# Hiding a Lookup Table in a Modulus Operation

Lookup tables are a tool found in every programmer’s tool belt.

Lookup tables let you pre-calculate a complex calculation in advance, store the results in a table (an array), and then during performance critical parts of your program, you access that table to get quick answers to the calculations, without having to do the complex calculation on the fly.

In this post I’ll show a way to embed a lookup table inside of a single (large) number, where you extract values from that lookup table by taking a modulus of that number with different, specific values.

This technique is slower and takes more memory than an actual lookup table, but it’s conceptually interesting, so I wanted to share.

Also, I stumbled on this known technique while working on my current paper. The paper will make this technique a bit more practical, and I’ll share more info as soon as I am able, but for now you can regard this as a curiosity 😛

Onto the details!

## 1 Bit Input, 1 Bit Output: Pass Through

Let’s learn by example and start with a calculation that takes in an input bit, and gives that same value for an output bit. It’s just a 1 bit pass through lookup table. $\begin{array}{c|c} \text{Input} & \text{Output} \\ \hline 0 & 0 \\ 1 & 1 \\ \end{array}$

To be able to convert that to something we can decode with modulus we have to solve the following equations: $x \% k_0 = 0 \\ x \% k_1 = 1$ $x$ is the number that represents our lookup table. $k_0$ and $k_1$ are the values that we modulus x against to get our desired outputs out.

It looks as if we have two equations and three unknowns – which would be unsolvable – but in reality, x is the only unknown. The k values can be whatever values it takes to make the equations true.

I wrote a blog post on how to solve equations like these in a previous post: Solving Simultaneous Congruences (Chinese Remainder Theorem).

You can also use this chinese remainder theorem calculator, which is handy: Chinese Remainder Theorem Calculator

The short answer here is that the k values can be ANY numbers, so long as they are pairwise co-prime to each other – AKA they have a greatest common divisor of 1.

If we pick 3 and 4 for k0 and k1, then using the chinese remainder theorem we find that x can equal 9 and the equations are true. Technically the answer is 9 mod 12, so 21, 33, 45 and many other numbers are also valid values of x, but we are going to use the smallest answer to keep things smaller, and more manageable.

So, in this case, the value representing the lookup table would be 9. If you wanted to know what value it gave as output when you plugged in the value 0, you would modulus the lookup table (9) against k0 (3) to get the output. If you wanted to know what value it gave as output when you plugged in the value 1, you would modulus the lookup table (9) against k1 (4) to get the output. The table below shows that it passes through the value in both cases like it should: $\begin{array}{c|c|c|c} \text{Input} & \text{Symbolic} & \text{Numeric} & \text{Output} \\ \hline 0 & x \% k_0 & 9 \% 3 & 0\\ 1 & x \% k_1 & 9 \% 4 & 1\\ \end{array}$

## 1 Bit Input, 1 Bit Output: Not Gate

Let’s do something a little more interesting. Let’s make the output bit be the reverse of the input bit. The equations we’ll want to solve are this: $x \% k_0 = 1 \\ x \% k_1 = 0$

We can use 3 and 4 for k0 and k1 again if we want to. Using the Chinese remainder theorem to solve the equations gives us a value of 4 for x. Check the truth table below to see how this works: $\begin{array}{c|c|c|c} \text{Input} & \text{Symbolic} & \text{Numeric} & \text{Output} \\ \hline 0 & x \% k_0 & 4 \% 3 & 1\\ 1 & x \% k_1 & 4 \% 4 & 0\\ \end{array}$

## 1 Bit Input, 1 Bit Output: Output Always 1

What if we wanted the output bit to always be 1 regardless of input bit? $x \% k_0 = 1 \\ x \% k_1 = 1$

Using 3 and 4 for our k values again, we solve and get a value of 1 for x. Check the truth table to see it working below: $\begin{array}{c|c|c|c} \text{Input} & \text{Symbolic} & \text{Numeric} & \text{Output} \\ \hline 0 & x \% k_0 & 1 \% 3 & 1\\ 1 & x \% k_1 & 1 \% 4 & 1\\ \end{array}$

Hopefully one bit input to one bit output makes sense now. Let’s move on (:

## 2 Bit Input, 1 Bit Output: XOR Gate

Things get a little more interesting when we bump the number of input bits up to 2. If we want to make a number which represents XOR, we now have 4 equations to solve. $x \% k_{00} = 0 \\ x \% k_{01} = 1 \\ x \% k_{10} = 1 \\ x \% k_{11} = 0$

In general we will have $2^N$ equations, where N is the number of input bits.

You might have noticed that I use subscripts for k corresponding to the input bits that the key represents. This is a convention I’ve found useful when working with this stuff. Makes it much easier to see what’s going on.

Now with four equations, we need 4 pairwise coprime numbers – no number has a common factor with another number besides 1.

Let’s pull them out of the air. Umm… 3, 4, 5, 7

Not too hard with only two bits of input, but you can see how adding input bits makes things a bit more complex. If you wanted to make something that took in two 16 bit numbers as input for example, you would need 2^32 co-prime numbers, since there was a total of 32 bits of input!

When we solve those four equations, we get a value of 21 for x.

Notice how x is larger now that we have more input bits? That is another added complexity as you add more input bits. The number representing your program can get very, very large, and require you to use “multi precision integer” math libraries to store and decode the programs, when the numbers get larger than what can be held in a 64 bit int.

Boost has a decent library for this, check out boost::multiprecision::cpp_int, it’s what I use. You can download boost from here: http://www.boost.org/doc/libs/1_59_0/more/getting_started/windows.html

Anyhow, let’s check the truth table to see if our values work: $\begin{array}{c|c|c|c} \text{Input} & \text{Symbolic} & \text{Numeric} & \text{Output} \\ \hline 00 & x \% k_{00} & 21 \% 3 & 0 \\ 01 & x \% k_{01} & 21 \% 4 & 1 \\ 10 & x \% k_{10} & 21 \% 5 & 1 \\ 11 & x \% k_{11} & 21 \% 7 & 0 \end{array}$

Woot, it worked.

