Young Tableau

Hi All,
The other day me, sanjeet, and pranjal were discussing a few questions ( or puzzles ) which are likely to be asked in technical interviews. One of the question that was asked to someone in google was about finding out a specific element in an n x n Young Tableau.

Let me first define, what a young tableau is. It is an m x n array with the following property.

for all j = 1 to n-1, A[p][i] <= A[p][i+1]
and for all , i = 1 to m-1 , A[i][q] <= A[i+1][q] It is easy to see that following statement holds for a young tableau.

1. For any sub-matrix of a young tableau, the top-left element is the smallest element within that sub-matrix.
2. For any sub-matrix of a young tableau, the bottom-right element is the largest element within that sub-matrix.

Now, the problem is to find out a given element within an n x n young tableau.

----
Method -1
----

Start with the top-right element of the tableau. Compare this element e with the element to be found x.
if e > x then
discard the column in which e appears. ( Because e is the first(smallest ) element of that column.
else
discard the row in which e appears.
end if.
Repeat this exercise until the element is found or the tableau is exhausted.

It is easy to see that time complexity of Method-1 is O(n). As at every step we will discard either row or column which will take at most 2n steps.

---------
Method - 2
---------
Do a binary search over the diagonal of the tableau. If we find the desired element x then we are done. If not, we will have two element u and v where u <> u is the bottom-right element and v is the top-left element. The elements that we will discard at each step would be at least half of the elements in the original matrix.
Here is why,

i^2 + (n-i)^2 function is minimum where i=n/2. With i=n/2 the value of the function is (n^2)/2.

Now, if we do the analysis with respect to number of elements.
T(n^2) = 2T((n^2)/4) + logn

or

T(m) = 2T(m/4) + 0.5 logm

Solution of this recurrence, turns out to be O(m^(0.5)) = O(n). So, there is no improvement over the complexity as compared to the first method. Besides, this second method fails when the sub-matrices that were discarded were not square. In which case, the remaining two sub-matrices will not be square either.
----
Method 3
------
Compare the central element e with the desired element x.
if e > x then discard the bottom-right quadrant of the tableau.
else discard the top-left quadrant of the tableau.

Now , we are left with three sub-matrices of the size n/2 x n/2.

The recurrence would be

T(n) = 3T(n/2) + O(1).

Solving it would give us n^(log_2 3) = n^1.585. It turns out that even though in Method-1 you discard n elements and in this method you discard (n^2)/4, the complexity does not favour Method-3. Anyways, it was a good exercise to do this analysis.

Necklace puzzle - yet another solution

Hi all,

I am posting here another solution to "Necklace Puzzle". The solution is by my friend Vaibhav Gupta

-------

Let A and B are the 2 necklaces with N gemstones. A and B has three types of
gemstones which are equal in number.

A = A[1] A[2] ... A[N]
B = B[1] B[2] ... B[N]

Where A[i] is the ith gemstone in Necklace A.
B[i] is the ith gemstone in Necklace B.

Let Color(A[i]) represents the type of the gemstone A[i].

We construct a bipartite graph G = (X,Y,E), with the following condition:

For each gemstone A[j] in A we create a node X_j in X. So |X| = N
Foe each gemstone B[j] in B we create a node Y_j in Y. So |Y| = N
E is the edges set from node in X to node in Y.

E = { (u,v) | u \in X and v \in Y and Color(u) = Color(v)}

|E| = (N^2) / 3 since each node has exactly N/3 edges.

We define weight to each edge in E:
Weight(e) = orientation of the string.
OR Mathematically
Weight((X_i,Y_j)) = i - j if i>=j
= N+i-j otherwise

0 <= Weight((X_i,Y_j)) <= N-1 since 1<=i<=N, 1<=j<=N So there are N possible values of weight.

Now we just need to prove that there are atleast N/3 edges of same weight.
This can be proved by pigon hole principle. There are (N^2) / 3 edges and
each edge has N possible values of weight so there must exist a set of
ceil(|E|/N) edges of same weight

|E| / N >= N/3

Hence proved.

