combination of numbers

Here's an example using the solver add-in

http://tinyurl.com/4doog

--
Regards,

Peo Sjoblom

(No private emails please)


"ErnestoMarti" <[email protected]>
wrote in message
news:[email protected]...
>
> undefined
> Hello!!!! The problem that I explain below is one that i have for a
> long time... hope someone can understand it...!!
> Is there any option in Excel, if I have a list of numbers (for example
> 20 different numbers), I want to know which combination of these
> numbers are the sum of "X" number???
>
>
> --
> ErnestoMarti
> ------------------------------------------------------------------------
> ErnestoMarti's Profile:
> http://www.excelforum.com/member.php...o&userid=26836
> View this thread: http://www.excelforum.com/showthread...hreadid=400817
>

Ernesto,

If Peo's Solver solution doesn't work (it usually won't with that many numbers), send me an email
(reply to this post, and take out the spaces and change the dot to . ), and I will send you a
workbook with a macro that can handle cases that Solver won't.

HTH,
Bernie
MS Excel MVP


"ErnestoMarti" <[email protected]> wrote in message
news:[email protected]...
>
> undefined
> Hello!!!! The problem that I explain below is one that i have for a
> long time... hope someone can understand it...!!
> Is there any option in Excel, if I have a list of numbers (for example
> 20 different numbers), I want to know which combination of these
> numbers are the sum of "X" number???
>
>
> --
> ErnestoMarti
> ------------------------------------------------------------------------
> ErnestoMarti's Profile: http://www.excelforum.com/member.php...o&userid=26836
> View this thread: http://www.excelforum.com/showthread...hreadid=400817
>

Here is a fast way (.4 sec) to find a set o 10 numbers (Num_Sel) out of
a set of 20 numbers (Bin1) that add up to (Total).
The first 10 columns of your Sheet1 are reserved for the several
thousand (Pcount) answers. Do not use these columns for the following
entries:
Setup: Tools > Options > General > R1C1 reference style
Create a 20 row vector named Bin1 and enter your 20 numbers.
Name a cell Numbers and enter
=COUNTA(Bin1)
Name a cell Num_Sel and enter 10
Name a cell CombTot and enter
=INT(COMBIN(Numbers,Num_Sel)/25)
Name a cell Total and enter your desired sum.
Name a cell Pcount.
Insert > Name > Define > Names > Results
Refers to:
=OFFSET(INDIRECT("Sheet1!C1:C"&(Num_Sel),0),1,,CombTot)
Copy the following macro into a VBA module and run it.
For other values of Num_Sel, expand or collapse the macro and modify
the spreadsheet formulas.

Option Explicit
Option Base 1

Sub sum_perm()
Dim i As Integer
Dim j As Integer
Dim k As Integer
Dim m As Integer
Dim n As Integer
Dim p As Long
Dim q As Integer
Dim r As Integer
Dim s As Integer
Dim t As Integer
Dim u As Integer
Dim v As Integer
Dim NN As Integer
Dim SS As Integer
Dim ix As Integer
Dim jx As Integer
Dim kx As Integer
Dim mx As Integer
Dim nx As Integer
Dim rx As Integer
Dim sx As Integer
Dim tx As Integer
Dim ux As Integer
Dim vx As Integer
Dim ns As Integer
Dim binx() As Variant
Dim accum() As Variant


Range("Results").ClearContents

NN = Range("Numbers")
p = Range("CombTot")
ns = Range("Num_Sel")
ReDim accum(p, ns)
ReDim binx(NN)
SS = Range("Total")


ReDim binx(NN)
For p = 1 To NN
binx(p) = Range("Bin1").Cells(p, 1)
Next p

p = 1

For i = 1 To NN
For j = i + 1 To NN
For k = j + 1 To NN
For m = k + 1 To NN
For n = m + 1 To NN
For r = n + 1 To NN
For s = r + 1 To NN
For t = s + 1 To NN
For u = t + 1 To NN
For v = u + 1 To NN
ix = binx(i)
jx = binx(j)
kx = binx(k)
mx = binx(m)
nx = binx(n)
rx = binx(r)
sx = binx(s)
tx = binx(t)
ux = binx(u)
vx = binx(v)
If (ix + jx + kx + mx + nx + rx + sx + tx + ux + vx) = SS Then
accum(p, 1) = ix
accum(p, 2) = jx
accum(p, 3) = kx
accum(p, 4) = mx
accum(p, 5) = nx
accum(p, 6) = rx
accum(p, 7) = sx
accum(p, 8) = tx
accum(p, 9) = ux
accum(p, 10) = vx
p = p + 1
End If
Next v
Next u
Next t
Next s
Next r
Next n
Next m
Next k
Next j
Next i

Range("Pcount") = p 'count of valid answers
Range("Results") = accum
End Sub

Herbert Seidenberg wrote...
>Here is a fast way (.4 sec) to find a set o 10 numbers (Num_Sel) out of
>a set of 20 numbers (Bin1) that add up to (Total).
....

What's magic about 10 out of 20 numbers?

>The first 10 columns of your Sheet1 are reserved for the several
>thousand (Pcount) answers. Do not use these columns for the following
>entries:
....

And you eat worksheet cells!

Why not create a new worksheet to store temporary results?

>Sub sum_perm()
....
>For i = 1 To NN
> For j = i + 1 To NN
> For k = j + 1 To NN
> For m = k + 1 To NN
> For n = m + 1 To NN
> For r = n + 1 To NN
> For s = r + 1 To NN
> For t = s + 1 To NN
> For u = t + 1 To NN
> For v = u + 1 To NN
....

Ah, brute force.

I've finally been tempted to do this myself. Brute force is
unfortunately necessary for this sort of problem, but there are better
control flows than hardcoded nested For loops.


Sub foo()
'This *REQUIRES* VBAProject references to
'Microsoft Scripting Runtime
'Microsoft VBScript Regular Expressions 1.0

Dim i As Long, n As Long, t As Double, u As Double
Dim x As Variant, y As Variant
Dim dv As New Dictionary, dc As New Dictionary
Dim re As New RegExp

re.Global = True
re.IgnoreCase = True

t = Range("A23").Value2 'target value - HARDCODED

For Each x In Range("A1:A20").Value2 'set of values - HARDCODED

If VarType(x) = vbDouble Then

If dv.Exists(x) Then
dv(x) = dv(x) + 1

ElseIf x = t Then
GoTo SolutionFound

ElseIf x < t Then
dc.Add Key:=Format(x), Item:=x
dv.Add Key:=x, Item:=1

End If

End If

Next x

For n = 2 To dv.Count

For Each x In dv.Keys

For Each y In dc.Keys
re.Pattern = "(^|\+)" & Format(x) & "(\+|$)"

If re.Execute(y).Count < dv(x) Then
u = dc(y) + x

If u = t Then
GoTo SolutionFound

ElseIf u < t Then
dc.Add Key:=y & "+" & Format(x), Item:=u

End If

End If

Next y

Next x

Next n

MsgBox Prompt:="all combinations exhausted", Title:="No Solution"

Exit Sub


SolutionFound:

If IsEmpty(y) Then
y = Format(x)
n = dc.Count + 1

Else
y = y & "+" & Format(x)
n = dc.Count

End If

MsgBox Prompt:=y, Title:="Solution (" & Format(n) & ")"

End Sub


The initial loop loads a dictionary object (dv) with the numeric values
from the specified range, storing distinct values as keys and the
number of instances of each distinct values as items.

It tracks combinations of values from the original set using a
dictionary object (dc) in which the keys are the symbolic sums (e.g.,
"1+2+3") and the items are the evaluated numeric sums (e.g., 6). It
uses a regex Execute call to ensure that each distinct value appears no
more times than it appears in the original set.

New combinations are added to dc only when their sums are less than the
target value. This implicitly eliminates larger cardinality
combinations of sums which would exceed the target value, thus
partially mitigating the O(2^N) runtime that's unavoidable from this
sort of problem.

FWIW, my test data in A1:A20 was

692
506
765
97
47
949
811
187
537
217
687
443
117
248
580
506
449
309
393
507

and the formula for my target value in A23 was

=A2+A3+A5+A7+A11+A13+A17+A19

which evaluates to 3775. When I ran the macro above, it returned the
solution

687+506+765+97+47+949+187+537

which is equivalent to

=A11+A2+A3+A4+A5+A6+A8+A9

Harlan Grove wrote...
....
>Sub foo()
....
> For n = 2 To dv.Count
>
> For Each x In dv.Keys
>
> For Each y In dc.Keys
....

This looping logic doesn't work. If there are no matching combinations,
this will run a LONG, LONG, LONG time.

Harlan:
Nothing magical about 10 and 20, I just picked them from
typical posts last month.
My 10 columns on Sheet1 do not list temporary results, but valid
solutions to the posed problem.
The length of the list depends on the interval of the 20 numbers,
on the count of numbers chosen and on the sum.
I ran your numbers and got the same result as you,
plus 97 more combinations of 8 out of your 20 numbers that equal 3775.
Herb

"Herbert Seidenberg" <[email protected]> wrote in message
news:[email protected]...
>
> Harlan:
> Nothing magical about 10 and 20, I just picked them from
> typical posts last month.
> My 10 columns on Sheet1 do not list temporary results, but valid
> solutions to the posed problem.
> The length of the list depends on the interval of the 20 numbers,
> on the count of numbers chosen and on the sum.
> I ran your numbers and got the same result as you,
> plus 97 more combinations of 8 out of your 20 numbers that equal
> 3775.
> Herb
>

Could Harlan's code be modified to list all unique combinations rather
than just one?

