The Penny Problem That Breaks Your Brain
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- เผยแพร่เมื่อ 15 มิ.ย. 2021
- Your mind is a massive collection of information, patterns, trivia, algorithms, and more -- and you have absolutely no idea how or when any of it is going to be useful. If you’ve got 50 stacks of 50 pennies, and you know one of those stacks is fake, do you have all the knowledge you need to find the most efficient measurement to find it?
Yes, you probably do. You just don’t know it. Part of the series is in the problem, but part of the series is inside you.
If you’ve ever thought about adding or subtracting consecutive numbers, mathematical patterns, or more advanced material like convergent and divergent series, you have all the ingredients you need for a solution to the penny brick problem that breaks your brain. The question is whether the two systems of thought in your brain are able to work together to find the answer.
From Euler to an 8-year old Carl Gauss to Ramanujan, the informational and mental tools to solve a simple penny problem likely exist somewhere in your head. And by harnessing the dual process theory of the brain, you can stumble on a mathematically elegant solution to a problem you never even knew you’d ever have to solve.
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What's in YOUR System 1?
The words, kevin sus
the contents of my refrigerator
Oh no I just finished the video I'm sorry for commenting sus before this is really serious
Hasbulla
Y=ax+b and a lot of addition, subtraktion, and the other 2 things that i cant remember
4:01 looks pretty mature for an 8 year old.
Obviously a joke, but the picture was made after he was 8.
@@lplatino6427 r/halfawhoosh
no
I'm a nine year old.
@@aashsyed1277 then name every Pewdiepie video 🔫
This seems like a lot more work than just weighing the stacks individually
In Real life, it's not speed, but in an video game or else, it's realy optimised
It may not be efficient in this situation, but it is more effecient when dealing with similar issues at a much larger scale
You're right Aaron. Even a computer would still have had to do 50 different computations with this method, (or 'x'number until a change occured) therefore not really changing the outcome.
The method he described IS great for finding the sum of a basic series from 1-N, for instance, 1-200 is easily 201x100, BUT you're correct that for this type of problem, or any similar problem, it would be equally effective to simply check them one by one instead of doing a complex algorithm AFTER adding parts of their sums one by one.
@@justcallmenoah5743It is only a different system. If you took a larger situation, 1000 boxes of pennies, you would always find it more efficient to weigh individually than to open, count each box down to a penny "without error", and stack them on a scale the size of your house.
Making it about that many pennies in rolls made me think about physically breaking open all those rolls and having to re-roll them. I thought they were supposed to remain in the roll. The solution they were looking for is such an inefficient way of dealing with actual pennies. It's completely ridiculous.
Quick note: The method mentioned at 2:00 wouldn't take max 15 tries. It's max 13.
If you check 4 groups and don't find the lighter one, then it's the 5th. No need to try it. Same with trying 9 from the group and not finding it.
Of course it'd be best to measure the last anyways, just for verification.
Nice
way to think about it
.
.
The reasoning is that you don't know if there is a fake or not
I learned that if I want to find the fake roll of pennies out of 50 rolls of pennies, that I SHOULD NOT UNROLL THEM. 👏🏻👏🏻👏🏻 Excellent lesson, sir.
Exactly. What good is it to learn which roll the fake pennies are from if they are no longer part of that roll. You then have to go through all of the loose pennies to find the fake ones. In order to put them back into the fake roll.
@jasonfrodge8742 this is incorrect. Since the weight difference identifies the exact roll, you would also know the exact loose stack that identified the roll.
@@JGeMcL ...but you also have some fake pennies in an undifferentiated mass of pennies on a scale.
@@jasonfrodge8742 just put them in an organized way so you can put them back easily
If you're able to stack them then you have them...if the shown example happens then you got yourself in a situation finding the 13 pennies in a pile of 1200 pennies.
From a fixed perspective it is really a nice mathematical trick...but from a realistic standpoint it's a total mess.
Unwrapping 50 rolls and counting the pennies and sorting them by stack is already very time consuming.
You need to be sure that one roll is fake, but if there're 50 fake pennies in the box that method is for the bin (we weighted it and we are 25 grams to light)
You could split the box in half to reduce the measurements and so on.
And actually you can literally do it with two measurements, one the whole box... empty the box and start filling it one by one, it needs to jump 128 g per roll, if it's not you got a fake in that roll.
Worst case is one in each roll😂
Ahh, the mind of a programmer: “How can I solve this simple problem? With this elegantly complicated solution
“
"Let me come up with the most complicated way of writing it"
simple find the missing ;
I got asked a question almost exactly like this in an interview. I wish I saw this video before then.
My mind immediately went to "are we allowed to take pennies out?" "do we have to measure in quantities of 50 pennies?"
Jokes on you, I can never face the dilemma because pennies don't exist in Canada
Same
Yes lol
Europe gang here
Nooooo penny dilemma
me being a penny collector 👁👄👁
My first instinct was to reach for binary search.
Guaranteed to find the right one in at most ceil(lg(50)) = 6 measurements (where lg is log base 2). Sometimes you may luck out on the last step be able to do it in 5.
And this way does not require unbundling the rolls.
Basically, take the set of rolls, split it in half and weigh one of those halves. If that half is less than the expected weight, then the fake roll is in that set. So split that one in half and repeat the process. If the half weighed is the expected weight, the fake roll is the other half. Split that one and repeat.
If you have a set with an odd number of rolls, just arbitrarily choose one of them to get the remainder roll. Just keep track of how many rolls are in each group.
Since 50 is not a power of 2, you will eventually get cases of 2 in one "half" and 1 in the other "half"
This is where the "luck out to get 5 measurements" comes in. If the 1 roll is lighter than expected, then you are done; you don't need to split that set of 2.
I feel like "at most 6 measurements" binary search gives the best _worst-case bound_ you can get if you aren't allowed to break up the rolls. I'd love to be proven wrong though.
It's true. There's no more information in the problem
Binary search was my #1 thought as well. I misinterpreted the problem initially - I thought one roll contained one counterfeit penny, not there was one roll of entirely counterfeit pennies. The solution to do it in one measurement is really elegant though, because it's a sneaky way to encode another piece of data (which roll the counterfeits are from) into a single measurement of one sample of all of the pennies!
You vastly underestimate my laziness if you think I'm opening the box, much less the rolls. Put it in the CT scanner
I've never seen someone put this much work into explaining a simple process with so much flamboyance to seem smarter than they are...
If you have a balance scale, you can do it faster. Split it into 3 equal groups (you'll have 2 left over. They can go sit in the corner and think about how they don't have a friend).
Balance 2 stacks of 16. If they are the same weight, you know the fake is in the final stack of 16 (or it's one of the two odd ones out). If they are different weights, then you know the fake is in whichever stack is lighter.
You now repeat with this stack of 16.
