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

“

"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?"

Same

Yes lol

Europe gang here

Nooooo penny dilemma

me being a penny collector 👁👄👁

Nice

way to think about it

.

.

The reasoning is that you don't know if there is a fake or not

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

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.

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 🔫

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.

Wtf spoilers

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.

'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.

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.

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.

Same

Yup!

@@ChayComas He saying Sus

He even played the music..

And then he called the counterfeit roll the _Impostor_

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.

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

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

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

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.

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...

That's modern Vsauce for you...

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.

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.

Why binary, and, for example, not one suggested in video? It worse in optimistic scenario but much better in the worst case.

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?

KZfaq 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.

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.

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

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.

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).

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.

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.

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.

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.

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!

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.

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.

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.

SUS PICIOUS

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

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

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

except the puzzle is how to find the fake stack with only one measurement. so ur solution is not the right answer for....how do we find the fake stack with one measurement

@@dagg310 I know that for the funky and arbitrary limitation that is true, but for real world testing speed a binary search is faster

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.

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.

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

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?

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)

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?

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.

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

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.

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.

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

Oh hi Carl!

Let my torque encode with distances the 50 weights by their lever arm input. Nice. Are you an experimental physicist, structural engineer or what?

LOL

Also, there's the chance you find the aberrant roll in your first measurement.

Ayo… it’s the son of da roach dogg

Is that really the RoachDogg's little boy?

so most missed the point then?

@@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🤔

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.

That’s not one measurement. Kevin’s method allows you to find the roll of fake pennies taking just one measurement, that’s the whole challenge of this problem. The problem isn’t just “find the fake pennies”

@@TheWiseGoat00 If you accidentally peek at the reading when you're carefully placing the unwrapped stacks on the scale you instantly explode.

@@nyororomeowelas1225 exactly, someone gets it