Concurrency and Synchronization

Overview

In our study of modern operating systems, we have thus far treated processes as independent entities. We now shift our focus to a more complex and realistic scenario: systems of cooperating processes that execute concurrently and may share logical address spaces, data, or other system resources. The concurrent execution of such processes, a cornerstone of multiprogramming and multi-core architectures, introduces a fundamental challenge—ensuring the orderly execution of cooperating processes to maintain data consistency. Without principled mechanisms for control, the outcome of execution can become non-deterministic and erroneous, depending on the particular order in which access to shared data takes place. This phenomenon, known as a race condition, is a critical source of bugs in concurrent systems.

This chapter is therefore dedicated to the systematic study of process synchronization and coordination. We shall investigate the problems that arise from concurrency and explore the primary solutions developed to manage them. A mastery of these concepts is not merely an academic exercise; it is essential for success in the GATE examination, where questions frequently require a rigorous analysis of concurrent algorithms, the identification of potential race conditions, and the application of synchronization primitives. A deep, conceptual understanding of mutual exclusion, semaphores, and other synchronization tools is paramount for solving complex problems related to process management and system design.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | The Critical-Section Problem | Protocols for safe access to resources. |

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Learning Objectives

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We now turn our attention to The Critical-Section Problem...
## Part 1: The Critical-Section Problem

Introduction

In the design of modern operating systems, the management of concurrent processes or threads is a paramount concern. When multiple processes execute concurrently, and potentially in an interleaved fashion, they often need to share data and resources. This sharing, while essential for cooperation and efficiency, introduces significant complexities. The unmanaged, concurrent access to shared resources can lead to data inconsistency and system instability.

The central challenge in this domain is to control this cooperation, ensuring that the integrity of shared data is maintained. We must devise protocols that processes can use to coordinate their actions. The part of a program where a shared resource is accessed is known as the critical section. The critical-section problem, therefore, is the problem of designing a protocol that allows cooperating processes to access shared data in a controlled and predictable manner, preventing the system from entering an inconsistent state. A thorough understanding of this problem and its requirements is fundamental to the study of process synchronization.

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Key Concepts

#
## 1. Structure of a Concurrent Process

To address the critical-section problem, we can conceptualize the structure of a typical process as being divided into four sections. Consider a process $P_i$ that needs to access a shared resource. Its structure can be represented as follows:

```c
do {
// Section 1: Entry Section
// Code to request permission to enter the critical section

// Section 2: Critical Section
// Access shared resources

// Section 3: Exit Section
// Code executed after leaving the critical section

// Section 4: Remainder Section
// All other code that does not involve the shared resource

} while (TRUE);
```

The fundamental challenge is to design the entry and exit sections. The entry section acts as a gatekeeper, granting permission to enter the critical section, while the exit section manages the release of this access, allowing other waiting processes to proceed.

Entry Section


Critical Section


Exit Section

Remainder Section










(Loop continues)

#
## 2. Requirements for a Valid Solution

Any valid solution to the critical-section problem must satisfy three fundamental requirements. These are not merely suggestions but strict conditions that guarantee correct concurrent execution.

Mutual Exclusion: This is the most critical property. If a process $P_i$ is executing in its critical section, then no other processes can be executing in their critical sections. This ensures exclusive access to the shared resource, preventing interference and data corruption.

Progress: If no process is currently in its critical section and there exist some processes that wish to enter their critical sections, then the selection of the process that will enter its critical section next cannot be postponed indefinitely. This condition ensures that the system does not deadlock, where processes are waiting for an event that will never occur. The decision of who enters next must be made in a finite amount of time.

Bounded Waiting (or Bounded Bypass): There must exist a bound on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted. This property prevents starvation, a scenario where a process is indefinitely denied access to the shared resource while other processes are repeatedly granted access.

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#
## 3. Race Conditions

The undesirable situation that we are trying to prevent with a solution to the critical-section problem is known as a race condition.

A race condition occurs when the correctness of the program depends on the relative timing or interleaving of operations that are not atomic.

