Question

Manish was on his way to an interview. On the way, he encountered his long lost cousin, Vijay, whom he hadn’t met in more than a decade. They started catching up on lost time. Manish learned that Vijay had 3 sons. When he asked about their ages, Vijay replied, “You’re going for an interview, right? Consider this a trial question. Figure out their ages from this: The product of the ages of my three sons is 36.” To this, Manish grumbled that he needed more information. Vijay, then, pointed to a sign board across the street that displayed the address of the area and said that the sum of the ages of his three children was equal to the last two digits of the pin code (zip code) of that area. Manish demanded still more information. Finally, Vijay said, “My eldest son wore a black shirt today. This is all I can tell you.”

What were the ages of the three children?

Solution

Say the ages are a, b and c. We know from clue 1 that a.b.c = 36, where “.” represents multiplication. First step in identifying such factors is to factorize 36 into its smallest factors – 1.2.2.3.3. The next step is to figure out how to group these 5 numbers into 3 groups. Note that we can have more 1s in the factorization. For example, what if 2 kids are 1 year old and 1 “kid” is 36 years old? So, to allow that to happen, we need another 1 in the factors. So the factors are 1.1.2.2.3.3. By trial and error, we recognize that the way to make 3 groups from 6 objects is by making groups of:

1 + 1 + 4 (call it Grouping Style 1, or GS1)

1 + 2 + 3 (call it Grouping Style 2, or GS2)

and 2 + 2 + 2 (call it Grouping Style 3, or GS3)

GS1 requires selecting 1 out of 6 for the first component, 1 out of remaining 5 for the second component and the remaining 4 automatically go into the third component. There are 6 ways to choose the first component, 5 ways to choose the second component and 1 way to chose the third component. This gives us a total of 6.5=30 combinations for GS1. Many of these will turn out to be identical. After some work, we can whittle down the GS1 groupings to the following:

1,1,36

1,2,18

1,3,12

2,2,9

2,3,6

3,3,4

GS2 requires selecting 1 out of 6 for the first component, 2 out of 5 for the second component and the remaining 3 automatically go into the third component. There are 6 ways to choose the first component, 5C2 = 10 ways to choose the second component, and 1 way to choose the third component. This gives us a total of 6.10=60 combinations for GS2. In reality it is easier to work it out by trial and error. After some work, and after recognizing and ignoring the combinations that we have already seen under GS1, we can whittle down the GS2 groupings to the following:

1,4,9

1,6,6

Similar analysis for GS3 gives us no new combinations.

Before we can use clue 2, we need to add up the ages in these combinations. Doing so, we get the following:

1+1+36=38

1+2+18=21

1+3+12=16

2+2+9=13

2+3+6=11

3+3+4=10

1+4+9=14

1+6+6=13

Notice that all the combinations give us unique totals, except for two combinations which both give us 13. The fact that the second clue did not suffice to answer the question indicates that the total of the ages must have been 13. The children could be aged (2,2 and 9) or (1,6 and 6). Any other value for the total age and the answer would have been clear after clue 2.

Clue 3 tells us of the existence of an eldest child. The color of the shirt is immaterial. In the (1,6 and 6) combination, there is no eldest child. There are 2 elder children, who are twins. In the (2,2 and 9) combination, there is an eldest child. Hence, that is the answer. The children are aged 2 years, 2 years and 9 years.

Intersecting Ramanujan’s trail

With the specific problem out of the way, let us think about a generalization to the problem. India’s best known mathematician, Srinivasa Ramanujan, hiked (given his genius, he probably breezed) along a mathematical thought process, probably in 1913, leading to his work on Integer Partitions. In attempting to solve this problem, I seem to have unknowingly stepped onto this trail briefly. Let me explain. Remember that we had to figure out how many ways could 6 objects be grouped into 3 groups. I listed these out as groupings with (1,1,4), (1,2,3) and (2,2,2) objects. There are 3 grouping styles possible, no more, no less. But as the number of total objects grows larger, or the number of groups to create changes, the number of such grouping styles are harder to figure out. At least they seem to follow no simple pattern. For example, to group 6 objects into 2 groups, there are also 3 ways – {5,1}, {4,2} and {3,3}. To group 7 objects into 2 groups, there are only 3 groupings – {6,1} and {5,2} and {4,3}. To group 7 objects into 3 groups, however, there are 4 groupings – {1,1,5}, {1,2,4}, {1,3,3}, and {2,2,3}.

The question is, is there a more general formula to figure this (the number of distinct ways to group N objects into G groups) out? And another question is – what is Integer Partitioning and how is that related to this problem?

So before going into the theory, let us try to do look at this problem from different angles, and we may see an opening to solving it. Let us use the following example – find the number of ways of grouping 7 objects into 2 groups. We discovered (through some mental enumerations, I admit) that there are three grouping styles – {6,1}, {5,2}, {4,3} – possible here. But now, let me draw your attention to a simple fact – notice that the sum of the numbers each of the groupings is equal to 7. Hardly a surprise, you say. There were 7 objects to begin with. And regardless of how we group them, the total object count is 7. Big deal! But it is often rewarding to look at problems from a different perspective. We now know that “grouping 7 objects into 2 groups” is essentially the same as “finding 2 positive integers that add up to 7″. How many ways are there to add up 2 positive integers to get a total of 7? 6+1=7 and 5+2=7 and 4+3=7. No more ways to do it. So, the number of ways to group N things into G groups is equal to the number of ways to add G positive integers to make N. Allow me create a short hand notation to identify this count. I will use P(N,G). P for partition, but really you can use any symbol. I think it only makes sense when N>=G. Otherwise P(N,G)=0.

P(N,G) = number of ways to add G positive integers to make N = numbers of ways to divide N objects into G groups

There are a couple of simple properties of P(N,G) which are easy to observe. There is only 1 way to divide N objects into N groups, and that is to assign 1 object per group. Similarly there is 1 way to divide N objects into 1 group, and that is to place all objects into 1 group. Symbolically,

P(N,N) = 1

P(N, 1) = 1

OK, I think we have beaten that one into submission (if not to death).

Note that we have only defined it. We we after a general formula for P(N,G). And we do not have that yet.

Now, let us visualize this slightly differently, meditate upon it a bit, and see if we can uncover some other properties about P(N,G). The following figure shows the objects, and groupings, visually. This style of representation, called, Ferrers Diagrams, uses one dot for one object. And it arranges the dots in each group along a column. There are 3 groupings shown – these are our friends {6,1}, {5,2} and {4, 3}. If we stare at this for some time we realize that the reason there are 2 columns is because we wanted 2 groups. This means there is at least 1 row, the first row, which has 2 dots in each of the groupings. Now, imagine that we remove those two dots from each of the groupings. What are we left with?