When you measure with a balance scale, there are 3 pieces of information you can get: Left is heavier. Right is heavier. They are the same. So it's faster than binary search.
I love how you went through all the trouble of doing this and then slammed every stack of pennies to the side and yelled out “NONE OF THEM ARE FAKE!”
Wtf spoilers
@@ninjaguyYT What do you expect from reading comments before watching the video?
Joke’s on you, counting and then taking all those pennies out is way more tedious than measuring 50 rolls.
I was thinking this. And rerolling them is worse.
especially when he didn't have to weigh them individually or even get precise weights lol. balance beam put 25 on one side and 25 on the other, discard the heaviest and go again, since you can't split 25 evenly you just weigh 12 and 12 if they weigh the same the one left out is fake otherwise repeat discarding the heavy and splitting the light again.
I think you missed the point of the video
@@Joe-Dead You ll eliminate heavier every time this way
@@Joe-Dead When you use balance instead of measuring the weight you dont go 50/50 but 1/3 to 1/3.
Plot twist: One of the stacks he got from the bank actually was fake, by sheer coincidence, and he will soon be arrested for fraud after buying gumballs.
'obviously, your honour, I had no idea they were fake as I got them from the bank'
'enter vsauce episode dated 17th June, the defendant was aware and intentionally mixed them in with the real ones knowing exactly how to seperate them again.'
LOL!
Gumballs with a penny??
@@v3rm1nentertainment6 The 25-cent machines are for snobby rich people.
Imagine putting 50 pennies in the coin slots for a gumball machine... good god, that's commitment.
A simpler way to do this with a single measurement is to have a long horizontal plank, pivoted at one end and with a single point of contact on a scale on the other end. Zero the scale, then line up all your rolls on the plank, evenly spaced so the first roll is at the pivot, the last roll is above the scale contact. If the pennies were all legit, the scale should read half the weight of pennies. The amount it reads below this value, divided by the absolute difference in weight of the pennies is the same as the normalised distance along the plank of the fake roll, i.e. if the scale reads the same value as it would if it was legit pennies, the fake pennies would be the first roll, at the pivot. If the scale reads the same difference as the full weight difference of the fake pennies, the fake pennies are the last roll, above the scale. If the scale reads half the difference, then it's the roll halfway along the plank.
I think part of the reason it's hard is that the answer is actually more labor intensive than than just measuring all of them.
The problem isn't stated correctly. It's presented as an efficiency problem. The final solution is not efficient -- it's work intensive, destructive, and prone to error. It does solve for minimizing the total number of measurements. My issue with this video is that he never says the goal is to minimize the total number of measurements. He states the problem incorrectly, then criticizes all of us for thinking about it the "wrong" way...
Yep. Life is about time and effort not number of steps. So our minds typically adjust and land on solutions that take less time or effort rather than less steps.
Most people's system 1 mind would correctly eliminate this method because it'll see that opening the nicely packed and easily handled rolls will make handling the resulting pennies harder and longer. Basically your system 1 didn't even allow your system 2 to think of this method because it failed a prerequisite requirement of being easy or timely.
Really this sort of method might have a few uses but I really fail to personally see one. In nearly every environment using this method would require more time and effort. Probably the only one is on computers and only if the algorithm is already made beforehand to make it plug and play.
it doesn’t even take less measurements… he still had Pennie’s from each stack…and had to put them all on the scale. So it’s a convoluted way to literally stack all 50 one at a time and just watch the number. He already did the math for how many pennies to remove. It’s not hard to do the same math for the consistent weight of each stack.
Counting is a measurement, so it doesn’t even hold up to the one measurement rule
@@mcfail3450 I thought the reference to system 1 and 2 thinking was confused and confusing. Saying that you have to 'rely on system 1' in order to use system 2 at all...well, okay, if by that he means that system 1 controls access to the kind of system 2 skills needed to solve the problem...then yes, but by that logic all system 2 thinking relies on system 1. And how did it make him 'think about how we think' in a new way?
Meh.
"Weighing each stack takes too long. Instead, open them all and count out the pennies to take."
Exactly
Real world application is obviously less efficient. But purpose a program could benefit from this on a much large scale in order to do less equations? Not totally sure.
O(n) vs O(n^2) operation
So, let's change the question a bit. What is the fastest way to determine the fake stack? Assume each measurement takes 10 seconds. Separating each coin from a stack takes 1 second. So to separate 1 stack would take 50 seconds. For ease of analysis, assume other operations (ex : unwrapping the coins from the stack, placing them on the scale etc) are instantaneous, ie, takes 0 seconds. Do we know the fastest way to do it?
If the measurement time is 1 minute instead of 10 seconds, would the fastest method be different? What if the measurement time is just 3 seconds instead? And finally, what if we just have 9 stacks of coins only?
Let's suppose that you get a continuous stream of some kind of input. Instead of weighing pennies, you do some much more expensive operation on each of these, like a checksum. If all is going correctly, all inputs should result in the same checksum.
Let's suppose the checksum operation is expensive to run, but has a quality that you can moosh inputs together somehow with some kind of additive operation, and the subsequent checksum operation is linear over this. Thus, you can run a single checksum on the mooshed together input much much cheaper than you can on individual inputs.
Furthermore, you can perform an operation that somehow "discards some fraction of" each input, and again the checksum is linear over this operation as well.
If those two things are true, then it pays to use this method, particularly because you don't have to store all these inputs in memory. As the inputs come in, you can take just the fraction of each according to its number in the sequence as shown in the movie, so that the 13th element gets 13/M of it saved, for M=the "size" of each input, and the 34th element gets 34/M etc. accumulated onto an accumulator. You do this immediately upon receiving the input, so no memory is needed beyond that single accumulation register.
Finally, at the end you run the checksum operation on the accumulation register and instantly find out not only if there was a problem, but which input had the problem.
There's quite a few constraints here, esp. re: needing to be linear operations. But if your problem meets these constraints, then this method is a perfect candidate for your problem.
TBH, I haven't figured out a good practical example yet but I'm working on it.
As for the pennies, maybe you are measuring something similar but only have access to a scale for a very limited amount of time. Maybe your neighbor is a grumpy old man who owns an accurate scale that's sitting out front in his driveway at the moment. So you prep your pennies by the method shown here, and put them in a single tote which you place on his scale. Then you dump the pennies into a bag you've brought along, measure the tote, and run the hell away before the old man catches you using his scale. You don't have your answer yet, but you've done all the measuring on the scale you need. You can compute the answer freely once you get back home.
Obviously the example here is to illustrate when a resource (in this case the scale) is under high contention or expensive to use per instance-of-use. The reason it's hard to think of non-ridiculous examples is because of the need for these operations to be linear and for the components in each input (each stack of pennies) to be identical when measured individually: i.e. to be a constant * a scalar.
