# A Blind Robot Beside an Infinite Wall

Let’s think about the following problem:

Consider a wall that stretches to infinitely in both directions. There is a robot at position $$0$$ and a door at position $$p\in\mathbb Z$$ along the wall $$(p\neq 0)$$. The robot would like to get to the door, but it knows neither $$p$$, nor the direction to the door. Furthermore, the robot cannot sense or see the door unless it stands right next to it. Give a deterministic algorithm that minimizes the number of steps the robot needs to take to get to the door.

This problem quite famous Continue reading A Blind Robot Beside an Infinite Wall

# Prime Counting Function and Chebyshev Bounds

The distribution of primes plays a central role in number theory. The famous mathematician Gauss had conjectured that the number of primes between $$1$$ and $$n$$ is roughly $$n/\log n$$. This estimation gets more and more accurate as $$n\to \infty$$. We use $$\pi(n)$$ to denote the number of primes between $$1$$ and $$n$$. So, mathematically, Gauss’s conjecture is equivalent to the claim

$\lim_{n\to\infty}\frac{\pi(n)}{n/\log n}=1$

# Multiplying Two Polynomials with Fast Fourier Transform

Polynomial multiplication is one of the most important problems in mathematics and computer science. It can be formally defined as follows:

You are given two polynomials roughly of equal size$A(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}$$B(x)=b_0+b_1x+\dots+b_{n-1}x^{n-1}$find a polynomial $C(x)=c_0+c_1x+\dots+c_{2n-2}x^{2n-2}$ such that $$A(x)\cdot B(x)=C(x)$$. At first, this may look like an easy problem, Continue reading Multiplying Two Polynomials with Fast Fourier Transform

No form of beauty is as flawless, as abstract, or as perennial as the mathematical beauty that derives from unassailable logical perfectness, for a mathematical proof is not earthly science, but a poetry written in logic.

# Generating All Subsets of Size k in Python

Suppose you are given a set $$S$$ with $$n$$ elements and you need to generate every subset of size $$k$$ from it. For instance, if $$S=\{3,4,5\}$$ and $$k=2$$, then the answer would be $$\{3,4\}$$, $$\{3,5\}$$, $$\{4,5\}$$. So, how would you do that in Python?

First of all, this is a really simple exercise. Stuff like this often comes up when someone writes a moderately large script. I wanna talk about this one in particular because it has a combinatorial solution. Continue reading Generating All Subsets of Size k in Python

# Lie Detectors and the Story of Halting Turing Machines

Is it possible to design a machine that detects lies? Sure, people have already built devices called polygraphs that monitor blood pressure, respiration, pulse etc. to determine if a person is giving false information. However, that is not same as “detecting lies” because polygraphs merely measure physical effects of telling lies. On the top of that, they are hella inaccurate. I am talking about REAL lie detectors, ie. machines that can instantly validate the statements themselves. Can we actually make such machines? (*insert new startup idea*) Continue reading Lie Detectors and the Story of Halting Turing Machines

ভোরের কুয়াশাস্নাত রোদের আঁচল
লুকিয়ে রেখো আমার কিছু ব্যর্থতা।
অর্ধরাতের জন্য পুরোনো জৌলুশটুকু বড্ড প্রয়োজন।
আজ আমার মন ভাল নেই।

# নৈশব্দ

সেই দিনগুলো ছিল উচ্ছল,
ছিল পলাশীর মত উত্তাল,
ছিল আকাশের তেজী রোদ্দুর,
নগরীর বেপরোয়া কোলাহল।
সেখানে ঘড়ি মেনে আসে রাত্রি
যেমন জাহাজের কোন যাত্রী
মৃত ডাহুকের মত নিশ্চুপ
জোছনার ফুল ঝরে টুপ টুপ
বুকের পাঁজরে আহত আবেগ
প্রেমিকার চোখে ভাসে অভিযোগ।
যদি দিগন্ত বেয়ে চুপিসারে
কোন ভালবাসা এসে কড়া নাড়ে
তবে সঁপে দিও তাকে যৌবন,
দিও অপরাধমাখা চুম্বন
যেভাবে সকালের ফোটা পলাশে
বসন্তের রঙ মেশে সহাসে।

(অসমাপ্ত)