The figure below shows the case where the 2 dots in the first row are removed. We are left with 5 dots. But more importantly notice the groupings that are left. {5}, {4,1} and {3,2}. These are some of the ways you can group 5 objects. This time, we do not necessarily group 5 objects into 2 groups – since we have a grouping, {5}, with only 1 group. It may be interesting to see how many ways there are in all to group 5 objects – not necessarily into 1 or 2 groups, but rather into any number of groups.

The following figure shows all the ways to group 5 objects.

1 group – {5}

2 groups – {4,1}, {3,2}

3 groups – {3,1,1}, {2,2,1}

4 groups – {2,1,1,1}

5 groups – {1,1,1,1,1}

Note that the number of ways to group 5 into 1 and 2 groups in this figure exactly matches the 3 groups in the previous figure. That is P(7,2) = P(5,1)+P(5,2)! That is a breakthrough. So maybe there is a way to break down any P(N,G) into smaller and smaller pieces and add up the totals of the smaller pieces. For example, P(N,G) = P(N-G,1)+P(N-G,2)+ … + P(N-G,G). Let’s try this for P(7,2).

P(7,2) = P(5,1) + P(5,2) = 1 + P(3,1) + P(3,2) = 1 + 1 + P(1,1) = 1 + 1 + 1 = 3

Not quite a closed form, but an recursive solution.

Integer Partitioning and a Generating Function for it

By the way, this is a good place to (finally) get into Integer Partitioning. In the figure above we listed out all the ways to group 5 objects. That is the Integer Partitioning of 5. I use P(5) to represent it.

P(5) = P(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5)

Though it is possible to use the recursive technique I described above to solve for P(5) term by term, there is a very interesting alternative way to figure it out. It is based on a near mind-bending technique, based on Generating Functions. It is powerful technique, used in many different places. I have not yet been able to pin down if the abstraction involved in using Generating Functions is pure genius or if it simply falls out of the mechanics of mathematics if only you start from the correct perspective. Let me attempt to convey how Generating Functions work in this specific situation.

We want to find out all the possible ways to group N objects. But instead of starting with this specific problem we turn the problem on its head and solve a much more general problem. Let us use a bunch of 1s, a bunch of 2s, a bunch of 3s etc. and see what we can add them up to. We do this exhaustively. That is, we we do not let any combination fall through the cracks. Then, we can count up all the different ways a bunch of addends give us the total N. For example, if N is 5, and we do this exhaustive (but structured) adding of all integers (repetitions allowed) and see how many combinations add to 5. We can be sure that none of the addends will be 6 or greater. Similarly, the number of addends will also be no greater than 5, because each addend is at least 1, and with 5 1s we already are at 5. So, the number of combinations we need to look at only needs to cover the addends 1 through 5, and no more than 5 of those addends.

This exhaustive search seems hard to do. But generating functions provide a structured method to approach this. Let us do it one step at a time. We will look at the potential contributions to the total, of each candidate addend one at a time.  Let us look at the addend 1. What totals can 1 contribute to? 1s can add up to 1 in 1 way, {1}. 1s can add up to 2 in 1 way, {1,1}. 1s can add up to 3 in 1 way {1,1,1} and so on. That is, 1 alone can add up to any N, AND, in only 1 way for each given N. In fact if there is no 1 in the addends, then 1 can contribute 0 to the total. In generating function terms this would be represented as follows:

What? Where did that come from? That is typically, always the reaction in my own mind when I see that. Essentially, the order of , which is k, indicates the total that the addends (remember we are only talking about the addends being 1) add up to. The coefficient of , which is 1, indicates the number of ways the addends can be added up to get to k. There is only 1 way to do this using only 1s. And there may be no 1s in the addends, and that is the term .

But what if 2s are allowed? There can be 1 two, or 2 twos, or 3 twos, etc. Thus the generating function based on contributions only from twos is as follows:

This basically means twos can add up to 0 or 2 or 4 or 6 or 8 etc.

Now, we know that we can have both 1s and 2s in the set of addends. In fact, we can have 1s, and 2s, and 3s, and 4s etc. Let us restrict the addends to 1s and 2s for the time being. There can be 1 one and 1 two, or 1 one and 2 twos or 2 ones and 1 two etc. How do we mathematically express these combinations? This is where the true import of the elegance of generating functions becomes clear. By allowing the contributions of the addends to be in the power of x position, products of and correctly give us the sum in the power’s position, .  Further, if there are multiple ways to add up to a given addend, they show up in the coefficient position. That is, as an example, say we want to figure out the Integer Partitions of any number N, using only the addends 1 and 2, we can multiply the individual addends’ generating polynomials.

This product of infinite polynomials will give us every possible way to add up 1s and 2s to get us to a specific total N. Say N=5, then we know we need not worry about and above in either polynomial.

And after some math, we end up with the necessary coefficient for , and that is the number of ways to get to 5 using only 1s and 2s.

Now, just extending this to allow contributions from the addends 3, 4, 5 etc., we get the full generating function for Integer Partitioning.

P(N) = coefficient of in

Whew! So we have that out of the way. What amazes me no end is how generating functions are able to hijack this polynomial product form to solve (or, more accurately, bring structure to) seemingly impossible scenarios that need counting. Notice that we still do not have a closed form for the Partition of Integers, P(N). Computers can be employed for the multiplication of terms and accumulation of coefficients.

A Closed Form Approximation

In 1918, Srinivasa Ramanujan and his advisor, G. Hardy, came up with a closed form , albeit a closed form for an approximation to P(N). With the current understanding of Integer Partitioning, this seems like an amazing accomplishment. This closed form was:

This produces pretty good approximations. For example, P(100) = 190,569,292, and this closed form approximates it to around 199 million.

