## Quantum Searching

Imagine a shuffled deck of 52 cards, and you are asked to find the ace of spades. The most natural thing to do is to take the top most card, see if it’s the ace of spades. If not, put it aside, take the top most and repeat the process until you find the ace of spades. If you are lucky, the ace of spades is the top most then we’re done. If you’re not so lucky, the ace of spades is at the bottom and it will take you 52 peeks before you find the card.

If we scatter the cards face down on the floor and randomly pick a card, then on the average, we will need $52/2$ peeks before we can find the ace of spades.

With quantum computing we can do even better!

For the sake of demonstration, let’s say we have 8 cards as shown in the figure below. We want to find the ace of spades.

Here are the steps:

1. Label each card from 0 to 7 in random order. The positions of the cards will represent the states of cards. The state $\left|6\right\rangle$ will therefore represent the ace of spades.

2. Let $\left|\phi\right\rangle$ be the superposition of states:

$\displaystyle \left|\phi\right\rangle = \frac{1}{2^{n/2}} \sum_{k=0}^{2^n-1} \left|x\right\rangle = \frac{1}{2^{n/2}} \Big( \left|0\right\rangle + \ldots + \left|7\right\rangle \Big) = \frac{1}{\sqrt{8}} \left[ \begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1\\ 1\\ 1\\ 1 \end{array} \right]$

3. Let $f$ be a function such that

$f(x) = \begin{cases} 1 & x \text{ corresponds to ace of spades}\\ 0 & \text{ otherwise} \end{cases}$

Define the operator $\mathbf{V}$

$\mathbf{V} = \mathbf{I} - 2\left|a\right\rangle \left\langle a\right|$

where a is the unique value where

$f(a) = 1$

In our example, the value of a is 6. Therefore,

$\mathbf{V} = \begin{bmatrix} 1.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 \\ 0.00 & 1.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 \\ 0.00 & 0.00 & 1.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 \\ 0.00 & 0.00 & 0.00 & 1.00 & 0.00 & 0.00 & 0.00 & 0.00 \\ 0.00 & 0.00 & 0.00 & 0.00 & 1.00 & 0.00 & 0.00 & 0.00 \\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 1.00 & 0.00 & 0.00 \\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & -1.00 & 0.00 \\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 1.00 \\ \end{bmatrix}$

Observe that the matrix element $\mathbf{V}_{6,6} = -1$

4. Define the operator W:

$\mathbf{W} = 2\left|\phi\right\rangle \left\langle\phi\right| - \mathbf{I} = \begin{bmatrix} -0.75 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 \\ 0.25 & -0.75 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 \\ 0.25 & 0.25 & -0.75 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & -0.75 & 0.25 & 0.25 & 0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & -0.75 & 0.25 & 0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.75 & 0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.75 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.75 \\ \end{bmatrix}$

5. Compute the operator $\mathbf{WV}$:

$\mathbf{WV} = \begin{bmatrix} -0.75 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.25 & 0.25 \\ 0.25 & -0.75 & 0.25 & 0.25 & 0.25 & 0.25 & -0.25 & 0.25 \\ 0.25 & 0.25 & -0.75 & 0.25 & 0.25 & 0.25 & -0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & -0.75 & 0.25 & 0.25 & -0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & -0.75 & 0.25 & -0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.75 & -0.25 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.75 & 0.25 \\ 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & 0.25 & -0.25 & -0.75 \\ \end{bmatrix}$

6. Multiply $\mathbf{WV}$ to $\left|\phi\right\rangle$ a number of times, about $\pi/4 \cdot 2^{n/2} = \pi/4 \cdot \sqrt{8} = 2.2 \approx 2$:

$\mathbf{WV}\left|\phi\right\rangle = \begin{bmatrix} 0.18 \\ 0.18 \\ 0.18 \\ 0.18 \\ 0.18 \\ 0.18 \\ 0.88 \\ 0.18 \\ \end{bmatrix}, \mathbf{(WV)^2}\left|\phi\right\rangle = \begin{bmatrix} -0.09 \\ -0.09 \\ -0.09 \\ -0.09 \\ -0.09 \\ -0.09 \\ 0.97 \\ -0.09 \\ \end{bmatrix}$

As you can see, the probability of getting the state $\left|6\right\rangle$ becomes very close to 1.

Making a measurement of the input register at this point will give us the state $\left|6\right\rangle$ with a probability very close to 1.

We have just demonstrated the quantum search algorithm!

## Why it works

Let $f$ be a function such that

$f(x) = \begin{cases} 1 & x \text{ the item we are looking for}\\ 0 & \text{ otherwise} \end{cases}$

Define the operator $\mathbf{U}_f$ whose action on an n qubit register and 1 qubit output register is

$\mathbf{U}_f(\left|x\right\rangle \otimes \left|y\right\rangle) = \left|x\right\rangle\otimes\left|y\oplus f(x)\right\rangle$

Prepare the n qubit input register and 1 qubit output register in the following state:

$\underbrace{\left|0\right\rangle\ldots \left|0\right\rangle}_{\text{n times}} \otimes \left|1\right\rangle$

Applying the Hadamard operator on the input and output qubits gives us a superposition of $N=2^n$ states:

$\begin{array}{rl} \displaystyle \mathbf{H}^{\otimes n}\otimes\mathbf{H}(\left|0\ldots 0\right\rangle\otimes \left|1\right\rangle )&= \mathbf{H}^{\otimes n}\left|0\ldots 0\right\rangle\otimes \mathbf{H}\left|1\right\rangle\\ &= \displaystyle \frac{1}{2^{n/2}}\sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \frac{1}{\sqrt{2}} \left(\left|0\right\rangle - \left|1\right\rangle\right) \end{array}$

Next apply the operator $\mathbf{U}_f$ to get

$\begin{array}{rl} \displaystyle \mathbf{U}_f \left[\frac{1}{2^{n/2}}\sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \frac{1}{\sqrt{2}} \left(\left|0\right\rangle - \left|1\right\rangle\right)\right] &= \displaystyle \frac{1}{\sqrt{2}} \frac{1}{2^{n/2}}\left[\sum_{x=0}^{2^n-1} \mathbf{U}_f \left|x\right\rangle \otimes \left|0\right\rangle - \sum_{x=0}^{2^n-1} \mathbf{U}_f \left|x\right\rangle \otimes \left|1\right\rangle\right] \\ &= \displaystyle \frac{1}{\sqrt{2}} \frac{1}{2^{n/2}}\left[\underbrace{\sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \left|0\oplus f(x) \right\rangle }_{\text{first underbrace}} - \underbrace{\sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \left|1\oplus f(x) \right\rangle}_{\text{second underbrace}}\right] \end{array}$

When $f(x) = 1$ for x=a, the first underbrace will expand to

$\displaystyle \sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \left|0\oplus f(x) \right\rangle = \left|0\right\rangle\otimes\left|0\right\rangle + \ldots + \underbrace{\left|a\right\rangle\otimes\left|1\right\rangle} + \ldots + \left|2^n-1\right\rangle \otimes \left|0\right\rangle$

The second underbrace will expand to

$\displaystyle \sum_{x=0}^{2^n-1} \left|x\right\rangle \otimes \left|1\oplus f(x) \right\rangle = \left|0\right\rangle\otimes\left|1\right\rangle + \ldots + \underbrace{\left|a\right\rangle\otimes\left|0\right\rangle} + \ldots + \left|2^n-1\right\rangle \otimes \left|1\right\rangle\\$

The underbraces in the above summations can be swapped so that we get

$\displaystyle \frac{1}{\sqrt{2}} \frac{1}{2^{n/2}} \left[ \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle \otimes \left|0\right\rangle - \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle \otimes \left|1\right\rangle\right]$

$\begin{array}{rl} &= \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle \otimes \frac{1}{\sqrt{2}} \left( \left|0\right\rangle - \left|1\right\rangle\right)\\ &= \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle \otimes \mathbf{H}\left|1\right\rangle \end{array}$

This means that the action of $\mathbf{U}_f$ leaves the output qubit unentangled with the input qubit. We can just ignore this moving forward and focus our attention to the input qubits.

We view the current state of the input qubit as the result of some operator $\mathbf{V}$ defined by

$\mathbf{V} \left|\phi\right\rangle = \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle$

where

$\left|\phi\right\rangle = \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} \left|x\right\rangle$

We can derive the expression of $\mathbf{V}$ by expanding $\mathbf{V} \left|\phi\right\rangle$, noting that at x=a, $f(x) = 1$:

$\begin{array}{rl} \mathbf{V} \left|\phi\right\rangle &= \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} (-1)^{f(x)} \left|x\right\rangle\\ &= \displaystyle \frac{1}{2^{n/2}} \left[\left|0\right\rangle + \ldots - \left|a\right\rangle +\ldots + \left|2^n-1\right\rangle\right]\\ &= \displaystyle \frac{1}{2^{n/2}} \left[\left|0\right\rangle + \ldots + \left|a\right\rangle +\ldots + \left|2^n-1\right\rangle - 2\left|a\right\rangle\right]\\ &= \displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} \left|x\right\rangle - \frac{2}{2^{n/2}} \left|a\right\rangle\\ &= \left|\phi\right\rangle - 2\left\langle a|\phi\right\rangle \left|a\right\rangle\\ &= \Big[\mathbf{I} - 2\left|a\right\rangle \left\langle a\right|\Big] \left|\phi\right\rangle\\ \end{array}$

which means

$\mathbf{V} = \mathbf{I} - 2\left|a\right\rangle \left\langle a\right|$

The two vectors $\left|\phi\right\rangle$ and $\left|a\right\rangle$ determine a plane P. Let $\left|a_{\perp}\right\rangle$ be the vector in the plane perpendicular to $\left|a\right\rangle$. We can write $\left|\phi\right\rangle$ as

$\left|\phi\right\rangle = \phi_1 \left|a_\perp\right\rangle + \phi_2\left|a\right\rangle$

Applying $\mathbf{V}$ to $\left|\phi\right\rangle$ gives us:

$\begin{array}{rl} \mathbf{V}\left|\phi\right\rangle &= \left(\mathbf{I} - 2\left|a\right\rangle \left\langle a\right| \right) \left( \phi_1 \left|a_\perp\right\rangle + \phi_2\left|a\right\rangle \right)\\ &= \phi_1 \left|a_\perp\right\rangle + \phi_2\left|a\right\rangle - 2\phi_2\left|a\right\rangle\\ &= \phi_1 \left|a_\perp\right\rangle - \phi_2\left|a\right\rangle \end{array}$

The effect of the operator $\mathbf{V}$ is therefore to reflect the vector $\left|\phi\right\rangle$ with respect to the vector $\left|a_\perp\right\rangle$.