I don't think there is a need to see each permutation, but certainly
unique combinations would be good.

Thanks,

Alan.

> plus 97 more combinations of 8 out of your 20 numbers that equal 3775.

Hi. Just gee wiz. Doesn't look like it, but I think there are 337
combinations that total 3775.
Ranging from

187, 449, 687, 692, 811, 949
to
97, 117, 187, 217, 248, 393, 443, 449, 507, 537, 580

--
Dana DeLouis
Win XP & Office 2003


"Herbert Seidenberg" <[email protected]> wrote in message
news:[email protected]...
> Harlan:
> Nothing magical about 10 and 20, I just picked them from
> typical posts last month.
> My 10 columns on Sheet1 do not list temporary results, but valid
> solutions to the posed problem.
> The length of the list depends on the interval of the 20 numbers,
> on the count of numbers chosen and on the sum.
> I ran your numbers and got the same result as you,
> plus 97 more combinations of 8 out of your 20 numbers that equal 3775.
> Herb
>

Dana:
Your first example has 6 numbers, the second 11.
Harlan and my criteria were 8 numbers.
Herb

"Dana DeLouis" <[email protected]> wrote in message
news:[email protected]...
>
> Hi. Just gee wiz. Doesn't look like it, but I think there are 337
> combinations that total 3775.
> Ranging from
>
> 187, 449, 687, 692, 811, 949
> to
> 97, 117, 187, 217, 248, 393, 443, 449, 507, 537, 580
>

Hi Dana,

That is amazing - how did you get all 337 answers?

The specific discussion above was restricted to only sets of a
particular size, but in the real world sets of *all* size would
normally be required.

Did you use code like Harlan's? If so, could you post it back here?

Thanks,

Alan.

Fixed the code. It's a bit more involved now. It should now list all
solutions in a new worksheet. (See my follow-up to Dana DeLouis in
another branch as to whether it misses some solutions).


'---- begin VBA code ----
Option Explicit


Sub foo()
Dim j As Long, k As Long, n As Long, p As Boolean
Dim s As String, t As Double, u As Double
Dim v As Variant, x As Variant, y As Variant
Dim dc1 As New Dictionary, dc2 As New Dictionary
Dim dcn As Dictionary, dco As Dictionary
Dim re As New RegExp

On Error GoTo CleanUp
Application.EnableEvents = False
Application.Calculation = xlCalculationManual

re.Global = True
re.IgnoreCase = True

t = Range("A23").Value2

Set dco = dc1
Set dcn = dc2

Call recsoln

For Each x In Range("A1:A20").Value2
If VarType(x) = vbDouble Then
If x = t Then
recsoln "+" & Format(x)

ElseIf dco.Exists(x) Then
dco(x) = dco(x) + 1

ElseIf x < t Then
dco.Add Key:=x, Item:=1
Application.StatusBar = dco.Count

End If

End If
Next x

n = dco.Count

ReDim v(1 To n, 1 To 2)

For k = 1 To n
v(k, 1) = dco.Keys(k - 1)
v(k, 2) = dco.Items(k - 1)
Next k

qsortd v, 1, n

For k = 1 To n
dcn.Add Key:="+" & Format(v(k, 1)), Item:=v(k, 1)
Next k

For k = 2 To n
dco.RemoveAll
swapo dco, dcn

For Each y In dco.Keys
p = False

For j = 1 To n
x = v(j, 1)
s = "+" & Format(x)
If Right(y, Len(s)) = s Then p = True

If p Then
re.Pattern = "\" & s & "(?=(\+|$))"

If re.Execute(y).Count < v(j, 2) Then
u = dco(y) + x

If u = t Then
recsoln y & s

ElseIf u < t Then
dcn.Add Key:=y & s, Item:=u
Application.StatusBar = dcn.Count

End If

End If

End If

Next j

Next y

Next k

If (recsoln() = 0) Then _
MsgBox Prompt:="all combinations exhausted", Title:="No Solution"

CleanUp:
Application.EnableEvents = True
Application.Calculation = xlCalculationAutomatic
Application.StatusBar = False

End Sub


Private Function recsoln(Optional s As String)
Static n As Long, ws As Worksheet, r As Range

If s = "" Then
recsoln = n

If n = 0 And r Is Nothing Then
Set ws = ActiveSheet
Set r = Worksheets.Add.Range("A1")
ws.Activate

Else
n = 0

End If

Else
r.Offset(n, 0).Value = s
n = n + 1
recsoln = n

End If

End Function


Private Sub qsortd(v As Variant, lft As Long, rgt As Long)
'ad hoc quicksort subroutine
'translated from Aho, Weinberger & Kernighan,
'"The Awk Programming Language", page 161

Dim j As Long, pvt As Long

If (lft >= rgt) Then Exit Sub

swap2 v, lft, lft + Int((rgt - lft + 1) * Rnd)

pvt = lft

For j = lft + 1 To rgt
If v(j, 1) > v(lft, 1) Then
pvt = pvt + 1
swap2 v, pvt, j
End If
Next j

swap2 v, lft, pvt

qsortd v, lft, pvt - 1
qsortd v, pvt + 1, rgt

End Sub


Private Sub swap2(v As Variant, i As Long, j As Long)
'modified version of the swap procedure from
'translated from Aho, Weinberger & Kernighan,
'"The Awk Programming Language", page 161

Dim t As Variant

t = v(i, 1)
v(i, 1) = v(j, 1)
v(j, 1) = t

t = v(i, 2)
v(i, 2) = v(j, 2)
v(j, 2) = t

End Sub


Private Sub swapo(a As Object, b As Object)
Dim t As Object

Set t = a
Set a = b
Set b = t

End Sub
'---- end VBA code ----

Dana DeLouis wrote...
....
>Hi. Just gee wiz. Doesn't look like it, but I think there are 337
>combinations that total 3775.
>Ranging from
>
>187, 449, 687, 692, 811, 949
>to
>97, 117, 187, 217, 248, 393, 443, 449, 507, 537, 580
....

I get the following as the final combination (smallest initial number,
vs your smallest final number).

+537+507+506+506+449+393+309+217+187+117+47

I also get only 247 combinations that sum to 3775, so our respective
sets of combinations differ in cardinality by a suspiciously round 90.
Here are my 247.