References

1. Joseph Laurendi, Partitions of Integers, 2005, http://www.artofproblemsolving.com/Resources/Papers/LaurendiPartitions.pdf

2. Wikipedia, http://en.wikipedia.org/wiki/Partition_%28number_theory%29

“Ammamma…Boost…”, I would ask her for a hot malt beverage, as she would get busy in her tiny kitchen after her short afternoon nap on the hard concrete floor, with a strategically placed pillow for her head. Ammamma means maternal grandmother in my mother tongue, Telugu. Amma is mom, and ammamma is, literally, momom. She would then get busy preparing the late afternoon coffees for the elders, starting with the eldest – Tatagaru, and Boost for the kids. Although I mostly saw her only over summer holidays, this particular aspect of her routine was probably eternal. In fact, all her routines were seemingly eternal yet inexplicably fresh every time. She would hunker down at the old, grime-laden, two-burner gas stove sitting on the floor of her tiny kitchen and with what seemed like an impossibly tiny collection of utensils, groceries and gadgets, came up with the most exquisite of dishes. Simple fare it always was, and she was not a great cook, but the taste of her cooking was earthy and heavenly. Vegetables of all manner were shallow fried. Coffee and Boost was not served before being poured several times, alternating between two tumblers to generate froth (steamed milk). “Boost tagutawa, Kishtappa, aain?”, she would ask. The “aain?” was kind of like Amitabh Bachchan’s pan-laden mouth confirming something – “aain?”.

Kishtappa was her pet name for me. And she had a way with all children. She was our story-teller come in-house-magician come mystery-keeper come stand-up-comedian come play-pal, all rolled into one. Above all, she was our friend. Maybe because she had only very little formal education (“Appatlo, SLC pass aiyyenu”, she would confirm – “I completed 8th grade in those days”.), or maybe because she was married off when only 9 years old, my grandmother never really got along as well with adults as she did with kids, and always seemed to have stamina for reliving the childhood which she did not get to truly enjoy. She had a way with kids. And with five daughters and three sons she never ran out of kids or grandkids or great-grandkids to enthrall.

Ammamma was the queen of her territory in my Tatagaru’s large house in Dondaparthy, in Visakhaptnam. Her territory included her tiny kitchen, the attached utility area (her bathing area), the verandah just outside the kitchen where we would all eat (sitting on the rough concrete floor), a step down area from the verandah where the dishes would be washed and which led to the back yard, the strip connecting these rooms in the back of the house to the front verandah of the house (at the head of this strip was where she would strategically place her pillow ever afternoon for a watchful nap), and, the sacred sanctum of her territory, the store room (ammamma kottu) which remained mostly under lock and key. Except at some strange hours at the night. This was truly ammamma’s territory. I have probably only caught a glimpse of the inside of this room a handful of times, and I think I have been lucky enough to step into the room only once or twice. The room was just a large walk-in pantry with wooden shelves loaded with boxes and bags of all sizes and descriptions. The occasional mice, cockroach, or a scorpion (none of which would faze the dauntless ammamma) along with the dim lighting in the room and ammamma’s reluctance to let kids enter added to the eeriness of that store room. On the rare occasion we kids found the room open and would gather enough courage to peek in, ammamma would bustle out form the dark confines of the room, shoo us kids away, and tell us kids to stay away from the ghoul-laden innards of the place. She would then promptly hand us 25 paise or 50 paise and ask us to go get ourselves some mint candy (straangu billalu) from the Nair kottu – Mr. Nair’s small shop which sold everything from candy to biscuits to soft-drinks (of which, goli-soda was a strange attraction) to cigarettes to stationery. It was a shop where it seemed you could find anything that could be held in one hand.

Ammamma had a very interesting relationship with her husband, my Tatagaru (respected grandfather). To us she seemed to never be in speaking terms with her husband. I have never seen her converse with Tatagaru. However, her day would revolve around him. Being a lawyer, Tatagaru would spent some hours of the day at the Vizag District Court. And the rest of the day, his clients would come see him at his elaborate home office. He was a simple man, often sporting only a dhoti (similar to Gandhiji), but his office was well endowed, with lots of Law books and legal proceedings, case studies etc. He had an office room, which was separated from the road outside by a small waiting room for clients to wait in if he had company. Ammamma would have to keep an eye on Tatagaru, for he was a man of few words and if he needed lunch or one of the many servings of coffee, she would prepare that. She also had to occasionally keep an eye out for important clients and make sure they were served coffee or water etc. She was particularly displeased with lady clients who would spend a lot of time in Tatagaru’s office and would keep an eye on them as well. The office room was always open and though there was nothing to be concerned about, I guess she was possessive after all. These lady clients were often from small villages, and were from farming families. They would be visiting Tatagaru in relation to some land-related case. Often they would wear only a saree without a blouse to go with it, and this seems to particularly displease ammamma, although she would never make a big fuss about it. At most it would be an extra visit to the office to remind Tatagaru, “Vadinchestaanu, kallu kadukkoni randi” (“I will serve” – likely lunch – “wash your feet and come”).

I felt that the best times ammamma had was with her grand kids. With her kids, she was probably too young herself. With her great-grand kids, she was too old. But the best times she had was with us grand kids. She would play and teach us card games and tricks, then she would tell us stories or read us stories from the children’s magazine “Chandamama”, or read us jokes from her Telugu magazines, such as “Swathi”, “Andhra Jyoti” or “Andhra Prabha”. She would play “ashta-chamma” (similar to Ludo) or some variant of knucklebones with us kids, and perpetually keep us enthralled.

Maybe it was childish innocence. More likely it was sagely wisdom. Ammamma was never one to get too emotionally attached or emotionally charged. Her actions were devoid of any scheming, but like I said, probably not out of innocence, but rather out of a sagely understanding of the futility of such schemes. She was happy-go-lucky. She might have been different when younger, but as she grew older, she became a person of caricaturable simplicity. Maybe that is why we kids loved her. We could understand her. She would get happy at the small things in life, while the elders were never truly happy with anything. Some small things I remember about her were the way she would comb her tiny swathe of hair very meticulously each day after her bath and after applying a generous portion of hair oil. She would neatly part her hair at the center and with a vigor exceeding her age she would straighten her hair on each side, before finally tying it up into a braid or a knot. That she managed to do all this with a tiny mirror, and a tiny comb, each less than a few inches in size, never ceased to amaze. She loved the cinema, and would take us kids and go watch movies, mostly mythological or allegoric ones. She took me to watch “Keelu Gurramu” (the Magic Horse) once. For someone who enjoyed mythological movies, she was not much into devout worship or religious rituals. She lived an uncomplicated, pragmatic life. Although some of her beliefs would be considered backward (such as resisting graduate education for her girls), her practical point of view was that it would get harder to find a good match if the girl were overly educated. Regardless, and thanks to the calming influence of Tatagaru, most of her sons and daughters got a good education, some going on to complete their Masters, and all others picking up Bachelors’ degrees.