Now, we want to reflect this vector $\mathbf{V}\left|\phi\right\rangle$ with respect to $\left|\phi\right\rangle$. To accomplish this, we define the operator $\mathbf{W}$ by

$\mathbf{W} = 2\left|\phi\right\rangle \left\langle\phi\right| - \mathbf{I}$

Apply this operator to $\mathbf{V}\left|\phi\right\rangle$:

$\begin{array}{rl} \mathbf{WV}\left|\phi\right\rangle &= \Big[ 2\left|\phi\right\rangle \left\langle\phi\right| - \mathbf{I} \Big]\mathbf{V}\left|\phi\right\rangle\\ &= 2\left|\phi\right\rangle \left\langle\phi\right|\mathbf{V}\left|\phi\right\rangle -\mathbf{V}\left|\phi\right\rangle \end{array}$

If we express $\mathbf{V}\left|\phi\right\rangle$ as a linear combination of $\left|\phi\right\rangle$ and a vector $\left|\phi_\perp\right\rangle$ in the plane P,

$\mathbf{V}\left|\phi\right\rangle = \alpha \left|\phi\right\rangle + \beta \left|\phi_\perp\right\rangle$

We have,

$\begin{array}{rl} \mathbf{WV}\left|\phi\right\rangle &= \Big[ 2\left|\phi\right\rangle \left\langle\phi\right| - \mathbf{I} \Big]\mathbf{V}\left|\phi\right\rangle\\ &= 2\left|\phi\right\rangle \left\langle\phi\right|\mathbf{V}\left|\phi\right\rangle -\mathbf{V}\left|\phi\right\rangle\\ &= 2\alpha\left|\phi\right\rangle - \alpha \left|\phi\right\rangle - \beta \left|\phi_\perp\right\rangle\\ &= \alpha\left|\phi\right\rangle - \beta \left|\phi_\perp\right\rangle \end{array}$

which demonstrates that $\mathbf{W}$ reflects $\mathbf{V}\left|\phi\right\rangle$ with respect to $\left|\phi\right\rangle$.

Therefore, the effect of $\mathbf{WV}$ on $\left|\phi\right\rangle$ is to rotate it by an angle $\gamma$ counter-clockwise.

We can compute this $\gamma$ by getting the inner product of $\mathbf{WV}\left|\phi\right\rangle$ and $\left|\phi\right\rangle$. First, let’s find the expression of $\mathbf{WV}\left|\phi\right\rangle$:

$\begin{array}{rl} \mathbf{WV}\left|\phi\right\rangle &= \mathbf{W} \left( \mathbf{I} - 2\left|a\right\rangle \langle a |\right) \left|\phi\right\rangle\\ &= \mathbf{W} \left(\left|\phi\right\rangle - 2 \underbrace{\langle a| \phi \rangle} \left|a\right\rangle \right) \end{array}$

The quantity $\langle a| \phi \rangle$ is

$\langle a| \phi \rangle = \displaystyle \langle a| \left( \frac{1}{2^{n/2}} \sum_{k=0}^{2^n-1} \left|x\right\rangle \right) = \frac{1}{2^{n/2}} = \cos \theta$

where $\theta$ is the angle between $\left|a\right\rangle$ and $\left|\phi\right\rangle$.

The complementary angle, $\rho = 90-\theta$ is the angle between $\left|\phi\right\rangle$ and $\left|a_\perp\right\rangle$. Using a well-known trigonometric identity, we can compute for the angle of $\rho$:

$\cos \theta = \sin \rho = \displaystyle \frac{1}{2^{n/2}}$

Since $\displaystyle \frac{1}{2^{n/2}}$ is very small if n is large,

$\sin \rho = \displaystyle \frac{1}{2^{n/2}} \approx \rho$

Continuing,

$\begin{array}{rl} \mathbf{WV}\left|\phi\right\rangle &= \mathbf{W} \left( \mathbf{I} - 2\left|a\right\rangle \langle a |\right) \left|\phi\right\rangle\\ &= \mathbf{W} \left(\left|\phi\right\rangle - 2 \cos\rho \left|a\right\rangle \right)\\ &= (2\left|\phi\right\rangle \langle \phi| - \mathbf{I}) \left(\left|\phi\right\rangle - 2 \cos\rho \left|a\right\rangle \right)\\ &= 2\left|\phi\right\rangle - \left|\phi\right\rangle - 2\cos\rho \left|\phi\right\rangle \langle \phi |a\rangle + 2 \cos\rho \left|a\right\rangle\\ &= \left|\phi\right\rangle - 4\cos^2\rho \left|\phi\right\rangle + 2 \cos\rho \left|a\right\rangle\\ &= (1- 4\cos^2\rho) \left|\phi\right\rangle + 2 \cos\rho \left|a\right\rangle \end{array}$

The inner product of $\mathbf{WV}\left|\phi\right\rangle$ and $\left|\phi\right\rangle$ is given by

$\begin{array}{rl} \langle\phi|\mathbf{WV}\left|\phi\right\rangle &= \langle\phi|\left[ (1- 4\cos^2\rho) \left|\phi\right\rangle + 2 \cos\rho \left|a\right\rangle\right]\\ &= (1- 4\cos^2\rho) + 2\cos^2\rho\\ &= 1- 2\cos^2\rho\\ &= \cos 2\rho \end{array}$

Therefore, the angle between these two vectors is $2\rho = \displaystyle \frac{1}{2^{n/2}}$. How many times do we have to apply the operator $\mathbf{WV}$ to get to $\pi/2$?

$\begin{array}{rl} m\cdot 2\rho &= \displaystyle \frac{\pi}{2}\\ \displaystyle m \frac{2}{2^{n/2}}&= \displaystyle \frac{\pi}{2}\\ m &=\displaystyle \frac{\pi\cdot 2^{n/2}}{4} \end{array}$

Therefore, we need to apply the operator $\mathbf{WV}$ $O(\sqrt{N})$ times to find our value (where $N=2^n$).

## RSA Encryption and Quantum Computing

We are going to crack the RSA by finding the period of the ciphertext. Let m be the message, c the ciphertext, e the encoding number and N=pq such that

$c=m^e \mod N$

From Period Finding and the RSA post, the period of a ciphertext is defined as the smallest number r such that

$c^r \equiv \mod N$

First we prepare $n=2n_0$ QBits at the state $\left|0\right\rangle$ where $n_0$ is the number of bits of $N=pq$. The number of bits can be computed using the formula:

$\lceil \log_2(N) \rceil$

We also prepare $n_0$ QBits for the output register.

So initially we have the following setup:

$\underbrace{\left|00\ldots 0\right\rangle}_{n\text{ times}} \otimes \underbrace{\left|00\ldots0\right\rangle}_{n_0\text{ times}}$

The output register will contain the output of the operator:

$\mathbf{U}_f \left|x\right\rangle\left|0\right\rangle = \left|x\right\rangle\left|f(x)\right\rangle$

where

$f(x) = c^x \mod N$

We now apply the Hadarmard gate to the input register and apply the operator $\mathbf{U}_f$:

$\displaystyle \frac{1}{2^{n/2}} \sum_{x=0}^{2^n-1} \left|x\right\rangle\left|f(x)\right\rangle$

The Hadamard gates transform the state into a superposition of all possible values of the period while the operator $\mathbf{U}_f$ transforms the output QBits as superposition of all possible values of $c^x\mod N$.

Next we make a measurement of the output QBits. This will give us a number $f_0$ which corresponds to all states whose value of $\left|x\right\rangle$ is of the form $x_0+kr$ since

$\begin{array}{rl} c^{x_0+kr} & \equiv c^{x_0}c^{kr} \mod N \\ & \equiv c^{x_0}(c^{r})^k \mod N \\ & \equiv c^{x_0} \mod N \end{array}$

since $(c^{r})^k \equiv 1 \mod N$ by definition of a period.

Therefore, after the measurement, the input state is now

$\displaystyle \left|\Psi\right\rangle = \frac{1}{\sqrt{m}} \sum_{k=0}^{m-1} \left|x_0 + kr\right\rangle$

where m is the smallest integer such that $x_0 + mr \ge 2^{n}$.

