Related
I want to implement a function which will return cartesian product of set, repeated given number. For example
input: {a, b}, 2
output:
aa
ab
bb
ba
input: {a, b}, 3
aaa
aab
aba
baa
bab
bba
bbb
However the only way I can implement it is firstly doing cartesion product for 2 sets("ab", "ab), then from the output of the set, add the same set. Here is pseudo-code:
function product(A, B):
result = []
for i in A:
for j in B:
result.append([i,j])
return result
function product1(chars, count):
result = product(chars, chars)
for i in range(2, count):
result = product(result, chars)
return result
What I want is to start computing directly the last set, without computing all of the sets before it. Is this possible, also a solution which will give me similar result, but it isn't cartesian product is acceptable.
I don't have problem reading most of the general purpose programming languages, so if you need to post code you can do it in any language you fell comfortable with.
Here's a recursive algorithm that builds S^n without building S^(n-1) "first". Imagine an infinite k-ary tree where |S| = k. Label with the elements of S each of the edges connecting any parent to its k children. An element of S^m can be thought of as any path of length m from the root. The set S^m, in that way of thinking, is the set of all such paths. Now the problem of finding S^n is a problem of enumerating all paths of length n - and we can name a path by considering the sequence of edge labels from beginning to end. We want to directly generate S^n without first enumerating all of S^(n-1), so a depth-first search modified to find all nodes at depth n seems appropriate. This is essentially how the below algorithm works:
// collection to hold generated output
members = []
// recursive function to explore product space
Products(set[1...n], length, current[1...m])
// if the product we're working on is of the
// desired length then record it and return
if m = length then
members.append(current)
return
// otherwise we add each possible value to the end
// and generate all products of the desired length
// with the new vector as a prefix
for i = 1 to n do
current.addLast(set[i])
Products(set, length, current)
currents.removeLast()
// reset the result collection and request the set be generated
members = []
Products([a, b], 3, [])
Now, a breadth-first approach is no less efficient than a depth-first one, and if you think about it would be no different from exactly what you're already doing. Indeed, and approach that generates S^n must necessarily generate S^(n-1) at least once, since that can be found in a solution to S^n.
I'm suppose to generate a maze using a binary matrix
when a 0 represents empty cell
and 1 a wall
I tried to use the DFS algorithm, the problem is the DFS refers to cells and walls between them(each cell has at most four walls)
" then selects a random neighbouring cell that has not yet been visited. The computer removes the 'wall' between the two cells and.."
I don't understand to analogy to the representation I've been asked to implement
anyone has any idea?
When asked to make a grid, I'd start by making a multi-dimensional array.
Have the outer array contain each row of your grid, and each nested array be the columns.
Depending if each cell needs to remember being visited, each slot of the array can contain a simple class or struct (depending on your language of choice). Else, they can simply contain an int or a bool.
A simple example could be:
var grid = [
[1, 1, 1],
[1, 0, 1],
[1, 1, 1]
];
Representing an empty cell in the middle, surrounded by walls.
The idea is to construct a pattern of cells, then build wall between them, but not when it has a edge of DFS tree.
Example DFS run (2D case, edge removed, can be inferred from order of nodes):
O
O
O
O
OO
OO
OO
(this branch is stuck so start a new one)
OOO
OO
...
OOO
OOO
OOO
Now construct the maze:
OXSOO
OXOXO
OOOXO
XXXXO
EOOOO
O -> cell
X -> wall
S -> start
E -> end
The wall can be either a block or just a plane, the topology is same.
I am trying to implement the Parallel Algorithm for Longest Common Subsequence Problem described in http://www.iaeng.org/publication/WCE2010/WCE2010_pp499-504.pdf
But i am having a problem with the variable C in Equation 6 on page 4
The paper refered to C on at the end of page 3 as
C as Let C[1 : l] bethe finite alphabet
I am not sure what is ment by this, as i guess it would it with the 2 strings ABCDEF and ABQXYEF be ABCDEFQXY. But what if my 2 stings is a list of objects (Where my match test for an example is obj1.Name = obj2.Name), what would my C be here? just a union on the 2 arrays?