Ammama visited Bhilai, the steel city that was my birth place, and where my father used to work, at least a couple of times. And for her, visiting Bhilai was a wonderful time off. A vacation. Once she had a cataract operation (I think it was her left eye) at Durg (a town near Bhilai) and she was very pleased with the outcome. The other eye, which was operated upon in Visakapatnam itself was never quite the same, she said. Regardless, ever since her eye operations we only remember seeing her with thick lensed spectacles, with a thick black frame. The only time we’d then see her without her glasses was during her afternoon naps, and even still, the glasses would be tucked neatly under her pillow, lest we kids stepped on her only pair while running around during our afternoon games.

Ammamma was born Adibhatla Venkata Ratnamma on May 5th, 1921, in the hamlet called Dimili Agraharam, near Elamanchili town in present day Andhra Pradesh. Her father was Sri Adibhatla Suryanarayana, who was an inspector in the Revenue Department, and her mother was Adibhatla Perindevamma. She had two elder sisters and an elder brother. When she was born she was grossly underweight (likely, very premature) and her mother had given up hope about her surviving. The newborn did not have enough strength to even suckle milk be it from the breast or bottle. The elder sisters would go and get milk from other nursing mothers in the village and feed the newborn using a cotton wick. This was the only way the sickly child would ingest food. After three months of such feeding the mother was finally convinced that ammamma would survive. She survived. And how.

The elder sisters died in their early twenties, probably during childbirth. Ammamma, though born sickly and underweight, outlived all her other siblings, and she led a blessed, healthy life. She was married to Sri Tata Sri Rama Murthy, my Tatagaru, on March 16th, 1930 in “Kanukurthi vari satram” in Vijayanagaram city. It was a 5-day long wedding, complete with city-tours aboard a pearl-encrusted palanquin. Henceforth she became Tata Venkata Ratnamma. Married to someone destined to be a renowned lawyer, blessed with loving children and grandchildren, living a exceptionally healthy life, managing to keep her distance from petty attachments while retaining the ability to stay happy, ammamma had a great life. Recently, she had her first great-great-grandchild. Her second daughter’s first daughter’s daughter had a daughter. Truly rare. If we go looking for any misfortunes in her life, the untimely loss of her eldest son-in-law is probably the only one.

Ammamma passed away about 4 hours ago, sometime between 5:15AM and 5:30 AM on September 2nd, 2010 (India time). She was admitted to the hospital in the early hours of September 1st when she seemed to have lost consciousness after a few days of minimized food intake. Her blood pressure was quite low. In the hospital, she regained consciousness, recognized people, recognized the doctors and asked about all her grandkids. The IV drip helped her improve her blood pressure to near normal levels, and the oxygen mask helped her weakened heart. She seemed to be on her way to a steady recovery. Finally, it was her kidneys that failed. A woman who was so strong in her life, by her 90th year had gotten really good at putting up a stern fight against and evading any illness. She was a throat cancer survivor. And just as she gave us all hope that this would just be another of her minor illnesses, which she would fight down handily, she pulled off her last magic trick. She decided to say goodbye without any drama, without any emotion, without any inconvenience to others. Kavita and I must have been at the ISKCON temple in Hillsborough, North Carolina chanting “Hare Krishna, Hare Krishna, Krishna Krishna, Hare Hare … Hare Rama, Hare Rama, Rama Rama, Hare Hare” right when, for the last time, Ammamma must have said her favorite prayer one last time – “Krishna Vasudeva”. On our way back, we got the call.

Ammamma left us on Krishnashtami (Lord Krishna’s birthday), and left us with happy memories.  There is not a single negative memory I have of her. That is probably how any grandson feels about his grandmother, and she was the only grandmother I knew. But still, I feel that she was special. She was different. Knowing her was a boon for me. And like Kavita reminded me, Ammamma was the only grandmother she knew as well. Kavita shares the same kind of happy memories with Maamma (as she called Ammamma) as I did. Ammamma taught me more than she will ever know. Or, maybe in her own way, she did know all along. Happy-go-lucky. Happy-went-lucky.

Acknowledgements: Some of the memories and Ammamma’s biography are based on recollections by my cousin, Prasad. Most of the pictures are from Anant.

One way to avoid (or, at least drastically reduce) this problem is to either send or store multiple copies of the information. Say you send the message “BILL JOHN” 3 times. One of the times the error converts this to “KILL JOHN”, but two of the times the message transmits successfully. Then you know that most likely the intended message was “BILL JOHN”. Similarly, your bank probably stores your account information of multiple storage locations to avoid this and other problems (such as, catastrophic data loss due to a fire).  However, redundancy is not very efficient, and may be overkill. Further, it may not be always possible to apply (say the communication takes a long time and repeat communication is not possible).

Another solution is to add a small amount of extra information to the message being communicated or stored to validate the correctness of the data. This is, in effect, redundancy, but is often a more space-efficient form of it. Its purpose is primarily to detect errors in the message data, not to handle catastrophic destruction of data (which is the primary purpose of full redundancy). The most common way to detect errors in transmitted data is by the use of parity bits. The data is first divided into blocks of bits, say 7 or 15-bit blocks. Next, for each block of bits a parity bit is added, thus making the block size 8 or 16 bits. For example, if the original data is “10110101000101010″, and the block size is 7 +1 parity bit, then the data is first extended to make it a multiple of 7. So, say the message is changed to “000010110101000101010″. Then, it is broken into 7-bit blocks – “0000101″, “1010100″, “0101010″. And finally, for each 7-bit block one more bit is added. Say, it is added to the right end. The parity bit is dependent on the data bits. The parity bit is set to a 1 or a 0 depending on the number of 1s in the message. If there are an odd number of 1s in the message, the parity bit is set to 1 to make the total parity of the 8-bit block even. I am assuming even parity in this example; the parity may also be set up to be odd. Going back to the example, if there are already an even number of 1s in the 7-bit data block the parity bit is set to 0. The recipient knows what the block size is, where the parity bit appears, and what kind of parity (Even or Odd Parity) is being used. This is part of a pre-established communication agreement. For example, “0000101″ has 2 1s, that is, an even number of 1s. The parity bit is set to 0 and appended to these 7 bits to make the transmission block, “00001010“. Upon transmission, say there is a bit flip and the message becomes “00101010″. The receiver can then figure out that the parity, which is supposed to be even is now odd, and therefore indicates an error. The receiver can then request a retransmission. Notice that the parity bit itself may be transmitted in error, and that triggers an error as well (even though in reality the data bits were all transmitted fine). The size of the block is set to be small enough so that the probability of a bit flip is very very low. More importantly, the probability of two flips is infinitesimally small. Notice that if there are two flips, however, a single bit parity scheme as described above will not work. There is then a need for a more involved parity scheme. If the channel or storage medium has a propensity for unidirectional flips (that is, say it can only flip a 1 to a 0, bit never a 0 to a 1), then a counter may be maintained along with the data to count the number of 1s in the data block. If the number of 1s changes, then the receiver can detect an error. It is a little tricky once you realize that the number of 1s counted must include the number of 1s in the counter itself! The counter bits need to be transmitted as well, and are subject to bit flips as well.

