4:45 - Brady - If I give you any number... Neil - Yes... Brady - ... can we always get to 1? Neil. Yes. So that's a very good question, and the answer is "No".
Fun fact: the name of this sequence is actually a pun: "Choix de Bruxelles" (Brussels Choice) is one letter away from "Choux de Bruxelles" (Brussels sprouts) in French.
Sometimes I feel like mathematicians just take numbers, smash them together like action figures, then write fanfiction about said action figure smash as their academic dissertation.
Sometimes even the most contrived maths turns out to be unexpectedly useful in some way, but even when it isn't, it's interesting enough in its own right to justify its own existence.
Just to be complete, find the first nonzero digit from the right. If it's a 5, double it to get 10. Turn the number formed from that last nonzero digit and every digit to the left to a 1, then turn every 100 to a 10 until you reach 10 itself.
@@danielyuan9862 Thanks a lot! There were several holes in his proof but this was a major one. It really bugged me till I thought: I can't be the only one, I'll search the comments. Bingo!
@@ReasonableForseeability None of the numbers in Numberphile need be useful. Some certainly are, of course, but this is Maths. Maths is abstract, and does not need a practical application.
Fascinating episode. Please, please make more of these videos with strange iterative rules that lead to amazing patterns. They are amazing to watch and this one actually struck a chord with me :)
It looks like he doesn't like book covers. He looks at either the white paper sides of the books, or brown paper titles. Maybe he finds all the different colours, fonts and font sizes too distracting?
@@pmcpartlan It should have been pointed out. It's definitely not self-evident (and I speak French reasonably well.) Animated sprouts would have been a welcome touch!
Dr Sloane is a hero of mine. I've got a few sequences on the OEIS, and I would totally lose my s**t if he talked about one of them on Numberphile. I fear none of them are interesting enough however.
This is a really nice operation. The iterative process reminds of Collatz' sequences, but instead we have 1) more choices, 2) much more can be proved about the sequence.
On intuition, I'd say any even base (except binary?) should have the 0/5 problem, but instead of 5 it would be your base/2. I'd be interested in odd bases, whether prime or composite.
@@odarkeq That can't be quite right, though, because in base 4, base/2 is 2, but you can get from 2 to 1 by halving. So you'd have to exclude any base that is itself a power of 2.
I don't think that the base being even has much impact; rather, any base with an _odd factor_ is gonna have the same kind of problem as base 10, where an odd factor of the base can't be halved and can't be removed by modifying substrings of the number. Bases that are powers of two also have a problem, though, in that starting from 1 you can only get to powers of two (e.g. in base 4 all you can do is go 1 -> 2 -> 10 -> 20 -> 100 -> 200 -> ...)
the digits of 0 and 5 can not be connected because you cant get to 5 without {odd number}0, witch ends in 0. and you cant get to a number that ends in 0 without and number that ends in 0 or 5
What can we say about sequences formed in that way: the third number is the linear combination of the two previous. Example: F(-1, 0) = 1, 0, -1, 0, 2, -2, -2, 2, 2, -2, -2, ..... F(1, -1) = 1, 0, 1, -1, 2, -3, 5, -8, 13, -21, Of course F(1,1) is the fibonacci sequence. The two starting numbers do not matter since F(a,b) is a two dimensional space and (I,0) and (0,1) is a base. Questions are: F(a, b) is periodic? It goes to infinity or minus infinity? It has subsequences bounded? etc.. It is asintotic a geometric sequence as Fibonacci sequence?
The reason why the alg. at the end is O(12*log(n)) = O(log(n)) is we get a geometric progression: After 12*log(n) steps, the # of digits halves, giving: 12*log(n) + 12*log(n)/2 + 12*log(n)/4 + ... = 12*log(n)*2 = total steps to get to 1.
The wording on the matchstick problem is a bit ambiguous because you're not really creating squares by removing matchsticks but rather reducing the number of squares from 10 to 4 without any restriction on the number of any other polygons nor that you even need to close all the shapes so you only need to remove 3 matchsticks to get 4 squares assuming you follow the rules from the last slide where the fifth square is the outer one. _ _ _ _ | _ _ | _ _ | | _ _ | _ | _ | This figure has the two large squares and the two little squares. You can also just remove the top row of 4 matchsticks and are left with just the 4 small squares on the botton. This means you can remove any other one or two of the matchsticks that don't make up a square and still have 4 squares and five fewer matchsticks, resolving the problem.
That numbers ending in 0 or 5 cannot get to 1 is very intuitive: any number ending with 5 can only be doubled and the new number will end in 0. Any number ending with 0 will end in 0 when doubled or 0 or 5 when halved.
A small clarification and a shortcut at the end. It takes 12 steps per pair of digits to go to 1. That is convert two digits into one digit. So in total it takes 12*k/2 for k digit number to have all its digits halved basically. You recursively then do it on a resulting number with half a digits (or half + 0.5 if there was an odd number of digits in the first place). That is 12*k/4. Then 12*k/8, etc. The total sum is
What if you are allowed to reverse the digits? Leading zero's get removed. That way you can do all of them to 1. For example 5 to 1: 5 -> 10 -> 01 -> 1. If you are allowed to add leading zero's, you can do it the other way around too...
This math series of favourite numbers bigger than 1 million is awesome! Spreading awareness of maths, thats what humans need, not a gaint wall between countries.