We will then apply the quantum Fourier transform to state $\left|\Psi\right\rangle$. The Quantum Fourier Transform is defined by its action on an arbitrary $\left|x\right\rangle$

$\displaystyle \mathbf{U}_{\mathbf{FT}}\left|x\right\rangle = \frac{1}{2^{n/2}} \sum_{y=0}^{2^n-1} e^{2\pi ixy/2^n}\left|y\right\rangle$

Applying the Fourier Transform to the state $\left|\Psi\right\rangle$,

$\begin{array}{rl} \displaystyle \mathbf{U}_{\mathbf{FT}} \frac{1}{\sqrt{m}} \sum_{k=0}^{m-1}\left|x_0 + kr\right\rangle &= \displaystyle \frac{1}{\sqrt{m}} \sum_{k=0}^{m-1}\mathbf{U}_{\mathbf{FT}} \left|x_0 + kr\right\rangle\\ &= \displaystyle \frac{1}{\sqrt{m}} \sum_{k=0}^{m-1} \frac{1}{2^{n/2}} \sum_{y=0}^{2^n-1} e^{2\pi i(x_0+kr)y/2^n}\left|y\right\rangle\\ &= \displaystyle \frac{1}{\sqrt{m}} \sum_{k=0}^{m-1} \frac{1}{2^{n/2}} \sum_{y=0}^{2^n-1} e^{2\pi ix_0y/2^n}e^{2\pi ikry/2^n}\left|y\right\rangle\\ &= \displaystyle \sum_{y=0}^{2^n-1} \underbrace{e^{2\pi ix_0y/2^n} \frac{1}{\sqrt{2^n m}} \sum_{k=0}^{m-1} e^{2\pi ikry/2^n}}_{\text{probability amplitude}}\left|y\right\rangle \end{array}$

Now if we make a measurement of this new state, the probability of getting a value of y is the probability amplitude multiplied by it’s complex conjugate:

$\displaystyle \Big[\underbrace{e^{2\pi ix_0y/2^n} } \frac{1}{\sqrt{2^n m}} \sum_{k=0}^{m-1} e^{2\pi ikry/2^n}\Big]\Big[\underbrace{e^{-2\pi ix_0y/2^n} } \frac{1}{\sqrt{2^n m}} \sum_{k=0}^{m-1} e^{-2\pi ikry/2^n}\Big]$

The factors in underbrace evaluates to 1 when multiplied and we’re left with

$\displaystyle \frac{1}{2^n m}\Big| \underbrace{\sum_{k=0}^{m-1} e^{2\pi ikry/2^n}}\Big|^2$

Using the formula

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

we can write the summation in underbrace as

$\displaystyle \Big(\sum_{k=0}^{m-1} \cos(2\pi kry/2^n)\Big)^2 + \Big(\sum_{k=0}^{m-1} \sin(2\pi kry/2^n)\Big)^2$

We’d like to know at what value of y the expression above is maximum. We can use the following formula to simplify above summations:

$\displaystyle \sum_{k=0}^{m-1} \cos(2\pi fk) = \frac{\cos(\pi f(m-1)) \sin(\pi fm)}{\sin(\pi f)}\\ \displaystyle \sum_{k=0}^{m-1} \sin(2\pi fk) = \frac{\sin(\pi f(m-1)) \sin(\pi fm)}{\sin(\pi f)}$

If we let

$f = ry/2^n$

this reduces the summations to:

$\displaystyle \Big(\sum_{k=0}^{m-1} \cos(2\pi kry/2^n)\Big)^2 + \Big(\sum_{k=0}^{m-1} \sin(2\pi kry/2^n)\Big)^2$

$\begin{array}{rl} &=\displaystyle \frac{\cos^2(\pi f(m-1)) \sin^2(\pi fm)}{\sin^2(\pi f)} + \frac{\sin^2(\pi f(m-1))\sin^2(\pi fm))}{\sin^2(\pi f)}\\ &= \displaystyle \frac{\sin^2(\pi fm)}{\sin^2(\pi f)}\Big[\underbrace{\cos^2(\pi f(m-1)) + \sin^2(\pi f(m-1))}_{\text{= 1}}\Big]\\ &= \displaystyle \frac{\sin^2(\pi fm)}{\sin^2(\pi f)} \end{array}$

Therefore, the probability of getting a specific y is

$p(y) = \displaystyle \frac{1}{2^nm} \displaystyle \frac{\sin^2(\pi fm)}{\sin^2(\pi f)}$

Here is a plot of the function $\sin^2(\pi fm)/\sin^2(\pi f)$ for m=4:

Notice that the graph attains its maximums when the values of f are integer values, that is when

$\displaystyle f = j= \frac{ry}{2^n} \text{ where } j= 1, 2, 3, 4, \ldots, r$

Since y is an integer value, this means that the value of y must be close to $j2^n/r$. We can use a theorem in Number theory which states that if $p/q$ is a rational number that approximates a real number x and

$\displaystyle \Big|\frac{p}{q} - x\Big| \le \frac{1}{2q}$

we can approximate x using continued fraction expansion of p/q. In fact, if we find a y that is within 1/2 of one of the $j2^n/r$ values, we have

$\displaystyle \left| y-j\frac{2^n}{r}\right| \le \displaystyle \frac{1}{2}\\$

and dividing both sides by $2^n$, we get

$\displaystyle \left| \frac{y}{2^n}-\frac{j}{r}\right| \le \displaystyle \frac{1}{2^{n+1}}$

which satisfies the theorem. We can therefore expand $y/2^n$ via continued fraction expansion and continue until we find a denominator $r$ that is less that $2^{n_0}$. We then test whether r could be our period if it satisfies

$\displaystyle c^r \equiv 1$

If it satisfies the above, then r is the period. It’s just a matter of computing $d^\prime$, the inverse of $e$ modulo r satisfying

$\displaystyle ed^\prime \equiv 1 \mod r$

Having computed $d^\prime$, we can then decrypt the ciphertext using

$m = c^{d^\prime} \mod N$

In our example in period finding, let’s suppose that the quantum computer gave us the following number

$y=1446311$

We can expand the fraction $y/2^n = 1446311/16777216$ as a continued fractions as follows:

$\displaystyle \frac{y}{2^n} = \frac{1446311}{16777216} = \frac{1}{11 + \displaystyle \frac{1}{1+\displaystyle\frac{1}{1+\displaystyle\frac{1}{1}}}} = \frac{5}{58}$

or

$\displaystyle \frac{y}{2^n} = \frac{1446311}{16777216} = \frac{1}{11 + \displaystyle \frac{1}{1+\displaystyle\frac{1}{1+\displaystyle\frac{1}{\displaystyle\frac{1}{6886}}}}} = \frac{34433}{399423}$

We can stop until $q=58$ since $q=58 < 2^{n_0} = 2^{12} = 4096$. So we use the first expansion where $q=58$ and testing it

$\begin{array}{rl} \displaystyle c^{58} \mod 3127 &= \displaystyle 794^{58} \mod 3127\\ &= \displaystyle 1 \mod 3127 \end{array}$

which confirms that $q=58$ is a period. Using $r=58$ in the equation

$\displaystyle ed^\prime = m\times r + 1$

and solving for $d^\prime$ (which is the inverse of e modulo 58) we get

$\displaystyle d^\prime = 25$

We can now get the original message using

$\displaystyle 0794^{25} \mod 3127 = 1907$

which agrees with our original message.

How lucky do we need to get in order for the state to collapse to one of these y values close to $j2^n/r$? Let’s evaluate the probability for

$y_j=\displaystyle j\frac{2^n}{r} \pm \delta_j$

where $\delta_j$ is the distance of the closest y to $j2^n/r$. Since $f = ry/2^n$, we have

$\begin{array}{rl} p(y_j) = \displaystyle \frac{1}{2^nm}\frac{\sin^2(\pi fm)}{\sin^2(\pi f)} &= \displaystyle \frac{1}{2^nm} \frac{\sin^2(\pi mr\displaystyle\frac{y}{2^n})}{\sin^2(\pi r\displaystyle\frac{y}{2^n}) }\\ &= \displaystyle \frac{1}{2^nm} \frac{\sin^2\Big[\displaystyle\frac{\pi mr}{2^n}(j\frac{2^n}{r} \pm \delta_j)\Big]}{\sin^2\Big[\displaystyle\frac{\pi r}{2^n}(j\frac{2^n}{r} \pm \delta_j)\Big] }\\ &= \displaystyle \frac{1}{2^nm} \frac{\sin^2\Big[\displaystyle j\pi m \pm \frac{\pi m r\delta_j}{2^n})\Big]}{\sin^2\Big[\displaystyle j\pi \pm \frac{\pi r\delta_j}{2^n}\Big] }\\ &= \displaystyle \frac{1}{2^nm} \frac{\sin^2\Big[\displaystyle \frac{\pi m r\delta_j}{2^n})\Big]}{\sin^2\Big[\displaystyle \frac{\pi r\delta_j}{2^n}\Big] } \end{array}$

We can approximate the denominator using the small angle approximation of sine:

$\sin(x) \approx x$

to get

$\displaystyle p(y_j) = \displaystyle \frac{1}{2^nm} \frac{\sin^2\Big[\displaystyle \frac{\pi m r\delta_j}{2^n})\Big]}{\left[\displaystyle \frac{\pi r\delta_j}{2^n}\right]^2}$

Since m is the smallest integer such that $x_0 + mr \ge 2^{n}$, then

$m = \displaystyle \left\lceil\frac{2^n}{r}\right\rceil$

We can use this to approximate

$\displaystyle m\frac{r}{2^n} \approx 1$

to get

$p(y_j) = \displaystyle \frac{1}{2^nm} \frac{\sin^2(\displaystyle \pi\delta_j)}{\left(\displaystyle \frac{\pi\delta_j}{m}\right)^2}$