In some situations, there may not be a second copy to rerequest the data upon detecting the error. Or in case of a communication channel, it may take too long to request a replay. And regardless of the application, it seems mathematically challenging to come up with a way to send a message such that not only is the error detected, it is also corrected by the receiver. This was the challenge that Richard Hamming took up an solved in an ingenious way. Here I would like to think through the same problem and come with a solution, which will then help us think through some of Hamming’s thoughts in the 1930s and 40s.  We will stick to the single bit error scenario. We can always reduce the block size sufficiently that only a single bit error can occur with any meaningful probability.

The first insight that led to the discovery of the Hamming code seems to be that some extra meta data (just like parity) needs to be transmitted in addition to the main data. In the presence of this extra data the receiver must be able to identify a bit flip in any bit in the transferred message (including a bit flip in the meta bits).  How do we uniquely identify a bit that flipped? We need to identify the position of the bit that flipped. In a message with d bits of data and p bits of meta data, the total number of transmitted bits is d+p. To identify a unique position where the flip may have occurred, we need log2(d+p) bits. The genius of Hamming was in recognizing that the bits being transmitted should be grouped into several groups such that each bit was a member of a unique set of groups. Further, the binary representation of the bit positions was the simplest way to identify the groups.

Say the message were 10 bits long. That is, d+p=10. The 10 positions can be represented as:

0001 – message bit 1
0010 – message bit 2
0011 – message bit 3
0100 – message bit 4
0101 – message bit 5
0110 – message bit 6
0111 – message bit 7
1000 – message bit 8
1001 – message bit 9
1010 – message bit 10

Notice that we start from 0001, and not 0000. This is because we need to make sure all numbers have 1 (or 0). This means to represent 8 positions (d+p=8), we actually need 4 bits. So the number of groups = log2(d+p+1). Further, notice that the binary representation of every message-bit-position has 1s in unique power-of-2 positions. This is obvious. After all, that is how we represent the positions uniquely. But the insight Hamming was able to draw was that each of the representations (left column above) with a 1 in a certain power-of-2 position could be clubbed together into a group. That is, in the example above, message bits with a 1 in the 2^0 position could be grouped into Club 0. Message bits with a 1 in the 2^1 position could be grouped into Club 1. Notice that some of the numbers which were members of Club 1 were also members Club 0 (for example message bit 3). Similarly, Club 2 and Club 3 could be formed. Further, exactly 4 clubs were needed to identify every bit (including the meta bits uniquely).

0 | 0 | 0 | 1 | – message bit 1  – Club 0
0 | 0 | 1 |0 | – message bit 2 – Club 1
0 | 0 | 1 | 1 | – message bit 3 – Club 0,1
0 | 1 | 0 | 0 | – message bit 4 – Club 2
0 | 1 | 0 | 1 | – message bit 5 – Club 0, 2
0 | 1 | 1 | 0 | – message bit 6 – Club 1, 2
0 | 1 | 1 | 1 | – message bit 7 – Club 0, 1, 2
1 | 0 | 0 | 0 | – message bit 8 – Club 3
1 | 0 | 0 | 1 | – message bit 9 – Club 0, 3
1 | 0 | 1 | 0 | – message bit 10 – Club 1, 3

Notice that no two bit-position-representations can belong to the same set of Clubs. After all, every bit position is represented uniquely in binary, and a 1 in the binary representation corresponds to a “key” to a certain club. No two binary representations are the same and so no two bit-positions can belong to the same club combination.

The groups/clubs thus formed are:

Club 0 – message bits 1, 3, 5, 7, 9
Club 1 – message bits 2, 3, 6, 7, 10
Club 2 – message bits 4, 5, 6, 7
Club 3 – message bits 8, 9, 10

Now, Hamming must have realized that he has created clubs with careful membership such that a unique member could be located if we knew which clubs he belonged to. So, a flipping bit can be located if that flip can identify all the clubs the bit belongs to. The clubs must start with some common property, and the flip must change that property for each club that it belongs to. This would help identify the clubs in which the flipped bit has membership, and by that unique combination, we can identify the flipped bit.

That unique property that Hamming gave each club was parity. Since no two clubs has the same membership, each club needed at least 1 unique parity bit. Since we want to keep the number of parity bits as small as possible to keep the message overhead as small as possible, the smallest solution would need as many parity bits as clubs. And we know that for transmitting d+n bits, we need log2(d+p+1) bits to represent the positions and, therefore, log2(d+p+1) clubs. This leads to the following observation:

The minimum overhead bits/parity bits = log2(d+p+1). But we assumed that the number of parity bits = p.

p >= log2(d+p+1)

Now we know how many groups, which bits make up the groups and the minimum number of parity bits. In the above example with d+p=10, p=4. Thus for 6 data bits we need to transmit 4 parity bits. But we must recognize that the parity bits cannot be simply lumped together into any casually selected positions. This is because without careful selection we may not have at least 1 parity bit per group. For example if the first 4 bit positions of a 10 bit message are used for parity, there would be no parity bits in Club 3. If we instead used the last 4 bits in a 10-bit message for parity, each club would have at least one parity bit, but all 3 of Club 3’s members would be parity bits. This creates a dependency on filling out the parity bits on the sender side. First parity bit 7 would have to be established based on Club 2 which has 3 data bits and only 1 parity bit. Then the remaining parity bits in Club 0 and Club 1 would have to be established, followed finally by the parity for Club 3. Hamming realized that it would be much easier to ensure that one parity bit belonged to 1 club, no more no less. All the members of a club are data bits except the one parity. This helps speed up setting the parity bit – there is no dependency and fixed order in which the parity bits need to be established. The bit positions in the message which belong t0 only 1 club are positions with a single 1 and all else 0s. These are also the power-of-two bit positions. Bit position 1, 2, 4 and 8. By making these bits parity bits, there is another advantage. At the receiving end, once parity is established for the various clubs, if it turns out that club 2 and club 1 are not maintaining parity, then to identify the bit position which flipped all that is needed is to calculate 2^2 + 2^1 = 6. Bit at position 6 is the only bit which is a member of Club 2 and 1 and no other club.

