Eive Impossible Questions

[R. Richard]

Question 1 is easily solvable. Questions 2,3 and 4 require only the methodology to be used to find the answer. Question 5 was solved only by Colleen Thomas. How can you do? (Answers later.)

1. You're in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You may inspect the room only once. How can you determine which switch is connected to which light bulb?

2. How many trees are there in Central Park?

3. How much does all the ice in a hockey rink weigh?

4. If you were building a new city with a projected population of 100,000, how many gas stations would you need?

5. You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

 

[DreamCloud]

Question 5 was the easiest. Assuming an equal-arm scale

1) Weigh 6 against 6. Whichever set is heavier contains the offending ball.

2) Weigh 3 against 3 from the heavy set. Whichever new set is heavier contains the offending ball.

3) Weigh any two balls from the heavy 3 set. If they are equal, the unweighed ball is the offending one. If not, the heavy ball is plainly identified on the scale.

 

[Bramblethorn]

Answers ROT-13ed:

Question 1 is easily solvable. Questions 2,3 and 4 require only the methodology to be used to find the answer. Question 5 was solved only by Colleen Thomas. How can you do? (Answers later.)

1. You're in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You may inspect the room only once. How can you determine which switch is connected to which light bulb?

Fjvgpu gjb ohyof ba. Yrnir gurz ba sbe n srj zvahgrf, gura fjvgpu bar bs gurz bss. Gur bar lbh whfg fjvgpurq bss jvyy or ubg gb gbhpu (nffhzvat lbh unir gur evtug xvaq bs ohyof).

5. You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

Nffhzvat guvf vf gur "jrvtu gjb guvatf ntnvafg bar nabgure" glcr bs fpnyr, abg na ryrpgebavp "qvfcynl gur jrvtug" fpnyr:

Fcyvg vagb guerr tebhcf bs sbhe. Jrvtu gjb tebhcf ntnvafg bar nabgure. Vs bar tebhc vf urnivre, xrrc gung tebhc; vs abg, xrrc gur guveq (hajrvturq) tebhc.

Frpbaq jrvtuvat: fcyvg gur tebhc lbh xrcg vagb gjb unyirf, jrvtu ntnvafg bar nabgure, xrrc gur urnivre unys (qbja gb gjb onyyf). Guveq jrvtuvat: fnzr ntnva jvgu gur gjb erznvavat.

Tvira guerr jrvtuvatf, lbh pna npghnyyl svaq gur urnil onyy bhg bs gjragl-frira, hfvat gur fnzr gevpx nf va gur svefg jrvtuvat nobir.

 

[Bramblethorn]

Here's another one then:

You have two identical glass balls. Dropped from a certain height (or above), they will always shatter; dropped from anything lower, they will survive undamaged.

You have a 36-floor building. You want to know what the maximum drop is that these balls can survive, to the nearest floor (it's possible that the answer is "less than 1 floor" or "more than 36 floors"). It doesn't matter if they both get smashed in the process, but obviously once you break both you can't do any more tests.

One way to do this would be to go to the first floor, drop a ball, retrieve it, then try from the second floor, then from the third, and so on, until a ball shatters. This could give you the answer in just one trial, but it might take as many as 36 trials (each drop = 1 trial).

What is the smallest number of trials in which you can guarantee finding the answer, to the nearest floor?

 

[49greg]

Question 5 was the easiest. Assuming an equal-arm scale

1) Weigh 6 against 6. Whichever set is heavier contains the offending ball.

2) Weigh 3 against 3 from the heavy set. Whichever new set is heavier contains the offending ball.

3) Weigh any two balls from the heavy 3 set. If they are equal, the unweighed ball is the offending one. If not, the heavy ball is plainly identified on the scale.

Very elegant, I don't think I ever would have gotten that one.

The first one, as stated, is easy. My way would be to turn two lights on, wait a few minutes and then turn one off. Go check, the light that is on is obvious, the light bulb that is warm is the one you turned off, and the cold one is the one you never turned on.

 

[R. Richard]

Very elegant, I don't think I ever would have gotten that one.

The first one, as stated, is easy. My way would be to turn two lights on, wait a few minutes and then turn one off. Go check, the light that is on is obvious, the light bulb that is warm is the one you turned off, and the cold one is the one you never turned on.

Very good! However, you might need a ladder to check the bulbs.

