How you can solve dice puzzles with polynomials

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[zachstar]

Slight error I just want to address here (and small spoiler ->). I should have said that the faces of the dice are required to be positive integers, at least that's what makes this problem have a unique answer. If you allow either die to have 0 on it's face then you get more solutions (for example, one die could be labelled 0, 1, 2, 3, 4, 5 and the other die would be 2, 3, 4, 5, 6, 7 and you'd get the exact same probability distribution). If you allow negative numbers on the faces you would have infinitely many solutions.

[abraham]

Zach has forgotten his engineering roots, he's become a pure mathematician ;(

[zi_t]

Never knew polynomials could be used like that… blew my mind

[Alexander-ohry]

I assume you mean the only pair of dice where only numbers >0 are allowed, because you could decrease all sides of one dice by 1 and increase the other dices sides

[DrTrefor]

Cool puzzle! Maybe my algebra is weak lol but I didn’t anticipate the unique factorization to jump into this

[nivolord]

To avoid the technical bits in the video, you can also do the following. Assume that 1, 2, 3, 4, 5, 6 are the sums of two other dice: a two-sided die, and a three sided die. We can get these sums with equal probability with the following two pairs of dice:
- 1, 2 and 0, 2, 4.
- 1, 4 and 0, 1, 2.
Hence, rolling all four dice together yields the right distribution of sums. So all we need to do is pair them differently:
-1, 2 + 0, 1, 2 yields 1, 2, 2, 3, 3, 4
-1, 4 + 0, 2, 4 yields 1, 2, 3, 4, 5, 6, 8
Just like the solution in the video.

[jonathanlevy]

I learned all the tools to answer the question and still fascinated by the elegance of it. Great riddle!

[aidanosullivan]

Love your videos Zack. Very interesting and engaging. Keep up the great work!

[JM-usfr]

Finally! Someone talked about Sicherman dice. I swear, every time numberphile does a dice video, I'm always expecting them to talk about Sicherman dice, and they never did.

[sunimod]

I just watched a veritasium video on clickbait and this is the most accurate title I've ever seen

[ithaca]

zach do you think you could check out how programming languages are essentially just fancy maths?

such as a for loop is almost / exactly the same as sigma in math depending on how you use it.

and how if statements can be represented by piecewise brackets

i think it would be amazing if you could cover that topic since all everything computers are, are just math machines

[Orengi-kun]

We studied generating functions on one of my introductory courses to Computer Science at college and they were the first thing that came to mind to solve the problem.

[shonsh]

thanks Zach for this 'outside the box' idia.
I realy enjoy your vidios.

In total, I found 5 solutions for this problem:

2 with 1 or more dots in each side (as shown):
(1, 2, 3, 4, 5, 6), (1, 2, 3, 4, 5, 6)
(1, 2, 2, 3, 3, 4), (1, 3, 4, 5, 6, 8)

3 with 0 or more dots in each side:
(0, 1, 2, 3, 4, 5), (2, 3, 4, 5, 6, 7)
(0, 1, 1, 2, 2, 3), (2, 4, 5, 6, 7, 9)
(0, 2, 3, 4, 5, 7), (2, 3, 3, 4, 4, 5)

[reidflemingworldstoughestm]

"Dice dice baby"
~ Vanilla Dice

[TeaBroski]

I was watching the last Veritasium's video and it's funny, it talks more to you than to me.

[aashsyed]

i love your videos! these are awesome! always hyped! :)

[alhdlakhfdqw]

really really awsome videos zach! thank you so much! subbed again :) please keep it up

[faulsname]

Dude you are so charismatic and likeable. Glad i found your channel.

[alejrandom]

4:58 here's how you can do such factorization:
Let P(x) = x1+x2+x3+x4+x5+x6
P(x) = x(x0+x1+x2+x3+x4+x5)
Note that
(x-1)P(x) = x(x6 -1)
Now x6 -1 is easier to factor as a difference of cubes
x6 -1 = (x3+1)(x3 -1) which can be factored as a sum of cubes and a difference of cubes
x6 -1 = (x+1)(x2 -x+1)(x-1)(x2+x+1)
Multiply by x to get (x-1)P(x)
x(x6 -1) = x(x+1)(x2 -x+1)(x-1)(x2+x+1)
Now we just divide by (x-1) to get P
P(x)=x(x+1)(x2 -x+1)(x2+x+1)
Square this and you get f(x)
f(x)=x^2*(x+1)^2*(x2 -x+1)^2*(x2+x+1)^2

[Jaylooker]

This is a very simplified version of the modularity theorem where coefficients from the elliptical curve of a Galois group representation correspond to coefficients of modular forms. The Langlands program hopes to generalize this.

In this case it is the 1-dimensional case with class field theory noting Kronecker-Weber’s theorem on abelian extensions, cyclomatic fields, and their cyclomatic polynomials. The generating function is derived from cyclic group. The cyclic group of the roll of dices determines the Cayley table and the generating function is the cyclomatic polynomial.