14:33 - you can get a tighter lower bound than 12 steps per digit. 1. Turn each pair of digits into 1 in 12 steps. That's 6 steps per digit. 2. Turn each triplet "111" into "1" in 8 steps: 122 -> 62 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1. That's 4 more steps per digit. Total is 10 steps per digit. I'm sure someone can do even better.
@@dcsignal5241 I tested using base 7 and found that 3, 5 and 6 connect; 1, 2 and 4 obviously connect; and that the two sets do not connect to each other. Numbers ending with 0 don't connect to any other type of number because 10 is an odd number in base 7 and divided by 2 is 3 and 1/2. Example of a path from 3 to 5 in base 7: 3 >> 6 >> 15 >> 113 >> 43 >> 23 >> 13 >> 5 If you can find a path from 1 to 3 that I couldn't, let me know. =]
@@Cuuniyevo It's easy to show that you can't go from the 1, 2, 4 set to the 3, 5, 6 set, ever. Since the operations are reversible, showing that you can't reach either set by doubling a number from the other set is enough. And that's obvious to prove, since you can trivially see that by doubling the digits themselves you never leave the sets.
I think it isn't fully explained, how you get to 5 from every number with a 5 or 0 at the end, because it can happen, that you have multipe zeros or fives at the end and then you can't argue like in the video: So if you have a number with a 5 at the end, just double the five and get a 10 at the end , so you can reduce the number without the zero to 1 and then devide 10 by 2 to get 5. If it has a 0 at the end, the number is divisible by 10, which means it is divisible by 2 and 5. So you halve the number. Because this new number still has to be divisible by 5, the last digit is either a 5 or 0. If it's a 0, just repeat that step until you get a 5 and then do the procedure above to finally get down to 5.
He says that any number ending in 0 or 5 can be brought back to any other number in 0 or 5, but then also says that for example reducing 117930 to 10 is possible because 11793 doesn't end in 0 or 5, suggesting that you couldn't, for example, reduce 117950 to 10 because 11795 ends in a 5
I was thinking about this operation in binary, because doubling and halving are fun in binary: it's just adding or removing 0's (i.e. 110 is 2 * 11, and 111 = 1110 / 2 in binary). This means, if I'm understanding correctly, this binary version of the operation groups the numbers by how many 1's are in its binary representation (zeroes can be added and removed freely, but it seems like ones cannot be created or destroyed). This has to do with the operation doubling and halving and the relationship of those actions to the base, i.e. if we were looking at tripling or dividing by 3, base 3 would have the same behavior. With that in mind, I find it interesting that there are only 2 "equivalence classes" in the base 10 case (multiples and non-multiples of 5) while there are infinitely many classes if the operation matches the base, so to speak. I'm confident this must relate to the prime factors of the base and how the operation matches one of those factors, but I haven't investigated enough to really see the connection. I wonder what happens if you multiply and divide by 3 in base 10 (the operation not matching a factor of the base) or by 2 in base 12 (since 2 is a factor in 12 2 times).
I feels like this should be easier than the Collatz conjecture since you can control what part you're modifying. … What about leading-zeroes; can you turn 1040 into 120 by halving 04? 🤨 Can a number that has a 5 or 0 _anywhere_ in it collapse to 1? 🤔
A couple years ago on another Neil Sloane Numberphile video, someone posted in the comments that Neil reminded them of Professor Farnsworth from Futurama. I have never been able to unsee that. Fateful commenter, if you're still around, I thank you.
It‘s interesting to see what happens in different bases! 12 is a base that has a lot more factors. You can’t escape a multiple of 6 (number ending in 0 or 6) [1/2 the base] You can escape a multiple of 4 (number ending in 4 or 8) [1/3 the base] by halving 4. You can’t escape a multiple of 3 (number ending in 3, 6, 9) [1/4 the base], which is pretty interesting! if the base is B, and n is an integer, I wonder if this only happens with numbers that are multiples of B/2^n, or maybe B/2n. In prime and odd bases, I suspect the only sets of numbers that get stuck in their own sets are ones that end in 0, 00, 000... What other interesting properties have you guys found?
I think something is missing. In the video you say you can pick any substring. What if I pick 02 from 20218? What should I do? Do I have 02 / 2 = 01 or 1? If 1 then 20218 -> 2118 and this is obviously not reversible. Otherwise you should forbid leading zeroes or force to keep them.
If you have a zero in the middle, e.g. 11011, when you double this with the number next to it (so the 01) would it become 11021, or 1121, seeing as 02 and 2 are just variants of the same number?
If you did this, the sequence would no longer be reversable. It seems to me like it would have to be illegal to transform sequences of digits starting with a '0'. You could of course play with the digits *after* the '0', so change 44014 to 44024 by doubling just the '1'. I guess this leaves the question of what would happen if you were allowed to add and remove 0's at will. This means 5 is solvable with 5->10->1, but maybe we could even figure out shortcuts for the other numbers?
@@Keldor314 Yeah. Adding and removing zeroes at will is a bit different, since this would allow, say 1000 > 1 in one step, whereas it would be impossible to do just with preceding zeroes. 5 can be gotten rid of as 10, but the zero still has to be dealt with as a preceding zero to something following, e.g. 54 > 104 > 18 > ... for instance. There's also the question of whether you could theoretically add infinite zeroes preceding, e.g. 15 > 100010. This seems unfair, so it does seem like messing with zero is off, but then it does seem like the issues with 5 and 0 are almost self-imposed? if you know what I mean.
