“If you don’t know where you are going, you might wind up someplace else.” [Yogy Berra]
Which rational number is a good proxy of π (3.1415926…)? Enter in cell A1 ‘=pi()', in cell B1 your maximal denominator (for example 10), and in cells C1:D1 ‘=sbNRN(A1,B1)' as array formula (with CTRL + SHIFT + ENTER). You will get in C1:D1 22 and 7. That means: 22/7 is the nearest rational number to π with a denominator not higher than 10. For 1000 in B1 you would get 355/113.
This algorithm does NOT necessarily find the nearest rational number to a given floating point number with a given maximal denominator and a maximal absolute error. The good message is, though, that it would then return a #NUM! error. In this case please try an individual maximal absolute error.
The author’s (Oliver Aberth) original intention was to support exact computation with rational numbers, for example solving a set of linear equations with rational coefficients.
Note: The last row in this graphic does not tell us that we have successfully squared the circle. We have reached (my) Excel’s limit of accuracy.
The fraction representations of the TEXT function are shown for comparison.
Example: =TEXT(PI(),"?/?") = “22/7”
Microsoft did not extend this representation for its 64-Bit version. It cannot get more accurate than PI() = “5419351/1725033”. With 64-Bit PI() = “245850922/78256779” would be more accurate, but in this case the absolute error is already less than 1e-15, of course.
A simple sample application you find at Quota Change as Fraction.
Excel can represent decimal numbers from -9.99999999999999E+307 to 9.99999999999999E+307. Excel’s 64-bit version can use integers of type LongLong from -9223372036854775808 to 9223372036854775807 which is about -1E+10 to 1E+10.
It is obvious that Aberth’s algorithm cannot calculate sufficiently accurate fractions for all available decimal numbers with Excel.
sbNRN - Compute nearest rational number to a given floating point number with a given maximal denominator
sbNRN(dFloat, lMaxDen, [dMaxErr])
sbNRN computes the nearest rational number to a given floating point number dFloat with a given maximal denominator lMaxDen and an optional maximal error dMaxErr.
dFloat - Floating point number for which you want to derive the nearest rational number
lMaxDen - Maximal denominator which you want to allow
dMaxErr - Optional - Maximal absolute error (absolute difference between input float and output rational number) which you want to allow
Oliver Aberth, Introduction to Precise Numerical Methods, ISBN 0-12-373859-8
In case you just need the relation to its power of 10, you can use the formula =IFERROR(–A2*10^(LEN(–A2)-SEARCH(",",–A2))&":"&10^(LEN(–A2)-SEARCH(",",–A2)),–A2&":1"):
Appendix – sbNRN Code
Please read my Disclaimer.
Option Explicit #If Win64 Then Function sbNRN(dFloat As Double, lMaxDen As LongLong, _ Optional dMaxErr As Double = -1#) As Variant #Else Function sbNRN(dFloat As Double, lMaxDen As Long, _ Optional dMaxErr As Double = -1#) As Variant #End If 'Computes nearest rational number to dFloat with a maximal denominator 'lMaxDen and a maximal absolute error dMaxErr and returns result as a 'variant Nominator / Denominator. 'See: Oliver Aberth, A method for exact computation with rational numbers, ' JCAM, vol 4, no. 4, 1978 'Source (EN): http://www.sulprobil.com/sbnrn_en/ 'Source (DE): http://www.bplumhoff.de/sbnrn_de/ 'Bernd Plumhoff V1.21 09-Oct-2020 Dim dB As Double #If Win64 Then Dim lA As LongLong, lSgn As LongLong Dim lP1 As LongLong, lP2 As LongLong, lP3 As LongLong Dim lQ1 As LongLong, lQ2 As LongLong, lQ3 As LongLong #Else Dim lA As Long, lSgn As Long Dim lP1 As Long, lP2 As Long, lP3 As Long Dim lQ1 As Long, lQ2 As Long, lQ3 As Long #End If If dMaxErr = -1# Then dMaxErr = 1# / (2# * CDbl(lMaxDen) ^ 2#) lSgn = Sgn(dFloat): dB = Abs(dFloat) lP1 = 0: lP2 = 1: lQ1 = 1: lQ2 = 0 Do While lMaxDen > lQ2 lA = Int(dB) lP3 = lA * lP2 + lP1: lQ3 = lA * lQ2 + lQ1 #If Win64 Then If Abs(dB - CDbl(lA)) < 1# / CLngLng("9223372036854775807") Then #Else If Abs(dB - CDbl(lA)) < 1# / 2147483647# Then #End If Exit Do End If dB = 1# / (dB - CDbl(lA)) lP1 = lP2: lP2 = lP3: lQ1 = lQ2: lQ2 = lQ3 Loop If lQ3 > lMaxDen Then lQ3 = lQ2: lP3 = lP2 If lQ2 > lMaxDen Then lQ3 = lQ1: lP3 = lP1 End If End If 'If absolute error exceeds 1/2Q^2 then Aberth's lemma p. 286 might not apply. 'But the user can override this and check the result himself. If Abs(dFloat - lSgn * lP3 / lQ3) > dMaxErr Then sbNRN = CVErr(xlErrNum) Else sbNRN = Array(lSgn * lP3, lQ3) End If End Function
Please read my Disclaimer.
sbNRN.xlsm [43 KB Excel file, open and use at your own risk]