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.

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Method -1
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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.

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Method - 2
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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.
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Method 3
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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

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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.

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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.

Cake cutting is NOT a piece of cake

Hi,
My Boss is out of india for a week so I am a bit relaxed now. Yesterday, I was just surfing the net and I recalled something about "Cake cutting". I did a bit of googling and found
this paper. The paper defines the problem formally and then give information about existing cake-cutting algorithms. Initially, I thought the paper might actually contain the algorithms but to my misfortune it does not. However, it shows how sorting can be reduced to the cake-cutting problem ( under certain conditions ). The most interesting piece of information I got from this paper is that it is not even easy ( in terms of complexity ) to cut a cake in a way so as to guarantee a positive share ( not even fair...just positive ) to every contender.

Well, for those of you who are baffled by the above text, let me refer to a lecture note. I am copying the text below in the fear of losing this nice piece of text in case it is removed from the original source.

There is also a very nice book named "Cake cutting algorithms : Be fair if you can" by Jack Robertson and William webb. I am saying this book is nice because I got very good reviews about it. The review says that it is lucid enough for a novice to try out. I wish to read it someday.

Anyways, it did not seem to cut a cake that difficult while celebrating b'days in the department :D ( I guess because we did not bother about the fair share ...:D)
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Lecture Notes on Cake Cutting & Fair Division, Thursday 9/18/2003, CS70

The cake-cutting problem:
We have a cake, and n people who want to split it amongst themselves.
However, each person might value different portions of the cake
differently. (I like flowers; you hate them. I hate icing; you
prefer it.) What's worse, we don't trust each other! What can we
do?

A protocol is *fair* for X if the following property is true:
If X follows the party, then X gets at least 1/n-th of the cake
(by X's measure), no matter how the other parties behave.
The protocol is *fair* if it is fair for all parties.

For n=2: cut-and-choose protocol
1. Alice cuts the cake into two equal pieces (equal by her measure)
2. Bob chooses whichever piece looks larger (by his measure)
3. Alice takes the remaining one
Theorem: Cut-and-choose is fair.
Proof: If Alice follows the protocol, she gets exactly 1/2 (by her
measure), no matter how Bob behaves.
Next Let's consider Bob. When it is time for him to choose,
he sees two pieces, one worth W and the other worth 1-W (by his measure).
It is guaranteed that either W >= 1/2 or 1-W >= 1/2 (if W < 1 =" W+1-W" 2 =" 1,">2?

For general n:
If n=2, use cut-and-choose. Otherwise:

Let first n-1 people divide the cake using a recursive
call to this procedure.
Then the nth person steps in and
asks each of the first n-1 people to divide her share into
n equal pieces (by her measure).
Finally, the nth person goes around and collects the
largest (in his view) of the n pieces from each of the other people.

Does each think they've got at least 1/n of the whole pile
(by their measure)?
For first n-1 people, yes, since they get
>= 1/(n-1) after recursive call, divide this into pieces each of
size >= 1/n(n-1), and then get n-1 of these pieces.
For last person, yes, since he gets at least x_i/n from the ith
person, if ith person's whole share is worth x_i. Also,
x_1 + ... + x_{n-1} = 1, so last person's total share is
x_1/n + ... + x_{n-1}/n >= 1/n. Therefore:
Theorem: This protocol is fair.
Proof: By induction on n. We've proved the base case (n=2).
The inductive step is exactly what appears in the previous paragraph. QED.

This general protocol can be implemented as a recursive program.
Q: What's the running time?
A: n! > 2^n ... which is exponential. yuck.

Q: Can it be done in time polynomial in n?



Puzzle: Find a fair cake-cutting protocol with running time polynomial in n.
One answer: A moving knife.
Move a knife slowly from left to right. When area to the left of the knife
covers at least 1/n-th of the cake by your measure, yell "Stop!". First person
to yell "Stop!" gets everything to the left of the knife, and we continue.

However, we might prefer to avoid using moving knifes.
Q: Why not use a moving knife?
A: hard to implement in a distributed system without time synchronization


Puzzle: Find a fair cake-cutting protocol with running time polynomial in n,
with no use of moving knives.
One solution:
Take the n-party moving knife algorithm, and translate it into
a non-moving-knife protocol by having each person cut at first
point they'd yell "Stop!". Take the smallest piece and give it to
the person who cut there. Then, continue dividing up the rest of the
cake up among the remaining n-1 parties.
Running time: O(n^2) basic operations

A protocol is *envy-free* for X if it has the following property:
if X follows the protocol, then X gets at least as much (by X's measure)
as anyone else gets (by X's measure), no matter how the other parties behave.
A protocol is *envy-free* if it is envy-free for all participants.

Q: Is 2-party cut-and-choose envy-free?
A: Yes (do you see why?)

Q: Is the recursive algorithm I gave last time envy-free for n=3?
(i.e., A+B cut-and-choose, then divide their pieces into 3 sub-pieces;
C gets to choose one sub-piece from A & one from B)
A: No.
(A+B might conspire to give whole cake to A; then C gets 1/3,
but A gets 2/3, not envy-free)

Q: Is the 3-party moving knife algorithm envy-free?
(i.e., move knife slowly to right until someone yells Stop!, repeat)
A: No.
(suppose A yells Stop! first. then B+C might conspire to give
all of remainder to B and none to C, not envy-free)

By the way, this is a non-zero-sum game.
Non-zero sum games often allow everyone to come away with better
than 1/n-th of the cake.

An example of "arbitrage" and a non-zero-sum game:
Alice believes that Gore will win the election with probability
5/8. Bob believes that Bush will win the election with probability
3/4.

Assuming that Alice and Bob are both willing to accept any bet
that gives them a positive expectation of winning, did you know
that there's a way to place bets with both of them so that you can
make money for certain?

Here's what you can do. Bet with Alice that you'll pay her $2 if
Gore wins and she'll pay you $3 otherwise. Alice agrees because
her expectation is: $2(5/8)-$3(3/8)=$1/8.

Bet with Bob that you'll pay him $2 if Bush wins, and he'll pay
you $3 otherwise. Bob agrees because his expectation is
$2(3/4)-$3(1/4)=$3/4.

Alice and Bob both believe they have positive expectation, but
you will win for certain: if either Bush or Gore wins, you will net
a dollar!
Credits: http://www.math.hmc.edu/funfacts/ffiles/30003.6.shtml


An envy-free cake-cutting protocol for n=3 people:
http://www.math.hmc.edu/funfacts/ffiles/30001.4-8.shtml

Problem: polynomial-time envy-free cake-cutting protocol for all n
(This might be open?)
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