If you are only allowing the deletion of preceding zeros then I don't think it changes that numbers ending in 5 or 0 can't get to 1 because the number at the end still stays the same and you always end up with a number ending in 0 or 1
To get rid of an inner 0, if the leading digit is odd, divide those two by 2, otherwise double the following digit until 6 or 8, then double the 3 digit group containing the 0. Either method works, though the second may take more steps depending on the number. The problem with trying to grab 0 with a following digit is that it isn’t a valid way of writing a number, so wouldn’t be allowed. And if it is considered a valid way of writing the number, it would be retained when doubling or halving in those contexts to maintain continuity. Therefore it wouldn’t help in getting down to 1 to do so. The arbitrary addition and subtraction of digits is not allowed, but it can fairly easily be removed anyway. In your example 11011 becomes 1511 by halving the 10. From there you can double the 1 following the 5 and reach 1521, half 152 to get to 761, then half the 76 to get 381, half that part again and get 191, double the trailing 1 to get 192, half the whole thing and get 96 to 48 to 24, 28, 56, 112, 16, 8, 4, 2, 1. You can probably find a faster route, even without arbitrarily removing intervening 0s.
Sure greedy algorithm will give you the maximum number you can get in next step, but I am doubt the greedy algorithm will always give you the maximum number it can get in finite step. Since sometimes small number can grow faster then the larger number, for example: 21 > 16, but 21->42, 16->112. This means you can choose grow the number slower at begin but possible make it grow even faster in longer steps.
When you can't double a digit 6 or higher, shouldn't you prioritize doubling a 3 or 4 digit, so that you guarantee the next resulting number has a digit 6 or 8 and you can increase the length by 1 again. This potentially grows faster overall, despite an individual step now growing as rapidly, because more operations will increase the length
There was a very vital bit of information missing: You can NOT take any sub-number and half it if it is even - it only allows sub-numbers that do not start with '0' - otherwise you could just take 1000000000000000000000006 and make it 16 in 1 step.
He didn't prove that the "greedy algorithm" is always best. He mentioned that there might be a "devious algorithm" that's better. It's also interesting that using the greedy algorithm, you get numbers whose only digits are 1, 2, 4, 6, and 8.
When he showed 117930 could get to 10, he didn't actually prove that *any* number divisible by 5 could get to 5: we know that 10 can't get to 1, which would mean it's still possible that you might not be able to get from 100 to 10 (since the strategy of ignoring the last zero and getting to 1 doesn't work). Thankfully, it turns out you are able, for example (and there's probably a quicker way): 100 -> 50 -> 25 -> 210 -> 220 -> 240 -> 280 -> 560 -> 1120 ->160 -> 80 -> 40 -> 20 -> 10. Because of that singularly pesky 25, we had to use the ones' digit, which is why you can't use a similar path to get from 10 to 1 (which would use the tenths' place). This is definitely enough to prove it entirely, since it also uses 50, and you can extend the whole thing by an arbitrary whole number power of ten. P.S. if anyone reads this and can find a quicker path from 100 to 10, let me know! You could splice in 25->45->90->180->280 for equal length, but I haven't found one of lesser length.
How to make any number ending in an odd digit other than 5 smaller: 1. Double the final digit 2. If that final digit was greater than 5, halve the final digit and halve the final two digits. 3. Otherwise, cut everything in half.
Regarding the biggest number in K steps, wouldn't it be possible to go infinitely big after the 4th step? 1 -> 2 -> 4 -> 8 -> 16 -> 100000...00003, that could lower the lower bound of certain numbers.
If we change the rule to multiply or divide multiples of 5 do we end up with 5 sets of numbers? I feel like the 2-ness 5-ness is base 10 related. I’d be interested to know what happens in other bases!
Also interesting to do it in other bases. In base 2, you get an infinite amount of classes where all numbers with the same amount of 1 digits belong to the same class. In base 8: 1, 2 and 4 are connected, and so are 3, 5, 6 and 7. I'm not yet sure if those two groups are also connected via some way, or if there exist other groups (but my guess would be no).
Woah. Thus reminds me of middle school on finding the prime numbers and squares in the fibonacci sequence and looking for palindromes in pi 3.(141) for instance. Good times
You are allowed to double a part of a number. From 26 to 212 for example. You kinda add a row to this number. In reverse logic. Why can't we substract a row? Can't we just be allowed to not write part of a number, if that number is basically nothing (zero at the end)? From 20 to 2. This way all numbers are connected.
According to an algorithm I wrote in Python; it takes 25 steps or less to get from any 3-digit number (that is not divisible by 5) to 1, 30 steps or less for 4-digit numbers, 34 steps or less for 5-digit numbers, 41 steps or less for 6-digit numbers. These by no means are optimal. I tried giving it some pseudo-random 100+ digit numbers and the number of steps would usually be between 1 and 4 times the number of digits.