If you can figure out how to open up these constraints, good examples will almost immediately flow.
he said "which one of you is looking a little", I said out loud "no don't" and he did it anyways
Same
Yup!
@@ChayComas He saying Sus
He even played the music..
And then he called the counterfeit roll the
_Impostor_
This is a fantastic method. Some are criticizing practicality while focusing on the pennies, but that's simply an example to illustrate the concept. Fundamentally, the solution is an elegant method of resolving a conditional problem in the fewest number of measurements by utilizing a mechanism that makes the information output of a single measurement represent a parameter only possible in one of the many options. It's interesting in that it doesn't feel like something that needs genius but is instead an answer that can be found through ingenuity and intuition given enough thought.
Exactly. This is the kind of thinking used to develope novel physics experiments
It's a bit deceptive because there was already an original measurement of how much the real and the fake pennies weight.
And you would never be able to tell them apart if the difference is along the instrument error.
Don't know if this is in comments somewhere already. But with the groups method (groups of 10 and then take the lightest and then measure all ten). You actually need to measure a maximum of 13 times. Since after 4 measurements you know which one should be lighter even though you're not measuring the 5th group. Same goes for the stacks. Only 9 measurements needed.
When that "single measurement" takes more effort than just measuring them all individual
Time complexity vs space complexity problem there
Exactly. The described method is good for something, just not this.
When that "single measurement" actually requires you to carefully count (measure) a certain amount of pennies 50x over, and THEN do the "single measurement" to find the answer.
@@AbjectPermanence a single measurement in this case meant a single usage of the scale.
No one said it should take less effort, just less measurements
Yeah, but my system 1 brain also tells me that separating all those Pennie’s is more work and time than just weighing each one until I find the fake. Laziness can be a beautiful thing.
Exactly. My system 1 brain just tells me to binary search the thing and call it a day.
You also can do it in a maximum of 5 weighings by splitting it in half w/o opening them. The odd one out isn't a problem, but a possible quick solution too
What bill gates said
@@Klust413 exactly what I though. Quickest way is to keep halving the stacks and weighing them.
This overly complicated solution also assumes that every penny inside of the bogus roll is fake. Yet this crucial presupposition wasn't even stated until three quarters of the way through the video...
So yes, opening the rolls and separating the pennies out to weigh is more work that just weighing each roll individually, *but* like this guy said, it's all about how you think about problems. This solution hadn't occurred to me, but now I'm looking forward to when I have a problem I can't solve, and this solution inspires a new way of thinking about the problem!
If the problem space is predicated on doing the least amount of work, as was proposed here, then it's unlikely people will come up with a solution that is more complex AND takes more work than equivalent simple solutions.
these are artificial problems made by humans and having certain parameters. in the real world, it is unapplicable since there are many more parameters
@@PhantomPhoton It's about minimizing a specific type of work, imagine to observe the reading you had to walk 1 mile away from where you placed the pennies on the scale. This shifts the real world work required so that all the work setting up the one measurement is much less work than the 9-11 miles you would have to walk to do the 5-6 measurements required to find it with a binary search so taking each additional measurement beyond the first is no longer trivial.
I know this solution from riddles and there are situations I'd consider it, like, when you have to find 1 out of 5 or so, up to 10, but with 50, I'd consider it only theoretically. Though I really like Gauss' idea of adding numbers 1 to 100 by simply doing 1+100 + 2+99 + ... = 50*101. Not very often, but sometimes I actually consider that kind of approach to a problem.
That seems a lot more complicated. I would've put all 50 stacks on the scale and remove one by one. The number on the scale should drop predictably and equally with each stack that's removed. The stack that removes a lesser amount is the fake.
Same as measuring each individually
@@avengemybreath3084 not necessarily. It's possible to know by he second one removed if it's lighter than the first
That would probably be the "quickest" way for sure since it doesn't take long to simply lift a stack off. His question was not what is the quickest way however, it was how to do it with just ONE measurement.
@@MrTomservo85 While the video did not explain this properly, the restriction of weighing only once means that you can only get one reading from the scale (and then you can assume it stops working). Your method fails in that situation.
@@MrTomservo85which also applies to doing it in reverse, you offer no benefit
At 1:53 he said there's a 2% chance you could do this in just 2 measurements. If you think about it though, there's 50 rolls total so there's literally a 2% chance of just picking the right one on the first try at random as well lol
I mean if it's 2%, then it's 50% because if its not 100%, it's 50%.
@@simplyoreo4899 and if its not 100% its 50%
2% to measure in one shot, sure, but a potential maximum of 50 measurements. Trade 2% to measure in 2 for a better maximum of 15 potential measurements seems worth it. I wonder, though, for each number of measurements n the odds of that being the number of measurements (for each method).
Mostly a neat application of an 8-year-old's work designed to show that off, then tweaked a bit to draw meaning from it as a demonstration of why we learn "pointless" things. I definitely plan to show this video to students asking why they'll need to know how many watermelons the elephant could squash if dropped from the sky; it's not about knowing how to do that, it's about demonstrating you have an expanse of understanding that allows you to have tools to solve problems like these.
@@ceribralboy4468you can change a method and have a chance of minimum 4 and maximum 8 or another method of minimum 5 to maximum 6.
Method 1 you group stacks in 5 groups, but instead of measuring them one by one after you have a correct stack you measure half of them and then again half, repeat till you get the correct one. Since you know that if the measured one have no fakes then it's in the other one. So worst case scenario it's in the last one group (4 measurements since last one don't need to be measured) and then you measure 5, halves them, 3, halves them 2 and 1 ( another 4 measurements). Second method you makes 2 groups instead of 5 and do same thing.
Yeah but you wouldn't know it's fake and also you can't continue to get closer to knowing. Very different scenarios
"the right process" is not unwrapping and counting down fifty stacks of pennies.
That's modern Vsauce for you...
2 years late, but ... your solution is a perfect illustration of the reasoning behind the first rule of optimization ("Don't do it!"). True, you managed to reduce the number of times you needed to weigh a bunch of pennies to its theoretical minimum, but at what cost? I dare say I could have done a binary search for the short roll, treating each roll as a unit, five times over before you were even ready to start loading pennies onto the scale.
Damn, that's actually pretty simple after knowing it. That kid was smart AF to figure it out the first time.
I love that doing this in one measurement still probably takes longer since you have to unravel all the stacks
Imagine doing all that and accidentally mixing the coins you took out. Now you know that the n-th stack is fake, but also you need to find the remaining n coins that you took out and are mixed with many others. Just stick with the first method lol
You don't have to mix them though
@@demerion @Guiorgy Potskhishvili Both of your comments are good. It can work but it's hard. This method obviously works for any number series with the same problem, it's just a handy demonstration. What else could he do to show this concept?
@@demerion 😒 seriously. What he said was just hypothetical. You may not need to, he's just saying if you did.