Having read and studied the paper, I can say that C is supposed to be an array holding the alphabet of your strings, where the alphabet size (and, thus, the size of C) is l.
By the looks of your question, however, I feel the need to go deeper on this, because it looks like you didn't get the whole picture yet. What is P[i,j], and why do you need it? The answer is that you don't really need it, but it's an elegant optimization. In page 3, a little bit before Theorem 1, it is said that:
[...] This process ends when j-k = 0 at the k-th step, or a(i) =
b(j-k) at the k-th step. Assume that the process stops at the k-th
step, and k must be the minimum number that makes a(i) = b(j-k) or j-k
= 0. [...]
The recurrence relation in (3) is equivalent to (2), but the fundamental difference is that (2) expands recursively, whereas with (3) you never have recursive calls, provided that you know k. In other words, the magic behind (3) not expanding recursively is that you somehow know the spot where the recursion on (2) would stop, so you look at that cell immediately, rather than recursively approaching it.
Ok then, but how do you find out the value for k? Since k is the spot where (2) reaches a base case, it can be seen that k is the amount of columns that you have to "go back" on B until you are either off the limits (i.e., the first column that is filled with 0's) OR you find a match between a character in B and a character in A (which corresponds to the base case conditions in (2)). Remember that you will be matching the character a(i-1), where i is the current row.
So, what you really want is to find the last position in B before j where the character a(i-1) appears. If no such character ever appears in B before j, then that would be equivalent to reaching the case i = 0 or j-1 = 0 in (2); otherwise, it's the same as reaching a(i) = b(j-1) in (2).
Let's look at an example:
Consider that the algorithm is working on computing the values for i = 2 and j = 3 (the row and column are highlighted in gray). Imagine that the algorithm is working on the cell highlighted in black and is applying (2) to determine the value of S[2,2] (the position to the left of the black one). By applying (2), it would then start by looking at a(2) and b(2). a(2) is C, b(2) is G, to there's no match (this is the same procedure as the original, well-known algorithm). The algorithm now wants to find the value of S[2,2], because it is needed to compute S[2,3] (where we are). S[2,2] is not known yet, but the paper shows that it is possible to determine that value without refering to the row with i = 2. In (2), the 3rd case is chosen: S[2,2] = max(S[1, 2], S[2, 1]). Notice, if you will, that all this formula is doing is looking at the positions that would have been used to calculate S[2,2]. So, to rephrase that: we're computing S[2,3], we need S[2,2] for that, we don't know it yet, so we're going back on the table to see what's the value of S[2,2] in pretty much the same way we did in the original, non-parallel algorithm.
When will this stop? In this example, it will stop when we find the letter C (this is our a(i)) in TGTTCGACA before the second T (the letter on the current column) OR when we reach column 0. Because there is no C before T, we reach column 0. Another example:
Here, (2) would stop with j-1 = 5, because that is the last position in TGTTCGACA where C shows up. Thus, the recursion reaches the base case a(i) = b(j-1) when j-1 = 5.
With this in mind, we can see a shortcut here: if you could somehow know the amount k such that j-1-k is a base case in (2), then you wouldn't have to go through the score table to find the base case.
That's the whole idea behind P[i,j]. P is a table where you lay down the whole alphabet vertically (on the left side); the string B is, once again, placed horizontally in the upper side. This table is computed as part of a preprocessing step, and it will tell you exactly what you will need to know ahead of time: for each position j in B, it says, for each character C[i] in C (the alphabet), what is the last position in B before j where C[i] is found (note that i is used to index C, the alphabet, and not the string A. Maybe the authors should have used another index variable to avoid confusion).
So, you can think of the semantics for an entry P[i,j] as something along the lines of: The last position in B where I saw C[i] before position j. For example, if you alphabet is sigma = {A, E, I, O, U}, and B = "AOOIUEI", thenP` is:
Take the time to understand this table. Note the row for O. Remember: this row lists, for every position in B, where is the last known "O". Only when j = 3 will we have a value that is not zero (it's 2), because that's the position after the first O in AOOIUEI. This entry says that the last position in B where O was seen before is position 2 (and, indeed, B[2] is an O, the one that follows A). Notice, in that same row, that for j = 4, we have the value 3, because now the last position for O is the one that correspnds to the second O in B (and since no more O's exist, the rest of the row will be 3).