For those wondering, the algorithm was basically: If the number is one digit long, use a look-up table for how to get to 1, else, go from right to left, looking for an even digit. Continue until either the end of the number is reached or a 1 is reached. Use the 'Choix de Bruxelles' to half the number starting at (including) the first even digit encountered, and either the 1 encountered or the end of the number (inclusive). If no even digit was found, use a lookup table for how to convert the right-most digit to a 6 (not resulting in a longer number, and there can be some variation here). Repeat this until 1 is reached. It works because; 1). excluding the case of one-digit numbers, it can only ever reduce a number's value (unless converting the right-most digit from 1, 3 or 5 to 6, but this will make the number divisible by 2 (because 16, 36, 56, 76, 96 all are divisible by 4), so its value will actually end up being lower after halving). 2). it can be applied to all positive integers not divisible by 5 (excluding the case of one-digit numbers).
I love your videos and all but what is the point? Can you add to the end of each video if there is an application/use for the math or if it is just a game?
You as a child in school multiplying integers in the way described at the beginning of the video. Teacher: "Bring out the dunce hat !" Professional Mathematicians:
What if there are leading zeros in the substring that you select? For example, can I start from 104, choose the 04, divide by 2 to get 02 = 2, and end up with 12? Similarly, can I take the 04, double it to get 08 = 8, and get 18?
![Kodak Black - Catch Fire [Official Music Video]](http://i.ytimg.com/vi/A4ch-olIuZo/mqdefault.jpg)
4:45 - Brady - If I give you any number...
Neil - Yes...
Brady - ... can we always get to 1?
Neil. Yes. So that's a very good question, and the answer is "No".
"Well yes, but actually no"
They had us in the first half not gonna lie
Well that's a typical professor way of answering... first praise your question then the answer
@@lucasmachain Saying "Yes" before praising the question wasn't super helpful though
yeah but no but no but yeah but yeah but but no but no but yeah...
Fun fact: the name of this sequence is actually a pun: "Choix de Bruxelles" (Brussels Choice) is one letter away from "Choux de Bruxelles" (Brussels sprouts) in French.
This man has gone through thousands of sequences and each one gets him excited like it's his first.
The OEIS conjecture: Neil Sloane has thought of any interesting sequence you come up with before.
Not the ones in less
I think we could all hope to find such passion in our lives
Placing in excitement rating after discovering a new sequence:
1, 1, 1, 1, 1, 1, ...
There is a Jeff Goldblum vibe happening here
Sometimes I feel like mathematicians just take numbers, smash them together like action figures, then write fanfiction about said action figure smash as their academic dissertation.
"Sometimes"? :D
That is 100% the accurate truth
I agree, and further, what’s the point?
That's quite possibly the best description I've ever seen of what mathematicians do.
Sometimes even the most contrived maths turns out to be unexpectedly useful in some way, but even when it isn't, it's interesting enough in its own right to justify its own existence.
In case anyone is wondering why numbers ending in 00 or 50 are connected to 10:
100 -> 50 -> 25 -> 45 -> 90 -> 180 -> 280 -> 560 -> 1120 -> 160 -> 80 -> 40 -> 20 -> 10
Just to be complete, find the first nonzero digit from the right. If it's a 5, double it to get 10. Turn the number formed from that last nonzero digit and every digit to the left to a 1, then turn every 100 to a 10 until you reach 10 itself.
Any stuff with 0s at the end can be halved until it shows 5 at the end.
@@danielyuan9862 Thanks a lot! There were several holes in his proof but this was a major one. It really bugged me till I thought: I can't be the only one, I'll search the comments. Bingo!
@@movax20h True, but how is this helpful?
@@ReasonableForseeability None of the numbers in Numberphile need be useful. Some certainly are, of course, but this is Maths. Maths is abstract, and does not need a practical application.
Find someone who loves you as much as Neil Sloane loves integer sequences. 😍
Make this integer sequence a date game.
So true
PC Filho as long as the number of people in your group does not end in a 0 or a 5, you will always find one :)
you'll find that person in a twelve steps program.
You'd be super-lucky then.
4:47 Brady: *Asks yes or no question*
Brussels Boi: "Yes! ...the answer is no."
LOL
Well yes but actually no
That is actually hilarious.
It's the right question but the answer os negative :)
??.
Fascinating episode. Please, please make more of these videos with strange iterative rules that lead to amazing patterns. They are amazing to watch and this one actually struck a chord with me :)
I second this motion
you could say that this is the 1st operation in a family of operations where the 2nd step is x3 or x1/3
A chord? Does it go through the center? If so, that'd be a diameter ;)
"Other numbers are available" nice one Brady
Yeah, that’s very funny.
_The number you have dialed is not available, please hang up and try again._
Best thing in all of the clip.
false..
That is a daunting stack of paper in the background...
I love all the different book labels. "UNIX" and "SIEVES" don't usually go together yet here they are atop the worktable of a mathematician.
@@Kapin05 Now we just need to figure out what a 'sieve' command would do to its standard input...
flau? didn't expect you here. i see you got a bit of a channel going. huh.
And the touch of "BRAZIL" and "BRAZIL 2"
It looks like he doesn't like book covers. He looks at either the white paper sides of the books, or brown paper titles. Maybe he finds all the different colours, fonts and font sizes too distracting?
I am very scared for that laptop in the right at 2:56
Well, you should be; after all, it is *based on* OSX, Unix, Mathematics and a lot more!
Hörmetjan Yiltiz Not ‘Mathematics’, Mathematica.
Although a little less interesting in binary, doing this in other bases leads to some new stuff (and all of it is as useful as lunar arithmetic).
What about doing this with multiplying and dividing by different numbers each time instead of 2? There's a lot of stuff that could be going on.