## 2 Bit Input, 2 Bit Output: OR, AND

What happens when we add another bit of output? Basically we just treat each output bit as it’s own lookup table. This means that if we have two output bits, we will have two numbers representing our program (one for each bit), and that this is true regardless of how many input bits we have.

Let’s make the left output bit ( $x_0$) be the OR of the input bits and the right output bit ( $x_1$) be the AND of the input bits.

That give us these two sets of equations to solve: $x_0 \% k_{00} = 0 \\ x_0 \% k_{01} = 1 \\ x_0 \% k_{10} = 1 \\ x_0 \% k_{11} = 1 \\ \\ x_1 \% k_{00} = 0 \\ x_1 \% k_{01} = 0 \\ x_1 \% k_{10} = 0 \\ x_1 \% k_{11} = 1 \\$

We can use the same coprime numbers for our k values as we used in the last section (3,4,5,7). Note that we use the same k values in each set of equations. This is intentional and required for things to work out!

If we solve each set of equations we get 141 for x0, and 120 for x1.

Let’s see if that worked: $\begin{array}{c|c|c|c} \text{Input} & \text{Symbolic} & \text{Numeric} & \text{Output} \\ \hline 00 & x_0 \% k_{00}, x_1 \% k_{00} & 141 \% 3, 120 \% 3 & 00 \\ 01 & x_0 \% k_{01}, x_1 \% k_{01} & 141 \% 4, 120 \% 4 & 10 \\ 10 & x_0 \% k_{10}, x_1 \% k_{10} & 141 \% 5, 120 \% 5 & 10 \\ 11 & x_0 \% k_{11}, x_1 \% k_{11} & 141 \% 7, 120 \% 7 & 11 \end{array}$

Hey, it worked again. Neat!

## Example Code

Now that we have the basics worked out, here is some sample code.

The lookup table takes in 8 bits as input, mapping 0..255 to 0…2pi and gives the sine of that value as output in a float. So it has 8 bits of input and 32 bits of output.

#include <vector>
#include <boost/multiprecision/cpp_int.hpp>
#include <stdint.h>
#include <string.h>
#include <memory>

typedef boost::multiprecision::cpp_int TINT;
typedef std::vector<TINT> TINTVec;

const float c_pi = 3.14159265359f;

//=================================================================================
void WaitForEnter ()
{
printf("nPress Enter to quit");
fflush(stdin);
getchar();
}

//=================================================================================
static TINT ExtendedEuclidianAlgorithm (TINT smaller, TINT larger, TINT &s, TINT &t)
{
// make sure A <= B before starting
bool swapped = false;
if (larger < smaller)
{
swapped = true;
std::swap(smaller, larger);
}

// set up our storage for the loop.  We only need the last two values so will
// just use a 2 entry circular buffer for each data item
std::array<TINT, 2> remainders = { larger, smaller };
std::array<TINT, 2> ss = { 1, 0 };
std::array<TINT, 2> ts = { 0, 1 };
size_t indexNeg2 = 0;
size_t indexNeg1 = 1;

// loop
while (1)
{
// calculate our new quotient and remainder
TINT newQuotient = remainders[indexNeg2] / remainders[indexNeg1];
TINT newRemainder = remainders[indexNeg2] - newQuotient * remainders[indexNeg1];

// if our remainder is zero we are done.
if (newRemainder == 0)
{
// return our s and t values as well as the quotient as the GCD
s = ss[indexNeg1];
t = ts[indexNeg1];
if (swapped)
std::swap(s, t);

// if t < 0, add the modulus divisor to it, to make it positive
if (t < 0)
t += smaller;
return remainders[indexNeg1];
}

// calculate this round's s and t
TINT newS = ss[indexNeg2] - newQuotient * ss[indexNeg1];
TINT newT = ts[indexNeg2] - newQuotient * ts[indexNeg1];

// store our values for the next iteration
remainders[indexNeg2] = newRemainder;
ss[indexNeg2] = newS;
ts[indexNeg2] = newT;

// move to the next iteration
std::swap(indexNeg1, indexNeg2);
}
}

//=================================================================================
void MakeKey (TINTVec &keys, TINT &keysLCM, size_t index)
{
// if this is the first key, use 3
if (index == 0)
{
keys[index] = 3;
keysLCM = keys[index];
return;
}

// Else start at the last number and keep checking odd numbers beyond that
// until you find one that is co-prime.
TINT nextNumber = keys[index - 1];
while (1)
{
nextNumber += 2;
if (std::all_of(
keys.begin(),
keys.begin() + index,
[&nextNumber] (const TINT& v) -> bool
{
TINT s, t;
return ExtendedEuclidianAlgorithm(v, nextNumber, s, t) == 1;
}))
{
keys[index] = nextNumber;
keysLCM *= nextNumber;
return;
}
}
}

//=================================================================================
void CalculateLookupTable (
TINT &lut,
const std::vector<uint64_t> &output,
const TINTVec &keys,
const TINT &keysLCM,
const TINTVec &coefficients,
)
{
// figure out how much to multiply each coefficient by to make it have the specified modulus residue (remainder)
lut = 0;
for (size_t i = 0, c = keys.size(); i < c; ++i)
{
// we either want this term to be 0 or 1 mod the key.  if zero, we can multiply by zero, and
// not add anything into the bit value!
if ((output[i] & bitMask) == 0)
continue;

// if 1, use chinese remainder theorem
TINT s, t;
ExtendedEuclidianAlgorithm(coefficients[i], keys[i], s, t);
lut = (lut + ((coefficients[i] * t) % keysLCM)) % keysLCM;
}
}

//=================================================================================
template <typename TINPUT, typename TOUTPUT, typename LAMBDA>
void MakeModulus (TINTVec &luts, TINTVec &keys, LAMBDA &lambda)
{
// to keep things simple, input sizes are being constrained.
// Do this in x64 instead of win32 to make size_t 8 bytes instead of 4
static_assert(sizeof(TINPUT) < sizeof(size_t), "Input too large");
static_assert(sizeof(TOUTPUT) < sizeof(uint64_t), "Output too large");

// calculate some constants
const size_t c_numInputBits = sizeof(TINPUT) * 8;
const size_t c_numInputValues = 1 << c_numInputBits;
const size_t c_numOutputBits = sizeof(TOUTPUT) * 8;

