How Game of Thrones Should Have Ended

This is a very unusual post for this blog. I hope my regular readers will bear with me.

I love Game of Thrones. I have read the books and the companion novels. I have been following the series for years. I am familiar with most of the fan theories too. Tonight the series finale aired. I didn’t like how the show ended. It felt rushed and very baffling.

I understand that I don’t get to question the showrunners’ creative decisions about the show. However, as a fan I can disagree with them. This post is about how I feel the show should have ended. It was written in haste, so please forgive the typos and the repeating sentences.

Before I begin, please note that this article contains major spoilers. Read at your own risk.

The Long Night

In my honest opinion, the army of the living should have lost this fight. They should have retreated back to Iron Islands. However, this goes too far from the show’s ending, so I shall not pursue this thread. I shall try to modify the show’s version of the story so that it makes some sense.

The battle plan of the army of the living SUCKED. I won’t even try to justify it.

First of all, the cavalry charge was stupid. There are other articles on the internet suggesting better ways to use the Dothraki, so I’ll not get to it.  Secondly, if the army of the living wanted to defend the castle,  they should have also dug a wider, deeper trench around winterfell. The troops should be positioned behind the trench. In that way, the wights would not be able to cross the trench just by dropping some bodies. They would eventually fill in the trench, but it would require more time while they take damage. And when they do cross the trench, they would face fresh troops awaiting for them.

They should have manned the wall from the beginning, not when the wights are pursuing them.

In the show, the army of the living had catapults, but they got overrun only after throwing a couple of vollies. A more realistic solution would be to position the catapults inside the castle walls so that they may function for longer. Also, throwing jars full of wildfire, instead of rocks, might have been more effective since wights are susceptible to fire. (In fact, it was out of character for Tyrion not to suggest this strategy, after all, he had destroyed King Stanis’s fleet in the Blackwater Bay with wildfire.)

Nobody should have taken shelter in the Crypt. That’s where the Starks bury their dead. Someone should have spotted that!

Brianne Should have died in Jamie’s arms. Her character arc was complete.

Sansa should have died. She was useless throughout the entire story. Too bad, north needs a ruler.

In the show, the living were overwhelmed by Night King’s army. Night King was coming for Bran. At that point, out of desperation, Bran should have tried a new strategy. As we have seen in the Hodor story, warging in the past can affect the future. Inspired from this, he would try to go back thousands of years, when Night King was a human, and try to warg into him to stop the army of the dead at present. This would freeze the dead and white walkers for am moment, which is enough for Jon to rush and and stab the Night King. That would fulfil Jon’s role as the Azor Ahai.

Again, it should have been Jon who killed Night King. Otherwise, his resurrection has no importance, storywise.

As for Bran, altering past should have serious side-effects, as the previous Three Eyed Raven said,

The past is already written. The ink is dry.

He could get stuck in the past for good. Then, he could become Bran the Builder.

Rhaegal’s Death

The overpowered Scorpion plot had more holes than Swiss cheese. Someone calculated that the bolts needed to have an initial velocity of 2000m/s to make it happen. Also, the dragons are practically invulnerable in the air according to the books.

The show should have given Euron a dragon horn instead. These horns were used in Old Valeriya and are presumably capable of controlling dragons. Dany should have spotted Iron fleet from far above. Unknowing that Euron has a dragon horn, she would proceed to destroy the Iron fleet. When she was close enough, Euron would blow the horn and take control of both dragons. However, Dany should regain control of Drogon soon, because of special bond with him. Rhaegal would go berserk, attacking Dany’s ships and pursuing Drogon. Dany would have no choice but to kill Rhaegal with Drogon. That would be a devastating blow for both Dany and Drogon. She would mourn for Rhaegal and become more revengeful and destroy Euron’s fleet with remaining strength.

Destruction of King’s Landing

Jaime should accompany Dany, not to save Cersei, but to kill her. Suspecting Dany’s plan to burn everyone in King’s Landing, Tyrion would ask Jaime to secretly go to the Red Keep and convince Cersei to surrender. Jaime would do so, while Dany would attack the City’s defenses.

