<p>First of all, it's important to note that Google's Sycamore has not (yet) processed any quantum algorithm, but instead has just proved that the quantum processor is capable of maintain quantum effects on all 53 qubits during operations. This is important because there's still <a href="https://en.wikipedia.org/wiki/Adiabatic_quantum_computation#D-Wave_quantum_processors" rel="nofollow noreferrer">controversy on if all quantum computers built to date exhibit real quantum advantages</a>.</p>

<p>In order for this processor to become a "practical" quantum computer, also <a href="https://www.youtube.com/watch?v=nycwc-wXuuw" rel="nofollow noreferrer">measurement physical errors must be lowered</a>.</p>

<p>Also, to make computations that can actually break current cryptography algorithms, the number of qubits must probably be raised, to around 1000 or more, all depending on how low the error rate will be.</p>

<p>And answering the second question, no, this result is in line with prospections on quantum computing, so the (<a href="https://medium.com/the-quantum-resistant-ledger/quantum-supremacy-and-the-case-for-quantum-security-today-in-blockchain-390fe55daab5" rel="nofollow noreferrer">optimistic</a>) <a href="https://arxiv.org/pdf/1710.10377.pdf" rel="nofollow noreferrer">estimations of around as early as 2027-2040 to break ECDS still maintain</a>. But all depends on current trends that nobody knows if can be maintained or not, as this is all based on cutting-edge engineering and mathematical advances.</p>

<p>Also note the link indicated by @ugam-kamat regarding <a href="https://crypto.stackexchange.com/questions/9480/assuming-a-1024qb-quantum-computer-how-long-to-brute-force-1024bit-rsa-256bit">practical mathematical implementation of cryptography breaking algorithms</a> that also points to a minimum of around a thousand usable qubits to break nowadays signatures. This would also raise the real number of needed qubits by some factor n>1 (n>>1?), as the previous noted increment was due to measurements errors which would still apply as the number of qubits raises.</p>

CC BY-SA 4.0
2019-11-03T22:21:36.023Z
2019-11-25T12:19:41.393Z
2019-11-25T12:19:41.393Z
circulosmeos
<p>Today, Google's Sycamore quantum computer has been <a href="https://www.nature.com/articles/s41586-019-1666-5" rel="nofollow noreferrer">confirmed</a> to process 53 qubits in parallel. </p>

<p>In practical terms, for non-quantum experts, how relevant is this breakthrough in terms of secp256k1 and SHA256? </p>

<p>Does this new development come earlier than expected or is it within Bitcoin's predicted quantum safety margin, by which we'd have a couple of decades before quantum computing threatens the cryptocurrency's "mathematical shields"?</p>

<p>Reference: <a href="https://www.nature.com/articles/d41586-019-03173-4" rel="nofollow noreferrer">Quantum computing takes flight</a></p>

CC BY-SA 4.0
2019-10-23T13:56:32.537Z
Related: [Assuming a 1024 qubit quantum computer, how long to brute force RSA, AES, ECC and SHA512?](https://crypto.stackexchange.com/questions/9480/assuming-a-1024qb-quantum-computer-how-long-to-brute-force-1024bit-rsa-256bit)
Ugam Kamat
<quantum-computing>
Google's Sycamore processor has parallel-processed 2^53 states. Is this a threat to Bitcoin?
Jose Fonseca
# Google's Sycamore processor has parallel-processed 2^53 states. Is this a threat to Bitcoin?

(Asked by: username_0 on 2019-10-23)

Today, Google's Sycamore quantum computer has been [confirmed](https://www.nature.com/articles/s41586-019-1666-5) to process 53 qubits in parallel.

In practical terms, for non-quantum experts, how relevant is this breakthrough in terms of secp256k1 and SHA256?

Does this new development come earlier than expected or is it within Bitcoin's predicted quantum safety margin, by which we'd have a couple of decades before quantum computing threatens the cryptocurrency's "mathematical shields"?

Reference: [Quantum computing takes flight](https://www.nature.com/articles/d41586-019-03173-4)

**Tags:** quantum-computing

> **Comments:**
> **username_1**: Related: [Assuming a 1024 qubit quantum computer, how long to brute force RSA, AES, ECC and SHA512?](https://crypto.stackexchange.com/questions/9480/assuming-a-1024qb-quantum-computer-how-long-to-brute-force-1024bit-rsa-256bit)  

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## 1 answers

### Answer by username_2 on 2019-11-03. Score: 1

First of all, it's important to note that Google's Sycamore has not (yet) processed any quantum algorithm, but instead has just proved that the quantum processor is capable of maintain quantum effects on all 53 qubits during operations. This is important because there's still [controversy on if all quantum computers built to date exhibit real quantum advantages](https://en.wikipedia.org/wiki/Adiabatic_quantum_computation#D-Wave_quantum_processors).

In order for this processor to become a "practical" quantum computer, also [measurement physical errors must be lowered](https://www.youtube.com/watch?v=nycwc-wXuuw).

Also, to make computations that can actually break current cryptography algorithms, the number of qubits must probably be raised, to around 1000 or more, all depending on how low the error rate will be.

And answering the second question, no, this result is in line with prospections on quantum computing, so the ([optimistic](https://medium.com/the-quantum-resistant-ledger/quantum-supremacy-and-the-case-for-quantum-security-today-in-blockchain-390fe55daab5)) [estimations of around as early as 2027-2040 to break ECDS still maintain](https://arxiv.org/pdf/1710.10377.pdf). But all depends on current trends that nobody knows if can be maintained or not, as this is all based on cutting-edge engineering and mathematical advances.

Also note the link indicated by @ugam-kamat regarding [practical mathematical implementation of cryptography breaking algorithms](https://crypto.stackexchange.com/questions/9480/assuming-a-1024qb-quantum-computer-how-long-to-brute-force-1024bit-rsa-256bit) that also points to a minimum of around a thousand usable qubits to break nowadays signatures. This would also raise the real number of needed qubits by some factor n>1 (n>>1?), as the previous noted increment was due to measurements errors which would still apply as the number of qubits raises.


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