// Generate the keys (coprimes)
TINT keysLCM;
keys.resize(c_numInputValues);
for (size_t index = 0; index < c_numInputValues; ++index)
MakeKey(keys, keysLCM, index);

// calculate co-efficients for use in the chinese remainder theorem
TINTVec coefficients;
coefficients.resize(c_numInputValues);
fill(coefficients.begin(), coefficients.end(), 1);
for (size_t i = 0; i < c_numInputValues; ++i)
{
for (size_t j = 0; j < c_numInputValues; ++j)
{
if (i != j)
coefficients[i] *= keys[j];
}
}

// gather all the input to output mappings by permuting the input space
// and storing the output for each input index
std::vector<uint64_t> output;
output.resize(c_numInputValues);
union
{
TINPUT value;
size_t index;
} input;
union
{
TOUTPUT value;
size_t index;
} outputConverter;

for (input.index = 0; input.index < c_numInputValues; ++input.index)
{
outputConverter.value = lambda(input.value);
output[input.index] = outputConverter.index;
}

// iterate through each possible output bit, since each bit is it's own lut
luts.resize(c_numOutputBits);
for (size_t i = 0; i < c_numOutputBits; ++i)
{
const size_t bitMask = 1 << i;
CalculateLookupTable(
luts[i],
output,
keys,
keysLCM,
coefficients,
);
}
}

//=================================================================================
int main (int argc, char **argv)
{
// Look up tables encodes each bit, keys is used to decode each bit for specific
// input values.
TINTVec luts;
TINTVec keys;

// this is the function that it turns into modulus work
typedef uint8_t TINPUT;
typedef float TOUTPUT;
auto lambda = [] (TINPUT input) -> TOUTPUT
{
return sin(((TOUTPUT)input) / 255.0f * 2.0f * c_pi);
};

MakeModulus<TINPUT, TOUTPUT>(luts, keys, lambda);

// show last lut and key to show what kind of numbers they are
std::cout << "Last Lut: " << *luts.rbegin() << "n";
std::cout << "Last Key: " << *keys.rbegin() << "n";

// Decode all input values
std::cout << "n" << sizeof(TINPUT) << " bytes input, " << sizeof(TOUTPUT) << " bytes outputn";
for (size_t keyIndex = 0, keyCount = keys.size(); keyIndex < keyCount; ++keyIndex)
{
union
{
TOUTPUT value;
size_t index;
} result;

result.index = 0;

for (size_t lutIndex = 0, lutCount = luts.size(); lutIndex < lutCount; ++lutIndex)
{
TINT remainder = luts[lutIndex] % keys[keyIndex];
size_t remainderSizeT = size_t(remainder);
result.index += (remainderSizeT << lutIndex);
}

TINT remainder = luts % keys[keyIndex];
std::cout << "i:" << keyIndex << " o:" << result.value << "n";
}

WaitForEnter();
return 0;
}


Here is some output from the program. The first is to show what the last (largest) look up table and key look like. Notice how large the look up table number is! Here it shows some sine values output from the program, using modulus against the large numbers calculated, to get the bits of the result out: ## How to Get Lots of Pairwise Co-Prime Numbers?

You can generate a list of pairwise coprimes using brute force. Have an integer that you increment, and check if it’s pairwise co-prime to the existing items in the list. If it is, add it to the list! Rinse and repeat until you have as many as you want.

That is the most practical way to do it, but there are two other interesting ways I wanted to mention.

The first way is using Fermat numbers. The Fermat numbers are an infinite list of pairwise co-prime numbers and are calculated as $2^{2^n}+1$ where n is an integer. Fermat numbers also have the benefit that you can get the nth item in the list without calculating the numbers that came before it. The only problem is that the numbers grow super huge very fast. The first 7 values are: 3, 5, 17, 257, 65537, 4294967297, 18446744073709551617. If Fermat numbers didn’t grow so quickly, they sure would be useful for things like this technique.

The second way is using something called Sylvester’s sequence. It too is an infinite list of pairwise co-prime numbers, and it too grows very large very quickly unfortunately. I also don’t believe there is a way to calculate the Nth item in the list directly. Every number is based on previous numbers, so you have to calculate them all from the beginning. No random access!

## Beyond Binary

In this post I showed how to work in binary digits, but there is no reason why you have to encode single bits in the lookup tables.

Instead of encoding 0 or 1 in each modulus “lookup table”, you could also perhaps store base 10 numbers in the tables and have 0-9. Or, maybe you encode a byte per lookup table.

Encoding more than one bit effectively makes both your input and your output smaller, which helps the algorithm do more with less.

Your keys will need to be larger though, since the keys have to be larger than the value you plan to store, and your resulting lookup table will be a larger number as well. It might make the technique more worth while though.

I’ll leave that as an exercise for you. If try it and find neat stuff, post a comment and let us know, or drop me an email or something. It’d be neat to hear if people find any practical usage cases of this technique 😛

## The End, For Now!

I want to point out that increasing the number of input bits in this technique is a pretty expensive thing to do, but increasing the number of output bits is a lot cheaper. It kind of makes sense in a way if you think about it. Input bits add information from the outside world that must be dealt with, while output bits are just fluff that can easily be diluted or concentrated by adding or removing bits that are associated with, and calculated from, the input bits.

Another problem you may have noticed with this technique is that if you have a really expensive calculation that you are trying to “flatten” into modulus math like this, that you have to run that calculation many, many times to know what values a lookup table would give you. You have to run it once per possible input to get every possible output. That is expected when making a lookup table, since you are paying a cost up front to make things faster later.

The paper I’m working on changes things a bit though. One of the things it does is it makes it so doing this technique only requires that you evaluate the function once, and it calculates all values simultaneously to give the end result that you can then do modulus against. It’s pretty cool IMO and I will share more details here as soon as I am able – and yes, i have actual working code that does that, believe it or not! I’m looking forward to being able to share it later on. Maybe someone will find some really cool usage case for it.

