Generating infinite sequences in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an infinite sequence starting from 1.

Kotlin

val numbers = generateSequence([1]) { it + 1 }

Drag options to blanks, or click blank then click option'

A-1

B0

Cnull

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0 or null will change the sequence start.

The sequence starts from 1, so the initial value must be 1.

2fill in blank

medium

Complete the code to generate an infinite sequence of even numbers starting from 2.

Kotlin

val evens = generateSequence([1]) { it + 2 }

Drag options to blanks, or click blank then click option'

A1

B0

C2

D-2

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 1 or 0 will not produce even numbers.

The sequence of even numbers starts at 2, so the initial value must be 2.

3fill in blank

hard

Fix the error in the code to generate an infinite sequence of odd numbers starting from 1.

Kotlin

val odds = generateSequence([1]) { it + 2 }

Drag options to blanks, or click blank then click option'

A0

B1

C2

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 as the start value will produce even numbers.

The sequence of odd numbers starts at 1, so the initial value must be 1.

4fill in blank

hard

Fill both blanks to generate the infinite sequence 1, 2, 5, 14, 41, … starting from 1.

Kotlin

val squares = generateSequence([1]) { it + [2] }

Drag options to blanks, or click blank then click option'

A1

B2

Cit * 2 - 1

Dit + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed increment like 2 will not generate the correct sequence.

The sequence starts at 1, and the next value is obtained by adding it * 2 - 1.

5fill in blank

hard

Fill all three blanks to generate an infinite Fibonacci sequence starting from 0 and 1.

Kotlin

val fibs = generateSequence(Pair([1], [2])) { Pair(it.second, it.first [3] it.second) } .map { it.first }

Drag options to blanks, or click blank then click option'

A0

B1

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or wrong starting numbers will break the Fibonacci sequence.

The Fibonacci sequence starts with 0 and 1, and each next number is the sum of the previous two.

Practice

1. What does an infinite sequence in Kotlin do?
sequence { yield(1); yieldAll(generateSequence(2) { it + 1 }) }

easy

A. Generates only the first two numbers and stops

B. Creates a fixed list of numbers from 1 to 10

C. Generates values endlessly, one at a time, starting from 1

D. Throws an error because sequences can't be infinite

Solution

1. Step 1: Understand sequence behavior

   The sequence starts by yielding 1, then continues with an infinite sequence starting at 2, increasing by 1 each time.

2. Step 2: Recognize infinite generation

   Since generateSequence with a lambda that adds 1 never returns null, it produces values endlessly.

3. Final Answer:

   Generates values endlessly, one at a time, starting from 1 -> Option C

4. Quick Check:

   Infinite sequence = endless values [OK]

Hint: Infinite sequences produce values one by one without stopping [OK]

Common Mistakes:

• Thinking infinite sequences create fixed lists
• Assuming infinite sequences cause errors immediately
• Believing sequences stop after a few values

2. Which of the following is the correct syntax to create an infinite sequence of even numbers starting from 0 in Kotlin?

easy

A. sequence { yield(0); yield(2); yield(4) }

B. sequenceOf(0, 2, 4, 6)

C. generateSequence(0, 2)

D. generateSequence(0) { it + 2 }

Solution

1. Step 1: Identify infinite sequence syntax

   generateSequence(seed) { nextValue } creates an infinite sequence by applying the lambda repeatedly.

2. Step 2: Check each option

   generateSequence(0) { it + 2 } correctly uses generateSequence(0) { it + 2 } to produce even numbers endlessly. Options B and D create finite sequences. generateSequence(0, 2) has incorrect syntax.

3. Final Answer:

   generateSequence(0) { it + 2 } -> Option D

4. Quick Check:

   Correct syntax = generateSequence(0) { it + 2 } [OK]

Hint: Use generateSequence(seed) { next } for infinite sequences [OK]

Common Mistakes:

• Using sequenceOf for infinite sequences
• Passing two arguments to generateSequence incorrectly
• Not providing a lambda for next values

3. What will be the output of this Kotlin code?

val seq = generateSequence(1) { it + 3 }
println(seq.take(4).toList())

medium

A. [1, 2, 3, 4]

B. [1, 4, 7, 10]

C. [1, 3, 6, 9]

D. [4, 7, 10, 13]

Solution

1. Step 1: Understand sequence generation

   The sequence starts at 1, then each next value adds 3: 1, 4, 7, 10, ...

2. Step 2: Apply take and toList

   take(4) takes first 4 values: 1, 4, 7, 10. toList() converts them to a list.

3. Final Answer:

   [1, 4, 7, 10] -> Option B

4. Quick Check:

   Sequence values = [1,4,7,10] [OK]

Hint: Add the step value repeatedly starting from seed [OK]

Common Mistakes:

• Adding 1 instead of 3 each step
• Confusing take with drop
• Expecting a finite sequence without take

4. Identify the error in this Kotlin code that tries to generate an infinite sequence of odd numbers:

val odds = generateSequence(1) { it + 2 }
println(odds.take(5).toList())

medium

A. No error; code runs correctly and prints odd numbers

B. Missing yield keyword inside generateSequence

C. The lambda should return null to stop sequence

D. Seed value should be 0 instead of 1

Solution

1. Step 1: Check generateSequence usage

   The code uses generateSequence(1) { it + 2 }, which correctly generates odd numbers starting at 1.

2. Step 2: Verify output and syntax

   take(5).toList() safely takes first 5 odd numbers. No syntax or runtime error exists.

3. Final Answer:

   No error; code runs correctly and prints odd numbers -> Option A

4. Quick Check:

   Correct infinite sequence = No error; code runs correctly and prints odd numbers [OK]

Hint: generateSequence needs only seed and next lambda, no yield [OK]

Common Mistakes:

• Thinking yield is needed inside generateSequence
• Expecting lambda to return null to continue
• Confusing seed value for even numbers

5. You want to generate an infinite sequence of Fibonacci numbers in Kotlin. Which code correctly creates this infinite sequence?

hard

A. Both A and B

B. sequence { var a=0; var b=1; while(true) { yield(a); val c=a+b; a=b; b=c } }

C. generateSequence(Pair(0, 1)) { Pair(it.second, it.first + it.second) }.map { it.first }

D. generateSequence(0) { it + it }

Solution

1. Step 1: Analyze generateSequence(Pair(0, 1)) { Pair(it.second, it.first + it.second) }.map { it.first }

   generateSequence(Pair(0, 1)) { Pair(it.second, it.first + it.second) }.map { it.first } uses generateSequence with pairs to generate Fibonacci pairs, then maps to first element, correctly producing Fibonacci numbers infinitely.

2. Step 2: Analyze sequence { var a=0; var b=1; while(true) { yield(a); val c=a+b; a=b; b=c } }

   sequence { var a=0; var b=1; while(true) { yield(a); val c=a+b; a=b; b=c } } uses a sequence builder with a while(true) loop yielding Fibonacci numbers step-by-step, also correct and infinite.

3. Step 3: Analyze generateSequence(0) { it + it }

   generateSequence(0) { it + it } doubles the number each time, not Fibonacci, so it's incorrect.

4. Final Answer:

   Both A and B -> Option A

5. Quick Check:

   Both methods generate Fibonacci infinite sequences [OK]

Hint: Use pairs or sequence builder with yield for Fibonacci [OK]

Common Mistakes:

• Using generateSequence without pairs for Fibonacci
• Trying to double numbers instead of summing previous two
• Not using yield in sequence builder for infinite values