+949+811+692+687+449+187
+949+811+765+580+506+117+47
+949+811+765+537+449+217+47
+949+811+765+537+309+217+187
+949+811+765+506+449+248+47
+949+811+765+506+309+248+187
+949+811+687+537+507+187+97
+949+811+537+507+506+248+217
+949+811+507+506+506+449+47
+949+811+507+506+506+309+187
+949+765+692+507+506+309+47
+949+765+687+580+449+248+97
+949+765+580+506+449+309+217
+949+692+687+537+506+217+187
+949+692+687+506+506+248+187
+949+692+580+537+507+393+117
+949+692+537+507+449+393+248
+949+687+580+507+506+449+97
+949+687+580+506+443+393+217
+949+580+537+506+506+449+248
+811+765+692+687+506+217+97
+811+765+692+507+443+309+248
+811+765+687+580+506+309+117
+811+765+687+537+449+309+217
+811+765+687+506+449+309+248
+811+692+687+506+449+443+187
+811+687+507+506+506+449+309
+765+692+580+537+449+443+309
+949+811+765+580+309+217+97+47
+949+811+765+506+393+187+117+47
+949+811+692+449+443+217+117+97
+949+811+687+537+309+248+187+47
+949+811+687+507+309+248+217+47
+949+811+580+537+506+248+97+47
+949+811+580+449+443+309+187+47
+949+811+580+449+393+309+187+97
+949+811+537+506+449+309+117+97
+949+765+687+580+443+187+117+47
+949+765+687+580+393+187+117+97
+949+765+687+537+506+187+97+47
+949+765+687+507+506+217+97+47
+949+765+687+449+443+248+187+47
+949+765+687+449+393+248+187+97
+949+765+580+537+443+217+187+97
+949+765+580+506+443+248+187+97
+949+765+580+506+393+248+217+117
+949+765+537+506+506+248+217+47
+949+765+506+449+393+309+217+187
+949+692+687+580+506+217+97+47
+949+692+580+507+443+309+248+47
+949+692+580+507+393+309+248+97
+949+687+580+537+449+309+217+47
+949+687+580+506+449+309+248+47
+949+687+537+507+443+248+217+187
+949+687+507+506+449+443+187+47
+949+687+507+506+449+393+187+97
+949+580+537+506+506+393+187+117
+949+580+507+506+506+443+187+97
+949+580+507+506+506+393+217+117
+949+537+506+506+449+393+248+187
+949+507+506+506+449+393+248+217
+811+765+692+537+449+217+187+117
+811+765+692+507+443+393+117+47
+811+765+692+506+449+248+187+117
+811+765+687+537+443+248+187+97
+811+765+687+537+393+248+217+117
+811+765+687+507+443+248+217+97
+811+765+687+506+449+393+117+47
+811+765+687+506+393+309+187+117
+811+765+580+506+506+443+117+47
+811+765+580+506+506+393+117+97
+811+765+580+449+443+393+217+117
+811+765+537+506+449+443+217+47
+811+765+537+506+449+393+217+97
+811+765+537+506+443+309+217+187
+811+765+506+506+449+443+248+47
+811+765+506+506+449+393+248+97
+811+765+506+506+443+309+248+187
+811+692+687+580+443+248+217+97
+811+692+687+506+506+309+217+47
+811+692+507+506+506+449+187+117
+811+687+580+537+506+309+248+97
+811+687+537+507+506+443+187+97
+811+687+537+507+506+393+217+117
+811+687+507+506+506+393+248+117
+811+537+507+506+506+443+248+217
+765+692+687+537+507+443+97+47
+765+692+580+537+443+393+248+117
+765+692+537+449+443+393+309+187
+765+692+507+506+506+443+309+47
+765+692+507+506+506+393+309+97
+765+692+507+449+443+393+309+217
+765+687+580+537+449+443+217+97
+765+687+580+506+449+443+248+97
+765+687+537+506+506+309+248+217
+765+580+506+506+449+443+309+217
+692+687+537+506+506+443+217+187
+692+580+537+507+506+443+393+117
+692+537+507+506+449+443+393+248
+687+580+537+506+506+449+393+117
+687+580+507+506+506+449+443+97
+949+811+765+393+309+217+187+97+47
+949+811+692+507+248+217+187+117+47
+949+811+580+443+393+248+187+117+47
+949+811+537+506+393+248+187+97+47
+949+811+507+506+393+248+217+97+47
+949+765+537+507+449+217+187+117+47
+949+765+507+506+449+248+187+117+47
+949+692+687+506+393+217+187+97+47
+949+692+580+537+449+217+187+117+47
+949+692+580+506+449+248+187+117+47
+949+692+507+443+393+309+248+187+47
+949+692+506+506+443+248+217+117+97
+949+687+580+537+443+248+187+97+47
+949+687+580+537+393+248+217+117+47
+949+687+580+507+443+248+217+97+47
+949+687+580+506+393+309+187+117+47
+949+687+537+449+443+309+187+117+97
+949+687+537+449+393+309+217+187+47
+949+687+507+449+443+309+217+117+97
+949+687+506+449+393+309+248+187+47
+949+580+537+506+449+393+217+97+47
+949+580+537+506+443+309+217+187+47
+949+580+537+506+393+309+217+187+97
+949+580+506+506+449+393+248+97+47
+949+580+506+506+443+309+248+187+47
+949+580+506+506+393+309+248+187+97
+949+580+449+443+393+309+248+217+187
+949+537+506+449+443+309+248+217+117
+811+765+692+580+449+217+117+97+47
+811+765+692+580+309+217+187+117+97
+811+765+692+449+309+248+217+187+97
+811+765+687+449+393+309+217+97+47
+811+765+580+506+443+309+217+97+47
+811+765+537+507+506+248+187+117+97
+811+765+506+506+443+393+187+117+47
+811+692+687+537+449+248+187+117+47
+811+692+687+507+449+248+217+117+47
+811+692+687+507+309+248+217+187+117
+811+692+687+443+393+248+217+187+97
+811+692+580+537+506+248+187+117+97
+811+692+580+507+506+248+217+117+97
+811+692+537+507+443+393+248+97+47
+811+692+507+506+449+309+217+187+97
+811+687+580+537+506+393+117+97+47
+811+687+580+449+443+393+248+117+47
+811+687+580+443+393+309+248+187+117
+811+687+537+506+449+393+248+97+47
+811+687+537+506+443+309+248+187+47
+811+687+537+506+393+309+248+187+97
+811+687+507+506+443+309+248+217+47
+811+687+507+506+393+309+248+217+97
+811+580+537+507+449+309+248+217+117
+811+580+537+506+506+443+248+97+47
+811+580+537+449+443+393+248+217+97
+811+580+506+449+443+393+309+187+97
+811+537+506+506+449+443+309+117+97
+811+537+506+506+449+393+309+217+47
+765+692+687+537+393+309+248+97+47
+765+692+687+507+506+217+187+117+97
+765+692+580+449+443+393+309+97+47
+765+692+537+506+506+248+217+187+117
+765+687+580+537+507+248+217+187+47
+765+687+580+506+443+393+187+117+97
+765+687+537+507+449+309+217+187+117
+765+687+537+506+506+443+187+97+47
+765+687+537+506+506+393+217+117+47
+765+687+537+449+443+393+217+187+97
+765+687+507+506+506+443+217+97+47
+765+687+507+506+449+309+248+187+117
+765+687+506+449+443+393+248+187+97
+765+580+537+507+506+449+217+117+97
+765+580+507+506+506+449+248+117+97
+765+580+506+506+443+393+248+217+117
+765+506+506+449+443+393+309+217+187
+692+687+580+537+449+309+217+187+117
+692+687+580+507+443+393+309+117+47
+692+687+580+506+506+443+217+97+47
+692+687+580+506+449+309+248+187+117
+692+687+537+507+506+393+309+97+47
+692+687+507+449+443+393+309+248+47
+692+580+537+507+443+393+309+217+97
+692+580+507+506+443+393+309+248+97
+687+580+537+506+449+443+309+217+47
+687+580+537+506+449+393+309+217+97
+687+580+506+506+449+443+309+248+47
+687+580+506+506+449+393+309+248+97
+687+507+506+506+449+443+393+187+97
+949+811+537+443+309+248+217+117+97+47
+949+765+580+507+309+217+187+117+97+47
+949+765+507+449+309+248+217+187+97+47
+949+692+580+449+309+248+217+187+97+47
+949+687+507+443+393+248+217+187+97+47
+949+580+537+507+506+248+187+117+97+47
+811+765+692+449+393+217+187+117+97+47
+811+765+687+507+309+248+187+117+97+47
+811+765+506+443+393+309+217+187+97+47
+811+692+687+580+309+248+187+117+97+47
+811+692+507+506+443+248+217+187+117+47
+811+692+507+506+393+248+217+187+117+97
+811+687+506+506+449+248+217+187+117+47
+811+580+537+507+449+443+187+117+97+47
+811+537+507+506+506+309+248+187+117+47
+811+537+507+449+393+309+248+217+187+117
+811+537+506+506+443+393+248+187+97+47
+811+507+506+506+443+393+248+217+97+47
+765+692+687+506+309+248+217+187+117+47
+765+692+580+506+506+248+217+117+97+47
+765+692+506+506+449+309+217+187+97+47
+765+687+580+537+449+309+187+117+97+47
+765+687+580+507+449+309+217+117+97+47
+765+687+506+506+393+309+248+217+97+47
+765+580+537+506+449+309+248+217+117+47
+765+537+507+506+449+443+217+187+117+47
+765+537+507+506+449+393+217+187+117+97
+765+507+506+506+449+443+248+187+117+47
+765+507+506+506+449+393+248+187+117+97
+692+687+580+507+443+248+217+187+117+97
+692+687+507+506+506+309+217+187+117+47
+692+687+506+506+443+393+217+187+97+47
+692+580+537+506+449+443+217+187+117+47
+692+580+537+506+449+393+217+187+117+97
+692+580+506+506+449+443+248+187+117+47
+692+580+506+506+449+393+248+187+117+97
+687+580+537+507+506+449+248+117+97+47
+687+580+537+507+506+309+248+187+117+97
+687+580+537+506+443+393+248+217+117+47
+687+580+506+506+443+393+309+187+117+47
+687+537+506+449+443+393+309+217+187+47
+687+506+506+449+443+393+309+248+187+47
+580+537+507+506+506+449+309+217+117+47
+580+537+506+506+449+443+393+217+97+47
+580+537+506+506+443+393+309+217+187+97
+949+506+449+443+393+309+248+217+117+97+47
+811+580+537+449+393+309+248+187+117+97+47
+811+580+507+449+393+309+248+217+117+97+47
+765+692+506+506+393+248+217+187+117+97+47
+765+687+507+449+393+309+217+187+117+97+47
+765+580+507+506+443+309+217+187+117+97+47
+765+537+506+449+393+309+248+217+187+117+47
+765+507+506+449+443+309+248+217+187+97+47
+692+687+580+449+393+309+217+187+117+97+47
+692+580+506+449+443+309+248+217+187+97+47
+687+537+507+506+449+393+248+187+117+97+47
+580+537+507+506+506+443+248+187+117+97+47
+580+537+507+449+443+393+248+217+187+117+97
+537+507+506+506+449+393+309+217+187+117+47

Presumably you calculated yours in Mathematica. Would you be willing to
share your code and your full set of combinations?

Thanks. I really like your code. I see where I went wrong. I was
treating 506 on one line as different than the 506 on another line. With
that in mind, I too get 247 unique solutions. Thanks for the catch. :>)
Actually, I was just trying to point out how surprising the total number of
combinations can be. I would have guessed maybe 2 or 3.
Just for fun, the totals that have the most combinations are 4265 and 4782.
Your program caught all 314 combinations. :>) I would have guessed 2,
maybe 3 at the most. Again, I just find it interesting that there are that
many. :>)

--
Dana DeLouis
Win XP & Office 2003


<....>
> I also get only 247 combinations that sum to 3775, so our respective
> sets of combinations differ in cardinality by a suspiciously round 90.
> Here are my 247.
>
> +949+811+692+687+449+187
> +949+811+765+580+506+117+47
> +949+811+765+537+449+217+47
etc...

<snip>

Harlan,

Below is a problem statement from years ago, and the code that solves it
relatively quickly - a few seconds. I tried your code to solve it, but my
machine locked up after a couple of minutes.

Perhaps there is something in Michel's code that might be of use in the
current application.

