Hi, I have a question which is not necessarily a ruby-related one. Anyway, I'm using Ruby for it. To make 5 by adding 2 more more natural numbers, there are different 6 ways. 1 + 1 + 1 + 1 + 1 2 + 1 + 1 + 1 2 + 2 + 1 3 + 1 + 1 3 + 2 4 + 1 In array form, you may express it like: [1,1,1,1,1] [2,1,1,1] [2,2,1] [3,1,1] [3,2] [4,1] Is there a general and efficient way to get the arrays? For example, set(5) #=> [[1,1,1,1,1], [2,1,1,1], [2,2,1], [3,1,1], [3,2], [4,1]] I came up with a mutual recursive solution but it's not very efficient. (Probably because it's recursive, which Ruby is not very good at.) I can't make it efficient for big numbers. I tried to use a cache but it didn't help much enough. def set n return [[1, 1]] if n == 2 result = [] (1...n).each do |i| subset = get_subset(n - i, i) subset.each do |j| result << [i] + j end end result end def get_subset n, max result = (n <= max) ? [[n]] : [] result += (set(n).select { |i| i.all? { |j| j <= max } }) end Do you know a good solution to this problem? Also, is there a mathematical or algorithmic term for this problem like combinations and permutations? Thanks. Sam

on 2007-01-23 06:10

on 2007-01-23 07:01

Sam Kong wrote: > 2 + 2 + 1 > [3,1,1] > (Probably because it's recursive, which Ruby is not very good at.) > end > Do you know a good solution to this problem? > Also, is there a mathematical or algorithmic term for this problem like > combinations and permutations? > > Thanks. > > Sam > I think it's called "partitions" and I think it's covered somewhere in Knuth's "Art Of Computer Programming". That's where I'd check first, anyhow. -- M. Edward (Ed) Borasky, FBG, AB, PTA, PGS, MS, MNLP, NST, ACMC(P) http://borasky-research.blogspot.com/ If God had meant for carrots to be eaten cooked, He would have given rabbits fire.

on 2007-01-23 07:04

On 1/22/07, Sam Kong <sam.s.kong@gmail.com> wrote: > 2 + 2 + 1 > [3,1,1] > (Probably because it's recursive, which Ruby is not very good at.) > end > Do you know a good solution to this problem? > Also, is there a mathematical or algorithmic term for this problem like > combinations and permutations? There are a few terms: partition function, number partitioning, integer partition,... There is a moderately large literature on the topic. Try googling. Wikipedia has some good introductory articles, like <http://en.wikipedia.org/wiki/Integer_partition>, and a number of references. > > Thanks. > > Sam > > > -- Stuart Yarus <syarus at gmail dot com>

on 2007-01-23 15:13

On Jan 22, 2007, at 11:10 PM, Sam Kong wrote: > To make 5 by adding 2 more more natural numbers, there are different 6 > ways. > > 1 + 1 + 1 + 1 + 1 > 2 + 1 + 1 + 1 > 2 + 2 + 1 > 3 + 1 + 1 > 3 + 2 > 4 + 1 > Is there a general and efficient way to get the arrays? You said you have a recursive solution. The first step to optimizing recursion is to make sure you aren't doing a bunch of unneeded work. For example, generating all the combinations and then removing duplicates is too much work. Instead, we want to make sure we recursively generate just one set of numbers: def partition(largest, rest = Array.new, &block) block.call([largest] + rest) (rest.first || 1).upto(largest / 2) do |i| partition(largest - i, [i] + rest, &block) end end partition(5) { |nums| p nums } # >> [5] # >> [4, 1] # >> [3, 1, 1] # >> [2, 1, 1, 1] # >> [1, 1, 1, 1, 1] # >> [2, 2, 1] # >> [3, 2] The optimization for recursion is to eliminate the recursion. You can always unroll recursive code into an iterative solution: def partition(num) partitions = [[num]] until partitions.empty? current = partitions.shift yield current largest = current.shift (current.first || 1).upto(largest / 2) do |i| partitions << Array[largest - i, i, *current.dup] end end end partition(5) { |nums| p nums } # >> [5] # >> [4, 1] # >> [3, 2] # >> [3, 1, 1] # >> [2, 2, 1] # >> [2, 1, 1, 1] # >> [1, 1, 1, 1, 1] I translated these examples from the book Higher-Order Perl and, in case you are curious, wrote more about them here: http://blog.grayproductions.net/articles/2006/01/3... chapters-4-and-5 Hope that helps. James Edward Gray II

