Re: [R] The 'subset matching' challenge
If you look for "subset sum problem", you will find relevant
information.
A starter:
http://en.wikipedia.org/wiki/Subset_sum_problem
Detlef
On Thu, 29 Oct 2009 15:47:22 +0100
Yvonnick Noël wrote:
> Dear all,
>
> The following problem just has been submitted to me by an accountant.
>
> In his new job, he has to close some old accounts. He has yearly
> amounts, and a list of products that have been bought over the years, at
> certain prices for which he has an exhaustive record. The problem is: He
> does not know what product was bought this or that year (don't ask). He
> does not want to find back the real story, but just write realistic
> accounts, for which the sum of a subset of product prices will give the
> exact yearly amount.
>
> Here is a real example from his data:
>
> # A list of 64 product prices
> products =
> c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,
> 53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
> 27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
> 59700,25908,829350,1198000,1031655)
>
> # Global amount
> amount = 4748652
>
> Now he wants to find all subsets of the 'product' vector which sums to
> 'amount'.
>
> I wrote the following code, which is clearly not optimal:
>
> # Create a matrix of subsets of size r among the integer set 1:n
> subsets <- function(n, r, v = 1:n) {
> if(r <= 0) vector(mode(v), 0)
> else if(r >= n) v[1:n]
> else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
> }
>
> # Main function
> find.amount = function(amount,products) {
>
> if(sum(products) cat("There is no solution.\n")
> return()
> }
>
> l = length(products)
> cat("\nThere are",l,"product prices\n\n")
> names(products) = paste("Product",1:l,sep="")
> products = sort(products)
>
> for(i in 2:l) {
>
> # If the sum of the i smallest prices is greater than amount, then stop
> if(sum(products[1:i])>amount) break
>
> # Look for matching subsets only in the case when the sum of i
> largest prices is greater than amount
> if(sum(rev(products)[1:i])>=amount) {
> # Generates all subsets of i indicies in 1:l
> subs = subsets(l,i)
> nl = nrow(subs)
> nc = ncol(subs)
>
> # Compute sums of corresponding price subsets
> sums = rowSums(matrix(products[subs],nl,nc))
>
> # Which ones match the global amount ?
> w = which(sums == amount)
> how.many = length(w)
> if(how.many>0) {
> cat("\n-->> There are",how.many,"solutions with",nc,"products :\n")
> for(j in 1:how.many) {
> print(products[subs[w[j],]])
> }
> }
> else cat("\n-->> There is no solution with",nc,"products.\n")
> }
> else cat("\n-->> There is no solution with",i,"products.\n")
> }
> }
>
>
> Then I can use these functions on a smaller example:
>
> > find.amount(4,c(1,1,1,1,2,2))
>
> and a number of matching subsets are provided. The problem is: This
> approach creates a whole matrix of subsets of r integers among 1:n,
> which rapidly gives huge matrices, and this is clearly not optimal for
> the real data provided above.
>
> Would anyone have a suggestion as to an alternative and more efficient
> strategy?
>
> Good luck,
>
> Yvonnick Noel
> University of Brittany, Rennes 2
> France
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] The 'subset matching' challenge
yvonnick noel wrote:
>
> Dear all,
>
> The following problem just has been submitted to me by an accountant.
>
> In his new job, he has to close some old accounts. He has yearly
> amounts, and a list of products that have been bought over the years, at
> certain prices for which he has an exhaustive record. The problem is: He
> does not know what product was bought this or that year (don't ask). He
> does not want to find back the real story, but just write realistic
> accounts, for which the sum of a subset of product prices will give the
> exact yearly amount.
>
> Here is a real example from his data:
>
> # A list of 64 product prices
> products =
> c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,
> 53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
> 27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
> 59700,25908,829350,1198000,1031655)
>
> # Global amount
> amount = 4748652
>
> Now he wants to find all subsets of the 'product' vector which sums to
> 'amount'.