Since $0 \le \delta_j \le 1/2$, we can see from the graph below that the line joining (0,0) and $(\pi/2,1)$, whose equation is $2x/\pi$ is less that the value of the sine function at that interval, that is,

$\displaystyle \frac{2x}{\pi} \le \sin(x)$

Using this inequality, we can get a lower bound of the probability of getting a specific y value to be

$\begin{array}{rl} p(y_j) = \displaystyle \frac{1}{2^nm} \frac{\sin^2(\displaystyle \pi\delta_j)}{\left(\displaystyle \frac{\pi \delta_j}{m}\right)^2} &\ge \displaystyle \frac{1}{2^nm} \left(\frac{\displaystyle \frac{2\pi\delta_j}{\pi}}{\displaystyle \frac{\pi \delta_j}{m}}\right)^2\\ &= \displaystyle \frac{1}{2^nm} \left( \frac{2m}{\pi}\right)\\ &= \displaystyle \frac{m}{2^n}\frac{4}{\pi^2}\\ &= \displaystyle \frac{1}{r}\frac{4}{\pi^2} \end{array}$

Since there are r of these $y_j$‘s, the probability of getting any one of these $y_j$‘s is therefore:

$p(y) = \displaystyle \frac{4}{\pi^2} \approx 0.405$

which means that there is 40% probability of the state to collapse to a value of y that will give us the correct period.

We have just seen how a quantum computer can be used to crack the RSA. We have exploited the fact that the measurement will give us a value of y that is near $j2^n/r$ with high probability. We then computed for the period using the continued fraction expansion of $y/2^n$. With the period computed, it’s becomes straightforward to decrypt the ciphertext.

## Period Finding and the RSA

In the previous post, we learned how to decrypt RSA by getting the factors of the big number N and computing for the inverse of e (the encoding number) modulo N. There is also another way to decrypt an RSA encrypted message. This is when you are able to get the period of the ciphertext. If c is the ciphertext, the period r is the smallest integer that satisfies:

$c^r \equiv 1 \mod N$

Once we get the period, we compute for $d^\prime$, the inverse of e modulo r:

$ed^\prime \equiv 1 \mod r$

The inverse can then be used to decrypt the ciphertext:

$m=c^{d^\prime}$

In our previous example, we encrypted the message

THIS IS A SECRET MESSAGE

using public key p=53, q=59, N=pq=3127 and e=7 and private key d=431. The “plain text” is

1907 0818 2608 1826 0026 1804 0217 0419 2612 0418 1800 0604

and the ciphertext is:

0794 1832 1403 2474 1231 1453 0268 2223 0678 0540 0773 1095

Let’s compute the period of the first block of our ciphertext:

$0794^r \equiv 1 \mod 3127$

Using the python script below, we can compute the period

for r in range(1,100):
p=pow(794,r,N)
if p == 1:
print "%d %d" % (r,p)


The result of running the above program gives r=58. We can then compute $d^\prime$ using the following equation:

$ed^\prime = m\times r + 1$

The above equation is satisfied when m=3 and $d^\prime = 25$. Using this value of $d^\prime$, we can compute for

$\begin{array}{rl} m&=0794^{25} \mod 3127 \\ &= 1907 \end{array}$

which gives us the original message!

However, unlike using the private key, you need to compute the period r and $d^\prime$ for every block of the ciphertext (unless the ciphertext is composed of only one block). However, that should not stop a cracker from deciphering all the blocks.

## How RSA Encryption Works

Alice and Bob live in different parts of the world. They want to communicate with each other but they don’t want anyone to know the messages they exchange. In order to protect the message, they need to encrypt their message. Bob comes up with a secret key that will allow both of them to encrypt and decrypt their messages so that when they send it via email they will be confident that no one can read the message if ever someone (like Eve) intercepted the message. However, they have a dilemma. How will Bob send the encryption key to Alice ? If he sends the key and Eve intercepts it, then Eve will be able to decrypt the messages both Alice and Bob exchange and know what they are up to.

Bob should be able to send the key to Alice encrypted so that Eve will not be able to read it. In order to do this, Bob will have to create a second key to encrypt the key he wants to send to Alice. The problem now is how will Alice decrypt the message (which is the encrypted key) if she does not have the second key?

This is where RSA encryption is used. Suppose we want to encrypt a message using RSA, what we’ll do is find 2 large prime numbers p and q and get their product N = pq. We will need another number e, which we will use to encode the message into a ciphertext. The set of numbers N and e is called the public key which Alice can send to Bob via email. Bob will use these numbers to encrypt the secret key before sending to Alice.

We will represent a textual message like “THIS IS A SECRET MESSAGE” into numbers. To accomplish this, we need to map letters into numbers like the following:

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline A & B & C & D & E & F & G & H & I & J & K & L & M & N & O & P & Q & R & S & T & U & V & W & X & Y & Z & \\\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12& 13& 14& 15& 16& 17& 18& 19& 20& 21& 22& 23& 24& 25& 26\\ \hline \end{tabular}$

Using the above mapping, we can write ‘THIS IS A SECRET MESSAGE’ as:

19 7 8 18 26 8 18 26 0 26 18 4 2 17 4 19 26 12 4 18 18 0 6 4

If a number is less than 10, we pad it with a zero to the left. The message then becomes:

19 07 08 18 26 08 18 26 00 26 18 04 02 17 04 19 26 12 04 18 18 00 06 04

To conserve some space, we can group the numbers into groups of 4:

1907 0818 2608 1826 0026 1804 0217 0419 2612 0418 1800 0604

Now for each $M$ number above, we will encode it using the formula:

$\displaystyle C = M^e \mod N$

What is this mod operation? The above says that we raise M to the exponent e, divide the result by N and get the remainder. For example, if M=10, e=7 and N=17 we have

$10^7=10000000$

Now divide the result by 17 and get the remainder:

$10000000 / 7 = 588235 \text { Remainder } 5$

Therefore

$10^7 \mod 17 = 5$

In python we can get the answer using the pow function:

>>> pow(10,7,17)
5


Let’s say we choose p = 53, q=59 and e = 7. This gives us $N=pq = 53\times 59 = 3127$. To encode 1907, we do

$\displaystyle C=1907^7 \mod 3127 = 0794$

The number 0794 is now the ciphertext. It is the number we give to the recipient of the message. We can use python to generate the ciphertext above.

for n in ("1907","0818","2608","1826","0026","1804","0217","0419","2612","0418","1800","0604"):
print pow(int(n),7,3127)


Doing this for all numbers we get:

0794 1832 1403 2474 1231 1453 0268 2223 0678 0540 0773 1095

When the recipient gets this message, she can decipher it using a key which she keeps private to herself. The key d is the inverse of e modulo $(p-1)(q-1)$. The key d is called the private key. We can retrieve the original message using the formula:

$\displaystyle M = C^d$

The number d can be calculated using the following formula:

$ed \mod (p-1)(q-1) \equiv 1$

Using python we can compute for d using the following program:

>>> p=53
>>> q=59
>>> e=7
>>> NN = (p-1)*(q-1)
>>> for d in range(1,NN):
...   x = e*d % NN
...   if x == 1:
...     print 'd = ', d
...
d =  431


Using d = 431 and applying the decipher formula to the first block, we get

$\displaystyle M = 0794^{431} \mod 3127 = 1907$

which is our original message!

We now apply this to the entire ciphertext

0794 1832 1403 2474 1231 1453 0268 2223 0678 0540 0773 1095

using the python program below:

for n in ("0794","1832","1403","2474","1231","1453","0268","2223","0678","0540","0773","1095"):
print pow(int(n),431,N)


we get

1907 0818 2608 1826 0026 1804 0217 0419 2612 0418 1800 0604

which is our original full message! It’s just a matter of mapping these numbers back to letters to get the message text.

Using this mechanism, Alice will send 2 numbers N and e to Bob which he will use to encrypt the secret key and send to Alice. When Alice receives the encrypted secret key, she will use her private key d to decrypt it and get the secret key. After that, they can now start using the secret key to encrypt messages between them.

## Parallel Computing in the Small: An Introduction to Quantum Computing

As I watched my one-month old baby sleeping, a thought ran through my mind. I wonder what technology would look like when she’s my age. There’s a lot of things going on simultaneously in technology today. I remember doing Artificial Intelligence/Machine Learning about 15 years ago. They used to be done in a university setting. They are now the new normal. Some technologies are quite recent like Big Data and Blockchain but they have become mainstream very quickly. I wonder what it will look like when my daughter grows up.

There is a technology I’m currently watching. It’s still in it’s infancy but this technology is really fascinating. It is a technology based on a remarkably counter-intuitive theory of the universe that governs the atoms and sub-atomic particles. It is called Quantum Computing. Given the high speed at which science fiction become realities, it’s possible that my baby girl will someday be programming on a quantum computer. But for now, we can only hope.

I have not seen a Quantum Computer and based on what I read so far, it’s still an engineering challenge. It might probably be a few more years before they are mass-produced but it is something to look forward to. But the good thing is, we can already talk about quantum algorithms! So let’s sharpen our pencils and get a few sheets of paper for we are going to talk about this new way of computing.