Oh man I love Neil
Yes actually
Nice Math discussion you got right there.
@Daniel Chang err?
Voi jonne
@@maikkelström voooii joonnneee
Sloane Incompleteness Theorem: There are sequences we may never know, and he's still gonna be excited about it.
"Choix de Brussels" is a pun on "Choux de Bruxelles (brussels sprouts)" BTW...Nicely done!
Haha, had I realized that I'd have filled the video with silly animated sprouts
@@pmcpartlan It should have been pointed out. It's definitely not self-evident (and I speak French reasonably well.) Animated sprouts would have been a welcome touch!
Thanks. I'm surprised it wasn't pointed out. Even dismayed.
I'm french and didn't even think of that.
@@josenobi3022 me neither weird pun btw
Woop woop, greetings from Brussels!
Hey - greetings to you too!
Ωραίος
All possible integer sequences are Neil's personal friends
*Integer sequences that have a rule
@@rana4410 Ya, with a rule
And are computable (there are uncountably infinite possible integer sequences, and countably many that are computable)
I see what you did there. ;)
@@Vaaaaadim urr=
When I first saw the title, I thought this was going to be a throwback to the Brussel Sprouts game.
Great video, I was very happy to see Neil
Seeing Neil speaking so passionately always cheers me up :)
Dr Sloane is a hero of mine. I've got a few sequences on the OEIS, and I would totally lose my s**t if he talked about one of them on Numberphile. I fear none of them are interesting enough however.
What are the sequences?
@@emuccino if you search my name on the OEIS they come up (but so do any that I have merely commented on too).
4:58 *Other numbers are available.
Thanks for clearing that up. Got a bit confused there for a second.
lol
This is a really nice operation. The iterative process reminds of Collatz' sequences, but instead we have 1) more choices, 2) much more can be proved about the sequence.
Neil is a treasure. I just love his enthusiasm.
This guy is precious. Favorite numberphile
I’d be interested to see how this game plays out in other bases
On intuition, I'd say any even base (except binary?) should have the 0/5 problem, but instead of 5 it would be your base/2.
I'd be interested in odd bases, whether prime or composite.
@@odarkeq That can't be quite right, though, because in base 4, base/2 is 2, but you can get from 2 to 1 by halving. So you'd have to exclude any base that is itself a power of 2.
and also, what about odd bases? There isn't an integer thats base/2 in odd bases. Maybe all the numbers are connected?
I don't think that the base being even has much impact; rather, any base with an _odd factor_ is gonna have the same kind of problem as base 10, where an odd factor of the base can't be halved and can't be removed by modifying substrings of the number. Bases that are powers of two also have a problem, though, in that starting from 1 you can only get to powers of two (e.g. in base 4 all you can do is go 1 -> 2 -> 10 -> 20 -> 100 -> 200 -> ...)
Just saying: in binary, only numbers with the same number of 1s (in binary) are connected.
"*other numbers are available" got a good laugh out of me
Professional mathematician: "you can't get 5 from this process."
Me: "oh, come on... What if you..."
my brain every time
I love the "other numbers are available" footnote
After a long time. I clicked as soon as the video was released.
never been so early, huh?
I... I kinda wanna get a giant sheet of paper, a sharpie... and sit in my yard and just.... THIS
Now, would you be a Numberphile or a Vi Hart?
the digits of 0 and 5 can not be connected because you cant get to 5 without {odd number}0, witch ends in 0. and you cant get to a number that ends in 0 without and number that ends in 0 or 5
Now I finally understand the rules for Numberwang.
You know the probability of numberphile uploading back to back tends to zero but when it happens it's great
It's like a pair of twin primes.
This reminds me of the collatz conjecture.
And this is not the first video that does remind me of the collatz conjecture.
Love the Hendrix shirt!
Give us more Neil. Think I have saved all his videos for my students 😍
What can we say about sequences formed in that way: the third number is the linear combination of the two previous. Example: F(-1, 0) = 1, 0, -1, 0, 2, -2, -2, 2, 2, -2, -2, ..... F(1, -1) = 1, 0, 1, -1, 2, -3, 5, -8, 13, -21, Of course F(1,1) is the fibonacci sequence. The two starting numbers do not matter since F(a,b) is a two dimensional space and (I,0) and (0,1) is a base. Questions are: F(a, b) is periodic? It goes to infinity or minus infinity? It has subsequences bounded? etc.. It is asintotic a geometric sequence as Fibonacci sequence?
It's like 6 Degrees of Kevin Bacon but with numbers.
A lot of things are. Collatz' conjecture centers around if numbers are connected to 1 through the process x -> 3x+1 if odd, otherwise x -> x/2.
CONGRATULATIONS on reaching 500,000,000 video views on RUclips!!
The reason why the alg. at the end is O(12*log(n)) = O(log(n)) is we get a geometric progression:
After 12*log(n) steps, the # of digits halves, giving:
12*log(n) + 12*log(n)/2 + 12*log(n)/4 + ... = 12*log(n)*2 = total steps to get to 1.
Small mistake at 15:27. The upper bound should be 12 times the number of *pairs* of digits in a number.
12(n/2+n/4+n/8+...)
=12n(1/2+1/4+1/8+...)
=12n(1)
=12n
@@lasithanirmitha321 Ah, I see what you mean.
