 'repeating decimal, unspecified denominator
tail_string_original = Mid(num, decimalposition + 1) 'digits after the decimal
delay_digits = -1 'the number of decimal place values removed from the tail, in case the repetition is delayed. i.e. 0.567777777...
Do 'loop through start points for the repeating digits
delay_digits = delay_digits + 1
If delay_digits >= Fix(Len(tail_string_original) / 2) Then
'Selection.TypeText Text:="Tried all starting points for the pattern, up to half way through the tail. None was found. I'll treat it as a terminating decimal." & vbCrLf
GoTo treat_as_terminating
End If
num_removed = Mid(num, 1, decimalposition) & Mid(num, decimalposition + 1 + delay_digits) 'original number with decimal values removed
tail_string_removed = Mid(num_removed, InStr(CStr(num_removed), ".") + 1)
repeat_digits = 0 'exponent on 10 for moving the decimal place over
'Selection.TypeText Text:="Searching " & num_removed & " for a pattern:" & vbCrLf
Do
repeat_digits = repeat_digits + 1
If repeat_digits = Len(tail_string_removed) - 1 Or repeat_digits >= 9 Then 'try removing a digit, incase the pattern is delayed
Exit Do
End If
tail_string_test = Mid(num_removed, decimalposition + 1 + repeat_digits)
'Selection.TypeText Text:=vbTab & "Comparing " & Mid(tail_string_removed, 1, Len(tail_string_removed) - repeat_digits) & " to " & tail_string_test & vbCrLf
If Mid(tail_string_removed, 1, Len(tail_string_removed) - repeat_digits) = tail_string_test Then
'Selection.TypeText Text:=num & ", " & Mid(tail_string_removed, 1, Len(tail_string_removed) - repeat_digits) & " vs " & tail_string_test & vbCrLf
GoTo foundpattern
End If
Loop
Loop 'next starting point for pattern
foundpattern:
If delay_digits = 0 Then 'found pattern right away
numerator = CLng(Mid(CStr(number), decimalposition + 1 + delay_digits, CInt(repeat_digits)))
'generate the denominator nines, same number of digits as the numerator
bottom = ""
For i_loop = 1 To repeat_digits
bottom = bottom & "9"
Next
denominator = CLng(bottom)
Else 'there were numbers before the pattern began
numerator = CLng(Mid(num, decimalposition + 1, delay_digits + repeat_digits)) - CLng(Mid(num, decimalposition + 1, delay_digits))
'i.e. x = 2.73232323232... delay_digits = 1, repeat_digits = 2, so numerator = 732 - 7 = 725
bottom = ""
For i_loop = 1 To repeat_digits
bottom = bottom & "9"
Next
For i_loop = 1 To delay_digits
bottom = bottom & "0"
Next
denominator = CLng(bottom)
'i.e. 990... 725/990 = 145/198 = 0.7323232...
End If
Else ' terminating decimal
treat_as_terminating:
'grab just the decimal trail
If denom = -1 Then
number = Math.Round(number, 8) 'reduce to fewer decimal places to avoid overload
'is it a whole number now?
If Fix(number) = number Then 'whole number
output = CStr(number)
GoTo all_done
End If
num = CStr(number)
numerator = CLng(Mid(num, decimalposition + 1))
denominator = 10 ^ (Len(num) - InStr(num, "."))
Else 'express as a fraction rounded to the nearest denom'th reduced
numerator1 = CDbl("0" & Mid(CStr(num), decimalposition))
numerator = CInt(Math.Round(numerator1 * denom))
denominator = CInt(denom)
End If
End If
'reduce the fraction if possible using Euclidean Algorithm
a = CLng(numerator)
b = CLng(denominator)
Dim t As Long
Do While b <> 0
t = b
b = a Mod b
a = t
Loop
gcd_ = a
numerator = numerator / gcd_
denominator = denominator / gcd_
whole_part = CLng(Mid(num, 1, decimalposition - 1))
'only write a whole number if the number is absolutely greater than zero, or will round to be so.
If whole_part <> 0 Or (whole_part = 0 And numerator = denominator) Then
'case where fraction rounds to whole
If numerator = denominator Then
'increase the whole by 1 absolutely
whole_part = (whole_part / Abs(whole_part)) * (Abs(whole_part) + 1)
End If
output = CStr(whole_part) & " "
End If
'if fraction rounded to a whole, it is already included in the whole number
If numerator <> 0 And numerator <> denominator Then
'negative sign may have been missed, if whole number was -0
If whole_part = 0 And number_ < 0 Then
numerator = -numerator
End If
output = output & CStr(numerator) & "/" & CStr(denominator) & " "
End If
If whole_part = 0 And numerator = 0 Then
output = "0"
End If
all_done:
If buildup = True Then 'build up the equation with a pretty fraction at the current selection range
Dim objRange As Range
Dim objEq As OMath
Dim AC As OMathAutoCorrectEntry
Application.OMathAutoCorrect.UseOutsideOMath = True
Set objRange = Selection.Range
objRange.Text = output
For Each AC In Application.OMathAutoCorrect.Entries
With objRange
If InStr(.Text, AC.Name) > 0 Then
.Text = Replace(.Text, AC.Name, AC.Value)
End If
End With
Next AC
Set objRange = Selection.OMaths.Add(objRange)
Set objEq = objRange.OMaths(1)
objEq.buildup
'Place the cursor at the end of the equation, outside of the OMaths object
objRange.OMaths(1).Range.Select
Selection.Collapse direction:=wdCollapseEnd
Selection.MoveRight Unit:=wdCharacter, count:=1
as_fraction = "" 'just a dummy return to make the function happy
Else 'just return a flat string value
as_fraction = output
End If
End Function
I shared an answer at this link : https://stackoverflow.com/a/57517128/11933717
It's also an iterative function, but unlike finding numerator and denominator in a nested loop, it just tests numerators only and so, should be faster.
Here is how it works :
It assumes that, based on the user input x, you want to find 2 integers n / m .
n/m = x , meaning that
n/x should give an almost integer m
Say one needs to find a fraction for x = 2.428571. Putting the int 2 aside for later, the algo starts by setting n and x and iterates n :
// n / x = m ( we need m to be an integer )
// n = 1 ; x = .428571 ;
1 / .428571 = 2.333335 (not close to an integer, n++)
2 / .428571 = 4.666671 (not close to an integer, n++)
3 / .428571 = 7.000007
At this point n = 3, we consider that m = 7.000007 is integer enough --based on some kind of accuracy the programmer decides-- and we reply the user
2.428571 = 2 + 3/7
= 14/7 + 3/7
= 17/7