Conway Checkers (proof) - Numberphile 

Vasiliy Sharapov You're looking for the real deal there. Stay Strong. *Insert appropiate emoji*

Holly Krieger also has magnificent handwriting

Yeah, there wasn't any "Okay let's just assume you know this formula" or "this would take too long to explain so I'm going to gloss over this one section". Very nicely done!

Yes, they should do more videos like this

IchBinKeinBaum ,

Communistic joke AF :D ... Communistic countries were "proforma" also free.

Nope, that's a joke about US, where they tell you how free you are, but, actually, you are bound by strict laws.

IchBinKeinBaum #lockherup

There was an old joke in the Soviet Union: In the Soviet Union, you have freedom of speech. In America, you have freedom after speech.

now THAT should be a shirt

Im quoting this given the first opportunity haha

This proof is given in Berlekamp, Conway and Guy's "Winning Ways", but the proof here is much clearer.

This has to be one of the best Numberphile videos made untill now.

How is it clear that moves are like powers of a number ? I understood the whole explanation, but not why remplacing cells with powers of X is relevant.

It's just because it lets you show what you want to show. There is no other explanation.

@@antonimaciag1259 sure it lets us show exactly what we want to show, but there was no indication on how you would come up with exactly that formula

I had no clue the video was this long!!!!

The longest is actually the interview of James Simons, I think!

Yea and i found it at 1.30am on a day before work. I'm sad I'm goto have to skip it and sleep

Agree. Wow!!!

You can never be too rich or too thin ;) or have enough Prof Z!

The longest on numberphile2 is an hour of coloring the collatz conjecture

Hahaha :D

Only true numberphile viewers understand

Well actually, if the first move took 1 second, then 2 seconds for the next move, then 3 etc. which makes sense because we have to keep reaching farther for each move. We get 1+2+3+4... And we'd win 1/12 seconds ago.

+Krekkertje Clever, very clever

I was thinking of supertasks too

there's all sorts of neat little numerical properties of phi. because of the relation x² = x + 1 , it is plain to see that the decimal expansion of phi and phi² are the exact same. The same is true for 1/phi through slight algebraic manipulation (divide everything by x) Personally I find it extremely cool how you can take the inverse or the square of a number with infinite decimals and have the exact same digits!

Yes also more simple - consider which of these numbers are the largest: 1/sqrt(2), sqrt(1/2), sqrt(2)/2 or (1/2)^(1/2)

all of them are the largest ;)

The golden ratio, e, pi, sqrt(2), ln(2), they show up everywhere!

@@skilz8098 Yeah, and π especially has the habit of appearing when you least expect it! Who would have said that the infinite sums of the reciprocals of the powers ≥ 2 of natural numbers (the zeta function ζ(x) for x ≥ 2) would contain π in it?

The main computational result from this proof is that the sum of the grid beneath the line is x^(n-5), for a center point on the nth line above the middle-line. So, just add copies of the grid for 3 dimensions. Based on the same calcuation as they did for the row, we conclude that the sum for a 3 dimensional lattice is x^(n-8). And, in general for an n-dimentional lattice, you get the sum below the line to be x^(n-3d+1). So, the methods of the proof lead to the following conjecture: Given a grid of dimension d, you can reach up to, via analogous rules, the (3d-2)th hyperrow above some dividing hyperplane. This proof suffices to show that you can't do any better than 3d-2. However, to show that you can actually get to the (3d-2)th hyperrow, you'd probably have to demonstrate an algorithm.

We know that you can't get 8 or more, but we don't know for sure that you can get 7.

So then the question is for which dimensions is it impossible to get to the (3d-2)th hyperrow?

Right. I'm inclined to think that you can always get to the (3d-2)th row, but that's just a conjecture: it requires a separate proof.

I need this answer!

The eureka moment if you will.

Happened at around 0:53 for me :/

About that same time I lost my sense of time and it felt like 5 minutes

ultear milkojohn it just didn't happen at all in my case..

Exactly!

That would be a game with an end but no beginning. Maybe we can call such thing a reverse supertask.