Bernie


'I was asked by a colleague to find the combination of certain numbers
'which will add up to a specific value. The numbers I was given were:
'
' 52.04;57.63;247.81;285.71;425.00;690.72;764.57;1485.00;1609.24;
' 3737.45;6485.47;6883.85;7309.33;12914.64;13714.11;14346.39;
' 15337.85;22837.83;31201.42;34663.07;321987.28
'
' (21 numbers in ascending order)
'
' I am trying to get a combination so that it adds up to 420422.19.
'
' On a sheet, put the following
' B1 Target 420422.19
' B2 number of parameters 21
' B3:B23 all parameters in descending order
' 321987.28
' 34663.07
' 31201.42
' 22837.83
' 15337.85
' 14346.39
' 13714.11
' 12914.64
' 7309.33
' 6883.85
' 6485.47
' 3737.45
' 1609.24
' 1485
' 764.57
' 690.72
' 425
' 285.71
' 247.81
' 57.63
' 52.04
' Start find_sol, it will put "1" or "0" in C3:Cx if you sum the
' parameters with a "1", you will have the best solution.
' It takes about 12 seconds on my very slow P133.
' The solution is
' 1 1 0 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 1 0 0
' Regards.
'
' Michel.
' Michel Claes <[email protected]>


Option Explicit

Global target As Double
Global nbr_elem As Integer
Global stat(30) As Integer
Global statb(30) As Integer
Global elems(30) As Double
Global best As Double

Sub store_sol()
Dim i As Integer
For i = 1 To nbr_elem
Cells(i + 2, 3) = statb(i)
Next i
End Sub

Sub copy_stat()
Dim i As Integer
For i = 1 To nbr_elem
statb(i) = stat(i)
Next i
End Sub

Sub eval(ByVal total As Double, ByVal pos As Integer)
If pos <= nbr_elem Then
stat(pos) = 0
eval total, pos + 1
stat(pos) = 1
eval total + elems(pos), pos + 1
Else
If (Abs(total - target) < Abs(target - best)) Then
best = total
copy_stat
End If
End If
End Sub

Sub find_sol()
Dim i As Integer
best = 0
target = Cells(1, 2)
nbr_elem = Cells(2, 2)
For i = 1 To nbr_elem
elems(i) = Cells(i + 2, 2)
Next i
eval 0, 1
store_sol
End Sub


"Harlan Grove" <[email protected]> wrote in message
news:[email protected]...
> Fixed the code.

Bernie Deitrick wrote...
>Below is a problem statement from years ago, and the code that solves it
>relatively quickly - a few seconds. I tried your code to solve it, but my
>machine locked up after a couple of minutes.
....
>' I am trying to get a combination so that it adds up to 420422.19.
>'
>' On a sheet, put the following
>' B1 Target 420422.19
>' B2 number of parameters 21
>' B3:B23 all parameters in descending order
>' 321987.28
>' 34663.07
>' 31201.42
>' 22837.83
>' 15337.85
>' 14346.39
>' 13714.11
>' 12914.64
>' 7309.33
>' 6883.85
>' 6485.47
>' 3737.45
>' 1609.24
>' 1485
>' 764.57
>' 690.72
>' 425
>' 285.71
>' 247.81
>' 57.63
>' 52.04
....

Problem with my code (1st revision) is using exact equality, killer for
fractional decimal values. It exhausted your data set without finding
any combination that summed to your target value. It took a few minutes
to do so on my machine.

I've modified it a bit in the last day and a half, in part to deal with
this. I'm sure you'll be thrilled to know it now finds the solution to
the problem above in a fraction of a second.

+321987.28+34663.07+22837.83+13714.11+12914.64+7309.33+3737.45+1609.24
+690.72+425+285.71+247.81

The macros you provided produce the single closest combination. Useful,
but not exactly the same as finding exact combinations (as rounded
decimals). Also, my revised code, in the absence of rounding error,
e.g., when all values are integers of 15 or fewer decimal digits,
produces all combinations summing to the target value. Modifying the
macros you provided to do the same would be a challenge.

And here's the revised code. It even has a user interface now!


'---- begin VBA code ----
Option Explicit


Sub findsums()
Const TOL As Double = 0.000001 'modify as needed
Dim c As Variant

Dim j As Long, k As Long, n As Long, p As Boolean
Dim s As String, t As Double, u As Double
Dim v As Variant, x As Variant, y As Variant
Dim dc1 As New Dictionary, dc2 As New Dictionary
Dim dcn As Dictionary, dco As Dictionary
Dim re As New RegExp

re.Global = True
re.IgnoreCase = True

On Error Resume Next

Set x = Application.InputBox( _
Prompt:="Enter range of values:", _
Title:="findsums", _
Default:="", _
Type:=8 _
)

If x Is Nothing Then
Err.Clear
Exit Sub
End If

y = Application.InputBox( _
Prompt:="Enter target value:", _
Title:="findsums", _
Default:="", _
Type:=1 _
)

If VarType(y) = vbBoolean Then
Exit Sub
Else
t = y
End If

On Error GoTo 0

Set dco = dc1
Set dcn = dc2

Call recsoln

For Each y In x.Value2
If VarType(y) = vbDouble Then
If Abs(t - y) < TOL Then
recsoln "+" & Format(y)

ElseIf dco.Exists(y) Then
dco(y) = dco(y) + 1

ElseIf y < t - TOL Then
dco.Add Key:=y, Item:=1

c = CDec(c + 1)
Application.StatusBar = "[1] " & Format(c)

End If

End If
Next y

n = dco.Count

ReDim v(1 To n, 1 To 3)

For k = 1 To n
v(k, 1) = dco.Keys(k - 1)
v(k, 2) = dco.Items(k - 1)
Next k

qsortd v, 1, n

For k = n To 1 Step -1
v(k, 3) = v(k, 1) * v(k, 2) + v(IIf(k = n, n, k + 1), 3)
If v(k, 3) > t Then dcn.Add Key:="+" & Format(v(k, 1)), Item:=v(k,
1)
Next k

On Error GoTo CleanUp
Application.EnableEvents = False
Application.Calculation = xlCalculationManual

For k = 2 To n
dco.RemoveAll
swapo dco, dcn

For Each y In dco.Keys
p = False

For j = 1 To n
If v(j, 3) < t - dco(y) - TOL Then Exit For

x = v(j, 1)
s = "+" & Format(x)
If Right(y, Len(s)) = s Then p = True

If p Then
re.Pattern = "\" & s & "(?=(\+|$))"
If re.Execute(y).Count < v(j, 2) Then
u = dco(y) + x

If Abs(t - u) < TOL Then
recsoln y & s

ElseIf u < t - TOL Then
dcn.Add Key:=y & s, Item:=u

c = CDec(c + 1)
Application.StatusBar = "[" & Format(k) & "] " &
Format(c)

End If
End If
End If
Next j
Next y

If dcn.Count = 0 Then Exit For
Next k

If (recsoln() = 0) Then _
MsgBox Prompt:="all combinations exhausted", Title:="No Solution"

CleanUp:
Application.EnableEvents = True
Application.Calculation = xlCalculationAutomatic
Application.StatusBar = False

End Sub


Private Function recsoln(Optional s As String)
Const OUTPUTWSN As String = "findsums solutions" 'modify to taste

Static r As Range
Dim ws As Worksheet

If s = "" And r Is Nothing Then
On Error Resume Next
Set ws = ActiveWorkbook.Worksheets(OUTPUTWSN)

If ws Is Nothing Then
Err.Clear
Application.ScreenUpdating = False
Set ws = ActiveSheet
Set r = Worksheets.Add.Range("A1")
r.Parent.Name = OUTPUTWSN
ws.Activate
Application.ScreenUpdating = False

Else
ws.Cells.Clear
Set r = ws.Range("A1")

End If

recsoln = 0

ElseIf s = "" Then
recsoln = r.Row - 1
Set r = Nothing

Else
r.Value = s
Set r = r.Offset(1, 0)
recsoln = r.Row - 1

End If

End Function


Private Sub qsortd(v As Variant, lft As Long, rgt As Long)
'ad hoc quicksort subroutine
'translated from Aho, Weinberger & Kernighan,
'"The Awk Programming Language", page 161

Dim j As Long, pvt As Long

If (lft >= rgt) Then Exit Sub

swap2 v, lft, lft + Int((rgt - lft + 1) * Rnd)

pvt = lft

For j = lft + 1 To rgt
If v(j, 1) > v(lft, 1) Then
pvt = pvt + 1
swap2 v, pvt, j
End If
Next j

swap2 v, lft, pvt

qsortd v, lft, pvt - 1
qsortd v, pvt + 1, rgt
End Sub


Private Sub swap2(v As Variant, i As Long, j As Long)
'modified version of the swap procedure from
'translated from Aho, Weinberger & Kernighan,
'"The Awk Programming Language", page 161

Dim t As Variant, k As Long

For k = LBound(v, 2) To UBound(v, 2)
t = v(i, k)
v(i, k) = v(j, k)
v(j, k) = t
Next k
End Sub


Private Sub swapo(a As Object, b As Object)
Dim t As Object

Set t = a
Set a = b
Set b = t
End Sub
'---- end VBA code ----

Hi. This doesn't apply here, so it's just for discussion..
In some optimization programs, it can sometimes be a good technique to
re-scale the problem. For financial data in some programs, one option would
be to multiply all data by 100 to make the numbers integers (working with
whole pennies). This doesn't work too well though with a spreadsheet and
Solver. However, there are some Solver problems that can benefit by working
with whole pennies.
So, another option in some programs might be to find a combination from
5204, 5763, 24781, 28571, 42500, ..etc

that sum to 42,042,219

--
Dana DeLouis
Win XP & Office 2003


> 'I was asked by a colleague to find the combination of certain numbers
> 'which will add up to a specific value. The numbers I was given were:
> '
> ' 52.04;57.63;247.81;285.71;425.00;690.72;764.57;1485.00;1609.24;
> ' 3737.45;6485.47;6883.85;7309.33;12914.64;13714.11;14346.39;
> ' 15337.85;22837.83;31201.42;34663.07;321987.28
> '
> ' (21 numbers in ascending order)
> '
> ' I am trying to get a combination so that it adds up to 420422.19.
> '
<<snip>>

"Dana DeLouis" wrote...
>Hi. This doesn't apply here, so it's just for discussion..
>In some optimization programs, it can sometimes be a good technique
>to re-scale the problem. For financial data in some programs, one
>option would be to multiply all data by 100 to make the numbers
>integers (working with whole pennies). This doesn't work too well
>though with a spreadsheet and Solver. However, there are some Solver
>problems that can benefit by working with whole pennies. So, another
>option in some programs might be to find a combination from 5204,
>5763, 24781, 28571, 42500, ..etc that sum to 42,042,219
....

