Algorithm for Shuffling a Linked List in n log n time - algorithm

Question

I'm trying to shuffle a linked list using a divide-and-conquer algorithm that randomly shuffles a linked list in linearithmic (n log n) time and logarithmic (log n) extra space.
I'm aware that I can do a Knuth shuffle similar to that could be used in a simple array of values, but I'm not sure how I would do this with divide-and-conquer. What I mean is, what am I actually dividing? Do I just divide to each individual node in the list and then randomly assemble the list back together using some random value?
Or do I give each node a random number and then do a mergesort on the nodes based on the random numbers?

Answer 1

What about the following? Perform the same procedure as merge sort. When merging, instead of selecting an element (one-by-one) from the two lists in sorted order, flip a coin. Choose whether to pick an element from the first or from the second list based on the result of the coin flip.
Edit (2022-01-12): As GA1 points out in the answer below, this algorithm doesn't produce a permutation uniformly at random.
Algorithm.
shuffle(list):
if list contains a single element
return list
list1,list2 = [],[]
while list not empty:
move front element from list to list1
if list not empty: move front element from list to list2
shuffle(list1)
shuffle(list2)
if length(list2) < length(list1):
i = pick a number uniformly at random in [0..length(list2)]
insert a dummy node into list2 at location i
# merge
while list1 and list2 are not empty:
if coin flip is Heads:
move front element from list1 to list
else:
move front element from list2 to list
if list1 not empty: append list1 to list
if list2 not empty: append list2 to list
remove the dummy node from list
The key point for space is that splitting the list into two does not require any extra space. The only extra space we need is to maintain log n elements on the stack during recursion.
The point with the dummy node is to realize that inserting and removing a dummy element keeps the distribution of the elements uniform.
Edit (2022-01-12): As Riley points out in the comments, the analysis below is flawed.
Analysis.
Why is the distribution uniform? After the final merge, the probability P_i(n) of any given number ending up in the position i is as follows. Either it was:
in the i-th place in its own list, and the list won the coin toss the first i times, the probability of this is 1/2^i;
in the i-1-st place in its own list, and the list won the coin toss i-1 times including the last one and lost once, the probability of this is (i-1) choose 1 times 1/2^i;
in the i-2-nd place in its own list, and the list won the coin toss i-2 times including the last one and lost twice, the probability of this is (i-1) choose 2 times 1/2^i;
and so on.
So the probability
P_i(n) = \sum_{j=0}^{i-1} (i-1 choose j) * 1/2^i * P_j(n/2).
Inductively, you can show that P_i(n) = 1/n. I let you verify the base case and assume that P_j(n/2) = 2/n. The term \sum_{j=0}^{i-1} (i-1 choose j) is exactly the number of i-1-bit binary numbers, i.e. 2^{i-1}. So we get
P_i(n) = \sum_{j=0}^{i-1} (i-1 choose j) * 1/2^i * 2/n
= 2/n * 1/2^i * \sum_{j=0}^{i-1} (i-1 choose j)
= 1/n * 1/2^{i-1} * 2^{i-1}
= 1/n
I hope this makes sense. The only assumption we need is that n is even, and that the two lists are shuffled uniformly. This is achieved by adding (and then removing) the dummy node.
P.S. My original intuition was nowhere near rigorous, but I list it just in case. Imagine we assign numbers between 1 and n at random to the elements of the list. And now we run a merge sort with respect to these numbers. At any given step of the merge, it needs to decide which of the heads of the two lists is smaller. But the probability of one being greater than the other should be exactly 1/2, so we can simulate this by flipping a coin.
P.P.S. Is there a way to embed LaTeX here?