on 2007-01-23 18:41

Hi James, James Edward Gray II wrote: > > 4 + 1 > block.call([largest] + rest) > # >> [1, 1, 1, 1, 1] > # >> [2, 2, 1] > # >> [3, 2] > > The optimization for recursion is to eliminate the recursion. You > can always unroll recursive code into an iterative solution: Right. But some recursions are harder to unroll than others.^^ > (current.first || 1).upto(largest / 2) do |i| > # >> [2, 2, 1] > # >> [2, 1, 1, 1] > # >> [1, 1, 1, 1, 1] > > I translated these examples from the book Higher-Order Perl and, in > case you are curious, wrote more about them here: > > http://blog.grayproductions.net/articles/2006/01/3... > chapters-4-and-5 Your weblog looks very helpful. I just bookmarked it. > > Hope that helps. Yes. It helped a lot. > > James Edward Gray II Thank you very much. Also, I thank others who replied to my question. Sam

on 2007-01-23 19:15

```
"Sam Kong" <sam.s.kong@gmail.com> wrote/schrieb
<1169528870.482644.83900@11g2000cwr.googlegroups.com>:
> Do you know a good solution to this problem?
I let you decide if mine is a good solution:
#\\\
$cache = []
def parts(s)
if s < 1
[]
else
$cache[s] ||
begin
a = []
s.downto(1) do |n|
k = s / n
left = [].fill(n, (0 .. (k-1)))
r = s - k * n
if (r > 0)
right = parts(r)
right.each do |elem|
a.push([left,elem])
end
else
a.push(left)
end
end
$cache[s] = a
end
end
end
def part(n)
parts(n).map{|x| x.flatten}
end
pp part(8)
#///
Regards
Thomas
```

on 2007-01-23 20:07

On Jan 23, 2007, at 11:40 AM, Sam Kong wrote: > Hi James, > > James Edward Gray II wrote: >> You can always unroll recursive code into an iterative solution: > > Right. But some recursions are harder to unroll than others.^^ You're not the first person to say that to me, but I still don't believe it. ;) Think of it this way. All recursion is doing for you is managing the call stack. You can always do that yourself. Just use an Array for the stack and put all the local variable in it. That's always works for unrolling recursion. Period. Frequently you can get away with much less too. In my example earlier in this thread, note that I didn't even need all the local values to unroll it. It just takes practice to think this way. James Edward Gray II

on 2007-01-23 20:35

James Edward Gray II wrote: > believe it. ;) > > It just takes practice to think this way. I'll take your words. Thank you very much for your advice.:-) > > James Edward Gray II Sam

on 2007-01-23 20:51

Hi Thomas, Thomas Hafner wrote: > if s < 1 > right = parts(r) > end > > def part(n) > parts(n).map{|x| x.flatten} > end > > pp part(8) > #/// Yes. That's much better than the code I posted. Actually I've already tried that. But the problem is that even the cached version is not fast enough for my purpose. James's non-recursive code is not fast enough either. Probably, the solution to my problem is not just algorithm. I need to find a mathematical formula. See http://home.att.net/~numericana/data/partition.htm What I want is not the array of partitions but only the size of partitions. For example, the number of different ways to make 1000 is 24061467864032622473692149727991. It's almost impossible to keep an array of that size. I'm satisfied with this thread of talks because it taught me some even if I couldn't find the answer. Thanks. Sam

on 2007-01-23 21:07

On Jan 23, 2007, at 2:50 PM, Sam Kong wrote: > Yes. That's much better than the code I posted. > partitions. > For example, the number of different ways to make 1000 is > 24061467864032622473692149727991. > It's almost impossible to keep an array of that size. > > I'm satisfied with this thread of talks because it taught me some even > if I couldn't find the answer. > > Thanks. > > Sam These are Catalan numbers which you can learn more about on Wikipedia [1] or MathWorld[2]. If you just need to know the number of solutions, both references give a formula. -Rob [1] http://en.wikipedia.org/wiki/Catalan_number [2] http://mathworld.wolfram.com/CatalanNumber.html Rob Biedenharn http://agileconsultingllc.com Rob@AgileConsultingLLC.com

