What do fractional bases have to do with cows?
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- čas přidán 15. 08. 2022
- An algorithm involving fractional bases and irrational numbers can tell you how many cows you have.
This video was made for the 2022 Summer of Math Exposition (#SoME2).
For more about negative bases, the Josephus problem, and a written derivation of the formula for the cow problem, see drive.google.com/file/d/14SE5...
thank you now i have many cows
This does pose the question: would you rather fight 10 blobs of mass 1 or 1 blob of mass 10?
10 blobs of mass 1 would essentially be the same as 1 blob of mass 10, except it can attack you from 10 angles at once
@@tastygold Sure, but 1 blob of mass 10, being one coherent blob, inherently coordinates better than 10 blobs of mass 1. Worth putting that into consideration
Bases can also be different in each digit. E.g., the least significant digit base could be 7. The next one could be 5. The next could be 3. You can get the original digits back out by dividing and using the remainder. I use this technique in computing to pack numbers together into less space when I don’t have nice powers of two. E.g., I need to store a number that can range from 0-8 (9 values), and a number that can range from 0-27 (28 values). Naively using 4 bits for the first and 5 bits for the second, we exceed 1 byte. But 28 * 9 = 252 which is
Multi-base numerals are also used in Hashcat's Mask Processor, but for different reasons. MP works like a traditional brute-force password cracker, but each char can have its own charset, and each charset can have different sizes.
Since the charset isn't treated like a set, but rather, a sequence of chars to test (an ordered array), each char in the charset has an implicit value equal to its index. This essentially means that the string is a numeral where the digits are (usually) alphabetical chars
@@Rudxain Yes. There are multiple representations for the same value many times. And projections from one problem space into another.
If your digits are base 2.4,10,6,10,6,10 you've invented the clock.
@@canaDavid1 I thought it was 12, 60, 60
If Φ is the Greek character designating the Golden Ratio, then I suggest μ (pronounced moo) for the Bovine Ratio.
This is brilliant
mu(μ) is already used to represent mean)
@@thegamingengine4537 most Greek letters have lots of usual meanings.
@@thegamingengine4537 meuh non, ce n'est pas un problème
The noise in the audio was a bit distracting (I think Audacity can clear up at least some of it out of the box), but this was a cool, nicely animated introduction to fractional bases and this Josephus problem constant.
I will definitely improve audio in the future - thanks for the recommendation. Great to hear!
The audio was also suuuuuuper quiet. I could bearly hear you on max volume. You have a good voice and great content. Can't wait to see what you make next! :)
@@adamscura9826 The audio was a normal volume when I was at like 3/4 volume so I don’t think it’s him
This was a lot of fun & a tricky topic too - I've never come across fractional bases.
Great job!
PS Enjoyed the Confucius quote!
Wow!! I would have never imagined that an apparently simple problem like the one of the cows can be studied via fractional bases. Thank you!! This video is amazing!
There's an interesting video about the sidenote presented at 3:10 on Jan Misali's channel about naming these bases.
Aha, great video - we didn't stumble across it in our research. So base 3/2 would be 'trivotbinary'.
For anyone who's interested:
czcams.com/video/7OEF3JD-jYo/video.html&ab_channel=janMisali
I'd expect that pairing two cows would produce zero calves. I'd pair a cow with a bull.
Yes, the narrator seems to be unaware that “cow” is not a species, but a female of any of several species, most often Bos taurus, “cattle”.
I think Minecraft cows are hermaphrodite.
Very interesting, I don't think I was able to really follow your explanation unfortunately but it definitely got me interested in other base representations
Any feedback on why you couldn't follow would be very useful. The script, animations, pacing...?
@@fauze5456 For me, the jump from alternate representations of 4 to the C(n) and R(n) was a little difficult to follow. I might have a little more time spent on this. But overall great explanation!
That's where I got lost too. Give me some time & restate what happened when we come upon a new and very novel conclusion.
And you said "make sure you understand the connection..." On one hand, an engaged listener/viewer has some responsibility to pause and ponder. On the other hand, it's your job to prevent me from being stranded at sea. I did not understand, and rewatching didn't really help.
A great start to your channel. I'm looking forward to seeing more :^)
13:12 "What is the growth rate if cows die?".
I'm afraid I don't follow your math well enough yet to use it in answer, but a branching diagram of the immortal case on lined paper with each line per generation could be an alternative, and could be adapted to illustrate which cows die when, and the consequences for subsequent generations.
