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Phil arrived at the Broadway theater. Mr. Smith, the director, had called him in for an emergency.

Phil: What is the problem, David?

David: Thank god you've arrived. We have a serious issue for tonight's show. Our electrical system has gone haywire and we need your help.

Phil: Electrical system? I'm not an electrical engineer. I don't know how I could possibly help.

David: Here's the problem: we managed to fix most of our lighting, but our biggest spotlights still have a serious issue. We have no idea what causes it, but it only started an hour ago, and the electrician said we would have to dismantle a large part of our rigging to fix it, which we don't have time for.

Phil: I still don't see where I come in...

David: Here's the problem:

Our 3 main spotlights that we don't use except for the finale are broken. They will work only for 6 seconds each before the fuse blows. The off-switches are broken, and the fastest we can replace the fuse when it blows is 10 seconds, and we need to start from the moment the light goes on, but your assistant is confident that it can always be done that fast. However, the rigging cannot support more than 1 person to replace a fuse at once, so they need to be replaced sequentially.

Our finale lasts 1 minute, and we need at least one light to be active during that entire period. If that's not possible, we need to know in advance how long we can keep those lights going before all the lights are without fuses. In theory, we can make the finale anywhere between 30 and 60 seconds long.

Phil: So, to summarize:

you only have 6 seconds per light, a fuse takes 10 seconds to replace, and the rigging can only support 1 person to replace fuses at a time. You wonder whether you can let the finale last the full 60 seconds or whether it needs to be shortened.

David: Correct. You're the best lighting expert we have. What's your verdict?

Can David still do the full finale? Or does he need to shorten it? And if so, how long can the finale last? Obviously, since this is a math problem, you need to "show your work".


Bonus: Friday the 12th of June, I will hand out a bounty of 50 reputation to the answer with the highest score that continues the dialog and gives the right answer.


Explanation for those that want the story behind this question

This core of this question is not about lighting. The original question I thought of was based around a mechanic from the video game World of Warcraft, more precisely the Priest's Angelic Feather ability:

Angelic Feather
40 yd range
Instant 10 sec recharge
3 Charges
Requires Priest
Requires level 30
Place a feather at the target location, granting the first ally to walk through it 60% increased movement speed for 6 sec.  Max 3 charges. Only 3 feathers can be placed at one time.

As you can see, one charge of the feather is equal in functionality to 1 light: once you activate it, it lasts 6 seconds, each charge has a separate 10 second cool-down period (called "recharge" here) and there are 3 charges. Charges in World of Warcraft have sequential cool-downs: they don't cool down at the same time, but one after another.

The reason I translated this to lighting instead of keeping it about the spell is because I assumed that people (especially those that don't play World of Warcraft) would understand the problem easier if I used more known terms like fuses and lighting than recharge, spells and duration.

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  • 1
    $\begingroup$ Is turn off and turn on instantaneous? If we choose to turn a light off while we turn another off at the same time does that satisfy the "at least one light to be active"? $\endgroup$ – LeppyR64 Jun 4 '15 at 11:28
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    $\begingroup$ @NateKerkhofs Suggestion: The off switches don't work, so each light is stuck on until the fuse blows after 6 seconds. There's a box of spare parts for each light, but for some reason it always takes 10 seconds to find and fit a new fuse. $\endgroup$ – LogicianWithAHat Jun 4 '15 at 13:49
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    $\begingroup$ What is this "cooling" people keep mentioning? I see no where in the puzzle indicating a "cooling". Only a "it takes 10 seconds to replace the fuse". $\endgroup$ – PiousVenom Jun 4 '15 at 14:24
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    $\begingroup$ @MyCodeSucks moving targets are hard to hit ;) $\endgroup$ – LeppyR64 Jun 4 '15 at 14:59
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    $\begingroup$ Lateral - Place a few fixed lights in the catwalk and have the prima donna pause in each one for 2 seconds each as she moves around the stage, extending the moving spotlight usage to 60 seconds. Or, better yet, park her in one location and don't use the follow spots. $\endgroup$ – Adam Davis Jun 4 '15 at 18:03
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This is the same solution as LeppyR64 and LogicianWithAHat, but I added a proof of optimality based on rand al'thor (who was working on a slightly different puzzle).

Phil: I'm afraid the best I can do is

36 seconds.

