Question: There are 3 light bulbs in a hidden room and 3 switches outside the room that correspond to those light bulbs. You do not know which switch affects which bulb and you cannot see inside of the room. You are allowed to go inside of the room only one time. How do you find out which switch corresponds to which bulb?
Answer: Turn on two switches and wait for a while. Then turn off one switch and go inside the room. The bulb that is still on corresponds to the switch that is still on. Touch the remaining bulbs. The hotter bulb is the switch that you turned off, and the remaining bulb is the switch that you never turned on.
Wednesday, July 23, 2008
Monday, June 23, 2008
9 Minutes
Question: You are given two hourglasses. One measures 4 minutes and one measures 7 minutes. How would you measure exactly 9 minutes?
Answer: we want to measure the 9 minutes right from the start.
Answer: we want to measure the 9 minutes right from the start.
| Step | Time | 4 minute timer | 7 minute timer |
|---|---|---|---|
| 1 | 0 min | Start | Start |
| 2 | 4 mins | Flip | 3 minutes left |
| 3 | 7 mins | 1 minute left | Flipt |
| 4 | 8 mins | Stop | Flip (1 minute left) |
| 5 | 9 mins | Stop | Stop |
Friday, May 23, 2008
Four People on a Rickety Bridge
Question: Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
Answer: The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.
Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins
Answer: The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.
Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins
Wednesday, April 23, 2008
How Strong is an Egg?
Question: You have two identical eggs. Standing in front of a 100 floor building, you wonder what is the maximum number of floors from which the egg can be dropped without breaking it. What is the minimum number of tries needed to find out the solution?
Answer: The easiest way to do this would be to start from the first floor and drop the egg. If it doesn’t break, move on to the next floor. If it does break, then we know the maximum floor the egg will survive is 0. If we continue this process, we will easily find out the maximum floors the egg will survive with just one egg. So the maximum number of tries is 100 that is when the egg survives even at the 100th floor.
Can we do better? Of course we can. Let’s start at the second floor. If the egg breaks, then we can use the second egg to go back to the first floor and try again. If it does not break, then we can go ahead and try on the 4th floor (in multiples of 2). If it ever breaks, say at floor x, then we know it survived floor x-2. That leaves us with just floor x-1 to try with the second egg. So what is the maximum number of tries possible? It occurs when the egg survives 98 or 99 floors. It will take 50 tries to reach floor 100 and one more egg to try on the 99th floor so the total is 51 tries. Wow, that is almost half of what we had last time.
Can we do even better? Yes we can . What if we try at intervals of 3? Applying the same logic as the previous case, we need a max of 35 tries to find out the information (33 tries to reach 99th floor and 2 more on 97th and 98th floor).
Interval – Maximum tries
1 – 100
2 – 51
3 – 35
4 – 29
5 – 25
6 – 21
7 – 20
8 – 19
9 – 19
10 – 19
11 – 19
12 – 19
13 – 19
14 – 20
15 – 20
16 – 21
So picking any one of the intervals with 19 maximum tries would be fine.
Can we do even better? Yes we can..
Instead of taking equal intervals, we can increase the number of floors by one less than the previous increment. For example, let’s first try at floor 14. If it breaks, then we need 13 more tries to find the solution. If it doesn’t break, then we should try floor 27 (14 + 13). If it breaks, we need 12 more tries to find the solution. So the initial 2 tries plus the additional 12 tries would still be 14 tries in total. If it doesn’t break, we can try 39 (27 + 12) and so on. Using 14 as the initial floor, we can reach up to floor 105 (14 + 13 + 12 + … + 1) before we need more than 14 tries. Since we only need to cover 100 floors, 14 tries is sufficient to find the solution.
Therefore, 14 is the least number of tries to find out the solution.
Answer: The easiest way to do this would be to start from the first floor and drop the egg. If it doesn’t break, move on to the next floor. If it does break, then we know the maximum floor the egg will survive is 0. If we continue this process, we will easily find out the maximum floors the egg will survive with just one egg. So the maximum number of tries is 100 that is when the egg survives even at the 100th floor.
Can we do better? Of course we can. Let’s start at the second floor. If the egg breaks, then we can use the second egg to go back to the first floor and try again. If it does not break, then we can go ahead and try on the 4th floor (in multiples of 2). If it ever breaks, say at floor x, then we know it survived floor x-2. That leaves us with just floor x-1 to try with the second egg. So what is the maximum number of tries possible? It occurs when the egg survives 98 or 99 floors. It will take 50 tries to reach floor 100 and one more egg to try on the 99th floor so the total is 51 tries. Wow, that is almost half of what we had last time.