Start off with 3 blobs on a line, the first 2 of which make out and the 3rd waits its turn. We represent the progeny of the first 2 with a blob on the next line, joined by two diagonal lines to its parents, so you have a V shape. But those 2 parental blobs also survive to the next generation, and this is symbolized by 2 vertical lines leading down from them to blobs on each side of the progeny blob, which now gives you an M shape. The singleton blob in that first 3 however simply has a vertical line leading down to a blob on the next line, symbolizing its survival to the next generation, and looks like an "I". However its patience is rewarded when as a blob on the next line it gets together with one of the 3 other blobs on that line.
So you should get MI on line 2, MM on line 3, MMM on line 4, MMMMI on line 5, etc, and the blobs where these letters rest on each line respectively should amount to 4, 6, 9, 13 and so on in accordance with your sequence. Hope all that's clear. What changes would you then make to kill off some of these cows?
I watched with wrapped attention, and as my grip of attention wained, I realized a sane person would gently yield his grasp, and realize reality was seeping away, leaving a perspective of complete and total confusion, with the phrase, "You are either entirely out of your mind, or hostage to a mathematical accounting system that has no empathy for any organic life forms." Knowing that your GRASP is well beyond the reach of what meare mortals could understand, I followed the rationality of your presentation to the very edge of the precipice where for self preservation I pulled back, and watched as the presentation slipped over the edge into the grey of the unknown. In my life (70 years) I've rarely visited this precipice, yet it is an experience of a lifetime, though druggies likely do it daily, or hourly, yet I intentionally shun it, as if it were yielding to the struggle to retain life. So much for my struggles. My sincere respects to the author of this video. To the audience, "DON'T DO DRUGS!!!"
Good video!
I'm once again reminded of how glad I am I played that old Rama game for the PC in the 90s, it might have frustrated the hell outta 8 year old me, but it also forced me to learn to do non-base 10 math at that wonderful age where you can just learn shit intuitively and keep it for life.
Dude this is so weird.
I remember watching a video much like this a while ago. I loved the channel that it was on but just can't find it anymore. It's nice to see this again.
P.s. if anyone knows the video please send it to me
Anytime i think about cows in any mathematics or scientific sense i remember that we should assume they are perfectly spherical
Amazing video!!
Very cool! Can this be used to study the Collatz conjecture? e.g. if we computed "r" for all starting values of the cow sequence (instead of just 2), is there some bound we could prove on r that would imply Collatz?
Most likely not. The problem is, knowing if 3n+1 would be even for any n is very hard
@@darkking571 3n+1 for any n in natural numbers is always even.
Edit: I messed up really badly here. Thank you Tom for correcting me. 3n+1 is even for any n that is odd. Which is what we would do i n the Collatz Conjecture.
@@austingarcia6060 Er, no. If n is even, 3n+1 is odd. Try n=2.
@@BridgeBum ah, yes, sorry. I was thinking in terms of the Collatz conjecture, 3x+1 is always odd if x is odd. And we only do 3x+1 on odd numbers. You are right. I'm not thinking right,
Impressive!
What an interesting proof!
However, this line: "depending on the starting value, Goodstein sequences can get larger than Graham's number" is a really trivial claim. If you pick Graham's number + 1 as your starting value, then the sequence already starts greater than that.
Some people need coffee in the morning, some need to throw cold water on their face, but me? Naw, i wake up by seeing Math problems on CZcams. Oh by the way, i learn math best in reverse. I go from complicated to simple, not simple to complicated. That way i know what's the bigger picture first, then all the tiny details later. Please upload more of those adorably cuuuute cows🐮
very well done
While I do think about different representations of numbers sometimes. I often end up in bases that aren't constant. So for example base_k where the first column is base_1, second base_2, etc etc or even use the primes instead of straight integers. This leads to great insights where you can represent some number in plenty of ways. And the amount of different representations usually says something about the number.
I never considered fractional bases.
I was a bit obsessed with the idea of fractional bases one day and made my own system for using them. I think it was tens of pages.
Wow, this is quite interesting, however I still have some difficulty wrapping my head around it all. Not a problem though. Thanks for your contribution to me learning something new 😊👍
Would base 3/2 be related to base 7 in any way? Like by a factor of 2 somewhere?
8:33 "For this, it makes more sense..."
Not to me. I easily understood why with the visual of the explanation before this. The explanation after was much more confusing, and even when I did understand at least where the numbers on the screen were coming from, it seemed further away from the cow problem than the previous explanation. Exactly the opposite effect from what you were apparently intending.
I felt the same way.
