Generating infinite sequences in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple infinite sequence with take

What is the output of this Kotlin code snippet?

Kotlin

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

A [1, 4, 7, 10, 13]

B [2, 4, 6, 8, 10]

C [1, 2, 3, 4, 5]

D [1, 3, 5, 7, 9]

💡 Hint

Look at how the sequence starts and how the lambda adds 2 each time.

The sequence starts at 1 and each next element adds 2, so the first five elements are 1, 3, 5, 7, 9.

🧠 Conceptual

intermediate

Understanding termination of infinite sequences

Which statement about Kotlin infinite sequences is true?

A Infinite sequences in Kotlin are stored fully in memory.

B An infinite sequence always runs forever and cannot be limited.

C You can limit an infinite sequence by using terminal operations like take or first.

D You must convert an infinite sequence to a list before using it.

💡 Hint

Think about how you can get a finite result from an infinite source.

Infinite sequences generate values lazily. Using functions like take(n) or first() stops the sequence after a finite number of elements.

❓ Predict Output

advanced

Output of a Fibonacci infinite sequence

What is the output of this Kotlin code?

Kotlin

val fibs = generateSequence(Pair(0, 1)) { Pair(it.second, it.first + it.second) }
println(fibs.take(7).map { it.first }.toList())

A [0, 1, 1, 2, 3, 5, 8]

B [1, 1, 2, 3, 5, 8, 13]

C [0, 1, 2, 3, 5, 8, 13]

D [1, 2, 3, 5, 8, 13, 21]

💡 Hint

The sequence generates pairs where the first is the current Fibonacci number.

The sequence starts with (0,1). Each next pair is (previous second, sum of previous first and second). Taking first elements gives Fibonacci numbers starting at 0.

❓ Predict Output

advanced

Terminating a sequence with null

What is the output of this Kotlin code?

Kotlin

val seq = generateSequence(1) { if (it < 5) it + 1 else null }
println(seq.toList())

A [1, 2, 3, 4, 5]

B Throws IllegalStateException due to null in sequence

C [1, 2, 3, 4]

D Throws NullPointerException

💡 Hint

Check what happens when the lambda returns null.

The lambda returns null when it >= 5, which signals the end of the sequence. toList() collects [1, 2, 3, 4, 5] and stops. No error occurs.

📝 Syntax

expert

Which option produces a valid infinite sequence of powers of 2?

Select the Kotlin code snippet that correctly generates an infinite sequence of powers of 2 starting at 1.

A val seq = generateSequence(1) { it + it }

B val seq = generateSequence(1) { it * 2 }

C val seq = generateSequence(1) { it xor 2 }

D val seq = generateSequence(1) { it ** 2 }

💡 Hint

Check Kotlin operators for multiplication and exponentiation.

Option B uses it * 2 which doubles the value each time, producing powers of 2. Option B also doubles but is equivalent to C. Option B uses ^ which is bitwise XOR in Kotlin, not exponentiation. Option B uses ** which is invalid syntax in Kotlin.