What if the cost of a single weigh is prohibitive somehow? Like you only had the power to do it once or twice?
@@alexortiz9777 Because that's a situation that ever happens in real life?
Put them side by side on a bar that pivots (like a teater totter) depending on the equilibrium point you will know which stack of pennies it is by knowing its deflection is proportional to the fake penny stack. All without removing the pennies from their wrapper.
That is genius, also very complicated math for the average person. You are obviously very intelligent.
✌️✝️
Alternatively, space them all out on a long lightweight bar and then weigh one end of the bar with the other end resting on something. The more it weighs, the closer your lighter fake penny roll is to the fulcrum. The less it weighs, the closer your lighter fake penny roll is to the scale. From there it's straight up math to determine what the weight should be if the lighter roll was at any specific point along the bar.
There are a lot of fun solutions to this problem.
This was really interesting, and amazing when you realize it works. But it might have been better to demonstrate with a much more modest number of stacks, say, 5 stacks. Then, for fun, people could actually try this at home. Although I don't know where you get light pennies. Maybe use some other medium, like aluminum washers vs. steel washers or something.
US Pennies before 1982 were mostly copper and weigh 3.11g each, pennies minted after 1982 are the mostly zinc ones discussed in the video and weigh 2.5g each. Pennies dated 1982 are mixed in that some are the old 3.11g composition of mostly copper and some are the new 2.5g composition of mostly zinc.
And prior to 1864 they were 4.67g each (1864 had a mix of 4.67g and 3.11g). Going even further back before 1857 they were 10g-13.5g each, but those would be very easy to notice since they were larger than a modern quarter, and closer to a half dollar in size.
Would probably be quicker to just do a binary search. Measure 25 stacks at a time. Take the lighter 25 then measure 12 and 13 (obviously needing to use the density of the stacks since one bunch has one more stack). And so on. 6 measurements tops and no preparation phase, right?
You could actually measure 12 and 12 too. If one group is lighter than the other, that one has the fake pennies. If they weigh the same, the leftover stack has the fake pennies.
If the issue was efficiency then yes, this was just talking about numbers of measurements and how it can be accomplished with one
But I totally agree with you that's where my head went first
You do not need to measure 12 and 13. You can measure 12 and 12. If both are equal, then leftover stack is the fake one. In worst case you need to measure maximum of 10 times. 25, 25, 12, 12, 6, 6, 3, 3, 1, 1
Yeah.
But they are two different optimization processes.
You are optimizing time, trying to minimize how long it takes to Discover the fake one.
Kevin is minimizing the use of the scale. I would like to know an scenario where his reasoning would be useful (in real world) but I can't think of one.
you could also take a balance, weigh 25 and 25 see which stack is heavier than eliminate the search down to 25, then you exclude 1 and you we 12 and 12, if they are equal then the one you excluded is the roll, otherwise, you've reduce it to 12, then 6 then 3 then you only need one more weighing to know. so in total you weighed a max of 4 weighings.
2 solutions, aside from that one which I didn't think was allowed since you didn't say group of pennies but said which wrapper, suggesting we were not allowed to open them. This would have been my first answer but decided on a second solution before I watched the rest, which is placing all 50 wraps about a circumference of a round object already balanced without them. The lighter stack will rise higher than all others and be easiest to see when the opposite stack on the other side of the wheel comes to rest closest to the table. Counter-balancing is easier if you don't want to unwrap them all.
Yeah but the solution required you to specifically "weigh" the pennies. There is a whole plethora of solutions if you remove that limitation
Counter balancing is one of the oldest ways of weighing objects.
"thinking, fast and slow" by Daniel Kahneman (a noble prize winner for economics) is all about system 1 and system 2. Great read.
Thanks for pointing out "further reading". I wish more videos have that. I'll take a look. Sounds like it would be interesting.
One of the best books I ever read
CORE A GAMING THOOO
Am I only person who felt their heart break when he splashed the pennies?
They weigh the same as when they are stacked.
@@TheMursk I don't think that's what the point was tho. It's the fact that he was stacking them up the same way that he took them out of each roll. That means even if he figured out which roll the fake pennies came out of, he wouldn't know which pennies he originally took out of that roll because he spilled them all over. In the end, it's all just means even more work for him.
While you were rambling on about systems and sequences, I went ahead and just weighed the rolls. It’s this one here, mate.
In real life I would check the weight of each roll separately because it’s much quicker than counting out the correct number of pennies from each roll. Also, with the 1 weigh solution you have found the fake roll (or what’s left of the roll), but you have also 1 or more fake pennies mixed up with over a thousand real pennies. If the challenge is to find all the fake pennies you have just made the problem harder.
If you mixed the pennies you're the one to blame ;-)
No because if the total measurement is down by say 25g, then it’s because of the 50 fake pennies from stack numbers 50 which weigh 2g each instead of 2.5g each, this solution is good if you only have 2 measurement to make like say you’re about to get executed unless you solve this in 1 weighing idk
I’d instinctively just bifurcate using groups of stacks. If I gave it a few seconds to ponder first, I’d realise I could trifurcate it and find the fakes in fewer measurements.
@@Tentin.Quarantino Ternary search gives 17/17/16 which need two measurements to narrow it to one of those, repeat this ternary search to 6/6/5 two measurements to narrow that down to one of those ternary search 2/2/2 two measurements to narrow that down to one of those 1/1/0 only one final measurement for 7 total in the worst case for ternary search for this case.
Binary search gives a worst case sequence of remaining rolls as 25 13 7 4 2 1 for a total of 6 measurements required in the worst case.
This is not a single weigh solution in my opinion. He's literally just weighing each penny and using the scale as a running total, that's it.
My idea was to do a sort of binary search: take 25 rolls, put them all on the scale at once, and check to see if one of them is lighter. If so, you cut the amount of rolls in half and try again, otherwise you use half of the _other_ rolls. This takes ceil(log_2(n)) tries for n rolls, so for 50 you would need 6 tries at max.
Precisely 6, no?
@@vaxrvaxr not always 6, no. If the number of rolls were a power of 2, then it would always take an exact amount, but since is not, you will often end up with odd numbers of stacks.
Lets say that your first measure shows which stack of 25 contains the counterfeit. You second measure would cut that into a stack of 12 and 13. Let's say it was in the 12. You third measure would split that into a stack of 6. Your fourth measure would split that into a stack of 3. 3 once again does not divide evenly so you would have to weigh a stack of 1 or 2. If the stack of 1 turns out to be the counterceit, then you've found it in 5 weighs. If the stack of 2 contains the counterfeit, then you need to do a 6th weigh. So binary search gurentees that to find an object in n objects, it will take at least log2(n) rounded down attempts and at most log2(n) attempts rounded up.