Another attempt on your question 13:12 "What is the growth rate if cows die?".
Start off as I did before with 3 blobs (representing 3 cows) on a line on a sheet of lined paper. Once again two of the blobs produce a third, which appears on the line below, but in this scenario the parents then promptly die. However the third blob, the one that doesn't yet get to reproduce, survives to the next generation. So there's now a total of just 2 blobs on the line below. These two produce a blob between them, and die. So now there's only 1 blob on the next line below. Being a singleton it survives to the next generation, so there's also just 1 blob on the next line. It then dies, so next line after that there are 0 blobs.
With these rules it's a negative growth rate. But notice how the population total changes. 3 2 1 1 0. It's the beginning of the Fibonacci sequence in reverse. If we start at any Fibonacci number like 13, we can have 10 (5 pairs) reproducing and 3 singletons. The rule is no singleton can couple with another singleton, they've got to wait till the next generation and couple with the progeny of a pair. This results in 8 the next generation. Then we can have 3 pairs reproducing and 2 singletons, resulting in 5 the next generation. This has 2 pairs reproducing and 1 singleton, resulting in 3, and so on as above.
What if you start at a different number of cows than 2, like 5? Do the sequences of numbers of cows ever connect? Do they behave differently in other bases, like 4/3?
7:07 isn't the equations on the right false cuz neglecting the remainder?
The [10] on the far right means that the digit in that spot is a 10. So 0.[10] = 0 + 10*(1/10), 0.9[10] = 0 + 9*(1/0) + 10*(1/100), etc.
@@fauze5456 I love the video, but that part absolutely needs explained because that confused me every second after 6:40. Without explaining that, I couldn't (and still don't) understand the core logic past 6:40.
Have you considered this problem for base transcendental numbers? Sounds like a good future project :D Keep up the good work!
What about Base i number system?
Yes, although just like binary is significantly more useful than base 1, the smallest complex/imaginary base you'd want to use is base 2i (quarter-imaginary base):
en.wikipedia.org/wiki/Quater-imaginary_base
Might go into more detail in a future video :)
I thought the graphic in the thumbnail had something to do with how if you breed two cows, you'll then have the original 2 cows plus a new baby cow. Then, you can only have one additional cow when you breed any 2 of the 3 cows, so then you'll have 4, and the 2 pairs can breed and make another pair.
how did you make these animations if you did not use manim?
Using a personal python library created for this video. It is very messy, but if I clean it up I'll put a GitHub link in the description.
For the text, it uses LaTeX and presumably a similar method of fading in the text to manim, hence the 3B1B look.
You haven't really solved the problem using fractional bases though, you've just pushed all the complicated part onto r. Without an analytic formula for r, I fail to see how the solution given at the end is in anymore of a closed form than the recursion relation at the start.
This is addressed in the conclusion. I’d implore you to watch until the end
@@fauze5456 Fair enough. I wasn't expecting something conceptually new in the conclusion so I had stopped then. I think I'm probably not alone that, it's standard writing advice to not add new ideas - especially important ones core to whole essay - in the conclusion.
mengy momento
based
At 3:25 you probably meant 10^{-2} for hundreds, instead of 10^{-1}
Ha, good catch! Thanks
*Everything* has to do with cows!
I don’t wait for the baby cow to grow up before I breed the parents again. This means the first generation of 2 cows actually produce several babies. Since the babies are born at different times, they don’t all join the next generation breeding population at the same time. The generations overlap as the eldest siblings start making the next generation of babies while the youngest are still babies themselves.
How the heck do I make a function out of this and plot it to see the curve? 🐄
And does this type of growth have any interesting properties? 🐄
like idek if this is exponential or something else 🐄
It should plot both the population of the adults and the babies, and it should be the most optimal graph (cows breed the instant they are able to, like when they grow up or when their cooldown expires, and then there are 2 of them) 🐄🐄
Is anyone here a fan of base 6, like jan Misali is?
This is some cursed shit
Unu bovino muĝas, mu
ĉu ĉi tio estas Esperanto por “one cow (female) moos, ‘moo’ “?
@@ValkyRiver Jes! :D Ĉu vi parolas E-on?
Ĝi estas aludo al tre fama infana kanto, _Dek bovinoj,_ pri kiu oni ofte faras memeojn.
@@ValkyRiver Ĉiam feliĉigas min trovi E-istojn en ne rilataj lokoj :D
"Memeo" signifas "meme" en la angla (kaj ĝi sonas kiel "mi pisas min" en la hispana, sed tio estas nerilata afero, mdr).