MPMSolution: How much fuel does the steam train need?

[@stefanwaal]

You shouldn't give participation points for people who gave it a go. You should give them Parker Points!

[@giadamoretti4819]

I WANT PARKER POINTS NOW

[@andymcl92]

Featured in a Matt Parker video: win ✔️ Got voice muted: win ✔️ Got cut before my shuffling got shown: 😭 ✖️ Thanks for more great work, Matt :)

[@andymcl92]

Also, the 2000 solution (the mode) was what you get if you assume discrete units of 100 fuel :)

[@JNCressey]

2:41, well, technically, 1733.33 isn't enough. you'll be 0.003333333... miles short. (that's almost 3 social distances)

[@cH3rtzb3rg]

If my math is correct, and assuming you optimize your drop-off stations for the amount of fuel you have, you'd be just 1/2100 miles short (no idea what a "mile" is, though). But I was actually struggling about whether I should round my solution up or down ...

[@1Resare]

I was at peace with answer ending with .33 given that 0.99 == 1.00 in 2 decimal precision and .33 is a third of that. kzbin.info/www/bejne/qIWUpHyMptShe5o

[@alandouglas2789]

JNCressey social distances? You Americans will use anything except metric

[@JNCressey]

@@alandouglas2789, it's just a joke, fam. I'm actually from the UK, and we're being recommended to keep more than a "social distance" (about 2 metres) away from other people in public.

[@variousthings6470]

@@JNCressey 1 social distance = 2 metres 1 square social distance = 4 square metres Area of Wales = 2.0798×10^10 m^2 So Wales is 5,199,500,000 square social distances

[@Imthefake]

parker proof: if i make a youtube video saying something and nobody writes me an email proving me wrong it must be true

[@andymcl92]

CF superpermutations :)

[@zinsch1]

Hey, I got featured in the video! If I had known, I would've put more effort into making the spreadsheet look nice. :)

[@Sahil-oq8ki]

same - i definitely would've used desmos properly if I had known it would be in the video!

[@SenselessUsername]

Jakob "Input"? Sounds suspicious...

[@daminox]

This MPMP thing is really cool. Maths teachers around the world should take a few minutes every week to show their students your newest puzzle and have them submit an answer. This segment should be way more popular than it is!

[@SimonTiger]

12:18 That's the Coding Train! Btw, you spoke with Daniel, a few years back.

[@SimonTiger]

I think you are, like, 2 degrees of separation with him. Link to his channel: kzbin.info/door/vjgXvBlbQiydffZU7m1_aw. I am also a personal friend of him. * Also, the graphics was done by Jonthan Feinberg (Coding Train, yell if I spelled that wrong).

[@SimonTiger]

Btw, I'm talking about Daniel, not whoever else there was in that animation.

[@spectralpiano3881]

When someone submits a correct answer after submitting an incorrect one, I think it would still be fair to give them 1000 points without any of the speed points. They gave it a double go after all :)

[@ascensionblade]

1000 is a perfect square, so I'm not sure if Matt would like that ;)

[@ohay12]

I got fourth place on this one! I initially found 2000, but paused on the submission page after seeing the decimal places note and eventually found the correct solution via Eugene's method. Great little puzzle!

[@squidygoo6613]

Matt in the video can you explain your process of solving the puzzle?

[@zinsch1]

Doing a spreadsheet helps. You can play around with different checkpoints to get a feel for how it plays out.

[@zacharybarbanell1064]

TBH I don't know how you would find the answer to this puzzle without finding a general solution

[@swerasnym]

@@zacharybarbanell1064 My thought process was the following: First I noticed that the train will travel an odd number of times over each part of the track. Then I realized that you want to maximize the amount of track where you travelled the fewest possible times, i.e. 500/T miles travelled T times thus you end up with a maximum of 500*(1+1/3+1/5+1/7+...+1/T) miles possible to travel when travelling a maximum of T times on any piece of track. The reason for the odd number of times is that you have to travel the whole rack the last time, but need to travel back over an area every time you drop coal off. With this in mind you can find that 500*(1+1/3+1/5) = 766.66 < 800 < 500*(1+1/3+1/5+1/7) = 838.01 so thus we need to use a full load of coal 3 times (1, 1/3, 1/5) e.g. 1500 units of coal and 7*(800 -766.66) = 7*33.33 = 233.33 for the last bit, ending up with 1733.33. This is basically what was animated in the fist animation shown. As a sanity check that it is possible to travel an infinite distance using this method we can notice that the odd numbers contains all primes > 2, and the known result by Euler that the sum of the reciprocals of the primes are divergent. en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

[@zacharybarbanell1064]

@@swerasnym RIght I get that, that's basically how I did it too, but my point is that in order to find the solution for 800 you had to find the general solution.

