Hi, Do you have ideas, how to find all those different combinations of integers (>0) that produce as a sum, a certain integer. i.e.: if that sum is 3, the possibilities are c(1,1,1), c(1,2), c(2,1) 4, the possibilities are c(1,1,1,1),c(1,1,2),c(1,2,1),c(2,1,1),c(2,2),c(1,3),c(3,1) etc. Best regards, Atte Tenkanen
Same sum, different sets of integers
5 messages · Atte Tenkanen, Jim Lemon, jim holtman +1 more
Hi Atte,
I'm not sure that this actually works, and it's very much a quick hack:
sums_x<-function(x,addends=1,depth=1) {
if(depth==1) {
addends<-rep(addends,x)
addlist<-list(addends)
} else {
addlist<-list()
}
lenadd<-length(addends)
while(lenadd > 2) {
addends<-c(addends[depth]+1,addends[-c(depth,depth+1)])
lenadd<-lenadd-1
if(sum(addends) == x) addlist[[length(addlist)+1]]<-addends
cat(depth,"-",addends,"\n")
if(lenadd > 2 && depth+1 < lenadd)
addlist<-c(addlist,(sums_x(x,addends=addends,depth=depth+1)))
}
return(addlist)
}
This doesn't return all the permutations of the addends, but it's all
the time I have to waste this morning.
Jim
On Thu, Apr 28, 2016 at 1:46 AM, Atte Tenkanen <attenka at utu.fi> wrote:
Hi, Do you have ideas, how to find all those different combinations of integers (>0) that produce as a sum, a certain integer. i.e.: if that sum is 3, the possibilities are c(1,1,1), c(1,2), c(2,1) 4, the possibilities are c(1,1,1,1),c(1,1,2),c(1,2,1),c(2,1,1),c(2,2),c(1,3),c(3,1) etc. Best regards, Atte Tenkanen
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This is not the most efficient, but gets the idea across. This is the largest sum I can compute on my laptop with 16GB of memory. If I try to set N to 9, I run out of memory due to the size of the expand.grid.
N <- 8 # value to add up to # create expand.grid for all combinations and convert to matrix x <- as.matrix(expand.grid(rep(list(0:(N - 1)), N))) # generate rowSums and determine which rows add to N z <- rowSums(x) # now extract those rows, sort and convert to strings to remove dups add2N <- x[z == N, ] strings <- apply(
+ t(apply(add2N, 1, sort)) # sort + , 1 + , toString + )
# remove dups
strings <- strings[!duplicated(strings)]
# remove leading zeros
strings <- gsub("0, ", "", strings)
# print out
cat(strings, sep = '\n')
1, 7 2, 6 3, 5 4, 4 1, 1, 6 1, 2, 5 1, 3, 4 2, 2, 4 2, 3, 3 1, 1, 1, 5 1, 1, 2, 4 1, 1, 3, 3 1, 2, 2, 3 2, 2, 2, 2 1, 1, 1, 1, 4 1, 1, 1, 2, 3 1, 1, 2, 2, 2 1, 1, 1, 1, 1, 3 1, 1, 1, 1, 2, 2 1, 1, 1, 1, 1, 1, 2 1, 1, 1, 1, 1, 1, 1, 1 Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it.
On Wed, Apr 27, 2016 at 11:46 AM, Atte Tenkanen <attenka at utu.fi> wrote:
Hi, Do you have ideas, how to find all those different combinations of integers (>0) that produce as a sum, a certain integer. i.e.: if that sum is 3, the possibilities are c(1,1,1), c(1,2), c(2,1) 4, the possibilities are c(1,1,1,1),c(1,1,2),c(1,2,1),c(2,1,1),c(2,2),c(1,3),c(3,1) etc. Best regards, Atte Tenkanen
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
I came up with this, using recursion. Short and should work for n
greater than 9 :)
Peter
sumsToN = function(n)
{
if (n==1) return(1);
out = lapply(1:(n-1), function(i) {
s1 = sumsToN(n-i);
lapply(s1, c, i)
})
c(n, unlist(out, recursive = FALSE));
}
sumsToN(4)
[[1]] [1] 4 [[2]] [1] 3 1 [[3]] [1] 2 1 1 [[4]] [1] 1 1 1 1 [[5]] [1] 1 2 1 [[6]] [1] 2 2 [[7]] [1] 1 1 2 [[8]] [1] 1 3
sumsToN(5)
[[1]] [1] 5 [[2]] [1] 4 1 [[3]] [1] 3 1 1 [[4]] [1] 2 1 1 1 [[5]] [1] 1 1 1 1 1 [[6]] [1] 1 2 1 1 [[7]] [1] 2 2 1 [[8]] [1] 1 1 2 1 [[9]] [1] 1 3 1 [[10]] [1] 3 2 [[11]] [1] 2 1 2 [[12]] [1] 1 1 1 2 [[13]] [1] 1 2 2 [[14]] [1] 2 3 [[15]] [1] 1 1 3 [[16]] [1] 1 4
On Wed, Apr 27, 2016 at 6:10 PM, jim holtman <jholtman at gmail.com> wrote:
This is not the most efficient, but gets the idea across. This is the largest sum I can compute on my laptop with 16GB of memory. If I try to set N to 9, I run out of memory due to the size of the expand.grid.
