## Posts Tagged ‘sorting’

### Puzzle : Nice puzzle on sorting!

August 11, 2010

Well,

The title is directly taken from a mail by Ramprasad, so is the puzzle :-). He in turn gives credit to Prof Meena for this nice puzzle.

Problem : Pasting the problem verbatim here from Ramprasad’s email.

Meena had given us this puzzle during this workshop and I found it
incredibly interesting.

Suppose we have 13 cards arranged in descending order: K Q J 10 9 8 7
6 5 4 3 2 1
At any move, you can take a substring of cards and move it elsewhere.
For example, 10 9 8 could be move to the right by 2 steps to get K Q
J 7 6 10 9 8 5 4 3 2 1. The goal is to minimize the number of such
moves.

It must be absolutely obvious that you can do it with 12 moves but the
very fact that I’m asking you this question means that you can do
better. How many moves do you need? ðŸ™‚

Now the obvious extension to this problem is to generalize this technique for any n.

Solution : Well, I got the solution myself, but not without the hint. And the hint is as follows.

See the snipped mailing conversation between me and Ramprasad.

> Is it the case that number of inversions should strictly decrease?

>

Actually inversions was my first guess towards proving the lower

bound. It is a very good guess but I do not know if it absolutely

necessary that inversions must go down at each step. But perhaps your

guess is right… maybe it must go down monotonically at each step.

[For those who do not know what an inversion is, a pair of indices

(i,j) is called an inversion if i < j and a_i > a_j ]

My first attempt towards the lower bound was that initially there are

n-choose-2 inversions and finally there are none. If we can show that

each move can remove only n inversions then we’d sort of have a good

lower bound. But unfortunately this is not true: take n n-1 n-2 … 1

and shift the first half and put it after the 2nd half to get n/2

n/2-1 … 1 n n-1 n-2… n/2+1. This process actually gets rid of

O(n^2) inversions and my proof attempt broke.

However there is a very similar parameter that can be used to argue

about the lower bound. And this parameter is called ‘descents’. These

are indices i such that a_i > a_{i+1}. Now clearly to begin with there

are (n-1) descents.

Claim: At every step, you can reduce the number of descents by at most

2. [prove it]

This immediately gives a (n-1)/2 lower bound. But do all steps

actually reduce the number of descents by 2? No! Take for example, the

first move. It can reduce the descents only by 1! And similarly the

last step. Thus you get the required lower bound of (n+1)/2.

And, this lower bound can be achieved!

```---
```

Well, there you go! It was sufficient for me to get the trick after this hint. The claim is easy to prove

hence I am leaving it for interested readers ( if any ðŸ™‚ ).

The sequence for 13 cards is as follows :

```K Q J 10 9 8 7 6 5 4 3 2 1
K 6 5 4 3 2 1 Q J 10 9 8 7
1 Q K 6 5 4 3 2 J 10 9 8 7
1 2 J Q K 6 5 4 3 10 9 8 7
1 2 3 10 J Q K 6 5 4 9 8 7
1 2 3 4 9 10 J Q K 6 5 8 7
1 2 3 4 5 8 9 10 J Q K 6 7
1 2 3 4 5 6 7 8 9 10 J Q K```

In seven steps!! Only, seven steps are required, as mentioned earlier by Ramprasad.

In general, for any decreasing sequence

n n-1 n-2 …. 2 1
First move would be to move the second half of the array just after n.

n n/2-1 n/2 -2 …. 1 n-1 n-2 …n/2

Now, we maintain the invariant that the subarray before and including n, does not have a descent.

On the right of n, find an element a_i such that a_i < a_(i+1) ( does not have a descent ).

Move these two elements to the left of n to maintain the invariant. In fact, one can put a_i at the a_i th

place from the left. Again, it is easy to prove the correctness of this strategy.
Copying verbatim interesting notes from Ramprasad :
Some general notes on this problem. The question of finding the
shortest set of moves to sort any given permutation is not known to be
in P, and not known to be NP-hard or anything either. There are
approximation algorithms for this known and that is how Meena asked me
this question. One of the participants of the workshop had worked on
this problem and we started discussing this when we spoke about his
problem.

One question I asked Meena was, is the decreasing sequence the worst
sequence? And turns out, surprisingly, the answer is no! People ran
some computer analysis and they figured a permutation on 15 and
another on 17 which took more than 8 and 9 respectively! But people
are making the conjecture that for sufficiently large n, the worst
sequence is the decreasing sequence [and some others also believe that
‘large n’ is any n > 17 ðŸ™‚ ].

— Saurabh Joshi