Fix acrolinx

Author: SoniaLopezBravo

learn-pr/quantum/estimate-resources-quantum-algorithms/includes/2-resource-estimation-in-quantum-computing.md

@@ -6,13 +6,13 @@ In this unit, you'll learn why the estimation of these resources is important in
 
 Quantum computers have the potential of *quantum advantage* to solve some scientifically and commercially valuable problems. For example, one of the top applications for quantum computing is to break encryption. The RSA encryption algorithm is based on the difficulty of factoring large numbers. A quantum computer can factor large numbers exponentially faster than a classical computer. So, the question is, how long does it take to run a quantum algorithm that breaks encryption on a real quantum computer? Or in other words, how long are my passwords secure?
 
-The truth is the required resources needed to run a quantum algorithm on a future scaled quantum computer vary for different computational scenarios based on the type of qubits, the error correction scheme, and other architecture design choices. The Azure Quantum Resource Estimator is a tool that helps you estimate the resources needed to run a quantum algorithm. 
+The truth is the required resources needed to run a quantum algorithm on a future scaled quantum computer vary for different computational scenarios based on the type of qubits, the error correction scheme, and other architecture design choices. The Azure Quantum Resource Estimator is a tool that helps you estimate the resources needed to run a quantum algorithm for a future scaled quantum computer. For example, the Azure Quantum Resource Estimator can estimate the resources needed to break a particular encryption algorithm. 
 
-The Azure Quantum Resource Estimator is a tool that helps you estimate the resources needed to run a quantum algorithm for a future scaled quantum computer. For example, the Azure Quantum Resource Estimator can estimate the resources needed to break a particular encryption algorithm. The diagram shows the estimated runtime and number of qubits needed to break different encryption algorithms for different qubit types. The diagram shows the following:
+The diagram shows the estimated runtime and number of qubits needed to break different encryption algorithms for different qubit types. The diagram shows the following:
 
-- Classical encryption algorithm, which are RSA (blue), Elliptic Curve (green), and Advanced Encryption Standard (red).
+- Classical encryption algorithms, which are RSA (blue), Elliptic Curve (green), and Advanced Encryption Standard (red).
 - Key strength, which is set to highest.
-- Qubit type, which are topological (circle) and superconducting (triangle).
+- Qubit types, which are topological (circle) and superconducting (triangle).
 - Qubit error rate, which is set to reasonable.
 
 

learn-pr/quantum/estimate-resources-quantum-algorithms/includes/3-azure-quantum-resource-estimator.md

@@ -30,7 +30,7 @@ For more information, see [Qubit parameters of the Azure Quantum Resource Estima
 
 ### Choose the QEC scheme
 
-Quantum error correction (QEC) is crucial for any quantum-computing platform to achieve truly scalable quantum computation. The set of operations a quantum computing platform permits is limited by physical constraints and might not match the operations prescribed in the algorithm. Even if the operations that the quantum computer offers match the operations in the algorithm, the accuracy to which the quantum computer can perform each operation is likely to be limited.
+Quantum error correction (QEC) is crucial for any quantum-computing platform to achieve truly scalable quantum computation. The set of operations a quantum computing platform permit is limited by physical constraints and might not match the operations prescribed in the algorithm. Even if the operations that the quantum computer offers match the operations in the algorithm, the accuracy to which the quantum computer can perform each operation is likely to be limited.
 
 The Azure Quantum Resource Estimator provides three predefined QEC schemes: two *surface code* protocols for gate-based and Majorana physical instruction sets, and the *Floquet code* protocol, which can be used only with a Majorana physical instruction set.
 
@@ -43,7 +43,9 @@ For more information, see [Quantum error correction schemes in the Azure Quantum
 
 ### Choose the error budget
 
-The total error budget sets the overall allowed error for the algorithm. The allowed error is the number of times the algorithm is allowed to fail. The value of the error budget must be between 0 and 1, and the default value is 0.001. The default value corresponds to 0.1 percent, and means that the algorithm is allowed to fail once in 1,000 executions. This parameter is highly specific to the application. For example, if you're running Shor’s algorithm for factoring integers, a large value for the error budget can be tolerated because you can check that the output is indeed the prime factors of the input. On the other hand, a smaller error budget might be needed for an algorithm solving a problem with a solution that can't be efficiently verified.
+The total error budget sets the overall allowed error for the algorithm. The allowed error is the number of times the algorithm is allowed to fail. The value of the error budget must be between 0 and 1, and the default value is 0.001. The default value corresponds to 0.1 percent, and means that the algorithm is allowed to fail once in 1,000 executions. 
+
+The error budget is highly specific to the application. For example, if you're running Shor’s algorithm for factoring integers, a large value for the error budget can be tolerated because you can check that the output is indeed the prime factors of the input. On the other hand, a smaller error budget might be needed for an algorithm solving a problem with a solution that can't be efficiently verified.
 
