README.md

josephus

This coding exercise solves a variant of the Josephus problem.

A circle of n elements is iteratively reduced to a single element by discarding every k-th element. The initial elements of the circle are labelled by their ordinal positions 1 to n in the direction of counting.

For example, suppose that n is 4 and k is 2. The initial elements in the circle can be represented as

1	2	3	4

A cursor is imagined to exist between the elements, initially located to the left of the first element. This extended circle can be visualized as:

*	1	2	3	4

where * signifies the cursor.

The first move for k = 2 is the advancement of the cursor by two positions:

1	2	*	3	4

and then the removal of the element before the cursor:

1	*	3	4

(Alternatively, we could advance the cursor by k - 1 positions and then remove the element after the cursor.)

Advancing two more positions:

1	3	4	*

Removing element before the cursor:

1	3	*

Now, since this is a circular data structure, this is equivalent to:

*	1	3

and when we advance two more positions (which happens to be the size of the remaining circle), we are back to:

1	3	*

i.e., nothing has changed.

After deleting the element before the cursor, the iterative reduction is complete:

1	*

Thus, Survivor(4, 2) = 1.

Two trivial cases stand out: k = 0 and k = 1.

When k = 0, the cursor never advances, so the highest numbered element is removed at each step. Hence, Survivor(n, 0) = 1.

When k = 1, the cursor advances over the lowest numbered element, which is then deleted. Hence Survivor(n, 1) = n.

It is also easy to show that Survivor(n, n) = Survivor(n - 1, n) = 1 + Survivor(n - 2, n)

Simulated Solution

To simulate a solution to this problem, start with a Circle of labelled ordinal positions from 1 to n. The simplest means of reducing this to a smaller problem parallels the example above by iterative reduction to the next smaller circle with special care to keep the cursor fixed before the first element of the Circle:

• let k' = k mod n be the effective step
• if k' is zero, generate a new Circle from the first n - 1 elements
• otherwise, move the first k - 1 elements to the end and skip the k-th element

This Circle is used by the NaiveSurvivor class to find the survivor index. The special cases noted above are used for testing, but not by NaiveSurvivor directly.

The problem with this solution is that the shuffling needed to keep the cursor at the beginning of the circular list is very expensive.

Nevertheless, this constitutes a reference implementation that can be used to verify the correctness of more complicated versions.

Improved Solution

If k ≤ n, we can perform multiple deletions in one step. Start by partitioning the elements of the circle into groups of size k and then drop the last element of each group. The tricky part occurs when the last group contains fewer than k elements.

FasterSurvivor uses this technique with Circle. For very large n, the performance difference is not as prounounced as expected:

Solver	Size	Step	Time
NaiveSurvivor	10000000	2	8732 ms
FasterSurvivor	10000000	2	4953 ms

Recursive Solution

Simulating the motion of a Circle to solve this problem is clearly overkill. Survivor(n, k) can be solved in terms of the next smaller problem by merely relabelling the indexes.

1	2	...	k-1	k	k+1	...	n
n-k+1	n-k+2	...	n-1	deleted	1	...	n-k

That is,

Survivor(n, k) = Survivor(n-1, k) + k,          if Survivor(n-1, k) <= n - k
Survivor(n, k) = Survivor(n-1, k) + k - n,      if Survivor(n-1, k) > n - k

or

Survivor(1, k) = 1
Survivor(n, k) = ((Survivor(n-1, k) + k - 1) mod n) + 1

This is a primitive recursive formula, so it is guaranteed to terminate; but it is not tail recursive. Consequently, this formula is expected to stack overflow prematurely. This can be avoided by converting the recursion to an iteration. After some manipulation and legerdemain, this reduces to

(1 to size).fold(1) {
  case (value, n) =>
    ((value + step - 1) mod n) + 1
}

which produces the same results as the reference implementation, but with much less time and bounded memory.

Solver	Size	Step	Time
NaiveSurvivor	10000000	2	8732 ms
FasterSurvivor	10000000	2	4953 ms
Survivor	10000000	2	178 ms

It's possible to optimize this even further by using the same trick that was used in the Improved Solution section above for k ≤ n. Now, all the multiples of k are deleted from the original circle, with some tricky left overs at the right edge. For example, with n = 11 and k = 4, the decoder ring looks like this:

1	2	3	4	5	6	7	8	9	10	11
4	5	6	-	7	8	9	-	1	2	3

The last (partial) group starts at index (n/k)k---8 in this case---where integer division is assumed. Thus,

Survivor(n, k) = Survivor(n - n/k, k) + (n/k)k,     if Survivor(n - n/k, k) <= n mod k

When the numbering wraps around on the bottom, we need to boost the output by one every k - 1 entries, starting at (n mod k) + 1; so if Survivor(n - n/k, k) > n mod k:

Survivor(n, k) = (Survivor(n - n/k, k) - (n mod k) - 1) + 1 + (Survivor(n - n/k, k) - (n mod k) - 1)/(k-1)
               = (k * ((Survivor(n - n/k, k) - (n mod k) - 1))/(k-1) + 1

which produces the same results as the reference implementation, but with much less time and bounded memory.

Solver	Size	Step	Time
NaiveSurvivor	10000000	2	8732 ms
FasterSurvivor	10000000	2	4953 ms
Survivor	10000000	2	178 ms
Survivor (revised)	10000000	2	2 ms

Now that's more like it!

Running

sbt "run size step"

where size is the initial size of the circle and step is the fixed number of positions to count off starting at zero. size should be positive. The resulting survivor position is 1-based.