I was actually thinking the same thing, but I didn’t realize it would take this few tries. Btw I love how you took the time to write out that whole answer 😂
If you used a balance scale you could do it in between 1 to floor(log_2(n)) weighs if you always compared groups of the same number of rolls, and left a random one out. If it any point you find that 2 groups balance evenly, then you know the extra one is fake. A group of rolls which is a power of 2 (e.g. 64), you'd weigh stacks of 32, 16, 8, 4, 2, and 1 each, so it would always take 6. For 63 you could weigh stacks of 31, 15, 7, 3, and 1, so it would take at most 5.
I was thinking the same thing. This would probably be the fastest way because log2 is the most efficient and counting hundreds of pennies individually and putting them on carefully just to get one measurement would take ages.
An interesting solution that could speed the first one a bit is: Adding to the pile a stack at a time, while measuring. This way you will encounter this number sequence as last digits: 8, 6, 4, 2, 0. As soon as the pattern changes, you know you added the false pile
Wouldn't that be counted as multiple measurements? Your solution is practical in real life but i doubt it's allowed through the problem's rules
So how does someone KNOW that one roll of pennies is fake before weighing them? Why would you have a fake roll of pennies?
@@petemcintire4339 they were all weighed before, and the result was off by a few grams? So you get tasked to find it, but your scale is faulty
@@MrZoolook Well, let's just consider this a riddle then, not a problem but a challenge, which makes using 1 measurement not a contraint but a rule, in reality this problem really wouldn't even work
Creative idea, but I already posted the Optimal solution in another comment... I can mathematically prove that a halves method works best on average and works out to 7.5 steps with minimal variance.
We need a system for quickly spotting videos that are 95% filler.
Yes Mr. Young. I’ve always believed every problem has one best solution, and it will be the simplest, most elegant one. I think you have it here.
Looking at the comments I'm not sure how many actually got the point of this video. It's not about finding the answer to the riddle of which is the fake stack. That is merely used as an example of how we need to all try to fill our brains with as much knowledge and experience as possible, even if we think it may not be of any use at the time or even ever at all, because if we have this info stored, then if one day we need that "tool" from our brain toolbox, it will be there to solve other, and often, much more consequential problems. Excellent video, great work, it really is eye opening. And also a confirmation that all this useless knowledge I have stored over the years is actually just the opposite. It's tools for my tool box that I can use in my life to better the world. Keep up the good work.
TH-cam commenters aren’t the brightest bunch as a general rule.
yeah that's exactly what I was thinking
It would have been far more useful to show the knowledge was useful and not a waste of time watching this video about nothing.
It would bring the point across better if it is actually better than the more intuitive solution most people would do. This is just over complicating the solution to a simple problem.
This is like watching that random tutorial video in your recommended, you may not have any use for it at the time, but, maybe in the future it will be just the thing you needed.
I thought of a completely different solution: you could put the stacks separated by a constant interval on top of a long ruler; support one end on a edge and the other edge on the scale; you will get different answers based on how far the fake stack is from the scale
Aw man, I thought I was the only one who thought of this.
I thought the solution was going to be something like that and was pretty disappointed that it turned out to be something that involved destructive manipulation of the stacks.
Opening the stacks seems like cheating.
@@majorgnu That's just a matter of principle. If it works, it is acceptable IMO (so is this method btw)
@@majorgnu I had the same felling
@@majorgnu It wouldn't be vsauce without at least some destructive manipulation.
I would have never figured this out on my own, but it makes perfect sense when explained to me.
Using this method you would still need to find the fake pennies from the lot you weighed.
I was sure this was going to be about binary search, until it wasn't.
That's what I was thinking too. And it would take way less than 15 measurements to get the right answer.
Me too, and I still think it would be easier than calculating all that
That's the entire point of the video. I'm a programmer (assuming you are too) and my first instinct was binary search, too. Our System 1 is loaded with information and algorithms that are applicable to writing software.
Actually this is like an optimized method for a particular situation if you think, binary search is the most optimized in a wider range, this one doesn't work in any other context, at this point you could imagine an array with all numbers from 0 to 9, binary search is efficient, but knowing the index equals the result would be faster, but wouldn't solve any other situation
@@heron3140 I think this would still be far inefficient because you're given 50 things and a scale. There would be a tremendous overhead of setting the problem up like this that grows faster than log2. The only case where this would be useful is if you could instantly set the pennies on the scale and had an incentive to use the scale as little as possible.
The irony of this solution is that it actually takes more setup and time to do it in technically less measurements than just brute forcing it. I would just calculate how much each stack should be if they're real, then as I'm loading the stacks onto the scale one at a time, and calculating the amount each one added to the total, once I hit the one that added less than it should I would know the solution. The solution he came up with is pretty neat though, mathematically speaking, it's just way too time consuming to be feasible.
It is the same thing when I notice that a task that could be automated, but then, in the end, I realize that the time it will take for me to develop the actual automation system is, at least, the same or greater than doing the task manually/traditional way.
I face this dilemma too often :(
Indeed, but that's because of the example he's using to try and explain it.
Isn't counting measuring?
if you code it in a computer it would take less cpu resources because it would only use each function once. 1 function to sort, 1 function to measure and 1 function for the output, rather than using the measure and output function multiple times until the stack is found.
This shows exactly why so many pupils have math problems. Our brains dont magically constrain themselves to artificial boundaries of a given math/logic problem so our problem solving thinking doesnt either. For this special problem, basically nobody would come up with the solution because we ourselves have certain expectations. We know it would be idiotic to count out penny stacks just to cut down the number of meeasurements from 6 to 1. We naturally wouldnt come up with a solution that involves breaking up the rolls either. All these "common sense" habits in our thinking make it possible for us to come up with solutions to real life situations which are infinitely more complex than most math problems, but which dont require perfect solutions, just efficient ones.
I wonder how long it took Carl to take out the garbage for his mom. Did he have to do the calculations first? What a maddening child.
that solution actually blew my mind tho. it’s brilliant
0:42 Amogus!! POGGERS
SUS
PICIOUS
This show has the massive budget of 2500 pennies.
Actually, they only had 600 pennies, the rest were rendered I'm CGI
@@ChayComas ohh... To save on production costs. Clever...
If you liked that, check out the recent bet between Steven Mould and Electroboom. They really upped the ante, putting a whopping TEN THOUSAND CENTS at stake.
@@ToyKeeper it was Canadian though
Does it actually, though? 50 are fake, after all
You tricked me. Here I thought this was a video about a cool math trick but instead you taught a valuable life lesson about continuing to learn and grow
There's another way to do it.
Have a long board balanced on a point (basically make a scale).
Lay your 50 stacks equally distant from one side to the other.
The scale will lean in a direction or another.
Based on the speed (or if you have a pressure sensor, based on that) you can tell which stack is the fake one (the further from the fulcrum, the more it will lean in the other direction) based on the "weight" of whichever side is leaning.
