Controlling distance of shuffling

Question

I have tried to ask this question before, but have never been able to word it correctly. I hope I have it right this time:

I have a list of unique elements. I want to shuffle this list to produce a new list. However, I would like to constrain the shuffle, such that each element's new position is at most d away from its original position in the list.

So for example:

L = [1,2,3,4]
d = 2
answer = magicFunction(L, d)

Now, one possible outcome could be:

>>> print(answer)
[3,1,2,4]

Notice that 3 has moved two indices, 1 and 2 have moved one index, and 4 has not moved at all. Thus, this is a valid shuffle, per my previous definition. The following snippet of code can be used to validate this:

old = {e:i for i,e in enumerate(L)}
new = {e:i for i,e in enumerate(answer)}
valid = all(abs(i-new[e])<=d for e,i in old.items())

Now, I could easily just generate all possible permutations of L, filter for the valid ones, and pick one at random. But that doesn't seem very elegant. Does anyone have any other ideas about how to accomplish this?

Answer 1

I am not sure how good it is, but maybe something like:

• create a list of same length than initial list L; each element of this list should be a list of indices of allowed initial indices to be moved here; for instance [[0,1,2],[0,1,2,3],[0,1,2,3],[1,2,3]] if I understand correctly your example;
• take the smallest sublist (or any of the smallest sublists if several lists share the same length);
• pick a random element in it with random.choice, this element is the index of the element in the initial list to be mapped to the current location (use another list for building your new list);
• remove the randomly chosen element from all sublists

For instance:

L = [ "A", "B", "C", "D" ]
i = [[0,1,2],[0,1,2,3],[0,1,2,3],[1,2,3]]
# I take [0,1,2] and pick randomly 1 inside
# I remove the value '1' from all sublists and since
# the first sublist has already been handled I set it to None
# (and my result will look as [ "B", None, None, None ]
i = [None,[0,2,3],[0,2,3],[2,3]]
# I take the last sublist and pick randomly 3 inside
# result will be ["B", None, None, "D" ]
i = [None,[0,2], [0,2], None]
etc.

I haven't tried it however. Regards.

Answer 2

Here are two sketches in Python; one swap-based, the other non-swap-based. In the first, the idea is to keep track of where the indexes have moved and test if the next swap would be valid. An additional variable is added for the number of swaps to make.

from random import randint

def swap(a,b,L):
  L[a], L[b] = L[b], L[a]

def magicFunction(L,d,numSwaps):
  n = len(L)
  new = list(range(0,n))
  for i in xrange(0,numSwaps):
    x = randint(0,n-1)
    y = randint(max(0,x - d),min(n - 1,x + d))
    while abs(new[x] - y) > d or abs(new[y] - x) > d:
      y = randint(max(0,x - d),min(n - 1,x + d))
    swap(x,y,new)
    swap(x,y,L)
  return L

print(magicFunction([1,2,3,4],2,3)) # [2, 1, 4, 3]
print(magicFunction([1,2,3,4,5,6,7,8,9],2,4)) # [2, 3, 1, 5, 4, 6, 8, 7, 9]

Using print(collections.Counter(tuple(magicFunction([0, 1, 2], 1, 1)) for i in xrange(1000))) we find that the identity permutation comes up heavy with this code (the reason why is left as an exercise for the reader).

---

Alternatively, we can think about it as looking for a permutation matrix with interval restrictions, where abs(i - j) <= d where M(i,j) would equal 1. We can construct a one-off random path by picking a random j for each row from those still available. x's in the following example represent matrix cells that would invalidate the solution (northwest to southeast diagonal would represent the identity permutation), restrictions represent how many is are still available for each j. (Adapted from my previous version to choose both the next i and the next j randomly, inspired by user2357112's answer):

n = 5, d = 2

Start:

0 0 0 x x
0 0 0 0 x
0 0 0 0 0
x 0 0 0 0
x x 0 0 0

restrictions = [3,4,5,4,3]  # how many i's are still available for each j

1.

0 0 1 x x  # random choice
0 0 0 0 x
0 0 0 0 0
x 0 0 0 0
x x 0 0 0

restrictions = [2,3,0,4,3] # update restrictions in the neighborhood of (i ± d)

2.