After meeting Jaime, Cersei would pretend to surrender. She would ring the bell and open the gates. When Dany’s troops get inside the walls, they would be greeted by common folk. Then the folk would suddenly attack Dany’s soldiers. It would become apparent that Cersei has hidden her soldiers among common folk and using them as a meat shield to slaughter Dany’s troops. There should also be archers in the buildings. Common folk would throw rocks at Dany and Drogon showing that she was undesirable. This would finally provoke Dany  and she would go on a rampage. She would also command Grey Worm to kill everyone who fought for Cersei.

On the other hand, seeing what Cersei had done, Jaime would finally choke Cersei, thus fulfilling Maggy the Frog’s prophecy.

The Ending

After witnessing the atrocities committed by Dany, Grey Worm and their army, Arya will finally kill Grey Worm and steal his face. Then she would get to Dany in the throne room and kill her. Seeing his mother dead, Drogon will burn Arya, melt the Iron Throne and fly away with Dany’s corpse.

Finally, Jon should become the king, with Tyrion becoming his hand. The post credit scene should show Night King is being reborn. (Implying that somehow the timeline has changed, which would provide the premise for HBO’s spin off.)

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 in Bangladesh because of Dr. Mohammad Kaykobad, a renowned Bangladeshi computer scientist. He maintains a list of interesting problems like this one to attract high school kids into problem solving.

Dr. Kaykobad’s formulation of this problem is quite ambiguous. He simply asks how the robot should get to the door, without specifying whether the algorithm should be deterministic. It matters as we shall see.

In this article, I shall provide a deterministic algorithm that takes at most \(9|p|\) steps for any \(p\). I shall also prove that any deterministic algorithm will take at least \(8|p|\) steps. I suspect this bound can be improved up to \(9|p|\), but I didn’t have time to think much.

My \(\leq 9|p|\) Deterministic Algorithm

for i = 0, 1, 2, ...:
    walk up to (-2)^i, but don't walk past p
    if you find the door: break loop
    else: get back to 0

Suppose \(p\) is found on \(i\)-th round. Then, \(|p| > 2^{i-2}\), otherwise \(p\) would have been found on \((i-2)\)-th round. Now, it is easy to see the robot takes \(2^1+\dots+2^{i}<2^{i+1}\) steps prior to \(i\)-th round. It takes \(|p|\) more steps to get to \(p\). Therefore, the total number of steps is at most \[|p|+2^{i+1}=|p|+8\cdot2^{i-2}<9|p|\] As \(|p|\to\infty\), total number of steps converges to \(9|p|\).

(Disclaimer: Finding a \(9|p|\) algorithm was a homework problem in my class 6.046. The algorithm and the runtime analysis shown above was part of the official solution. My algorithm was the same, but my runtime analysis was much uglier.)

Other Deterministic Algorithms

In this section, we shall prove that any deterministic algorithm for this problem will take at least \(8|p|\) steps.

In any algorithm, the robot has to walk some steps to the left/right, then turn back and walk some more steps, then turn again, and so on. Without loss of generality, we may assume the optimum algorithm takes first step to the right.

Consider the the optimum deterministic algorithm \(OPT\) with minimum number of turns \(n\). Assume that \(\{a_i\}_{i=1}^n\) denotes the number of steps \(OPT\) takes during the \(i\)-th turn. Define \(A_i=\sum_{j=1}^i(-1)^{j+1}a_j\) to be the position of the robot after \(i\) steps. Also define \(A_0=0\). Suppose \(OPT\) takes fewer than \(8|p|\) steps for all \(p\).

Claim 1: \(a_n>\cdots a_i>\cdots a_1 > 0\) for \(n> i > 1\).

Proof: If the robot takes \(0\) step at any turn, we can simply ignore that turn and merge the steps of the next turn with the steps of the previous turn. Hence, \(a_1,a_i,a_n > 0\).