Recall that building P is a preprocessing step necessary if you want to easily find the value of k that makes the recursion from equation (2) stop. It should make sense by now that P[i,j] is the k you're looking for in (3). With P, you can determine that value in O(1) time.
Thus, the C[i] in (6) is a letter of the alphabet - the letter that we are currently considering. In the example above, C = [A,E,I,O,U], and C[1] = A, C[2] = E, etc. In equaton (7), c is the position in C where a(i) (the current letter of string A being considered) lives. It makes sense: after all, when building the score table position S[i,j], we want to use P to find the value of k - we want to know where was the last time we saw an a(i) in B before j. We do that by reading P[index_of(a(i)), j].
Ok, now that you understand the use of P, let's see what's happening with your implementation.
About your specific case
In the paper, P is shown as a table that lists the whole alphabet. It is a good idea to iterate through the alphabet because the typical uses of this algorithm are in bioinformatics, where the alphabet is much, much smaller than the string A, making the iteration through the alphabet cheaper.
Because your strings are sequences of objects, your C would be the set of all possible objects, so you'd have to build a table P with the set of all possible object instance (nonsense, of course). This is definitely a case where the alphabet size is huge when compared to your string size. However, note that you will only be indexing P in those rows that correspond to letters from A: any row in P for a letter C[i] that is not in A is useless and will never be used. This makes your life easier, because it means you can build P with the string A instead of using the alphabet of every possible object.
Again, an example: if your alphabet is AEIOU, A is EEI and B is AOOIUEI, you will only be indexing P in the rows for E and I, so that's all you need in P:
This works and suffices, because in (7), P[c,j] is the entry in P for the character c, and c is the index of a(i). In other words: C[c] always belongs to A, so it makes perfect sense to build P for the characters of A instead of using the whole alphabet for the cases where the size of A is considerably smaller than the size of C.
All you have to do now is to apply the same principle to whatever your objects are.
I really don't know how to explain it any better. This may be a little dense at first. Make sure to re-read it until you really get it - and I mean every little detail. You have to master this before thinking about implementing it.
NOTE: You said you were looking for a credible and / or official source. I'm just another CS student, so I'm not an official source, but I think I can be considered "credible". I've studied this before and I know the subject. Happy coding!
I have a list of elements, each one identified with a type, I need to reorder the list to maximize the minimum distance between elements of the same type.
The set is small (10 to 30 items), so performance is not really important.
There's no limit about the quantity of items per type or quantity of types, the data can be considered random.
For example, if I have a list of:
5 items of A
3 items of B
2 items of C
2 items of D
1 item of E
1 item of F
I would like to produce something like:
A, B, C, A, D, F, B, A, E, C, A, D, B, A
A has at least 2 items between occurences
B has at least 4 items between occurences
C has 6 items between occurences
D has 6 items between occurences
Is there an algorithm to achieve this?
-Update-
After exchanging some comments, I came to a definition of a secondary goal:
main goal: maximize the minimum distance between elements of the same type, considering only the type(s) with less distance.
secondary goal: maximize the minimum distance between elements on every type. IE: if a combination increases the minimum distance of a certain type without decreasing other, then choose it.
-Update 2-
About the answers.
There were a lot of useful answers, although none is a solution for both goals, specially the second one which is tricky.
Some thoughts about the answers:
PengOne: Sounds good, although it doesn't provide a concrete implementation, and not always leads to the best result according to the second goal.
Evgeny Kluev: Provides a concrete implementation to the main goal, but it doesn't lead to the best result according to the secondary goal.
tobias_k: I liked the random approach, it doesn't always lead to the best result, but it's a good approximation and cost effective.
I tried a combination of Evgeny Kluev, backtracking, and tobias_k formula, but it needed too much time to get the result.
Finally, at least for my problem, I considered tobias_k to be the most adequate algorithm, for its simplicity and good results in a timely fashion. Probably, it could be improved using Simulated annealing.