[@1Resare]

@@zacharybarbanell1064 I think finding a general solution, does not necessary mean that you've found an optimal solution. Finding an optimal solution may require one to disregard a number of general solutions, until major contributors are identified and solution developed with those in mind.

[@Pulsar77]

I'm one of those who entered 1933.33 and then found the correct solution. Lesson learned :) Great puzzle.

[@jw41538]

I got 2000 in my head and then solved the next day on paper, as I assumed it was all about speed points.

[@Pulsar77]

@@jw41538 Same. I won't be doing that again, from now on I triple-check before I submit anything :)

[@AlRoderick]

Bonus round: minimum amount of fuel required if you need to make the whole round trip without refueling at party Town. Is it twice as much? More? Less? Because functionally this is similar to how one might do something like a polar expedition, where you'd be establishing base camps and storing food and supplies at each of them on the idea that you'd pick them up for the return trip while you're coming back but you're constantly lightening the load as you go.

[@1Resare]

It will take 14 trips and consume 6639.06 units of fuel. Segments will be at 4.96, 24.19, 45.03, 67.75, 92.75, 120.53, 151.78, 187.5, 229.16, 279.16, 341.66, 425, and 550 miles Extra bonus, how much fuel (minimum) do you need to rescue a second train from a party town (if the party town has no fuel) and the first train has to be back to the depot to clear the tracks.

[@dominiquefortin5345]

That would mean depositing 1733,333 fuel at the 800 mark and using that fuel to come back at the 0 mark. In that case, you would need 42239,44 fuel total and there would be 84 segment on your way to the 800 mark and 3 segment on you way back.

[@boomiTech]

Code train by Daniel Shiffman! at 12:22

[@SenselessUsername]

What stressed me out was the mixed-up rainbow in Partytown... By now you should know the colours of rainbows out-to-in from NHS rainbows in windows. (Although, "If we've learned one thing from Covid-19, it's that kids are sh#t at drawing rainbows".)

[@konstantinkh]

I did eventually come up with general solution that matches what's been described in the video, but at the time of submission, I was still at a point where I had some thoughts on that, but not a full algorithm nor general formula. What I did have is a list of linear constraints and a linear objective function. So I did what any software engineer with computational math background would do with such information. I converted it into a linear programming problem and stuck it into a solver. While that only works for specific distance, 800 miles in this case, it does guarantee that the solution I got was optimal within numerical error, which was way better than 2 decimal places, so I was happy enough submitting it. The full, general solution, along with proof that you can take this to arbitrary distance at an approximately exponential cost came to me the next day.

[@dominiquefortin5345]

Using the relation Xn-1 = Xn - (Cap*n - Sn)/(2*n - 1) where Xn-1 and Xn are positions so that Xn-1 < Xn, Sn is the stock pile at position Xn, Cap is the train’s capacity (here it is 500) and n is the number of trips forward between Xn-1 and Xn. Starting from the end with the values Xn = 800 and Sn = 0, we find the first Xn-1 < Xn by increasing n starting at 1 and we find Sn-1 = Sn + (2*n - 1)*(Xn - max(0; Xn-1)). Rinse and repeat until Xn-1

[@abcdefg9213]

Regular people Wrong solution for this problem Right solution for the current one Me Right solution for this problem Rushed a wrong one for the next

[@philipG2337]

No guarantees for the correctness of the closed form, though. ;) Also: The fuel you need to get x Coal Capacities far can be approximated quite well by e^(2x-2). see www.desmos.com/calculator/i0sedbfilq

[@Macieks300]

Thanks for sending a proof! You should get like 1000 bonus points for your solution.

