Given a list L of an even number (2k) of elements, I'm looking for an algorithm to produce a list of 2k-1 sublists with the following properties:
each sublist includes exactly k 2-combinations (pairs where the order does not matter) of elements from L,
each sublist includes every elements from L exactly once, and
the union of all elements from all sublists is exactly the set of all possible 2-combinations of the elements from L.
For example, if the input list is L = [a, b, c, d], we have k = 2 with 3 sublists, each including 2 pairs. A possible solution would look like [[ab, cd], [ac, bd], [ad, bc]]. If we ignore the ordering for all elements in the lists (think of all lists as sets), it turns out that this is also the only solution for k = 2.
My aim now is not only to find a single solution but all possible solutions. As the number of involved combinations grows pretty quickly, it would be nice to have all results be constructed in a clever way instead of generating a huge list of candidates and removing the elements from it that don't satisfy the given properties. Such a naïve algorithm could look like the following:
Find the set C of all 2-combinations for L.
Find the set D of all k-combinations for C.
Choose all sets from D that union equals L, call the new set D'.
Find the set E of all (2k-1)-combinations for D'.
Choose all sets from E that union is the set C, and let the new set be the final output.
This algorithm is easy to implement but it's incredibly slow for bigger input lists. So is there a way to construct the result list more efficently?
Edit: Here is the result for L = [a,b,c,d,e,f] with k = 3, calculated by the above algorithm:
[[[ab,cd,ef],[ac,be,df],[ad,bf,ce],[ae,bd,cf],[af,bc,de]],
[[ab,cd,ef],[ac,bf,de],[ad,be,cf],[ae,bc,df],[af,bd,ce]],
[[ab,ce,df],[ac,bd,ef],[ad,be,cf],[ae,bf,cd],[af,bc,de]],
[[ab,ce,df],[ac,bf,de],[ad,bc,ef],[ae,bd,cf],[af,be,cd]],
[[ab,cf,de],[ac,bd,ef],[ad,bf,ce],[ae,bc,df],[af,be,cd]],
[[ab,cf,de],[ac,be,df],[ad,bc,ef],[ae,bf,cd],[af,bd,ce]]]
All properties are satisfied:
each sublist has k = 3 2-combinations,
each sublist only includes each element once, and
the union of all 2k-1 = 5 sublists for one solution is exactly the set of all possible 2-combinations for L.
Edit 2: Based on user58697's answer, I improved the calculation algorithm by using the round-robin tournament scheduling:
Let S be the result set, starting with an empty set, and P be the set of all permutations of L.
Repeat the following until P is empty:
Select an arbitrary permutation from P
Perform full RRT scheduling for this permutation. In each round, the arrangement of elements from L forms a permutation of L. Remove all these 2k permutations from P.
Add the resulting schedule to S.
Remove all lists from S if the union of their sublists has duplicate elements (i.e. doesn't add up to all 2-combinations of L).
This algorithm is much more performant than the first one. I was able to calculate the number of results for k = 4 as 960 and k = 5 as 67200. The fact that there doesn't seem to be an OEIS result for this sequence makes me wonder if the numbers are actually correct, though, i.e. if the algorithm is producing the complete solution set.
It is a round-robin tournament scheduling:
A pair is a match,
A list is a round (each team plays with some other team)
A set of list is an entire tournament (each team plays each other team exactly once).
Take a look here.
This was an interesting question. In the process of answering it (basically after writing the program included below, and looking up the sequence on OEIS), I learned that the problem has a name and rich theory: what you want is to generate all 1-factorizations of the complete graph K2k.
Let's first restate the problem in that language:
You are given a number k, and a list (set) L of size 2k. We can view L as the vertex set of a complete graph K2k.
For example, with k=3, L could be {a, b, c, d, e, f}
A 1-factor (aka perfect matching) is a partition of L into unordered pairs (sets of size 2). That is, it is a set of k pairs, whose disjoint union is L.
For example, ab-cd-ef is a 1-factor of L = {a, b, c, d, e, f}. This means that a is matched with b, c is matched with d, and e is matched with f. This way, L has been partitioned into three sets {a, b}, {c, d}, and {e, f}, whose union is L.