# Quantum Computing For Programmers Part 2: Multiple Qubits

In part 1 (Quantum Computing For Programmers Part I: One Qubit) we looked at the basics of quantum computing and how to deal with a single qubit. Here we’ll talk about the interesting things that happen when you involve multiple qubits!

## Multiple Qubit Probabilities & Possibilities

In the last post we used the analogy of a coin to describe a qubit. The coin can be either heads or tails, or flipping in the air, waiting to become either a heads or tails when it lands. The same is true of how a qubit works, where it can either be a 0 or a 1 or some superposition of both, only deciding which it is when observed. Furthermore, just like a real coin, there can be a bias towards becoming one value or the other.

We represented a coin by having a probability for each possible state (heads or tails), but represented a qubit by having an AMPLITUDE for each possible state (0 or 1), where you square an amplitude to get the probability.

Going back to the coin analogy, how would probabilities work if we had two coins?

The four possible outcomes with two coins are:
Tails/Tails

Each coin individually could either be sitting on the table heads or tails, or up in the air, waiting to become either a heads or a tails, with some probability of becoming a heads or a tails.

To figure out the probability of each possible outcome, you just multiply the probability of each coin’s outcome together.

For instance, let’s say that the first coin (coin A) is already sitting on the table as a heads, and the second coin (coin B) is flipping in the air, with a 1/3 chance of becoming heads and a 2/3 chance of becoming tails. Here is how you would calculate the probability of each possible state: $\begin{array}{c|c|c|c} \text{Outcome A/B} & \text{Coin A Probability} & \text{Coin B Probability} & \text{Outcome Probability (A*B)}\\ \hline heads / heads & 100\% & 33\% & 33\% \\ heads / tails & 100\% & 67\% & 67\% \\ tails / heads & 0\% & 33\% & 0\% \\ tails / tails & 0\% & 67\% & 0\% \\ \end{array}$

Qubits actually work the same way! Using the same values as the coins, let’s say qubit A is a 0, and qubit B has a 1/3 chance of becoming a 0, and a 2/3 chance of becoming a 1. Converting those probabilities to amplitudes you could get $A=[1,0], B=[\frac{1}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}]$. $\begin{array}{c|c|c|c|c} \text{Outcome AB} & \text{Qubit A Amplitude} & \text{Qubit B Amplitude} & \text{Outcome Amplitude(A*B)} & \text{Outcome Probability}\\ \hline 00 & 1 & 1/\sqrt{3} & 1/\sqrt{3} & 33\% \\ 01 & 1 & \sqrt{2}/\sqrt{3} & \sqrt{2}/\sqrt{3} & 67\% \\ 10 & 0 & 1/\sqrt{3} & 0 & 0\%\\ 11 & 0 & \sqrt{2}/\sqrt{3} & 0 & 0\% \\ \end{array}$

Note that in both cases, the probabilities add up to 100% like you’d expect.

In the case of qubits, the resulting vector is $[\frac{1}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}, 0, 0]$, which is still normalized, and represents the amplitudes for the 4 possible states of those two qubits: 00, 01, 10 and 11.

When working with one qubit (or coin), there are 2 possible outcomes 0 or 1 (heads or tails). When working with two qubits (or coins), there are four possible outcomes 00, 01, 10, 11 (heads/heads, heads/tails, tails/heads, tails/tails). When working with three qubits or coins, there are eight possible outcomes: 000,001,010 … 111.

With both coins and qubits there are $2^N$ possibilities, where $N$ is the number of qubits or coins you have.

If you had 8 qubits, you’d have to use a 256 dimensional vector to describe the possibilities, since $2^8$ is 256. When performing quantum computing with 8 qubits, you only have to deal with the 8 qubits. When simulating quantum computing on a regular, classical computer, you have to deal with the 256 amplitudes. This kind of gives a glimpse at how quantum computers can be faster than regular computers at certain things. There is an economy of scale working against us on regular computers simulating quantum computing.

The method we used to combine the probabilities of single qubits into an amplitude vector representing multiple qubits is called the Kronecker product. That’s just a fancy way of saying we have to multiply everything from the first vector by everything from the second vector, to get a third vector that is bigger than the first two. You’ll see it represented like this: $A \otimes B$ and while it’s similar to the “outer product” and even uses the same symbol (!), it gives a slightly different result versus if you did an outer product and then vectorized the matrix.

The Kronecker product of vectors works like this: $\begin{bmatrix} A_1 \\ A_2 \\ ... \\ A_M \\ \end{bmatrix} \otimes \begin{bmatrix} B_1 \\ B_2 \\ ... \\ B_N \\ \end{bmatrix} = \begin{bmatrix} A_1 B_1 \\ A_1 B_2 \\ ... \\ A_1 B_N \\ A_2 B_1 \\ A_2 B_2 \\ ... \\ A_2 B_N \\ ... \\ A_M B_1 \\ A_M B_2 \\ ... \\ A_M B_N \\ \end{bmatrix}$

Let’s show an example involving two qubits.

The first qubit has a 75% chance of being heads so it’s amplitude is $[\sqrt{3}/2,1/2]$. The second qubit has a 2/3 chance of being heads so has an amplitude of $[\sqrt{2}/\sqrt{3}, 1/\sqrt{3}]$.

To calculate the amplitude vector representing these two qubits, we do a kronecker product: $\begin{bmatrix} \cfrac{\sqrt{3}}{2} \\ \cfrac{1}{2} \\ \end{bmatrix} \otimes \begin{bmatrix} \cfrac{\sqrt{2}}{\sqrt{3}} \\ \cfrac{1}{\sqrt{3}} \\ \end{bmatrix} = \begin{bmatrix} \cfrac{\sqrt{3}}{2}\cfrac{\sqrt{2}}{\sqrt{3}} \\ \cfrac{\sqrt{3}}{2}\cfrac{1}{\sqrt{3}} \\ \cfrac{1}{2}\cfrac{\sqrt{2}}{\sqrt{3}} \\ \cfrac{1}{2}\cfrac{1}{\sqrt{3}} \\ \end{bmatrix}$

If you simplify that, you get: $\begin{bmatrix} \cfrac{1}{\sqrt{2}} \\ \cfrac{1}{2} \\ \cfrac{1}{\sqrt{6}} \\ \cfrac{1}{2\sqrt{3}} \\ \end{bmatrix}$ $\begin{bmatrix} \cfrac{1}{\sqrt{2}} & \cfrac{1}{2} & \cfrac{1}{\sqrt{6}} & \cfrac{1}{2\sqrt{3}} & \end{bmatrix}$

Squaring those values to get the probabilities of each state, we get the below, which you might notice adds up to 100%, meaning the vector is still normalized! $\begin{bmatrix} 50\% & 25\% & 17\% & 8\% \end{bmatrix}$

This generalizes for larger numbers of qubits, so you could use the Kronecker product to combine a vector describing 4 qubits with a vector describing 3 qubits, to come up with a vector that describes 7 qubits (which would have 128 amplitudes in it, since 2^7 = 128!).