In classical computing, a bit has a value that is either one or zero. In quantum computing, a bit can now have a value that is a superposition of 1 and 0. We call this a QBit. The values 1 and 0 are now symbolized as $\left| 1\right\rangle$ and $\left| 0\right\rangle$ respectively. These are unit vectors in what is known as a Hilbert space. They are also known as kets, derived from the word bracket which we’ll talk more of later. An example of a vector space that you are familiar with is the color wheel. Each color is actually a combination of red, blue and green. For example, the color yellow is a combination of red and green. It’s also the same as with the qubit. A general QBit vector is of the form

$\displaystyle \left| \Psi \right\rangle = a\left| 0\right\rangle + b\left| 1\right\rangle$

such that $\mid a\mid^2 + \mid b\mid^2 = 1$. The numbers a and b are complex numbers and the symbol $\mid a\mid^2$ is defined as

$\displaystyle \mid a\mid^2 = a^*a$

where $a^*$ is the complex conjugate of a. A QBit vector represents the state of the QBit. We shall use the term vector and state interchangeably moving forward.

However, you won’t be able to see them in superposition state. If you want to know the state of a qubit, you will have to measure it. But when you measure it, it will collapse to one of either $\left| 0\right\rangle$ or $\left| 1\right\rangle$ but you will have a clue as to what that state will be. The coefficients a and b will give you the probability of the result of the measurement: it will be $\mid a\mid^2$ in state $\left| a\right\rangle$ and $\mid b\mid^2$ in state $\left| b\right\rangle$.

If the state of a QBit is a superposition of basis states $\left| 0\right\rangle$ and $\left| 1\right\rangle$, does this mean our computation is also random? Does it mean we get different answers every time ?

It turns out that we can get a definite answer! Let’s illustrate this by a simple quantum computation.

Suppose a I have a number between 0 and 1 million. What is my number? You can guess my number and I will only tell you if you got it correctly or not. I won’t tell you if my number is higher or lower than your guess. How many tries do you think you need in order to guess my number? I’m sure our tries will be many. In the worst case, you’ll need about 1 million tries.

For a quantum computer, you only need one try to guess it and that’s what we’re going to see. However, bear with me since we have to build up our arsenal of tools first before we can even talk about it. Let’s start with how a QBit looks like as a matrix.

The basis QBits $\left| 0\right\rangle$ and $\left| 1\right\rangle$ can be written as 1×2 matrices:

$\displaystyle \left| 0 \right\rangle = \begin{bmatrix} 1\\ 0 \end{bmatrix}$

$\displaystyle \left| 1 \right\rangle = \begin{bmatrix} 0\\ 1 \end{bmatrix}$

Using this representation, we can write a general QBit as:

$\displaystyle \mid\Psi\rangle = a\mid 0\rangle + b\mid 1\rangle = a \begin{bmatrix} 1\\ 0 \end{bmatrix} + b \begin{bmatrix} 0\\ 1 \end{bmatrix} = \begin{bmatrix} a\\ b \end{bmatrix}$

So now we know that every QBit is a linear combination of ket vectors $\left| 0\right\rangle$ and $\left| 1\right\rangle$. These ket vectors have a corresponding “bra” vectors $\langle 0 \mid$ and $\langle 1\mid$. In matrix representation, the bra basis vectors are defined as:

$\displaystyle \langle 0\mid = \begin{bmatrix} 1 & 0 \end{bmatrix}$

and

$\displaystyle \langle 1\mid = \begin{bmatrix} 0 & 1 \end{bmatrix}$

For a general QBit $\mid\Psi\rangle = a\mid 0\rangle + b\mid 1\rangle$, the corresponding bra vector is

$\displaystyle \langle \Psi \mid = a^*\langle 0\mid + b^*\langle 1\mid = \begin{bmatrix} a^* & b^* \end{bmatrix}$

There is an operation between bra and ket vectors of the basis states called inner product which is defined as:

$\begin{array}{rl} \displaystyle \langle 0\mid 0 \rangle &= 1\\ \displaystyle \langle 1\mid 1 \rangle &= 1\\ \displaystyle \langle 0\mid 1 \rangle &= \langle 1\mid 0\rangle = 0 \end{array}$

Using the above rules, we can define the inner product of any two QBits $\mid \Psi\rangle = a\mid 0\rangle + b\mid 1\rangle$ and $\mid\Phi\rangle = c\mid 0\rangle + d\mid 1\rangle$ as

$\begin{array}{rl} \displaystyle \langle \Psi\mid\Phi\rangle &= \big(a^*\langle 0\mid + b^*\langle 1\mid\big)\big(c\mid 0\rangle + d\mid 1\rangle\big)\\ &= \displaystyle a^*c \langle 0\mid 0\rangle + a^*\langle 0\mid 1\rangle + b^*c\langle 1\mid 0\rangle + b^*d\langle 1\mid 1\rangle\\ &= a^*c + b^*d \end{array}$

Using the above definition of an inner product, the norm of a vector is defined as

$\displaystyle \mid\left|\Psi\right\rangle\mid = \displaystyle \sqrt{\langle \Psi \mid \Psi \rangle} =\displaystyle \sqrt{a^*a + b^*b}$

Geometrically, the norm is the length of the ket $\mid \Psi\rangle$.

Quantum computations are done using quantum operators. They are also called quantum gates. Operators are linear mappings that take a state to another state. That is, an operator $\mathbf{A}$ acts on $a\left| 0 \right\rangle + b\left| 1\right\rangle$ to give another vector $c\left| 0 \right\rangle + d\left| 1\right\rangle$. We write this as

$\displaystyle \mathbf{A}(a\left| 0\right\rangle + b\left| 1\right\rangle) = c\left| 0\right\rangle + d\left| 1\right\rangle$

Since the basis vectors $\left| 0\right\rangle$ and $\left| 1\right\rangle$ are also vectors, operators can also operate on them. An example of a very important operator and it’s action on the basis vectors is the Hadamard operator:

$\begin{array}{rl} \displaystyle \mathbf{H} \left| 0 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1\right\rangle)\\ \displaystyle \mathbf{H} \left| 1 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle - \left| 1\right\rangle) \end{array}$

By knowing the action of an operator on the basis vectors, we can determine the matrix representation of an operator. For the Hadamard operator, the matrix representations are:

$\displaystyle \mathbf{H} \left| 0 \right\rangle = \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1\right\rangle) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1 \end{bmatrix}$

and

$\displaystyle \mathbf{H} \left| 1 \right\rangle = \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle - \left| 1\right\rangle) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ -1 \end{bmatrix}$

Therefore the matrix representation of the Hadamard operator is

$\displaystyle \mathbf{H}= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$

The identity operator is defined as the operator that maps a vector into itself and is symbolized as $\mathbf{I}$:

$\displaystyle \mathbf{I}\left|\Psi\right\rangle = \left|\Psi\right\rangle$

The NOT operator is defined as

$\begin{array}{rl} \mathbf{X}\left|0\right\rangle &= \left|1\right\rangle\\ \mathbf{X}\left|1\right\rangle &= \left|0\right\rangle \end{array}$

What it does is transform the $\left|0\right\rangle$ to $\left|1\right\rangle$ and vice versa, just like the classical bits.

Operators are linear. We can use linearity to derive the resulting vector when an operator acts on a general QBit:

$\displaystyle \mathbf{A}(a\left| 0\right\rangle + b\left| 1\right\rangle) = a\mathbf{A} \left| 0\right\rangle + b\mathbf{A} \left| 1\right\rangle)$

If $\mathbf{A}$ is the Hadamard operator, that is, $\mathbf{A} = \mathbf{H}$, then the above operation becomes:

$\displaystyle \mathbf{H}(a\left| 0\right\rangle + b\left| 1\right\rangle) = a\mathbf{H} \left| 0\right\rangle + b\mathbf{H} \left| 1\right\rangle)$

Since we know that:

$\begin{array}{rl} \displaystyle \mathbf{H} \left| 0 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1\right\rangle)\\ \displaystyle \mathbf{H} \left| 1 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle - \left| 1\right\rangle) \end{array}$

We can then write this as:

$\begin{array}{rl} \displaystyle \mathbf{H}(a\left| 0\right\rangle + b\left| 1\right\rangle) &= \displaystyle a\mathbf{H} \left| 0\right\rangle + b\mathbf{H} \left| 1\right\rangle)\\ &= \displaystyle a\frac{1}{\sqrt{2}}\big(\left| 0\right\rangle + \left| 1\right\rangle\big) + b\frac{1}{\sqrt{2}}\big(\left| 0\right\rangle - \left| 1\right\rangle\big)\\ &= \displaystyle \frac{a+b}{\sqrt{2}} \left| 0\right\rangle + \frac{a-b}{\sqrt{2}} \left| 1\right\rangle \end{array}$

Let see what the operators look like when they operate on bra vectors. Let’s start with the Hadamard operator acting on a general QBit above.

$\begin{array}{rl} \displaystyle \mathbf{H} \mid 0 \rangle &= \displaystyle \frac{1}{\sqrt{2}}(\mid 0 \rangle + \mid 1\rangle) \rightarrow \langle 0\mid \mathbf{H^\dagger} = \displaystyle \frac{1}{\sqrt{2}}(\langle 0 \mid + \langle 1\mid)\\ \displaystyle \mathbf{H} \mid 1 \rangle &= \displaystyle \frac{1}{\sqrt{2}}(\mid 0 \rangle - \mid 1\rangle) \rightarrow \langle 1\mid \mathbf{H^\dagger} = \displaystyle \frac{1}{\sqrt{2}}(\langle 0 \mid - \langle 1\mid) \end{array}$

where we symbolize $\mathbf{H^\dagger}$ as the operator corresponding to the Hadamard operator acting on bras.