Answer 2

Code
Up shuffle approach
This (lua) version is improved from foxcub's answer to remove the need of dummy nodes.
In order to slightly simplify the code in this answer, this version suppose that your lists know their sizes. In the event they don't, you can always find it in O(n) time, but even better: a few simple adaptation in the code can be done to not require to compute it beforehand (like subdividing one over two instead of first and second half).
function listUpShuffle (l)
local lsz = #l
if lsz <= 1 then return l end
local lsz2 = math.floor(lsz/2)
local l1, l2 = {}, {}
for k = 1, lsz2 do l1[#l1+1] = l[k] end
for k = lsz2+1, lsz do l2[#l2+1] = l[k] end
l1 = listUpShuffle(l1)
l2 = listUpShuffle(l2)
local res = {}
local i, j = 1, 1
while i <= #l1 or j <= #l2 do
local rem1, rem2 = #l1-i+1, #l2-j+1
if math.random() < rem1/(rem1+rem2) then
res[#res+1] = l1[i]
i = i+1
else
res[#res+1] = l2[j]
j = j+1
end
end
return res
end
To avoid using dummy nodes, you have to compensate for the fact that the two intermediate lists can have different lengths by varying the probability to choose in each list. This is done by testing a [0,1] uniform random number against the ratio of nodes popped from the first list over the total number of node popped (in the two lists).
Down shuffle approach
You can also shuffle while you subdivide recursively, which in my humble tests showed slightly (but consistently) better performance. It might come from the fewer instructions, or on the other hand it might have appeared due to cache warmup in luajit, so you will have to profile for your use cases.
function listDownShuffle (l)
local lsz = #l
if lsz <= 1 then return l end
local lsz2 = math.floor(lsz/2)
local l1, l2 = {}, {}
for i = 1, lsz do
local rem1, rem2 = lsz2-#l1, lsz-lsz2-#l2
if math.random() < rem1/(rem1+rem2) then
l1[#l1+1] = l[i]
else
l2[#l2+1] = l[i]
end
end
l1 = listDownShuffle(l1)
l2 = listDownShuffle(l2)
local res = {}
for i = 1, #l1 do res[#res+1] = l1[i] end
for i = 1, #l2 do res[#res+1] = l2[i] end
return res
end
Tests
The full source is in my listShuffle.lua Gist.
It contains code that, when executed, prints a matrix representing, for each element of the input list, the number of times it appears at each position of the output list, after a specified number of run. A fairly uniform matrix 'show' the uniformity of the distribution of characters, hence the uniformity of the shuffle.
Here is an example run with 1000000 iteration using a (non power of two) 3 element list :
>> luajit listShuffle.lua 1000000 3
Up shuffle bias matrix:
333331 332782 333887
333377 333655 332968
333292 333563 333145
Down shuffle bias matrix:
333120 333521 333359
333435 333088 333477
333445 333391 333164

Answer 3

You can actually do better than that: the best list shuffle algorithm is O(n log n) time and just O(1) space. (You can also shuffle in O(n) time and O(n) space by constructing a pointer array for the list, shuffling it in place using Knuth and re-threading the list accordingly.)
Complexity proof
To see why O(n log n) time is minimal for O(1) space, observe that:
With O(1) space, updating the successor of an arbitrary list element necessarily takes O(n) time.
Wlog, you can assume that whenever you update one element, you also update all the other elements (leaving them unchanged if you wish), as this also takes just O(n) time.
With O(1) space, there are at most O(1) elements to choose from for the successor of any element you're updating (which specific elements these are will obviously depend on the algorithm).
Therefore, a single O(n) time update of all the elements could result in at most c^n different list permutations.
Since there are n! = O(n^n) = O(c^(n log n)) possible list permutations, you require at least O(log n) updates.
Linked-list data structure (because Python)
import collections
class Cons(collections.Sequence):
def __init__(self, head, tail=None):
self.head = head
self.tail = tail
def __getitem__(self, index):
current, n = self, index
while n > 0:
if isinstance(current, Cons):
current, n = current.tail, n - 1
else:
raise ValueError("Out of bounds index [{0}]".format(index))
return current
def __len__(self):
current, length = self, 0
while isinstance(current, Cons):
current, length = current.tail, length + 1
return length
def __repr__(self):
current, rep = self, []
while isinstance(current, Cons):
rep.extend((str(current.head), "::"))
current = current.tail
rep.append(str(current))
return "".join(rep)
Merge-style algorithm
Here is an O(n log n) time and O(1) space algorithm based on iterative merge sort. The basic idea is simple: shuffle the left half, then the right half, then merge them by randomly selecting from the two lists. Two things worth noting:
By making the algorithm iterative rather than recursive, and returning a pointer to the new last element at the end of every merge step, we reduce the space requirement to O(1) while keeping the time cost minimal.
To make sure that all possibilities are equally likely for all input sizes, we use probabilities from the Gilbert–Shannon–Reeds model riffle shuffle when merging (see http://en.wikipedia.org/wiki/Gilbert%E2%80%93Shannon%E2%80%93Reeds_model).
import random
def riffle_lists(head, list1, len1, list2, len2):
"""Riffle shuffle two sublists in place. Returns the new last element."""
for _ in range(len1 + len2):
if random.random() < (len1 / (len1 + len2)):
next, list1, len1 = list1, list1.tail, len1 - 1
else:
next, list2, len2 = list2, list2.tail, len2 - 1
head.tail, head = next, next
head.tail = list2
return head
def shuffle_list(list):
"""Shuffle a list in place using an iterative merge-style algorithm."""
dummy = Cons(None, list)
i, n = 1, len(list)
while (i < n):
head, nleft = dummy, n
while (nleft > i):
head = riffle_lists(head, head[1], i, head[i + 1], min(i, nleft - i))
nleft -= 2 * i
i *= 2
return dummy[1]
Another algorithm
Another interesting O(n log n) algorithm that produces not-quite-uniform shuffles involves simply riffle shuffling the list 3/2 log_2(n) times. As described in http://en.wikipedia.org/wiki/Gilbert%E2%80%93Shannon%E2%80%93Reeds_model, this leaves only a constant number of bits of information.