David: Hmm, well that's not great. But tell me, how do you do it:

Phil: Here's a schedule, light-by-light.

Light 1:

On from 0-6 secs. Cooling from 0-10 secs. On from 18-24 secs.

Light 2:

On from 6-12 secs. Cooling from 10-20 secs. On from 24-30 secs.

Light 3:

On from 12-18 secs. Cooling from 20-30 secs. On from 30-36 secs.

That gets you the lighting I promised.

David: I see. But our prima donna won't be happy. Maybe I should give rand al'thor's brother-in-law a call -- I'm not going to waste money on one of his lights, but maybe he can come up with a better way to use the ones I have.

Phil: No, actually I'm quite sure he can't do any better.

David: And how are you so confident?

Phil: Think about it like this:

Imagine each light has six units of coolness in it. When you turn it on, it burns the coolness at a rate of one unit / second, for six seconds. When you work on a light, it adds coolness at a rate of .6 units / second.

Got that so far?

David: No, wait. That doesn't make sense.

What if you turn on a light and start working on it at the same time? After 6 seconds it's used up the original 6 units, but you've put in another 3.6. So in your way of thinking about it, the light still has plenty of coolness left... but what I said was, the light will turn off after 6 seconds, even if you are working on it!

Phil: I see. Well, I can explain that...

When you're fixing a light, the light can't use any of the coolness you put into it, until you have put in all 6 units. So, let's agree that the coolness you put in while you are working on a light is only potential coolness: you put in potential coolness at a rate of .6 units per second. At the end of ten seconds, there are 6 units of coolness in the light, and 6 units of potential coolness are automatically converted to actual coolness.

David: So, in the scenario I was worried about,

After 6 seconds the light has used up all 6 units of actual coolness, and it has 3.6 units of potential coolness. So the light will turn off, and you can't turn it back on until it has 6 units of coolness, so the potential coolness can convert to actual coolness.

Phil: Exactly.

David: So why can't you get any more time?

Phil:

Well, let's think about the total coolness of all the lights, actual plus potential, all added together. At start the lights are all OK, so there are 18 units of coolness. Now whenever a light is on, the light is using 1 unit/sec, and you can only replenish .6 units/sec, so the total coolness goes down by at least .4 units/sec.

David: Wait... I'm confused again.

The lights use actual coolness, but you're replenishing potential coolness... and then potential coolness will convert to actual coolness...

Phil: Don't worry about that.

We're counting the total coolness, potential and actual all added together. So when potential converts to actual, it won't affect our total.

David: I see.

Phil:

Anyway, to turn on a light the light has to have 6 units of coolness, so certainly you have to have at least 6 units of coolness total. But starting at 18 units, and losing .4 units a second, you know that after 30 seconds...

David:

After 30 seconds you have lost 12 units of coolness, so you have exactly 6 units left!

Phil:

Exactly. Which means that after the 30 second mark, you won't have enough to turn on a light. No matter how you plan it, the latest you can turn on a light is 30 seconds... and that means the light will go out by 36.

David: So you really did squeeze every last second! Phil, you saved my day.

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  • $\begingroup$ Nice work with the spoilers and such. I think this one is actually worth getting the bonus more than the one I planned originally. $\endgroup$ – Nzall Jul 2 '15 at 7:55
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rand al'thor had the answer to the initial question, where the lights had to wait to be off before repairing.

Phil: I cannot do the full finale, but I can give you

36 seconds of continuous light!

A light takes 10 seconds to repair and provides 6 seconds of light. All lights are available at the beginning.

Light 1: On at 0, Off at 6, Available at 10
Light 2: On at 6, Off at 12, Available at 20
Light 3: On at 12, Off at 18, Available at 30
Light 1: On at 18, Off at 24, Available at 40
Light 2: On at 24, Off at 30, Available at 50
Light 3: On at 30, Off at 36, Available at 60
Light 1: Not Available at 36
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  • $\begingroup$ This is optimal. After 36 seconds, there will have been 6 fuse breakings (one after every 6 seconds) and at most 3 fuse repairs, so at this point, all fuses will be broken. $\endgroup$ – Mike Earnest Jun 5 '15 at 3:42
  • $\begingroup$ @MikeEarnest The puzzle doesn't place any limitation on the number of fuse repairs. You might be confusing the puzzle's definition with the source of the inspiration for it, where there was such a limitation. $\endgroup$ – mechalynx Jun 5 '15 at 15:07
  • $\begingroup$ @ivy_lynx he means that the fuses in all three lights will be broken. $\endgroup$ – LeppyR64 Jun 5 '15 at 15:08
  • $\begingroup$ @LeppyR64 Right, sorry, I interpreted the comment differently. $\endgroup$ – mechalynx Jun 5 '15 at 15:09
  • $\begingroup$ Can you start changing Light 1's fuse at a time of -4? $\endgroup$ – Joel Rondeau Jun 5 '15 at 15:43
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Phil: It's impossible, I'm afraid.