Can we do even better? Yes we can . What if we try at intervals of 3? Applying the same logic as the previous case, we need a max of 35 tries to find out the information (33 tries to reach 99th floor and 2 more on 97th and 98th floor).
Interval – Maximum tries
1 – 100
2 – 51
3 – 35
4 – 29
5 – 25
6 – 21
7 – 20
8 – 19
9 – 19
10 – 19
11 – 19
12 – 19
13 – 19
14 – 20
15 – 20
16 – 21
So picking any one of the intervals with 19 maximum tries would be fine.
Can we do even better? Yes we can..
Instead of taking equal intervals, we can increase the number of floors by one less than the previous increment. For example, let’s first try at floor 14. If it breaks, then we need 13 more tries to find the solution. If it doesn’t break, then we should try floor 27 (14 + 13). If it breaks, we need 12 more tries to find the solution. So the initial 2 tries plus the additional 12 tries would still be 14 tries in total. If it doesn’t break, we can try 39 (27 + 12) and so on. Using 14 as the initial floor, we can reach up to floor 105 (14 + 13 + 12 + … + 1) before we need more than 14 tries. Since we only need to cover 100 floors, 14 tries is sufficient to find the solution.
Therefore, 14 is the least number of tries to find out the solution.
Sunday, March 23, 2008
12 Identical Balls Problem
Question: You are give 12 identical looking balls. One of them is fake (could be heavier or lighter) than the rest of the 11 (all the others weight exactly the same). You a provided with a simple mechanical balance and you are restricted to only 3 uses. Find the fake ball.
Answer: For convenience sake, let’s name the balls 1-12. This question is quite tricky. If you want to warm up your skills, you can try this easier 8 ball problem first.
Ok, let’s get on with it.
First we weigh {1,2,3,4} on the left and {5,6,7,8} on the right. There are three scenarios which can arise from this.
If they balance, then we know 9, 10, 11 or 12 is fake. Weigh {8, 9} and {10, 11} (Note: 8 is surely not fake)
If they balance, we know 12 is the fake one. Just weigh it with any other ball and figure out if it is lighter or heavier.
If {8, 9} is heavier, then either 9 is heavy or 10 is light or 11 is light. Weigh {10} and {11}. If they balance, 9 is fake (heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {8, 9} is lighter, then either 9 is light or 10 is heavy or 11 is heavy. Weigh {10} and {11}. If they balance, 9 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
Phwww, that was confusing enough but we are not done yet.
If {1,2,3,4} is heavier, we know either one of {1,2,3,4} heavier or one of {5,6,7,8} is lighter but it is guarantees that {9,10,11,12} are not fake. This is where it gets really tricky, watch carefully. Weigh {1,2,5} and {3,6,9} (Note: 9 is surely not fake).
If they balance, then either 4 is heavy or 7 is light or 8 is light. Following the last step from the previous case, we weigh {7} and {8}. If they balance, 4 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {1,2,5} is heavier, then either 1 is heavy or 2 is heavy or 6 is light. Weigh {1} and {2}. If they balance, 6 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {3,6,9} is heavier, then either 3 is heavy or 5 is light. Weigh {5} and {9}. They won’t balance. If {5} is lighter, 5 is fake (lighter). If they balance, 3 is fake (heavier).
If {5,6,7,8} is heavier, it is the same situation as if {1,2,3,4} was heavier. Just perform the same steps using 5,6,7 and 8. Unless maybe you are too lazy to try and reprocess the steps, then you continue reading the solution. Weigh {5,6,1} and {7,2,9} (Note: 9 is surely not fake).
If they balance, then either 8 is heavy or 3 is light or 4 is light. Following the last step from the previous case, we weigh {3} and {4}. If they balance, 8 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {5,6,1} is heavier, then either 5 is heavy or 6 is heavy or 2 is light. Weigh {5} and {6}. If they balance, 2 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {7,2,9} is heavier, then either 7 is heavy or 1 is light. Weigh {1} and {9}. If they balance, 7 is fake (heavier). If they don’t balance then 1 is fake (lighter).
