# Combinatorics and permutations? Help?

Hi,
I have a problem involving combinatorics and permutations and my brain
hurts.

The problem: I have a hash of products in input. I then build an array
of equivalent products for each input product. I end up with an array
such as this:
{
:input_prod_1 => [:p1_1, :p1_2, … , p1_x],
:input_prod_2 => [:p2_1, :p2_1, …, p2_y],

:input_prod_m => [:pm_1, :pm_1, …, pm_z]
}

I need all the unique combinations of each product from the first group
with each of the second. For example:
given: [ [:p1, :p2], [:p3, p4] ]
I need to obtain: [ [:p1, :p3], [:p1, :p4], [:p2, :p3], [:p2, :p4] ]

Given: [ [:p1], [:p2, :p3, :p4] ]
I need: [ [:p1, :p2], [:p1, :p3], [:p1, :p4] ]

The number of input products is in the order of tens (max); the same
goes for the number of equivalent products. This means the total number
of partitions is pretty large and I need to weed out the equivalent ones
soon/fast enough to reduce the total number.

I’ve been making attempts with the permutation gem all morning; looks
like it’s doing the “right thing”, but frankly my maths skills are
lacking and I can’t really figure out how to use it (should I?).

Also: order is not important for the result arrays.

This is easy, right? :-/

On Mar 11, 5:06 am, David P. [email protected] wrote:

I need all the unique combinations of each product from the first group with each of the second. For example:
given: [ [:p1, :p2], [:p3, p4] ]
I need to obtain: [ [:p1, :p3], [:p1, :p4], [:p2, :p3], [:p2, :p4] ]

Given: [ [:p1], [:p2, :p3, :p4] ]
I need: [ [:p1, :p2], [:p1, :p3], [:p1, :p4] ]

p [[1,2],[3,4]].inject([[]]){|old,lst|
lst.inject([]){|new,e| new + old.map{|c| c.dup << e}}}

On Tue, 11 Mar 2008 20:40:11 +0900, William J. wrote:

p [[1,2],[3,4]].inject([[]]){|old,lst|
lst.inject([]){|new,e| new + old.map{|c| c.dup << e}}}

awesome. Totally rocks.

given: [{“a” => [1, 2]}, {“b” => [3, 4]}]

I need: [{“a” => 1, “b” => 3}, {“a” => 1, “b” => 4}, {“a” => 2, “b” =>
3}, {“a” => 2, “b” => 4}]

Any ideas?

Yup, the following method does nearly the same thing, except it takes
arrays. You could either use this method for inspiration, or convert
your
desired input into my array of arrays input format and then convert the
output back to your desired format. Hope that helps.

def selections(arrays, results=[Array.new])
return results if arrays.empty?

new_selection_choices = arrays.shift
results_with_new_selections = []

results.each do |prev_selection_set|
new_selection_choices.each do |selection|
results_with_new_selections <<
prev_selection_set.clone.push(selection)
end
end

return selections(arrays, results_with_new_selections)
end

selections([[1, 2], [3, 4]]) #=> [[1, 3], [1, 4], [2, 3], [2, 4]]

I have a similar challenge:

given: [{“a” => [1, 2]}, {“b” => [3, 4]}]

I need: [{“a” => 1, “b” => 3}, {“a” => 1, “b” => 4}, {“a” => 2, “b” =>
3}, {“a” => 2, “b” => 4}]

Any ideas?

Thanks.
S

Sorry took so long to respond. Your solution worked as expected. Thanks!

SB

Josh C. wrote:

On Sun, Feb 7, 2010 at 9:29 PM, Sergio B.
[email protected]wrote:

## S

Posted via http://www.ruby-forum.com/.

Here is a solution I have come up with (http://gist.github.com/297937).
I
don’t know how you want all edge cases to be handled, I made my best
guesses, you can see them in the tests.

On Sun, Feb 7, 2010 at 9:29 PM, Sergio B.
[email protected]wrote:

## S

Posted via http://www.ruby-forum.com/.

Here is a solution I have come up with (http://gist.github.com/297937).
I
don’t know how you want all edge cases to be handled, I made my best
guesses, you can see them in the tests.