Ractor for true parallelism in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Ractor communication code?

Consider this Ruby code using Ractors. What will it print?

Ruby

r = Ractor.new do
  (0...5).map { |i| i * 2 }
end

result = r.take
puts result.inspect

A [0, 2, 4, 6, 8]

B nil

C 0

D RactorError

💡 Hint

Remember that Ractor.new runs the block in parallel and take receives the result.

The Ractor runs the block which returns an array of doubled numbers from 0 to 4. The take method receives this array, so the output is [0, 2, 4, 6, 8].

🧠 Conceptual

intermediate

Which statement about Ractors is true?

Choose the correct statement about Ruby Ractors.

A Ractors share all instance variables between them by default.

B Ractors can directly modify each other's local variables.

C Ractors communicate only by message passing and do not share memory.

D Ractors run on the same thread and do not provide parallelism.

💡 Hint

Think about how Ractors achieve true parallelism safely.

Ractors do not share memory to avoid race conditions. They communicate only by passing messages, which makes them safe for parallel execution.

🔧 Debug

advanced

What error does this Ractor code raise?

What error will this code produce when run?

Ruby

r1 = Ractor.new do
  @counter = 0
  @counter += 1
  @counter
end

puts r1.take

A NameError

B Ractor::IsolationError

C NoMethodError

D No error, outputs 1

💡 Hint

Instance variables inside a Ractor are local to it.

The instance variable @counter is local to the Ractor and can be used normally. The code runs without error and outputs 1.

❓ Predict Output

advanced

What is the output of this Ractor message passing example?

What will this Ruby code print?

Ruby

r = Ractor.new do
  msg = Ractor.receive
  msg * 3
end

r.send(7)
puts r.take

A 21

B 7

C Ractor::ClosedError

D nil

💡 Hint

Look at how the Ractor receives and processes the message.

The Ractor receives the number 7, multiplies it by 3, and returns 21. The take method prints 21.

🚀 Application

expert

How many items are in the resulting array after this Ractor code runs?

Given this Ruby code using multiple Ractors, how many items will the final array contain?

Ruby

ractors = (0...3).map do |i|
  Ractor.new(i) do |index|
    Array.new(index + 2) { index }
  end
end

results = ractors.map(&:take)
final = results.flatten
puts final.size

A 6

B 9

C 3

D 7

💡 Hint

Count how many elements each Ractor creates and sum them.

The Ractors create arrays of sizes 2, 3, and 4 (because index goes 0,1,2). So total elements = 2 + 3 + 4 = 9. But index starts at 0, so arrays are of sizes 2,3,4 respectively. Actually, 0+2=2, 1+2=3, 2+2=4. So total is 2+3+4=9. The code prints 9.

Practice

1. What is the main purpose of using Ractor in Ruby?

easy

A. To run code truly in parallel on multiple CPU cores

B. To create graphical user interfaces

C. To handle database connections

D. To write single-threaded scripts

Solution

1. Step 1: Understand Ractor's role

   Ractor is designed to enable true parallel execution in Ruby, using multiple CPU cores.

2. Step 2: Compare options

   Options A, B, and C do not relate to parallelism or CPU core usage.

3. Final Answer:

   To run code truly in parallel on multiple CPU cores -> Option A

4. Quick Check:

   Ractor = true parallelism [OK]

Hint: Ractor means real parallel tasks in Ruby [OK]

Common Mistakes:

• Confusing Ractor with threads
• Thinking Ractor is for UI or database
• Assuming Ractor runs code sequentially

2. Which of the following is the correct way to create a new Ractor in Ruby?

easy

A. Ractor.new { puts 'Hello from Ractor' }

B. Thread.new { puts 'Hello from Ractor' }

C. Process.spawn { puts 'Hello from Ractor' }

D. Fiber.new { puts 'Hello from Ractor' }

Solution

1. Step 1: Identify Ractor creation syntax

   Ractor is created using Ractor.new with a block of code to run in parallel.

2. Step 2: Eliminate other options

   Thread, Process, and Fiber are different concurrency tools, not Ractor.