0 0 1 x x 
0 0 0 0 x  
0 0 0 0 0
x 0 0 0 0
x x 0 1 0  # random choice

restrictions = [2,3,0,0,2] # update restrictions in the neighborhood of (i ± d)

---

3.

0 0 1 x x
0 0 0 0 x
0 1 0 0 0  # random choice
x 0 0 0 0
x x 0 1 0

restrictions = [1,0,0,0,2] # update restrictions in the neighborhood of (i ± d)

only one choice for j = 0 so it must be chosen

4.

0 0 1 x x  
1 0 0 0 x  # dictated choice
0 1 0 0 0
x 0 0 0 0
x x 0 1 0

restrictions = [0,0,0,0,2] # update restrictions in the neighborhood of (i ± d)

Solution:

0 0 1 x x
1 0 0 0 x
0 1 0 0 0
x 0 0 0 1  # dictated choice
x x 0 1 0

[2,0,1,4,3]

Python code (adapted from my previous version to choose both the next i and the next j randomly, inspired by user2357112's answer):

from random import randint,choice
import collections

def magicFunction(L,d):
  n = len(L)
  restrictions = [None] * n
  restrict = -1
  solution = [None] * n
  for i in xrange(0,n):
    restrictions[i] = abs(max(0,i - d) - min(n - 1,i + d)) + 1
  while True:
    availableIs = filter(lambda x: solution[x] == None,[i for i in xrange(n)]) if restrict == -1 else filter(lambda x: solution[x] == None,[j for j in xrange(max(0,restrict - d),min(n,restrict + d + 1))])
    if not availableIs:
      L = [L[i] for i in solution]
      return L
    i = choice(availableIs)
    availableJs = filter(lambda x: restrictions[x] <> 0,[j for j in xrange(max(0,i - d),min(n,i + d + 1))])
    nextJ = restrict if restrict != -1 else choice(availableJs)
    restrict = -1
    solution[i] = nextJ
    restrictions[ nextJ ] = 0
    for j in xrange(max(0,i - d),min(n,i + d + 1)):
      if j == nextJ or restrictions[j] == 0:
        continue
      restrictions[j] = restrictions[j] - 1
      if restrictions[j] == 1:
        restrict = j

print(collections.Counter(tuple(magicFunction([0, 1, 2], 1)) for i in xrange(1000)))

Using print(collections.Counter(tuple(magicFunction([0, 1, 2], 1)) for i in xrange(1000))) we find that the identity permutation comes up light with this code (why is left as an exercise for the reader).

Answer 3

This is going to be long and dry.

I have a solution that produces a uniform distribution. It requires O(len(L) * d**d) time and space for precomputation, then performs shuffles in O(len(L)*d) time¹. If a uniform distribution is not required, the precomputation is unnecessary, and the shuffle time can be reduced to O(len(L)) due to faster random choices; I have not implemented the non-uniform distribution. Both steps of this algorithm are substantially faster than brute force, but they're still not as good as I'd like them to be. Also, while the concept should work, I have not tested my implementation as thoroughly as I'd like.

Suppose we iterate over L from the front, choosing a position for each element as we come to it. Define the lag as the distance between the next element to place and the first unfilled position. Every time we place an element, the lag grows by at most one, since the index of the next element is now one higher, but the index of the first unfilled position cannot become lower.

Whenever the lag is d, we are forced to place the next element in the first unfilled position, even though there may be other empty spots within a distance of d. If we do so, the lag cannot grow beyond d, we will always have a spot to put each element, and we will generate a valid shuffle of the list. Thus, we have a general idea of how to generate shuffles; however, if we make our choices uniformly at random, the overall distribution will not be uniform. For example, with len(L) == 3 and d == 1, there are 3 possible shuffles (one for each position of the middle element), but if we choose the position of the first element uniformly, one shuffle becomes twice as likely as either of the others.