Let S (called C in the question) denote the set of all pairs of elements of L. (In terms of the complete graph, if L is its vertex set, S is its edge set.) Note that S contains (2k choose 2) = k(2k-1) pairs. So for k = 0, 1, 2, 3, 4, 5, 6…, S has size 0, 1, 6, 15, 28, 45, 66….
For example, S = {ab, ac, ad, ae, af, bc, bd, be, bf, cd, ce, cf, de, df, ef} for our L above (k = 3, so |S| = k(2k-1) = 15).
A 1-factorization is a partition of S into sets, each of which is itself a 1-factor (perfect matching). Note that as each of these matchings has k pairs, and S has size k(2k-1), the partition has size 2k-1 (i.e., is made of 2k-1 matchings).
For example, this is a 1-factorization: {ab-cd-ef, ac-be-df, ad-bf-ce, ae-bd-cf, af-bc-de}
In other words, every element of S (every pair) occurs in exactly one element of the 1-factorization, and every element of L occurs exactly once in each element of the 1-factorization.
The problem asks to generate all 1-factorizations.
Let M denote the set of all 1-factors (all perfect matchings) of L. It is easy to prove that M contains (2k)!/(k!2^k) = 1×3×5×…×(2k-1) matchings. For k = 0, 1, 2, 3, 4, 5, 6…, the size of M is 1, 1, 3, 15, 105, 945, 10395….
For example, for our L above, M = {ab-cd-ef, ab-ce-df, ab-cf-de, ac-bd-ef, ac-be-df, ac-bf-de, ad-bc-ef, ad-be-cf, ad-bf-ce, ae-bc-df, ae-bd-cf, ae-bf-cd, af-bc-de, af-bd-ce, af-be-cd} (For k=3 this number 15 is the same as the number of pairs, but this is just a coincidence as you can from the other numbers: this number grows much faster than the number of pairs.)
M is easy to generate:
def perfect_matchings(l):
if len(l) == 0:
yield []
for i in range(1, len(l)):
first_pair = l[0] + l[i]
for matching in perfect_matchings(l[1:i] + l[i+1:]):
yield [first_pair] + matching
For example, calling perfect_matchings('abcdef') yields the 15 elements ['ab', 'cd', 'ef'], ['ab', 'ce', 'df'], ['ab', 'cf', 'de'], ['ac', 'bd', 'ef'], ['ac', 'be', 'df'], ['ac', 'bf', 'de'], ['ad', 'bc', 'ef'], ['ad', 'be', 'cf'], ['ad', 'bf', 'ce'], ['ae', 'bc', 'df'], ['ae', 'bd', 'cf'], ['ae', 'bf', 'cd'], ['af', 'bc', 'de'], ['af', 'bd', 'ce'], ['af', 'be', 'cd'] as expected.
By definition, a 1-factorization is a partition of S into elements from M. Or equivalently, any (2k-1) disjoint elements of M form a 1-factorization. This lends itself to a straightforward backtracking algorithm:
start with an empty list (partial factorization)
for each matching from the list of perfect matchings, try adding it to the current partial factorization, i.e. check whether it's disjoint (it should not contain any pair already used)
if fine, add it to the partial factorization, and try extending
In code:
matching_list = []
pair_used = defaultdict(lambda: False)
known_matchings = [] # Populate this list using perfect_matchings()
def extend_matching_list(r, need):
"""Finds ways of extending the matching list by `need`, using matchings r onwards."""
if need == 0:
use_result(matching_list)
return
for i in range(r, len(known_matchings)):
matching = known_matchings[i]
conflict = any(pair_used[pair] for pair in matching)
if conflict:
continue # Can't use this matching. Some of its pairs have already appeared.
# Else, use this matching in the current matching list.
for pair in matching:
pair_used[pair] = True
matching_list.append(matching)
extend_matching_list(i + 1, need - 1)
matching_list.pop()
for pair in matching:
pair_used[pair] = False
If you call it with extend_matching_list(0, len(l) - 1) (after populating known_matchings), it generates all 1-factorizations. I've put the full program that does this here. With k=4 (specifically, the list 'abcdefgh'), it outputs 6240 1-factorizations; the full output is here.
