Fibonacci function/command in excel?...tia sal

[Guest]

Fibonacci function/command in excel?...tia sal

Greetings all,

Is thier a Fibonacci function or command built into excel?

TIA

 

[JE McGimpsey]

No.

 

[Dana DeLouis]

If you would like a non-looping custom function, this is good to n=73 (all
digits). Not sure if you need to go up to Excel vba's limit of about n=136
or so.

Function Fibonacci(n As Long)
'// Good to n=73
With WorksheetFunction
Fibonacci = Round((.Power((1 + Sqr(5)) / 2, n) - .Power(2 / (1 + Sqr(5)), n)
* .Power(-1, n)) / Sqr(5), 0)
End With
End Function

HTH :>)

 

[Kaak]

Put this in module and you an use it in your sheet.

Function Fibonacci(i)

If i = 1 Or i = 2 Then

tempsum = 1

Else

t1 = 1
t2 = 1

For j = 3 To i

ts = t1 + t2
t1 = t2
t2 = ts

Next

tempsum = ts

End If

Fibonacci = tempsum

End Function

 

[Dana DeLouis]

Hi. For a loop, the following is an interesting alternative.

Function Fibonacci(n)
Dim v, t, x
v = [{1,1;1,0}]
t = v
For x = 2 To CInt(n)
v = WorksheetFunction.MMult(v, t)
Next x
Fibonacci = v(2, 1)
End Function

 

[Dana DeLouis]

Oops! Old notes! This does it with 1 less loop. :>)

Function Fibonacci(n)
Dim v, t, x
v = [{1,1;1,0}]
t = v
For x = 2 To CInt(n - 1)
v = WorksheetFunction.MMult(v, t)
Next x
Fibonacci = v(1, 1)
End Function

--
Dana DeLouis
Win XP & Office 2003


<snip>

 

[Tushar Mehta]

Nice application of MMULT but a bit of overkill, don't you think?

The following will do just fine:

Function FibonacciNumber(ByVal N As Long)
Dim I As Long, X0 As Variant, X1 As Variant
X0 = 1: X1 = 1
For I = 3 To N Step 2
X0 = CDec(X0 + X1)
X1 = CDec(X0 + X1)
Next I
FibonacciNumber = IIf(N Mod 2 = 1, X0, X1)
End Function
Sub testIt2()
MsgBox FibonacciNumber(100)
End Sub

Without the CDec piece and with X0 and X1 declared as longs it works
fine upto FibonacciNumber(46)

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

 

[Dana DeLouis]

Hi Tushar. Yes, it's definitely not too efficient. I just thought it was
interesting. :>)
I noticed that your example can't squeeze out the largest number
possible...139 because X1 will overflow inside the loop. Perhaps as an
idea, stop the loop just prior to X1 overflowing, and go from there.
Perhaps:

Function FibonacciNumber(ByVal n As Long)
Dim I As Long, X0 As Variant, X1 As Variant
X0 = CDec(1): X1 = X0
For I = 3 To (n - 1) Step 2
X0 = X0 + X1
X1 = X0 + X1
Next I
FibonacciNumber = IIf(n Mod 2 = 1, X0 + X1, X1)
End Function

?FibonacciNumber(139)
50095301248058391139327916261

Just for gee-wiz, there are additional neat techniques. For example, if N
is an even number, one can cut the number of loops in half again.

Function Fibonacci_Even(ByRef N As Long)
Dim X As Variant
Dim Y As Variant
Dim j As Long
Dim Half As Long

'// For Even numbers only...
If N Mod 2 = 1 Then Exit Function

Select Case N
Case Is < 2, Is > 138: Fibonacci_Even = CVErr(9) 'Subscript out of range
Case 2: Fibonacci_Even = 1
Case Else
Half = N / 2
X = CDec(1): Y = X
For j = 3 To Half - 1 Step 2
X = X + Y
Y = X + Y
Next j

If Half Mod 2 = 0 Then
Fibonacci_Even = Y * (2 * X + Y)
Else
Fibonacci_Even = (X + Y) * (X + 3 * Y)
End If
End Select
End Function

?Fibonacci_Even(138)
30960598847965113057878492344

--
Dana DeLouis
Win XP & Office 2003

Nice application of MMULT but a bit of overkill, don't you think?

The following will do just fine:

Function FibonacciNumber(ByVal N As Long)
Dim I As Long, X0 As Variant, X1 As Variant
X0 = 1: X1 = 1
For I = 3 To N Step 2
X0 = CDec(X0 + X1)
X1 = CDec(X0 + X1)
Next I
FibonacciNumber = IIf(N Mod 2 = 1, X0, X1)
End Function
Sub testIt2()
MsgBox FibonacciNumber(100)
End Sub

Without the CDec piece and with X0 and X1 declared as longs it works
fine upto FibonacciNumber(46)

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

Oops! Old notes! This does it with 1 less loop. :>)

Function Fibonacci(n)
Dim v, t, x
v = [{1,1;1,0}]
t = v
For x = 2 To CInt(n - 1)
v = WorksheetFunction.MMult(v, t)
Next x
Fibonacci = v(1, 1)
End Function

 

[Tushar Mehta]

Oh, yes, it was interesting. But, then, so are many of your posts.

I realized the limitation of the code I shared while writing it but
figured...wtf...

The alternative that I thought of was

X0=1: X1=1
for i=3 to n
Xtemp=CDec(X0+X1): X0=X1: X1=Xtemp
next i
FibonacciNumber=X1

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

 

[Dana DeLouis]

Hi. I know this really doesn't apply here, but I thought you may find the
following topic interesting. I know that MMult was a bit of "overkill", but
certain math programs take advantage of that with a more efficient version.
The fastest algorithms look at the bit pattern of the number in binary form
and choose 1 of 2 simple operations. They then use a more efficient form of
MMult. For example, 128 (2^7) can be calculated in 7 loops.

Sub Fib_128()
Dim v, t, j

'// Note: Log(128)/Log(2) = 7
v = Array(CDec(1), CDec(1), CDec(0))
For j = 1 To 7
t = v(1) * v(1)
v = Array(v(0) * v(0) + t, v(1) * (v(0) + v(2)), t + v(2) * v(2))
Next j
Debug.Print v(1)
End Sub

Returns: 251728825683549488150424261

The real speed comes from wanting to do Fibonacci (16384) (ie 2^14) where
you only need to loop 7 more times.
Excel can't do that directly, of course. The real code is written slightly
more efficiently then that above.
Anyway, just gee-wiz. :>) The op probably only wanted Fibonacci (10).
:>)