The matrix representation of the Hadamard operator acting on bras is therefore:

$\mathbf{H^\dagger} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} = \mathbf{H}$

In general, the matrix representation of $\mathbf{H^\dagger}$ is the conjugate transpose of the matrix of $\mathbf{H}$. In simple terms, this means take the transpose of $\mathbf{H}$ and apply the complex conjugates to each element. It’s amazing that the Hadamard operator is equal to its conjugate transpose. The types of operators that behave that way are called Hermitian operators.

Let $\mathbf{A}$ be an operator and $\left| \Psi\right\rangle$ a vector and let $\Phi = \mathbf{A}\left|\Psi\right\rangle$, then the norm of $\Phi$ is

$\displaystyle \langle\Phi\mid\Phi\rangle = \langle \Psi\mid \mathbf{A^\dagger}\mathbf{A}\mid \Psi\rangle = \langle\Psi\mid\Psi\rangle$

This means that

$\displaystyle \mathbf{A^\dagger}\mathbf{A} = \mathbf{I}$

The kind of operator that satisfies the above relationship is called Unitary.

Now that we have gained knowledge of some of the tools, we can now describe how a quantum computation is set up. We need QBits for the input and QBits for the output. The simplest setup is a one QBit input and one QBit output. Let $\left| x\right\rangle$ and $\left| y\right\rangle$ be the input and output QBits,respectively. The general state of this quantum system is a superposition of the following states:

$\left| 0 \right\rangle \otimes \left| 0 \right\rangle, \left| 0 \right\rangle \otimes \left| 1 \right\rangle, \left| 1 \right\rangle \otimes \left| 0 \right\rangle, \left| 1 \right\rangle \otimes \left| 1 \right\rangle$

which could also be written as

$\left|00\right\rangle, \left|01\right\rangle, \left|10\right\rangle, \left|11\right\rangle,$

or, if we take them as binary representation of numbers, we can write them concisely as

$\left|0\right\rangle, \left|1\right\rangle,\left|2\right\rangle,\left|3\right\rangle,$

Using the above basis vectors, we can write a general 2-QBit system as:

$\displaystyle \left|\Psi\right\rangle = c_1\left|0\right\rangle + c_1\left|1\right\rangle + c_1\left|2\right\rangle + c_1\left|3\right\rangle = \sum_{i=0}^{2^n-1} c_i\left|i\right\rangle$

where the numbers $c_i$ are complex and n=2. The summation is up to $2^n-1$ since the maximum number n QBits can represent is $2^n-1$. The summation on the right also gives us the general form of a QBit for an n-QBit system.

As we saw earlier, the quantum operators (or gates) are the ones used to compute from a given QBit inputs. Let $\mathbf{U}_f$ represent a unitary operator implementing the function f. The function f is a function that takes the numbers from the domain $\{0,1\}$ to $\{0,1\}$, that is,

$\displaystyle f: \{0,1\}\rightarrow \{0,1\}$

For a two-QBit system, the form of quantum computation is

$\displaystyle \mathbf{U}_f\left|x\right\rangle \left|y\right\rangle = \left|x\right\rangle \left|x\oplus f(x)\right\rangle$

where the symbol $\oplus$ represent bitwise OR operation. The table below will give you the results of the bitwise OR operation for every combination of values in $\{0,1\}$:

$\begin{array}{cc} 0 \oplus 0 &= 0\\ 0 \oplus 1 &= 1\\ 1 \oplus 0 &= 1\\ 1 \oplus 1 &= 0 \end{array}$

A simple but important example of a unitary operator acting on two QBits is the $\mathbf{C_{\text{NOT}}}$ gate, defined as:

$\displaystyle \mathbf{C_{\text{NOT}}}\left|x\right\rangle\left|y\right\rangle = \left|x\right\rangle\left|y \oplus x\right\rangle$

Here is the complete action of the $\mathbf{C_{\text{NOT}}}$:

$\begin{tabular}{c|c|c} \text{Input} & \text{Output} & \text{Result}\\ 0 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{tabular}$

If you study the table above, you’ll notice that if the input QBit is in the state $\left|0\right\rangle$, the output QBit stays the same. However, if the input QBit is in $\left|1\right\rangle$, the output QBit is flipped. The input QBit in this case controls the flipping of the second QBit. The input QBit is said to be the control bit. That’s why it’s called a $\mathbf{C_{\text{NOT}}}$ which stands for Controlled-NOT gate.

Let’s now go back to our original problem. There is a positive number $a$ that is less than $2^n$ . We want to find this number using quantum computation. We are given a black box operator $\mathbf{U}_f$ whose action on a bra is

$\displaystyle \mathbf{U}_f \left|x\right\rangle\left|y\right\rangle = \left|x\right\rangle\left| y \oplus f(x)\right\rangle$

where

$\displaystyle f(x) = x_0 a_0 \oplus x_1 a_1 \oplus \cdots x_{n-1}a_{n-1} = x \cdot a$

where $a$ is the unknown number we are going to guess and $x_ia_i$ are the i-th bits of x and a.

So how many times do we have to apply $\mathbf{U}_f$ in order to guess the number $a$?

To solve this, we need $n$ input QBits and 1 output QBit. We will initialize our QBits to $\left|0\right\rangle$. So the initial configuration is

$\displaystyle \underbrace{\left|000...0\right\rangle}_{\text{n number of zeroes}}\otimes\left|1\right\rangle$

Notice how we separate the input and output qubits.

Next we apply the Hadamard operator to each QBit:

$\displaystyle \mathbf{H}^{\otimes n}\otimes \mathbf{H} \left|000...0\right\rangle\left|1\right\rangle = \mathbf{H}^{\otimes n} \left|000...0\right\rangle\otimes \mathbf{H}\left|1\right\rangle$

We have seen the Hadamard operator earlier. For the sake of convenience we’ll repeat it here:

$\begin{array}{rl} \displaystyle \mathbf{H} \left| 0 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1 \right\rangle)\\ \displaystyle \mathbf{H} \left| 1 \right\rangle &= \displaystyle \frac{1}{\sqrt{2}}(\left| 0 \right\rangle - \left| 1\right\rangle) \end{array}$

We can generalize this by letting x be either 0 or 1:

$\begin{array}{rl} \displaystyle \mathbf{H}\left| x\right\rangle &= \frac{1}{\sqrt{2}}\big(\left|0\right\rangle + (-1)^x \left|1\right\rangle \end{array}$

We can use this to compute the action on two QBits and generalize:

$\begin{array}{rl} \displaystyle \mathbf{H}^{\otimes 2}\left| x\right\rangle &= \mathbf{H}^{\otimes 2}\left| x_0x_1\right\rangle = \mathbf{H}\left| x_0\right\rangle \otimes \mathbf{H}\left| x_1\right\rangle\\ &= \displaystyle \frac{1}{\sqrt{2}}\big(\left|0\right\rangle + (-1)^{x_0} \left|1\right\rangle \otimes \frac{1}{\sqrt{2}}\big(\left|0\right\rangle + (-1)^{x_1} \left|1\right\rangle\\ &= \displaystyle \big(\frac{1}{\sqrt{2}}\big)^2 \big( \left|00\right\rangle + (-1)^{x_0} \left|01\right\rangle + (-1)^{x_1}\left|10\right\rangle + (-1)^{x_1 + x_0}\left|11\right\rangle\\ &= \displaystyle \big(\frac{1}{\sqrt{2}}\big)^2 \big( (-1)^{0\cdot x_1 + 0\cdot x_0} \left|00\right\rangle + (-1)^{0\cdot x_1 + 1\cdot x_0} \left|01\right\rangle + (-1)^{1\cdot x_1 + 0\cdot x_0}\left|10\right\rangle + (-1)^{1\cdot x_1 + 1\cdot x_0}\left|11\right\rangle\\ &= \displaystyle \big(\frac{1}{\sqrt{2}}\big)^2 \big( (-1)^{0\cdot x_1 + 0\cdot x_0} \left|0\right\rangle + (-1)^{0\cdot x_1 + 1\cdot x_0} \left|1\right\rangle + (-1)^{1\cdot x_1 + 0\cdot x_0}\left|2\right\rangle + (-1)^{1\cdot x_1 + 1\cdot x_0}\left|3\right\rangle\\ &= \displaystyle \big(\frac{1}{\sqrt{2}}\big)^2 \sum_{y=0}^{2^2-1} (-1)^{x\cdot y}\left|y\right\rangle \end{array}$

In general,

$\displaystyle \mathbf{H}^{\otimes n}\left| x\right\rangle = \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{x\cdot y}\left|y\right\rangle$

Using the above formula, the action of $\mathbf{H}^{\otimes n}$ on $\left|0\right\rangle$ is

$\displaystyle \mathbf{H}^{\otimes n} \left|0\right\rangle = \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \left|y\right\rangle$

The next step is to apply our operator $\mathbf{U}_{f}$ which was defined above:

$\begin{array}{rl} \displaystyle \mathbf{U}_f\mathbf{H}^{\otimes n}\left| 0\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big) &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \mathbf{U}_f\left|y\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)\\ &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \frac{1}{\sqrt{2}} \big( \mathbf{U}_f\left|y\right\rangle \otimes \left|0\right\rangle - \mathbf{U}_f\left|y\right\rangle \otimes \left|1\right\rangle\big)\\ &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \frac{1}{\sqrt{2}} \big( \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle - \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle\big)\\ \end{array}$

