I'm trying to build a pair generator. It takes a list of six names, and generates pairs for the week (5 days), with as few replications as possible.
The minimum I've gotten my replicated pairs down to is 2 (so I have found 5 days of pairs, i.e. 15 total pair combinations, with only 2 identical sets).
My method:
# Start with individuals in an array
[1, 2, 3, 4, 5, 6]
# Bisect the array
[1, 2, 3]
[4, 5, 6] => yields pair combinations [1, 4], [2, 5], [3, 6]
# Move the lower of the bisected arrays along
[1, 2, 3]
[6, 4, 5] => yields pair combinations [1, 6], [2, 4], [3, 5]
# Move along once more
[1, 2, 3]
[5, 6, 4] => yields pair combinations [1, 5], [2, 6], [3, 4]
# Since there are no more unique pair combinations, bisect each array again
(Array 1) [1, 2]
(Array 1) [3] => yields pair combination [1, 3] with 2 'spare'
(Array 2) [4, 5]
(Array 2) [6] => yields pair combination [4, 6] with 6 'spare'
=> 'spare' pair combination [2, 6] is a replication
# Move the lower of the bisected arrays along
(Array 1) [1, 2]
(Array 1) [3] => yields pair combination [2, 3] with 1 'spare'
(Array 2) [4, 5]
(Array 2) [6] => yields pair combination [5, 6] with 4 'spare'
=> 'spare' pair combination [1, 4] is a replication
This process above gives us 13 unique pairs, and then 2 that are non-unique. Every day of the week is covered, but we replicate.
Is there any way to do this more efficiently/to avoid the replication?
This is a round robin tournament where every player plays every other player. Line up the players like below to form the pairs 1 4, 2 5 and 3 6:
123
456
fix player 1, rotate the remaining players:
142
563
to produce pairs 1 5, 4 6 and 2 3. Keep rotating:
154
632
165
324
136
245
I think you're simply after the built-in combination method which returns an enumerator. You can use .to_a to turn it into an array of unique combinations.
[1, 2, 3, 4, 5, 6].combination(2).to_a
# => [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5], [2, 6], [3, 4], [3, 5], [3, 6], [4, 5], [4, 6], [5, 6]]
This is called a 1-factorization. One 1-factorization of the complete graph on 6 vertices {0,1,2,3,4,oo} is to let the schedule on day i be {{oo,i},{i+1,i+4},{i+2,i+3}} where all of the numbers i+j are reduced mod 5.
Related
The input is a length n array, containing distinct integers from 1 to n.
We will perform the following algorithm. There are no choices to be made in the algorithm: the only goal is counting the number of reversals this algorithm will perform.
Algorithm is like below:
While(array not sorted in ascending order) do:
Find all the maximal contiguous subarrays which are in decreasing order,
and reverse each of them (simultaneously).
We want to find the total number of times reverse will be called in the above algorithm, in an optimal way (i.e. as efficiently as possible, so direct simulation is probably too slow).
Example 1:
4,3,1,2 // 1 reversal: [4, 3, 1]
1,3,4,2 // 1 reversal: [4, 2]
1,3,2,4 // 1 reversal: [3, 2]
1,2,3,4
// Reverse is called 3 times in above example
Example 2:
5 3 4 2 1 ---> 2 reversals: [5, 3], [4, 2, 1]
3 5 1 2 4 ---> 1 reversal: [5, 1]
3 1 5 2 4 ---> 2 reversals: [3, 1], [5, 2]
1 3 2 5 4 ---> 2 reversals: [3, 2], [5, 4]
1 2 3 4 5 ---> sorted
Total of 7 reversals
Note that O(n^2) reversals may be necessary to sort a sequence in this way, so a direct simulation can take that many steps to run, but there may be a way to count them faster than O(n^2).
If n = 2k, then the sequence k+1, k+2, ... , 2k, 1, 2, 3, ..., k will require k^2 reversals:
Reversals to be performed are surrounded by brackets
5, 6, 7, [8, 1], 2, 3, 4 // 1 reversal
5, 6, [7, 1], [8, 2], 3, 4 // 2 reversals
5, [6, 1], [7, 2], [8, 3], 4 // 3 reversals
[5, 1], [6, 2], [7, 3], [8, 4] // 4 reversals
1, [5, 2], [6, 3], [7, 4], 8 // 3 reversals
1, 2, [5, 3], [6, 4], 7, 8 // 2 reversals
1, 2, 3, [5, 4], 6, 7, 8 // 1 reversal
1, 2, 3, 4, 5, 6, 7, 8
Total of 16 = (8/2)^2 reversals.
Things which I tried:
Trying to convert it to recursion.