If it's perfectly balanced, then the one in the middle (in case of an odd number of stacks) is the fake.
Nice idea. What do you think about my comment from today? I didn't read much comments but stopped at your "board", because i thought, you had the same idea as me, but mine is a bit different
While this will theoritically work, in practice, placing equally distant 25 item from center is nearly impossible, a minor or slightly off (especially near the edge) position will cause it to disrupt the balance hugely, so unfortunately this won't work unless in very precise laboratory condition which most people won't have the capability to do the setup. Also the stack must be perfectly vertical because if the stack has a coin that is slightly off position or the stack slightly leaned (not perfect 90 degree line) the balance will be off (and will be off by far if it happens at the edge).
The outside the box part of this is realizing you can open the wrappers up and split out individual pennies. If these were instead, say, 50 gold bars and one was fake then the method in the video wouldn't work.
However I think you could still do only a single measurement on 50 gold bars to see which one was fake, assuming you know the real weight of a gold bar, by putting the bars on a balance beam at different points and weighing a counterbalance against that. By having each bar at a different length on the beam the amount the fake bar is off will be scaled by whatever the distance is of that bar from the fulcrum (e.g. if the fake is 4 units out it will "weigh" half as much as if it's 8 units out on the beam). So if, like in the video, you already know that the fake bar differs from a real bar by x pounds, then you could do this single measurement and the difference will be a multiple of x, with that multiple telling you which bar is the fake.
I could be mistaken but I believe this would only work if the bars and counterweight were single points, which they cant be physically
@@willmueller3637 Physical objects tend to project weight as if they projected the force only from their center of mass. It should work, as long as the distribution of mass in the gold bars is uniform and the gold bars are placed so they wouldn't have any torque.
@@willmueller3637 I was thinking about that.
But imagine that somehow You attach the bars to a string and hang the string on to the beam.
I think it could work like that because strings are thin and would be as close as possible to a point.
I thought of this same idea
That is actually really smart!
oddly enough, although he put a lot of focus on Gauss' discovery, it actually has nothing to do with the solution. The solution was simply based on the fact that by including a unique number of pennies from each roll the difference in mass would correspond to a specific number of fake pennies identifying which roll was fake. Gauss' formula provided a shortcut to counting how many total pennies would be being weighed, but doesn't actually have anything to do with the result, you could have done the same thing and simply added up how many pennies you took from each stack manually.
Thank you
@@VOTVRe but wether he calculated it by Gauss or added the numbers by hand, it makes no difference as it happens before the measurement. It doesn't have any effect on amount of measurements needed, and "finding it in one measurement" was the whole point.
They didn't explain the solution very well. Kinda got distracted. Oh well now the solution is in my system 1.
Agreed. The Gauss method of counting is cool, but is totally separate to the exponential addition trick he uses to solve the problem.
That makes so much more sense.
I also missed at the start of the video that one FULL roll of pennies was fake and the goal was not to find which roll(s) had 1 or X amount of fake pennies. So I was very, very confused.
Bro this was absolutely brilliant and absolutely so simple an 8 year old would have had to think it up. This actually made a hole in my brain and forced new connections.
I love this solution. Actually doing it is not elegant at all, but the solution itself is very elegant.
The other thing is this doesn't even translate well into a practical problem using just math. How would you do this problem mathematically, and not mechanically? How do you "take 1 from the first stack, 2 from the second stack, etc etc" mathematically without having the weight of each individual penny already? And that defeats the whole purpose of the problem as you've already weighed each individual penny, and can immediately identify the fake ones. I get that the point here is that you can use statistical analysis in a huge data set to identify outliers, but in any real world example, the individual data points aren't all exactly the same, and don't immediately tell you what class that data point falls into (legit or fake penny)
Gauss: and that's why my first employer, the moneylender, fired me.
Just one problem: The one-weighing solution is more difficult than the original solution.
It’s not about which one takes less time. It’s about thinking outside the box. The simple solution isn’t always available.
@@DragonsinGenesisPodcast It's a poor analogy.
Working that out doesn’t make you smart.
@@filonin2 right, I don't doubt this principle might be useful in some context but here
my system one is telling me "maybe we can do something by unrolling... nah nah nah, fk that noise"
Impractical example but for a useful concept. It is how you think about things that is important.
Align all the rolls of pennies on a rigid beam. One end of the beam on a fulcrum, the other end on the scale. If the fakes are nearer the scale, the scale will read a lighter weight, if the fakes are nearer the fulcrum, the scale will read a heavier weight. Using structural shear calculations for beams allows one to determine the location of the fakes.
You can actually take one single measurement without breaking any roll.
Put the rolls beside each other in a row on a board of known weight and the same length as the row. Weigh one end of the board. The weight discrepancy from the ideal will tell you the position of the fake roll, by the virtue of center of mass. The closer to the scale the fake one is, the larger the discrepancy.
(I am sorry if someone else has already pointed this out; I have not been able to read all 5033 comments.)
It seriously didn't occur to me that we could split these rolls of pennies. I just assumed it was an artificial problem, and the roll with counterfeit pennies had an arbitrary number of counterfeits.
This is a valuable lesson on checking your biases and assumptions before attempting to solve a problem.
Except the was no way to check your bias here.
Just because the weight of the fake coins was different does not mean it was consistently different. Some could be too heavy, some too light and some could get the weight just right.
Riddles like this require you to think out of the box to an exact point and no farther.
@@JoseMartinez-ll7vo Yes, when talking about the entire roll, it would average itself out. But you might not be able to tell if you weight each penny individually.
i didn't understand sh!t about that gauss method. He went from solving the problem to solving why geniuses are geniuses. Idk, it made me quite uncomfortable
@@picapica8266 yes he scrambled the information.......all reall pennies weigh the same, all fake pennies the same but less than the real pennies, each roll has 50
if you remove 1 from roll 1, 2 from roll 2 and so on......when put the rolls on the scale you can calcufugure from the weight how many fake pennies are weighed, thus telling you which roll contained the fake pennies.........if it weighs exactly how much it should with all real pennies then the roll number is 50.....you removed all 50 from roll 50, if the weight says it is missing 25 fake pennies then it is roll 25
2:03 At worst it would take 13, since you don’t have to measure the last stack or the last roll. If the rest of them check out, then you can infer the fake is in the one you didn’t measure.
Same with doing them 1 by 1, it would take at most 49.
Thanks, I also realized that
That's assuming the premise that one of them is actually fake is correct. Which in this case, it wouldn't have been.
50 / 2
25/2
13/2
7/2
4/2
2/2
that works out at 7 times I think.