If we want a uniform distribution over valid shuffles, we need to make a weighted random choice for the position of each element, where the weight of a position is based on the number of possible shuffles if we choose that position. Done naively, this would require us to generate all possible shuffles to count them, which would take O(d**len(L)) time. However, the number of possible shuffles remaining after any step of the algorithm depends only on which spots we've filled, not what order they were filled in. For any pattern of filled or unfilled spots, the number of possible shuffles is the sum of the number of possible shuffles for each possible placement of the next element. At any step, there are at most d possible positions to place the next element, and there are O(d**d) possible patterns of unfilled spots (since any spot further than d behind the current element must be full, and any spot d or further ahead must be empty). We can use this to generate a Markov chain of size O(len(L) * d**d), taking O(len(L) * d**d) time to do so, and then use this Markov chain to perform shuffles in O(len(L)*d) time.

Example code (currently not quite O(len(L)*d) due to inefficient Markov chain representation):

import random

# states are (k, filled_spots) tuples, where k is the index of the next
# element to place, and filled_spots is a tuple of booleans
# of length 2*d, representing whether each index from k-d to
# k+d-1 has an element in it. We pretend indices outside the array are
# full, for ease of representation.

def _successors(n, d, state):
    '''Yield all legal next filled_spots and the move that takes you there.

    Doesn't handle k=n.'''
    k, filled_spots = state
    next_k = k+1

    # If k+d is a valid index, this represents the empty spot there.
    possible_next_spot = (False,) if k + d < n else (True,)

    if not filled_spots[0]:
        # Must use that position.
        yield k-d, filled_spots[1:] + possible_next_spot
    else:
        # Can fill any empty spot within a distance d.
        shifted_filled_spots = list(filled_spots[1:] + possible_next_spot)
        for i, filled in enumerate(shifted_filled_spots):
            if not filled:
                successor_state = shifted_filled_spots[:]
                successor_state[i] = True
                yield next_k-d+i, tuple(successor_state)
                # next_k instead of k in that index computation, because
                # i is indexing relative to shifted_filled_spots instead
                # of filled_spots


def _markov_chain(n, d):
    '''Precompute a table of weights for generating shuffles.

    _markov_chain(n, d) produces a table that can be fed to
    _distance_limited_shuffle to permute lists of length n in such a way that
    no list element moves a distance of more than d from its initial spot,
    and all permutations satisfying this condition are equally likely.

    This is expensive.

    '''

    if d >= n - 1:
        # We don't need the table, and generating a table for d >= n
        # complicates the indexing a bit. It's too complicated already.
        return None

    table = {}
    termination_state = (n, (d*2 * (True,)))
    table[termination_state] = 1

    def possible_shuffles(state):
        try:
            return table[state]
        except KeyError:
            k, _ = state
            count = table[state] = sum(
                    possible_shuffles((k+1, next_filled_spots))
                    for (_, next_filled_spots) in _successors(n, d, state)
            )
            return count

    initial_state = (0, (d*(True,) + d*(False,)))
    possible_shuffles(initial_state)
    return table

def _distance_limited_shuffle(l, d, table):
    # Generate an index into the set of all permutations, then use the
    # markov chain to efficiently find which permutation we picked.

    n = len(l)

    if d >= n - 1:
        random.shuffle(l)
        return

    permutation = [None]*n
    state = (0, (d*(True,) + d*(False,)))
    permutations_to_skip = random.randrange(table[state])

    for i, item in enumerate(l):
        for placement_index, new_filled_spots in _successors(n, d, state):
            new_state = (i+1, new_filled_spots)
            if table[new_state] <= permutations_to_skip:
                permutations_to_skip -= table[new_state]
            else:
                state = new_state
                permutation[placement_index] = item
                break
    return permutation

class Shuffler(object):
    def __init__(self, n, d):
        self.n = n
        self.d = d
        self.table = _markov_chain(n, d)
    def shuffled(self, l):
        if len(l) != self.n:
            raise ValueError('Wrong input size')
        return _distance_limited_shuffle(l, self.d, self.table)
    __call__ = shuffled

¹We could use a tree-based weighted random choice algorithm to improve the shuffle time to O(len(L)*log(d)), but since the table becomes so huge for even moderately large d, this doesn't seem worthwhile. Also, the factors of d**d in the bounds are overestimates, but the actual factors are still at least exponential in d.

