The Math Behind the Bulb Switcher Challenge: How Many Bulbs Will Remain On After N Rounds?by@okyrcheuskaya

# The Math Behind the Bulb Switcher Challenge: How Many Bulbs Will Remain On After N Rounds?

4mApril 27th, 2023

This article describes several solutions and approaches to the "Bulb Switcher" problem. The problem is a classic brain teaser that requires a bit of mathematical intuition and careful analysis to solve. Reading this article will help you understand the problem better and provide you with a useful tool to solve similar problems in the future.

If you're interested in solving challenging math problems and want to improve your logical thinking and programming skills, then you should read this article on the "Bulb Switcher" problem.

This problem is a classic brain teaser that requires a bit of mathematical intuition and careful analysis to solve. The article provides a clear explanation of the problem and walks through the steps needed to arrive at the solution. Reading this article will help you understand the problem better and provide you with a useful tool to solve similar problems in the future.

This article describes several solutions and approaches, making it a pretty useful resource for anyone interested in the “Bulb Switcher” problem.

## Description

There are `n` bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.

On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the `ith` round, you toggle every `i` bulb. For the `nth` round, you only toggle the last bulb.

Return the number of bulbs that are on after `n` rounds.

## The first solution

The solution to this problem involves finding the pattern in the bulb switching process. After the first round, all bulbs are on. After the second round, every other bulb is turned off. After the third round, every third bulb is toggled. This pattern continues until the nth round, where only the nth bulb is toggled.

It is observed that for any given round, the bulbs that will be on are those with an odd number of factors. For example, the number 6 has factors 1, 2, 3, and 6. In the third round, bulbs 1, 2, 3, and 6 will be toggled. Bulbs 1 and 6 will end up on, since they have an odd number of factors, while bulbs 2 and 3 will end up off, since they have an even number of factors.

Therefore, to solve the problem, we need to count the number of integers between 1 and n that have an odd number of factors. This can be done efficiently using the square root property: every integer that is a perfect square has an odd number of factors. Therefore, the number of bulbs that will be on after n rounds is the number of perfect squares less than or equal to n.

fun bulbSwitch(n: Int): Int {

var count = 0

for (i in 1..n) {

if (i * i <= n) {

count++

} else {

break

}

}

return count

}

## The second solution

fun bulbSwitcher(n: Int): Int {

return kotlin.math.sqrt(n.toDouble()).toInt()

}

This solution takes advantage of the fact that the number of times a bulb gets toggled is equal to the number of factors it has. For example, bulb 6 gets toggled on rounds 1, 2, 3, and 6, so it has 4 factors. A bulb will only be on at the end if it has an odd number of factors, because toggling a bulb twice cancels out the effect of toggling it once.

The only numbers that have an odd number of factors are perfect squares. So the number of bulbs that will be on at the end is equal to the number of perfect squares less than or equal to `n`. We can calculate this by taking the square root of `n` and rounding down to the nearest integer.

The first solution uses a boolean array to represent the status of each bulb, where `true` means the bulb is on and `false` means the bulb is off. It then iterates through the rounds and toggles the status of the bulbs according to the given rules. Finally, it counts the number of bulbs that are on and returns the count.

The second solution also uses a boolean array to represent the status of each bulb, but instead of iterating through the rounds and toggling the status of the bulbs, it calculates the number of factors for each bulb. If the number of factors is odd, the bulb is on; otherwise, it is off. It then counts the number of bulbs that are on and returns the count.

Both solutions use the same data structure and return the same output, but the second solution has a different approach to determining the status of each bulb by counting its factors, while the first solution toggles the status of each bulb based on the given rules.

In general, the article provides a good example of how to approach a problem like the bulb switcher and how to find efficient solutions using different approaches. It also shows the importance of analyzing the problem and understanding its underlying mathematical concepts to find the most optimized solution.

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