For general case :

For 3 gemstone types
( a^2 + b^2 + c^2 ) / (a + b + c)

Where a is the number of gemstone of type 1.
b is the number of gemstone of type 2.
c is the number of gemstone of type 3.

For m gemstone types:
(N_1^2 + N_2^2 + N_3^2 ... + N_m^2) / (N_1+N_2+N_3 ..+N_m)

where N_i is the number of gemstone of type i.

---------

Quote by John Von Neumann

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is"
-- John Von Neumann

I saw this quote as saket's status message in google talk.

Necklace Theorem - Solutions to puzzles

Hi All,
So I am back again after some gap. This post will describe the solutions to the two puzzle that I posed as part of my prequel post "Necklace Theorem". For the sake of clarity I am copying the puzzles again over here.

1) Given a point set of size n in 2D, prove that there would exist a pair of orthogonal lines which would divide the point set in such a way that in each quadrant number of points are at most n/4.
As shown in the figure on the left, let us consider a baseline B. The circle represent the point set we are interested in. Refer to the prequel of this post and convince yourself that it is possible to have a line in any given direction which divides the point set into two equal half. Hence, it is also possible to have two orthogonal lines both of which will divide the point set into two equal half. ( When I say, two equal half, I mean that division in two portion satisfying the constraint "less or equals n/2" ). As both p and q divides the point set into two equal half, the region which is captured on top-left and bottom-right will be equal ( call it x, or to be more precise "less or equal x" ). Similarly, tom-right and bottom-left would be equal ( call it y). Let us have a function f(x,y) = (x-y) = v. By rotating p and q , 90 degrees we will have function value exactly negative, as x and y would swap their places. From intermediate value theorem we can argue that there will be a configuration where x = y = n/4.

2) Given a point set of size n in 2D, prove that there would exist 3 concurrent lines which would divide the point set in such a way that in each segment the number of points are at most n/6.

With respect to the baseline B, we can have two lines p and q, each dividing the point set into equal half and region trapped as shown in figure would be equal to n/6 on each side. ( n/6 is written outside the circle to make it readable ). Now, as shown in the prequel post, we can have a line r which will divide the two point sets ( each of size n/3 ) into two equal half. This line r need not be concurrent with p and q. Let the triangle trapped between these three lines is equal to A. As we rotate this configuration 180 degree the area will become -A. Again, from intermediate value theorem we can state that area will be 0 at some configuration making r concurrent to p and q. Thus, each segment will have size n/6.

Necklace Puzzle

Hi All,
So I am here back again with a new interesting puzzle ( courtesy Vaibhav Gupta ). I am copying his mail verbatim below followed by the solution that I have in mind.
---------- Puzzle ------
NECKLACE PUZZLE #2

There are two circular necklaces with same number of three types of gemstones,
but the gemstones may be strung in different order along the necklace.

Now the two necklaces are placed on one top of the other so that the
gems are aligned
one above the other. Count the number of locations where the pair of
gems (one from
the lower necklace, one from the upper) is of the same type.
Since one necklace can be rotated relatively to the other, there are
many orientations
in which the count of matching gems can be taken.

Prove that there is at least one orientation where the number of
matches is at least
a certain number v.

For the special case of equal numbers of each kind of gem, v = N/3,
where N = total number of gems in one necklace (and 3 kinds of gems).

1. prove this for the special case.
2. find, with proof, value of v in the general case (with m kinds of
gems not necessarily equal in number).

some clarifications:
1. the necklace is circular i.e. its a simple closed loop. so given
two necklaces RGB and RBG,
the second can be rotated wrt first to give three orientations (but
this small example
is unusual because the number of matches is same in all 3 orientations):

RGB
RBG 1 matches

RGB
GRB 1 matches

RGB
BGR 1 matches

so v = 1

2. "there is at least one orientation where the number of matches is
at least v"
label the orientations from 1 to N. denote the number of matches for
orientation i as m[i].
then " there exists an i such that m[i] >= v" is the more formal way
of saying this.