 

[R. Richard]

Here's another one then:

You have two identical glass balls. Dropped from a certain height (or above), they will always shatter; dropped from anything lower, they will survive undamaged.

You have a 36-floor building. You want to know what the maximum drop is that these balls can survive, to the nearest floor (it's possible that the answer is "less than 1 floor" or "more than 36 floors"). It doesn't matter if they both get smashed in the process, but obviously once you break both you can't do any more tests.

One way to do this would be to go to the first floor, drop a ball, retrieve it, then try from the second floor, then from the third, and so on, until a ball shatters. This could give you the answer in just one trial, but it might take as many as 36 trials (each drop = 1 trial).

What is the smallest number of trials in which you can guarantee finding the answer, to the nearest floor?

You go to the 18th floor and try one ball. If the ball breaks, you go to the first floor and work your way up, one floor at a time, a maximum of 18 trials.
I the ball survives the 18th floor, you go to the ((36+16)/2 \ 27th floor and try again. The method depends on what's called a binary search and the precise number of trials depends on the rounding process used for the binary search. Simple.

 

[LaRascasse]

Here's another one then:

You have two identical glass balls. Dropped from a certain height (or above), they will always shatter; dropped from anything lower, they will survive undamaged.

You have a 36-floor building. You want to know what the maximum drop is that these balls can survive, to the nearest floor (it's possible that the answer is "less than 1 floor" or "more than 36 floors"). It doesn't matter if they both get smashed in the process, but obviously once you break both you can't do any more tests.

One way to do this would be to go to the first floor, drop a ball, retrieve it, then try from the second floor, then from the third, and so on, until a ball shatters. This could give you the answer in just one trial, but it might take as many as 36 trials (each drop = 1 trial).

What is the smallest number of trials in which you can guarantee finding the answer, to the nearest floor?

11 tries. Skipping 6 floors each try with the first ball.

Let "n" be the number of trials to find the threshold floor and "k" be the gap between floors when you're dropping the balls.

Basically if k is 4, you drop the ball from floors 4, 8, 12 until it breaks (say on 12). Now you know that your threshold floor is somewhere between 9 and 12 (because it survived on 8 and not on 12). Then you try on 9, 10 and 11 and voila, you have your floor aka no. 11.

Total no. of tries in the above example = 6 (floors 4, 8, 12, 9, 10, 11)

n = 36/k + (k-1) in the worst case. The worst case being when the threshold floor is 35th, meaning if k = 4, your tries go floors 4, 8, 12, 16, 20, 24, 28, 32, 36 (ball 1 breaks), 33, 34, 35. 12 tries.

With a bit of Newton magic (or Liebnitz, if you are so opined)

dn/dk = (-36/k^2) + 1

Now for minima of n, dn/dk = 0

so (-36/k^2) + 1 = 0

solving we get k = 6.

Hence, n is minimum when k is 6. Putting that back in our original equation.

n = 11.

 

[49greg]

Very good! However, you might need a ladder to check the bulbs.

That would depend on the height of the ceiling, in a normal house I can usually reach up and easily touch the ceiling with with my fingertips. Bulbs hanging down are even less of a problem.

 

[LaRascasse]

My turn

There is a long hallway containing 1000 doors. Each door is numbered 1-1000 and has a guard wearing the number of his door. All the doors are initially closed. All the guards in sequence go about toggling (opening if closed and closing if open) all the doors whose numbers are a multiple of their own.

So guard #1 opens every door (every door is a multiple of 1). Then guard #2 closes every alternate door. Guard #3 goes to every third door and alters the state of it. So on and so forth.

How many doors remain open after all 1000 guards have had their turn? (for bonus points - which ones?)

 

[JKendallDane]

11 tries. Skipping 6 floors each try with the first ball.

Let "n" be the number of trials to find the threshold floor and "k" be the gap between floors when you're dropping the balls.

Basically if k is 4, you drop the ball from floors 4, 8, 12 until it breaks (say on 12). Now you know that your threshold floor is somewhere between 9 and 12 (because it survived on 8 and not on 12). Then you try on 9, 10 and 11 and voila, you have your floor aka no. 11.

Total no. of tries in the above example = 6 (floors 4, 8, 12, 9, 10, 11)

n = 36/k + (k-1) in the worst case. The worst case being when the threshold floor is 35th, meaning if k = 4, your tries go floors 4, 8, 12, 16, 20, 24, 28, 32, 36 (ball 1 breaks), 33, 34, 35. 12 tries.