3. Final Answer:

   Ractor.new { puts 'Hello from Ractor' } -> Option A

4. Quick Check:

   Use Ractor.new to create Ractors [OK]

Hint: Ractor.new starts a parallel task [OK]

Common Mistakes:

• Using Thread.new instead of Ractor.new
• Confusing Process.spawn with Ractor
• Using Fiber.new which is cooperative, not parallel

3. What will be the output of this Ruby code using Ractor?

r = Ractor.new { 5 * 5 }
result = r.take
puts result

medium

A. RactorError

B. 5 * 5

C. nil

D. 25

Solution

1. Step 1: Understand Ractor block execution

   The block { 5 * 5 } runs inside the Ractor and returns 25.

2. Step 2: Use take to get the result

   r.take receives the value returned by the Ractor block, which is 25.

3. Final Answer:

   25 -> Option D

4. Quick Check:

   Ractor returns 25, take receives it [OK]

Hint: Ractor block returns value, take fetches it [OK]

Common Mistakes:

• Expecting string '5 * 5' instead of calculation
• Confusing take with yield
• Assuming nil or error without communication

4. Identify the error in this Ractor code snippet:

r = Ractor.new do
  Ractor.yield 10
  20
end
puts r.take
puts r.take

medium

A. Ractor.new requires arguments

B. Second r.take call will cause Ractor::Closed error

C. No error, code runs fine

D. Calling Ractor.yield inside Ractor block is invalid

Solution

1. Step 1: Understand Ractor.yield and r.take

   Ractor.yield sends a value to the main thread, and r.take receives it. The first r.take gets 10.

2. Step 2: Analyze second r.take call

   The block returns 20 after yield, which is the final value. The second r.take gets 20. After that, the Ractor is closed. A third r.take would cause error, but here only two calls are made, so no error.

3. Step 3: Check the code carefully

   Both r.take calls are valid; no error occurs.

4. Final Answer:

   No error, code runs fine -> Option C

5. Quick Check:

   Ractor.yield and take work twice safely [OK]

Hint: Ractor.yield sends values, take receives them sequentially [OK]

Common Mistakes:

• Thinking Ractor.yield is invalid inside Ractor
• Expecting error on second take call
• Assuming Ractor.new needs arguments

5. You want to sum numbers from 1 to 100 using two Ractors running in parallel, each summing half the range. Which code correctly implements this?

hard

A. r1 = Ractor.new { (1..100).sum }
r2 = Ractor.new { (1..100).sum }
puts r1.take + r2.take

B. r1 = Ractor.new { (1..50).sum }
r2 = Ractor.new { (51..100).sum }
puts r1.take + r2.take

C. r1 = Thread.new { (1..50).sum }
r2 = Thread.new { (51..100).sum }
puts r1.value + r2.value

D. r1 = Ractor.new { (1..50).reduce(:+) }
r2 = Ractor.new { (51..100).reduce(:+) }
puts r1.take * r2.take

Solution

1. Step 1: Split the range correctly

   We want two Ractors, one summing 1 to 50, the other 51 to 100.

2. Step 2: Check the operations and combining results

   r1 = Ractor.new { (1..50).sum } r2 = Ractor.new { (51..100).sum } puts r1.take + r2.take sums each half and adds results correctly. r1 = Ractor.new { (1..100).sum } r2 = Ractor.new { (1..100).sum } puts r1.take + r2.take sums full range twice (wrong). r1 = Thread.new { (1..50).sum } r2 = Thread.new { (51..100).sum } puts r1.value + r2.value uses Thread, not Ractor. r1 = Ractor.new { (1..50).reduce(:+) } r2 = Ractor.new { (51..100).reduce(:+) } puts r1.take * r2.take multiplies sums instead of adding.

3. Final Answer:

   r1 = Ractor.new { (1..50).sum } r2 = Ractor.new { (51..100).sum } puts r1.take + r2.take -> Option B

4. Quick Check:

   Split range, sum halves, add results [OK]

Hint: Split work, sum halves in Ractors, then add [OK]

Common Mistakes:

• Using Thread instead of Ractor
• Summing full range twice
• Multiplying sums instead of adding