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Todays IPG seminar had Fritz Eisenbrand (the Disctete Opt chair, Math department EPFL) talking about Diameter of Polyhedra:Limits of Abstraction. I don’t think I followed the topic too well, but this is a share of what I understood.

The topic is about a convex geometric problem on the diameter of a polyhedra. The question of whether the diameter of a polyhedron is polynomial or not seemed to be a longstanding open problem. The largest diameter ${\Delta_{u}(d,n)}$ of a ${d}$ dimensional polyhedron with ${n}$ facets has known upper and lower bounds.

${n-d+\lfloor d/5 \rfloor \le \Delta_{u}(d,n) \le n^{\log d +1}}$.

The lower bound is due to Klee and Walkup and upper bound to Kalai and Kleitman. These bounds also hold good for combinatorial abstractions of the 1-skeleton of non-degenerate polyhedra (Polyhedron here is called non-degenrate). What Fritz and his colleagues have done is to look into the gap between these known lower and upper bounds. Apparently, the gap is wide and they have made some progress to get a super linear lower bound ${\Delta_{u}(d,n) \le \Omega\left(n^{3/2}\right)}$ if ${d}$ is allowed to grow with ${n}$.

The way they showed this bound is by establishing the bound for the largest diemeter of a graph in a base abstraction family. Let us say, the abstraction family of connected graphs be denoted by ${\mathcal{B}_{d,n}}$.The largest diameter of a graph in ${\mathcal{B}_{d,n}}$ is denoted by ${D(d,n)}$. They find that,${D(d,n) =\Omega\left(n^{3/2}\right)}$ and then using the fact that ${\Delta_{u}(d,n) \le D(d,n)}$, they conclude the bound ${\Delta_{u}(d,n) \le \Omega\left(n^{3/2}\right)}$

I have not had a chance to see their paper yet. I must say, the proof was not all that within my grab during the talk. However it appeared that it is based on some layering and combinatorics. He said some applications to covering problem, in particular disjoint covering design which I didn’t follow that well. Sometimes I get the feeling that I am a little dumb to grasp these ideas during a talk. I wonder whether others understand it very well on a first shot presentation. I have put it in my agenda (among the millions of other papers to read) to see through this problem and proof, one day! His presentation was very clear and legible though.

A few years ago, during undergrad days, myself and  friend Ramani during our lazy 75 paise mini canteen tea outing, were discussing a small riddle. It was motivated from a real world experience from our computer center in NIT Calicut (REC Calicut). In REC those days, we students almost exclusively used rubber slippers (Yes, those Paraqon brand which used to cost 20 rupees or so), usually called by the name ‘chappels’. With that, we were not only comfortable while walking and running around, but we’re equally at ease playing cricket and badminton with the very same foot support; and many other things too, including jogging. Those thin hard rubber slippers used to last an year or more without giving much trouble, other than perhaps an occasional tearing of the rubber tie. In all, we were at peace with that.

But there was an issue, not exclusively for this brand, but for chappals in general (shoes were a luxury of sort in the campus;atleast it wasnt very common). Not for everyone though! If and only if you were fancied of visiting the computer center! Well, computer center wasn’t all that fanciful then, since we were provided with only graphics less Unix terminals (no colour monitors!). You might wonder, huh! what age am I talking about? Besides, Internet and Emails were only taking shape then. Chats and browsing were not quite there yet;Unless you felt a touch inferior to the computer wizkid around, that was not a compelling centre de visite. As, ‘would be‘ electronics and communication engineers we had that occasional inferiority complex!. Computer center was air conditioned and was strictly slippers free. We were expected to keep our valuable slippers outside (no clock room luxury! well that was not a necessity either) before entering to that cooler room, filled with monochromatic terminals. Since most of the chappals dropped outside were alike (in size and also sometimes color) there was a good chance that at the time return, we ended up with a different pair of slippers (Some folks found happy for themselves by a visit to the computer center, just for a pair change, often to an improved lot!).  Sometimes, we ended up having differently colored ones, say left foot white and right foot blue. That wasn’t a problem socially either, as long as you stayed within the campus. It was socially accepted within the walls!