First, you don't have a well-defined optimization problem yet. If you want to maximized the minimum distance between two items of the same type, that's well defined. If you want to maximize the minimum distance between two A's and between two B's and ... and between two Z's, then that's not well defined. How would you compare two solutions:
A's are at least 4 apart, B's at least 4 apart, and C's at least 2 apart
A's at least 3 apart, B's at least 3 apart, and C's at least 4 apart
You need a well-defined measure of "good" (or, more accurately, "better"). I'll assume for now that the measure is: maximize the minimum distance between any two of the same item.
Here's an algorithm that achieves a minimum distance of ceiling(N/n(A)) where N is the total number of items and n(A) is the number of items of instance A, assuming that A is the most numerous.
Order the item types A1, A2, ... , Ak where n(Ai) >= n(A{i+1}).
Initialize the list L to be empty.
For j from k to 1, distribute items of type Ak as uniformly as possible in L.
Example: Given the distribution in the question, the algorithm produces:
F
E, F
D, E, D, F
D, C, E, D, C, F
B, D, C, E, B, D, C, F, B
A, B, D, A, C, E, A, B, D, A, C, F, A, B
This sounded like an interesting problem, so I just gave it a try. Here's my super-simplistic randomized approach, done in Python:
def optimize(items, quality_function, stop=1000):
no_improvement = 0
best = 0
while no_improvement < stop:
i = random.randint(0, len(items)-1)
j = random.randint(0, len(items)-1)
copy = items[::]
copy[i], copy[j] = copy[j], copy[i]
q = quality_function(copy)
if q > best:
items, best = copy, q
no_improvement = 0
else:
no_improvement += 1
return items
As already discussed in the comments, the really tricky part is the quality function, passed as a parameter to the optimizer. After some trying I came up with one that almost always yields optimal results. Thank to pmoleri, for pointing out how to make this a whole lot more efficient.
def quality_maxmindist(items):
s = 0
for item in set(items):
indcs = [i for i in range(len(items)) if items[i] == item]
if len(indcs) > 1:
s += sum(1./(indcs[i+1] - indcs[i]) for i in range(len(indcs)-1))
return 1./s
And here some random result:
>>> print optimize(items, quality_maxmindist)
['A', 'B', 'C', 'A', 'D', 'E', 'A', 'B', 'F', 'C', 'A', 'D', 'B', 'A']
Note that, passing another quality function, the same optimizer could be used for different list-rearrangement tasks, e.g. as a (rather silly) randomized sorter.
Here is an algorithm that only maximizes the minimum distance between elements of the same type and does nothing beyond that. The following list is used as an example:
AAAAA BBBBB CCCC DDDD EEEE FFF GG
Sort element sets by number of elements of each type in descending order. Actually only largest sets (A & B) should be placed to the head of the list as well as those element sets that have one element less (C & D & E). Other sets may be unsorted.
Reserve R last positions in the array for one element from each of the largest sets, divide the remaining array evenly between the S-1 remaining elements of the largest sets. This gives optimal distance: K = (N - R) / (S - 1). Represent target array as a 2D matrix with K columns and L = N / K full rows (and possibly one partial row with N % K elements). For example sets we have R = 2, S = 5, N = 27, K = 6, L = 4.
If matrix has S - 1 full rows, fill first R columns of this matrix with elements of the largest sets (A & B), otherwise sequentially fill all columns, starting from last one.
For our example this gives:
AB....
AB....
AB....
AB....
AB.
If we try to fill the remaining columns with other sets in the same order, there is a problem:
ABCDE.
ABCDE.
ABCDE.
ABCE..
ABD
The last 'E' is only 5 positions apart from the first 'E'.
Sequentially fill all columns, starting from last one.
For our example this gives:
ABFEDC
ABFEDC
ABFEDC
ABGEDC
ABG
Returning to linear array we have:
ABFEDCABFEDCABFEDCABGEDCABG
Here is an attempt to use simulated annealing for this problem (C sources): http://ideone.com/OGkkc.
I believe you could see your problem like a bunch of particles that physically repel eachother. You could iterate to a 'stable' situation.