# শহর

প্রতিটি ক্লান্ত দিনের শেষে
পরমকাঙ্ক্ষিত বিকেলের মত
তোমাকে আমার প্রয়োজন।

প্রতিটি প্রিয় ছুটির দিনে
আলস্যমাখা দিবানিদ্রার মতো
তোমাকে আমার প্রয়োজন।

একবার এসো না এই শহরে!
তোমাকে আমার বড্ড প্রয়োজন।

# A Special Case of Zsigmondy’s Theorem

Theorem (Zsigsmondy):

1. For two coprime positive integers $$a$$ and $$b$$ and for any positive integer $$n$$, $$a^n-b^n$$ has a prime divisor that does not divide $$a^k-b^k$$ for all positive integers $$k< n$$ with the following exceptions:
• $$n = 2$$ and $$a+b$$ is a power of two.
• $$n=6, a=2, b=1$$.
2. For two coprime positive integers $$a$$ and $$b$$ and for any positive integer $$n$$, $$a^n+b^n$$ has a prime divisor that does not divide $$a^k+b^k$$ for all positive integers $$k< n$$ with the following exception:
• $$n=3, a=2, b=1$$.

This theorem is very helpful in solving many olympiad number theory problems. However, its use is often frowned upon, as this theorem is quite hard to prove using only elementary mathematics. (I am aware of the proof using Cyclotomic Polynomials, but that’s not “elementary enough” in my opinion.)

In olympiad mathematics, one can get away in most of the cases by just using this theorem for a fixed pair of $$n$$ and $$k$$. This modification allows us to provide a very simple proof using LTE. Here I’ll restate this particular case, and then prove the first part. The second part can be proven analogously.

Theorem (Zsigsmondy Special Case):

1. For two coprime positive integers $$a$$ and $$b$$ and any two positive integers $$n$$ and $$k$$ with $$k<n$$, $$a^n-b^n$$ has a prime divisor that does not divide $$a^k-b^k$$ with the following exception:
• $$n = 2, k=1$$ and $$a+b$$ is a power of two.
2. For two coprime positive integers $$a$$ and $$b$$ and any two positive integers $$n$$ and $$k$$ with $$k<n$$, $$a^n+b^n$$ has a prime divisor that does not divide $$a^k+b^k$$ with the following exception:
• $$n=3, k=1,a=2, b=1$$.

Proof of 1:

Suppose $$a^n-b^n$$ and $$a^k-b^k$$ share same set of prime divisors. This implies $$a^n-b^n$$ and $$a^{\gcd(n,k)}-b^{\gcd(n,k)}$$ also share same set of prime divisors. Assuming $$A=a^{\gcd(n,k)}$$ and $$B=b^{\gcd(n,k)}$$, we could follow the rest of this argument and get to a contradiction. So, without loss of generality, we may assume $$k=1$$.

Now we consider two cases:

Case 1: $$n$$ is a power of $$2$$.
If $$n=2$$ and $$a+b$$ is a power of two, we arrive at one of the listed exceptions. Now note that,
\begin{align*}&\gcd(a+b, a^2+b^2)\\ =&\gcd(a+b, (a+b)^2-a^2-b^2)\\ =&\gcd(a+b,2ab)\\ =&\gcd(a+b,2)\end{align*}
This implies both $$a+b$$ and $$a^2+b^2$$ can’t be powers of two unless $$a+b=2\implies a=b=1$$. Hence at least one of them must have an odd prime divisor. Furthermore, this prime divisor does not divide $$a-b$$ since $$\gcd(a-b, a^2+b^2)=\gcd(a-b,2)$$ (following the previous steps). However, if $$n>2$$, then $$a^2+b^2|a^n-b^n$$, implying that odd prime will divide $$a^n-b^n$$. This is a contradiction.

Case 2: $$n=2^md$$ with $$d>1$$ being odd.
Without loss of generality, we may assume $$n>1$$ is odd, since $$a^d-b^d|a^n-b^n$$ and it is sufficient to show that $$a^d-b^d$$ has a prime divisor that does not divide $$a-b$$.  From LTE, $$v_p(a-b)+v_p(n)=v_p(a^n-b^n)$$ for each odd prime $$p|a-b$$. Furthermore, $\frac{a^n-b^n}{a-b}\equiv na^{n-1}\equiv 1\pmod 2$ implying $$v_2(a-b)=v_2(a^n-b^n)$$. Therefore, we can conclude $\frac{a^n-b^n}{a-b}=\prod_{p|\gcd(n,a-b)}p^{v_p(n)}\leq n$ which is impossible for $$n>1$$. This raises another contradiction and we are done.