UNCRACKABLE? The Collatz Conjecture - Numberphile

Professor David Eisenbud on the infamous Collatz Conjecture, a simple problem that mathematicians may not be "ready" to crack.
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NUMBERPHILE

Videos by Brady Haran

[scottanderson]

“Seven seems to be an odd number.”

That’s why he’s a maths professor.

[Eleni_E]

I'm reminded of the wikipedia phenomenon where, if you click the first link in almost any article that isn't in parenthesis or italics, and keep doing that long enough, you will eventually end up at philosophy.

[jalepenopvp]

This guy has legit the most calming and comforting voice I’ve ever heard

[Croxmata]

And as we learned, Brady was most likely to choose 7 out of all numbers from 1 to 10.

[simonreinsperger]

"Wow... 16, very even number!"

[badhombre]

This must be the formula that banks use, to calculate fees, that reduce my account to 1.

[andlabs]

My first introduction to the problem was one of the example in former Bell Labs Unix co-pioneer Jon Bentley's book "Programming Pearls", which is a collection of columns on the theory and practice of software design he wrote for seminal comp sci journal Communications of the ACM over the years, expanded and with exercises. One of the exercises in one of the early chapters was this conjecture (in the Second Edition it's Column 4 Question 5).

In the back of the book is a collection of hints. Here is the hint for this question. It has stuck with me ever since I first read it:

"If you solve this problem, run to the nearest mathematics department and ask for a Ph.D."

[JimCullen]

Man I chose 4 initially. Worst possible number…

[finchisneat]

My friend in high school told me about this nearly 20 years ago, so cool to see a video about it!

[Rock]

what about the 360 noscope conjecture tho

[zacherylunnonvesty]

choose a number from 1-10
"42"

[itamarmedved]

Hi, I want to offer a seemingly simple answer, though probably not new/right:
1. Any even number you continuously Divide by two will reach one.
2. What we need to show is that the division action is more dominant than the multiplication.
3. You can see that after every ‘up’ there is a ‘down’, but after a ‘down’ you can have either an ‘up’ or another ‘down’. Statistically this gives you 3 ‘up’s for every 5 ‘downs’.
4. Now, let’s say that the probability was 1 ‘up’ for 2 ‘downs’, in that case it would be easy to see why the division action is dominant. Since what you get from two successive division (/4) is more powerful than what you get from one multiplication (*3).
5. Now you can ask yourself what should the probability be in order for the actions to cancel each other out? Let’s assume extremely high numbers, so the +1 factor can be ignored. The answer is that the division (/2) should be one and a half times more probable than the multiplication (*3), in other words that for every 3 multiplications there should be 4.5 divisions.
6. Since we now that the probability exceeds that, then we can say that when reaching very high numbers the ‘down’ action will surely be more prominent.
7. Since we can see that what goes up will surely come down, but, what comes down will not surely come up, since it can get trapped in the 1-4 loop, then given enough time you will always get stuck in the loop.
8. Problem solved!

[esotericVideos]

The first part of every numberphile video gets me excited to try to solve a problem, the second part convinces me that there's no point in trying because tons of people have already tried. :/

[nimets]

it is interesting to look at the problem with binary numbers. dividing by 2 is just binary shifting to the right, because lowest bit is 0 for even number. 3n+1 is shifting the initial number to the left, adding the original number and then add 1. the middle path is reached when the highest bit is 1 and the rest is 0. they then can be reduced to "1" by shifting to the right.

[ljfaag]

I found a counterexample, but the comment section is too small to contain it :P

[yashrawat]

*" Don't Judge a b̶o̶o̶k̶ problem by its c̶o̶v̶e̶r̶ statement "*
*- Collatz Conjecture*

[garrett]

I was messing around on my calculator and I think I found a similar problem.  It has 3 rules. If even dived by 2, If divisible by 3 dived by 3, IF the number isn't divisible by 2 or 3 the multiply by 5 and add 1.  do this and It always seems to get stuck at the loop 6, 3, 1, 6, 3, 1.  Or depending on If you divided by 3 or 2 first 6, 2, 1, 6, 2, 1.  This works regardless of the number.

[airmusic]

"I bet the person who found it thought maybe he was on to something..."

I'm willing to bet the guy who found the tree for 63, 728, 127 wasn't sitting down and doing it 😂

[simonmultiverse]

The cover of the book is taking a short cut. It has an arrow going from 1 to 2 so it's showing two steps n-> (3n+1)/2 (combining the multiplication and the division by 2).

[MiaVilleneuve]

When he said any fourth grader could understand it, I was like, that's probably an exaggeration. Then he explained it and sure enough I remembered reading about it in a children's math book called "Math For Smarty Pants" in elementary school.