Suppose \(a_{i-1}-a_{i} \ge 0\) for some \(i\). This means at \(i\)-th step, the robot backtracked to somewhere between \(A_{i-1}\) and \(A_{i-2}\). Then, at the next step, it must go past \(A_{i-1}\), otherwise, \((i+1)\)-th step will revisit some previously visited vertices and it can be removed by walking to \((a_{i-1}-a_i+a_{i+1})\) at \(i\)-th step. This is a contradiction to the minimality of \(OPT\). Hence, we must have \((a_{i-1}-a_i)+a_{i+1} > a_{i-1}\). However, in this case, we can get a better algorithm by taking \(a_{i-1}-a_i+a_{i+1}\) steps at the \((i-1)\)-th turn, eliminating both \(a_i\) and \(a_{i+1}\). This is another contradiction. Hence, \(a_{i-1}-a_{i}< 0\).

\(a_{n}> a_{n-1}\) must be true, otherwise the \(n\)-th step is redundant.

Claim 2: \(a_i=|A_i|+|A_{i-1}|\) for all \(i\ge 1\). Furthermore, \(|A_{i+2}|-|A_{i}|=a_{i+2}-a_{i+1}>0\).

Proof: Trivial to check for odd and even \(i\). This claim also implies \(|A_{i+2}|>0\).

Claim 3: For \(i\ge 2\),
\[\sum_{j=1}^{i}a_j=|A_{i}|+2\sum_{j=1}^{i-1}|A_j|\]

Proof: Apply claim 3 for every \(a_j\).

Claim 4: \(|A_i| \leq 2.5|A_{i-1}|\) for \(i\ge 6\).

Proof: Suppose \(i\) is odd since the other case is analogous. Consider \( p= -|A_{i-1}|- 1 < A_{i-1}\) . The robot does not find \(p\) until at least \((i+1)\)-th turn, since \(|A_{i-1}|>|A_{i-3}|>\cdots>|A_2|\) (claim 2). Thus,
\[
\left(\sum_{j=1}^{i}a_j\right)+a_{i}+1 < 8(|A_{i-1}|+1) \] The left hand side can be thought as follows. First, the robot walks to \(A_{i}\). After \(a_{i}\) more steps in the opposite direction, it reaches \(A_{i-1}\) again. Then it takes 1 more step to reach \(p\). Rewrite the left hand side using claim 2 and 3 to get \[ 1+|A_{i-1}|+2\sum_{j=1}^{i}|A_j|< 8|A_{i-1}|+8 \] It simplifies to \[ 1+2|A_{i}|+2\sum_{j=1}^{i-2}|A_j|< 5|A_{i-1}|+8 \] \[ \implies |A_{i}|< 2.5|A_{i-1}|+3.5-\sum_{j=1}^{i-2}|A_j|<2.5|A_{i-1}| \] The last inequality follows from the fact that \(i\ge 6\) and \(|A_j| > 0\). (Q.E.D)

Notice that claim 4 can be strengthened iteratively.

Claim 5: Suppose we have proven \(|A_i|\leq b|A_{i-1}|\) for some \(b\) and for all sufficiently large \(i\). Let \(b’=(2.5-1/b-1/b^2)\). We claim that \(|A_k|\le b’|A_{k-1}|\) for all sufficiently large \(k\).

Proof: Pick a sufficiently large \(i\) such that \(|A_{i-2}|\leq b|A_{i-3}|\) We proceed like in the previous proof to get
\[
|A_{i}|< 2.5|A_{i-1}|+3.5-\sum_{j=1}^{i-2}|A_j|\qquad(1) \] Notice that \[|A_{i-2}|+|A_{i-3}|\ge \left( \frac{1}{b}+\frac{1}{b^2} \right)|A_{i-1}| \] Since \(i\) is sufficiently large, the right hand side of (1) is at most \[(2.5-1/b-1/b^2)|A_{i-1}|=b'|A_{i-1}|\] Hence, \(|A_{i}| < b' |A_{i-1}|\) . (Q.E.D.)

