# New quantum computer

Then’s a one-judgment summary of what amount a computer is an amount computer is a type of computer that uses amount mechanics so that it can perform certain kinds of calculations more efficiently than a regular computer can.

There’s a lot to unpack in this judgment, so let me walk you through what it’s exactly using a simple illustration.

To explain what an amount computer is, I’ll need to first explain a little about regular(non-quantum) computers.

Now, what about amount computers?

An amount computer doesn’t use bits to store information. Rather, it uses a commodity called cubits.

Each cubit can’t only be set to 1 or 0, but it can also be set to 1 and 0. But what does that mean exactly?

This is going to be a kindly artificial illustration. But it’s still going to help understand how quantum computers work.

A simple illustration of understanding how quantum computers work

Now, suppose you’re running a trip agency, and you need to move a group of people from one position to another.

Suppose that you have reserved 2 hacks for this purpose, and you want to figure out who gets into which hack.

Also, suppose then that you’re given information about whose musketeers with whom, and who’s adversaries with whom.

Then, let’s say that

Alice and Becky are musketeers

Becky and Chris are adversaries

And suppose that your thing then’s to divide this group of 3 people into two hacks to achieve the following two objects

Maximize the number of friend dryads that partake in the same auto

Minimize the number of adversary dryads that partake in the same auto

Okay, so this is the introductory premise of this problem. Let’s first suppose how we’d break this problem using a regular computer.

Working this problem with a regular computer

To break this problem with a regular, non-quantum computer, you’ll need first to figure out how to store the applicable information with bits.

Let’s mark the two hacks Taxi# 1 and hack# 0.

Also, you can represent who gets into which auto with 3 bits.

For illustration, we can set the three bits to 0, 0, and 1 to represent

Alice gets into hack# 0

Becky gets into hack# 0

Chris gets into hack# 1

Since there are two choices for each person, there are 2 × 2 × 2 = 8 ways to divide this group of people into two buses.

Then’s a list of all possible configurations

A| B| C

0| 0| 0

0| 0| 1

0| 1| 0

0| 1| 1

1| 0| 0

1| 0| 1

1| 1| 0

1| 1| 1

Computing the score for each configuration now, using a regular computer, how would we determine which configuration is the stylish result?

To do this, let’s define how we can cipher the score for each configuration. This score will represent the extent to which each result achieves the two objects I mentioned before

Maximize the number of friend dryads that partake in the same auto

Minimize the number of adversary dryads that partake in the same auto

Let me explain simply the score of a given configuration) = (# friend dryads participating in the same auto)-(# adversary dryads participating in the same auto)

For illustration, suppose that Alice, Becky, and Chris all get into hack# 1.

In this case, there’s only one friend brace participating in the same auto — Alice and Becky.

Still, two adversary dryads are participating in the same auto — Alice and Chris, and Becky and Chris.

Working the problem

With all of this setup, we can eventually go about working on this problem.

With a regular computer, to find the stylish configuration, you’ll need to go through all configurations to see which one

Achieves the loftiest score.

So, you can suppose about constructing a table like this

A| B| C| Score

0| 0| 0|-1

0| 0| 1| 1<- bone

of the stylish results

0| 1| 0|-1

0| 1| 1|-1

1| 0| 0|-1

1| 0| 1|-1

1| 1| 0| 1<- the other stylish result

1| 1| 1|-1

As you can see, there are two correct results then — 001 and 110, both achieving a score of 1.

This problem is fairly simple. It snappily becomes too delicate to break with a regular computer as we increase the number of people in this problem.

We saw that with 3 people, we need to go through 8 possible configurations.

What if there are 4 people? In that case, we ’ll need to go through 2 * 2 * 2 * 2 = 16 configurations.

With n people, we’ll need to go through (2 to the power of n) configurations to find the stylish result.

So, if there are 100 people, we’ll need to go through

2 ¹⁰⁰

= 10 ³⁰ = one million configurations

This is simply insolvable to break with a regular computer.

working this problem with an amount of computer

How would we go about working this problem with several computers?

To suppose about that, let’s go back to the case of dividing 3 people into two hacks.

As we saw before, there were 8 possible results of this problem

A| B| C

0| 0| 0

0| 0| 1

0| 1| 0

0| 1| 1

1| 0| 0

1| 0| 1

1| 1| 0

1| 1| 1

With a regular computer, using 3 bits, we were suitable to represent only one of these results at a time for illustration, 001.

still, with several computers, using 3 qubits, we can represent all 8 of these results at the same time.

There are debates as to what it means exactly, but then’s the way I suppose about it.

When you set it to both 0 and 1, it’s kind of like creating two resemblant worlds. ( Yes, it’s strange, but just follow on then.)

In one of those resemblant worlds, the qubit is set to 0. In the other one, it’s set to 1.

Now, what if you set the alternate qubit to 0 and 1, too? also, it’s kind of like creating 4 resemblant worlds.

the two qubits are set to 00. In the alternate bone, they’re 01. In the third one, they’re 10. In the fourth bone, they’re 11.

also, if you set all three qubits to both 0 and 1, you’d be creating 8 resemblant worlds — 000, 001, 010, 011, 100, 101, 110, and 111.

This is a strange way to suppose, but it’s one of the correct ways to interpret how the qubits bear in the real world.

Now, when you apply some kind of calculation on these three qubits, you’re applying the same calculation in all of those 8 resemblant worlds at the same time.

So, rather than going through each of those implicit results successionally, we can cipher the scores of all results at the same time.

With this particular illustration, in the proposition, your amount computer would be suitable to find one of the stylish results in many milliseconds. Again, that’s 001 or 110 as we saw before

A| B| C| Score

0| 0| 0|-1

0| 0| 1| 1<- bone

of the stylish results

0| 1| 0|-1

0| 1| 1|-1

1| 0| 0|-1

1| 0| 1|-1

1| 1| 0| 1<- the other stylish result

1| 1| 1|-1

In reality, to break this problem, you would need to give your amount computer two effects

All implicit results are represented with qubits

A function that turns each implicit result into a score. In this case, this is the function that counts the number of friend dyads and adversary dyads participating in the same auto.

Given these two effects, your amount computer will spear out one of the stylish results in many milliseconds. In this case, that’s 001 or 110 with a score of 1.

Now, in the proposition, an amount computer can find one of the stylish results every time it runs.

still, in reality, there are crimes when running several computers. So, rather than changing the stylish result, it might find the alternate-stylish result, the third-stylish result, and so on.

These crimes come more prominent as the problem becomes more and more complex.

So, in practice, you’ll presumably want to run the same operation on an amount of computers dozens of times or hundreds of times. Also, pick the stylish result out of the numerous results you get.

How several computer scales Indeed with the crimes I mentioned, the number of computers doesn’t have the same scaling issue a regular computer suffers from.

When there are 3 people we need to divide into two buses, the number of operations we need to perform on an amount computer is 1. This is because an amount computer computes the score of all configurations at the same time.

When there are 4 people, the number of operations is still 1.

When there are 100 people, the number of operations is still 1. With a single operation, an amount computer computes the scores of all 2 ¹⁰⁰

= 10 ³⁰ = one million configurations at the same time.

As I mentioned before, in practice, it’s presumably stylish to run your amount computer dozens of times or hundreds of times and pick the stylish result out of the numerous results you get.

still, it’s much better than running the same problem on a regular computer and having to repeat the same type of calculation one million times.

belting up

D- Wave lately launched a pall terrain for interacting with several computers.

still, it’s presumably the easiest way to do so, If you’re an investor and would like actually to try using the amount computer.