With a bit of Newton magic (or Liebnitz, if you are so opined)

dn/dk = (-36/k^2) + 1

Now for minima of n, dn/dk = 0

so (-36/k^2) + 1 = 0

solving we get k = 6.

Hence, n is minimum when k is 6. Putting that back in our original equation.

n = 11.

It was stuff like this that caused me to paperclip a $100 bill to my Quantitative Analysis II final when I handed it in my senior year of college.

 

[Bramblethorn]

You go to the 18th floor and try one ball. If the ball breaks, you go to the first floor and work your way up, one floor at a time, a maximum of 18 trials.
I the ball survives the 18th floor, you go to the ((36+16)/2 \ 27th floor and try again. The method depends on what's called a binary search and the precise number of trials depends on the rounding process used for the binary search. Simple.

Recursive bisection is optimal if you have unlimited balls, but not here. LR's answer is pretty good...

11 tries. Skipping 6 floors each try with the first ball.

Let "n" be the number of trials to find the threshold floor and "k" be the gap between floors when you're dropping the balls.

Basically if k is 4, you drop the ball from floors 4, 8, 12 until it breaks (say on 12). Now you know that your threshold floor is somewhere between 9 and 12 (because it survived on 8 and not on 12). Then you try on 9, 10 and 11 and voila, you have your floor aka no. 11.

Total no. of tries in the above example = 6 (floors 4, 8, 12, 9, 10, 11)

n = 36/k + (k-1) in the worst case. The worst case being when the threshold floor is 35th, meaning if k = 4, your tries go floors 4, 8, 12, 16, 20, 24, 28, 32, 36 (ball 1 breaks), 33, 34, 35. 12 tries.

With a bit of Newton magic (or Liebnitz, if you are so opined)

dn/dk = (-36/k^2) + 1

Now for minima of n, dn/dk = 0

so (-36/k^2) + 1 = 0

solving we get k = 6.

Hence, n is minimum when k is 6. Putting that back in our original equation.

n = 11.

...but it's still not quite optimal. I'll leave this up a little while and see if anybody can improve on 11!

 

[Handley_Page]

I think I have just realised why I never made it as a computer programmer.

 

[Bramblethorn]

There is a long hallway containing 1000 doors. Each door is numbered 1-1000 and has a guard wearing the number of his door. All the doors are initially closed. All the guards in sequence go about toggling (opening if closed and closing if open) all the doors whose numbers are a multiple of their own.

So guard #1 opens every door (every door is a multiple of 1). Then guard #2 closes every alternate door. Guard #3 goes to every third door and alters the state of it. So on and so forth.

How many doors remain open after all 1000 guards have had their turn? (for bonus points - which ones?)

ROT-13'ed again:

Guvegl-bar. N qbbe jvyy or gbttyrq nf znal gvzrf nf vg unf cbfvgvir-vagrtre snpgbef (vapyhqvat bar). Nal ahzore gung vf abg n cresrpg fdhner unf na rira ahzore bs snpgbef: rirel snpgbe terngre guna gur fdhner ebbg unf n pbeerfcbaqvat snpgbe gung'f yrff guna gur fdhner ebbg, naq ivpr irefn, fb vg jvyy or gbttyrq na rira ahzore bs gvzrf naq raq hc pybfrq ntnva. Nal cresrpg fdhner unf na bqq ahzore bs snpgbef, orpnhfr gur fdhner ebbg vf vgfrys na vagrtre snpgbe naq vf vgf bja pbhagrecneg, fb cresrpg-fdhner-ahzorerq qbbef trg gbttyrq na bqq ahzore bs gvzrf naq raq hc bcra.

Fb gur fdhner-ahzorerq qbbef (naq ab bguref) raq hc bcra. Ynetrfg cresrpg fdhner yrff guna 1000 vf guvegl-bar fdhnerq.

 

[Handley_Page]

There is a long hallway containing 1000 doors. Each door is numbered 1-1000 and has a guard wearing the number of his door. All the doors are initially closed. All the guards in sequence go about toggling (opening if closed and closing if open) all the doors whose numbers are a multiple of their own.

So guard #1 opens every door (every door is a multiple of 1). Then guard #2 closes every alternate door. Guard #3 goes to every third door and alters the state of it. So on and so forth.

How many doors remain open after all 1000 guards have had their turn? (for bonus points - which ones?)