It was at this point that I fed the sequence 1, 6, 6240 into OEIS, and discovered OEIS A000438, sequence 1, 1, 6, 6240, 1225566720, 252282619805368320,…. It shows that for k=6, the number of solutions ≈2.5×1017 means that we can give up hope of generating all solutions. Even for k=5, the ≈1 billion solutions (recall that we're trying to find 2k-1=9 disjoint sets out of the |M|=945 matchings) will require some carefully optimized programs.
The first optimization (which, embarrassingly, I only realized later by looking closely at trace output for k=4) is that (under natural lexicographic numbering) the index of the first matching chosen in the partition cannot be greater than the number of matchings for k-1. This is because the lexicographically first element of S (like "ab") occurs only in those matchings, and if we start later than this one we'll never find it again in any other matching.
The second optimization comes from the fact that the bottleneck of a backtracking program is usually the testing for whether a current candidate is admissible. We need to test disjointness efficiently: whether a given matching (in our partial factorization) is disjoint with the union of all previous matchings. (Whether any of its k pairs is one of the pairs already covered by earlier matchings.) For k=5, it turns out that the size of S, which is (2k choose 2) = 45, is less than 64, so we can compactly represent a matching (which is after all a subset of S) in a 64-bit integer type: if we number the pairs as 0 to 44, then any matching can be represented by an integer having 1s in the positions corresponding to elements it contains. Then testing for disjointness is a simple bitwise operation on integers: we just check whether the bitwise-AND of the current candidate matching and the cumulative union (bitwise-OR) of previous matchings in our partial factorization is zero.
A C++ program that does this is here, and just the backtracking part (specialized for k=5) does not need any C++ features so it's extracted out as a C program here. It runs in about 4–5 hours on my laptop, and finds all 1225566720 1-factorizations.
Another way to look at this problem is to say that two elements of M have an edge between them if they intersect (have a pair (element of S) in common), and that we're looking for all maximum independent set in M. Again, the simplest way to solve that problem would still probably be backtracking (we'd write the same program).
Our programs can be made quite a lot more efficient by exploiting the symmetry in our problem: for example we could pick any matching as our first 1-factor in the 1-factorization (and then generate the rest by relabelling, being careful not to avoid duplicates). This is how the number of 1-factorizations for K12 (the current record) was calculated.
A note on the wisdom of generating all solutions
In The Art of Computer Programming Volume 4A, at the end of section 7.2.1.2 Generating All Permutations, Knuth has this important piece of advice:
Think twice before you permute. We have seen several attractive algorithms for permutation generation in this section, but many algorithms are known by which permutations that are optimum for particular purposes can be found without running through all possibilities. For example, […] the best way to arrange records on a sequential storage […] takes only O(n log n) steps. […] the assignment problem, which asks how to permute the columns of a square matrix so that the sum of the diagonal elements is maximized […] can be solved in at most O(n3) operations, so it would be foolish to use a method of order n! unless n is extremely small. Even in cases like the traveling salesrep problem, when no efficient algorithm is known, we can usually find a much better approach than to examine every possible solution. Permutation generation is best used when there is good reason to look at each permutation individually.
This is what seems to have happened here (from the comments below the question):
I wanted to calculate all solutions to run different attribute metrics on these and find an optional match […]. As the number of results seems to grow quicker than expected, this is impractical.
Generally, if you're trying to "generate all solutions" and you don't have a very good reason for looking at each one (and one almost never does), there are many other approaches that are preferable, ranging from directly trying to solve an optimization problem, to generating random solutions and looking at them, or generating solutions from some subset (which is what you seem to have done).
Further reading
Following up references from OEIS led to a rich history and theory.
On 1-factorizations of the complete graph and the relationship to round robin schedules, Gelling (M. A. Thesis), 1973
On the number of 1-factorizations of the complete graph, Charles C Lindner, Eric Mendelsohn, Alexander Rosa (1974?) -- this shows that the number of nonisomorphic 1-factorizations on K2n goes to infinity as n goes to infinity.
E. Mendelsohn and A. Rosa. On some properties of 1-factorizations of complete graphs. Congr. Numer, 24 (1979): 739–752
E. Mendelsohn and A. Rosa. One factorizations of the complete graph: A survey. Journal of Graph Theory, 9 (1985): 43–65 (As long ago as 1985, this exact question was studied well-enough to need a survey!)
