Category Archives: Problem Solving

Counting Cycles

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During one of my reading projects in 2015, I read about the Enigma cipher machine. While reading about it, I came to know that the number of possible keys of this machine is 7, 156, 755, 732, 750, 624, 000. One can see the counting procedure at pp. 22 of this document. But the counting procedure was not found to be satisfactory by most members of the audience (during my presentation). My failure to convince the audience that the counting procedure was correct, lead to my distrust in the counting arguments in general. Many times, I still, find the counting procedures controversial.

So, in an attempt to regain the trust, I will present two counting procedures for counting the number of cycles of length r when n objects (colours/beads/numbers) are given.

Procedure A: Using multiplication principle
Step 1: Choose r objects from the n choices.
Step 2: Arrange the selected r objects in a cyclic order.

  1. Since the Step 1 and Step 2 are independent of each other but should be performed together, we will multiply the results (i.e. use the multiplication principle). From Step 1 we will get \binom{n}{r} and from Step 2, we will get (r-1)! as per the circular permutation formula. Hence we get:

\displaystyle{\# r-\text{cycles from } n \text{ objects} =\binom{n}{r}\times(r-1)! = \frac{n!}{r (n-r)!}}

Procedure B: Using division principle
Step 1: Permute r of the n objects.
Step 2: Realise the mistake that you counted the permutations r extra times because these circular permutations of objects are equivalent since the circle can be rotated.

Since in Step 2 we want to correct the overcounting mistake of Step 1 performed for different objects simultaneously, we will divide the result of Step 1 by the result of Step 2. From Step 1 we will get ^n P_r and from Step 2 we will get r. Hence we get:

\displaystyle{\# r-\text{cycles from } n \text{ objects} =\frac{^n P_r}{r} = \frac{n!}{r (n-r)!}}

I am still not happy with the Procedure B, so if you have a better way of stating it please let me know.

When the intuition is correct

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Before starting college I read Paul Zeitz’s The Art and Craft of Problem Solving (a must read book along with the books by Arthur Engel and Terence Tao) and after the first example to illustrate Psychological Strategies he writes (pp. 15):

Just because a problem seems impossible does not mean that it is impossible. Never admit defeat after a cursory glance. Begin optimistically; assume that the problem can be solved. Only after several failed attempts should try to prove impossibility. If you cannot do so, then do not admit defeat. Go back to the problem later.

And today I will share a problem posed by August Ferdinand Möbius around 1840:

Problem of the Five Princes:
There was a king in India who had a large kingdom and five sons. In his last will, the king said that after his death the sons should divide the kingdom among themselves in such a way that the region belonging to each son should have a borderline (not just a point) in common with the remaining four regions. How should the kingdom be divided?

The hint is in the title of this blog post. The solution is easy, hence I won’t discuss it here. The reader is invited to write the solution as a comment to this post.

Ulam Spiral

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Some of you may know what Ulam’s spiral is (I am not describing what it is because the present Wikipedia entry is awesome, though I mentioned it earlier also). When I first read about it, I thought that it is just a coincidence and is a useless observation. But a few days ago while reading an article by Yuri Matiyasevich, I came to know about the importance of this observation. (Though just now I realised that Wikipedia article describes is clearly, so in this post I just want to re-write that idea.)

It’s an open problem in number theory to find a non-linear, non-constant polynomial which can take prime values infinitely many times. There are some conjectures about the conditions to be satisfied by such polynomials but very little progress has been made in this direction. This is a place where Ulam’s spiral raises some hope. In Ulam spiral, the prime numbers tend to create longish chain formations along the diagonals. And the numbers on some diagonals represent the values of some quadratic polynomial with integer coefficients.

ulam_spiral_by_splatbang-d5b0yfj

Ulam spiral consists of the numbers between 1 and 400, in a square spiral. All the prime numbers are highlighted. ( Ulam Spiral by SplatBang)

Surprisingly, this pattern continues for large numbers. A point to be noted is that this pattern is a feature of spirals not necessarily begin with 1. For examples, the values of the polynomial x^2+x+41 form a diagonal pattern on a spiral beginning with 41.

 

Finite Sum & Divisibility

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I wish to discuss a small problem from The USSR Olympiad Problem Book (problem 59) about the finite sum of harmonic series. The problem asks us to prove that

\displaystyle{\sum_{k=2}^{n} \frac{1}{k}}  can never be an  integer for any value of n.

I myself couldn’t think much about how to prove such a statement. So by reading the solution, I realised that how a simple observation about parity leads to this conclusion.

Firstly, observe that among the natural numbers from 2 to n there is exactly one natural number which has the highest power of 2 as its divisor. Now, while summing up the reciprocals of these natural numbers we will get a fraction as the answer. In that fraction, the denominator will be an even number since it’s the least common multiple of all numbers from 2 to n. And the numerator will be an odd number since it’s the sum of (n-2) even numbers with one odd number (corresponding to the reciprocal of the number with the highest power of 2 as the factor). Since under no circumstances an even number can completely divide an odd number, denominator can’t be a factor of the numerator. Hence the fraction can’t be reduced to an integer and the sum can never be an integer.

Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

  • \displaystyle{\sum_{p}\frac{1}{p}}   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

  • \displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}, where s is any complex number with \text{Re}(s)>1.

The outline of proof,   when s is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html

Solving Logarithmic Equations

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While reading John Derbyshire’s Prime Obsession I came across the following statement (clearly explained on pp. 74):

Any positive power of \log(x) eventually increases more slowly than any positive power of x.

It is easy to prove this (existence) analytically, by taking derivative to compare slopes. But algebraically it implies that (for example):

There are either no real solution or two real solutions of the equation
\log(x) = x^\varepsilon
for any given \varepsilon>0.

Now the question that arises is “How to find this x?” I had no idea about how to solve such logarithmic equations, so I took help of Google and discovered this Mathematic.SE post. So, we put \log(x)=y and re-write the equation as:

y=e^{y\varepsilon}

Now to be able to use Lambert W function (also called the product logarithm function) we need to re-write the above equation, but I have failed to do so. 

But using WolframAlpha I was able to solve \log(x)=x^2 to get x=e^{\frac{-W(-2)}{2}} (which is an imaginary number, i.e. no real solution of this equation) but I was not able to figure out the steps involved. So if you have any idea about the general method or the special case of higher exponents, please let me know.

Combinatorial Puzzle

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This is a continuation of previous post:

How many distinct numbers can be formed by using four 2s and the four arithmetic operations +,-,\times, \div.

For example:
1 = \frac{2+2}{2+2}=\frac{2}{2}\times\frac{2}{2}
2 = 2+\frac{2-2}{2}=\frac{2}{2}+\frac{2}{2}
3 = 2+2 - \frac{2}{2}
4 = 2+2+2-2 = (2\times 2) + (2-2)
(note that some binary operations do not make sense without parenthesis)

I have no idea about how to approach this problem (since I am not very comfortable with combinatorics). So any help will be appreciated.

Edit[29 May 2017]: This problem has been solved in the comments below.