We start with \(b=2.5\) and repeatedly apply claim 5. We find the following values for \(b\):

\[
\begin{array}{cl}
\text{iter.} & b\\
1 & 1.94\\
2 & 1.71883\\
3 & 1.57973\\
4 & 1.46627\\
5 & 1.35287\\
6 & 1.21445\\
7 & 0.99857
\end{array}
\]

We observe that \(|A_{k}|\leq .999 |A_{k-1}|\) for all sufficiently large \(k\). However, that’s a contradiction, since we know \(|A_{k}| > |A_{k-2}|\). Hence, \(OPT\) has to take at least \(8|p|\) steps in the worst case.

Probabilistic Algorithms

The \(8|p|\) bound does not hold for probabilistic algorithms. In fact, it can be shown that my first algorithm takes \(7|p|\) steps in expectation when the first step is taken in a random direction.

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

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

This conjecture (currently known as the Prime Number Theorem) is notoriously difficult to prove. The first proof appeared almost a century after it was stated.

However, there are much simpler methods to bound \(\pi(n)\) above and below. Today I shall show such a bound that gives the correct order of magnitude for \(\pi(n)\) up to some constants. In particular, I shall prove that

\[(1+2\log 4)\frac{n}{\log n}\ge \pi(n)\ge \frac{\log 2}{2}\frac{n}{\log n}\]

for \(n\ge 12\). This proof technique is originally due to nineteenth century mathematician Pafnuty Chebyshev. He used a more complicated version of this technique to show that

\[1.1\frac{n}{\log n}\ge \pi(n)\ge 0.92\frac{n}{\log n}\]

for sufficiently large \(n\)s.

The Lower Bound

Suppose \(n\) is even, ie. \(n=2k\). Let us consider the prime factorization of \(\binom{2k}{k}\). All of its prime factors are between \(1\) and \(2k\). Therefore,

\[\binom{2k}{k}=p_1^{a_1}\cdots p_{\pi(2k)}^{a_{\pi(2k)}}\]

Claim: For every prime \(p_i\), \(p_i^{a_i}\leq 2k\)

Proof: We shall use this very simple identity \[\lfloor a+b\rfloor-\lfloor a\rfloor-\lfloor b\rfloor\leq1\]

It is well-known that the largest exponent of a prime dividing \(r!\) is

\[\left\lfloor\frac{r}{p_i}\right\rfloor+\left\lfloor\frac{r}{p_i^2}\right\rfloor+\cdots\]

Since \(\binom{2k}{k}=\frac{(2k)!}{(k!)^2}\), we have the following relationship

\begin{align}
a_i=&\left\lfloor\frac{2k}{p_i}\right\rfloor-2\left\lfloor\frac{k}{p_i}\right\rfloor+\left\lfloor\frac{2k}{p_i^2}\right\rfloor-2\left\lfloor\frac{k}{p_i^2}\right\rfloor+\cdots \\
&\leq 1+1+\cdots
\end{align}

Now there are as many \(1\)s as there are powers of \(p_i\leq 2k\). Hence, \(p_i^{a_i}\leq 2k\).

(Q.E.D)

So, we can conclude,

\[\binom{2k}{k}=p_1^{a_1}\cdots p_{\pi(2k)}^{a_{\pi(2k)}}\leq (2k)^{\pi(2k)}\]

It is easy to prove \(\binom{2k}{k}\ge 2^k\), hence

\[(2k)^{\pi(2k)}\ge 2^k\implies \pi(2k)\ge \frac{\log 2}{2}\cdot\frac{2k}{\log 2k}\]

The same bound can be derived for \(n=2k+1\) using \(\binom{2k+1}{k+1}\).  Therefore, the lower bound is

\[\boxed{\pi(n)\ge \frac{\log 2}{2}\cdot\frac{n}{\log n}}\]

The Upper Bound

Claim: For all \(n\ge 2\), \[\prod_{p\leq n\\ p\ \text{prime}}p\leq 4^n\]

Proof: We induct on \(n\). The base case trivially holds. \(n=2k-1\to 2k\) also holds because the product on the left hand side does not increment. So, we only need to prove \(n=2k\to 2k+1\)
\begin{align}
\prod_{p \leq 2k+1\\ p\ \text{prime}}p &= \prod_{p \leq k+1\\ p\ \text{prime}}p \cdot\prod_{k+2\leq p \leq 2k+1\\ p\ \text{prime}}p\\
&\leq 4^{k+1}\cdot \binom{2k+1}{k}\quad (1)
\end{align}
The first part of the product is \(\leq 4^{k+1}\) because of the inductive hypothesis. The second part is just the product of all the primes in between \(k+2\) and \(2k+1\). It’s easy to see that they all divide \(\binom{2k+1}{k}\), hence their product is \(\leq \binom{2k+1}{k}\).