Let’s break this down a bit. Since the value of $f(y)$ is either 0 or 1, we have 2 cases:

When $f(y) = 0$,

$\displaystyle \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle = \left|y\right\rangle \otimes \left|0\right\rangle$
$\displaystyle \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle = \left|y\right\rangle \otimes \left|1\right\rangle$

which means

$\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \frac{1}{\sqrt{2}} \big( \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle - \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle\big) = \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \left|y\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)$

When $f(y) = 1$,

$\displaystyle \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle = \left|y\right\rangle \otimes \left|1\right\rangle$
$\displaystyle \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle = \left|y\right\rangle \otimes \left|0\right\rangle$

which means

$\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \frac{1}{\sqrt{2}} \big( \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle - \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle\big) = \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1) \left|y\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)$

Combining these 2 we have

$\begin{array}{rl} \displaystyle \mathbf{U}_f\mathbf{H}^{\otimes n}\left| 0\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big) &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} \frac{1}{\sqrt{2}} \big( \left|y\right\rangle \otimes \left|0\oplus f(y)\right\rangle - \left|y\right\rangle \otimes \left|1\oplus f(y)\right\rangle\big)\\ &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big) \end{array}$

Finally, getting the Hadamard action on this gives:

$\begin{array}{rl} \displaystyle \mathbf{H}^{\otimes n}\otimes\mathbf{H}\Big[\mathbf{U}_f\mathbf{H}^{\otimes n}\left| 0\right\rangle \otimes \frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)\Big] &=\mathbf{H}^{\otimes n} \mathbf{U}_f\mathbf{H}^{\otimes n}\left| 0\right\rangle \otimes \mathbf{H}\Big[\frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)\Big]\\ &= \underbrace{\mathbf{H}^{\otimes n} \Big[\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \Big]}_{A} \otimes \underbrace{\mathbf{H}\Big[\frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)\Big]}_{B} \end{array}$

Expanding underbrace B gives us:

$\begin{array}{rl} \displaystyle \mathbf{H}\Big[\frac{1}{\sqrt{2}} \big( \left|0\right\rangle - \left|1\right\rangle\big)\Big] &= \displaystyle \frac{1}{\sqrt{2}}\big(\mathbf{H}\left|0\right\rangle - \mathbf{H}\left|1\right\rangle\big)\\ &= \displaystyle \frac{1}{\sqrt{2}}\Big[\frac{1}{\sqrt{2}} \big( \left|0\right\rangle + \left|1\right\rangle \big) - \frac{1}{\sqrt{2}}\big( \left|0\right\rangle - \left|1\right\rangle\big) \Big]\\ &= \displaystyle \frac{1}{2}\Big[ 2\left|1\right\rangle \Big]\\ &= \left|1\right\rangle \end{array}$

Expanding underbrace A above:

$\begin{array}{rl} \mathbf{H}^{\otimes n} \Big[\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \Big] &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)}\mathbf{H}^{\otimes n} \left|y\right\rangle \end{array}$

We have found earlier that

$\displaystyle \mathbf{H}^{\otimes n}\left| y\right\rangle = \frac{1}{2^{n/2}}\sum_{x=0}^{2^n-1} (-1)^{x\cdot y}\left|x\right\rangle$

Substituting this to the above and noting that

$\displaystyle f(y) = y\cdot a = \sum_{i=0}^{n-1} y_ia_i \mod 2$,

we get:

$\begin{array}{rl} \mathbf{H}^{\otimes n} \Big[\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \Big] &= \displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)}\mathbf{H}^{\otimes n} \left|y\right\rangle \\ &= \displaystyle \frac{1}{2^{n}}\sum_{y=0}^{2^n-1} \sum_{x=0}^{2^n-1} (-1)^{f(y)}(-1)^{y\cdot x}\left|x\right\rangle\\ &= \displaystyle \frac{1}{2^{n}}\sum_{y=0}^{2^n-1} \sum_{x=0}^{2^n-1} (-1)^{y\cdot a}(-1)^{y\cdot x}\left|x\right\rangle\\ &= \displaystyle \frac{1}{2^{n}}\sum_{x=0}^{2^n-1} \underbrace{\sum_{y=0}^{2^n-1} (-1)^{(a+x)\cdot y}}\left|x\right\rangle\\ \end{array}$

where we have interchanged the order of the summation. To simplify the term with the underbrace, notice that

$\begin{array}{rl} \displaystyle \Big[1+ (-1)^{z_0}\Big]\Big[1+(-1)^{z_1}\Big] &= 1 + (-1)^{z_0} + (-1)^{z_1} + (-1)^{z_0+z_1}\\ &= \displaystyle (-1)^{0\cdot z_0 + 0\cdot z_1} + (-1)^{1\cdot z_0 + 0\cdot z_1} + (-1)^{0\cdot z_0 + 1\cdot z_1} + (-1)^{1\cdot z_0+ 1\cdot z_1}\\ &= \displaystyle(-1)^{0\cdot z} + (-1)^{1\cdot z} + (-1)^{2\cdot z} + (-1)^{3\cdot z}\\ &= \displaystyle \sum_{y=0}^{2^2-1} (-1)^{y\cdot z} \end{array}$

We can therefore write the factor in underbrace as

$\begin{array}{rl} \mathbf{H}^{\otimes n} \Big[\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \Big] &= \displaystyle \frac{1}{2^{n}}\sum_{x=0}^{2^n-1} \underbrace{\sum_{y=0}^{2^n-1} (-1)^{(a+x)\cdot y}}\left|x\right\rangle\\ &= \displaystyle \frac{1}{2^{n}}\sum_{x=0}^{2^n-1} \prod_{i=0}^{n-1} \Big[ \underbrace{1 +(-1)^{(a_i+x_i)}}\Big] \left|x\right\rangle\\ \end{array}$

Notice that

$\displaystyle 1 +(-1)^{(a_i+x_i)} = \begin{cases} 0 & x\ne a\\ 2 & x=a \end{cases}$

that is, the product $\prod_{i=0}^{n-1} \Big[1 + (-1)^{(a_i+x_i)}\Big] = 0$ unless $x=a$. Therefore

$\begin{array}{rl} \mathbf{H}^{\otimes n} \Big[\displaystyle \frac{1}{2^{n/2}}\sum_{y=0}^{2^n-1} (-1)^{f(y)} \left|y\right\rangle \Big] &= \displaystyle \frac{1}{2^{n}}\sum_{x=0}^{2^n-1} \prod_{i=0}^{n-1} \Big[ \underbrace{1 +(-1)^{(a_i+x_i)}}\Big] \left|x\right\rangle\\ &= \displaystyle \frac{1}{2^{n}} 2^n \left|a\right\rangle\\ &= \left|a\right\rangle \end{array}$

Therefore, the quantum computation gives the answer in one application of $\mathbf{U}_f$:

$\displaystyle \mathbf{H}^{\otimes n+1}\mathbf{U}_f \mathbf{H}^{\otimes n+1} \left|0\right\rangle = \left| a\right\rangle \left| 1\right\rangle$

So what just happened? We began with a superposition of states and finished with the correct answer in one application of the black box operator $\mathbf{U}_f$. That’s the power of Quantum Computing!

## Prototyping Parallel Programs on OpenShift

In the previous posts, we were introduced to parallel computing and saw how it can speed up computation and by what factor (depending on the algorithm). We also introduced Chapel, a language for parallel computation, developed by Cray. Using Chapel, we first developed a very simple parallel algorithm to compute the sum of a list of numbers. Then we learned how to implement parallel matrix multiplication. In this post, we will talk about how to run our parallel programs using a cluster of machines. The source code referenced here can be found on github.

To run our parallel program, we will need to set up a bare-metal server or a Virtual Machine (VM), install Chapel and clone it to N instances. A shared filesystem will be mounted on a pre-defined directory on each instance. This filesystem will allow all instances to load data from the same location. Any output from each instance will also be written to this filesystem (although we must ensure they write to different files or directories to avoid corruption of data).

Here is how our setup will look like:

We will also need to setup the following:

• SSH server – this will run as an ordinary user using port 2222. The Chapel platform will connect to port 2222 on each machine.

We will need to export the following environment variables to enable parallel computation across different machines. The purpose of those variables can be found here. The variable GASNET_SSH_SERVERS is a space-separated list of IP addresses or hostnames of our nodes. In our setup, we will use the following:

export CHPL_COMM=gasnet
export CHPL_LAUNCHER=amudprun
export GASNET_SSH_SERVERS="node00 node01 node10 node11"


where node00, node01, node10, node11 are hostnames of our machines/nodes.

## How to Run a Parallel Program

We will run our parallel matrix_multiplication on our cluster. To launch the program, we will login to one of the nodes and change directory to /home/chapel containing the mat_mul binary and issue the following command:

chpl mat_mul -nl 4


## Running in OpenShift

If having to provision a bare-metal or VM environment for parallel computing is costly, we can turn to containers for a much cheaper and faster way to provision. To make our lives easier, we will use OpenShift Origin platform to run our parallel computing environment.

In the succeeding sections, we assume that we have a working installation of OpenShift. See the instructions here on how to set up OpenShift on VirtualBox or here to install a single-node OpenShift in AWS.

### Containerizing the “Node”

We will have to build a container image of the bare metal or VM we mentioned above. The format we will use is Docker. First, we clone the project from github:

git clone https://github.com/corpbob/chapel-openshift-demo.git


Change directory to chapel-openshift-demo and build the image:

docker built -t chapel .