Basic pseudo-code:
force( x, y ) = 0 if x.type==y.type
1/distance(x,y) otherwise
nextposition( x, force ) = coined?(x) => same
else => x + force
notconverged(row,newrow) = // simplistically
row!=newrow
row=[a,b,a,b,b,b,a,e];
newrow=nextposition(row);
while( notconverged(row,newrow) )
newrow=nextposition(row);
I don't know if it converges, but it's an idea :)
I'm sure there may be a more efficient solution, but here is one possibility for you:
First, note that it is very easy to find an ordering which produces a minimum-distance-between-items-of-same-type of 1. Just use any random ordering, and the MDBIOST will be at least 1, if not more.
So, start off with the assumption that the MDBIOST will be 2. Do a recursive search of the space of possible orderings, based on the assumption that MDBIOST will be 2. There are a number of conditions you can use to prune branches from this search. Terminate the search if you find an ordering which works.
If you found one that works, try again, under the assumption that MDBIOST will be 3. Then 4... and so on, until the search fails.
UPDATE: It would actually be better to start with a high number, because that will constrain the possible choices more. Then gradually reduce the number, until you find an ordering which works.
Here's another approach.
If every item must be kept at least k places from every other item of the same type, then write down items from left to right, keeping track of the number of items left of each type. At each point put down an item with the largest number left that you can legally put down.
This will work for N items if there are no more than ceil(N / k) items of the same type, as it will preserve this property - after putting down k items we have k less items and we have put down at least one of each type that started with at ceil(N / k) items of that type.
Given a clutch of mixed items you could work out the largest k you can support and then lay out the items to solve for this k.
I apologize for not have the math background to put this question in a more formal way.
I'm looking to create a string of 796 letters (or integers) with certain properties.
Basically, the string is a variation on a De Bruijn sequence B(12,4), except order and repetition within each n-length subsequence are disregarded.
i.e. ABBB BABA BBBA are each equivalent to {AB}.
In other words, the main property of the string involves looking at consecutive groups of 4 letters within the larger string
(i.e. the 1st through 4th letters, the 2nd through 5th letters, the 3rd through 6th letters, etc)
And then producing the set of letters that comprise each group (repetitions and order disregarded)
For example, in the string of 9 letters:
A B B A C E B C D
the first 4-letter groups is: ABBA, which is comprised of the set {AB}
the second group is: BBAC, which is comprised of the set {ABC}
the third group is: BACE, which is comprised of the set {ABCE}
etc.
The goal is for every combination of 1-4 letters from a set of N letters to be represented by the 1-4-letter resultant sets of the 4-element groups once and only once in the original string.
For example, if there is a set of 5 letters {A, B, C, D, E} being used
Then the possible 1-4 letter combinations are:
A, B, C, D, E,
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE,
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE,
ABCD, ABCE, ABDE, ACDE, BCDE
Here is a working example that uses a set of 5 letters {A, B, C, D, E}.
D D D D E C B B B B A E C C C C D A E E E E B D A A A A C B D D B
The 1st through 4th elements form the set: D
The 2nd through 5th elements form the set: DE
The 3rd through 6th elements form the set: CDE
The 4th through 7th elements form the set: BCDE
The 5th through 8th elements form the set: BCE
The 6th through 9th elements form the set: BC
The 7th through 10th elements form the set: B
etc.
* I am hoping to find a working example of a string that uses 12 different letters (a total of 793 4-letter groups within a 796-letter string) starting (and if possible ending) with 4 of the same letter. *
Here is a working solution for 7 letters:
AAAABCDBEAAACDECFAAADBFBACEAGAADEFBAGACDFBGCCCCDGEAFAGCBEEECGFFBFEGGGGFDEEEEFCBBBBGDCFFFFDAGBEGDDDDBE
Beware that in order to attempt exhaustive search (answer in VB is trying a naive version of that) you'll first have to solve the problem of generating all possible expansions while maintaining lexicographical order. Just ABC, expands to all perms of AABC, plus all perms of ABBC, plus all perms of ABCC which is 3*4! instead of just AABC. If you just concatenate AABC and AABD it would cover just 4 out of 4! perms of AABC and even that by accident. Just this expansion will bring you exponential complexity - end of game. Plus you'll need to maintain association between all explansions and the set (the set becomes a label).