It’s easy to show \(\binom{2k+1}{k}<4^k\). Substituting this bound back to (1) gives us

\[\prod_{p \leq 2k+1\\ p\ \text{prime}}p \leq 4^{k+1}\cdot 4^k=4^{2k+1}\quad (\text{Q.E.D.})\]

Now take a number \(2\leq m\leq n\). Consider the primes that are between \(m\) and \(n\). There are \(\pi(n)-\pi(m)\) of them, each of which is \(\ge m\). Consequently,

\[m^{\pi(n)-\pi(m)}\leq \prod_{m\leq p \leq n\\ p\ \text{prime}}p\leq \prod_{1\leq p \leq n\\ p\ \text{prime}}p\leq 4^n\]

Now take log of both sides

\[(\pi(n)-\pi(m))\log m\leq n\log 4\]

Rewrite this as

\[\pi(n)\leq \pi(m)+\frac{n\log 4}{\log m}\leq m+\frac{n\log 4}{\log m}\]

Set \(m=\frac{n}{\log n}\) and simplify the expression

\[\pi(n)\leq \frac{n}{\log n}\left(1+\frac{\log 4\cdot\log n}{\log(n/\log n)}\right)\]

Easy to show \(\frac{\log n}{\log(n/\log n)}\leq 2\) for \(n\ge 12\). (In fact you can show much tighter bound for sufficiently large \(n\).) Hence, the upper bound is

\[\boxed{\pi(n)\leq \left(1+2\log 4\right)\frac{n}{\log n}}\]

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, since you already know how to multiply polynomials. You just sum the product of every two terms, right?

def multiply(A, B):
	"""
	A = list of coefficients of A(x)
	B = list of coefficients of B(x)
	""" 
	n = len(A) - 1
	# C can have 2*n-2 degree at most
	C = [0 for i in range(2*n-1)]
	for i in range(n):
		for j in range(n):
	    	# adding a_i*b_j to c_{i+j}
	    	C[i+j] += A[i]*B[j]

It definitely works, but has asymptotic runtime of \(O(n^2)\). That’s quite bad. Can we do it faster?

Turns out we can! We can in fact multiply two polynomials in \(O(n\log n)\) time. How is that possible? It all comes down to looking at the problem from another perspective.

Change of Perspective

So far we only saw polynomials as a list of coefficients. However, a polynomial of degree \(n\) can also be viewed as a curve in \(n+1\) dimensional space. It is quite possible define it in terms of the points it passes through. For instance, take the line \(P(x)=2x+1\) in 2d space. We only need to know two points on this line, say \((0, 1)\) and \((1, 3)\), to identify it uniquely. To determine a polynomial with degree \(n\), It has been proven that you need to know its values at \(n+1\) distinct points. This is called the Point Value Representation of Polynomials.

A polynomial \(P(x)\) in this form can be saved like \(\{0 : P(0), \dots, n : P(n)\}\), although you can evaluate \(P(x)\) at any complex point. (This will come handy later)

Note that multiplication is really cheap in this form. Given
\[A(x)=\{0:A(0),\dots (2n-2): A(2n-2)\}\] and \[B(x)=\{0:B(0),\dots (2n-2): B(2n-2)\}\] we can compute \(C(i)=A(i)\cdot B(i)\) at every point \(i\) from \(0\) to \(2n-2\). For instance, \(C(0)=A(0)\cdot B(0)\).This costs \(O(n)\) time. So, we are happy.

Wait, are we? We do not know the coefficients of \(C(x)\)! Evaluation at arbitrary points is very costly in point value form. So, we could not, for instance, raise \(C(x)\) to some arbitrary power. What if we could find a way to go back and forth between coefficient representation and point value representation? We could get the best of both worlds.