Via papers of Dinitiz:
D. K. Garnick and J. H. Dinitz, On the number of one-factorizations of the complete graph on 12 points, Congressus Numerantium, 94 (1993), pp. 159-168. They announced they were computing the number of nonisomorphic 1-factorizations of K12. Their algorithm was basically backtracking.
Jeffrey H. Dinitz, David K. Garnick, Brendan D. McKay: There are 526,915,620 nonisomorphic one-factorizations of K12 (also here), Journal of Combinatorial Designs 2 (1994), pp. 273 - 285: They completed the computation, and reported the numbers they found for K12 (526,915,620 nonisomorphic, 252,282,619,805,368,320 total).
Various One-Factorizations of Complete Graphs by Gopal, Kothapalli, Venkaiah, Subramanian (2007). A paper that is relevant to this question, and has many useful references.
W. D. Wallis, Introduction to Combinatorial Designs, Second Edition (2007). Chapter 10 is "One-Factorizations", Chapter 11 is "Applications of One-Factorizations". Both are very relevant and have many useful references.
Charles J. Colbourn and Jeffrey H. Dinitz, Handbook of Combinatorial Designs, Second Edition (2007). A goldmine. See chapters VI.3 Balanced Tournament Designs, VI.51 Scheduling a Tournament, VII.5 Factorizations of Graphs (including its sections 5.4 Enumeration and Tables, 5.5 Some 1-Factorizations of Complete Graphs), VII.6 Computational Methods in Design Theory (6.2 Exhaustive Search). This last chapter references:
[715] How K12 was calculated ("orderly algorithm"), a backtracking -- the Dinitz-Garnick-McKay paper mentioned above
[725] “Contains, among many other subjects related to factorization, a fast algorithm for finding 1-factorizations of K2n.” ("Room squares and related designs", J. H. Dinitz and S. R. Stinson)
[1270] (P. Kaski and P. R. J. Östergård, One-factorizations of regular graphs of order 12, Electron. J. Comb. 12, Research Paper 2, 25 pp. (2005))
[1271] “Contains the 1-factorizations of complete graphs up to order 10 in electronic form.” (P. Kaski and P. R. J. Östergård, Classification Algorithms for Codes and Designs, Springer, Berlin, 2006.)
[1860] “A survey on perfect 1-factorizations of K2n” (E. S. Seah, Perfect one-factorizations of the complete graph—A survey, Bull. Inst. Combin. Appl. 1 (1991) 59–70)
[2107] “A survey of 1-factorizations of complete graphs including most of the material of this chapter.” W. D. Wallis, One-factorizations of complete graphs, in Dinitz and Stinson (ed), Contemporary Design Theory, 1992
[2108] “A book on 1-factorizations of graphs.” W. D. Wallis, "One-Factorizations", Kluwer, Dordrecht, 1997
Some other stuff:
*Factors and Factorizations of Graphs by Jin Akiyama and Mikio Kano (2007). This looks like a great book. “Frank Harary predicted that graph theory will grow so much that each chapter of his book Graph Theory will eventually expand to become a book on its own. He was right. This book is an expansion of his Chapter 9, Factorization.” There's not much about this particular topic (1-factorizations of complete graphs), but there is a proof in Chapter 4 (Theorem 4.1.1) that K2n always has a 1-factorization.
Papers on special types of 1-factorizations:
[Symmetry Groups Of] Some Perfect 1-Factorizations Of Complete Graphs, B. A. Anderson, 1977 (1973). Considers 1-factorizations that are in fact "perfect", having the property that the union of any two 1-factors (matchings) is a Hamiltonian cycle. (There's one up to isomorphism for K2k k ≤ 5, and two for K12.)
On 4-semiregular 1-factorizations of complete graphs and complete bipartite graphs.
Low Density MDS Codes and Factors of Complete Graphs -- also about perfect 1-factorizations
Self-invariant 1-Factorizations of Complete Graphs and Finite Bol Loops of Exponent 2
See also OEIS index entry for [sequences related to tournaments].