I was think about that. It'd be possible to promt for users inputs of
scaling values and rounding tolerance. Setting the former to 100 and the
latter to 0 would scale monetary amounts to integers and only accept exact
equality. But it'd also allow for other sorts of problems.

Hi

I have a similar problem as mentioned above.

Say I have 10 numbers, and some will be duplicated.

eg 1,2,2,5,5,7,8,9,10,12

How can I produce a list of combinations of say 5 numbers that add up to say
30.

I want my list to include ALL combinations

ie 1 2 8 9 10 will appear twice as it will use a different number 2.

I think I need Dana's macro mentioned previously but please enlighten me.

Hope my ramblings make sense - Many thanks.

Mike

"Dana DeLouis" wrote:

> Thanks. I really like your code. I see where I went wrong. I was
> treating 506 on one line as different than the 506 on another line. With
> that in mind, I too get 247 unique solutions. Thanks for the catch. :>)
> Actually, I was just trying to point out how surprising the total number of
> combinations can be. I would have guessed maybe 2 or 3.
> Just for fun, the totals that have the most combinations are 4265 and 4782.
> Your program caught all 314 combinations. :>) I would have guessed 2,
> maybe 3 at the most. Again, I just find it interesting that there are that
> many. :>)
>
> --
> Dana DeLouis
> Win XP & Office 2003
>
>
> <....>
> > I also get only 247 combinations that sum to 3775, so our respective
> > sets of combinations differ in cardinality by a suspiciously round 90.
> > Here are my 247.
> >
> > +949+811+692+687+449+187
> > +949+811+765+580+506+117+47
> > +949+811+765+537+449+217+47
> etc...
>
> <snip>
>
>
>

Mike__ wrote...
....
>Say I have 10 numbers, and some will be duplicated.
>
>eg 1,2,2,5,5,7,8,9,10,12
>
>How can I produce a list of combinations of say 5 numbers that add up to say
>30.
>
>I want my list to include ALL combinations
>
>ie 1 2 8 9 10 will appear twice as it will use a different number 2.
>
>I think I need Dana's macro mentioned previously but please enlighten me.
....

Dana never provided the code he used to produce his results. I
speculated that he used Mathematica to generate all nonempty
combinations, then summed each of them. If so, that's a relatively
simple operation in Mathematica because it includes built-in means to
generate combinations and sum the combinations. It's not as easy in
Excel.

Worst case, this class of problem requires checking all 2^N
combinations. It's a practical necessity to eliminate unnecessary
branches and reduce unnecessary duplication from the iterative process.
That's why my macro doesn't produce multiple identical combinations
when there are duplicate numbers in the original set. Doing so requires
additional overhead that grows with the number of combinations in each
iterative step.

If you took the output from my macro, you have the distinct
combinations that sum to the target value. Use Data > Text to Columns
to split those into separate columns. If your original set were in
J5:J14 and the parsed (Data > Text to Columns) distinct combinations
were in L5:Q24, you could calculate the number of instances in K5:K24
using the following formulas.

K5 [array formula]:
=PRODUCT(IF(COUNTIF($J$5:$J$14,L5:Q5),
COUNTIF($J$5:$J$14,L5:Q5)/COUNTIF(L5:Q5,L5:Q5)))

Select K5 and fill down into K6:K24.

The distinct combinations of your original data that sum to 30 are

12 10 8
10 8 7 5
12 10 7 1
12 9 8 1
12 9 7 2
12 8 5 5
9 8 7 5 1
10 9 8 2 1
10 9 7 2 2
10 9 5 5 1
10 8 5 5 2
12 10 5 2 1
12 9 5 2 2
12 8 7 2 1
12 7 5 5 1
9 8 5 5 2 1
9 7 5 5 2 2
10 8 7 2 2 1
10 7 5 5 2 1
12 8 5 2 2 1

and the number of instances of each using the col K formulas above are

1
2
1
1
2
1
2
2
1
1
2
4
2
2
1
2
1
1
2
2

Macros are the only way to generate the necessary combinations with
some efficiency. Formulas are more efficient counting the instances of
each of the distinct combinations in the solution set.

Harlan

I'm way out of my comfort zone for knowing what i'm taking about but could I
ask just a couple more questions.

In my example I'm only interested in 5 numbers pulled out of the master list

ie in the example below only these combnations are relevant
> 9 8 7 5 1
> 10 9 8 2 1
> 10 9 7 2 2
> 10 9 5 5 1
> 10 8 5 5 2
> 12 10 5 2 1
> 12 9 5 2 2
> 12 8 7 2 1
> 12 7 5 5 1

And these aren't

> 12 10 8
> 10 8 7 5
> 12 10 7 1
> 12 9 8 1
> 12 9 7 2
> 12 8 5 5
> 9 8 5 5 2 1
> 9 7 5 5 2 2
> 10 8 7 2 2 1
> 10 7 5 5 2 1
> 12 8 5 2 2 1

If it not too much hassle could you post a macro for me that will do this
for me.
I've simplified my actual real life problem by saying its 5 from 20. But its
actually 11 from about 150. Will the macro able to handle this or is there
too much number crunching involved ?

Thanks very much for your assistance so far.

Mike

"Harlan Grove" wrote:

> Mike__ wrote...
> ....
> >Say I have 10 numbers, and some will be duplicated.
> >
> >eg 1,2,2,5,5,7,8,9,10,12
> >
> >How can I produce a list of combinations of say 5 numbers that add up to say
> >30.
> >
> >I want my list to include ALL combinations
> >
> >ie 1 2 8 9 10 will appear twice as it will use a different number 2.
> >
> >I think I need Dana's macro mentioned previously but please enlighten me.
> ....
>
> Dana never provided the code he used to produce his results. I
> speculated that he used Mathematica to generate all nonempty
> combinations, then summed each of them. If so, that's a relatively
> simple operation in Mathematica because it includes built-in means to
> generate combinations and sum the combinations. It's not as easy in
> Excel.
>
> Worst case, this class of problem requires checking all 2^N
> combinations. It's a practical necessity to eliminate unnecessary
> branches and reduce unnecessary duplication from the iterative process.
> That's why my macro doesn't produce multiple identical combinations
> when there are duplicate numbers in the original set. Doing so requires
> additional overhead that grows with the number of combinations in each
> iterative step.
>
> If you took the output from my macro, you have the distinct
> combinations that sum to the target value. Use Data > Text to Columns
> to split those into separate columns. If your original set were in
> J5:J14 and the parsed (Data > Text to Columns) distinct combinations
> were in L5:Q24, you could calculate the number of instances in K5:K24
> using the following formulas.
>
> K5 [array formula]:
> =PRODUCT(IF(COUNTIF($J$5:$J$14,L5:Q5),
> COUNTIF($J$5:$J$14,L5:Q5)/COUNTIF(L5:Q5,L5:Q5)))
>
> Select K5 and fill down into K6:K24.
>
> The distinct combinations of your original data that sum to 30 are
>
> 12 10 8
> 10 8 7 5
> 12 10 7 1
> 12 9 8 1
> 12 9 7 2
> 12 8 5 5
> 9 8 7 5 1
> 10 9 8 2 1
> 10 9 7 2 2
> 10 9 5 5 1
> 10 8 5 5 2
> 12 10 5 2 1
> 12 9 5 2 2
> 12 8 7 2 1
> 12 7 5 5 1
> 9 8 5 5 2 1
> 9 7 5 5 2 2
> 10 8 7 2 2 1
> 10 7 5 5 2 1
> 12 8 5 2 2 1
>
> and the number of instances of each using the col K formulas above are
>
> 1
> 2
> 1
> 1
> 2
> 1
> 2
> 2
> 1
> 1
> 2
> 4
> 2
> 2
> 1
> 2
> 1
> 1
> 2
> 2
>
> Macros are the only way to generate the necessary combinations with
> some efficiency. Formulas are more efficient counting the instances of
> each of the distinct combinations in the solution set.
>
>

Mike,

There are approximately 5.94 x10^23 possible combinations of 11 numbers out of 150 numbers,
compared to 1.8 x 10^6 when you are dealing with 5 out of 20. If the 20 number problem took .001
second, the 150 number problem would take 10 million years....

HTH,
Bernie
MS Excel MVP

> I've simplified my actual real life problem by saying its 5 from 20. But its
> actually 11 from about 150. Will the macro able to handle this or is there
> too much number crunching involved ?