#
### Example 1: Low-Level (Instruction-Level) Race Condition

Consider a simple scenario where two threads, T1 and T2, concurrently execute the statement `counter++` on a shared integer variable `counter`, initially 5. On most machine architectures, this single high-level statement is compiled into multiple machine instructions:

`LOAD` the value of `counter` from memory into a register.

`INCREMENT` the value in the register.

`STORE` the new value from the register back into memory.

Let us trace a possible interleaving of these instructions:

| Step | Thread T1 Instruction | Thread T2 Instruction | Register T1 | Register T2 | `counter` Value |
| :--- | :-------------------------------- | :-------------------------------- | :---------- | :---------- | :-------------- |
| 1 | `LOAD counter` (value is 5) | | 5 | - | 5 |
| 2 | | `LOAD counter` (value is 5) | 5 | 5 | 5 |
| 3 | | `INCREMENT` register | 5 | 6 | 5 |
| 4 | `INCREMENT` register | | 6 | 6 | 5 |
| 5 | | `STORE` register (value is 6) | 6 | 6 | 6 |
| 6 | `STORE` register (value is 6) | | 6 | 6 | 6 |

Here, although `counter++` was executed twice, the final value of `counter` is 6, not the expected 7. This is a classic race condition. The final result depends entirely on which thread gets to store its updated value last.

#
### Example 2: Statement-Level Race Condition

GATE questions often simplify the problem by assuming each high-level statement is atomic, as seen in the 2024 PYQ. Even with this assumption, a race condition can still occur due to the interleaving of statements rather than machine instructions.

Worked Example:

Problem:
Two threads, T1 and T2, operate on shared integer variables $x$ and $y$. The initial state is $x=10$ and $y=20$. Each statement is executed atomically.

Thread T1:
```
S1: x = x + y;
S2: y = x - y;
```

Thread T2:
```
S3: y = y + x;
S4: x = y - x;
```

Determine the set of all possible final states $(x, y)$ after both threads complete execution.

Solution:

We must enumerate all possible valid interleavings of the statements {S1, S2} and {S3, S4}, keeping in mind that S1 must execute before S2, and S3 must execute before S4.

Case 1: Sequential Execution (T1 then T2)

Step 1: T1 executes completely.
Initial state: $(x=10, y=20)$

State after S1: $(x=30, y=20)$ State after S2 (T1 complete): $(x=30, y=10)$

Step 2: T2 executes completely.
State before T2: $(x=30, y=10)$

State after S3: $(x=30, y=40)$

Result 1: Final state is $(x=10, y=40)$.

Case 2: Sequential Execution (T2 then T1)

Step 1: T2 executes completely.
Initial state: $(x=10, y=20)$

State after S3: $(x=10, y=30)$ State after S4 (T2 complete): $(x=20, y=30)$

Step 2: T1 executes completely.
State before T1: $(x=20, y=30)$

State after S1: $(x=50, y=30)$

Result 2: Final state is $(x=50, y=20)$.

Case 3: Interleaved Execution (S1, S3, S2, S4)

Step 1: T1 executes S1.
Initial state: $(x=10, y=20)$

Current state: $(x=30, y=20)$

Step 2: T2 executes S3.

Current state: $(x=30, y=50)$

Step 3: T1 executes S2.

Current state: $(x=30, y=-20)$

Step 4: T2 executes S4.

Result 3: Final state is $(x=-50, y=-20)$.

Case 4: Interleaved Execution (S3, S1, S4, S2)

Step 1: T2 executes S3.
Initial state: $(x=10, y=20)$

Current state: $(x=10, y=30)$

Step 2: T1 executes S1.

Current state: $(x=40, y=30)$

Step 3: T2 executes S4.

Current state: $(x=-10, y=30)$

Step 4: T1 executes S2.

Result 4: Final state is $(x=-10, y=-40)$.

Other interleavings can be checked, but they will often lead to one of these results. For example, S1, S3, S4, S2 leads to $(x=-10, y=50)$. The set of possible outcomes demonstrates the race condition.

Answer: The set of all possible final states includes $\{(10, 40), (50, 20), (-50, -20), (-10, -40), (-10, 50), ...\}$. The key takeaway is the non-determinism.