AMS feature column: Mathematics and Sports (April 2010) -- despite the overly broad name, is quite related.
Suppose you have a bunch of sets, whereas each set has a couple of subsets.
Set1 = { (banana, pineapple, orange), (apple, kale, cucumber), (onion, garlic) }
Set2 = { (banana, cucumber, garlic), (avocado, tomato) }
...
SetN = { ... }
The goal now is to select one subset from each set, whereas each subset must be conflict free with any other selected subset. For this toy-size example, a possible solution would be to select (banana, pineapple, orange) (from Set1) and (avocado, tomato) (from Set2).
A conflict would occur, if one would select the first subset of Set1 and Set2 because the banana would be contained in both subsets (which is not possible because it exists only once).
Even though there are many algorithms, I was unable to select a suitable algorithm. I'm somehow stuck and would appreciate answers targeting the following questions:
1) How to find a suitable algorithm and represent this problem in such a way that it can be processed by the algorithm?
2) How a possible solution for this toy-size example may look like (any language is just fine, I just want to get the idea).
Edit1: I was thinking about simulated annealing, too (return one possible solution). This could be of interest to minimize, e.g., the overall cost of selecting the sets. However, I could not figure out how to make an appropriate problem description that takes the 'conflicts' into account.
This problem can be formulated as a generalized exact cover problem.
Create a new atom for each set of sets (Set1, Set2, etc.) and turn your input into an instance like so:
{Set1, banana, pineapple, orange}
{Set1, apple, kale, cucumber}
{Set1, onion, garlic}
{Set2, banana, cucumber, garlic}
{Set2, avocado, tomato}
...
making the Set* atoms primary (covered exactly once) and the other atoms secondary (covered at most once). Then you can solve it with a generalization of Knuth's Algorithm X.
Looking at the list of sets, I had the image of a maze with multiple entrances. The task is akin to tracing paths from top to bottom that are free of subset-intersections. The example in Haskell picks all entrances, and tries each path, returning those that succeed.
My understanding of how the code works (algorithm):
For each subset in the first set, pick each subset in the next set where the intersection of that subset with each of the subsets in the accumulated result is null. If there are no subsets matching the criteria, break that strain of the loop. If there are no sets left to pick from, return that result. Call the function recursively for all chosen subsets (and corresponding accumulating-results).
import Data.List (intersect)
import Control.Monad (guard)
sets = [[["banana", "pineapple", "orange"], ["apple", "kale", "cucumber"], ["onion", "garlic"]]
,[["banana", "cucumber", "garlic"], ["avocado", "tomato"]]]
solve sets = solve' sets [] where
solve' [] result = [result]
solve' (set:rest) result = do
subset <- set
guard (all null (map (intersect subset) result))
solve' rest (result ++ [subset])
OUTPUT:
*Main> solve sets
[[["banana","pineapple","orange"],["avocado","tomato"]]
,[["apple","kale","cucumber"],["avocado","tomato"]]
,[["onion","garlic"],["avocado","tomato"]]]
I have a symmetric matrix like shown in the image attached below.
I've made up the notation A.B which represents the value at grid point (A, B). Furthermore, writing A.B.C gives me the minimum grid point value like so: MIN((A,B), (A,C), (B,C)).
As another example A.B.D gives me MIN((A,B), (A,D), (B,D)).
My goal is to find the minimum values for ALL combinations of letters (not repeating) for one row at a time e.g for this example I need to find min values with respect to row A which are given by the calculations:
A.B = 6
A.C = 8
A.D = 4
A.B.C = MIN(6,8,6) = 6
A.B.D = MIN(6, 4, 4) = 4
A.C.D = MIN(8, 4, 2) = 2
A.B.C.D = MIN(6, 8, 4, 6, 4, 2) = 2
I realize that certain calculations can be reused which becomes increasingly important as the matrix size increases, but the problem is finding the most efficient way to implement this reuse.
Can point me in the right direction to finding an efficient algorithm/data structure I can use for this problem?
You'll want to think about the lattice of subsets of the letters, ordered by inclusion. Essentially, you have a value f(S) given for every subset S of size 2 (that is, every off-diagonal element of the matrix - the diagonal elements don't seem to occur in your problem), and the problem is to find, for each subset T of size greater than two, the minimum f(S) over all S of size 2 contained in T. (And then you're interested only in sets T that contain a certain element "A" - but we'll disregard that for the moment.)