Mike__ wrote...
....
>In my example I'm only interested in 5 numbers pulled out of the master list
....
>If it not too much hassle could you post a macro for me that will do this
>for me.

It's not the hassle, it's the point that if you can't modify the code,
how could you understand it? If you can't understand it, why would you
rely on it?

It's easy enough to extract the solution combinations with only 5
numbers from the exhaustive list, then calculate their respective
numbers of instances.

Finaly, I hate taking general code an making it overly particular.
However, if you don't want combinations of more than 11 numbers, change
the

For k = 2 To n

loop to

For k = 2 To 11

>I've simplified my actual real life problem by saying its 5 from 20. But its
>actually 11 from about 150. Will the macro able to handle this or is there
>too much number crunching involved ?

It may take a LONG TIME, but if there's pronounced variance in your 150
number set, then many branches of combinations should be eliminated
quickly. However, it's impossible to say for sure whether my macro
would fail or not. Worst case, you won't have sufficient memory to
store the intermediate combinations.

Still, post your data and your target sum. If you don't post any
description of the numbers, they'd just be numbers.

Bernie Deitrick wrote...
>There are approximately 5.94 x10^23 possible combinations of 11
>numbers out of 150 numbers, compared to 1.8 x 10^6 when you are
>dealing with 5 out of 20. If the 20 number problem took .001
>second, the 150 number problem would take 10 million years....

Only if the target value were approximately equal to a random sum
of 11 of the largest half of the numbers would there be anything
close to this number of combinations actually generated. If the
target value were approximately equal to 11 times the average of
the 150 numbers, combinations involving more than a few of the
numbers below the 30th percentile or above the 70th percentile
would have been ruled out in shorter combinations. This assumes
a reasonable variance in the 150 numbers.

Also, if 80% of that 0.001 second runtime were overhead, it'd
only take a few million years. I suspect the OP would have run out
of RAM in his lifetime.

"Harlan Grove" <[email protected]> wrote in message
news:[email protected]...
> Only if the target value were approximately equal to a random sum
> of 11 of the largest half of the numbers would there be anything
> close to this number of combinations actually generated. If the
> target value were approximately equal to 11 times the average of
> the 150 numbers, combinations involving more than a few of the
> numbers below the 30th percentile or above the 70th percentile
> would have been ruled out in shorter combinations. This assumes
> a reasonable variance in the 150 numbers.

Harlan,

I was guessing that the same sort of population statistics applied to both the 150 number set as the
20 number set, and that of the ~1sec (I may have blinked and mis-timed the routine), 99.9% was
overhead. (I didn't actually time it.) Still, even if the actual calc took only 1E-9 sec, it would
still take 10 years.... on a machine with unlimited memory. But I think we can agree that the
problem won't be solved on his PC anytime soon. And thanks, by the way, for the code. Works
sweetly.

Bernie

Bernie Deitrick wrote...
....
> . . . Still, even if the actual calc took only 1E-9 sec, it would
>still take 10 years.... on a machine with unlimited memory. But I
>think we can agree that the problem won't be solved on his PC
>anytime soon. . . .

I agree. This sort of problem is usually too big to be handled.
There's a small chance that the OP's data has a bit more variance
than the sample posted, in which case there's some likelihood that
huge swaths of combinations would be eliminated early on. Still, I'd
figure Excel would need to churn through billions of combinations,
and that'd probably take days.

BTW, COMBIN(150,11) returns 1.48852E+16. You claimed 5.94171E+23,
which is PERMUT(150,11). Since addition is commutative, there's no
difference between the sum of one permutation and another. So it's
only necessary to check distinct combinations. That drops your
original 10 million year runtime estimate down to the order of one
year. Any appreciable elimination of smaller cardinality combinations
would drop the runtime to the order of a single day or several hours.
At that order runtime, memory would be the limiting factor.

These are the actual numbers I'm working with at the moment

A C 10
A D 20
A L 8
A S 1
AH 4
A Y 2
B M 8
C J 9
C P 11
C R 14
D K 6
D B 7
D D 8
D Z 0
D M 13
D V 9
E H 11
E D 13
F L 17
F Q 11
G H 12
G M 9
H P 4
J B 10
J C 10
J D 3
J P 15
J R 7
J S 0
J T 10
JA 0
J Z 2
JF 1
K D 3
K N 9
K P 6
L B 2
L M 0
L R 12
LT 7
M B 1
M Z 5
M D 9
M V 18
N H 10
O M 5
P C 9
P Z 11
P F 6
R E 1
R F 6
R K 2
R v 5
S A 3
S E 4
S K 3
S W 13
T H 17
T R 4
T S 7
T Z 15
U E 5
W G 12
W R 15


My aim is to find out all the combinations using 11 numbers that add up to
153.

If it would take far too long to run I could simplify it as below.

As 153 is at the top range,I don't believe that there would be many
combinations compared to the maximum amount possible.

It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
and probably the lower numbers and duplicated numbers can be removed if its
still too unwieldy.

My data and total will change periodically -although my total will still be
very high- so would it be possible to type my total in cell A1 and my data
below it and then run the macro?

Thanks for all your help

MIke.


"Harlan Grove" wrote:

> Mike__ wrote...
> ....
> >In my example I'm only interested in 5 numbers pulled out of the master list
> ....
> >If it not too much hassle could you post a macro for me that will do this
> >for me.
>
> It's not the hassle, it's the point that if you can't modify the code,
> how could you understand it? If you can't understand it, why would you
> rely on it?
>
> It's easy enough to extract the solution combinations with only 5
> numbers from the exhaustive list, then calculate their respective
> numbers of instances.
>
> Finaly, I hate taking general code an making it overly particular.
> However, if you don't want combinations of more than 11 numbers, change
> the
>
> For k = 2 To n
>
> loop to
>
> For k = 2 To 11
>
> >I've simplified my actual real life problem by saying its 5 from 20. But its
> >actually 11 from about 150. Will the macro able to handle this or is there
> >too much number crunching involved ?
>
> It may take a LONG TIME, but if there's pronounced variance in your 150
> number set, then many branches of combinations should be eliminated
> quickly. However, it's impossible to say for sure whether my macro
> would fail or not. Worst case, you won't have sufficient memory to
> store the intermediate combinations.
>
> Still, post your data and your target sum. If you don't post any
> description of the numbers, they'd just be numbers.
>
>

Mike__ wrote...
>These are the actual numbers I'm working with at the moment
....
>My aim is to find out all the combinations using 11 numbers that add up to
>153.

I was affraid of something like this. Your largest 9 numbers alone add
up to 144, and that 9th largest number is the first of 3 13s. There are
6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
8 numbers, one of the 13s, one of the 9s and one of the 0s.

Off the top of my head, I'd guess the number of combinations of 11
numbers from your set of just 64 that'd sum to 153 would number on the
order of thousands, and it'd require trillions of combinations to
determine with certainty.

>If it would take far too long to run I could simplify it as below.
>
>As 153 is at the top range,I don't believe that there would be many
>combinations compared to the maximum amount possible.

You're wrong. Intuition is useless for this sort of problem.

>It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
>and probably the lower numbers and duplicated numbers can be removed if its
>still too unwieldy.

Again, intuition is useless. 20 isn't needed.
SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
there are 4 instances of

18,17,17,15,15,15,14,13,13,13,3

SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
6 9s, so there are 18 instances of

17,17,15,15,15,14,13,13,13,12,9

SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
different combinations of these 9 numbers. There are 3 12s and 4 4s, so
36 combinations of

18,17,17,15,15,15,14,13,13,12,4

4 11s and 4 5s, so 48 combinations of

18,17,17,15,15,15,14,13,13,11,5

5 10s and 4 6s, so 90 combinations of

18,17,17,15,15,15,14,13,13,10,6

6 9s and 4 7s, so 72 combinations of

18,17,17,15,15,15,14,13,13,9,7

and 3 8s, so 9 combinations of

18,17,17,15,15,15,14,13,13,8,8

SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
6s, so 144 instances of

18,17,17,15,15,15,14,13,12,11,6

That's enough for me. As I said, the number of combinations of 11
numbers that sum to 153 is on the order of thousands, and most of those
won't include 20. To repeat, INTUITION IS USELESS.

FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
that sum to 153. Given the repeat counts for most of your 64 numbers,
I'd guess there were around 4,000 combinations in total.

"Harlan Grove" wrote:

> Mike__ wrote...
> >These are the actual numbers I'm working with at the moment
> ....
> >My aim is to find out all the combinations using 11 numbers that add up to
> >153.
>
> I was affraid of something like this. Your largest 9 numbers alone add
> up to 144, and that 9th largest number is the first of 3 13s. There are
> 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> 8 numbers, one of the 13s, one of the 9s and one of the 0s.
>
> Off the top of my head, I'd guess the number of combinations of 11
> numbers from your set of just 64 that'd sum to 153 would number on the
> order of thousands, and it'd require trillions of combinations to
> determine with certainty.
>
> >If it would take far too long to run I could simplify it as below.
> >
> >As 153 is at the top range,I don't believe that there would be many
> >combinations compared to the maximum amount possible.
>
> You're wrong. Intuition is useless for this sort of problem.
>
> >It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
> >and probably the lower numbers and duplicated numbers can be removed if its
> >still too unwieldy.
>
> Again, intuition is useless. 20 isn't needed.
> SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> there are 4 instances of
>
> 18,17,17,15,15,15,14,13,13,13,3
>
> SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> 6 9s, so there are 18 instances of
>
> 17,17,15,15,15,14,13,13,13,12,9
>
> SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> 36 combinations of
>
> 18,17,17,15,15,15,14,13,13,12,4
>
> 4 11s and 4 5s, so 48 combinations of
>
> 18,17,17,15,15,15,14,13,13,11,5
>
> 5 10s and 4 6s, so 90 combinations of
>
> 18,17,17,15,15,15,14,13,13,10,6
>
> 6 9s and 4 7s, so 72 combinations of
>
> 18,17,17,15,15,15,14,13,13,9,7
>
> and 3 8s, so 9 combinations of
>
> 18,17,17,15,15,15,14,13,13,8,8
>
> SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> 6s, so 144 instances of
>
> 18,17,17,15,15,15,14,13,12,11,6
>
> That's enough for me. As I said, the number of combinations of 11
> numbers that sum to 153 is on the order of thousands, and most of those
> won't include 20. To repeat, INTUITION IS USELESS.
>
> FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> that sum to 153. Given the repeat counts for most of your 64 numbers,
> I'd guess there were around 4,000 combinations in total.
>
>

Have I lost my last message ?