>
> I wrote the following code, which is clearly not optimal:
>
> # Create a matrix of subsets of size r among the integer set 1:n
> subsets <- function(n, r, v = 1:n) {
> if(r <= 0) vector(mode(v), 0)
> else if(r >= n) v[1:n]
> else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
> }
>
> # Main function
> find.amount = function(amount,products) {
>
> if(sum(products) cat("There is no solution.\n")
> return()
> }
>
> l = length(products)
> cat("\nThere are",l,"product prices\n\n")
> names(products) = paste("Product",1:l,sep="")
> products = sort(products)
>
> for(i in 2:l) {
>
> # If the sum of the i smallest prices is greater than amount, then
> stop
> if(sum(products[1:i])>amount) break
>
> # Look for matching subsets only in the case when the sum of i
> largest prices is greater than amount
> if(sum(rev(products)[1:i])>=amount) {
> # Generates all subsets of i indicies in 1:l
> subs = subsets(l,i)
> nl = nrow(subs)
> nc = ncol(subs)
>
> # Compute sums of corresponding price subsets
> sums = rowSums(matrix(products[subs],nl,nc))
>
> # Which ones match the global amount ?
> w = which(sums == amount)
> how.many = length(w)
> if(how.many>0) {
> cat("\n-->> There are",how.many,"solutions with",nc,"products
> :\n")
> for(j in 1:how.many) {
> print(products[subs[w[j],]])
> }
> }
> else cat("\n-->> There is no solution with",nc,"products.\n")
> }
> else cat("\n-->> There is no solution with",i,"products.\n")
> }
> }
>
>
> Then I can use these functions on a smaller example:
>
> > find.amount(4,c(1,1,1,1,2,2))
>
> and a number of matching subsets are provided. The problem is: This
> approach creates a whole matrix of subsets of r integers among 1:n,
> which rapidly gives huge matrices, and this is clearly not optimal for
> the real data provided above.
>
> Would anyone have a suggestion as to an alternative and more efficient
> strategy?
>
> Good luck,
>
> Yvonnick Noel
> University of Brittany, Rennes 2
> France
>
>
I believe this is the "knapsack problem".
RSiteSearch("knapsack") might help a little bit.
http://xkcd.com/287/
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__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] The 'subset matching' challenge
Dear all,
The following problem just has been submitted to me by an accountant.
In his new job, he has to close some old accounts. He has yearly
amounts, and a list of products that have been bought over the years, at
certain prices for which he has an exhaustive record. The problem is: He
does not know what product was bought this or that year (don't ask). He
does not want to find back the real story, but just write realistic
accounts, for which the sum of a subset of product prices will give the
exact yearly amount.
Here is a real example from his data:
# A list of 64 product prices
products =
c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,
53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
59700,25908,829350,1198000,1031655)
# Global amount
amount = 4748652
Now he wants to find all subsets of the 'product' vector which sums to
'amount'.
I wrote the following code, which is clearly not optimal:
# Create a matrix of subsets of size r among the integer set 1:n
subsets <- function(n, r, v = 1:n) {
if(r <= 0) vector(mode(v), 0)
else if(r >= n) v[1:n]
else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
}
# Main function
find.amount = function(amount,products) {
if(sum(products)amount) break
# Look for matching subsets only in the case when the sum of i
largest prices is greater than amount
if(sum(rev(products)[1:i])>=amount) {
# Generates all subsets of i indicies in 1:l
subs = subsets(l,i)
nl = nrow(subs)
nc = ncol(subs)
# Compute sums of corresponding price subsets
sums = rowSums(matrix(products[subs],nl,nc))
# Which ones match the global amount ?
w = which(sums == amount)
how.many = length(w)
if(how.many>0) {
cat("\n-->> There are",how.many,"solutions with",nc,"products :\n")
for(j in 1:how.many) {
print(products[subs[w[j],]])
}
}
else cat("\n-->> There is no solution with",nc,"products.\n")
}
else cat("\n-->> There is no solution with",i,"products.\n")
}
}
Then I can use these functions on a smaller example:
> find.amount(4,c(1,1,1,1,2,2))
and a number of matching subsets are provided. The problem is: This
approach creates a whole matrix of subsets of r integers among 1:n,
which rapidly gives huge matrices, and this is clearly not optimal for
the real data provided above.
Would anyone have a suggestion as to an alternative and more efficient
strategy?
Good luck,
Yvonnick Noel
University of Brittany, Rennes 2
France
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