This is what Fast Fourier Transform allows us to do.

Before I move on, please note a couple of things.

  • Prior to multiplication, we must evaluate both \(A(x)\) and \(B(x)\) at at least \(2n-1\) points even though each of them have degree \(n\). This is because \(C(x)\) has degree \(2n-2\).
  • It is not necessary to evaluate the polynomials exactly at \(1,2,\dots,2n-2\). There is nothing special about these numbers. However, it is necessary that we evaluate both \(A(x)\) and \(B(x)\) at the same set of points.

Coefficients to Point Value Representation

The only way to convert a polynomial with degree \(n\) to point value representation is to evaluate it at \(n+1\) distinct points. However, polynomial evaluation at a single point stills costs \(O(n)\) in coefficient representation. For instance, you can evaluate \(A(p)\) at any point \(p\) like this: \[A(p)=a_0+p(a_1+p(a_2+\cdots+pa_{n-1}))\] Then for \(n+1\) points, we would end up with \(O(n^2)\) runtime. However, you can save a lot of this cost by algebraic manipulation. Let me give you an example.

Suppose we want to compute \(P(x)=2+3x-4x^2+2x^3+5x^4\) at \(1, 2, -1, -2\). We can evaluate the odd powers and even powers separately since even powers of negative numbers is same as that of positive numbers. To be more precise, define \(Q(x)=2-4x+5x^2\) and \(R(x)=3x+2x^2\).

Note that we only kept the order of the coefficients from \(P(x)\), but not the associated powers \(x\). This gives an algebraic relationship as follows:
\[P(x)=Q(x^2)+xR(x^2)\] Now, note that \(Q((-1)^2)=Q(1^2), Q((-2)^2)=Q(2^2)\) and so on. Therefore, we will only need to compute \(Q(1), Q(4), R(1), R(4)\) in order to evaluate \(P(x)\) at four points. Both \(Q(x)\) and \(R(x)\) has half the degree as \(P(x)\), hence we’re saving almost half the work.

In order for trick to work, we must select our list of points very carefully. If the polynomial were much bigger, we would have to do that odd-even split multiple times. Every time we would like to have numbers with opposite signs, so that when squared, they give the same value. You can easily see that no initial set of real numbers can satisfy such requirements. This is why we choose roots of unity. In particular, we choose \(m=2^k\)th roots of unity such that \(2n\leq m\leq 4n\). (This is essentially the smallest power of two that’s greater than \(2n\)). They have this magic property that if \(w\) is a \(2^k\)th root, then so is \(-w\) regardless of which \(k\) you choose.

Point Value to Coefficient Representation

If you know some linear algebra, you might notice that evaluating a polynomial at \(n+1\) point essentially defines a linear transformation, which can be represented by a matrix. Due to special nature of roots of unity, the inverse matrix consists of conjugates of previously used roots scaled by a factor of \(1/n\). I won’t show it here in details, but if you are interested you should check out this blogpost.

In English, the equations tell us that inverse of Fast Fourier Transform is another Fast Fourier Transform where every point of polynomial evaluation must be replaced by its conjugate, and the final result has to be scaled by a factor of \(1/n\).

Time Complexity

Fast Fourier Transform is essentially a divide-and-conquer algorithm. The divide step corresponds to odd-even coefficient split. Each of the resulting subproblems has cost \(T(n/2)\). The combining step has cost \(O(n)\). (There are \(2n-2\) points and you have to compute the algebraic identity at each point.) So, the recurrence relationship should be
\[T(n)=2T\left(\frac{n}{2}\right)+O(n)\] Using master theorem, \(T(n)=O(n\log(n))\). Since FFT is the most expensive step in this approach, runtime of the entire process is dominated by \(O(n\log n)\). Therefore, polynomial multiplication also costs \(O(n\log n)\).

Summary

  • Given \(A(x)\) and \(B(x\) in coefficient representation, convert them to point value representation with Fast Fourier Transform. (\(O(n\log n)\)))
  • Compute \(C(x)=A(x)\cdot B(x)\). (\(O(n)\))
  • Go back to coefficients representation with Inverse Fast Fourier Transform. (\(O(n\log n)\)))

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.
[ –Adib Hasan 😛 ]

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.