[@philipG2337]

@@Macieks300 I didn't proof it to be fair, matt twisted that a bit. I even said that it is not a formal proof. But what i did find, is that the fuel needed can be well approximated by an exponential. I thought he would focus more on that part if my submission was chosen, but I guess a diagram is more photogenic ¯\_(ツ)_/¯

[@KuraIthys]

Since I'm more inclined towards physics and engineering than pure mathematics... What the formulation of this puzzle makes me think of is this: If you've got a train that runs on coal or diesel (electricity doesn't work, but then it wouldn't in general since battery trains aren't really a thing yet), there is a fuel capacity in the locomotive itself (or any associated tender), but of course, a train can pull a load, and this can include coal hoppers or fuel tanks... That means your fuel limit is flexible, from the capacity of the locomotive itself, up to the weight limit that the locomotive can reasonably pull (obviously that weight limit isn't simple to define either, since it would vary with speed, the braking equipment, if there's any hills along the route, etc) However, if we simplify and assume the train can pull x tonnes, then every extra wagon of fuel we take along is y tonnes less we can take with us. (some for the weight of the extra wagon, and separately for however much fuel we load into it.) If we're just trying to get ourselves (the train crew) to party town then that won't matter. We just load up however much fuel it will take to get us there. However, if we have to take some cargo or passengers, then we get a related, but different problem. (assuming the maximum load the train can pull isn't simply so amazingly huge that we can easily take anything the whole distance. - though of course, fuel usage is likely related to weight hauled as well, just to confuse matters further) Now the question becomes if we can load some mixture of fuel and cargo/passengers, assuming we can't get our entire load of cargo/passengers to the destination in one go, what is the optimal approach? Do we take passengers/cargo part of the way then go back for more? Do we first do a run to move nothing but fuel somewhere partway, then go back and take a mixed load of passengers/cargo partway, refuelling somewhere? Anyway, totally overthinking it... But it's actually an interesting variant of the problem to think about... If you can easily take extra fuel along to cover the entire distance, but also need to take something else along that limits the amount of fuel you can take for a given trip...

[@aliceanderson5154]

I remember seeing this problem illustrated in a film. A group lost in a desert had water and ostrich eggs. They had to work out how to deposit caches of water to get to a rescue station. Might have been "Flight of the Phoenix".

[@abafaba]

The "hint" of decimal places psyched me out to the point I never got to an answer and forgot to get an guess in before the due time. 1850

[@1Resare]

Just realized that there is, in fact, a limit to how far you can go using an optimal method. It is 47.5 times the fuel size. For any distance further than that the first refueling stop would have to be closer than a plank length away from the depot.

[@Macieks300]

I sent a nice spreadsheet where you can input any distance and any capacity and it gives you the answer. It gave 1733.33333 for 500 and 800 so I think it was correct.

[@MrMaelstrom07]

Matt, please do a followup on the 3-sided coin. What did you find?

[@gregchamblin1965]

I didn't know the challenge videos would be on this channel until I found the solution video.

[@bbgun061]

He's doing some on this channel and some on his other, which is mildly annoying.

[@gregchamblin1965]

@@bbgun061 Well I know that now! LOL

[@Hwd405]

Hi Matt, I appear to have been split in two on the spreadsheet! Both Hwd405 and Hwd405 A are me (I assume? Presumably I'm not facing identity theft)

[@kane2742]

Maybe you used different email addresses for the submissions (or meant to use the same one, but made a typo one time)?

[@Hwd405]

Kane Casolari typo sounds likely, I'm sure I used the same address though because I've been receiving emails about them from the same address since the beginning

[@Tondadrd]

You can post the workings out on the website. Automatically, just check it is either jpg, link to a KZbin video or some other non security material. And maybe scan through them to check it looks like a solution :o.

[@t71024]

I'm one of the lucky 288 but I'd still like to see your explanation just to assure myself I did it correctly and it wasn't just a fluke. I did it with a 5-row spreadsheet but I was too lazy to figure out where to email it.

[@asafattia9000]

Why dont you explain the solution? I'd love to see that

[@Booone008]

What's the formula to compute the weighted scores on the scoreboard? Even the full spreadsheet only shows points1, points2 and weighted result without the formula.

[@Booone008]

Currently it seems to be (0.8 * points1) + points2, but how will the weighting change for future puzzles? 0.8^age?

[@Quantris]

New from the makers of the Parker Square: Proof by KZbin

[@alfiestoppani]

I spent ages on this problem, now I realise I misunderstood it. I thought we were finding the minimum tank limit, not total fuel use. Damn.

[@WillMoff0]

i wish this wasnt the first video of this kind to pop up in my suggested :(

[@antisocialbob968]

Your point about only taking the first entry is a fair one. However my suggestion would be to give people say 3 chances like the Mathsbombe by the University of Manchester. (maybe a penalty of 100 speed points could be applied too)

[@antisocialbob968]

Ps. this also could tell the user they are wrong which would mean they can carry on working on the problem not stop thinking they are correct.

