List of Problem Solving books in Maths

Here's a complete list of problem solving books that I like and recommend -

1. Art and craft Of Problem Solving

2. Mathematical Circles

3. Techniques Of Problem Solving

4. Winning solutions

5. Arthur Engel

6. Solving Mathematical Problems A Personal Perspective by Terence Tao

7. Springer - Problem Books in Mathematics

Conversation with Prof. Partha

The article (paper) http://www.freewebs.com/profpartha/publications/futile.pdf

My email to the prof when I read this at 6 a.m. (I was very sleepy) :

Dear Sir,

I am third year undergraduate student at Dept of Mathematics, IIT KGP. And the first thing I would like to say is that the Department Sucks!
I apologize for the language but this is the best I can do considering the enormous amount of frustration I have right now.

I love maths (mainly pure maths) and I am pretty good at it, well at least good enough to understand and appreciate the beauty and depth of this subject. I was up all night going through the awesome library of math books that I have created at my room, solving some olympiad problems and now I'll sleep till the evening missing all my lectures. Is it the right thing to do? Well if the "professors" who teach us maths have no love for the subject or teaching then I think that it is. And I even have real doubts about the competence of some profs as well. Ask them a problem and they'll fumble and they'll ask for some time till the next class and even then it seems like they just mugged up a solution.

It's pathetic! I want to complain, I want to shout at the administration, at the profs, abuse them! But I can't, because I fear they'll ruin my only chance of escape - going for a PhD in maths at a better place, a place where the people teach because they love teaching and because they are good at it.

I am sorry for bothering you with this mail, but after reading your article I felt that you are one of those people who can understand my frustration and I felt like venting.


--
Anurag Bishnoi,
Third Year (undergraduate) ,
Department of Mathematics ,
IIT Kharagpur


P.S. : When I talked abt the professors here, I was talking about the majority. Since I do have a few good profs who are really good teacher and mathematicians like Prof V.K. jain, Prof. Pratima Panigrahi, Prof. Abhiji Das (Computer Science dept).



His reply at 8 am :
I am happy to hear the views on the side of the fence. I have always
seen maths like what you are experiencing. That is why I was never a
good student of maths. But one day, it dawned on me that there is a
better way to look at things. YES, stop blaming your teachers. It is
just a mental frame of mind. There is no such thing as a bad teacher,
and there is no such thing as a great teacher either. It is all within
us, students. Teachers are not GODs. So you cant expect them to know
everything. Besides, they are constrained by the system which imposes
unrealistic goals and targets (explained in my paper). We can overcome
all this, by trust. Trust yourself. Keep up the effort. You will
succeed. How do you explain an average student of maths (me) becoming
a teacher and crusader or maths ?

Cheer up. You will succeed. Never mind the frustrations you are facing.

All the best,

partha



And then later ...

our mail is now on my blog. I hope some teacher will see it and find a way out.

http://partha-the-prof.blogspot.com/2011/04/lament-of-student.html

partha

Ternary Numeral System

First some wiki links to get familiar with the base 3 numerical system -

Ternary numeral System : base 3 system as most of the people imagine it to be

Balanced ternary : an interesting form with several advantages.

I will discuss here, the "two" problems that got me interested in "ternary" numeral system -

note : now that I have mentioned ternary, these problems are mere exercises.

Problem 1 :

"""

A 40 kg stone breaks into 4 pieces. Determine the weight of those 4 pieces such that using them and a "2-pan" balance you can measure all the weights from 1kg to 40kg.

"""

Solution : This can be easily solved by playing around a bit. Here's one line of thought -

It's a good guess to start with that exactly 1 piece must weigh 1 kg. And then to measure 2kg, lets take a piece weighing 3kg. So till now we can measure weights from 1 to 4. What should be the 3rd piece. We can check one by one weights from 5 onwards, which is the weight that gives longest sequence of weights starting from 5. We can easily see that if we take the 3rd piece to be 9kg, then we can measure all the weights from 1 to 13. So the 4th piece is 40 - 13 = 27kg.

And now it can be easily verified 1,3,9,27 solves the problem.

Generalizing the above solution suggests that powers of 3 have some "nice property" which can be used in this "2-pan" balance problem. Let's see why is it so.

Some formalism will be useful here -

set the 4 pieces be x1,x2,x3,x4. The constraints are

a) x1 + x2 + x3 + x4 = 40

b) a1*x1 + a2*x3 + a3*x3 + a4*x4 gives all the numbers from 1 to 40 with ai's in {-1,0,1}

it is the second constraint that suggests base 3 system here. The coefficients are 3 consecutive numbers, -1, 0 and 1. Using euclidean division we can prove that any number n can be written as a0 + a1*3 + a2*3^2 ... where ai is in {r-1, r, r + 1} for some integer r. For r = 1 we get the standard base 3 representation and with r = 0 we get the balanced ternary system.

So if we take 4 weights {1,3,9,27} using them and a balanced ternary system we can represent all the numbers from -40 to +40. And as it should be clear that using a 2-pan balance is same balanced ternary system, we know that we can measure all the weights from 1 to 40.

Problem 2 :

"""

You have eight balls all of the same size. Seven of them weigh the same, and one

of them weighs slightly more. How can you find the ball that is heavier by using a 2-pan balance and only two weighings?

"""

Solution : Thinking for a while gives the following heuristic -

Take two groups of 3 balls say A,B. Weigh them, if A = B then the heavy ball is in remaining 2 balls which can be found in another weighing. Else, say A > B. Now, we need to find the heavy ball from a group of 3 balls using 1 weighing. Even this is quite easy as you can pick 2 balls, compare them and if they are equal the one that is left is the answer else the heavier one is.

Let's Generalize!

What is the minimum number of weighings that is required for identifying a the heavy balls from N balls.

Say f(N) is the solution.

Playing around a bit gives this values for f

f(2) = f(3) = 1

f(4) = f(5) = f(6) = f(7) = f(8) = f(9) = 2

If you observe carefully, then f(N) is the number of "trits" ( :P ) in N. (N > 1)

which is actually clear if you see the recurrence relation

f(N) = f(N/3) where N/3 = floor(N/3)

and f(3) = 1

hence f(N) = floor( log3(N) + 1) = number of trits in N

Being Human

1. Always in search for a physical/emotional thing that generates an urge to live.

2. Always have another person to care for, in the fear of having no one when you need to be cared.

3.

...

Hypocricy

I am not a Hypocrite, I am just absent minded.

We live most of our life in hibernation.
the only times that we have been awake are those whch we can recall by closing our eyes.

continuous and the discrete

it seems that the continuous is just in our mind, our thought an our logic while the nature is discrete............