 For more information, see [Error budget in the Azure Quantum Resource Estimator](/azure/quantum/overview-resources-estimator#error-budget).
 

learn-pr/quantum/estimate-resources-quantum-algorithms/includes/4-get-started-with-azure-quantum-resource-estimator.md

@@ -55,7 +55,7 @@ To get started with the Resource Estimator, you estimate the required resources
 
 ## Estimate the quantum algorithm
 
-Now, run the Resource Estimator to estimate the physical resources for the `RandomBit` operatio. If you don't specify anything, the Resource Estimator uses the default assumptions, that is the `qubit_gate_ns_e3` qubit model, the `surface_code` error correction code, and 0.001 error budget.
+Now, run the Resource Estimator to estimate the physical resources for the `RandomBit` operation. If you don't specify anything, the Resource Estimator uses the default assumptions, that is the `qubit_gate_ns_e3` qubit model, the `surface_code` error correction code, and 0.001 error budget.
 
 1. Add a new cell and copy the following code:
 

learn-pr/quantum/explore-entanglement/includes/2-what-is-entanglement.md

@@ -21,12 +21,12 @@ $$\ket{\phi}=\frac1{\sqrt2}(\ket{0_A 0_B}+ \ket{1_A 1_B})$$
 > [!NOTE]
 > In Dirac notation, $\ket{0_A 0_B}= |0\rangle_\text{A} |0\rangle_\text{B}$. The first position corresponds to the first qubit, and the second position corresponds to the second qubit.
 
-The global system $\ket{\phi}$ is in a superposition of the states $|00\rangle$ and $|11\rangle$. If you measure both qubits, only two outcomes are possible: $\ket{00}$ and $\ket{11}$, and each has the same probability of $\frac{1}{2}$.
+The global system $\ket{\phi}$ is in a superposition of the states $\ket{00}$ and $\ket{11}$. If you measure both qubits, only two outcomes are possible: $\ket{00}$ and $\ket{11}$, and each has the same probability of $\frac{1}{2}$.
 
 But what is the individual state of qubit $A$? And of qubit $B$? If you try to describe the state of qubit $A$ without considering the state of qubit $B$, you would fail. Subsystems $A$ and $B$ are **entangled**, which means that they are correlated, and cannot be described independently.
 
 > [!TIP]
-> If you're familiar with algebra and Dirac notation, a good exercise is to try to modify the $\ket{\phi}$ state to get something like the state of qubit $A$ times the state of qubit $B$. If you try to expand the parenthesis, get the common factor, etc, you'll see that it's not possible.
+> If you're familiar with algebra and Dirac notation, a good exercise is to try to modify the $\ket{\phi}$ state to get something like the state of qubit $A$ times the state of qubit $B$. You'll see that it's impossible to do so.
 
 The quantum state $\ket{\phi}$ is a special entangled state, called **Bell state**. There are four Bell states. 
 
@@ -37,21 +37,21 @@ $$\ket{\psi^{-}}=\frac1{\sqrt2}\ket{01} - \frac1{\sqrt2}\ket{10} $$
 
 ## Using entanglement as a resource
 
-At this point, you might be wondering what's the big deal about entanglement?
+At this point, you might be wondering: what's the big deal about entanglement?
 
 When two particles are entangled, subsystems are correlated and cannot be described independently. But here's the interesting part: **the measurement outcomes are also correlated.** That is, whatever operation happens to the state of one qubit in an entangled pair, also affects to the state of the other qubit.
 
 For example, consider the $\ket{\phi^{+}}$ state,
 
 $$\ket{\phi^{+}}=\frac1{\sqrt2}\ket{00} + \frac1{\sqrt2}\ket{11}$$
 
-If you measure both qubits, you get either $|00\rangle$ or $|11\rangle$ with equal probability. There's zero probability of obtaining the states $|01\rangle$ and $|10\rangle$. 
+If you measure both qubits, you get either $\ket{00}$ or $\ket{11}$ with equal probability. There's zero probability of obtaining the states $\ket{01}$ and $\ket{10}$. 
 
 But what happens if you measure only one qubit? 
 
-If you measure only the qubit  $A$  and you get the $|0\rangle$ state, this means that the global system collapses to the state $\ket{00}$. This is the only possible outcome, since the probability of measuring $|01\rangle$ is zero.
+If you measure only the qubit  $A$  and you get the $\ket{0}$ state, this means that the global system collapses to the state $\ket{00}$. This is the only possible outcome, since the probability of measuring $\ket{01}$ is zero.
 
-Therefore, without measuring the qubit $B$ you can be positive that the second qubit is also in $|0\rangle$ state. The measurement outcomes are correlated because the qubits are entangled.
+Therefore, without measuring the qubit $B$ you can be positive that the second qubit is also in $\ket{0}$ state. The measurement outcomes are correlated because the qubits are entangled.
 
 Entanglement can exist between two particles even if they are separated by large distances. This correlation is stronger than any classical correlation, and it is a key resource for quantum information processing tasks such as quantum teleportation, quantum cryptography, and quantum computing.