# I need a more efficient algorithm for this problem

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

Sam K. 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§

If God had meant for carrots to be eaten cooked, He would have given
rabbits fire.

On 1/22/07, Sam K. [email protected] 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

On Jan 22, 2007, at 11:10 PM, Sam K. 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 }

# >> [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 }

# >> [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/2006/01/31/iterators-
chapters-4-and-5

Hope that helps.

James Edward G. II

“Sam K.” [email protected] wrote/schrieb
[email protected]:

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®
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

Hi James,

James Edward G. II wrote:

4 + 1
block.call([largest] + rest)

# >> [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|
``````

# >> [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/2006/01/31/iterators-
chapters-4-and-5

I just bookmarked it.

Hope that helps.

Yes. It helped a lot.

James Edward G. II

Thank you very much.
Also, I thank others who replied to my question.

Sam

On Jan 23, 2007, at 11:40 AM, Sam K. wrote:

Hi James,

James Edward G. 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 G. II

James Edward G. II wrote:

believe it. It just takes practice to think this way.

Thank you very much for your advice. James Edward G. II

Sam

On Jan 23, 2007, at 2:50 PM, Sam K. 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

 or MathWorld. If you just need to know the number of
solutions, both references give a formula.

-Rob

Hi Rob,

Rob B. wrote:

 or MathWorld. If you just need to know the number of
solutions, both references give a formula.

-Rob

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

Hi Thomas,

Thomas H. wrote:

if s < 1
right = parts®
end

def part(n)
parts(n).map{|x| x.flatten}
end

pp part(8)
#///

Yes. That’s much better than the code I posted.
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 1/24/07, Sam K. [email protected] 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

Hi Sam,

You might also find
interesting.

Regards,
Sean

“Sam K.” [email protected] wrote/schrieb
[email protected]:

Yes. That’s much better than the code I posted.

Unfortunately my ``solution’’ of
[email protected] 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
#///

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 Jan 24, 1:00 am, “Martin DeMello” [email protected] wrote:

On 1/24/07, Sam K. [email protected] 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 Jan 24, 3:45 am, Thomas H. [email protected] 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