References:
I found this video of a class, held in University of New South Wales, Australia, and taught by Professor Richard Buckland, to be very useful in helping me understand how Hamming Codes work. The Hamming Code material starts at minute 8:00. http://www.youtube.com/watch?v=kE7V7UI4jpk

भीडों के धक्कों में खुश रहने वाले,
तमाशों के रंगों को सच कहने वाले,
औरों की बातों को अपना बता कर,
चुराये खयालात अकड से जता कर,
जो डरते हैं पर डर जताते नहीं हैं,
शर्मिन्दा हैं, सर झुकाते नहीं हैं ।
न जाने के उनमे वो खुद ही कहां हैं ।
ढूंढें तो औरों कि परछाइयां हैं ।
जो खुद से, खुदी से, खुदा से ख़फा हैं,
बेखबर, बेहुनर, बेकस, बेवफा हैं ।
बदल दो ये मक्सद, ये मंज़र, ये मंज़िल ।
कफस से कदम पर कफन ही कज़ा है ।

Transliterated to English:
bheedon ke dhakkon mein khush rehne waale,
tamaashon ke rangon ko sach kehne waale,
auron ki baaton ko apna bataa kar,
churaaye khayaalaat akad se jataa kar,
jo darte hian par dar jataate nahin hain,
sharminda hian, sar jhukaate nahin hain,
na jaane ke unme vo khud hi kahaan hain.
dhoonde to auron ki parchchaaiyaan hain.
jo khud se, khudi se, khuda se khafaa hain,
bekhabar, behunar, bekas, bewafaa hain.
badal do ye maksad, ye manzar, ye manzil,
kafas se kadam par kafan hi kazaa hai.

Meaning in English:
(Note: I present the intended meaning. If the actual meaning of my words turns out different, especially if it changes the intention, please let me know.)
Badal = Change (as a command)

Those who are happy jostling amidst crowds,
Those who say that the colors of worldly parades are true,
Those who claim others’ words as their own,
Those who show off stolen ideas,
Those who are afraid, but do not show their fear,
Those who are ashamed, but don’t let their heads hang,
They do not know who they really are within,
Upon looking, they find that they are only the shadows of others.
Those who are unhappy with themselves, their sense of self-respect, and God,
They are unware, unskilled, friendless and unfaithful.
Change such ambition, such scenery, such a goal.
By destiny, one step from this cage is death.

http://www.justiceharvard.org/

It contains some fascinating discussions on morals, philosophy, rights and justice. Professor Sandel has a very interesting teaching style, where he almost helps the students discover right vs. wrong, rather than just teaching it to them. Also, he has an extraordinary delivery style – careful and sincere. It is clear that he is actively participating in the discussion himself; he tailors the material such that it conveys all the crucial points while at the same time allowing the journey to these crucial points to be shaped by the students in the classroom. The class itself is comprised of over a thousand students, hanging on to every word from the teacher, and is a sight to behold.

A must watch. More accurately, a must think.

The mathematical constant shows up in strange places. Moreover, its significance is not as easy to grasp as that of the other famous constant, , because there is no easy physical object in whose context to imagine it. For example, is the ratio of the circumference to the diameter of a circle. Yes, it is irrational, but if you can get over that mystery (or ignore it for the time being), it is straightforward to imagine what is. Every circle does seem to have a certain circleness, which makes them all look the same. It is intuitively not hard to agree with the hunch that every circle has an unchanging ratio between the circumference and the diameter; and it makes sense to keep that ratio handy and give it a name.

The constant is considerably more elusive. It appears, at first, to be a number you would not go hunting after. You just happen to stumble upon in during one of your mathematical excursions; it seems interesting enough that you then pick it up and put in in your pocket for some potential use later. After stumbling upon the same thing along other mathematical excursions, in hindsight, it does seem to be something rather useful. Something you should have gone looking for.

Jacob Bernoulli, sometime in the late 1600s, stumbled upon this constant when he was trying to calculate the maximum compound interest that can be earned on an investment by compounding continuously.

If you invest 1 dollar in an account that earns 100% annual interest, then after 1 year the value of your investment will be 2 dollars assuming the compounding (interest calculation) happens once during the year. Assuming you do not touch the money and let it grow, by the end of 2 years, the money would have doubled again and you’ll have 4 dollars in all. Now, imagine the bank decided to compound every 6 months instead of once a year. The annual interest rate remains the same, 100%. 50% is applied at the end of 6 months and another 50% is applied at the end of the year. The 1 dollar grows to 1.50 after the first 6 months. For the second half of the year 1.50 dollars grow in value. At the end of the first year, the total value in your account is 1.50 + (1.50*50%), that is, 2.25 dollars. Not bad! Just by compounding every 6 months you made 2.25 instead of 2. Now you start thinking (as, perhaps, did Jacob Bernoulli). What if the bank compounded every 3 months? What if the bank compounded every month? Every week? Every day? Every second? Every picosecond? A similar question is what Jacob Bernoulli asked himself in the late 1600s. It turns out that this calculation results in an infinite series, which when added up closes in on 2.718281…This means if your bank was really generous and compounded continuously, your 1 dollar would become 2.718281 dollars at the end of the year. Quite a swing there – between 2 and 2.71, wouldn’t you say? And this given that the interest rate is unchanged between the two scenarios. (Aside: In general, pay some attention to how often your investments are being compounded, not just to the the annual rate of return.)

That number, 2.718281…, shows up in such seemingly disparate scenarios that it may be hard to see the connection between those scenarios. I recall that the way this constant was introduced to me in school was entirely different from the discussion above. I was taught that , where is a constant, is the only function whose derivative is equal to itself. That is, the slope of the curve is the value of the curve. Wow, I thought. That is quite a curve. I now feel that though that was correct, it may not have been the best way to introduce ; it does not say why is the only function that satisfies this property. Further, it is not explained why a function that does satisfy this property should even be of the basic form , where a is a constant.

I decided therefore to think through two things. First, I wanted to understand why the only function whose slope at a given is the same as its value. I wanted to approach this problem by using the knowledge of this special property, and then trying to come up with the function which would fit this requirement. Second, I wanted to tie together the two seemingly different arenas where e shows up – continuous compounding and as the base of the curve whose slope is equal to its value. There are other areas where shows up also. For example, if the probability of winning a wager is 1 in , and a man places bets, then the probability that he wins at least one bet is . This question is related closely to the hat check problem (also known as Derangements). also shows up in Euler’s Identity. I will need to do more digging before I attempt to understand, intuitively, why shows up in those scenarios so I may not tackle these in much depth in this article.