[@Idran]

I have to admit, it still wasn't clear how to submit anything _but_ the answer. The email address callout in this video helped for future reference, and I see it on the main puzzle page now, but it'd be helpful if that was included alongside the Google Forms submission on the individual puzzle pages too.

[@andymcl92]

I wonder if that's so it's harder for spambots to find. I imagine the range of files sent means the spam filter has to be pretty open so stuff doesn't get missed

[@Idran]

@@andymcl92 There's almost no kind of obfuscation you can use nowadays to keep spambots away from email but still let humans know how to use it, so spambot obfuscation doesn't really achieve anything anyway But also having a link on one page but not another wouldn't even do that much, since spambots just spider entire websites, they don't stop at individual pages

[@andymcl92]

@@Idran Ah, sorry, I missed that it's on the puzzle's page and didn't read your comment properly. I thought you meant it only featured onscreen in the video. I take it back!

[@Macieks300]

@@Idran The e-mail was in the video description. I didn't have a hard time finding it.

[@TheCrewdy]

2000 was the answer I gave which I think is the best option if you only stop to drop off fuel at exactly multiples of 100 - It did not even cross my mind that a train can stop if it isn't at a multiple of 100. In any case, it's best to be fashionably late to Party Town so 2000 is probably the socially optimal solution.

[@KuraIthys]

Lol. That's a pretty weird assumption to make, given how arbitrary the distances between train stops can be in reality. If you tried this with a real train... Well, you'd get a huge number of complicating factors about loading the train and such. And you almost certainly couldn't leave fuel in any arbitrary location you felt like. But the places you COULD drop fuel certainly wouldn't be multiples of 100... Constrained, yes. But not in any such neat numerical pattern.

[@calmelbourne]

I just used solved the integral for continuous amounts of fuel on the Wikipedia page for the Jeep Problem.

[@tompov227]

The coding train artwork!!!

[@jeffreyblack666]

The problem with you assuming it would have been fine to keep submitting is how many people just didn't bother submitting because of them? What is the point in trying when I am already 1000 points behind and am unlikely to get any significant amount of speed points?

[@franciscowagnerdemoura3912]

The solution 1733.33 is the most efficient way for the train to travel 800 miles. Is the reciprocal true? Like, is 800 miles the max distance the train can travel with 1733.33 units of fuel?

[@cheaterman49]

12:24 I'm surprised you never heard of the Coding Train Matt, I'm sure you'd love it :-) I sure did recognize the graphics and Daniel's surname hehe.

[@johnchessant3012]

Damn, I foolishly assumed that you could only drop fuel at 100, 200, 300, etc.; that's how I got the answer of 2,000.

[@andreacolongo8094]

Me too dude! Same mistake

[@TheCrewdy]

@@andreacolongo8094 I think a few of us were on that track.

[@fpth848]

Yeah, I did that too

[@perry4754]

Dang, I got close. I think I had my math wrong, found a solution, fixed the math, but then double check that my solution was truly optimal. I ended up making stops at 50, 83.33, 133.33, and 300

[@Pulsar77]

Here's my solution, which has a step-by-step explanation: www.docdroid.net/tDUhrnN/steam-train-puzzle-pdf

[@severnsevern580]

excellent.

[@bobzhang9966]

thank you so much for sharing your excellent solution Pulsar77!

[@ChrisBryantMusic]

Is there a way to solve it in a continuous rather than discrete way? By dumping and picking up fuel at every infinitesimal location with some continuous mass distribution

[@1Resare]

This is what the first illustration showed. The black lines drawn on the outward journey were representing numerous continuous layers of fuel, not discrete pickup locations. Number of layers is still discrete though, since the train cannot use partial fuel for movement.

[@ChrisBryantMusic]

@KZbinr Oh, got it, I understand your solution now! The way Matt explained it didn’t seem to have the same interpretation. If I understand you correctly, each black line is a uniform distribution of fuel that you lay down as you move forward. They turn red as they are permanently depleted. Right?

[@1Resare]

@@ChrisBryantMusic Yes :-) The fuel consumption on the new segments is a multiple of the layers needed for that segment, that is why the full tank gets used up faster for segments where more layers are needed. Having a train that can run on empty or keep the tank full, as long as there is a previously laid layer available to burn, helped me to think of fuel requirements during an individual segment, since at that point fuel to get to and from the segment would not come from the onboard storage.