Is this a Fibonachi thing ?

 

[pbbj]

Another version of the first one. You have 13 balls and a scale. One of the balls is either lighter or heavier than the others(the others are all the same weight).

How do you find out which one, only using the scale 3 times?

 

[LaRascasse]

Recursive bisection is optimal if you have unlimited balls, but not here. LR's answer is pretty good...



...but it's still not quite optimal. I'll leave this up a little while and see if anybody can improve on 11!

Yeah, I see that. My bad, should've tried dynamic programming instead of trying calculus.

I'll leave the answer open to others who want to try. The trick is basically to reduce the problem to two similar sub-problems until you hit a trivial solution and then work your way back up.

 

[LaRascasse]

ROT-13'ed again:

Guvegl-bar. N qbbe jvyy or gbttyrq nf znal gvzrf nf vg unf cbfvgvir-vagrtre snpgbef (vapyhqvat bar). Nal ahzore gung vf abg n cresrpg fdhner unf na rira ahzore bs snpgbef: rirel snpgbe terngre guna gur fdhner ebbg unf n pbeerfcbaqvat snpgbe gung'f yrff guna gur fdhner ebbg, naq ivpr irefn, fb vg jvyy or gbttyrq na rira ahzore bs gvzrf naq raq hc pybfrq ntnva. Nal cresrpg fdhner unf na bqq ahzore bs snpgbef, orpnhfr gur fdhner ebbg vf vgfrys na vagrtre snpgbe naq vf vgf bja pbhagrecneg, fb cresrpg-fdhner-ahzorerq qbbef trg gbttyrq na bqq ahzore bs gvzrf naq raq hc bcra.

Fb gur fdhner-ahzorerq qbbef (naq ab bguref) raq hc bcra. Ynetrfg cresrpg fdhner yrff guna 1000 vf guvegl-bar fdhnerq.

Qvat! Qvat! Qvat! naq jr unir n jvaare.

 

[Over_Red]

Question 1 is easily solvable. Questions 2,3 and 4 require only the methodology to be used to find the answer. Question 5 was solved only by Colleen Thomas.

That's funny, I couldn't solve number 1, but I figured out the answer to question 5 in a few moments. I guess everyone thinks differently.

#2:

Is there some special method? I suppose I'd just walk around and count them, marking them somehow with string or tape to show that I counted them.

#3:

To determine the weight of the ice, you just need to multiply the area of a hockey rink by the thickness of the ice. This gives you the volume of ice. Using the average density of ice will then let you determine the mass of the ice. Then multiply that mass by the acceleration due to gravity, and you have the weight of the ice.

#4

Hmm, this is a weird question. It depends on the layout of the city and how big each gas station is. It also depends on where you put the gas stations, if the price of gas can support a more concentrated number or a less concentrated number...

Anyway, gas stations are private or corporate owned enterprises, so you wouldn't really have to plan for them other than to allow building permits for interested companies. Seems like a problem that solves itself. If there's too many gas stations, they'll be weeded out by the competition, and the reverse is true as well; too few gas stations means opportunity for more to move in.

#5

I came to a slightly different method on this one than what was already given.

First, put 6 balls on either side of the scale. The group of 6 containing the heavier ball will tip down. Isolate that group.

Pick 4 balls at random from the remaining 6. Place 2 on either side of the scale. The group containing the heavier ball will tip down. If all 4 are the same weight, the scale won't tip, indicating that the heaviest ball is one of the remaining 2 not placed on the scale.

At this point there are only 2 balls left, so they just need to be compared. The scale will tip down for the heaviest.

Another version of the first one. You have 13 balls and a scale. One of the balls is either lighter or heavier than the others(the others are all the same weight).

How do you find out which one, only using the scale 3 times?

Use the same exact method as above; just take 12 of the balls at random and go. If by chance the two groups are equal during the first weigh-in, then the 13th ball you didn't pick is the one with an alternate weight.

 

[Bramblethorn]

Another version of the first one. You have 13 balls and a scale. One of the balls is either lighter or heavier than the others(the others are all the same weight).

How do you find out which one, only using the scale 3 times?

Ooh, haven't seen this one before... tricky, but I think I have it.

Use the same exact method as above; just take 12 of the balls at random and go. If by chance the two groups are equal during the first weigh-in, then the 13th ball you didn't pick is the one with an alternate weight.