on 2007-01-23 22:23

Hi Rob, Rob Biedenharn wrote: > These are Catalan numbers which you can learn more about on Wikipedia > [1] or MathWorld[2]. If you just need to know the number of > solutions, both references give a formula. > > -Rob > > [1] http://en.wikipedia.org/wiki/Catalan_number > [2] http://mathworld.wolfram.com/CatalanNumber.html Thank you for the information. That's almost what I was looking for. Catalan numbers are for combinatorics not for partitions in that the order of numbers is significant in combinatorics. Anyway, the documents will give me hints. And at least, now I know that Mathematica provides the function. I can install Mathematica if I need to. I'm trying to solve math problems in projecteuler.net . They are very helpful when you want to apply your coding skills to solving problems. (Similar to ruby quiz but more math-oriented.) Some problems require math knowledge, though. (Brute force won't help.) It's fun. I hope more Ruby programmers will join it and challenge. It shows staticstics which compares languages. There are not many ruby users there yet. Thanks again. Sam

on 2007-01-24 10:00

On 1/24/07, Sam Kong <sam.s.kong@gmail.com> wrote: > > Thank you for the information. > That's almost what I was looking for. > Catalan numbers are for combinatorics not for partitions in that the > order of numbers is significant in combinatorics. You want the Stirling Numbers of the second kind, I think http://en.wikipedia.org/wiki/Stirling_number martin

on 2007-01-24 11:18

Hi Sam, You might also find http://www.ics.uci.edu/~eppstein/PADS/IntegerPartitions.py interesting. Regards, Sean

on 2007-01-24 12:45

"Sam Kong" <sam.s.kong@gmail.com> wrote/schrieb <1169581542.741753.78250@13g2000cwe.googlegroups.com>: > Yes. That's much better than the code I posted. Unfortunately my ``solution'' of <dfvh84-viu.ln1@faun.hafner.nl.eu.org> is none, because it does not report all combinations, sorry. Here's a better approach: #\\\ $cache = {} def parts(s,u) u0 = [s,u].min if u0 < 1 [] else i = (s-1)*s/2+u0-1 $cache[i] || begin a = [] u0.downto(1) do |n| r = s - n if (r > 0) parts(r,n).each do |elem| a.push([n, elem]) end else a.push([n]) end end $cache[i] = a a end end end def part(n) parts(n,n).map{|x| x.flatten} end #/// > Actually I've already tried that. > But the problem is that even the cached version is not fast enough for > my purpose. On my PC evaluation of parts(45) takes 11.53 seconds without cache, and 2.96 seconds with cache. But it's definitely not appropriate for calculating parts(1000). But you've already mentioned, that you need only the number of combinations, not the combinations itself. Many years ago at University I've been told that it's often better to solve the problem directly. Solving an arbitrary intermediate problem can make things worse or even impossible. Here the intermediate problem is ``calculate the list of combinations''. Some others are (numerics): - Don't calculate the inverse of a matrix. The original problem is probably a linear system of equations. Just solve that. - Never calculate the characteristic polynomial just to find the roots. Solve the Eigenvalue-problem directly. Regards Thomas

on 2007-01-24 17:20

On Jan 24, 1:00 am, "Martin DeMello" <martindeme...@gmail.com> wrote: > On 1/24/07, Sam Kong <sam.s.k...@gmail.com> wrote: > > > > > Thank you for the information. > > That's almost what I was looking for. > > Catalan numbers are for combinatorics not for partitions in that the > > order of numbers is significant in combinatorics.You want the Stirling Numbers of the second kind, I think > > http://en.wikipedia.org/wiki/Stirling_number Wow, finally I've got what I've been looking for. Thanks. Sam

on 2007-01-24 17:24

On Jan 24, 3:45 am, Thomas Hafner <tho...@hafner.NL.EU.ORG> wrote: > u0 = [s,u].min > parts(r,n).each do |elem| > end > calculating parts(1000). But you've already mentioned, that you need > roots. Solve the Eigenvalue-problem directly. Yes, you said it right. I didn't know that the combination grows that fast. Now I know what's right approach to the solution. Thanks. Sam