The kronecker product also generalizes to matrices, which we will talk about shortly.

To be able to work with multiple qubits in a quantum circuit, you need to represent all qubits involved in a single vector. Doing this, while being the correct thing to do, also allows entanglement to happen, which we will talk about next!

## Entanglement – Simpler Than You Might Think!

Time to demystify quantum entanglement!

Quantum entanglement is the property whereby two qubits – which can be separated by any distance, such as several light years – can be observed and come up with the same value (0 or 1). Whether they are 0s or 1s is random, but they will both agree, when entangled in the way that causes them to agree (you could alternately entangle them to always disagree).

Interestingly, in 1965 John Bell proved that the mechanism that makes this happen is NOT that the qubits share information in advance. You can read more about that here – it’s near the end: Ars Technica – A tale of two qubits: how quantum computers work.

A common misconception is that this means that we can have faster than light (instantaneous) communication across any distance. That turns out not to be true, due to the No-Communication-Theorem (Wikipedia).

Interestingly though, you can use separated entangled qubits in separated quantum circuits to do something called Quantum Pseudo Telepathy (Wikipedia), which lets you do some things that would otherwise be impossible – like reliably winning a specially designed game that would otherwise be a game of chance.

I don’t yet understand enough about Quantum Pseudo Telepathy to see why it isn’t considered communication. I also have no idea how entanglement is actually “implemented” in the universe, but nobody seems to (or if they do, they aren’t sharing!). How can it be that two coins flipped on opposite sides of the galaxy can be guaranteed to both land with the same side facing up?

Despite those mysteries, the math is DEAD SIMPLE, so let me share it with you, it’ll take like one short paragraph. Are you ready?

Quantum computing works by manipulating the probabilities of the states of qubits. If you have two qubits, there are four possible sets of values when you observe them: 00, 01, 10, 11. If you use quantum computing to set the probabilities of the 01 and 10 states to a 0% chance, and set the probabilities of the 00 and 11 states to a 50% chance each, you’ve now entangled the qubits such that when you observe them, they will always agree, because you’ve gotten rid of the chances that they could ever DISAGREE. Similarly, if instead of setting the 01 and 10 states to 0%, you set the probability of the 00 and 11 states to 0%, you’d have entangled qubits which would always disagree when you observed them, because you’ve gotten rid of the chances that they could ever AGREE.

That’s all there is to it. Strange how simple it is, isn’t it? The table below shows how the only possible outcomes are that the qubits agree, but it is a 50/50 chance whether they are a 0 or a 1: $\begin{array}{c|c|c} \text{Outcome} & \text{Amplitude} & \text{Probability}\\ \hline 00 & 1/\sqrt{2} & 50\% \\ 01 & 0 & 0\% \\ 10 & 0 & 0\% \\ 11 & 1/\sqrt{2} & 50\% \\ \end{array}$

Entanglement isn’t limited to just two qubits, you can entangle any number of qubits together.

Entanglement has a special mathematical meaning. If you can represent a state of a group of qubits by a kronecker product, they are not entangled. If you CAN’T represent a state of a group of qubits by a kronecker product, they ARE entangled. These two things – entanglement and lack of kronecker product factorability (made that term up) – are the same thing.

As an example, what two vectors could you use the kronecker product on to get the entangled two qubit state $1/\sqrt{2}(|00\rangle+|11\rangle)$ (or in vector form $[1/\sqrt{2}, 0, 0, 1/\sqrt{2}]$)? You’ll find there aren’t any! That state is the entangled state where the two qubits will always have the same value when you observe them.

Entangled qubits are nothing special in quantum circuits. You don’t have to take any special precautions when working with them. They are an interesting byproduct of quantum computing, and so basically are something neat that comes out of your quantum circuit, but they don’t require any extra thought when using them within your circuits. Don’t worry about them too much (:

## Multi Qubit Gates

Let’s have a look at some multi qubit gates! (These are again from Wikipedia: Quantum Gate)

### Swap Gate

Given two qubits, with possible states $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, this gate swaps the amplitudes (probabilities) of $|01\rangle, |10\rangle$ and leaves the probabilities of the states $|00\rangle$ and $|11\rangle$ alone.

That might seem weird, but the probability of the $|00\rangle$ state is the probability of the first qubit being 0 added to the probability of the second qubit being 0. If those probabilities swap, they still add to the same value, so this probability is unaffected. It’s the same situation for the $|11\rangle$ state.

Here’s the matrix for the swap gate: $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

### Square Root of Swap Gate

I don’t really understand a usage case for this gate myself, but it apparently does a half way swap between two qubits. That once again modifies the probabilities of the $|01\rangle, |10\rangle$ states, but leaves state $|00\rangle$ and $|11\rangle$ alone again, for the same reason as the swap gate.

Wikipedia says that if you have this gate, and the single qubit gates, that you can do universal quantum computation. In other words, you can build a fully functional quantum computer using just this and the single qubit gates.

Here’s the matrix: $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2}(1+i) & \frac{1}{2}(1-i) & 0 \\ 0 & \frac{1}{2}(1-i) & \frac{1}{2}(1+i) & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

### Controlled Not Gate

The controlled not (CNOT) gate flips the value of the second qubit if the first qubit is true. This is a logic / flow control type of gate.

Interestingly, this gate is also useful for creating entangled qubits, which you’ll be able to see lower down!