First of all, note that if you have n letters, that this amounts to asking Omega(2^n) questions, roughly one for each subset. (Excluding the zero- and one-element subsets and those that don't include "A" saves you n + 1 sets and a factor of two, respectively, which is allowed for big Omega.) So if you want to store all these answers for even moderately large n, you'll need a lot of memory. If n is large in your applications, it might be best to store some collection of pre-computed data and do some computation whenever you need a particular data point; I haven't thought about what would work best, but for example computing data only for a binary tree contained in the lattice would not necessarily help you anything beyond precomputing nothing at all.
With these things out of the way, let's assume you actually want all the answers computed and stored in memory. You'll want to compute these "layer by layer", that is, starting with the three-element subsets (since the two-element subsets are already given by your matrix), then four-element, then five-element, etc. This way, for a given subset S, when we're computing f(S) we will already have computed all f(T) for T strictly contained in S. There are several ways that you can make use of this, but I think the easiest might be to use two such subset S: let t1 and t2 be two different elements of T that you may select however you like; let S be the subset of T that you get when you remove t1 and t2. Write S1 for S plus t1 and write S2 for S plus t2. Now every pair of letters contained in T is either fully contained in S1, or it is fully contained in S2, or it is {t1, t2}. Look up f(S1) and f(S2) in your previously computed values, then look up f({t1, t2}) directly in the matrix, and store f(T) = the minimum of these 3 numbers.
If you never select "A" for t1 or t2, then indeed you can compute everything you're interested in while not computing f for any sets T that don't contain "A". (This is possible because the steps outlined above are only interesting whenever T contains at least three elements.) Good! This leaves just one question - how to store the computed values f(T). What I would do is use a 2^(n-1)-sized array; represent each subset-of-your-alphabet-that-includes-"A" by the (n-1) bit number where the ith bit is 1 whenever the (i+1)th letter is in that set (so 0010110, which has bits 2, 4, and 5 set, represents the subset {"A", "C", "D", "F"} out of the alphabet "A" .. "H" - note I'm counting bits starting at 0 from the right, and letters starting at "A" = 0). This way, you can actually iterate through the sets in numerical order and don't need to think about how to iterate through all k-element subsets of an n-element set. (You do need to include a special case for when the set under consideration has 0 or 1 element, in which case you'll want to do nothing, or 2 elements, in which case you just copy the value from the matrix.)
Well, it looks simple to me, but perhaps I misunderstand the problem. I would do it like this:
let P be a pattern string in your notation X1.X2. ... .Xn, where Xi is a column in your matrix
first compute the array CS = [ (X1, X2), (X1, X3), ... (X1, Xn) ], which contains all combinations of X1 with every other element in the pattern; CS has n-1 elements, and you can easily build it in O(n)
now you must compute min (CS), i.e. finding the minimum value of the matrix elements corresponding to the combinations in CS; again you can easily find the minimum value in O(n)
done.
Note: since your matrix is symmetric, given P you just need to compute CS by combining the first element of P with all other elements: (X1, Xi) is equal to (Xi, X1)
If your matrix is very large, and you want to do some optimization, you may consider prefixes of P: let me explain with an example
when you have solved the problem for P = X1.X2.X3, store the result in an associative map, where X1.X2.X3 is the key
later on, when you solve a problem P' = X1.X2.X3.X7.X9.X10.X11 you search for the longest prefix of P' in your map: you can do this by starting with P' and removing one component (Xi) at a time from the end until you find a match in your map or you end up with an empty string
if you find a prefix of P' in you map then you already know the solution for that problem, so you just have to find the solution for the problem resulting from combining the first element of the prefix with the suffix, and then compare the two results: in our example the prefix is X1.X2.X3, and so you just have to solve the problem for
X1.X7.X9.X10.X11, and then compare the two values and choose the min (don't forget to update your map with the new pattern P')
if you don't find any prefix, then you must solve the entire problem for P' (and again don't forget to update the map with the result, so that you can reuse it in the future)
This technique is essentially a form of memoization.