"Mike__" wrote:

>
>
> "Harlan Grove" wrote:
>
> > Mike__ wrote...
> > >These are the actual numbers I'm working with at the moment
> > ....
> > >My aim is to find out all the combinations using 11 numbers that add up to
> > >153.
> >
> > I was affraid of something like this. Your largest 9 numbers alone add
> > up to 144, and that 9th largest number is the first of 3 13s. There are
> > 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> > 8 numbers, one of the 13s, one of the 9s and one of the 0s.
> >
> > Off the top of my head, I'd guess the number of combinations of 11
> > numbers from your set of just 64 that'd sum to 153 would number on the
> > order of thousands, and it'd require trillions of combinations to
> > determine with certainty.
> >
> > >If it would take far too long to run I could simplify it as below.
> > >
> > >As 153 is at the top range,I don't believe that there would be many
> > >combinations compared to the maximum amount possible.
> >
> > You're wrong. Intuition is useless for this sort of problem.
> >
> > >It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
> > >and probably the lower numbers and duplicated numbers can be removed if its
> > >still too unwieldy.
> >
> > Again, intuition is useless. 20 isn't needed.
> > SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> > there are 4 instances of
> >
> > 18,17,17,15,15,15,14,13,13,13,3
> >
> > SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> > 6 9s, so there are 18 instances of
> >
> > 17,17,15,15,15,14,13,13,13,12,9
> >
> > SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> > different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> > 36 combinations of
> >
> > 18,17,17,15,15,15,14,13,13,12,4
> >
> > 4 11s and 4 5s, so 48 combinations of
> >
> > 18,17,17,15,15,15,14,13,13,11,5
> >
> > 5 10s and 4 6s, so 90 combinations of
> >
> > 18,17,17,15,15,15,14,13,13,10,6
> >
> > 6 9s and 4 7s, so 72 combinations of
> >
> > 18,17,17,15,15,15,14,13,13,9,7
> >
> > and 3 8s, so 9 combinations of
> >
> > 18,17,17,15,15,15,14,13,13,8,8
> >
> > SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> > different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> > 6s, so 144 instances of
> >
> > 18,17,17,15,15,15,14,13,12,11,6
> >
> > That's enough for me. As I said, the number of combinations of 11
> > numbers that sum to 153 is on the order of thousands, and most of those
> > won't include 20. To repeat, INTUITION IS USELESS.
> >
> > FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> > that sum to 153. Given the repeat counts for most of your 64 numbers,
> > I'd guess there were around 4,000 combinations in total.
> >
> >

Aaaaargh

After spending 20 mins listing out exactly what my data consists of, I find
its been lost in the ether.

If I just could summarize, perhaps you could tell me if its worth continuing.

You told me earlier that my data would create about 4000 possible
combinations.

However what I didnt mention is that my raw data is split into 4 distinct
groups and my 11 numbers that will total 153 must have 1 from Group A and 4
from Group B and any combination from Group C & D as long as I finish with 11
numbers. So this will reduce the combinations down to a more workable amount.
(hopefully)

Is it worth me telling you which data is in which group, could you program a
model that i could sort my data into the separate groups,or has it just got
too complicated?

I'll understand if its the latter.

Mike

"Mike__" wrote:

> Have I lost my last message ?
>
> "Mike__" wrote:
>
> >
> >
> > "Harlan Grove" wrote:
> >
> > > Mike__ wrote...
> > > >These are the actual numbers I'm working with at the moment
> > > ....
> > > >My aim is to find out all the combinations using 11 numbers that add up to
> > > >153.
> > >
> > > I was affraid of something like this. Your largest 9 numbers alone add
> > > up to 144, and that 9th largest number is the first of 3 13s. There are
> > > 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> > > 8 numbers, one of the 13s, one of the 9s and one of the 0s.
> > >
> > > Off the top of my head, I'd guess the number of combinations of 11
> > > numbers from your set of just 64 that'd sum to 153 would number on the
> > > order of thousands, and it'd require trillions of combinations to
> > > determine with certainty.
> > >
> > > >If it would take far too long to run I could simplify it as below.
> > > >
> > > >As 153 is at the top range,I don't believe that there would be many
> > > >combinations compared to the maximum amount possible.
> > >
> > > You're wrong. Intuition is useless for this sort of problem.
> > >
> > > >It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
> > > >and probably the lower numbers and duplicated numbers can be removed if its
> > > >still too unwieldy.
> > >
> > > Again, intuition is useless. 20 isn't needed.
> > > SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> > > there are 4 instances of
> > >
> > > 18,17,17,15,15,15,14,13,13,13,3
> > >
> > > SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> > > 6 9s, so there are 18 instances of
> > >
> > > 17,17,15,15,15,14,13,13,13,12,9
> > >
> > > SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> > > different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> > > 36 combinations of
> > >
> > > 18,17,17,15,15,15,14,13,13,12,4
> > >
> > > 4 11s and 4 5s, so 48 combinations of
> > >
> > > 18,17,17,15,15,15,14,13,13,11,5
> > >
> > > 5 10s and 4 6s, so 90 combinations of
> > >
> > > 18,17,17,15,15,15,14,13,13,10,6
> > >
> > > 6 9s and 4 7s, so 72 combinations of
> > >
> > > 18,17,17,15,15,15,14,13,13,9,7
> > >
> > > and 3 8s, so 9 combinations of
> > >
> > > 18,17,17,15,15,15,14,13,13,8,8
> > >
> > > SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> > > different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> > > 6s, so 144 instances of
> > >
> > > 18,17,17,15,15,15,14,13,12,11,6
> > >
> > > That's enough for me. As I said, the number of combinations of 11
> > > numbers that sum to 153 is on the order of thousands, and most of those
> > > won't include 20. To repeat, INTUITION IS USELESS.
> > >
> > > FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> > > that sum to 153. Given the repeat counts for most of your 64 numbers,
> > > I'd guess there were around 4,000 combinations in total.
> > >
> > >

Can you post your macro that would create about 4000 possibilities ?

Then I can leave you in peace !

Mike



"Mike__" wrote:

> Aaaaargh
>
> After spending 20 mins listing out exactly what my data consists of, I find
> its been lost in the ether.
>
> If I just could summarize, perhaps you could tell me if its worth continuing.
>
> You told me earlier that my data would create about 4000 possible
> combinations.
>
> However what I didnt mention is that my raw data is split into 4 distinct
> groups and my 11 numbers that will total 153 must have 1 from Group A and 4
> from Group B and any combination from Group C & D as long as I finish with 11
> numbers. So this will reduce the combinations down to a more workable amount.
> (hopefully)
>
> Is it worth me telling you which data is in which group, could you program a
> model that i could sort my data into the separate groups,or has it just got
> too complicated?
>
> I'll understand if its the latter.
>
> Mike
>
> "Mike__" wrote:
>
> > Have I lost my last message ?
> >
> > "Mike__" wrote:
> >
> > >
> > >
> > > "Harlan Grove" wrote:
> > >
> > > > Mike__ wrote...
> > > > >These are the actual numbers I'm working with at the moment
> > > > ....
> > > > >My aim is to find out all the combinations using 11 numbers that add up to
> > > > >153.
> > > >
> > > > I was affraid of something like this. Your largest 9 numbers alone add
> > > > up to 144, and that 9th largest number is the first of 3 13s. There are
> > > > 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> > > > 8 numbers, one of the 13s, one of the 9s and one of the 0s.
> > > >
> > > > Off the top of my head, I'd guess the number of combinations of 11
> > > > numbers from your set of just 64 that'd sum to 153 would number on the
> > > > order of thousands, and it'd require trillions of combinations to
> > > > determine with certainty.
> > > >
> > > > >If it would take far too long to run I could simplify it as below.
> > > > >
> > > > >As 153 is at the top range,I don't believe that there would be many
> > > > >combinations compared to the maximum amount possible.
> > > >
> > > > You're wrong. Intuition is useless for this sort of problem.
> > > >
> > > > >It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
> > > > >and probably the lower numbers and duplicated numbers can be removed if its
> > > > >still too unwieldy.
> > > >
> > > > Again, intuition is useless. 20 isn't needed.
> > > > SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> > > > there are 4 instances of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,13,3
> > > >
> > > > SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> > > > 6 9s, so there are 18 instances of
> > > >
> > > > 17,17,15,15,15,14,13,13,13,12,9
> > > >
> > > > SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> > > > different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> > > > 36 combinations of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,12,4
> > > >
> > > > 4 11s and 4 5s, so 48 combinations of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,11,5
> > > >
> > > > 5 10s and 4 6s, so 90 combinations of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,10,6
> > > >
> > > > 6 9s and 4 7s, so 72 combinations of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,9,7
> > > >
> > > > and 3 8s, so 9 combinations of
> > > >
> > > > 18,17,17,15,15,15,14,13,13,8,8
> > > >
> > > > SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> > > > different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> > > > 6s, so 144 instances of
> > > >
> > > > 18,17,17,15,15,15,14,13,12,11,6
> > > >
> > > > That's enough for me. As I said, the number of combinations of 11
> > > > numbers that sum to 153 is on the order of thousands, and most of those
> > > > won't include 20. To repeat, INTUITION IS USELESS.
> > > >
> > > > FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> > > > that sum to 153. Given the repeat counts for most of your 64 numbers,
> > > > I'd guess there were around 4,000 combinations in total.
> > > >
> > > >

Hi. I can't add anything to the problem of using a fix size (ie 11 numbers)
from a larger size. Although I could have sworn I've read a technique
somewhere.
However, I thought I'd share an interesting technique for testing one's
program for selecting numbers that total a specific value.
Suppose one wanted to test a program with a large number of solutions.
One technique is to use a sequence of data 1,2,...n without duplicates.
For example, suppose we want to test our program to see if it will find all
combinations that total 153 from the numbers 1 thru 20.