[@1Resare]

By the way, this analogy works with arbitrary distances. To get to 2000 miles for example, your first segment would require 837 layers (if you include initial depletion layer), thus that segment can have a maximum length of (500 fuel capacity / 837 layers) = 0.597 miles, but you would start the journey with only 211 miles worth of fuel. This is because the first segment would be only 0.252 miles long, rest is covered by sum (500 / (2n-1)), n=1 to 418 Reason why the number of layers needed is increased by two for each segment (thus 2n in calculations) is that you have two additional trips to get to the end of the segment and back, in order to lay down the fuel needed later on.

[@loreleihillard5078]

That's the method I used and landed at 1733.33 I ended up starting from the 300 mark and working out how much fuel it would require if I started a set distance beforehand and wanted 500 miles of fuel to be dropped off at that point, then repeating until I got back to the starting point. then with a little help from limits, I ended up at the solution that everyone else got, and what I saw as the lower bound. Didn't actually come up with a way to do the trip in a finite amount of moves bc I was worn out and I was up at night until like 5am (as Matt said, time zones are not so kind) but I got the answer I was looking for

[@michaelcastelpoggi1735]

Didn't make it on this time. Just more motivation for the next puzzle!

[@kayleighlehrman9566]

I'm here for ProofTube

[@FiniteJest]

Just change your first name to Maths so that the title of this series is Maths Parker’s Maths Problems. It just rolls off the tongue.

[@henrikwannheden7114]

Just a stab in the dark here, not a mathematician at all: What about dropping and refilling continuously as you chug along? Would this be beneficial, detrimental, or not really effect anything at all?

[@d1rcwill]

Like... for every 1 unit of distance you travel, you also drop off 1 unit (or 2, 3, n) of fuel? That's functionally the same as stopping and dropping off large chunks of fuel. If the efficiency of the train changed with the load... THEN it might change things. At that point I think you would be looking at something that resembles rocket fuel calculations.

[@alexbanks9510]

Ok but can we actually have a step by step solution? I've looked through others and the only ones that do step by step get 1933.33

[@loreleihillard5078]

Leg 0: start at 0 with 1733.33 fuel Leg 1: Travel from 0 miles -> 33.33 miles 7 times delivering 1500 fuel and using remaining 233.33 fuel for traveling ( 4 forward trips, 3 return trips) Leg 2: Travel from 33.33 miles -> 133.33 miles 5 times delivering 1000 fuel and using remaining 500 fuel for travel (3 forward trips, 2 return trips) Leg 3: Travel from 133.33 miles -> 300 miles 3 times delivering 500 fuel and using remaining 500 fuel for travel (2 forward trips, 1 return trip) Leg 4: Travel from 300 miles -> 800 using the remaining 500 fuel

[@1Resare]

