## Solution of 100 door puzzle....microsoft interview puzzle

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Solution:

This problem is designed to seem overwhelming. You don’t have time to draw a diagram of 100 lockers and count 100 passes through them. Even if you did, solving the problem that way wouldn’t illustrate any skill or intuition, so there must be some trick that can be used to determine how many doors will be open. You just have to figure out what that trick is.
It’s unlikely that you’re going to be able to intuit the solution to this problem by just staring at it. What can you do? Although it’s not practical to solve the entire problem by brute force, solving a few lockers in this manner is reasonable. Perhaps you’ll notice some patterns you can apply to the larger problem.
Start by choosing an arbitrary locker, 12, and determining whether it will end open or closed. On which passes will you toggle locker 12? Obviously on the first pass, when you toggle every locker, and on the twelfth pass when you start with 12. You don’t need to consider any pass after 12 because those will all start farther down the hall. This leaves passes 2 through 11. You can count these out: 2, 4, 6, 8, 10, 12 (you toggle on pass 2); 3, 6, 9, 12 (on 3); 4, 8, 12 (on 4); 5, 10, 15 (not on 5); 6, 12 (on 6); 7, 14 (not on 7), and so on. Somewhere in the middle of this process, you will probably notice that you toggle locker 12 only when the number of the pass you’re on is a factor of 12. If you think about this, it makes sense: When counting by n, you hit 12 only when some integer number of n’s add to 12, which is another way of saying that n is a factor of 12. Though it seems simple in retrospect, this probably wasn’t obvious before you worked out an example.
The factors of 12 are 1, 2, 3, 4, 6, and 12. Correspondingly, the operations on the locker door are open, close, open, close, open, close. So locker 12 will end closed. The solution seems to have something to do with factors. Primes are numbers with unique factor properties. Perhaps it would be instructive to investigate a prime numbered locker. You might select 17 as a representative prime. The factors are 1 and 17, so the operations are open, close. It ends closed just like 12. Apparently primes are not necessarily different from nonprimes for the purposes of this problem.
What generalizations can you make about whether a locker ends open or closed? All lockers start closed and alternate between being open and closed. So lockers are closed after the second, fourth, sixth, and so on, times they are toggled - in other words, if a locker is toggled an even number of times, then it ends closed; otherwise, it ends open. You know that a locker is toggled once for every factor of the locker number, so you can say that a locker ends open only if it has an odd number of factors.
The task has now been reduced to finding how many numbers between 1 and 100 have an odd number of factors. The two you’ve examined (and most others, if you try a few more examples) have even numbers of factors. Why is that? If a number i is a factor of n, what does that mean? It means that i times some other number j is equal to n. Of course, because multiplication is commutative (i × j = j × i), that means that j is a factor of n, too, so the number of factors is usually even because factors tend to come in pairs. If you can find the numbers that have unpaired factors, you will know which lockers will be open. Multiplication is a binary operation, so two numbers will always be involved, but what if they are both the same number (that is, i = j)? In that case, a single number would effectively form both halves of the pair and there would be an odd number of factors. When this is the case, i × i = n. Therefore, n would have to be a perfect square. Try a perfect square to check this solution. For example, for 16, the factors are 1, 2, 4, 8, 16; operations are open, close, open, close, open - as expected, it ends open.
Based on this reasoning, you can conclude that only lockers with numbers that are perfect squares end open. The perfect squares between 1 and 100 (inclusive) are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. So 10 lockers would remain open.
Similarly, for the general case of k lockers, the number of open lockers is the number of perfect squares between 1 and k, inclusive. How can you best count these? The perfect squares themselves are inconvenient to count because they’re unevenly spaced. However, the square roots of the perfect squares greater than zero are the positive integers. These are very easy to count: The last number in the list of square roots gives the number of items in each list. For example, the square roots of 1, 4, 9, 16, and 25 are 1, 2, 3, 4, and 5; the last number in the list of square roots is the square root of the largest perfect square and is equal to the number of perfect squares. You need to find the square root of the largest perfect square less than or equal to k.
This task is trivial when k is a perfect square, but most of the time it won’t be. In these cases, the square root of k will be a noninteger. If you round this square root down to the nearest integer, then its square is the largest perfect square less than k-just what you were looking for. The operation of rounding to the largest integer less than or equal to a given number is often called floor. Thus, in the general case of k lockers, there will be floor(sqrt(k)) lockers remaining open.
The key to solving this problem is trying strategies to solve parts of the problem even when it isn’t clear how these parts will contribute to the overall solution. Although some attempts, such as the investigation of prime numbered lockers, may not be fruitful, others are likely to lead to greater insight about how to attack the problem, such as the strategy of calculating the result for a single locker. Even in the worst case, where none of the things you try lead you closer to the final solution, you show the interviewer that you aren’t intimidated by difficult problems with no clear solution and that you are willing to keep trying different approaches until you find one that works.

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#### 1 comment

Anonymous
4 July 2013 at 18:06

Really very nice and informative blog. The technique described here is really very useful.

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