One technique to calculate the number of solutions is to use the Generation
Function 1+z^i.
Excel can not do this of course, so we use a math program.
So, the real generation function goes out to our maximum value in our data
range. We take the product as i goes from 1 to 20.

gf = Product[1 + z^i, {i, 20}]

(1 + z)*(1 + z^2)*...(1 + z^19)*(1 + z^20)

We then look at the series expansion of this function, and pick the
coefficient associated with our z^153 term.

Timing[Coefficient[Series[gf, {z, 0, 153}], z, 153]]

{0.*Second, 3288}

So, we immediately see that the solution is 3288.
Given the numbers 1 thru 20, there are 3288 combinations that total 153.
And then we test our vba program to see if it finds all of them.

Stated another way, the above problem can also be worded from number theory
as "Given the number of Partitions of the integer 153 (of which there are
54,770,336,324), how many have unique solutions who's maximum value is 20?
Note that integer partitions have duplicates. For example, the partitions
of the number 4 are:
{4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}

Anyway, I just thought this might be interesting to share.
--
Dana DeLouis
Win XP & Office 2003


"Harlan Grove" <[email protected]> wrote in message
news:[email protected]...
> Mike__ wrote...
>>These are the actual numbers I'm working with at the moment
> ...
>>My aim is to find out all the combinations using 11 numbers that add up to
>>153.
>
> I was affraid of something like this. Your largest 9 numbers alone add
> up to 144, and that 9th largest number is the first of 3 13s. There are
> 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> 8 numbers, one of the 13s, one of the 9s and one of the 0s.
>
> Off the top of my head, I'd guess the number of combinations of 11
> numbers from your set of just 64 that'd sum to 153 would number on the
> order of thousands, and it'd require trillions of combinations to
> determine with certainty.
>
>>If it would take far too long to run I could simplify it as below.
>>
>>As 153 is at the top range,I don't believe that there would be many
>>combinations compared to the maximum amount possible.
>
> You're wrong. Intuition is useless for this sort of problem.
>
>>It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
>>and probably the lower numbers and duplicated numbers can be removed if
>>its
>>still too unwieldy.
>
> Again, intuition is useless. 20 isn't needed.
> SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> there are 4 instances of
>
> 18,17,17,15,15,15,14,13,13,13,3
>
> SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> 6 9s, so there are 18 instances of
>
> 17,17,15,15,15,14,13,13,13,12,9
>
> SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> 36 combinations of
>
> 18,17,17,15,15,15,14,13,13,12,4
>
> 4 11s and 4 5s, so 48 combinations of
>
> 18,17,17,15,15,15,14,13,13,11,5
>
> 5 10s and 4 6s, so 90 combinations of
>
> 18,17,17,15,15,15,14,13,13,10,6
>
> 6 9s and 4 7s, so 72 combinations of
>
> 18,17,17,15,15,15,14,13,13,9,7
>
> and 3 8s, so 9 combinations of
>
> 18,17,17,15,15,15,14,13,13,8,8
>
> SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> 6s, so 144 instances of
>
> 18,17,17,15,15,15,14,13,12,11,6
>
> That's enough for me. As I said, the number of combinations of 11
> numbers that sum to 153 is on the order of thousands, and most of those
> won't include 20. To repeat, INTUITION IS USELESS.
>
> FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> that sum to 153. Given the repeat counts for most of your 64 numbers,
> I'd guess there were around 4,000 combinations in total.
>

I know this is an old thread, but I came across the generating function for
the following question if anyone is interested:

Given the numbers 1 - 20 (no duplicates), how many groups of 11 numbers
total 153?

gf = Product[x*y^j + 1, {j, 1, 20}]

There is no Series expansion in this version, but one would still want to
use a math program instead of Excel to do this.
m=20, n=11, t=153

Coefficient[Coefficient[Collect[Product[x*y^j+1,{j,m}],x],x,n],y,t]

As a function, there are 72 combinations of 11 numbers that total 153

Timing[Fx[20, 11, 153]]

{0.016*Second, 72}

Given the numbers 1-64, there are about 3.7 million combinations of 11
numbers that total 153. Timing seems slow, so I'm sure there's a more
efficient algorithm.

Timing[Fx[64, 11, 153]]

{2.5*Second, 3,699,726}

--
Dana DeLouis
Win XP & Office 2003

<snip>

Harlan Grove wrote:
> Mike__ wrote...
> >These are the actual numbers I'm working with at the moment
> ...
> >My aim is to find out all the combinations using 11 numbers that add up to
> >153.
>
> I was affraid of something like this. Your largest 9 numbers alone add
> up to 144, and that 9th largest number is the first of 3 13s. There are
> 6 9s and 4 0s, so there are 72 (=3*6*4) combinations using the largest
> 8 numbers, one of the 13s, one of the 9s and one of the 0s.
>
> Off the top of my head, I'd guess the number of combinations of 11
> numbers from your set of just 64 that'd sum to 153 would number on the
> order of thousands, and it'd require trillions of combinations to
> determine with certainty.
>
> >If it would take far too long to run I could simplify it as below.
> >
> >As 153 is at the top range,I don't believe that there would be many
> >combinations compared to the maximum amount possible.
>
> You're wrong. Intuition is useless for this sort of problem.
>
> >It almost certainly uses AD 20 so perhaps I can look for 10 that total 133
> >and probably the lower numbers and duplicated numbers can be removed if its
> >still too unwieldy.
>
> Again, intuition is useless. 20 isn't needed.
> SUM(18,17,17,15,15,15,14,13,13,13) equals 150, and there are 4 3s, so
> there are 4 instances of
>
> 18,17,17,15,15,15,14,13,13,13,3
>
> SUM(17,17,15,15,15,14,13,13,13,12) equals 144, and there are 3 12s and
> 6 9s, so there are 18 instances of
>
> 17,17,15,15,15,14,13,13,13,12,9
>
> SUM(18,17,17,15,15,15,14,13,13) equals 137, and there are 3 13s, so 3
> different combinations of these 9 numbers. There are 3 12s and 4 4s, so
> 36 combinations of
>
> 18,17,17,15,15,15,14,13,13,12,4
>
> 4 11s and 4 5s, so 48 combinations of
>
> 18,17,17,15,15,15,14,13,13,11,5
>
> 5 10s and 4 6s, so 90 combinations of
>
> 18,17,17,15,15,15,14,13,13,10,6
>
> 6 9s and 4 7s, so 72 combinations of
>
> 18,17,17,15,15,15,14,13,13,9,7
>
> and 3 8s, so 9 combinations of
>
> 18,17,17,15,15,15,14,13,13,8,8
>
> SUM(18,17,17,15,15,15,14,13) equals 124, and there are 3 13s, so 3
> different combinations of these 8 numbers. There are 3 12s, 4 11s and 4
> 6s, so 144 instances of
>
> 18,17,17,15,15,15,14,13,12,11,6
>
> That's enough for me. As I said, the number of combinations of 11
> numbers that sum to 153 is on the order of thousands, and most of those
> won't include 20. To repeat, INTUITION IS USELESS.
>
> FWIW, my macro came up with 828 *distinct* combinations of 11 numbers
> that sum to 153. Given the repeat counts for most of your 64 numbers,
> I'd guess there were around 4,000 combinations in total.

hi..sorry to jump in here but I think youse may be able to help me..I
have a column in excel that can list about 2,000 values (positive and
negative) to 2.d.p. I need excel to tell me which of these values add
up to a given value (x), it can be to the nearest whole number. I was
informed about using solver but it cant handle this many values..can
you help? can I use some sort of macro here?

Hi, Mel. The only thing I've ever seen like this is here:
http://www.mrexcel.com/pc09.shtml
************
Anne Troy
www.OfficeArticles.com

"mellowe" <[email protected]> wrote in message
news:[email protected]...
> hi..sorry to jump in here but I think youse may be able to help me..I
> have a column in excel that can list about 2,000 values (positive and
> negative) to 2.d.p. I need excel to tell me which of these values add
> up to a given value (x), it can be to the nearest whole number. I was
> informed about using solver but it cant handle this many values..can
> you help? can I use some sort of macro here?
>