-Vaibhav.
------------------
Well, indeed an interesting puzzle. Intially for a few minutes I thought whether it can be proved by contradiction or not. I tried to see whether I am able to hit a contradiction by imagining that "v is always less than N/3". Within a few minutes I realized that this is not going to work. As I remembered the lessons from Prof Sundar Vishvanathan , I thought of trying to prove it by induction. Prof Sundar always used to say that most of the proofs can be given by the principle of mathematical induction. This time I did not bank too much on this options as I learnt it from my earlier experience as I mentioned in my previous post. The solution of this proof also along the line which was taught to me by Prof Sundar.

The proof here makes use of the fact that if expected value of some random variable is a then there exist a possibility where the random variables has a values at least a.

Let us assume that first necklace is fixed and the second necklace is rotated with respect to the first one. Let us define a random variable X such that
X_i = 1 ( if ith gem of first necklace matches with the second one )
X_i = 0 Otherwise.

The ordering in the second necklace can be anything so assuming a uniform distribution there.
Calculating the expected value,
E(X_i) = Pr(X_i).X_i = (Number of gems having the same colour as ith gem/Total number of gems)
For 3 different kind of gems
E(X) = Sum(i=1 to N) ( E(X_i) ) = R^2/N + B^2 /N + G^2/N.

For the special case where R = B = G = N/3 we will have
E(X) = N/3. As the expected value is N/3 we can say that there is at least one configuration of second necklace which would make the number of matching gems at least N/3.

When there are m kinds of gems, not necessarily equal in number, let C[i] denote the number of gems of the same kind as i, where i varies from 1 to m.
Therefore, for general case the expected value is
E(X) = Sum (i = 1 to n ) ( C[i]^2/N )

Well, a nice puzzle indeed. As a closing remark, I have the solutions to the puzzle that I mentioned in my previous post. It will require some figures to be embedded to explain the solution in a lucid way. I will post it soon as I get the time.

Necklace Theorem

Hi all,
As mentioned in my last post, after lunch talks can be very fruitful at times. Describing a yet another fruitful chat with my friend Sambuddha Roy.

The other day he gave me an interesting problem to think about, which is formerly known as, as he told me, necklace theorem. Two thieves have stolen queen's very precious necklace. There are three kinds of gems in the necklace called R, G and B. All are even. The problem is to divide the necklace in such a way that both the thieves get equal amount of gems of each kind. The necklace can have any interleaving of the gems. ( e.g. RRGBGGRBBBG..... ) Prove that it is always sufficient to divide the necklace in equal share by at most 3 cuts.

The proof makes use of Ham Sandwich Theorem. For brevity, I will quote it as HST now on. HST for 2-dimension says that given any two point sets in a plane, there always exist a line which will cut both these point sets simultaneously into half. To say it more formally, call the point sets R and B. Then one side of line will contain at most |R|/2 points and at most |B|/2 points. The same constraints will be followed on the other side too.

If there is only 1 point set R. For a given direction we can always have a line which will divide R in two. The proof is based on intermediate value theorem. It says that for a continuous function if f(x1) = 0 and f(x2)=1 then there exist a z \in (x1,x2) such that f(z) = 1/2.

Now that we have two point sets. What we can do is we get a line for point set R. Now we evaluate how this line L behaves with respect to B. Let us define diff(L) = number of blue gems on the left of L - number of blue gems on the right of L. For initial line L, let the value of diff(L) be b. What we can do is rotate this line a bit. We know that for any given direction, we have a corresponding line L which will divide R equally. We will again evaluate wrt B. We keep on doing this until we have rotated the line for 180 degree. At this point value of diff(L) would be -b. From intermediate value theorem, we can state that there will be a line where diff(L) will be 0.

How about, Ham Sandwich theorem in 3-dimension?? Well, my friend has given me the proof as an exercise. The statement is that given 3 point sets R, G, B in R^3 ( or 3-dim space ) there always exist a plane which will divide all 3 point sets simultaneously into equal portion.

Now, we need to reduce Necklace Theorem to HST in 3D. Starting from left to right, assign every gem an index i. Plot a parametric curve ( t, t^2, t^3 ) using these indices. HST assures us that we will have a plane ax+by+cz+d=0 which will divide R, G, B equally. This plane will divide the parametric curve at , at most 3 points as at + bt^2 + ct^3 + d = 0 will have at most 3 roots. Now, it is easy to map these 3 intersection points on the curve to 3 cuts on the necklace.