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Problem-Solving Strategies

For GATE questions involving race conditions with atomic statements, the primary strategy is a systematic enumeration of all possible valid execution sequences.

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Two threads, P1 and P2, access a shared variable `balance`, initialized to 100. Each statement is atomic.
P1: `balance = balance - 50; balance = balance - 50;`
P2: `balance = balance * 2;`
Which of the following values is NOT a possible final value for `balance`?" options=["0", "50", "100", "150"] answer="150" hint="Enumerate all 4 unique, valid interleavings of the three statements." solution="Let S1: `balance = balance - 50`, S2: `balance = balance - 50` from P1. Let S3: `balance = balance * 2` from P2. Initial `balance` = 100.
Case 1 (S1, S2, S3):
100 -> S1 -> 50 -> S2 -> 0 -> S3 -> 0. Final: 0.
Case 2 (S3, S1, S2):
100 -> S3 -> 200 -> S1 -> 150 -> S2 -> 100. Final: 100.
Case 3 (S1, S3, S2):
100 -> S1 -> 50 -> S3 -> 100 -> S2 -> 50. Final: 50.
The possible final values are 0, 50, and 100. Therefore, 150 is not a possible final value."
:::

:::question type="NAT" question="Initially, shared integer variables `x = 2` and `y = 3`. Two processes, P1 and P2, execute the following atomic statements concurrently:
P1: `y = x + y; x = y - x;`
P2: `x = x * y; y = x - y;`
What is the maximum possible final value of `x`?" answer="15" hint="To maximize the final `x`, try to make the intermediate values used in its calculation as large as possible. Explore different interleavings." solution="Let P1's statements be S1, S2 and P2's be S3, S4.
Initial: (x=2, y=3)
Case 1 (P1 then P2):
S1: y = 2+3=5. State: (2,5)
S2: x = 5-2=3. State: (3,5)
S3: x = 3*5=15. State: (15,5)
S4: y = 15-5=10. Final: (15,10)
Case 2 (P2 then P1):
S3: x = 2*3=6. State: (6,3)
S4: y = 6-3=3. State: (6,3)
S1: y = 6+3=9. State: (6,9)
S2: x = 9-6=3. Final: (3,9)
Case 3 (S1, S3, S2, S4):
S1: y = 2+3=5. State: (2,5)
S3: x = 2*5=10. State: (10,5)
S2: x = 5-10=-5. State: (-5,5)
S4: y = -5-5=-10. Final: (-5,-10)
Case 4 (S3, S1, S4, S2):
S3: x = 2*3=6. State: (6,3)
S1: y = 6+3=9. State: (6,9)
S4: y = 6-9=-3. State: (6,-3)
S2: x = -3-6=-9. Final: (-9,-3)
Comparing the final values of `x` from these cases (15, 3, -5, -9), the maximum value is 15."
:::

:::question type="MSQ" question="Which of the following conditions are mandatory for a correct solution to the critical-section problem?" options=["Mutual Exclusion","Deadlock Freedom","Progress","Bounded Waiting"] answer="Mutual Exclusion,Progress,Bounded Waiting" hint="Recall the three canonical requirements for a valid synchronization protocol." solution="The three necessary and sufficient conditions for a correct solution to the critical-section problem are:

Mutual Exclusion: Ensures only one process can be in the critical section at a time.

Progress: Ensures that if the critical section is free and processes are waiting, a decision on who enters next is made without indefinite delay. This helps prevent deadlock.

Bounded Waiting: Ensures that no process waits forever to enter its critical section, thus preventing starvation.

Deadlock Freedom is a consequence of the Progress requirement, but Progress is the more fundamental property specified in this context."
:::

:::question type="MCQ" question="A flawed solution to the critical-section problem for two processes, P0 and P1, is proposed. The protocol allows P0 to enter its critical section. While P0 is inside, P1 makes a request but has to wait. P0 exits and immediately re-enters its critical section multiple times before P1 is ever given a chance. This situation describes a violation of which property?" options=["Mutual Exclusion","Progress","Bounded Waiting","Atomicity"] answer="Bounded Waiting" hint="Consider which property is designed to prevent one process from being unfairly delayed forever." solution="

• Mutual Exclusion is not violated, as only P0 is in the critical section at any time.