Why is the only function whose slope at a given is the same as its value?

Before I looked at this particular issue and attempted to uncover the function which gave me this special property, I decided to start small. I decided to look at some other canonical functions that might have been interesting. For example, “What is a function whose value, , is equal to ?”, “What is a function whose value is equal to ?”, “What is a function whose value is equal to ?”. The first question is easy. It defines . The curve for this is a straight line at a angle through the origin. The second one is easy as well. It defines . A parabola passing through the origin, (1,1) and (-1,1). The third question is the easiest of all, but the one that gets us closest to the question we are going after. It defines ! That is silly. Any function will fit. Notice, however, that in the third question the value of the function is dependent on the value of the function and not directly on . This starts to highlight the significance of the question we are trying to answer. We want to discover a function, . There is something quirky going on here. The right hand side is the derivative of the left hand side. So which comes first? It seems like you cannot really know the slope at , until you define . But then, you need the slope of to define because its value is the slope of . It is a bit confusing. The trick is to realize that this function is continuous. So the value of the slope of will be pretty close to the value of the slope of , the slope of the curve just prior to x. (This recursion also vaguely points to an upcoming infinite series.)

Now, let us try to find the function which satisfies the interesting property, . In words, this means the rate of growth of this function at a given point is equal to the value of the function at that point. More on that in a bit. Let us use the fact that the function is continuous and therefore, , the slope of the function at is given by the following.

Since we know that the function we are trying to uncover has the special property that , we can substitute instead of in the previous equation, to get the following.

This can be rewritten as

which in turn can be rewritten as

I will substitute with (this is unfortunately one of those things which you do after you have worked out the steps and realize that doing something like this may make things a little cleaner to understand).

So we get,

Swapping the right and left sides, we get the following.

The above equation represents in terms of . Using steps similar to the ones shown above if we express in terms of we get the following.

Extending this process times we get the following.

If is large enough that then the above equation becomes very interesting.

Notice that is still not a power to any base. So is still not an exponential curve, at least it does not look so. It does seem like it may expand into a sum of polynomials with the largest polynomial being of degree . A polynomial, however large in degree, grows slower than an exponential curve (as long as the degree is finite). However, note that is not a finite integer. It is an infinitely large number because is infinitesimally small. So there is still hope that this sum of polynomials may catch up to an exponential curve. Now, we apply a trick which changes the nature of this curve. A sum of polynomials becomes an exponential curve, and the goes up to a “position of power”. We simply reapply the equation to the and the term in the above equation to get the following.

Now, say we represent by a new variable, . Then the above equation can be written as follows.

which in turn can be written as

And now we start seeing the exponential form emerge. The is the power to which must be raised to evaluate .

This term, , when expanded out (remember that is infinitely large) hones in on a constant, 2.718281. And that is why, a special name is given to this term (). The right hand side thus becomes .

This takes the general form , where , the value of the function at . In other words we have proved that a function that satisfies the property that the curve’s value at any point is equal to the slope of the curve at that point follows the general form of an exponential curve.

How is continuous compounding related to the base of the exponential curve whose slope is equal to its value?

Now that we understand the general structure of the exponential curve and its special property, slope equals value, let us revisit the issue of continuous compounding. How is that related to this curve? It is intuitive once we realize that the rate at which money grows at any point in time is directly proportional to the amount of money in the account at that point in time. This is the same as saying the slope of the curve at a given point is directly proportional to the value of the function at that point. The interest rate plays the role of tempering the proportionality. For example, if the interest rate is 100%, the value of the continuously compounding account after units of time ( years) is . Here is the amount in the account at the beginning (value of the function at ). If the interest rate is 8%, then the value at time years is . Once we see that the two phrases, “slope of a curve is proportional to its value” and “rate of growth of something is proportional to the amount of that something”, are saying the same thing, it is clearer that these two ways of defining or discovering are equivalent.

Other places shows up

Another place shows up is called Derangements, as in the opposite of arrangement. A popular example used to explain this issue is the hat check problem. Say there are people who check in at a party, and the butler takes their hats as they enter the party. There are boxes to hold the hats, one per guest. The boxes have names on them to identify which guest’s hat should go into it. The butler, however, does not know the names of the guests. He puts the hats into those boxes randomly. When the number of hats (and boxes) is large, the probability that no hat is in the right box is . I have been able to work this out mathematically, but I still have not been able to understand this intuitively. Maybe the reason shows up here has nothing to do with the properties of . Maybe it just so happens that the same summation, which adds up to e, , shows up here.

Solution to Hat Check Problem: An Outline

The probability that all hat are in wrong boxes = 1 – probability that at least one hat is in the right box

= 1 – (prob. that the hat is in the right box OR prob that the hat is the right box OR … OR probability that the hat is in the right box)

Note that for the hat to be in the right box the first i-1 hats have to avoid that box. Therefore,

Probability that the hat is in the right box = (probability that the hat has not taken up the right box of the hat).(probability that the hat has not taken up the right box of the hat)…(probability that the hat has not taken up the right box of the hat).(the hat does take up the right box)

Using this, the probability that all hat are in wrong boxes

where,

Putting this back in the equation for the probability we are looking for, we get,

now, replacing by ,

The second term goes to as increases. So we are left with,

I do not know why shows up here. But, well, it does. Yet another place where shows up with unabated vigor is when representing complex numbers, and, of course, in that famous Euler’s equation that brings together 5 of the most interesting quantities in mathematics, . I will go into exploring that another time.

References and Acknowledgments

My discussions with my friend Mani, and my investigations on the internet led to this write up. I found the Wikipedia site very useful to tie together the seemingly unconnected ways in which the constant shows up in real life. The BetterExplained.com website has a very good illustrated tutorial for those who want to understand the concept of in simple terms.

Wikipedia.com – http://en.wikipedia.org/wiki/E_(mathematical_constant)

BetterExplained.com – http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/

Reference for the population figure: http://www.google.com/publicdata?ds=wb-wdi&met=sp_pop_totl&tdim=true&dl=en&hl=en&q=world+population

Figure1: Basic plan showing the material required

We have two 4′x8′ (4 feet by 8 feet) raised beds, which we use for vegetable plants. Kavita has been asking me to either buy or build a trellis for her climbing plants (cucumbers, tomatoes and eventually some types of squash and gourds). I read several websites online and decided to build a simple trellis using PVC piping. It took one trip to the local Home Depot, and then about 2 hours of work. The cost for the material was under $10 (I already had all the tools needed).