Phew!!! It never occurred to me that this proof will use such weird things. Induction based method was too tempting for me to let me think in some different direction. But the story is not over yet. The proof given above is not constructive. So if we were to find an algorithm to find these 3 cuts we might have to explore all possibilities taking O(n^3) time. So another exercise would be to try to find more efficient algorithm to find these cuts.

The generalization of HST is that given k point sets in R^k there would be always be a hyperplane which would cut all these k point sets simultaneously into equal portion. Yet another generalization one can think of is that, what if there are more thieves??? Given q number of thieves and k kind of gems, what would be the upper bound on the number of cuts required?? The answer is due to dold's theorem ( Neither sambuddha nor I have read the proof ). It say that for such situation at most k(q-1) cuts would be necessary.

What if there are 3-point sets in two dimension?? Definitely a line would not suffice. An exercise would be to prove that two rays originating from an apex would be sufficient. Give an algorithm for finding them out.

Two more puzzles are :
1) Given a point set of size n in 2D, prove that there would exist a pair of orthogonal lines which would divide the point set in such a way that in each quadrant number of points are at most n/4.

2) Given a point set of size n in 2D, prove that there would exist 3 concurrent lines which would divide the point set in such a way that in each segment the number of points are at most n/6.

That's enough to tease our brain for a while :D.

After lunch talks

Hi,
It's been a long pause on my blog. I would say nothing significant to write. Even in this post not much significant but few small interesting pieces of interesting discussion.

Sometimes we go to IRL foyer after lunch and have some discussion. At times, those discussions are fruitful too. By fruitful, I mean it adds something in your knowledge. For example, today I was chatting with my friend Sambuddha . He is a very good narrator, I must admit. He was talking about his BTech advisor Manindra Agrawal who has been awarded with a Godel Prize.

He narrated an incident when Prof Manindra was out to present his work on a conference. There were a few Fields medalist , who were unable to grasp the idea of a young talent like Prof Manindra taking away the Godel prize on the same problem that they once tried to tackle. As Prof Manindra was presenting his talks, these fellows tried to interrupt him in order to embarrass him every now and then. At one point, a question was posed on such a trivial thing that Prof Manindra could not resist but say "I don't know about the undergraduates in US, but I can assure you that any undergraduate student in India would be able to understand this". Needless to say, no further interruptions occurred during his presentation :-).

There were also a few things that I came to know like a complexity class ( NP^NP ) as well as
game of Hex, Secretary Problem and things like that.

Sometimes, chatting can be more brain stimulating than the work :-)

Sudoku Solver

It is very unlikely for me to post twice in a day, yet I am back again. I had a little idle time at my disposal so I thought to put it to some use.

A couple of months back I read in Times of India about world Sudoku championship which was to happen in March 2007. In addition to the day to day sudoku puzzles they included certain variation of the puzzle too, to encourage readers to take part as well as to stimulate their brain :D.

Sudoku took over my mind during those days and I was completely engrossed in reading about Sudoku as well as solving them at websudoku . It occurred to me why not write a program to solving the puzzle? So I took up the task and was ready with a small Sudoku Solver .

Of course, for certain inputs it does not give the solution. I pondered over the completeness and soundness issues of the initial sudoku board. I don't know whether it is appropriate to use these terms which are often used and associated with algorithms. Let me define it this way. Completeness means that the initial information given in the board should be sufficient to solve the puzzle by using only inferences ( No guessing game/No coin tossing/ No breaking tie). In other words, there must exist at most one solution that can be derived from the initial configuration of the board. Soundness means that the information should not be conflicting, or say, at least one solution exist that can be derived from the initial information. I discussed the matter with my mathematics teacher Mr Motilal Panigrahi at GCET .

We could not arrive at any definitive proofs or conditions to ensure the soundness and completeness, but it was nice to chat with him over these things years after I left the college.