Answer 4

Here's an adaptation of @גלעד ברקן's code that takes only one pass through the list (in random order) and swaps only once (using a random choice of possible positions):

from random import choice, shuffle

def magicFunction(L, d):
  n = len(L)
  swapped = [0] * n     # 0: position not swapped, 1: position was swapped
  positions = list(xrange(0,n))         # list of positions: 0..n-1
  shuffle(positions)                    # randomize positions
  for x in positions:
    if swapped[x]:      # only swap an item once
      continue
    # find all possible positions to swap
    possible = [i for i in xrange(max(0, x - d), min(n, x + d)) if not swapped[i]]
    if not possible:
      continue
    y = choice(possible)        # choose another possible position at random
    if x != y:
      L[y], L[x] = L[x], L[y]   # swap with that position
      swapped[x] = swapped[y] = 1       # mark both positions as swapped
  return L

Here is a refinement of the above code that simply finds all possible adjacent positions and chooses one:

from random import choice

def magicFunction(L, d):
  n = len(L)
  positions = list(xrange(0, n))        # list of positions: 0..n-1
  for x in xrange(0, n):
    # find all possible positions to swap
    possible = [i for i in xrange(max(0, x - d), min(n, x + d)) if abs(positions[i] - x) <= d]
    if not possible:
      continue
    y = choice(possible)        # choose another possible position at random
    if x != y:
      L[y], L[x] = L[x], L[y]   # swap with that position
      positions[x] = y
      positions[y] = x
  return L

Answer 5

In short, the list that should be shuffled gets ordered by the sum of index and a random number.

import random
xs = range(20) # list that should be shuffled
d = 5          # distance
[x for i,x in sorted(enumerate(xs), key= lambda (i,x): i+(d+1)*random.random())]

Out:

[1, 4, 3, 0, 2, 6, 7, 5, 8, 9, 10, 11, 12, 14, 13, 15, 19, 16, 18, 17]

Thats basically it. But this looks a little bit overwhelming, therefore...

The algorithm in more detail

To understand this better, consider this alternative implementation of an ordinary, random shuffle:

import random
sorted(range(10), key = lambda x: random.random())

Out:

[2, 6, 5, 0, 9, 1, 3, 8, 7, 4]

In order to constrain the distance, we have to implement a alternative sort key function that depends on the index of an element. The function sort_criterion is responsible for that.

import random

def exclusive_uniform(a, b):
    "returns a random value in the interval  [a, b)"
    return a+(b-a)*random.random()

def distance_constrained_shuffle(sequence, distance,
                                 randmoveforward = exclusive_uniform):
    def sort_criterion(enumerate_tuple):
        """
        returns the index plus a random offset,
        such that the result can overtake at most 'distance' elements
        """
        indx, value = enumerate_tuple
        return indx + randmoveforward(0, distance+1)

    # get enumerated, shuffled list
    enumerated_result = sorted(enumerate(sequence), key = sort_criterion)
    # remove enumeration
    result = [x for i, x in enumerated_result]
    return result

With the argument randmoveforward you can pass a random number generator with a different probability density function (pdf) to modify the distance distribution.

The remainder is testing and evaluation of the distance distribution.

---

Test function

Here is an implementation of the test function. The validatefunction is actually taken from the OP, but I removed the creation of one of the dictionaries for performance reasons.

def test(num_cases = 10, distance = 3, sequence = range(1000)):
    def validate(d, lst, answer):
        #old = {e:i for i,e in enumerate(lst)}
        new = {e:i for i,e in enumerate(answer)}
        return all(abs(i-new[e])<=d for i,e in enumerate(lst))
        #return all(abs(i-new[e])<=d for e,i in old.iteritems())


    for _ in range(num_cases):
        result = distance_constrained_shuffle(sequence, distance)
        if not validate(distance, sequence, result):
            print "Constraint violated. ", result
            break
    else:
        print "No constraint violations"


test()

Out:

No constraint violations

---

Distance distribution

I am not sure whether there is a way to make the distance uniform distributed, but here is a function to validate the distribution.