Create the chapel project on OpenShift. We use the admin user in this example:

oc login -u admin
oc new-project chapel


We need to know the ip address of the docker-registry service in OpenShift. We execute the following command:

[root@openshift chapel-openshift-demo]# oc describe svc docker-registry -n default
Name:			docker-registry
Namespace:		default
Labels:			docker-registry=default
Selector:		docker-registry=default
Type:			ClusterIP
IP:			172.30.1.1
Port:			5000-tcp	5000/TCP
Endpoints:		172.17.0.7:5000
Session Affinity:	ClientIP
No events.


Line 7 (highlighted) of the output gives us the IP address of the docker-registry service. We will use this to push the Chapel image we just created to the registry. First we login to the docker registry:

docker login -u admin -p $(oc whoami -t) 172.30.1.1:5000  Tag the chapel image using the following format <registry_ip>:<port>/<project_name>/<image_name>:<version>  For our example, we use the following command to tag the image: docker tag chapel 172.30.1.1:5000/chapel/chapel:v0.1  We can now push the image to the internal docker registry: docker push 172.30.1.1:5000/chapel/chapel  ### Importing the Chapel Template In the previous section, we built the container image of Chapel and pushed it to the private docker registry. In this section, we will import a template that will do the following: • Set up the Chapel containers inside OpenShift • Create a file that will dynamically generate the variable GASNET_SSH_SERVERS containing the IP addresses of the Chapel pods that will be used in the parallel computation. The name of the template is chapel.yml. Import the template using the command oc create -f chapel.yml  We need to give the default service account the view role so that it can read the IP addresses of the pods associated with chapel. To do this, execute the command: oc policy add-role-to-user view system:serviceaccount:chapel:default  After this we can now create the chapel application: oc new-app chapel  This will automatically trigger a deployment of 4 chapel instances with a shared volume mounted at /home/chapel/mnt. The “hook-post” pod is a separate instance of the chapel image that will execute the following commands echo "export GASNET_MASTERIP=\$MY_NODE_IP" > /home/chapel/mnt/exports && \
echo "export GASNET_SSH_OPTIONS=\"-p 2222\"" >> /home/chapel/mnt/exports && \
for pod in oc get pods -l app=chapel|grep chapel|awk '{print $1}'; \ do \ oc describe pod$pod |grep ^IP:|awk '{print $2}'; \ done| \ awk 'BEGIN { x="" } \ {x = x$1" "} \
END {print "export GASNET_SSH_SERVERS=\""x"\""}' >> \
/home/chapel/mnt/exports


The output of the above command is a file named exports and looks like the below:

### Running the Sample Program

We now go to the web console-> Applications -> Pods. Select any of the pods and click Terminal. In the /home/chapel directory, there is a file named run-test.sh. This file contains the following commands:

export GASNET_SPAWNFN=S
source /home/chapel/mnt/exports

./hello6-taskpar-dist -nl $*  The commands above executes the pre-compiled chapel binary hello6-taskpar-dist which was compiled when we built the container image earlier. Executing this file gives us: # the parameter 4 tells chapel to use 4 pods to execute the command. sh-4.2$ ./run-test.sh 4
Warning: Permanently added '[172.17.0.16]:2222' (ECDSA) to the list of known hosts.
Warning: Permanently added '[172.17.0.18]:2222' (ECDSA) to the list of known hosts.
Warning: Permanently added '[172.17.0.15]:2222' (ECDSA) to the list of known hosts.
Warning: Permanently added '[172.17.0.17]:2222' (ECDSA) to the list of known hosts.
Hello, world! (from locale 0 of 4 named chapel-1-fgg4b)
Hello, world! (from locale 2 of 4 named chapel-1-pt3v9)
Hello, world! (from locale 1 of 4 named chapel-1-rg668)
Hello, world! (from locale 3 of 4 named chapel-1-rw2rc)


### Running the Parallel Matrix Multiplication

Copy the file mat_mul.chpl to the pod and compile.

chpl mat_mul.chpl -o mnt/mat_mul


The command above will place the resulting binary inside the directory /home/chapel/mnt. This will be accessible from all pods.

Finally, execute the parallel matrix multiplication:

./run.sh mnt/mat_mul 4


## Conclusion

I have not tried this in a real production environment or even on a bare-metal installation of OpenShift. This seems to be a promising use of OpenShift but I still have to find out.

## Implementing Parallel Algorithms Part 2

In the previous post, we implemented a very simple parallel program to add a set of numbers. In this post, we will implement parallel matrix multiplication.

We have shown a parallel algorithm to multiply 2 big matrices using message passing. The algorithm involved block sub-matrices to be passed from node to node and multiplied within a node until the answer is found.

There will be 2 square matrices A and B. In our example, the dimension of both A and B is 4×4. We will distribute the matrices evenly to 4 nodes.

$A = \left[ \begin{array}{cc|cc} a_{00} & a_{01} & a_{02} & a_{03}\\ a_{10} & a_{11} & a_{12} & a_{13}\\ \hline a_{20} & a_{21} & a_{22} & a_{23}\\ a_{30} & a_{31} & a_{32} & a_{33} \end{array} \right], B = \left[ \begin{array}{cc|cc} b_{00} & b_{01} & b_{02} & b_{03}\\ b_{10} & b_{11} & b_{12} & b_{13}\\ \hline b_{20} & b_{21} & b_{22} & b_{23}\\ b_{30} & b_{31} & b_{32} & b_{33} \end{array} \right]$

In this example, node 00 will have the following matrices:

$A_{00}=\begin{bmatrix} a_{00} & a_{01}\\ a_{10} & a_{11} \end{bmatrix}, B_{00}= \begin{bmatrix} b_{00} & b_{01}\\ b_{10} & b_{11} \end{bmatrix}$

const n = 4;
var vec = 1..n;
var blockSize = 2;

var A: [vec, vec] real;
var B: [vec, vec] real;
var C: [vec, vec] real;

coforall loc in Locales {
on loc {
var i = loc.id/2;
var j = loc.id%2;
var istart = i*blockSize;
var iend = istart + blockSize;
var jstart = j*blockSize;
var jend = jstart + blockSize;

for (r,s) in {istart + 1..iend, jstart + 1..jend} {
B(r,s) = r+s;
A(r,s) = 2*r + s;
}
}
}


Each node has limited memory physically exclusive to itself. In order for node A to have access to the contents of the memory of another node B, node B should pass the data to node A. Fortunately, Chapel can use a library called GASNet that allows each node to have a global view of the memory of all nodes participating in the computation.

In the code above, each node loads its own data. However, the GASNet library allows each node to access the matrix elements loaded by the other nodes. Consequently, we are able to reference the sub-matrix held by each node without doing fancy message passing. The algorithm is then a straightforward implementation of

$\displaystyle \mathbf{C}_{ij}=\sum_{k=0}^{2} \mathbf{A}_{ik} \mathbf{B}_{kj}$

where $\mathbf{A_{ij}}$, $\mathbf{B_{ij}}$ and $\mathbf{C_{ij}}$ are submatrices of $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$, respectively.

Below is the straightforward implementation of parallel block multiplication:

coforall loc in Locales {
on loc {
var i = loc.id/2;
var j= loc.id%2;
var istart = i*blockSize;
var iend = istart + blockSize;
var jstart = j*blockSize;
var jend = jstart + blockSize;
var r = { istart + 1..iend, jstart + 1..jend };
ref W = C[r].reindex( { 1..2,1..2 });

coforall k in 0..1 {
var U=get_block_matrix(A[vec,vec],i,k,blockSize);
var V=get_block_matrix(B[vec,vec],k,j,blockSize);
var P = mat_mul(U,V);
coforall (s,t) in { 1..2,1..2 } {
W(s,t) += P(s,t);
}
}
}
}


The procedure get_block_matrix will return the sub-matrix given the (i,j)th index and the block size.

proc get_block_matrix(A: [?D], i:int, j:int , blockSize:int) {
var r = { i*blockSize+1 .. i*blockSize
+  blockSize, j*blockSize
+ 1 .. j*blockSize + blockSize };
return A[r];
}


The procedure mat_mul will return the matrix product of two sub-matrices:

proc mat_mul(A: [?D1], B: [?D2]) {
var D3 = { 1..2, 1..2 };
var C: [D3] real;
var AA = A.reindex({1..2,1..2});
var BB = B.reindex({1..2,1..2});

for row in 1..2 {
for col in 1..2 {
var sum:real = 0;
for k in 1..2 {
sum += AA(row,k) * BB(k,col);
}
C(row,col) = sum;
}
}
return C;
}

writeln(C[vec,vec]);


To run this code, we need to set the following environment variables:

source \$CHPL_HOME/util/setchplenv.bash

export CHPL_COMM=gasnet
export CHPL_LAUNCHER=amudprun

export GASNET_SSH_SERVERS="127.0.0.1 127.0.0.1 127.0.0.1 127.0.0.1"


Compiling and running this program gives the output:

#Compile
chpl mat_mul.chpl -o mat_mul

# Run using 4 nodes
./mat_mul -nl 4

A=
3.0 4.0 5.0 6.0
5.0 6.0 7.0 8.0
7.0 8.0 9.0 10.0
9.0 10.0 11.0 12.0
B=
2.0 3.0 4.0 5.0
3.0 4.0 5.0 6.0
4.0 5.0 6.0 7.0
5.0 6.0 7.0 8.0
C=
68.0 86.0 104.0 122.0
96.0 122.0 148.0 174.0
124.0 158.0 192.0 226.0
152.0 194.0 236.0 278.0