@@troodon1096 Then at worst it would take 14. Check all 5 groups (worst case) to confirm one group is off before proceeding, then you only need to check 9 of the 10 rolls to confirm the last one is fake (worst case). Likewise you could check 4 of the groups, then check all 10 rolls in the final one that's assumed to contain a fake. This option is worse though, as you would find out there's no fake later on than in the other order, but the amount of tests to find a fake if one exists are the same.
I came to the comments to see if anyone noticed lol
You could split the rolls into two groups, set them equally in rows on either side of a balance beam and measure the force acting on the heavier side. It will be proportional to the distance from the pivot and hence you will be able to determine the distance from the pivot on the lighter side will be the dodgy roll. Much quicker that unpacking them all...
Pretty cool visualization of a conceptual problem. Love it
I think you could solve this by placing all of the rolls on a board, spaced out by prime intervals, and then find the center of mass by balancing it.
Or on a wheel and well lubricated axle. The light coin-roll will rise to the top automatically
Why
I think they can be spaced out with equal intervals and it would still work.
@@joseville I think you are right.
@@joseville but... but... primes?
Kevin brings the wisdom! The more nodes (diverse basic knowledge and experience), the more opportunity for connections leading to insight. Wonderful presentation.
Yep. The true value of education even though you may not use most of what you learn. Now if they would couple that with critical and skeptical thinking....
That's one cool job to have, never disappointed to check Vsauce channels from time to time
I'd just measure half the stacks then half again until I found it
Numbers are fun, but this breaks down on a practical scale.
I think my system 1 is knowing how subtle differences can create imbalances and my system2 is using that to concieve of placing the rolls sidebyside on a ruler and measuring the weight of both sides. With perfect measurements, placing, composition etc the delta in weights could be used with geometry to calculate which is roll is lighter.
This is the best analogy/visualisation and even motivation I've heard regarding learning and information. Damn. Great job Vcauce
Saw this problem on an episode of Columbo. They phased a little differently though. You were measuring gold coins and a penny scale that could be used once.
Came here to see if anyone would say this! We had a similarly phrased problem on the test to enter the maths faculty (idk how to say it in english), with the king measuring gold coins on a scale that can be used once
Sorry sir, just one more thing...
Oh the memories of watching the show with my dad
_"Saw this problem on an episode of Columbo."_
YES! I was just going to say that. In fact, I may go ahead and say it anyway as a separate original comment.
The episode in question is "The Bye-Bye Sky High I.Q. Murder Case" from 1977, starring Theodore Bikel as a murderous accountant, Sorrell Booke ("Boss Hogg" from the Dukes of Hazzard) as his victim, and a then-unknown Jamie Lee Curtis as a waitress.
Why could the scale only be used once though?
I guess my intuition ruled out unpacking the pennys, because it would mean much more work than doing multiple measurments. I wonder If I would have gotten to the solution when the Problem was framed in wa way where unpacking was actually easier than measuring multiple times.
btw. Good luck finding the fake pennys in that collapsed pile^^
So, this problem is about inserting a "signal" into the input of the black-box, and then read the output, assessing it according to the knowledge of the previously established pattern.
This reminds me of using a lock-in amplifier that we used in our research-lab, decades ago. Mix your weak analog signal from an experiment with a known sine-wave with a very precise frequency. Then, mix your received signal, which is loaded with noise, with the same frequency and filter it. The sought signal appears out of the sea of noise in a crystal-clear quality.
The premise of this particular problem is that we are allowed to do only one measurement, whilst dealing with 50 individual components. So, the answer is to impose a specific pattern onto those 50 components, thereby rendering them unique within the previously uniform distribution of 50 equal components. The output "signal" will carry the pattern of the input, identifying the one and only component that decided the outcome.
I would never have known that opening the penny rolls was even allowed. Overall the work involved in this process is actually LESS efficient than just doing 50 measurements, unless the problem is that you have a dying scale that can only take 1 more measurement.
Math problems allow us to think creatively. The practicality of solutions is less important then the problem solving skills you develop. Word problems like this are simplified toy versions of complex problems. For instance the internet is a glorious mess that's half bodge half space age technology but it works.
Agreed. The problem is poorly defined.
@@JD-tp2lj no, it wasn't
That's the difference between a mathematician and an engineer (in theory/education; in practice it's the difference between a mathematician and literally anyone else). One will spend lots of time to devise a way to spend lots of time on the problem at hand. The other one will just get the tedious work done fastest.
Yeah it’s not optimal but still really cool
We don’t have pennies in our currency
Our currency is backed by cheese
Why are you everywhere?
@@juanromannavarro1303 Why are you everywhere?
your currency is act- * gunshot *
2 wholes and 2 holes
But your people can't afford cheese
Very well put. I feel like I've got so much junk in my System 1 and maybe they'll come into use one day.
Without watching, here is my first solution: roll them all into each other. If they are the same mass, they will simply stop moving they collide. If they are not, one will push the other with slightly more velocity. At maximum, you will need to try this 25 times.
You're better off using some sort of binary tree. Measure one half of the pennies, and then the other half. The one measurement with a lower weight has the counterfeit roll in it. So you take that set, divide it in half, and do it again. It'll only take like 5 or 6 measurements to identify the counterfeit roll.
Exactly what I thought, it would just take a couple minutes lmfao. Honestly, it would probably take less time than that whole system took.
@@saihajmangat2995 point of the video isnt the system to figure out the stack of counterfeit pennies. its about the accumulation of knowledge to one day use it to figure out a much more complex problem.
That’s not scalable to a large dataset though. That’s where this kind of stuff comes into play and matter when we’re talking about billions or trillions of stacks of pennies.
Einstein, do you know why going by this method he suggested to divide them on 5 parts? BECAUSE YOU CANNOT DIVIDE 25 COINS ON 2 EQUAL PARTS!
Binary tree are too big words for someone who failed preschool math.
@@Qwex1992 You don't have to divide 25 in 2 equal parts. I never said you had to divide 25 in 2 equal parts. There's plenty of ways you can use a binary tree-like system to measure 25 coins.
I understand that comprehensive reading is not your strongest point. And neither is thinking before you post.
The realization of the solution as you started taking pennies out of the series was SOOOO satisfying. Great video.
Weigh them on a balance [you place each of them at x times a fixed distance from the centre ] and measure the moment of torsion vs expected.
This is an example of overthinking a problem. It's more complex to make the system "easier" than to just go with what experience tells you. Life lesson, folks. Fancy methods aren't always the best.
Fing it in one step: swing all the rolls together by applying equal force, and you will get the real one out
I approached this from a rotational dynamics perpective. If the fake stack has a different mass then it also has a different moment of inertia meaning that, if the dimensions are the same as the other stacks, the rate at which the fake stack rotates will be different than that of the others. You could place them all at the same position on an inclined plane and just observe which one does not roll with the others when they are all let go at the same instance.😅 It is a bit impractical but you would know immediately.
That sounds like the best method, wonder how that'd work out
It will roll more quickly given the same force but the force from gravity will also be less due to being lighter.