Sherlock Holmes

Lately, I have been reading famous "Sherlock Holmes" stories. What a nice piece of literature it is! Sir Arthur Conan Doyle is truly a master of creating writings that hold the reader till the end. It really takes a true genius to create such a timeless character like Sherlock Holmes.

Though, some of the inferences employed by Sherlock Holmes in unleashing the truth seems to be unrealistic, I especially admire the literature for the flair with which the author wrote it. The description of a scene creates enough impact on your mind to force it to weave a virtual world around as if you are witnessing the entire series of events. Author has put to use his vocabulary so perfectly that it does not drive away a reader with average vocabulary and yet maintaining enough variations to make it far from banal. I wish I could attain at least 1/20th of author's writing skill.

Now I know, why Sir Arthur Conan Doyle was awarded knighthood. :-)

Bitwise fiasco

Yesterday was one of the worst day. It was the day of Bitwise 2007 . I participated in this contest with Pranjal as my team mate. I had really high hopes and a bit of arrogance as I could claim 26th rank in this prestigious contest last time amongst 2700 participating teams. Bitwise 2006 was my first official participation in a programming contest. Having done so well in my debut, I was dreaming to climb further up the ladder with Pranjal ( who secured 8th rank in Bitwise 2006 ) along the side.

The arrogance had fogged the part of the brain which does rational thinking, and I ended up starting my day without any kind of warm up practice for the contest. I had not even revised any relevant material.

Bitwise 2k7 started a little late than the announced time of 1300hrs. And as usual, I thought to attempt the "easier" problems ( 100 points ) first. It was based on graph theory where one was required to find out the vertice(s) having the minimum distance to all other vertices in a given tree. Initially, I started with an O(n^3) implementation and when I found the response of the system to be "Time Limit Exceeded" ( which was kind of obvious with such a naive approach to the problem ), I improved it to O(n^2). Even though, I tried harder I could neither change the complexity nor the response of the system. After 2-3 failed attempts I decided to concentrate on some other problem and come back to this one at a later time.

Problem 4, had two parts, the initial part was to decide whether a certain door k out of n will be open or closed having gone through a certain toggling scheme. It was kind of easy. The next part was a variation of the classical Tower of Hanoi problem. No matter how hard I tried, my algorithm was not matching the given sample input/output. I asked the admins to look at the problem again for any possible mistake in the statement or sample input/output. But this time they were not as quick in response as they were for the last event. After around 10-15 questions fired by different teams, they answered only one. The overall contest also had attained height of mismanagement with problem statements being updated several times after the contest timer was started. Having tried for 2-3 hours on Problem 4, to my utter surprise, I found out that many other teams were cribbing about the correctness of its problem statement. I checked the rankings and was shocked to see that none of the teams ( not even those who were sitting at Top 10 ) could solve that problem. So I gave up on this problem too.

Then, I started yet another problem having to do with dynamic programming, coins and denominations. It was a variation of the classic denomination problem. Not that it was too hard to do it, I was kind of getting low because of failures and hunger. We went out to have a dinner at a small restaurant just opposite to IIT where we discussed yet another problem involving prime number and modular arithmetic. I thought to give it a shot after coming back, but that damn "Time Limit Exceeded" was not getting out of my sight.

Finally, we decided to give up as we were still sitting at zero where other Top 50 teams begged more than 800 points. I had never been so dejected in a long time. I was literally ashamed of myself.

Today morning I read a couple of articles on Modular Arithmetic and
Fermat's little theorem and realized why my implementation was not fast enough to defeat that "Time Limit Exceeded" demon. For Problem 1 ( The one with the vertices ) I learned that the middle vertice(s) of the longest path in the tree always have minimum sum of distance to all other vertices. How to find the longest path in a tree? Well, start from any vertex A and find out the furthest vertex B from it. Now start with B and find out the furthest vertex C from B. Path B to C is the longest path and its middle vertice(s) are the ones that we needed to find out. Running time?? O(n)

The bottom line is that I have got rusty. My programming skills and algorithm designing skills have fallen down. It seems I have not been doing much brain stimulating work for a long time. May be this failure is an indication for me to start honing my skills and in addition to giving exercise to my body I should give enough exercise to my brain as well.