Here is the matrix: $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$

The controlled not gate is also referred to as the quantum XOR gate. To see why, check out the truth table below for this gate: $\begin{array}{c|c} \text{Input} & \text{Output} \\ \hline 00 & 00 \\ 01 & 01 \\ 10 & 11 \\ 11 & 10 \\ \end{array}$

If you look at the right most bit of the outputs, you’ll notice that it’s the XOR of the two input bits!

All quantum gates need to be reversible, and having these two qubits be the output of a hypothetical quantum XOR gate allows that to happen. If you look at the right most bit of the inputs, you’ll notice that it is also the XOR of the two output bits. It’s bidirectional, which is kind of weird and kind of interesting 😛

### Generalized Control Gate

You can actually convert any single qubit gate into a controlled gate. That makes a 2 qubit gate which only does work on the second qubit if the first qubit is true.

How you do that is you make a 4×4 identity matrix, and make the lower 2×2 sub-matrix into the single qubit matrix you want to use.

In other words, if your single qubit matrix is this: $\begin{bmatrix} U_{00} & U_{01} \\ U_{10} & U_{11} \\ \end{bmatrix}$

Then the controlled version of the matrix would be this: $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & U_{00} & U_{01} \\ 0 & 0 & U_{10} & U_{11} \\ \end{bmatrix}$

Pretty simple right?

### Toffoli Gate

The Tofolli gate is a 3 qubit gate that is also known as the CCNOT gate or controlled controlled not gate. It flips the third qubit if the first two qubits are true.

It’s matrix looks pretty uninteresting: $\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix}$

This is also known as the quantum AND gate. If you look at the truth table below, you’ll see that when the input has the right most qubit of 0, that in the output, the right most qubit will be the AND value of the two left qubits. We need the three bits of input mapping to three bits of output like this so that the gate is reversible. $\begin{array}{c|c} \text{Input} & \text{Output} \\ \hline 000 & 000 \\ 001 & 001 \\ 010 & 010 \\ 011 & 011 \\ 100 & 100 \\ 101 & 101 \\ 110 & 111 \\ 111 & 110 \\ \end{array}$

### Fredkin Gate

The Fredkin gate is a controlled swap gate. Here’s the matrix: $\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$

Just like we talked about how to make a generalized controlled gate acting on two qubits, you should be able to notice that this gate is just a specific version of a generalized controlled 3 qubit gate. You take an 8×8 identity matrix, and make the lower right 4×4 matrix be the 2 qubit gate that you want to add a control on. In this case, the lower right 4×4 matrix is just the swap gate we mentioned earlier in the post (:

## Circuit To Entangle Qubits

The circuit to entangle qubits is pretty simple, so let’s start with that as the first quantum circuit we look at. It takes in two qubits. The first qubit is put through a Hadamard gate, and then both qubits are put through a controlled not gate. That circuit looks like this (image courtesy of Quantum Circuit Simulator): The value you plug in for the second qubit determines what type of entanglement you get out. Setting the second qubit to 0, you will get entangled qubits which always agree when they are observed. Setting it to 1, you will get entangled qubits which always disagree when they are observed. The first qubit controls whether the phase of each state matches or mismatches. Note that these are the four Bell States (Wikipedia: Bell State). $\begin{array}{c|c|c} \text{Input} & \text{Output In Ket Notation} & \text{Output As Vector} \\ \hline 00 & \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) & [1/\sqrt{2},0,0,1/\sqrt{2}] \\ 01 & \frac{1}{\sqrt{2}}(|01\rangle+|10\rangle) & [0,1/\sqrt{2},1/\sqrt{2},0] \\ 10 & \frac{1}{\sqrt{2}}(|00\rangle-|11\rangle) & [1/\sqrt{2},0,0,-1/\sqrt{2}] \\ 11 & \frac{1}{\sqrt{2}}(|01\rangle-|10\rangle) & [0,1/\sqrt{2},-1/\sqrt{2},0]\\ \end{array}$

In a quantum circuit, you can’t just apply a gate to an individual quabit at a time though. You have to make the matrix of your gate such that it applies to all qubits, applying the “identity” matrix to the qubits you don’t want to be affected by the gate.

So, how do we make a matrix that applies the Hadamard gate to qubit 1, and identity to qubit 2? You use the kronecker product!

Since we want to apply the Hadamard matrix to qubit 1 and identity to qubit 2, we are going to calculate $H \otimes I$ (If we wanted to apply the Hadamard gate to qubit 2 and identity to qubit 1 we would calculate $I \otimes H$ instead). $H \otimes I = 1/\sqrt{2}* \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} = 1/\sqrt{2}* \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ \end{bmatrix}$

If you notice, the result of the kronecker product is just every value in the left matrix, multiplied by every value in the right matrix. Basically, the result is a 2×2 grid of identity matrices, where each of those identity matrices is multiplied by the corresponding value from the same cell in the left matrix. Since the left matrix has a 1 in all cells except the lower right, the same is true of the result… it’s a positive identity matrix in each cell, except the lower right one, which is a negative identity matrix. Hopefully that makes sense, it’s a lot easier than it sounds…

The second gate in the quantum circuit is the CNOT gate. We can actually multiply the gate we just made by the CNOT gate to represent the full quantum circuit as a single matrix. This uses regular matrix multiplication. $1/\sqrt{2}* \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ \end{bmatrix} * \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} = 1/\sqrt{2}* \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2} & 0 & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 1/\sqrt{2} & 0 & 0 & -1/\sqrt{2} \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ \end{bmatrix}$

Lets plug some values into the circuit and see what comes out!

Lets start by plugging in 00. Our input qubits are $[1, 0]$ and $[1, 0]$. The kronecker product of those two qubit vectors is $[1, 0, 0, 0]$. Now let’s multiply that vector by the matrix of our quantum circuit. $\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} * \begin{bmatrix} 1/\sqrt{2} & 0 & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 1/\sqrt{2} & 0 & 0 & -1/\sqrt{2} \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2} & 0 & 0 & 1/\sqrt{2} \end{bmatrix}$

Comparing this to the table showing how our input maps to output, we got the right answer! If you plugged in the other values listed, you would get the rest of the bell state entanglements.