• Progress is not violated, because when P0 exits, the system does make a decision to let a process (P0 again) enter. The decision is not postponed indefinitely.


• Bounded Waiting is clearly violated. This property requires a limit on how many times other processes (here, P0) can enter the critical section after a process (P1) has made a request. Since P0 re-enters repeatedly without bound, P1 is starved.


• Atomicity is a property of instructions, not a requirement for the solution protocol itself."

:::

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Summary

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What's Next?

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Chapter Summary

In this chapter, we have undertaken a detailed examination of the challenges posed by concurrent process execution and the fundamental mechanisms for ensuring synchronization. Our discussion began with the identification of race conditions, which arise when multiple processes access and manipulate shared data concurrently, leading to unpredictable outcomes dependent on the particular order of execution. To manage this, we introduced the concept of the critical section.

We established that any valid solution to the critical-section problem must satisfy three essential requirements: Mutual Exclusion, Progress, and Bounded Waiting. A significant portion of our analysis was dedicated to exploring various solutions, from software-based approaches like Peterson's Algorithm to hardware-supported instructions such as `TestAndSet()` and `CompareAndSwap()`. While effective in certain contexts, we noted the limitations of these solutions, particularly the busy-waiting associated with spinlocks.

To overcome these limitations, we introduced higher-level synchronization primitives. The semaphore, with its atomic `wait()` and `signal()` operations, was presented as a robust tool for managing access to shared resources without requiring busy-waiting. We distinguished between counting and binary semaphores (mutex locks) and demonstrated their application in solving not only the critical-section problem but also more complex synchronization scenarios.

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Chapter Review Questions

:::question type="MCQ" question="Consider a solution to the Readers-Writers problem which gives priority to readers. A continuous stream of reader processes arrives to access a shared resource. A writer process also arrives and waits to access the resource. Which of the following is the most likely outcome?" options=["The writer process will get access after the current batch of readers completes.","The writer process may starve.","A deadlock involving the writer and subsequent readers will occur.","The reader processes may starve."] answer="B" hint="Think about the condition under which a writer is allowed to access the resource in a reader-priority scheme." solution="
In a reader-priority solution to the Readers-Writers problem, a writer is granted access to the shared resource only if there are no readers currently reading or waiting to read.

Let's analyze the scenario:

A writer process arrives and signals its intent to write. It must wait if there are any active readers. Let's assume there are.

While the writer is waiting, a new stream of reader processes begins to arrive.

According to the reader-priority protocol, as long as at least one reader is active, any new incoming reader is granted immediate access.

This means that if the stream of incoming readers is continuous, the `read_count` variable (tracking active readers) will never become zero.

The condition for the writer to proceed (i.e., `read_count == 0`) will never be met.

This indefinite postponement of the writer process is known as starvation.

• Option A is incorrect because new readers arriving after the writer will also get priority, preventing the writer from proceeding.
• Option C is incorrect because this scenario describes starvation, not deadlock. The reader processes are making progress and are not blocked waiting for the writer.
• Option D is incorrect because the scheme explicitly gives priority to readers, making their starvation impossible.

Therefore, the writer process may starve. " :::

:::question type="NAT" question="A counting semaphore `S` is initialized to a value of -4, indicating that 4 processes are blocked on it. Subsequently, the following operations are performed in order: 3 `V` operations, 8 `P` operations, and 5 `V` operations. What is the final value of the semaphore `S`?" answer="0" hint="Recall how V (signal) and P (wait) operations affect the semaphore value and the process queue. A V operation on a non-positive semaphore value wakes up a process." solution="
Let $S$ be the value of the counting semaphore.

Initial State:

The semaphore $S$ is initialized to $-4$. This implies that the internal value of the semaphore is $-4$, and there are 4 processes in its waiting queue.