Think of it in terms of how many "secondsworth" of light you have available. You start off with 3 cool lamps - that's 18 seconds' worth. For the first 6 seconds, you're using one lamp to light the stage, so there's nothing to cool. Then you have 12 seconds' worth of light left.

It's going to take you 10 seconds to earn another 6 seconds' worth, and you need to use up 10 seconds' worth of light in that time, leaving you with only 8 seconds' worth.

Now you move the cooling system to another light, but by the time you get that cool, 10 seconds will have passed and you only had 8 seconds' worth of light to play with, so 2 seconds must have been dark.

The most light you can get is 24 seconds, which is less than the 30-second minimum you mentioned.

If you like, my brother-in-law could do you a good deal on a fourth spotlight...

David: I called you in to advise me, not to advertise at me! Everyone knows your brother-in-law's lights are dodgy! Thanks for the advice; we'll have to either cut out some of the primadonna's finale or let her do some of it in darkness.

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  • $\begingroup$ I just realized that I made a fundamental mistake in my question. you can already start cooling the lamps DURING their lighted period. it still takes 10 seconds to cool down, and they can still work only 6 seconds, but you don't need to wait for the lamp to overheat before you can start cooling. $\endgroup$ – Nzall Jun 4 '15 at 11:27
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Phil: I don't know if I can get you your full minute, but:

I think I can do 36 seconds
Turn lamp 1 on for 6 seconds, whilst cooling it
Turn lamp 2 on for 6 seconds - continue cooling lamp 1 for 4 seconds, then switch the cooling device to lamp 2 for 2 seconds
Turn lamp 3 on for 6 seconds - continue cooling lamp 2.
Turn lamp 1 on for 6 seconds - coll lamp 2 for another 2 seconds, then switch the cooler to lamp 3 for 4 seconds
Turn lamp 2 on for 6 seconds. Once this is done, lamp 3 should be cool, but you don't have time to cool anything else.
Turn lamp 3 on for 6 seconds. That should give you a total of 36 seconds of light

Are you sure you're ok with cutting the performance down by that much? I'll sit down with the calculations and see if I can get you a little longer...

David: That's fantastic! If you can get a few more seconds I'd greatly appreciate it, but at least now we know that the show can go on!

Phil: I'll do what I can. In the meantime, here's an diagram that should help explain it to any confused techs. Green cells represent a second of a light being on, blue a second of the cooling mechanism

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David: I can't quite get you the full minute - 45 seconds is the best I can do, and here's how:

Use the cooling system on the first light even as you use it. If it would cool it from critical in 10 seconds, and heat to critical in 6 seconds without cooling, then with cooling it should last 15 seconds. We have three lights, so we can switch twice and get 45 seconds in total. If you can boost the cooling system just a bit more, and get it to cool in 8.5 seconds, that will be enough to get the full minute. Any chance of that?


The reasoning here is simple.

It cools by $\frac1{10}$ of the difference between "cool" and "critical" each second, and heats by $\frac16$ of the difference when operating. So the rate of heating with the cooling system is $\frac16 - \frac1{10} = \frac{10 - 6}{60}= \frac1{15}$, and thus one light will last 15 seconds. After the 15 seconds, the light will be at critical, and it won't cool by much without the cooling system, so it will effectively stay at critical, making that light no longer usable.

