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Mark Spinelli
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What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices of the test graph, and $\pi_i(G)$ being another adjacency matrix having the $i$th permutation applied to $G$, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary wayinterpretation of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary way of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, $G$ being an adjacency matrix on $N$ vertices of the test graph, and $\pi_i(G)$ being another adjacency matrix having the $i$th permutation applied to $G$, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary interpretation of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

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Mark Spinelli
  • 14.4k
  • 2
  • 24
  • 78

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary way of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary way of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {N!}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary way of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.

Source Link
Mark Spinelli
  • 14.4k
  • 2
  • 24
  • 78

What do you mean that we cannot "represent an input to this oracle as a bitstring"?

For example we could have the basis states in our Hilbert space be the adjacency matrices over $N$ vertices, with $G$ being one of these states, while $\pi_i\in S_N$ being permutations of these vertices.

I claim it's easy to prepare the state:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|i\rangle,\tag 1$$

for example where each $i$ is written as a number in the factoradic number system.

With $\pi_i\in S_N$, and $G$ being an adjacency matrix on $N$ vertices, I then also claim that it's easy to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|G\rangle|\pi_i(G)\rangle,\tag 2$$

by applying the $i$th permutation to the vertices on the test graph $G$. If $G$ is given as an adjacency matrix on $N$ vertices in, say, some canonical form, and $\pi_i$ is a permutation, then we can easily come up with classical code, and hence with a quantum circuit, to permute the $N$ vertices in the adjacency matrix to find another adjacency matrix.

The problem, though, is that we cannot then easily disentangle the first register from the second register to prepare:

$$\frac{1}{\sqrt {2^{N!}}}\sum_{i=1}^{N!}|\pi_i(G)\rangle,\tag 3$$

because we need to uncompute the garbage that's picked up along the way, while computing $\pi_i(G)$. For, if we had such a state, we could solve graph isomorphism in quantum polynomial time.


Note I'm using "easy" as synonymous with "polynomially", and not necessarily as meaning easy in the plain and ordinary way of effortlessly or uncomplicated; it might indeed be actually be challenging to engineer the actual snippet of code to do the permutations, and to then convert the code into a quantum circuit.