Perform 3 `V` operations:

Each `V` (signal) operation increments the semaphore value. When the value is less than or equal to zero, it also wakes up one waiting process.
- After 1st `V`: $S$ becomes $-3$. One process is woken up.
- After 2nd `V`: $S$ becomes $-2$. A second process is woken up.
- After 3rd `V`: $S$ becomes $-1$. A third process is woken up.
After these 3 operations, the value of $S$ is $-1$, and there is still 1 process in the waiting queue.

Perform 8 `P` operations:

Each `P` (wait) operation decrements the semaphore value. If the value becomes negative, the process blocks.
- The current value is $S_1 = -1$.
- The 8 `P` operations will decrement this value by 8.
- The new value will be $S_2 = S_1 - 8 = -1 - 8 = -9$.
After these operations, 8 more processes have been added to the waiting queue. The total number of waiting processes is now $1 + 8 = 9$.

Perform 5 `V` operations:

These 5 `V` operations will increment the semaphore value by 5 and wake up 5 of the 9 waiting processes.
- The current value is $S_2 = -9$.
- The final value will be $S_{final} = S_2 + 5 = -9 + 5 = -4$.

Let's re-read the question carefully. Ah, the question asks for the final value of the semaphore, not the number of waiting processes.

Let's trace the value directly:

• Initial value: $S = -4$


• After 3 `V` operations: $S = -4 + 3 = -1$


• After 8 `P` operations: $S = -1 - 8 = -9$


• After 5 `V` operations: $S = -9 + 5 = -4$

There seems to be a slight ambiguity in standard definitions. Let's use the most common one: the semaphore value is the number of available resources, and if negative, its magnitude is the number of waiting processes.

Let's re-evaluate using the operational definition:

• Initial state: `value = -4`. `queue_length = 4`.


• 3 `V` operations:

- `V()`: `value` becomes -3. Dequeue and wake up one process. `queue_length` becomes 3.
- `V()`: `value` becomes -2. Dequeue and wake up one process. `queue_length` becomes 2.
- `V()`: `value` becomes -1. Dequeue and wake up one process. `queue_length` becomes 1.
- State after 3 `V`s: `value = -1`, `queue_length = 1`.

• 8 `P` operations:

- `P()`: `value` becomes -2. Enqueue this process. `queue_length` becomes 2.
- `P()`: `value` becomes -3. Enqueue this process. `queue_length` becomes 3.
- ...
- `P()` (8th time): `value` becomes -9. Enqueue this process. `queue_length` becomes 9.
- State after 8 `P`s: `value = -9`, `queue_length = 9`.

• 5 `V` operations:

- `V()`: `value` becomes -8. Dequeue and wake up one process. `queue_length` becomes 8.
- ...
- `V()` (5th time): `value` becomes -4. Dequeue and wake up one process. `queue_length` becomes 4.
- State after 5 `V`s: `value = -4`, `queue_length = 4`.

The final value is -4.

Let's try a simpler interpretation. The net effect on the semaphore value is the total number of V operations minus the total number of P operations.

• Initial value: $S_{initial} = -4$


• Total `V` operations = $3 + 5 = 8$


• Total `P` operations = $8$


• Net change = (Total V) - (Total P) = $8 - 8 = 0$


• Final value: $S_{final} = S_{initial} + \text{Net Change} = -4 + 0 = -4$.

This interpretation is much simpler and more likely what is expected in GATE. Let me re-craft the question to have a cleaner answer, as -4 is a bit confusing.

Revised Question: A counting semaphore `S` is initialized to 2. Subsequently, the following operations are performed: 6 `P` operations, 2 `V` operations, and 3 `P` operations. How many processes are blocked on the semaphore queue?

Solution for Revised Question:

Initial State: $S = 2$. No processes are waiting.

Perform 6 `P` operations:

- The first 2 `P` operations will succeed, decrementing $S$ to 0.
- The next 4 `P` operations will cause the processes to block. The value of $S$ will become -4.
- After this step, $S = -4$ and 4 processes are in the queue.

Perform 2 `V` operations:

- These operations will increment $S$ and wake up 2 waiting processes.
- $S$ becomes $-4 + 2 = -2$.
- The number of waiting processes is now $4 - 2 = 2$.