You will need to take time to understand the ratio of (1 / (2n-1)), as soon as you do, solution will follow. This ratio describes a decreasing distance a train can travel if it needs to provide fuel for n number of trips. So, if you only need to worry about 1 trip, than the train can travel (1 / (2*(1) - 1)), in other words (1/(2-1)), aka 1 mile per unit of fuel. This means that if the trip length is lower than the range of a full tank, than your trip length will equal to your minimum fuel requirements. What if the distance is further than the range of a full tank of fuel? Your send trip efficiency will drop to (1 / (2 * (2) - 1)), in other words (1 / (4-1) ), aka 1/3 miles per unit of fuel. This is because during the second trip, the train will have to travel a 3rd of a way, drop of a 3rd worth of fuel for the other trip and return a 3rd of a way back. Notice how your "used" fuel increases by 2, thus "2n" term in the calculations. This is because it now takes an additional round trip (to the refueling point & back) in order to have that fuel be available for the final trip. In order to calculate the distance we can travel so far, after two trips, we can use a mathematical sum operation, as in: Range * sum ( 1 / (2n - 1) ) for n = 1 to 2 In our problem, this would mean that after two trips we can cover: 500 * ( 1 + 1/3 ) = 666.66 miles Had our target been within 600 miles (as original party town was), we could stop right there, we know we would need to take two trips to get to the party town with 66.66 miles of fuel still in the tank. Had we taken on the full 1000 miles worth of fuel and maximized the second trip where the fuel efficiency is at 1/3, we would only save 66.66 miles worth of fuel (as our other trip has efficiency of 1 mile per unit of fuel). On the other hand, reducing the trip with 1/3 fuel requirements by 66.66 miles, would gain us 66.66 * 3 = 200 miles worth of fuel savings, so it makes sense to maximize the segments with higher fuel efficiency and minimize the one segment with the minimum fuel efficiency. Now the question becomes, which trip is which? How do we know where to drop off the fuel? Math says that the 1st trip is the most efficient, so it makes sense that the first trip is the first time the train leaves the station, right? Sadly, no. Imagine that the train started on the 1st trip with the highest fuel efficiency. That means that it can travel 500 miles on a full tank of fuel, if you end up at the 500 miles away from the fuel depot, how are you going to get back? On the other hand, if you end up at the party town, you are done, no matter how far you are away from the start. Because of this, we count our trips backwards from the goal. Our first, and most efficient, trip is the one closest to the party town. Here we do not care about a return trip and we can burn all of the fuel available to us to get to the goal. Our second trip, on the other hand, has to worry about getting back for the refuel AND drop off the fuel to use on the 1st trip. You can think of it as credit. You can make one run, non stop, from start to finish, but you will end up using more fuel than you've got. Your 2nd, 3rd, 4th, etc... trip will need to settle your bill and deposit the fuel you've borrowed to make the first trip. Ok, I've covered the "2n" part of the (1/(2n - 1)) ratio earlier on, but what about that negative one in the denominator, that works in our favor? That represents that we don't plan to make a return trip when we reach our destination. You subtract that trip not taken from every step of the way, since it would not require any fuel. There was a "Bonus round" question by Alexander Roderick which asked how much fuel one would need to make a round trip to the party town and back, if we could not refuel at the party town itself. To solve this, we would simply get rid of the "negative one" credit within our calculations and redo all of the steps using (1 / 2n) ratio. By the way, 1 in the numerator of the ratio represents a full tank of fuel, in case it is not clear. This way solution is the same, no matter what range the fuel tank has. Going back to the original problem, the "Final" refuel stop will be at the ( 800 party town location ) - (500 fuel range * (1 (full tank) / (2 * 1 (1st stop) - 1 (no return trip) ) = 800 - 500 * 1 = at the 300th mile. The second fuel stop will be (300 (n-1 stop) - (500 fuel range * (1 (full tank) / (2 * 2 (2nd stop) - 1 (no return trip) ) = 300 - (500 / 3) = 133.33rd mile marker. The third stop will be (133.33 (n-1 stop) - (500 fuel range * (1 (full tank) / (2 * 3 (3rd stop) - 1 (no return trip) ) = 300 - (500 / 5) = 33.33rd mile marker. Finally our last stop will be (33.33 (n-1 stop) - (500 fuel range * (1 (full tank) / (2 * 4 (4th stop) - 1 (no return trip) ) = 300 - (500 / 7) = -38.09 mile marker. The refuel location can be generalized by function f(x) which is equal to (800 target distance) - (500 full tank range) * sum (1 / (2n - 1) ) for n = 1 to x Wait a minute, our 4th stop has overshot our starting point, this means that we don't need a full tank in order to reach the 33.33rd mile and can take just (33.33 * (2 * 4 (4th stop) - 1 (no return trip)) = 233.33 miles worth of fuel. Generalized, the fuel for the trip should equal a minimum between (full tank) and the (distance to the (n-1) refuel point * (2n-1)). On every trip, besides our initial departure from the fuel depot, our fuel use will be equal to the full tank of fuel worth, whether we are burning it to travel, or dropping it off for later. On the shortest segment though, we may end up incurring fuel savings, as our fuel tank will be sufficient to meet the fuel needs without the full load.

[@achtsekundenfurz7876]

> There was a "Bonus round" question (...) if we could not refuel at the party town itself. > To solve this, we would (...) redo all of the steps using (1 / 2n) ratio. That sounds like a minor change, but would take a fuel depot at 550, which would be very expensive to build up. A similar fuel plan was used once in real life, for a mission in the Falklands war IIRC. (I'm not counting space missions; those use delta v instead of distance.) Another interesting question would be to plan a round trip if we could refuel at both ends. In that case, it would make sense to put some excess fuel on the way to Partytown for the return leg, somewhere close to zero, since that would be more efficient than getting the whole fuel for the return leg from Partytown.

[@NashvilleMonkey1000]

simply combine the rocket equation with a binary/halves algorithm.