Anyway, coming back to the riddle we were busy conjecturing on. We wanted to automate a clock room. The idea then would be to just deposit the chappals there at random. The clock room work automatically. Upon asking (at the time of return, say) it will select a pair at random and give it to you. Sorry, you cant have a choice. Just accept and hope for the best. We asked the questions:

1) What is the probability that everyone gets their own chappals

2) What is the probability that none of them get their submitted pairs

Assume $n$ number of  people (and hence $n$ pairs). We can assume that, a pair is a single entity (say both left and right slippers are tied and submitted as one) . This simplified the problem to $n$ people $n$ slipper scenario. A simplistic model assumeed that all $n$ people submit their slippers at the same time. We wanted to build that great randomized clocker machine! And we wanted that to work for any $n$, which means, the algorithm had to be implementable and to work well in expectation!

We had thought and pondered about it for a while, then. In the end, we had found that the first one is easy, but the second one a little harder to generalize for beyond $n=10$ or something.  As busy undergrads, we left the problem after an hour of discussion, probably until we had finished sipping the tea. Aside, we were busy with many other extra curricular activities including a 3 hour daily cricket match at the lush green international hostel ground. The megadeth team, as we proudly grouped ourselves, the electronics and communication batch hardly missed those cricket matches. We were electronics engineers and had taken pride in ourselves by not really bothered to ask any fellow discrete math or combinatorics folks! That perhaps helped in some sense.  Ramani found management more interesting than those technical details of counting. I am sure he took the right career. Anyway…too much digressing already!

Now, it turns out that, the very same problem is akin to a well known problem in combinatorics. It is called the Hatcheck lady problem. It is fairly easy to solve it using the inclusion exclusion principle. The proof outline is shown below. As I type, memory fetches that discussion,  sitting leg-folded on the cement bench at the REC mini-canteen, perhaps an occasional cool breeze around too.

The inclusion exclusion principle is the following:

$\lvert \bigcup_{i=1}^{n} A_{i} \rvert=\displaystyle\sum_{i=1}^{n}{\lvert A_{i}\rvert}-\displaystyle\sum_{1\le i_{1}

$+\displaystyle\sum_{1\le i_{1}

$+(-1)^{n-1}{\lvert A_{1}\cap A_{2}\cap A_{3}\cap\ldots\cap A_{n} \rvert}$

The Hatchek lady problem can be stated with a similar story as the random clocker machine. (From Harris, Mossinghoff, Hirst’s book on Combinatorics and Graph Theory)

A lazy professor gives a quiz to a class of $n$ students, then collects the papers, shuffles them, and redistribute them randomly to the class for grading. The professor would prefer that no student receives his or her own paper to grade. What is the probability that this occurs? This indeed is an equivalent statement of the well known Hatcheck lady problem (I guess the exact name come from a hatcheck lady who collects hats and absentmindedly return them)

For Hatcheck lady problem, the probability $P(n)=\frac{D(n)}{n!}$.

$D(n)=n!-\lvert A_{1}\cup A_{2}\ldots\cup A_{n}\rvert=n!-\frac{n!}{1!}+\frac{n!}{2!}-\ldots+(-1)^{n}\frac{n!}{n!}$

$= n!-\displaystyle\sum_{k=1}^{n}{(-1)^{k-1}\binom{n}{k}(n-k)!}=n!-\displaystyle\sum_{k=1}^{n}{(-1)^{k-1}\frac{n!}{k!}}$

$P(n)= 1-\displaystyle\sum_{k=1}^{n}{(-1)^{k-1}\frac{1}{k!}}$

When $n$ gets larger and larger it converges asymptotically to a constant!

$\displaystyle\lim_{n\to\infty} P(n)=\displaystyle\lim_{n\to\infty}{\displaystyle \sum_{k=1}^{n}{\frac{1}{k!}}}=\frac{1}{e}$

The popular documentary on this fascinating mathematical prodigy of 20th century is now available on you tube. Personally, while watching the video, the cam river and the row boat brought a touch of nostalgia! I have heard mountains of stories about Paul Erdős. This documentary is a must watch for not only mathematicians and mathematically curious guys (or guys like me who are just curious about mathematics, mathematicians and mathematical minds or for that matter about anything in this world!), but for everyone interested to know about such an extra ordinary mind of our times.  What a fascinating experience it would have been to listen to one of his lectures live. Now this youtube brought the gap down to finite length/time reality.