Your best bet is to use one of known efficient De Bruijn constuctors and try to see if you can put your set-equivalence in there. Check out
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.14.674&rep=rep1&type=pdf
and
http://www.dim.uchile.cl/~emoreno/publicaciones/FINALES/copyrighted/IPL05-De_Bruijn_sequences_and_De_Bruijn_graphs_for_a_general_language.pdf
for a start.
If you know graphs, another viable option is to start with De Bruijn graph and formulate your set-equivalence as a graph rewriting. 2nd paper does De Bruijn graph partitioning.
BTW, try VB answer just for A,B,AB (at least expansion is small) - it will make AABBAB and construct ABBA or ABBAB (or throw in a decent language) both of which are wrong. You can even prove that it will always miss with 1st lexical expansions (that's what AAB, AAAB etc. are) just by examining first 2 passes (it will always miss 2nd A for NxA because (N-1)xA+B is in the string (1st expansion of {AB}).
Oh and if we could establish how many of each letters an optimal soluton should have (don't look at B(5,2) it's too easy and regular :-) a random serch would be feasible - you generate candidates with provable traits (like AAAA, BBBB ... are present and not touching and is has n1 A-s, n2 B-s ...) and random arrangement and then test whether they are solutions (checking is much faster than exhaustive search in this case).
Cool problem. Just a draft/psuedo algo:
dim STR-A as string = getall(ABCDEFGHIJKL)
//custom function to generate concat list of all 793 4-char combos.
//should be listed side-by-side to form 3172 character-long string.
//different ordering may ultimately produce different results.
//brute-forcing all orders of combos is too much work (793! is a big #).
//need to determine how to find optimal ordering, for this particular
//approach below.
dim STR-B as string = "" // to hold the string you're searching for
dim STR-C as string = "" // to hold the sub-string you are searching in
dim STR-A-NEW as string = "" //variable to hold your new string
dim MATCH as boolean = false //variable to hold matching status
while len(STR-A) > 0
//check each character in STR-A, which will be shorted by 1 char on each
//pass.
MATCH = false
STR-B = left(STR-A, 4)
STR-B = reduce(STR-B)
//reduce(str) is a custom re-usable function to sort & remove duplicates
for i as integer = 1 to len((STR-A) - 1)
STR-C = substr(STR-A, i, 4)
//gives you the 4-character sequence beginning at position i
STR-C = reduce(STR-C)
IF STR-B = STR-C Then
MATCH = true
exit for
//as long as there is even one match, you can throw-away the first
//letter
END IF
i = i+1
next
IF match = false then
//if you didn't find a match, then the first letter should be saved
STR-A-NEW += LEFT(STR-B, 1)
END IF
MATCH = false //re-init MATCH
STR-A = RIGHT(STR-A, LEN(STR-A) - 1) //re-init STR_A
wend
Anyway -- there could be problems at this, and you'd need to write another function to parse your result string (STR-A-NEW) to prove that it's a viable answer...
I've been thinking about this one and I'm sketching out a solution.
Let's call a string of four symbols a word and we'll write S(w) to denote the set of symbols in word w.
Each word abcd has "follow-on" words bcde where a,...,e are all symbols.
Let succ(w) be the set of follow-on words v for w such that S(w) != S(v). succ(w) is the set of successor words that can follow on from the first symbol in w if w is in a solution.
For each non-empty set of symbols s of cardinality at most four, let words(s) be the set of words w such that S(w) = s. Any solution must contain exactly one word in words(s) for each such set s.
Now we can do a reasonable search. The basic idea is this: say we are exploring a search path ending with word w. The follow-on word must be a non-excluded word in succ(w). A word v is excluded if the search path contains some word w such that v in words(S(w)).
You can be slightly more cunning: if we track the possible "predecessor" words to a set s (i.e., words w with a successor v such that v in words(s)) and reach a point where every predecessor of s is excluded, then we know we have reached a dead end, since we'll never be able to obtain s from any extension of the current search path.
