**4. Non-Adaptive Sorting: Batcher's Algorithm**

The weighing algorithm we found was non-adaptive. We
were able to spell out in advance exactly which side each of the balance each
coin would be.

The weighing problem is a special case of the sorting
problem where we have the special case that all but one coin or key has the
same weight, while in the usual sorting problem, all weights are different.

So we can ask in the sorting problem: can we find a
non-adaptive algorithm for it: one for which there are a fixed set of
comparisons; the keys are fed in unsorted, and come out after a predetermined
set of operations, all sorted.

We will produce just such an algorithm, in which the basic
operation is a "compare and switch": given two keys x and y, located
in boxes a and b, we compare x and y and put the larger of the two in a and the
smaller in b.

We can describe such an **operation** by setting

New a = max(
old a, old b), New b = old a + old b
– New a.

An algorithm of this kind consists of a sequence of
operations every one of which has this form, so it can be described completely
by describing which pairs (a,b) are to be treated in this way in each step.

There is a very simple and slick way to do this, called
Batcher's algorithm after its discoverer. It is based on the following simple
and wonderful fact.

**If you have a list
of keys arranged from left to right, and you sort the left and right halves of
the list separately, and then sort the keys in even positions on the list, and
in odd positions separately, **

** **

** Then
all you need do is compare and switch each even key from the left with the odd
key immediately to its right, and you will have completely sorted the whole
list.**

Consider the following example: we start with 8 keys
ordered as follows

1 7 3 4 5 2 8 6

Sorting the first and last halves separately changes this
to

**1**
3 **4**
7 ** 2** 5** 6** 8

Sorting the odd and
even placed halves separately and in place changes this to

**1**
3 ** 2** 5 **4**
7 ** 6** 8,

and you will notice that
comparison and switching the second and third, fourth and fifth and sixth and
seventh finishes the sorting job,

**First we will ask,
why is this so?** Then we will see how to exploit this fact to create a
sorting algorithm.

Consider a key in an even position, say the 2kth after this
first last and odd even sorts, and suppose when we compare and switch the
smaller key goes to the left.

If we have 2m keys all together, then we know that there
are k-1 even position keys to its left and m-k even position keys to its right.

Each even position key had a unique odd position key to its
immediate left that had to be smaller than it after the first round of left
right sorts: there are therefore at least k odd position keys that are smaller
than the key in position 2k which makes 2k-1 smaller keys all together, and so after a complete sort this key
belongs at position 2k or to its right.

Similarly each odd position key has an even position key
that is larger immediately to its right after the left right sorts, so that it
must, after a complete sort be in its present position or to its left.

If we were to do a similar count of keys that belong to the
right of the 2kth key we find that each even position key except the top key on
the left and the top key on the right has an odd key to its right.

If the keys were completely sorted we would know the exact
sorted position of the key in the 2kth position and the same count would differ
by exactly one; each even position key has an odd larger key except for the
last key.

We can conclude that the count of keys to the right of the
2kth key in our case is at least 2m-2k-1 instead of the 2m-2k we would have if
the list were completely sorted.

The extra one comes from the fact that in our case there is
a top key on the left and on the right, while in the sorted case there is only
one rather than two top keys.

This means our 2kth position key after the left right and
odd even sorts is either where it belongs or one place to the left of where it
belongs.

An odd position key is similarly either where it belongs or
one place to the right of it.

Thus comparing and switching each even key with the odd key
to its right will finish the sorting job,

**What algorithm can we make from this fact?**

This fact tells us how to sort 2m keys if we can sort m
keys. We can sort left and right sets of m keys, then odd and even, and then do
the last round of even odd comparisons and switches.

The algorithm is even easier than this sounds, because
after we sort the left and right halves of the keys, the left and right halves
of both the even and the odd keys are already sorted, so we do not have sort
lefts and rights in sorting the even and odd keys..

Thus if we write the algorithm schematically as

**Sort Algorithm for 2m keys = Sort left half and Sort
right half; Then Merge the two halves**,

we can describe the Merge
step as

**Merge 2m keys =
Merge m odd keys and m even keys. Then compare and switch each even key
with odd key to its right.**

** **

This is a complete description of the algorithm but it
leaves me at least a bit confused as to what it actually does. So let us dig a
bit deeper.

A comparison and switch of two keys can be described by an
ordered pair of key positions which can be described by a directed edge of a
directed graph. The edge will point from the position that is to get the
smaller key to the one that gets the larger one.

Suppose for example, we want to sort our initially unsorted
keys into pairs. The algorithm for this is not what we have been describing but
it is easier. You compare and switch them.

Suppose now we want to sort four keys in positions p1 p2 p3
and p4

We first sort the left and right pairs which can be
described by the edges (p1, p2) and
(p3, p4).

Then we merge by sorting the odds and evens with edges

(p1, p3) and (p2, p4)

and complete the job with
(p2,p3).

We will sort 8 keys this way, and leave the 16 and 32 cases
to you.

To sort 8 we first apply the sort 4 algorithm to the first
half and second half of our keys, separately. This takes the same three rounds
of comparison switches just described.

Sort pairs

(p1, p2) (p3, p4) (p5, p6) (p7, p8)

Merge to fours

(p1, p3) ( p2,
p4) (p5, p7) (p6, p8)

(p2, p3) (p6, p7)

Now we repeat the merge 4
steps now on odds and evens separately

(p1, p5) (p2, p6) (p3, p7) (p4, p8)

(p3, p5) (p4, p6)

and finally compare and
switch evens with next odds:

(p2, p3), (p4, p5), (p6, p7)

Notice that repeating steps on evens and odds, has the
effect of doubling the distance between the front and back of each edge and
making each edge into two parallel overlapping edges.

One nice feature of this algorithm which is perhaps not so
nice for you, is that it is easy to implement this algorithm on a spreadsheet,
where you can put the original keys in
the first column, put each round of comparison switches in subsequent columns
(using say b1= max(a1,a2), b2=a1-b1+a2 for a comparison and switch). You can
then watch your keys get sorted,

Exercises: 1. How many rounds of comparison switches (not using
the same key twice in any one round) are required to sort 2^{k} keys?
How many comparison switches are used in this algorithm? You need not get the
answer exactly, but will get extra credit for an exact answer.

2. Draw a graph of each of the
rounds needed to sort 16 keys here.

3. Set up a spreadsheet that
will Batcher's algorithm for 32 keys,

hint: you can do this with
entering only a small number of entries and doing judicious copying.

4. Exactly how many
comparison switches are needed here for 16 keys? Compare that with the number
needed with say insert or simple merge sort.

This sort is very nice in that there is no excess movement
of keys, no wastage of space, easy to do in parallel, but unfortunately for
huge sorts it takes too many rounds.

There is an incredibly complicated algorithm (discovered by
Ajtai, Komlos and Szemeredi) that accomplishes the same task with cnlog n
comparison and switches, but unfortunately the c here is very large.