I have never seen Erdős. Now that he is no more warrants any thoughts anyway.  In a away I am lucky this semester to attend courses of another prolific mathematician of this era Janos Pach. Interestingly, Janos Pach is one of the few living mathematicians with Erdős number 1.  His lectures on Graph theory as well as the one on geometrical graph theory are truly fascinating.

Anyway, if you have not seen the documentary yet, here is the link. It is a must watch. I bet, you wouldnt miss the time. On many occasions, the Cam river and its slow movement etches something in the backdrop of those days.  Didnt I like that place?

While trying to understand the Luby transform (LT) code, I stumbled upon the well known coupon collector’s problem. It took a while for me to figure out the connection, but as it turned out, there is a stronger connection between these two.  In LT parlance, if we were to use only degree one packets (that is, packets sent and collected as it is) what is the expected number of packets to be collected (when collected randomly, one at a time) such that all the required packets are collected atleast once. For illustration let us say we have $n$ information packets at the transmitter. The receiver collectes these one by one at random. How many packets on the average, we need to collect until we have collected all of the $n$ different information packets. Remember we are collecting the packets randomly (On the other hand, if we were to collect things deterministically, we just need to collect $n$ packets to get all these $n$, when done without replacement).

Assume that there are $n$ distinct coupon types.  We have, a large pool of these coupons at disposal. Every time you go to the shop you collect a coupon picked uniformly at random from the pool.  The picked coupon has equal probability of  being any of the $n$ types.  Naturally, some of the collected coupons (over multiple visits to the shop) may be of the same type. The question asked is this:  Suppose the coupon collector aims to have coupons of all types.  How many (number of visits) coupons he  has to collect till he possess all the $n$ distinct types of coupons?

In expectation, the coupon collector should make  $n \log(n) + O(1)$ visits to the shop in order to have atleast one copy of all $n$ distinct types of coupons . This coupon collector problem can sound a little confusing to a fresh reader. For simplicity sake we can assume that, there are $n$ differently coloured coupons at the shop. The question then is, on average (i.e., expectation) how many times one needs to visit (each visit fetch a coupon) the shop so that all coloured coupons are fetched atleast once.

There are $n$ different type of coupons.  The coupon collector collects a coupon upon each visit. The collected coupon is among the $n$ types, picked uniformly at random (from a set of possibly large pool of coupons) .  Since the coupon is drawn uniformly at random, there is a non zero probability that some of the collected coupons over multiple visits may be of the same type.  Suppose that at some stage, the coupon collector has $r$ different type of coupons collected.  The probability that his next visit fetch a new coupon type (not of the $r$ types he already have in the kitty) is $p_r=\frac{n-r}{n}$.  So, the expected number of coupons to be collected to fetch a new coupon type is $\frac{n}{n-r}$.  Let us denote this number by $E\left[N_r\right]$.

The expected value $E\left[N_i\right]=\frac{1}{p_i}=\frac{n}{n-i}$. From this we can compute the expected value of $N$. In other words, $E[N]$, the expected number of coupons to be collected (i.e, number of visits to the shop!) so that, the he would have all the different $n$ types of coupons is:

$E[N]=\displaystyle \sum_{i=1}^{n-1} {\frac{n}{n-i}}=n\sum_{i=1}^{n-1}{\frac{1}{i}}=nH(n)=n\log(n)+O(1)$

So, what is the trouble? This number $n\log(n)$ is prohibitively high a number to be acceptable (as decoding time of $n\log (n)$ is significantly higher than the wishful linear time $n$!). So, simply using degree $1$ is not a good idea. This is why Luby went ahead and identified some smarter distribution like Soliton (and its variants proposed later on, such as robust soliton and then the recent raptor codes by Amin).