def distance_distribution(maxdistance = 3, sequence = range(3000)):
    from collections import Counter

    def count_distances(lst, answer):
        new = {e:i for i,e in enumerate(answer)}
        return Counter(i-new[e] for i,e in enumerate(lst))    

    answer = distance_constrained_shuffle(sequence, maxdistance)
    counter = count_distances(sequence, answer)

    sequence_length = float(len(sequence))

    distances = range(-maxdistance, maxdistance+1)
    return distances, [counter[d]/sequence_length for d in distances]

distance_distribution()

Out:

([-3, -2, -1, 0, 1, 2, 3],
 [0.01,
  0.076,
  0.22166666666666668,
  0.379,
  0.22933333333333333,
  0.07766666666666666,
  0.006333333333333333])

Or for a case with greater maximum distance:

distance_distribution(maxdistance=9, sequence=range(100*1000))

Answer 6

This is a very difficult problem, but it turns out there is a solution in the academic literature, in an influential paper by Mark Jerrum, Alistair Sinclair, and Eric Vigoda, A Polynomial-Time Approximation Algorithm for the Permanent of a Matrix with Nonnegative Entries, Journal of the ACM, Vol. 51, No. 4, July 2004, pp. 671–697. http://www.cc.gatech.edu/~vigoda/Permanent.pdf.

Here is the general idea: first write down two copies of the numbers in the array that you want to permute. Say

1   1
2   2
3   3
4   4

Now connect a node on the left to a node on the right if mapping from the number on the left to the position on the right is allowed by the restrictions in place. So if d=1 then 1 on the left connects to 1 and 2 on the right, 2 on the left connects to 1, 2, 3 on the right, 3 on the left connects to 2, 3, 4 on the right, and 4 on the left connects to 3, 4 on the right.

1 - 1
  X 
2 - 2
  X
3 - 3
  X
4 - 4

The resulting graph is bipartite. A valid permutation corresponds a perfect matching in the bipartite graph. A perfect matching, if it exists, can be found in O(VE) time (or somewhat better, for more advanced algorithms).

Now the problem becomes one of generating a uniformly distributed random perfect matching. I believe that can be done, approximately anyway. Uniformity of the distribution is the really hard part.

What does this have to do with permanents? Consider a matrix representation of our bipartite graph, where a 1 means an edge and a 0 means no edge:

1 1 0 0
1 1 1 0
0 1 1 1
0 0 1 1

The permanent of the matrix is like the determinant, except there are no negative signs in the definition. So we take exactly one element from each row and column, multiply them together, and add up over all choices of row and column. The terms of the permanent correspond to permutations; the term is 0 if any factor is 0, in other words if the permutation is not valid according to the matrix/bipartite graph representation; the term is 1 if all factors are 1, in other words if the permutation is valid according to the restrictions. In summary, the permanent of the matrix counts all permutations satisfying the restriction represented by the matrix/bipartite graph.

It turns out that unlike calculating determinants, which can be accomplished in O(n^3) time, calculating permanents is #P-complete so finding an exact answer is not feasible in general. However, if we can estimate the number of valid permutations, we can estimate the permanent. Jerrum et. al. approached the problem of counting valid permutations by generating valid permutations uniformly (within a certain error, which can be controlled); an estimate of the value of the permanent can be obtained by a fairly elaborate procedure (section 5 of the paper referenced) but we don't need that to answer the question at hand.

The running time of Jerrum's algorithm to calculate the permanent is O(n^11) (ignoring logarithmic factors). I can't immediately tell from the paper the running time of the part of the algorithm that uniformly generates bipartite matchings, but it appears to be over O(n^9). However, another paper reduces the running time for the permanent to O(n^7): http://www.cc.gatech.edu/fac/vigoda/FasterPermanent_SODA.pdf; in that paper they claim that it is now possible to get a good estimate of a permanent of a 100x100 0-1 matrix. So it should be possible to (almost) uniformly generate restricted permutations for lists of 100 elements.

There may be further improvements, but I got tired of looking.

If you want an implementation, I would start with the O(n^11) version in Jerrum's paper, and then take a look at the improvements if the original algorithm is not fast enough.

There is pseudo-code in Jerrum's paper, but I haven't tried it so I can't say how far the pseudo-code is from an actual implementation. My feeling is it isn't too far. Maybe I'll give it a try if there's interest.

Answer 7

My idea is to generate permutations by moving at most d steps by generating d random permutations which move at most 1 step and chaining them together.