N <- 8 # value to add up to # create expand.grid for all combinations and convert to matrix x <- as.matrix(expand.grid(rep(list(0:(N - 1)), N))) # generate rowSums and determine which rows add to N z <- rowSums(x) # now extract those rows, sort and convert to strings to remove dups add2N <- x[z == N, ] strings <- apply(
+ t(apply(add2N, 1, sort)) # sort + , 1 + , toString + )
# remove dups
strings <- strings[!duplicated(strings)]
# remove leading zeros
strings <- gsub("0, ", "", strings)
# print out
cat(strings, sep = '\n')
1, 7 2, 6 3, 5 4, 4 1, 1, 6 1, 2, 5 1, 3, 4 2, 2, 4 2, 3, 3 1, 1, 1, 5 1, 1, 2, 4 1, 1, 3, 3 1, 2, 2, 3 2, 2, 2, 2 1, 1, 1, 1, 4 1, 1, 1, 2, 3 1, 1, 2, 2, 2 1, 1, 1, 1, 1, 3 1, 1, 1, 1, 2, 2 1, 1, 1, 1, 1, 1, 2 1, 1, 1, 1, 1, 1, 1, 1 Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Wed, Apr 27, 2016 at 11:46 AM, Atte Tenkanen <attenka at utu.fi> wrote:
Hi, Do you have ideas, how to find all those different combinations of integers (>0) that produce as a sum, a certain integer. i.e.: if that sum is 3, the possibilities are c(1,1,1), c(1,2), c(2,1) 4, the possibilities are c(1,1,1,1),c(1,1,2),c(1,2,1),c(2,1,1),c(2,2),c(1,3),c(3,1) etc. Best regards, Atte Tenkanen
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Thanks for the suggestions, all of you! I first began to think about using somehow permutations of the gtools-package. I will continue utilizing Peter's solution. My purpose is to divide basic musical rhythm units (whole note, half note, quarter note etc durations) to meaningful entities in algorithmic composition. I use R for that. Usually, there are - in the composition algorithm - random number generators, which "composes" different versions of music. Those RNG's (sample in R) can be - for their part - conducted by using different constraints and probabilities. Here I draw rhythms. For example, if my quarter note (1/4) is 1024 ticks in MIDI format, I may like to split it into quintuples, ie. five units, using ties differently: 1024*c(1/5, 4/5), 1024*c(2/5, 3/5),... This way I can produce and output rich and still usable rhythms to music notation software, keep the rhythmic entities playable and readable over barlines. Yours, Atte 28.4.2016, 5.00, Peter Langfelder kirjoitti:
I came up with this, using recursion. Short and should work for n
greater than 9 :)
Peter
sumsToN = function(n)
{
if (n==1) return(1);
out = lapply(1:(n-1), function(i) {
s1 = sumsToN(n-i);
lapply(s1, c, i)
})
c(n, unlist(out, recursive = FALSE));
}
sumsToN(4)
[[1]] [1] 4 [[2]] [1] 3 1 [[3]] [1] 2 1 1 [[4]] [1] 1 1 1 1 [[5]] [1] 1 2 1 [[6]] [1] 2 2 [[7]] [1] 1 1 2 [[8]] [1] 1 3
sumsToN(5)
[[1]] [1] 5 [[2]] [1] 4 1 [[3]] [1] 3 1 1 [[4]] [1] 2 1 1 1 [[5]] [1] 1 1 1 1 1 [[6]] [1] 1 2 1 1 [[7]] [1] 2 2 1 [[8]] [1] 1 1 2 1 [[9]] [1] 1 3 1 [[10]] [1] 3 2 [[11]] [1] 2 1 2 [[12]] [1] 1 1 1 2 [[13]] [1] 1 2 2 [[14]] [1] 2 3 [[15]] [1] 1 1 3 [[16]] [1] 1 4 On Wed, Apr 27, 2016 at 6:10 PM, jim holtman <jholtman at gmail.com> wrote:
This is not the most efficient, but gets the idea across. This is the largest sum I can compute on my laptop with 16GB of memory. If I try to set N to 9, I run out of memory due to the size of the expand.grid.
N <- 8 # value to add up to # create expand.grid for all combinations and convert to matrix x <- as.matrix(expand.grid(rep(list(0:(N - 1)), N))) # generate rowSums and determine which rows add to N z <- rowSums(x) # now extract those rows, sort and convert to strings to remove dups add2N <- x[z == N, ] strings <- apply(
+ t(apply(add2N, 1, sort)) # sort + , 1 + , toString + )
# remove dups
strings <- strings[!duplicated(strings)]
# remove leading zeros
strings <- gsub("0, ", "", strings)
# print out
cat(strings, sep = '\n')
1, 7 2, 6 3, 5 4, 4 1, 1, 6 1, 2, 5 1, 3, 4 2, 2, 4 2, 3, 3 1, 1, 1, 5 1, 1, 2, 4 1, 1, 3, 3 1, 2, 2, 3 2, 2, 2, 2 1, 1, 1, 1, 4 1, 1, 1, 2, 3 1, 1, 2, 2, 2 1, 1, 1, 1, 1, 3 1, 1, 1, 1, 2, 2 1, 1, 1, 1, 1, 1, 2 1, 1, 1, 1, 1, 1, 1, 1 Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Wed, Apr 27, 2016 at 11:46 AM, Atte Tenkanen <attenka at utu.fi> wrote:
Hi, Do you have ideas, how to find all those different combinations of integers (>0) that produce as a sum, a certain integer. i.e.: if that sum is 3, the possibilities are c(1,1,1), c(1,2), c(2,1) 4, the possibilities are c(1,1,1,1),c(1,1,2),c(1,2,1),c(2,1,1),c(2,2),c(1,3),c(3,1) etc. Best regards, Atte Tenkanen
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
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______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.