The wording on the matchstick problem is a bit ambiguous because you're not really creating squares by removing matchsticks but rather reducing the number of squares from 10 to 4 without any restriction on the number of any other polygons nor that you even need to close all the shapes so you only need to remove 3 matchsticks to get 4 squares assuming you follow the rules from the last slide where the fifth square is the outer one.
_ _ _ _
| _ _ | _ _ |
| _ _ | _ | _ |
This figure has the two large squares and the two little squares. You can also just remove the top row of 4 matchsticks and are left with just the 4 small squares on the botton.
This means you can remove any other one or two of the matchsticks that don't make up a square and still have 4 squares and five fewer matchsticks, resolving the problem.
I could genuinely listen to Neil talk about interesting sequences for hours. I want this at feature film length
Getting to 7 is easy. Once you hit 12, then 14 (double the 2), then 7 (halving the 14).
That numbers ending in 0 or 5 cannot get to 1 is very intuitive: any number ending with 5 can only be doubled and the new number will end in 0. Any number ending with 0 will end in 0 when doubled or 0 or 5 when halved.
A small clarification and a shortcut at the end. It takes 12 steps per pair of digits to go to 1. That is convert two digits into one digit. So in total it takes 12*k/2 for k digit number to have all its digits halved basically. You recursively then do it on a resulting number with half a digits (or half + 0.5 if there was an odd number of digits in the first place). That is 12*k/4. Then 12*k/8, etc. The total sum is
What if you are allowed to reverse the digits? Leading zero's get removed. That way you can do all of them to 1. For example 5 to 1: 5 -> 10 -> 01 -> 1. If you are allowed to add leading zero's, you can do it the other way around too...
Amazing Video!!
arya maroo wait how did you do that 30 minutes before the video was uploaded
_Wait, that's illegal._
Im new to Numberphile and really njoying it. 1 question; Whats with the brown paper? Everybody I've watched so far uses it for their work.
This math series of favourite numbers bigger than 1 million is awesome!
Spreading awareness of maths, thats what humans need, not a gaint wall between countries.
I really enjoy listening to Prof Sloane.
14:33 - you can get a tighter lower bound than 12 steps per digit.
1. Turn each pair of digits into 1 in 12 steps. That's 6 steps per digit.
2. Turn each triplet "111" into "1" in 8 steps: 122 -> 62 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1. That's 4 more steps per digit.
Total is 10 steps per digit. I'm sure someone can do even better.
Even better: use 111 > 112 > 16 > 8 > 4 > 2 > 1 = 6 steps.
WHY is Neil SO GOOD!
"K could be 11511 I still haven't ruled it out" the guy is checking every single number
we could add a new rule where erasing a zero counts as one step, so all numbers are connected and could go down to 1.
As this goes both ways, that means that we also must have a rule to "add one 0".
Surely all you need to do is use in a number base which is odd.
@@dcsignal5241 I tested using base 7 and found that 3, 5 and 6 connect; 1, 2 and 4 obviously connect; and that the two sets do not connect to each other. Numbers ending with 0 don't connect to any other type of number because 10 is an odd number in base 7 and divided by 2 is 3 and 1/2.
Example of a path from 3 to 5 in base 7: 3 >> 6 >> 15 >> 113 >> 43 >> 23 >> 13 >> 5
If you can find a path from 1 to 3 that I couldn't, let me know. =]
@@Cuuniyevo It's easy to show that you can't go from the 1, 2, 4 set to the 3, 5, 6 set, ever. Since the operations are reversible, showing that you can't reach either set by doubling a number from the other set is enough. And that's obvious to prove, since you can trivially see that by doubling the digits themselves you never leave the sets.
I love this man. So much energy.
I think it isn't fully explained, how you get to 5 from every number with a 5 or 0 at the end, because it can happen, that you have multipe zeros or fives at the end and then you can't argue like in the video:
So if you have a number with a 5 at the end, just double the five and get a 10 at the end , so you can reduce the number without the zero to 1 and then devide 10 by 2 to get 5.
If it has a 0 at the end, the number is divisible by 10, which means it is divisible by 2 and 5. So you halve the number. Because this new number still has to be divisible by 5, the last digit is either a 5 or 0. If it's a 0, just repeat that step until you get a 5 and then do the procedure above to finally get down to 5.
Tip: the step counts for every integer is A323454 on the OEIS
Small typo from description: 'Neil Sloane founded the runs the OEIS' should be: '... founded and runs the OEIS'
He says that any number ending in 0 or 5 can be brought back to any other number in 0 or 5, but then also says that for example reducing 117930 to 10 is possible because 11793 doesn't end in 0 or 5, suggesting that you couldn't, for example, reduce 117950 to 10 because 11795 ends in a 5
I was thinking about this operation in binary, because doubling and halving are fun in binary: it's just adding or removing 0's (i.e. 110 is 2 * 11, and 111 = 1110 / 2 in binary). This means, if I'm understanding correctly, this binary version of the operation groups the numbers by how many 1's are in its binary representation (zeroes can be added and removed freely, but it seems like ones cannot be created or destroyed). This has to do with the operation doubling and halving and the relationship of those actions to the base, i.e. if we were looking at tripling or dividing by 3, base 3 would have the same behavior.