Answer 4

I'd say, that foxcub's answer is wrong. To prove that I will introduce a helpful definition for a perfectly shuffled list (call it array or sequence or whatever you want).
Definition: Assume we have a List L containing the elements a1, a2 ... an and the indexes 1, 2, 3..... n. If we expose the L to a shuffle operation (to which internals we have no access) L is perfectly shuffled if and only if by knowing indexes of some k (k< n) elements we can't deduce the indexes of remaining n-k elements. That is the remaining n-k elements are equally probable to be revealed at any of the remaining n-k indexes.
Example: if we have a four element list [a, b, c, d] and after shuffling it, we know that its first element is a ([a, .., .., ..]) than the probability for any of the elements b, c, d to occur in, let's say, the third cell equals 1/3.
Now, the smallest list for which the algorithm does not fulfil the definition has three elements. But the algorithm converts it to a 4-element list anyway, so we will try to show its incorrectness for a 4-element list.
Consider an input L = [a, b, c, d]Following the first run of the algorithm the L will be divided into l1 = [a, c] and l2 = [b, d]. After shuffling these two sublists (but before merging into the four-element result) we can get four equally probable 2-elements lists:
l1shuffled = [a , c] l2shuffled = [b , d]
l1shuffled = [a , c] l2shuffled = [d , b]
l1shuffled = [c , a] l2shuffled = [b , d]
l1shuffled = [c , a] l2shuffled = [d , b]
Now try to answer two questions.
1. What is the probability that after merging into the final result a will be the first element of the list.
Simply enough, we can see that only two of the four pairs above (again, equally probable) can give such a result (p1 = 1/2). For each of these pairs heads must be drawed during first flipping in the merge routine (p2 = 1/2). Thus the probability for having a as the first element of the Lshuffled is p = p1*p2 = 1/4, which is correct.
2. Knowing that a is on the first position of the Lshuffled, what is the probability of having c (we could as well choose b or d without loss of generality) on the second position of the Lshuffled
Now, according to the above definition of a perfectly shuffled list, the answer should be 1/3, since there are three numbers to put in the three remaining cells in the list
Let's see if the algorithm assures that.
After choosing 1 as the first element of the Lshuffled we would now have either:
l1shuffled = [c] l2shuffled = [b, d]
or:
l1shuffled = [c] l2shuffled = [d, b]
The probability of choosing 3 in both cases is equal to the probability of flipping heads (p3 = 1/2), thus the probability of having 3 as the second element of Lshuffled, when knowing that the first element element of Lshuffled is 1 equals 1/2. 1/2 != 1/3 which ends the proof of the incorrectness of the algorithm.
The interesting part is that the algorithm fullfils the necessary (but not sufficient) condition for a perfect shuffle, namely:
Given a list of n elements, for every index k (<n), for every element ak: after shuffling the list m times, if we have counted the times when ak occured on the k index, this count will tend to m/n by probability, with m tending to infinity.

Answer 5

Here is one possible solution:
#include <stdlib.h>
typedef struct node_s {
struct node_s * next;
int data;
} node_s, *node_p;
void shuffle_helper( node_p first, node_p last ) {
static const int half = RAND_MAX / 2;
while( (first != last) && (first->next != last) ) {
node_p firsts[2] = {0, 0};
node_p *lasts[2] = {0, 0};
int counts[2] = {0, 0}, lesser;
while( first != last ) {
int choice = (rand() <= half);
node_p next = first->next;
first->next = firsts[choice];
if( !lasts[choice] ) lasts[choice] = &(first->next);
++counts[choice];
first = next;
}
lesser = (counts[0] < counts[1]);
if( !counts[lesser] ) {
first = firsts[!lesser];
*(lasts[!lesser]) = last;
continue;
}
*(lasts[0]) = firsts[1];
*(lasts[1]) = last;
shuffle_helper( firsts[lesser], firsts[!lesser] );
first = firsts[!lesser];
last = *(lasts[!lesser]);
}
}
void shuffle_list( node_p thelist ) { shuffle_helper( thelist, NULL ); }
This is basically quicksort, but with no pivot, and with random partitioning.
The outer while loop replaces a recursive call.
The inner while loop randomly moves each element into one of two sublists.
After the inner while loop, we connect the sublists to one another.
Then, we recurse on the smaller sublist, and loop on the larger.
Since the smaller sublist can never be more than half the size of the initial list, the worst case depth of recursion is the log base two of the number of elements. The amount of memory needed is O(1) times the depth of recursion.
The average runtime, and number of calls to rand() is O(N log N).
More precise runtime analysis requires an understanding of the phrase "almost surely."

Answer 6

Bottom up merge sort without compares. while merging don't do any comparison just swap the elements.

Answer 7

You could traverse over the list, randomly generating 0 or 1 at each node.
If it is 1, remove the node and place it as the first node of the list.
If its is 0, do nothing.
loop this until you reach the end of the list.