This only works if the moment of inertia is not linearly proportional to the mass like the force due to gravity is.
@@puddlejumper3259 The acceleration due to gravity does not depend on the mass of the object. mass would effect the friction with the incline plan.
but how far would the need to go no make a noticeable difference the release would not be 100% in sink and the mass difference is small.
There's also a way to do it using a scale.
Have a long plank, balanced in the middle on a fulcrum, and then lay the stacks equally distant from the center on both sides.
It will lean one way or another, and based on how fast it's leaning (or if you have a sensor, or the plank is suspended by some springs) you can calculate which stack it is, based on the amount of force generated.
Why does he look like he's having a panic attack during the whole video
that process feels like it would take more time than to just measure half, then to measure half of the failed set, then to measure half of that set until you get to 1.
The time it took to get the pennies out of the things you could’ve measured all fifty stacks
"Which one of you... Is looking a li'l..."
"Don't say it, please don't do it"
" *S U S ?* "
LOL
I got half way there, but I hadn't yet thought of the consequitve numbers method. I was just trying to brute force it in my head.
My Algebra 1 Math teacher actually quizzed us on this exact same problem, if only I was able to watch this video before.
I didn't know that taking apart all the rolls to find the fake was an option, since you're essentially doing destructive testing and mixing the coins at that point. If you have a way to keep the coins separate on the scale in order to rebuild the rolls, then this might have a purpose. But it's obviously much easier just to weigh the complete stacks a few times.
I mean, you could always weigh with the rolls instead of without, that way you can separate everything easily on the table
I agree. I'm not fond of riddles where the solution breaks strongly implied rules. That happened to me a lot in school- I'd have the correct solution but dismiss it because I thought it broke the rules.
@@moonbeamstry5321 The moment he started with the unwrapping I was like "WAIT! That's against the rules!". With this "cheating" allowed you technically have to not weigh at all, because you could just take a coin from each roll, melt them down and check their metal content and find out which ones are the fakes. Or get an expert to evaluate the stamping etc
@@moonbeamstry5321 Thing is: You say this rule was "strongly implied", but in fact, there is no way to solve this problem while sticking to both the "real" rule (just one meassurement) and your implied rule, which is not a rule at all.
@@Chaosghoul why did you assume you couldnt take them out the rolls? I didnt get that implication at all. I'm genuinely curious I watched twice to try and fine it.
When you said the false stack can be identified in one measurement I immediately thought at Archimedes law of buoyancy, I can place all stacks in a tub filled with a liquid dense enough to make the stacks float, the stack that floats differently is the false one
@@Imperial_Squid Well, the report between the density of the object and the density of the fluid is equal with the report between the weight of the object and the weight of the displaced fluid. He was looking after a weight difference through his method, so I guess I can still find the false stack through Archimedes law, but now that I think about it maybe I should put each stack in it's own tub, not all at once🤔
Like, there is a faster way to do it. If you take the number of stacks, split them into two groups of equal size call them groups one and two, and calculate how much such a group should weight, you can weigh half of the stacks in a single measurement. We will call that result A. If result A is less, then the fakes are in that group, if not then you know that the other group has the fakes. You then repeat the process with group 3 and 4 coming from the stack that got split. If it is odd, you measure the left over stack separately, and since you know the weight of a single stack, if it is the odd one out, you solved it in 2 weights. If it isn't, you know the weight that stacks 3 and 4 should be. Weight one of them, that is result B and it will tell you if you weighted the fake stack or not. Repeat this process until you have measured all the stacks in individual groups down the fake stack. This should result in finding the stack in less weights then by measuring n split stacks since the faulty stack has either been split off and weighed separately, or has taken exactly how many halvings it takes to get to 1 stack remaining. Since there are 50 stacks, the first weighing will be a stack of 25. From there, it will split into two stacks of 12 and a single roll left over. That is 3 weightings to get to 6. Then 4 weightings to get to 3, and then at most 2 more weightings gets you the roll. The minimum is 2 weightings, and the maximum is 6.
I used to work in a resort hotel in Florida. One of my tedious jobs stacking the drinking glasses on a table. I would stack one on top of two on top of three on top of four... you get the idea. When doing this, I discovered that if you multiply the number of glasses on the bottom by 1/2 the number of glasses, and then added one half, you get the number of glasses in the stack. So, 10 times 1/2 10 plus 1/2 10 or (10x5) + 5 = 55. By the way, don't stack the glasses 10 high in a resort hotel, the managers tend to get upset.
I was waiting the moment he says the balance only has a 99% accuracy making the hole measurement process false 😅
the weight of each penny is probably different enough to throw the numbers way off.
@luuk indeed it was a poor choice of word
Real pennies weigh 25% more than fake pennies. 99% accuracy is more than good enough.
A) it’s not about the percentage (the scale could be more or less precise: I simply put 99% because I would assume it’s close)
B) but for the sake of the example: If we assume the 13th stack is the false one and the scale is 99% accurate : you would assume the total weight is 3187,5 minus 13 times 0,5g so 3181g total. But the confidence interval would be [3149,19; 3212,81]*. With this interval of confidence you could literally assume every stack is the false one.
* because we don’t have the standard deviation of the measurement we (I) can’t calculate what the exact confidence interval is but for the sake of the example a simply took 0,99 and 1,01 times the measurement (I know this isn’t how it’s properly done but it help to see how big 99%accuracy or 1% error can affect this problem
My method was like this: put all of the 50 coin stacks on the scale then start removing them one by one till you find the fake one.
That IS one measurement isn't it!
Nice ;)
this is not one measurement, the moment you take one roll of and compare you have measured twice
the method kevin uses has an estimate and then he measures the rolls once comparing the estimate to that.
@@richardgurney1844 it could be a maximum of 49 weighings.
Here's the component you forgot with a "real world" elegant solution: time. The simplest and easiest solution is often the best. It would take maybe 2 minutes to measure 50 rolls, how long did it take you to organize, unwrap, select a specific number of coins, stack the pennies, then do your calculations? So what if you did it in one measurement. The first method did it faster.
Also, there's the chance you find the aberrant roll in your first measurement.
The actual important lesson here is how the framing of a question with misleading visual cues ("stacks" of pennies that are in paper rolls) can mislead an audience and obscure potential solutions.
0:37 time stamp for amogus joke
3:55 you heard it, we all heard it
It was seeming like you wanted to keep the stacks whole (not break the wrapper) while also singling out the fake stack. Later knowing that you could divide the stacks did make the conclusion much more obvious.
I think the one problem I have with this is that from a practical sort of view use a classic weighing scale (the ones Gods of justice use) and dividing the stacks into progressively smaller groups of 2 would from a practicality stand point take far less time to do in discovering the answer, taking only 5 weighings over the course of a few minutes