That doesn't work, because this time around you don't know whether the odd ball out is lighter or heavier - so if the first two groups of 6 aren't equal, you don't immediately know which one to eliminate.

I think this works:

First weighing: Weigh #1-4 against #5-8. If equal, the odd ball is somewhere in 9-13, otherwise it's not.

Case 1: if 1-4 = 5-8, then we know the odd ball is somewhere in 9-13 and we also know 1-8 are normal. Second weighing: 1-3 against 9-11.

Case 1.1: if 1-3 = 9-11, then the odd ball is either 12 or 13. Third weighing: #1 against #12; if equal, then the odd ball is #13, otherwise it's #12.

Case 1.2: if 9-11 is lighter than 1-3, we know the odd ball is in 9-11 and we know it's lighter than the others. Third weighing: #9 vs #10, odd ball is the lighter of the two, or if they're equal then it's #11.

Case 1.3: if 9-11 is heavier than 1-3, then just take 1.2 and replace "lighter" with "heavier" throughout.

Case 2: 1-4 is lighter than 5-8, so balls 9-13 are all normal. Second weighing: (1-2 + 5-7) against 9-13.

Case 2.1: 1-2+5-7 lighter than 9-13: we know #9-13 are normal, therefore the odd ball is light, so it must be either 1 or 2. Third weighing: 1 vs 2, lighter one is the odd one out.

Case 2.2: 1-2 + 5-7 heavier than 9-13: tells us the odd ball is heavy, so it must be one of 5-7. Third weighing: 5 vs 6, heavier one is odd one out, if they're both equal then it's #7.

Case 2.3: 1-2 + 5-7 = 9-13: tells us the odd one out must be either one of 3-4 (and lighter than normal) or 8 (and heavier). Third weighing: 3 vs 4, lighter one is odd one out, if they're both equal then it's #8.

Case 3: 1-4 is heavier than 5-8: same approach as case 2 but switch "light" and "heavy" throughout.

 

[Handley_Page]

How big is an ice rink ?

 

[Zeb_Carter]

Question 1 is easily solvable. Questions 2,3 and 4 require only the methodology to be used to find the answer. Question 5 was solved only by Colleen Thomas. How can you do? (Answers later.)

1. You're in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You may inspect the room only once. How can you determine which switch is connected to which light bulb?

2. How many trees are there in Central Park?

3. How much does all the ice in a hockey rink weigh?

The same as it weighs when not in an ice rink.

4. If you were building a new city with a projected population of 100,000, how many gas stations would you need?



5. You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

...

 

[Zeb_Carter]

Here's another one then:

You have two identical glass balls. Dropped from a certain height (or above), they will always shatter; dropped from anything lower, they will survive undamaged.

You have a 36-floor building. You want to know what the maximum drop is that these balls can survive, to the nearest floor (it's possible that the answer is "less than 1 floor" or "more than 36 floors"). It doesn't matter if they both get smashed in the process, but obviously once you break both you can't do any more tests.

One way to do this would be to go to the first floor, drop a ball, retrieve it, then try from the second floor, then from the third, and so on, until a ball shatters. This could give you the answer in just one trial, but it might take as many as 36 trials (each drop = 1 trial).

What is the smallest number of trials in which you can guarantee finding the answer, to the nearest floor?

Once you have broken both balls, the answer is obvious.

According to Schrodinger the balls would have survived the drop from one level down from where they shattered.

 

[litmlove]

2. How many trees are there in Central Park?

Satellite photography with an image recognition algorithm. Google suggestions the answer is 26,000 trees in 843 acres, but that is an approximation from an official NYC website,

3. How much does all the ice in a hockey rink weigh?

Let the ice melt then measure the water. Measuring the ice depth & the rink's area is too much like an estimate.

4. If you were building a new city with a projected population of 100,000, how many gas stations would you need?

I would plan for gas stations to be within 10 minutes drive of everyone.

initially one gas station would be built in the city centre at the inconvenience of everyone. How many more gas stations are actually built depends on the population at the time and how much convenience I can afford.

 

[Hypoxia]

4. If you were building a new city with a projected population of 100,000, how many gas stations would you need?

If you're a socialist central planner, you have an established formula already, and you'll impose it despite actual needs and desires. If you're a free-market libertarian, you'll let corporate interests bid on desirable locations. If you're an ordinary pol, you'll take whatever bribes are offered for as many fuel points as possible. See, that's easy.