Code to follow after the weekend, with a bit of luck...
Here is my proposal. I'll admit upfront this is a performance and memory hog.
This may be overkill, but have a class We'll call it UniqueCombination This will contain a unique 1-4 char reduced combination of the input set (i.e. A,AB,ABC,...) This will also contain a list of possible combination (AB {AABB,ABAB,BBAA,...}) this will need a method that determines if any possible combination overlaps any possible combination of another UniqueCombination by three characters. Also need a override that takes a string as well.
Then we start with the string "AAAA" then we find all of the UniqueCombinations that overlap this string. Then we find how many uniqueCombinations those possible matches overlap with. (we could be smart at this point an store this number.) Then we pick the one with the least number of overlaps greater than 0. Use up the ones with the least possible matches first.
Then we find a specific combination for the chosen UniqueCombination and add it to the final string. Remove this UniqueCombination from the list, then as we find overlaps for current string. rinse and repeat. (we could be smart and on subsequent runs while searching for overlaps we could remove any of the unreduced combination that are contained in the final string.)
Well that's my plan I will work on the code this weekend. Granted this does not guarantee that the final 4 characters will be 4 of the same letter (it might actually be trying to avoid that but I will look into that as well.)
If there is a non-exponential solution at all it may need to be formulated in terms of a recursive "growth" from a problem with a smaller size i.e to contruct B(N,k) from B(N-1,k-1) or from B(N-1,k) or from B(N,k-1).
Systematic construction for B(5,2) - one step at the time :-) It's bound to get more complex latter [card stands for cardinality, {AB} has card=2, I'll also call them 2-s, 3-s etc.] Note, 2-s and 3-s will be k-1 and k latter (I hope).
Initial. Start with k-1 result and inject symbols for singletons
(unique expansion empty intersection):
ABCDE -> AABBCCDDEE
mark used card=2 sets: AB,BC,CD,DE
Rewriting. Form card=3 sets to inject symbols into marked card=2.
1st feasible lexicographic expansion fires (may have to backtrack for k>2)
it's OK to use already marked 2-s since they'll all get replaced
but may have to do a verification pass for higher k
AB->ACB, BC->BCD, CD->CED, DE->DAE ==> AACBBDCCEDDAEEB
mark/verify used 2s
normally keep marking/unmarking during the construction but also keep keep old
mark list
marking/unmarking can get expensive if there's backtracking in #3
Unused: AB, BE
For higher k may need several recursive rewriting passes
possibly partitioning new sets into classes
Finalize: unused 2-s should overlap around the edge (that's why it's cyclic)
ABE - B can go to the begining or and: AACBBDCCEDDAEEB
Note: a step from B(N-1,k) to B(N,k) may need injection of pseudo-signletons, like doubling or trippling A
B(5,2) -> B(5,3) - B(5,4)
Initial. same: - ABCDE -> AAACBBBDCCCEDDDAEEEB
no use of marking 3-sets since they are all going to be chenged
Rewriting.
choose systematic insertion positions
AAA_CBBB_DCCC_EDDD_AEEE_B
mark all 2-s released by this: AC,AD,BD,BE,CE
use marked 2-s to decide inserted symbols - totice total regularity:
AxCB D -> ADCB
BxDC E -> BEDC
CxED A -> CAED
DxAE B => DBAE
ExBA C -> ECBA
Verify that 3-s are all used (marked inserted symbols just for fun)
AAA[D]CBBB[E]DCCC[A]EDDD[B]AEEE[C]B
Note: Systematic choice if insertion point deterministically dictated insertions (only AD can fit 1st, AC would create duplicate 2-set (AAC, ACC))
Note: It's not going to be so nice for B(6,2) and B(6,3) since number of 2-s will exceede 2x the no of 1-s. This is important since 2-s sit naturally on the sides of 1-s like CBBBE and the issue is how to place them when you run out of 1-s.
B(5,3) is so symetrical that just repeating #1 produces B(5.4):
AAAADCBBBBEDCCCCAEDDDDBAEEEECB