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  • $\begingroup$ As mentioned in the question, the cooling system and overheating don't affect each other. Cooling a light always takes 10 seconds, and a light always overheats in 6 seconds. Cooling a light that is on does not make it overheat later. $\endgroup$ – Nzall Jun 4 '15 at 13:06
  • $\begingroup$ So what if you have the light on for 4 seconds, and then switch? I think there's a problem with exact definition of what's going on, here. I assumed that, so long as the temperature is sub-critical, the light will continue to work. It takes 6 seconds to go from cool to critical, without the cooling system. If it always turns off after 6 seconds, then does that mean that the light will have to be off for 4 seconds more after that before it's usable again, if the cooling system were directed at it? $\endgroup$ – Glen O Jun 4 '15 at 13:14
  • $\begingroup$ I'm currently working on a clarification in my question that will explain why it doesn't work like that. Short answer: this is not the original problem, it's a translation of a game mechanic, and some elements turn out somewhat weird while translating. I translated it because it would be easier to explain to most people how it works using lighting than using the original mechanics from the game. $\endgroup$ – Nzall Jun 4 '15 at 13:18
  • $\begingroup$ @NateKerkhofs What game? I'm interested. $\endgroup$ – LeppyR64 Jun 4 '15 at 13:27
  • $\begingroup$ @LeppyR64 World of warcraft. I explained some more in the question. $\endgroup$ – Nzall Jun 4 '15 at 13:29
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I think your spotlight analogy does not match up very well to the World of Warcraft scenario, for two reasons:

1) It seems electrically hazardous to change a fuse on a light that's still "hot"
2) If I knew I was going to blow a fuse after 6 seconds of use, wouldn't it make sense to shut off the light after 5 seconds and use a different light?

Anyhow, as a retired WoW player, thinking of the scenario in that context makes more sense to me. The WoW solution depends on whether you can "pre-drop" feathers before starting. If yes:

T = -30: drop feather A
T = -20: drop feather B
T = -10: drop feather C
T = 0: consume feather A, drop feather D (2 charges left)
T = 6: consume feather B, drop feather E (1.6 charges left)
T = 12: consume feather C, drop feather F (1.2 charges left)
T = 18: consume feather D, drop feather G (0.8 charges left)
T = 24: consume feather E, drop feather H (0.4 charges left)
T = 30: consume feather F, drop feather I (0 charges left)
T = 36: consume feather G (0.6 charges left)
T = 42: consume feather H, drop feather J (0.2 charges left)
T = 48: consume feather I (0.8 charges left)
T = 54: consume feather J, drop feather K (0.4 charges left)
T = 60: consume feather K, drop feather L (0 charges left)
T = 66: consume feather L (0.6 charges left)
T = 70: drop feather M (0 charges left)
T = 72: consume feather M (0.2 charges left)
T = 78: timer of feather M expires and we are still 2 seconds before we can drop another feather. Done.

If you're not allowed to drop feathers before starting the timer, then it works as follows:
T = 0: drop and consume feather A (2 charges left)
T = 6: drop and consume feather B (1.6 charges left)
T = 12: drop and consume feather C (1.2 charges left)
T = 18: drop and consume feather D (0.8 charges left)
T = 24: drop and consume feather E (0.4 charges left)
T = 30: drop and consume feather F (0 charges left)
T = 36: feather F runs out and we still have 4 seconds before we can drop again. Done.

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  • $\begingroup$ Yeah, I originally used an overheating example instead of fuses, but that lead to something that almost violated the laws of of physics (or maybe even completely) and people interpreting overheating and cooling down as interacting with each other. But fuses lead to similar issues. I think I should have just stuck to the original question I wanted to ask, namely the one you answered. $\endgroup$ – Nzall Jun 4 '15 at 19:55
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If we assume that everything is done perfectly(so that we can cooldown a light immediately after we done cooling the previous one) we can certainly get at least 45 seconds of show. Here's how:

Start with light 1 and cool it at the same time. Assume that light has 60HP, so each second of work will consume 10HP and each second of cooldown will add 6HP. So that leaves us with a constant HP loss of 4/sec. Because the light has 60HP, it will take 15 seconds to get down to 0($60/4=15$).
Then use second light exactly the same till it's 30 seconds total.
And last use the third light for another 15 seconds which will sum up to 45 seconds in total.

I'll update my answer if I find a way to get more time

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  • $\begingroup$ I like your answer and +1'd it but it seems like it's incorrect in the same way that @glen-o solution is. $\endgroup$ – LeppyR64 Jun 4 '15 at 13:17

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