[@achtsekundenfurz7876]

I can't believe how far I had to scroll to find a comment about the rocket equation. Forget distance, put velocity there, and there you have it. Its not rocket science, people! *. . . WAIT.*

[@zacharybarbanell1064]

Hey I made it into the top 200! Nice

[@NetAndyCz]

My thinking for this puzzle was, it is probably most fuel efficient to do 500 miles in a row, but getting 500 miles of fuel to 300 miles point will require probably over 1000 miles of fuel and I am too lazy to count that to two decimal places and there will be probably some fractions:p

[@rtpoe]

I wonder what a histogram / distribution of the answers would look like.

[@ten.seconds]

I was convinced that there can't be any subdivision of 100 in the solution. I was wrong. I also wrote a program that works with integer steps only, so I guess that a mini Sapir-Whorf situation happened with me.

[@bbgun061]

I tried three times, with initial stops at 200, 100, and 50 miles. 200 was the worst, 100 and 50 were equal. I suspected there was a more optimal solution, but I had no clue how to start finding it.

[@poohstix16]

I broke my brain similarly, considering the possibility of dropping and retrieving continuous amount of coal, thereby needing calculus to solve (so my thinking went). And I wasn't up for that this week.

[@kfftfuftur]

77th - 4529 points

[@Nathan0A]

now do the same problem, but the train is marginally more efficient the lower it runs on fuel, and each individual fuel stockpile has a minimum fuel quantity

[@SoleaGalilei]

Thanks for the solution! May I ask why you often refer to submitters as "they" even when the name seems to be obviously male or female? Is it something you do on purpose or just a habit when talking about people you don't personally know?

[@kevinxiehk2909]

11:33 You could call that a "Parker Solution"

[@cheaterman49]

Philip you're crazy xD excellent work! And Matt, I think your German pronunciation is very good? But I guess only Philip can tell how we're supposed to say his last name, hehe.

[@fdagpigj]

I got to 2000 because I limited myself to solving it by playing around with physical objects and thus didn't want to bother with fractions, even though I suspected there might be a better solution. Should've tried using some actual maths or at least code, I guess.

[@ditzfough]

1:43 thought i was watching a Turbo Tax commercial....

[@haroerhaktak2613]

I feel like we're being used in some sort of study. Or will be eventually.

[@severnsevern580]

i tried a few scenarios, all of them leading to the answer 2,000 consistently. interesting.

[@martinlamb6122]

I got 2000 since I was working under the assumption that fuel was divided into integer sections.

[@DontMockMySmock]

In hindsight, I got really close to the solution, but I thought I was going in circles and gave up. Oh well!

[@TrimutiusToo]

Oh well I am bad at solving this kinds of puzzles it turns out... i get railroaded into certain type of solution and then cannot optimize... pun intended...

[@TheCrewdy]

I'm sure you'll be back on track next time

[@bobzhang9966]

should be 1600?

[@bobzhang9966]

1st round: go 100 and dump 100 (delivery 100 with coal cost 200); 2nd round: go 200 and dump 200 (delivery 200, cost 400, pick up 100 along the way); 3rd round: repeat 1st round: last round go full 800 pick 100+200 along the way, so the cost will be 800 plus previous 1 - 3 round 800)...1600 is the correct answer?

[@andermium]

@@bobzhang9966 That's awfully convincing. But I finally found the problem. You cannot pick up the 100+200 anymore. The 100 you pick up after using 100 of your fuel and the 200 you pick up after using 200 of your fuel. That's right. But if you pick up the 100 at the first stop, you only have room for 100 fuel more at the second stop. Our fuel capacity is not high enough to ride from the 200 to the 800 in one go

[@bobzhang9966]

@@andermium yes, i got your point, thank you Anderium! i will try again to see if it's possible to do 100, 100, 100 dump for example, i contact you if i could, thanks again! happy holiday...:)

[@HavasiP]

I work in a train yard. I was gonna do a nice video driving a real train to party town. Didn't have time though :(

[@d.lawrencemiller5755]

Oh my god this is the nerdiest shit I've ever seen I love this.