## Unentangling Qubits

All gates are reversible, so all circuits are reversible. To get the reverse circuit for a given matrix, you just get the inverse of the matrix. Since quantum gates (and circuits) are unitary matrices, taking the inverse of one of these matrices just means taking the conjugate transpose of the matrix. In other words, you take the transpose of the matrix, and then just negate the imaginary component of any complex numbers. In this example, there are no imaginary numbers, so you just take the transpose of the matrix.

Since our circuit matrix is this: $\begin{bmatrix} 1/\sqrt{2} & 0 & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 1/\sqrt{2} & 0 & 0 & -1/\sqrt{2} \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ \end{bmatrix}$

That means that the inverse must be this, since this is the (conjugate) transpose. $\begin{bmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 0 & -1/\sqrt{2} \\ 1/\sqrt{2} & 0 & -1/\sqrt{2} & 0 \\ \end{bmatrix}$

Let’s try it out by putting the output from last section in and seeing what comes out the other side: $\begin{bmatrix} 1/\sqrt{2} & 0 & 0 & 1/\sqrt{2} \end{bmatrix} * \begin{bmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 0 & -1/\sqrt{2} \\ 1/\sqrt{2} & 0 & -1/\sqrt{2} & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}$

It worked! We got our original input back.

## Making Matrices for More Complex Circuits

We talked about how to make single qubit gates apply to specific qubits by using the kronecker product with identity to move the gate to the right location.

If you have 4 qubits and you want to apply some single qubit gate (say Hadamard) to the 3rd qubit, you would just do this to get the matrix: $I \otimes I \otimes H \otimes I$

What if you want to do complex things with multi qubit gates though? Like what if you want to do a controlled not gate where you want the 2nd qubit to flip it’s value if the 4th qubit was true?

The answer is actually pretty simple… you use swaps gates to flip the values of the qubits so that your qubit values line up with the inputs of the gate, then you do the gate, and finally undo all the swaps to get the qubit values back to the right positions.

We need to swap qubits so that the 4th qubit value is in the 1st qubit slot, and the 2nd qubit value is in the 2nd qubit slot. We can do that with the following swaps: Once we have done those swaps, we can do our controlled not, then do the swaps in reverse order to return the qubits to their original positions.

Here’s what the circuit looks like: You’ll see the simpler version in circuit diagrams, but at least now you’ll know how to make things that look like this: Below is the mathematical way that you would get the matrix representing this gate. $I$ is the identity matrix, $S$ is the swap gate, and $C$ is the controlled not gate. $M = \\ (I \otimes I \otimes S) * \\ (I \otimes S \otimes I) * \\ (S \otimes I \otimes I) * \\ (I \otimes S \otimes I) * \\ (C \otimes I \otimes I) * \\ (I \otimes S \otimes I) * \\ (S \otimes I \otimes I) * \\ (I \otimes S \otimes I) * \\ (I \otimes I \otimes S) \\$

## Code

Here is some simple C++ to show both the entanglement circuit and the more complicated controlled not gate we described.

#include <stdio.h>
#include <vector>
#include <complex>
#include <assert.h>

typedef std::complex<float> TComplex;

//=================================================================================
struct SComplexVector
{
public:
TComplex& Get (size_t i) { return v[i]; }

const TComplex& Get (size_t i) const { return v[i]; }

size_t Size() const
{
return v.size();
}

void Resize (size_t s)
{
v.resize(s);
}

void Print () const
{
printf("[");
for (size_t i = 0, c = v.size(); i < c; ++i)
{
if (i > 0)
printf(", ");
const TComplex& val = Get(i);
if (val.imag() == 0.0f)
{
if (val.real() == 0.0f)
printf("0");
else if (val.real() == 1.0f)
printf("1");
else
printf("%0.2f", val.real());
}
else
printf("%0.2f + %0.2fi", val.real(), val.imag());
}
printf("]n");
}

std::vector<TComplex> v;
};

//=================================================================================
struct SComplexMatrix
{
public:
TComplex& Get (size_t x, size_t y)
{
return v[y*Size() + x];
}

const TComplex& Get (size_t x, size_t y) const
{
return v[y*Size() + x];
}

// For an MxM matrix, this returns M
size_t Size () const
{
size_t ret = (size_t)sqrt(v.size());
assert(ret*ret == v.size());
return ret;
}

// For an MxM matrix, this sets M
void Resize(size_t s)
{
v.resize(s*s);
}

void Print() const
{
const size_t size = Size();

for (size_t y = 0; y < size; ++y)
{
printf("[");
for (size_t x = 0; x < size; ++x)
{
if (x > 0)
printf(", ");

const TComplex& val = Get(x, y);
if (val.imag() == 0.0f)
{
if (val.real() == 0.0f)
printf("0");
else if (val.real() == 1.0f)
printf("1");
else
printf("%0.2f", val.real());
}
else
printf("%0.2f + %0.2fi", val.real(), val.imag());
}
printf("]n");
}
}

std::vector<TComplex> v;
};

//=================================================================================
static const SComplexVector c_qubit0 = { { 1.0f, 0.0f } };  // false aka |0>
static const SComplexVector c_qubit1 = { { 0.0f, 1.0f } };  // true aka |1>

// 2x2 identity matrix
static const SComplexMatrix c_identity2x2 =
{
{
1.0f, 0.0f,
0.0f, 1.0f,
}
};

// Given the states |00>, |01>, |10>, |11>, swaps the |01> and |10> state
// If swapping the probabilities of two qubits, it won't affect the probabilities
// of them both being on or off since those add together.  It will swap the odds of
// only one of them being on.
static const SComplexMatrix c_swapGate =
{
{
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
}
};

// Controlled not gate
// If the first qubit is true, flips the value of the second qubit
static const SComplexMatrix c_controlledNotGate =
{
{
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f,
0.0f, 0.0f, 1.0f, 0.0f
}
};

// Takes a pure |0> or |1> state and makes a 50/50 superposition between |0> and |1>.
// Put a 50/50 superposition through and get the pure |0> or |1> back.
// Encodes the origional value in the phase information as either matching or
// mismatching phase.
{
{
1.0f / std::sqrt(2.0f), 1.0f / std::sqrt(2.0f),
1.0f / std::sqrt(2.0f), 1.0f / -std::sqrt(2.0f)
}
};