Using stack to solve it.
Read about inversion but not able to solve this.
Read on codeforces about the thread but was not relevant to this post.
Are there any algorithm that solves the following problem in time less than O(n!), like polynomial time?
Otherwise, for this problem, does not anyone have found any polynomial time algorithm, like NP problems?
Input: n (number of elements)
Output: a list of all combinations of two, where, from top of the list, each unit of combinations of n/2 must have all elements.
Example 1
Input: n=4
Output:
[0, 1], [2, 3],
[0, 2], [1, 3],
[0, 3], [1, 2]
Example 2
Input: n=8
Output:
[0, 1], [2, 3], [4, 5], [6, 7],
[0, 2], [1, 3], [4, 6], [5, 7],
[0, 3], [1, 2], [4, 7], [5, 6],
[0, 4], [1, 5], [2, 6], [3, 7],
[0, 5], [1, 4], [2, 7], [3, 6],
[0, 6], [1, 7], [2, 4], [3, 5],
[0, 7], [1, 6], [2, 5], [3, 4]
P.S.
The following answer does not meet the requirements.
The first two (= n/2) pairs ([0, 1], [0, 2]) do not have "3", so the answer does not meet the condition where "0" and "1", "2", "3" must be in the first two pairs.
>>> n=4
>>> for i in range(0, n-1):
... for j in range(i+1,n):
... print( [i, j] )
...
[0, 1]
[0, 2]
[0, 3]
[1, 2]
[1, 3]
[2, 3]
As I said in my comment, this appears to be a type of (relaxed) Sports League Scheduling problem. If I understand what you are asking for, it can be summarized as follows:
Given a positive even integer N generate a set of n/2 "rounds" of pairings with the following qualities:
A pairing is a pair of two different integers [a, b] such that a and b are integers from 0..n-1 and a < b.
A round consists of n/2 pairings, such that every element from 0..n-1 appears exactly once in a pairing in the round, and
All pairings are unique across all rounds (that is no pairing ever appears more than once in the complete solution).
Assuming that this is a correct formulation of your problem, then the answer is
Yes, this can be done in O(n^2).
Further, not only can it be done, there exists a simple method to solve it for any even N:
For the first round, make n-1 pairs, filling in the first element of the pairs with the integers from 0 to (n/2)-1 going left-to-right. This how it would look for N=8:
[0, ], [1, ], [2, ], [3, ]
Then, fill in the second elements with (n/2) to n-1, but going right-to-left:
[0, 7], [1, 6], [2, 5], [3, 4]
This completes your first round.
For the next round, copy the first round, but keeping 0 in the same place, move the remaining left-side elements up the list, and the right-side elements down the list. When an element reaches the end of the list, reverse direction and swap them from first elements to second elements (or vice-versa):
----------------------->
[0, 7], [1, 6], [2, 5], [3, 4]
<-----------------------
Becomes
----------------------->
[0, 6], [7, 5], [1, 4], [2, 3]
<-----------------------
Now you just continue this process until you have N/2 rounds:
[0, 7], [1, 6], [2, 5], [3, 4]
[0, 6], [7, 5], [1, 4], [2, 3]
[0, 5], [6, 4], [7, 3], [1, 2]
[0, 4], [5, 3], [6, 2], [7, 1]
Finally swap any pairings where the first element happens to be greater than the second:
[0, 7], [1, 6], [2, 5], [3, 4]
[0, 6], [5, 7], [1, 4], [2, 3]
[0, 5], [4, 6], [3, 7], [1, 2]
[0, 4], [3, 5], [2, 6], [1, 7]
If you check this solution you will find that it fulfills all of the constraints. This solutions works for any even value of N and obviously runs in O(n^2) time.
Yes, this problem can be solved in quadratic time. It is not too hard to explicitly construct these pairings.
It is quite helpful to consider a regular (n-1)-gon with one additional point in the middle. Then take the lines through one of the (n-1) endpoints and the midpoint and choose the pairs given by the symmetry of this line.
I have a range of numbers R = (1..n). I also have another character 'a'. I want to generate strings of length L (L > n + 2) that have all the numbers in the same order, but go through every repeated permutation of 'a' to fill the length L. For example, if n = 3, and L = 7, then some valid strings would be :
"123aaaa",
"1a23aaa",
"1aa2a3a",
"aaaa123"
while the following strings would be invalid:
"213aaaa", # invalid, because 1,2,3 are not in order
"123a", #invalid, because length < L
"1123aaa", # invalid because a number is repeated
I am currently doing this, which is way too inefficient:
n = 3
L = 7
all_terms = (1..n).to_a + Array.new(L - n, 'a')
all_terms.permutation.each do |permut|
if(valid_permut? permut) # checks if numbers are in their natural order
puts permut.join
end
end
How do I directly generate valid strings more efficiently?