With that in mind, I find it interesting that there are only 2 "equivalence classes" in the base 10 case (multiples and non-multiples of 5) while there are infinitely many classes if the operation matches the base, so to speak. I'm confident this must relate to the prime factors of the base and how the operation matches one of those factors, but I haven't investigated enough to really see the connection. I wonder what happens if you multiply and divide by 3 in base 10 (the operation not matching a factor of the base) or by 2 in base 12 (since 2 is a factor in 12 2 times).
I feels like this should be easier than the Collatz conjecture since you can control what part you're modifying. … What about leading-zeroes; can you turn 1040 into 120 by halving 04? 🤨 Can a number that has a 5 or 0 _anywhere_ in it collapse to 1? 🤔
No and yes (unless it's at the end).
@@awebmate That's being overly harsh. They both involve halving even numbers, enlarging odd numbers, and possibly reaching 1 eventually.
after watching all (/lots of) numberphile videos during the last weeks..
I just realised that I was not subscribed!
I hope you have not only fixed that, but also bashed the bell. 🛎
Is this useful in any way or just a thing to play around with?
This is kind of similar to the Collatz conjecture just with different operations.
A couple years ago on another Neil Sloane Numberphile video, someone posted in the comments that Neil reminded them of Professor Farnsworth from Futurama. I have never been able to unsee that. Fateful commenter, if you're still around, I thank you.
This is one of the most interesting people I Have ever seen in my life.
It‘s interesting to see what happens in different bases!
12 is a base that has a lot more factors.
You can’t escape a multiple of 6 (number ending in 0 or 6) [1/2 the base]
You can escape a multiple of 4 (number ending in 4 or 8) [1/3 the base] by halving 4.
You can’t escape a multiple of 3 (number ending in 3, 6, 9) [1/4 the base], which is pretty interesting! if the base is B, and n is an integer, I wonder if this only happens with numbers that are multiples of B/2^n, or maybe B/2n.
In prime and odd bases, I suspect the only sets of numbers that get stuck in their own sets are ones that end in 0, 00, 000...
What other interesting properties have you guys found?
Why is it n/10 < n < 10n?
I wonder, what they do for this paper which they use in the videos. This Channel is running for years, there would be house full of papers
Example is the easiest way to get to 1 from 1551:
1551, 11101, 11201, 1601, 801, 802, 804, 808, 408, 204, 104, 52, 26, 16, 8, 4, 2, 1
Took 17 steps
I think something is missing. In the video you say you can pick any substring. What if I pick 02 from 20218? What should I do? Do I have 02 / 2 = 01 or 1? If 1 then 20218 -> 2118 and this is obviously not reversible. Otherwise you should forbid leading zeroes or force to keep them.
If you have a zero in the middle, e.g. 11011, when you double this with the number next to it (so the 01) would it become 11021, or 1121, seeing as 02 and 2 are just variants of the same number?
If you did this, the sequence would no longer be reversable. It seems to me like it would have to be illegal to transform sequences of digits starting with a '0'. You could of course play with the digits *after* the '0', so change 44014 to 44024 by doubling just the '1'.
I guess this leaves the question of what would happen if you were allowed to add and remove 0's at will. This means 5 is solvable with 5->10->1, but maybe we could even figure out shortcuts for the other numbers?
@@Keldor314 Yeah. Adding and removing zeroes at will is a bit different, since this would allow, say 1000 > 1 in one step, whereas it would be impossible to do just with preceding zeroes.
5 can be gotten rid of as 10, but the zero still has to be dealt with as a preceding zero to something following, e.g. 54 > 104 > 18 > ... for instance.
There's also the question of whether you could theoretically add infinite zeroes preceding, e.g. 15 > 100010. This seems unfair, so it does seem like messing with zero is off, but then it does seem like the issues with 5 and 0 are almost self-imposed? if you know what I mean.
If you are only allowing the deletion of preceding zeros then I don't think it changes that numbers ending in 5 or 0 can't get to 1 because the number at the end still stays the same and you always end up with a number ending in 0 or 1
Adding a zero would be equivalent to multiplying by 10 (or multiplying by n in base n). Removing a zero would be equivalent to dividing by 10.
To get rid of an inner 0, if the leading digit is odd, divide those two by 2, otherwise double the following digit until 6 or 8, then double the 3 digit group containing the 0. Either method works, though the second may take more steps depending on the number. The problem with trying to grab 0 with a following digit is that it isn’t a valid way of writing a number, so wouldn’t be allowed. And if it is considered a valid way of writing the number, it would be retained when doubling or halving in those contexts to maintain continuity. Therefore it wouldn’t help in getting down to 1 to do so. The arbitrary addition and subtraction of digits is not allowed, but it can fairly easily be removed anyway. In your example 11011 becomes 1511 by halving the 10. From there you can double the 1 following the 5 and reach 1521, half 152 to get to 761, then half the 76 to get 381, half that part again and get 191, double the trailing 1 to get 192, half the whole thing and get 96 to 48 to 24, 28, 56, 112, 16, 8, 4, 2, 1. You can probably find a faster route, even without arbitrarily removing intervening 0s.
Sure greedy algorithm will give you the maximum number you can get in next step, but I am doubt the greedy algorithm will always give you the maximum number it can get in finite step.
Since sometimes small number can grow faster then the larger number, for example: 21 > 16, but 21->42, 16->112.
This means you can choose grow the number slower at begin but possible make it grow even faster in longer steps.
Is the no 5s or 0s thing dependent on the base though. How does this game work in other bases?