We decided to build one and test it out before getting carried away and building more. We decided that we would roughly want the trellis to be 4 feet wide by 5 feet high. At the Home Depot we did some quick calculations based on the basic design we had in mind and came up with a total of about 29 feet of PVC tube. The calculation is shown in Figure 1.

Figure 2: Materials and Tools for the project (4-way, + shaped, PVC connector missing)

The PVC pipes are sold in 10′ pieces. We got 3 pieces. We also got some string (I tried polypropylene string since I did not know any better, we’ll see how that works out).

Figure 2 shows most of the material and tools. The one caveat is, since I took this picture after completing the project, the one 4-way 1/2″ PVC pipe connector used at the center of the frame is missing from the picture. I had extra connectors of the other type, so I could use them for the picture. Also, one other thing that is missing from the picture is a power drill and drill bits. I used a 3/16″ drill bit to drill evenly spaced holes in the pipes to draw the string through, to create a framework.

Figure 3: Trellis plan with the stringing shown

Figure 4: Framework ready, stringing is yet to be completed

The spacing between the holes and how the trellis is supposed to look eventually is shown in Figure 3.

The thing that took the most time was measuring and marking the PVC pipes, cutting them to the right size with the saw, then measuring and marking the locations for the holes for the string to go through and then drilling the holes with the power drill. Since the PVC pipe keeps rolling about, making it stable before drilling is important. I just used an old rag to wrap around the pipe in order to hold it somewhat still.

Once the pieces were all ready, putting the trellis together took less than 10 minutes. There was no need for glue, since the connectors fit quite snugly. Figure 4 shows the trellis laid out on the lawn (with only one piece of string drawn through, Kavita will work on getting the rest of the mesh this evening).

We are not sure how well this will hold up, how long it will last etc. I will update the post with some pictures on the trellis in action, later.

Light, at the end of the tunnel

Dr. K. Ramamurthy (my dad’s uncle) wrote the following in response to this issue.

This gave rise to some thoughts which� I present here.

I agree that educated incompetence is a real phenomenon. I see it at work, at school, and even within our family. Typically, it manifests itself as a minor irritant with small repercussions; however, on occasion, it can bloat into a dangerous phenomenon, affecting life-changing decisions we make.

The people who think that they know (or, more accurately, who have been conditioned to think they ought to know) feel this pressure to pretend that they know. They may not know, but their ego does not let them accept or project their ignorance.

Now, take the case of a person who is formally uneducated. That person may not have any degrees, but may have sufficient common sense and, indeed, an abundant supply of humility. Unfortunately, lack of formal education, brings lack of confidence in addition to the abundance of humility. Abundance of formal education brings abundance of confidence (over-confidence, in most cases) and a lack of humility. The best kind of education encourages the right amount of confidence, and the right amount of humility.

That begs the question, “What is the fundamental problem with formal education?”.

Formal educational arrangements often cannot give one-on-one attention that is necessary to truly explore the limits of an individual’s potential. The result of this fundamental clash between individual potentials and group educational-arrangements is a lowering of the education bar. Further, in order to smoothly impart knowledge to a group, and then to smoothly measure the efficiency of that process, a common canonical framework (the box) must be constructed. All formal education builds within the confines of this framework. This explains why, in exams, typically, all questions have a reasonably fixed answer. Mathematical problems, typically, only provide precisely the data necessary to solve the problem. No unnecessary data is provided. In Chemistry lab, only the reagents and equipment necessary to prove the workings of a certain reaction are provided. To me it is not at all surprising that relying only on formal education as a means to educating yourself is bound to teach you a canonical view of the world and its problems. It teaches you to construct the boundaries of your sandbox of thoughts before trying to solve a problem. It teaches you that when you have a problem you just need to use the resources readily visible within this sandbox. It does not, typically, teach you where to look for resources to solve a problem. It does not teach you how to get rid of the sandbox. And most significantly it does not teach you how to look *for* problems. Problems are assumed to be handed to you, with your job being restricted to looking for solutions. This is a fundamental problem with all forms of formal education. And, if you think about it, this formal educational structure is a direct result of the simplifications that *must* be made for any, large-scale, practical, hand-off of human understanding from one generation to the next. That is, I am not blaming the formal education. I am saying it is doing exactly what it was meant to do. To identify and solve real-life problems, it helps to know how to solve a canonical, artificially-created, abstract versions of the real-life problem. Formal education signed up to distill real-life problems into canonical problems and teach us how to proceed from that point further. It did not sign up for more.

Now, the responsibility of the second half of education – the ability to identify a real-life problem (before it is handed over to you on a platter), the ability to whittle down a real-life problem into a canonical form, the ability to identify parameters that affect the solution to this problem from a much larger selection than you are used to dealing with – lies with the individual. Even if we could, in a perfect society, provide one-on-one attention to each student, human brains are not very good at communication of complex thought. The only brain that can efficiently sift through the deluge of complex thoughts that arise inside it, it the very same brain where these thoughts arise. This ability to learn by looking inwards has been called meditation, self-awareness, reflection, wisdom, and, common sense. This is an ability that cannot be taught by formal educational tools. It is, therefore, a resource available to the educated and the uneducated in equal measure. The uneducated person, probably enjoys it in purer, undiluted proportions. The formally educated person may, unfortunately, end up suspicious of such free-ranging thoughts arising in his or her mind and proceed to quell them. This may explain why educated people often appear to be unimaginative and conforming – thinking “inside-the-box” out of habit.

Maybe this makes a case for a statutory warning to be legally enforced on all formal educational fora – FORMAL EDUCATION, HOWEVER ADVANCED, IS ONLY RESPONSIBLE FOR A SMALL FRACTION OF YOUR OVERALL EDUCATION

(PS: By the way, be careful not to interpret this to mean that everything your maid-servant says is wise.)

I read some more interesting insights form Dr. Ramamurthy a few emails down the chain.

And here is my response:

In fact, rational thinking is a necessary condition to allow out of the box thinking. It is not a sufficient condition though. A bit of inspired (some may even call it non-rational, creative or artistic) thinking becomes necessary as well. This is that spark of subconscious creativity that conscious rational thinking can then give form to.

All this talk of rational vs creative, boxed-in vs out-of-the box thinking reminds me of the following incident which happened recently.

"No, K. The problem is not that I use only my left brain. The problem is that I can only use the brain that is left."