That's how I tried to find the arrangements in the main video, but it guess it would require an algorithm which can establish a new configuration to pop a new cell, given a configuration. That surely would involve recursivity which makes it even harder to compute, let's find a better way. (I will not show you the way)

I'm sure there exists a path you can take that would do that. If you wanted to fill the entire board (not just below the line), you could follow a Hilbert Curve (en.wikipedia.org/wiki/Hilbert_curve ), which would cover the entire board in checkers. There probably exists some path which does the same as the Hilbert Curve but staying below the line, if you wanted to only cover that segment.

@@jarredallen3228 the real question would be: could you find a path that leaves the top of the board empty?

It would be impossible to end up with no checkers on the starting side with finitely many moves

Incidentally, there is at least one other mathematical game where the golden ratio appears. It's called Wythoff's game. In Wythoff's game, two players have two piles of (not necessarily equally large) coins in front of them. A turn consists of removing any number of coins from either pile, or removing the same number of coins from both piles. The winner is the person who takes the last coin. It turns out that the winning strategy is to keep the ratio of the sizes of the two piles as close to the golden ratio as possible. Phi is one of those numbers that has a habit of popping up when you least expect it.

All math talk is dirty talk to me

That accent does to me what it does to Jamie Lee Curtis in A Fish Called Wanda.

Could you elaborate?

No, if you did win the game it doesn't mean that you've made a finite number of moves

You also don’t show that it’s possible to win just by showing the sum at the beginning is equal to the sum at the end, all you show is that you can’t exclude that possibility using this method. But you can obviously come up with several initial and final configurations in which the sums are the same and yet there is no way to move from one configuration to the other (for example, if no two pieces are adjacent).

make your first move at time t_0 = 0 make the nth move at time t_n = t_(n-1) + 1/2^n you are done with infinite moves at t = 1

It's a little more subtle. What's it's saying is that in order to have enough "energy" to get to row number 5, you need to have infinitely many checkers underneath the line. In fact, the numbers work out such that you need ALL of the checkers underneath the line. If you don't use some of them, you don't have enough energy to get to row number 5. Therefore, it's impossible to do in finitely many moves.

SirFloIII. The moves are countable, so each move has a positive integer associated with it. The first move is associated with 1, the second with 2, etc. Which positive integer is associated with the move which first reaches row 5? Hint: it's a rhetorical question.

You came here from Vsauce2 right, WRONG

@@AminGhomati maybe?

I saw this before, but this time, from vsauce 2

Me.Right?WROOONG!!!!!!

or is it?

Also, the problem (conway checkers) was clearly not specifially engeneered to produce such a miracle, it all happened in purely a natural way. Mind bending.

Right? All of my infinite checkers keep leaking out through it!

OriginalPiMan , oh cool!

That would make the unit more impractical than it already is. We do not need that really. It is a mathematically neat concept, but units of measurement should stay on the practical side of things. After all, they're made for measurements, not abstract thinking.

That would be gold.

Alistair Shaw i think a lot of people have come up with interesting ideas that lead to cool maths, but not many others will pursue them and make them popular. Conway is such a big name that people will devote careers to solving his ideas

bluekeybo my point wasnt that other people havent done created lots of important and cool maths. Far from. Nor was it an overt celebration of conway himself, although he does in many ways deserve it. No its more that cool maths arises from both interesting and trivial places.

You might like the book on mathematical analysis of games which he wrote with Elwyn Berlekamp and Richard K Guy, "Winning Ways". I have had it for twenty years or so and read bits of it many times, but never come close to mastering it. In one game I used to play at school, Fox and Geese (though we called it Fox and Hounds), it turns out that the geese have an advantage of one plus the reciprocal of the largest possible infinite number!

His study of games actually led to the development of a new set of numbers! Look up "surreal numbers;" they're incredibly interesting!

Ethan Smoller thats exactly my point

Umm… who are you talking to?

Ian Barton , Ha ha ha.

well obviously he did it by already starting in the second row

@Ian Barton@@Tracequaza Underrated comment + underrated reply

Watched again. Mind blown!

Don't you like maffs Angelov ?