[@andypyne]

Don't we need to know the weight of the train because train+fuel = 500 miles, so for journeys with less fuel on board (like a return back to the fuel cache) fewer miles will be required for the same amount of fuel

[@KuraIthys]

Once you go down that path you realise the purpose of a train is to pull a load. At that point you realise that the weight of whatever cargo (or passengers) you have to take along matters. that you can haul extra fuel above and beyond what the locomotive can contain (fuel wagons or coal hoppers depending on type of train), and now you have extra factors such as is it better to do your initial trips carrying only fuel for the later legs, or is it better to take some of the cargo partway as well, and in general you've now got a VASTLY more complicated problem with multiple interdependent factors... If the locomotive can pull 500 tonnes, how much of that should be fuel and how much cargo? What is the deadweight of whatever carriages you are using for each given type of cargo or fuel? Do you have an infinite supply of wagons, or would you need to haul the empty wagons back with you to fill up again? What does the cargo you need to move weigh? How many pieces can you break it up into, if any? And that's only some of the many complications that you run into if you make this problem more in line with reality... (but notice that the moment you bring in reality you realise that the fuel load the train can carry isn't a fixed maximum, but inter-related with the total amount of weight the train can carry, which in turn depends on how much other stuff you are being asked to carry...)

[@frognik79]

Do you get 1 million points if you solve the 3 sided coin problem?

[@theod0r3]

There is a Theo 1 just above me in the league table (I am Theodore K.) with just 2 more points. This is really weird!!!

[@livedandletdie]

Troll answer is 0 fuel just move party town to the train, no fuel wasted.

[@tbpotn]

Got this wrong and the Scrabble one as well. Guess ill take a bit more time next time haha.

[@jw41538]

that scrabble one was no fun at all.

[@Tfin]

@@jw41538 Why was it no fun? There aren't that many possibilities.

[@EldeNova]

I didn't participate in this, but I think .33 should not be included because they would not actually make the trip...

[@oak3001]

Yay! I got participation points! (probably)

[@Macieks300]

you can check it on the website

[@stepheneefting7989]

how the hell do you get 1733 actually... i ran through my method 4 times and i got 2000 4 times

[@Krebzonide]

Did your method include equations?

[@stepheneefting7989]

@@Krebzonide no What i did was use 100 and 200 as stockpiles to get 300 on 300 and that gave me 2000

[@Krebzonide]

@@stepheneefting7989 The reason your method gets you 2000 each time is because your method is not optimal.

[@christianbarnay2499]

@@stepheneefting7989 I think your issue is you limited your calculations to using stockpiles at positions that are multiples of 100. The solution is given with the train animation at 05:05. Stockpiles are at 33.33, 133.33 and 300. The general solution uses the sequence 1+1/3+1/5+1/7+... Counting stockpiles from Party Town to King's Cross: - The 1st stockpile is 1 full tank from Party Town. - The 2nd is 1/3 of a tank from the 1st. - The 3rd is 1/5 of a tank from the 2nd. - And so on until you reach King's Cross. One optimal way of using these stockpiles is this: You make as many two-way journeys as there are stockpiles, going one stockpile further each time and taking a small amount in each intermediate stockpile to fully refuel on the way out and refuel just what you need to reach the next stockpile on the way in. And then you do the final one-way journey to Party Town using what's remaining in each stockpile to fully refuel. There can be further optimisation if the stockpile closest to King's Cross is really close but this is the general idea.

[@Lioness99a]

@@christianbarnay2499 Clearly I'm being rather dim today - I still don't get it and none of the links in the description or comments have cleared it up! :/ I get where you are stopping but not how much fuel goes where. How much fuel do you actually drop at each point? Can someone spell it out for me please in numbers, not just the fractions..!

[@ifer1280]

You could instead only accept the last answer

[@GrayBlood1331]

I take it that matt isn't familiar with dan shiffman

[@robertshort9487]

So I thought the intention was to figure it out in Your head. (Found the original one today. So too late for this one). I got 1800. I didn't think of using excel or anything like that ...

Proof by KZbin :D :D

[@steamedup2]

Silly me, I thought were talking about real steam..srry..

[@HeathLedgersChemist]

Zee, Matt? Really?

[@TheGreatSteve]

Eugene can't tell a steam train from a Diesel/electric.

[@AspieGamer13]

>_< I think I forgot to re-multiply by 100 on my answer (Either way, I was still off slightly)

[@covle9180]

Surprise coding train!

[@roberteospeedwagon3708]

first

[@nanigopalsaha2408]

It is awful to hear a person from UK say zee.

[@belg4mit]

He's Australian, not British

[@nanigopalsaha2408]

@@belg4mit He lives in Godalming, UK

[@KuraIthys]

@@nanigopalsaha2408 And that still doesn't change the fact that he's Australian. ESPECIALLY not when your comment is related to accent.