The problem is equivalent to: select n elements from index 0 to L - 1, fill these with 1 to n accordingly, and fill the rest with some constant character.
In your example, it's taking 3 elements from 0..6:
(0..6).to_a.combination(3).to_a
=> [[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 1, 5], [0, 1, 6], [0, 2, 3], [0, 2, 4],
[0, 2, 5], [0, 2, 6], [0, 3, 4], [0, 3, 5], [0, 3, 6], [0, 4, 5], [0, 4, 6], [0, 5, 6],
[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5], [1, 3, 6], [1, 4, 5],
[1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6],
[3, 4, 5], [3, 4, 6], [3, 5, 6], [4, 5, 6]]
Every subarray here represents a possible result. For example, [0, 2, 3] corresponds to '0a12aaa', [3, 5, 6] corresponds to 'aaa0a12', etc. The code for this conversion is straight-forward.
You can model this as all possible interleavings of two strings, where relative order of the input elements is preserved. Here's a recursive solution. It works by choosing an element from one list, and prepending it to all possible subproblems, then doing it again where an element is chosen from the second list instead, and combining the two solution sets at the end.
# Returns an array of all possible interleaving of two strings
# Maintains relative order of each character of the input strings
def interleave_strings_all(a1, a2)
# Handle base case where at least one input string is empty
return [a1 + a2] if a1.empty? || a2.empty?
# Place element of first string, and prepend to all subproblems
set1 = interleave_strings_all(a1[1..-1], a2).map{|x| a1[0] + x}
# Place element of second string and prepend to all subproblems
set2 = interleave_strings_all(a1, a2[1..-1]).map{|x| a2[0] + x}
# Combine solutions of subproblems into overall problem
return set1.concat(set2)
end
if __FILE__ == $0 then
l = 5
n = 3
a1 = (1..n).to_a.map{|x| x.to_s}.join()
a2 = 'a' * (l - n)
puts interleave_strings_all(a1, a2)
end
The output is:
123aa
12a3a
12aa3
1a23a
1a2a3
1aa23
a123a
a12a3
a1a23
aa123
I have this array
a = [[1,2,3,4,5],[2,3,6],[2,8,9]]
How can I get the intersection of all the arrays in a? It should be the same as a[0]&a[1]&a[2], which is:
[2]
a = [[1 ,2, 3, 4, 5], [2, 3, 6], [2, 8, 9]]
a.inject(:&)
# => [2]
How can I go from this:
for number in [1,2] do
puts 1+number
puts 2+number
puts 3+number
end
which will return 2,3,4 then 3,4,5 -> 2,3,4,3,4,5. This is just an example, and clearly not the real use.
Instead, I would like it to return 2,3 3,4 4,5 -> 2,3,3,4,4,5. I would like each of the puts to be iterated for each of the possible values of number; In this case 1 and 2 are the two possible values of 'number', before moving on to the next puts.
One way to do this is to create two lists, [2,3,4] and [3,4,5] and then use the zip method to combine them like [2,3,4].zip([3,4,5]) -> [2,3,3,4,4,5].
zip is good. You should also look at each_cons:
1.9.2p290 :006 > [2,3,4].each_cons(2).to_a
=> [[2, 3], [3, 4]]
1.9.2p290 :007 > [2,3,4,5,6].each_cons(2).to_a
=> [[2, 3], [3, 4], [4, 5], [5, 6]]
1.9.2p290 :008 > [2,3,4,5,6].each_cons(3).to_a
=> [[2, 3, 4], [3, 4, 5], [4, 5, 6]]
Because each_cons returns an Enumerator, you can use a block with it, as mentioned in the documentation for it, or convert it to an array using to_a like I did above. That returns the array of arrays, which can be flattened to get a single array:
[2,3,4,5].each_cons(2).to_a.flatten
=> [2, 3, 3, 4, 4, 5]
From the ri docs:
Iterates the given block for each array of consecutive elements. If no
block is given, returns an enumerator.
e.g.:
(1..10).each_cons(3) {|a| p a}
# outputs below
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 10]
Maybe not the most readable code but you could use inject on the first range to create an array based on the summed up second range.
(1..3).inject([]){|m,n| (1..2).each{|i| m<<n+i }; m }
=> [2, 3, 3, 4, 4, 5]
This might be a little more readable
res=[]
(1..3).each{|r1| (1..2).each{|r2| res<<r1+r2 } }
[1, 2, 3].each { |i| [1, 2].each { |y| puts i + y } }