Before I show it to you, I highly recommend you think about it on your own.

Solution

Notice that if you convert the set to a list, every element is assigned to a unique index between \(0\) and \(n-1\). So, it suffices to generate all the subsets of size \(k\) from \(\{0,1,\ldots,n-1\}\).

Now, every such subset falls into one of two categories: either it contains \(n-1\) or it doesn’t.

If \(n-1\) belongs to the subset, the rest of its elements form a subset of size \((k-1)\) in \(\{0,1,\ldots,n-2\}\).

If \(n-1\) doesn’t belong to the subset, then that subset is also a subset of \(\{0,1,\ldots,n-2\}\).

This is essentially a recursion. So, we are done.

Pretty neat, eh?

def gen_subset(n, k):
    if n >= 0 and n >= k >= 0:
        if n == 0 or k == 0: 
            yield set()
        elif n == k:
            # returns {0, 1, ..., n-1}
            yield set(range(n))
        else:
            # ksubsets without n-1
            yield from gen_subset(n-1, k)
            # ksubsets with n-1
            for subset in gen_subset(n-1, k-1):
                subset.add(n-1)
                yield subset
    else: 
        raise ValueError("Illegal Parameters")

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*)

Well, turns out if we ever could make such machines, they would become incredibly handy tools for scientists and mathematicians. That’s because they could babble all sorts of scientific and mathematical conjectures to such a machine, if it beeped, they would know those conjectures were false. Their conversation might go like this:

Crazy Physicist: Higgs Boson exists.
Machine: *Says Nothing*
Crazy Physicist: Ureka! I was right all along.

Computer Scientist: P=NP #changemymind
Machine: *beep*
Computer Scientist: OMG. P=/=NP. I just became a millionaire.

Clearly, you can see how these lie detectors would trivialize almost every known problem of science and mathematics. Then, a very legit question would be why nobody even tried to build such a machine. Indeed, humankind has spared far more effort after making junks like philosopher stones and elixir of life.

Turns out there is something fundamentally wrong about the design of the lie detectors. Before I get to the details, let me tell you a cool life hack. If you ever meet people that claim to be oracles, do this to them:

You: Will I offer you $10?

Whatever the oracle says, do the opposite.

This clearly shows the oracles cannot predict the future. The same thing is true for our lie detectors. Here is why:

Suppose you brought your friend Joe to show him the lie detector you bought. Now do the following:

You: I’m gonna hand Joe $10.

Whatever the lie detector says, do the opposite.

Therefore, just as above, our precious lie detectors cannot exist. However, they did have a purpose. They helped me to show you a proof technique called Cantor’s Diagonal Argument. It is a really nifty tool to prove many advanced statements in computability theory. Here is the general idea:

  • First you assume a machine with utterly unbelievable functions exists.
  • Then you come up for a statement or input for this machine using its own functions.
  • Finally, show that existence of such an input is contradictory to the existence of the machine itself.

As I’m running out of time, I’ll give you just one example of this technique. First proven by Alan Turing, it’s still one of the most famous proofs in computability theory.

Halting Turing Machine (HTM) is a special type of Turing machine that can instantly determine whether another Turing Machine halts on a specific input. In English, this means HTM is a cool app that can tell you whether one of the apps in your smartphone is gonna become unresponsive forever for a specific environment condition inside your smartphone. HTM would be a handy app, right? If HTM tells you an app is gonna be unresponsive beforehand, you’ll save time by not installing that crappy app. However, like all good things in computability theory, turns out HTM cannot exist either. This is because what if an evil developer created an app like this?

  • Give my app’s code and current environment condition of the smartphone to the HTM
  • Whatever the HTM says about my app, do the opposite inside the app.

Hence HTM is unable to predict what this evil dev’s app will do, and that’s why HTM cannot exist.

That’s it for today. I hope you enjoyed this blogpost. If you want to learn Computability Theory for real, I highly recommend Michael Sipser’s book.

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।