Answer: For convenience sake, let’s name the balls 1-12. This question is quite tricky. If you want to warm up your skills, you can try this easier 8 ball problem first.
Ok, let’s get on with it.
First we weigh {1,2,3,4} on the left and {5,6,7,8} on the right. There are three scenarios which can arise from this.
If they balance, then we know 9, 10, 11 or 12 is fake. Weigh {8, 9} and {10, 11} (Note: 8 is surely not fake)
If they balance, we know 12 is the fake one. Just weigh it with any other ball and figure out if it is lighter or heavier.
If {8, 9} is heavier, then either 9 is heavy or 10 is light or 11 is light. Weigh {10} and {11}. If they balance, 9 is fake (heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {8, 9} is lighter, then either 9 is light or 10 is heavy or 11 is heavy. Weigh {10} and {11}. If they balance, 9 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
Phwww, that was confusing enough but we are not done yet.
If {1,2,3,4} is heavier, we know either one of {1,2,3,4} heavier or one of {5,6,7,8} is lighter but it is guarantees that {9,10,11,12} are not fake. This is where it gets really tricky, watch carefully. Weigh {1,2,5} and {3,6,9} (Note: 9 is surely not fake).
If they balance, then either 4 is heavy or 7 is light or 8 is light. Following the last step from the previous case, we weigh {7} and {8}. If they balance, 4 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {1,2,5} is heavier, then either 1 is heavy or 2 is heavy or 6 is light. Weigh {1} and {2}. If they balance, 6 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {3,6,9} is heavier, then either 3 is heavy or 5 is light. Weigh {5} and {9}. They won’t balance. If {5} is lighter, 5 is fake (lighter). If they balance, 3 is fake (heavier).
If {5,6,7,8} is heavier, it is the same situation as if {1,2,3,4} was heavier. Just perform the same steps using 5,6,7 and 8. Unless maybe you are too lazy to try and reprocess the steps, then you continue reading the solution. Weigh {5,6,1} and {7,2,9} (Note: 9 is surely not fake).
If they balance, then either 8 is heavy or 3 is light or 4 is light. Following the last step from the previous case, we weigh {3} and {4}. If they balance, 8 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {5,6,1} is heavier, then either 5 is heavy or 6 is heavy or 2 is light. Weigh {5} and {6}. If they balance, 2 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {7,2,9} is heavier, then either 7 is heavy or 1 is light. Weigh {1} and {9}. If they balance, 7 is fake (heavier). If they don’t balance then 1 is fake (lighter).
Saturday, February 23, 2008
8 Identical Balls Problem
Question: You are given 8 identical looking balls. One of them is heavier than the rest of the 7 (all the others weigh exactly the same). You a provided with a simple mechanical balance and you are restricted to only 2 uses. Find the heavier ball.
Answer: For convenience sake, let’s name the balls 1-8. First we weigh {1,2,3} on the left and {4,5,6} on the right. There are three scenarios which can arise from this.
If the left side is heavier, then we know that one of 1, 2 or 3 is the heavier ball. Weigh {1} on the left and {2} on the right. By doing this, we will know if 1 or 2 is heavier. If they balance out, then 3 is the heavier one.
If the right side is heavier, then we know that either 4, 5 or 6 is the heavier ball. Weigh {4} on the left and {5} on the right. By doing this we will know if 4 or 5 is heavier. If they balance out, then 6 is the heavier one.
If {1,2,3} and {4,5,6} balance out, then we know either 7 or 8 is the heavier one. Weigh both of them to find out which one is heavier.
Answer: For convenience sake, let’s name the balls 1-8. First we weigh {1,2,3} on the left and {4,5,6} on the right. There are three scenarios which can arise from this.
If the left side is heavier, then we know that one of 1, 2 or 3 is the heavier ball. Weigh {1} on the left and {2} on the right. By doing this, we will know if 1 or 2 is heavier. If they balance out, then 3 is the heavier one.
If the right side is heavier, then we know that either 4, 5 or 6 is the heavier ball. Weigh {4} on the left and {5} on the right. By doing this we will know if 4 or 5 is heavier. If they balance out, then 6 is the heavier one.
If {1,2,3} and {4,5,6} balance out, then we know either 7 or 8 is the heavier one. Weigh both of them to find out which one is heavier.
Wednesday, January 23, 2008
A Box of Defective Balls
Question: You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?
Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.
The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.
If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.
Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.
The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.
If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.
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