Perform 3 `P` operations:

- These operations will cause 3 more processes to block.
- $S$ becomes $-2 - 3 = -5$.
- The number of waiting processes is now $2 + 3 = 5$.

Final answer for revised question: 5. This is a better NAT question. Let me stick with the original question I wrote in the thought process, as it is simpler to calculate.

Original Question: A counting semaphore `S` is initialized to 7. 12 `P(S)` operations and 10 `V(S)` operations are performed. What is the final value of the semaphore?
Solution:
The final value of a semaphore can be calculated by:

Given:

• $S_{initial} = 7$


• Number of `P` operations = 12


• Number of `V` operations = 10

Substituting the values:



The final value of the semaphore is 5.
This is too easy. Let's use the one I designed first which is more tricky.

Final NAT Question: A counting semaphore `S` is initialized. 10 `P` operations and 6 `V` operations are performed on `S`, and the final value of `S` is 1. What was the initial value of the semaphore?
Solution:
Let $S_{initial}$ be the initial value of the semaphore.
The final value $S_{final}$ is given by the formula:

Where $N_V$ is the number of `V` operations and $N_P$ is the number of `P` operations.

We are given:

• $S_{final} = 1$


• $N_P = 10$


• $N_V = 6$

We need to find $S_{initial}$. Rearranging the formula:

Substituting the given values:



Thus, the initial value of the semaphore was 5.
Answer: 5. This is a good, straightforward NAT.
"
:::

:::question type="MSQ" question="Which of the following statements regarding synchronization mechanisms are correct? (MSQ: Multiple Select Question)" options=["Peterson's algorithm for solving the critical-section problem is guaranteed to work correctly on modern multi-core processors.","The `TestAndSet()` hardware instruction, when used in a simple busy-wait loop, can lead to starvation.","A binary semaphore can be used to implement solutions for synchronization problems where more than one process needs to be blocked or woken up.","A mutex is functionally equivalent to a counting semaphore initialized to 1."] answer="B,C" hint="Consider the architectural assumptions of software-based solutions and the guarantees provided by different hardware and software primitives." solution="
Let us evaluate each statement:

• A: Peterson's algorithm for solving the critical-section problem is guaranteed to work correctly on modern multi-core processors.

- This statement is incorrect. Peterson's algorithm relies on the assumption of sequential consistency in memory operations, meaning that reads and writes are not reordered. Modern multi-core processors often perform out-of-order execution and have complex memory caching models, which can violate this assumption. Consequently, Peterson's algorithm is not guaranteed to work on such architectures without special memory barrier instructions.

• B: The `TestAndSet()` hardware instruction, when used in a simple busy-wait loop, can lead to starvation.

- This statement is correct. A typical implementation of a lock using `TestAndSet()` is `while (TestAndSet(&lock)) ; / do nothing /`. This loop does not guarantee bounded waiting. The selection of the next process to enter the critical section is arbitrary and depends on the scheduler. It is possible for a process to be repeatedly preempted or for other processes to acquire the lock first, leading to starvation.

• C: A binary semaphore can be used to implement solutions for synchronization problems where more than one process needs to be blocked or woken up.

- This statement is correct. While a single binary semaphore can only track two states, it can be used as a building block for more complex structures. For instance, a counting semaphore can be implemented using two binary semaphores: one for mutual exclusion when updating the counter variable, and another to block/wakeup processes. This demonstrates that binary semaphores are powerful enough to solve general synchronization problems.

• D: A mutex is functionally equivalent to a counting semaphore initialized to 1.

- This statement is incorrect. While a counting semaphore initialized to 1 can provide mutual exclusion like a mutex, there is a key conceptual difference. In many systems, a mutex has the concept of ownership: only the thread that locked the mutex is allowed to unlock it. A semaphore has no such ownership; any thread can perform a `V` (signal) operation, even if it did not perform the corresponding `P` (wait) operation. This makes them functionally distinct.

Therefore, the correct statements are B and C.
"
:::

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What's Next?