We can generate permutations which move at most 1 step quickly by the following recursive procedure: consider a permutation of {1,2,3,...,n}. The last item, n, can move either 0 or 1 place. If it moves 0 places, n is fixed, and we have reduced the problem to generating a permutation of {1,2,...,n-1} in which every item moves at most one place.

On the other hand, if n moves 1 place, it must occupy position n-1. Then n-1 must occupy position n (if any smaller number occupies position n, it will have moved by more than 1 place). In other words, we must have a swap of n and n-1, and after swapping we have reduced the problem to finding such a permutation of the remainder of the array {1,...,n-2}.

Such permutations can be constructed in O(n) time, clearly.

Those two choices should be selected with weighted probabilities. Since I don't know the weights (though I have a theory, see below) maybe the choice should be 50-50 ... but see below.

A more accurate estimate of the weights might be as follows: note that the number of such permutations follows a recursion that is the same as the Fibonacci sequence: f(n) = f(n-1) + f(n-2). We have f(1) = 1 and f(2) = 2 ({1,2} goes to {1,2} or {2,1}), so the numbers really are the Fibonacci numbers. So my guess for the probability of choosing n fixed vs. swapping n and n-1 would be f(n-1)/f(n) vs. f(n-2)/f(n). Since the ratio of consecutive Fibonacci numbers quickly approaches the Golden Ratio, a reasonable approximation to the probabilities is to leave n fixed 61% of the time and swap n and n-1 39% of the time.

To construct permutations where items move at most d places, we just repeat the process d times. The running time is O(nd).

Here is an outline of an algorithm.

arr = {1,2,...,n};
for (i = 0; i < d; i++) {
  j = n-1;
  while (j > 0) {
    u = random uniform in interval (0,1)
    if (u < 0.61) { // related to golden ratio phi; more decimals may help
      j -= 1;
    } else {
      swap items at positions j and j-1 of arr // 0-based indexing
      j -= 2;
    }
  }
}

Since each pass moves items at most 1 place from their start, d passes will move items at most d places. The only question is the uniform distribution of the permutations. It would probably be a long proof, if it's even true, so I suggest assembling empirical evidence for various n's and d's. Probably to prove the statement, we would have to switch from using the golden ratio approximation to f(n-1)/f(n-2) in place of 0.61.

There might even be some weird reason why some permutations might be missed by this procedure, but I'm pretty sure that doesn't happen. Just in case, though, it would be helpful to have a complete inventory of such permutations for some values of n and d to check the correctness of my proposed algorithm.

Update

I found an off-by-one error in my "pseudocode", and I corrected it. Then I implemented in Java to get a sense of the distribution. Code is below. The distribution is far from uniform, I think because there are many ways of getting restricted permutations with short max distances (move forward, move back vs. move back, move forward, for example) but few ways of getting long distances (move forward, move forward). I can't think of a way to fix the uniformity issue with this method.

import java.util.Random;
import java.util.Map;
import java.util.TreeMap;

class RestrictedPermutations {
  private static Random rng = new Random();

  public static void rPermute(Integer[] a, int d) {
    for (int i = 0; i < d; i++) {
      int j = a.length-1;
      while (j > 0) {
        double u = rng.nextDouble();
        if (u < 0.61) { // related to golden ratio phi; more decimals may help
          j -= 1;
        } else {
          int t = a[j];
          a[j] = a[j-1];
          a[j-1] = t;
          j -= 2;
        }
      }
    }
  }

  public static void main(String[] args) {
    int numTests = Integer.parseInt(args[0]);
    int d = 2;
    Map<String,Integer> count = new TreeMap<String,Integer>();
    for (int t = 0; t < numTests; t++) {
      Integer[] a = {1,2,3,4,5};
      rPermute(a,d);
      // convert a to String for storage in Map
      String s = "(";
      for (int i = 0; i < a.length-1; i++) {
        s += a[i] + ",";
      }
      s += a[a.length-1] + ")";
      int c = count.containsKey(s) ? count.get(s) : 0;
      count.put(s,c+1);
    }
    for (String k : count.keySet()) {
      System.out.println(k + ": " + count.get(k));
    }
  }

}