When you can't double a digit 6 or higher, shouldn't you prioritize doubling a 3 or 4 digit, so that you guarantee the next resulting number has a digit 6 or 8 and you can increase the length by 1 again. This potentially grows faster overall, despite an individual step now growing as rapidly, because more operations will increase the length
There was a very vital bit of information missing:
You can NOT take any sub-number and half it if it is even - it only allows sub-numbers that do not start with '0' - otherwise you could just take 1000000000000000000000006 and make it 16 in 1 step.
He didn't prove that the "greedy algorithm" is always best. He mentioned that there might be a "devious algorithm" that's better.
It's also interesting that using the greedy algorithm, you get numbers whose only digits are 1, 2, 4, 6, and 8.
I started watching numberphile so I can understand some of the stuff my brother tells me. Good shit!
*other numbers are available
When he showed 117930 could get to 10, he didn't actually prove that *any* number divisible by 5 could get to 5: we know that 10 can't get to 1, which would mean it's still possible that you might not be able to get from 100 to 10 (since the strategy of ignoring the last zero and getting to 1 doesn't work). Thankfully, it turns out you are able, for example (and there's probably a quicker way): 100 -> 50 -> 25 -> 210 -> 220 -> 240 -> 280 -> 560 -> 1120 ->160 -> 80 -> 40 -> 20 -> 10. Because of that singularly pesky 25, we had to use the ones' digit, which is why you can't use a similar path to get from 10 to 1 (which would use the tenths' place). This is definitely enough to prove it entirely, since it also uses 50, and you can extend the whole thing by an arbitrary whole number power of ten.
P.S. if anyone reads this and can find a quicker path from 100 to 10, let me know! You could splice in 25->45->90->180->280 for equal length, but I haven't found one of lesser length.
Man I love proof by contradiction. It's oddly satisfying idk why 8:35
How to make any number ending in an odd digit other than 5 smaller:
1. Double the final digit
2. If that final digit was greater than 5, halve the final digit and halve the final two digits.
3. Otherwise, cut everything in half.
8:40 Doubling the last 2 1’s gives you ”22”, not ”12”.
Regarding the biggest number in K steps, wouldn't it be possible to go infinitely big after the 4th step? 1 -> 2 -> 4 -> 8 -> 16 -> 100000...00003, that could lower the lower bound of certain numbers.
Eric Angelini is from Brussels, last time I checked still Belgium, and the flag shown is from France. :)
There are no french flag in this video
If we change the rule to multiply or divide multiples of 5 do we end up with 5 sets of numbers? I feel like the 2-ness 5-ness is base 10 related. I’d be interested to know what happens in other bases!
This number theory kind of math is soooo much my favourite!
Also interesting to do it in other bases. In base 2, you get an infinite amount of classes where all numbers with the same amount of 1 digits belong to the same class. In base 8: 1, 2 and 4 are connected, and so are 3, 5, 6 and 7. I'm not yet sure if those two groups are also connected via some way, or if there exist other groups (but my guess would be no).
This has Collatz conjecture vibes
Woah. Thus reminds me of middle school on finding the prime numbers and squares in the fibonacci sequence and looking for palindromes in pi 3.(141) for instance. Good times
You are allowed to double a part of a number. From 26 to 212 for example. You kinda add a row to this number.
In reverse logic. Why can't we substract a row? Can't we just be allowed to not write part of a number, if that number is basically nothing (zero at the end)? From 20 to 2.
This way all numbers are connected.
According to an algorithm I wrote in Python; it takes 25 steps or less to get from any 3-digit number (that is not divisible by 5) to 1, 30 steps or less for 4-digit numbers, 34 steps or less for 5-digit numbers, 41 steps or less for 6-digit numbers. These by no means are optimal. I tried giving it some pseudo-random 100+ digit numbers and the number of steps would usually be between 1 and 4 times the number of digits.
For those wondering, the algorithm was basically:
If the number is one digit long, use a look-up table for how to get to 1, else, go from right to left, looking for an even digit. Continue until either the end of the number is reached or a 1 is reached. Use the 'Choix de Bruxelles' to half the number starting at (including) the first even digit encountered, and either the 1 encountered or the end of the number (inclusive). If no even digit was found, use a lookup table for how to convert the right-most digit to a 6 (not resulting in a longer number, and there can be some variation here). Repeat this until 1 is reached.
It works because; 1). excluding the case of one-digit numbers, it can only ever reduce a number's value (unless converting the right-most digit from 1, 3 or 5 to 6, but this will make the number divisible by 2 (because 16, 36, 56, 76, 96 all are divisible by 4), so its value will actually end up being lower after halving). 2). it can be applied to all positive integers not divisible by 5 (excluding the case of one-digit numbers).
I would watch tv everyday if this man was the host of every show !!
"Yes. That's a no." to can any number be brought to one.
I'm sure I heard Brady's head melt at that moment!
I love your videos and all but what is the point? Can you add to the end of each video if there is an application/use for the math or if it is just a game?
@4:54 "Other numbers available" 😂😂😂
Really like your videos!
You as a child in school multiplying integers in the way described at the beginning of the video.
Teacher: "Bring out the dunce hat !"
Professional Mathematicians:
What if there are leading zeros in the substring that you select? For example, can I start from 104, choose the 04, divide by 2 to get 02 = 2, and end up with 12? Similarly, can I take the 04, double it to get 08 = 8, and get 18?