//=================================================================================
void WaitForEnter ()
{
printf("nPress Enter to quit");
fflush(stdin);
getchar();
}

//=================================================================================
SComplexVector KroneckerProduct (const SComplexVector& a, const SComplexVector& b)
{
const size_t aSize = a.Size();
const size_t bSize = b.Size();

SComplexVector ret;
ret.Resize(aSize*bSize);

for (size_t i = 0, ic = aSize; i < ic; ++i)
{
for (size_t j = 0, jc = bSize; j < jc; ++j)
{
size_t n = i * bSize + j;
ret.Get(n) = a.Get(i)*b.Get(j);
}
}
return ret;
}

//=================================================================================
SComplexMatrix KroneckerProduct (const SComplexMatrix& a, const SComplexMatrix& b)
{
const size_t aSize = a.Size();
const size_t bSize = b.Size();

SComplexMatrix ret;
ret.Resize(aSize*bSize);

for (size_t ax = 0; ax < aSize; ++ax)
{
for (size_t ay = 0; ay < aSize; ++ay)
{
const TComplex& aValue = a.Get(ax, ay);

for (size_t bx = 0; bx < bSize; ++bx)
{
for (size_t by = 0; by < bSize; ++by)
{
const TComplex& bValue = b.Get(bx, by);

size_t nx = ax*bSize + bx;
size_t ny = ay*bSize + by;

ret.Get(nx,ny) = aValue * bValue;
}
}
}
}

return ret;
}

//=================================================================================
SComplexMatrix operator* (const SComplexMatrix& a, const SComplexMatrix& b)
{
assert(a.Size() == b.Size());
const size_t size = a.Size();

SComplexMatrix ret;
ret.Resize(size);

for (size_t nx = 0; nx < size; ++nx)
{
for (size_t ny = 0; ny < size; ++ny)
{
TComplex& val = ret.Get(nx, ny);
val = 0.0f;
for (size_t i = 0; i < size; ++i)
val += a.Get(i, ny) * b.Get(nx, i);
}
}

return ret;
}

//=================================================================================
SComplexVector operator* (const SComplexVector& a, const SComplexMatrix& b)
{
assert(a.Size() == b.Size());
const size_t size = a.Size();

SComplexVector ret;
ret.Resize(size);

for (size_t i = 0; i < size; ++i)
{
TComplex& val = ret.Get(i);
val = 0;
for (size_t j = 0; j < size; ++j)
val += a.Get(j) * b.Get(i, j);
}

return ret;
}

//=================================================================================
int main (int argc, char **argv) {

// 2 qubit entanglement circuit demo
{
// make the circuit
const SComplexMatrix H1 = KroneckerProduct(c_hadamardGate, c_identity2x2);
const SComplexMatrix circuit = H1 * c_controlledNotGate;

// display the circuit
printf("Entanglement circuit:n");
circuit.Print();

// permute the inputs and see what comes out when we pass them through the circuit!
SComplexVector input = KroneckerProduct(c_qubit0, c_qubit0);
SComplexVector output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(c_qubit0, c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(c_qubit1, c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(c_qubit1, c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();
}

// 4 qubit demo: flip the second qubit if the fourth qubit is true
{
// make the circuit
const SComplexMatrix cnot4qubit = KroneckerProduct(KroneckerProduct(c_controlledNotGate, c_identity2x2), c_identity2x2);
const SComplexMatrix swap12 = KroneckerProduct(KroneckerProduct(c_swapGate, c_identity2x2), c_identity2x2);
const SComplexMatrix swap23 = KroneckerProduct(KroneckerProduct(c_identity2x2, c_swapGate), c_identity2x2);
const SComplexMatrix swap34 = KroneckerProduct(KroneckerProduct(c_identity2x2, c_identity2x2), c_swapGate);
const SComplexMatrix circuit =
swap34 *
swap23 *
swap12 *
swap23 *
cnot4qubit *
swap23 *
swap12 *
swap23 *
swap34;

// display the circuit
printf("nFlip 2nd qubit if 4th qubit true circuit:n");
circuit.Print();

// permute the inputs and see what comes out when we pass them through the circuit!
SComplexVector input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit0), c_qubit0), c_qubit0);
SComplexVector output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit0), c_qubit0), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit0), c_qubit1), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit0), c_qubit1), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit1), c_qubit0), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit1), c_qubit0), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit1), c_qubit1), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit0, c_qubit1), c_qubit1), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit0), c_qubit0), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit0), c_qubit0), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit0), c_qubit1), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit0), c_qubit1), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit1), c_qubit0), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit1), c_qubit0), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit1), c_qubit1), c_qubit0);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();

input = KroneckerProduct(KroneckerProduct(KroneckerProduct(c_qubit1, c_qubit1), c_qubit1), c_qubit1);
output = input * circuit;
printf("ninput:");
input.Print();
printf("output:");
output.Print();
}

WaitForEnter();

return 0;
}


Here is the first part of the output, which shows the results of the entangling circuit. You can see that the input gave the expected output Bell states: Below is the second part of the output, which is the circuit that flips the 2nd qubit if the 4th qubit is true.

Each entry in the input and output vectors is the amplitude (probability) that the state it represents is true. The states start at the left with 0000, then 0001, then 0010 and continue until the very right most value which represents 1111.

If you look at the second input/output pair the input is [0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0], which means it has a 100% chance of the 4 qubits being 0001 (aka the qubits represent the number 1). The output vector is [0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0], which means that it has a 100% chance of being 0101 (aka the qubits represent the number 5). Since the input did have the 4th bit set (4th from the left), it flipped the 2nd bit. So, you can see that it worked!

If you check all the input/output pairs, you’ll see that they all follow this rule.

We used only whole, real numbers, no using fractional probabilities, or imaginary amplitudes. What it does in those situations is a little bit harder to get intuition for, but rest assured that it does “the right thing” in those situations as well. ## Next Up

Now that we have the basics of quantum computing down pretty well, it’s time to analyze a couple quantum algorithms to see how they work!