What is the maximum number of pieces of pizza one can get by making n straight cuts through?

What is the maximum number of pieces of pizza one can get by making n straight cuts through?

Assuming it is a continuous pizza: the answer is {}

define a slice of pizza, you could make infinite cuts with just this information, or I guess you get as many slices as there are *insert smallest kind of particle you can get from this super god knife* in a pizza

many

The result is supposed to be a function of n

For example, with n = 0, the function should return 1
With n = 1, the function should return 2
With n = 2, the function should return 4 (it is also possible to get 3 pieces, but this is not the maximum)

I didn't read the n heh

f(n)=2n; for n>0; if n=0: 1;

The amount of single atoms it contains, if not then into smaller particles, I guess

Physics realm doesn't exactly works like pure mathematics, you forget this fundamental.

But you can slice 3 times and end up with 8 pieces if you just move one of the cuts sideways a bit. 2 of the slices will be tiny but that shouldn't matter.

I cant be bothered to draw it because I'm on my phone. But the answer is not trivial.

if you are allowed to move the pieces around then it's 2^n if you just lay every slice on top of each other after slicing

Well that's true, but I wasn't talking about moving the slices between cuts, I was talking about making off-axis cuts.

I assume in OPs problem you cant move the slices between cuts either

Your solution only works for n < 3

No, you're not allowed to move the pieces, and the cuts are straight

It's a mathematical question, not a trick question

This also goes for (and also "the amount of single atoms it contains" is definitely not the solution for small n)

Oh then I am too retarded to get to 8 it seems, I only get 7

Nah I think it is 7, my bad.

That's if you assume that only convex pizzas exist.

If you bake an "W" shaped pizza then you can divide it into 5 pieces with a single cut.

That's clearly not the problem being discussed here

Is anyone still guessing/calculating? Otherwise I will post the solution

>With n = 2, the function should return 4 (it is also possible to get 3 pieces, but this is not the maximum)

This is a bit weird because if you think about it, to get 4 pieces with n=2 you would have to move your knife across one slice... and then go through the other one. Isn't that to cuts?

But whatever, if you count it like that then think about.

At n=3 I could line up all the pizzas in a line and then just run my knife along them and literally get 8 pizzas.

Then do the same and 16 pizzas.

And this is the max so I guess 2n.

What exactly do you mean by a cut straight through? Is the idea that each cut must intersect the other cuts but they dont all need to cross at the same point?

No, it's not about moving the pizza

You have a pizza in front of you and you make two cuts
If they intersect, you get 4 pieces
If you cut in a way that they don't intersect, you get 3 pieces

I mean that you don't make any curved cuts, they are completely straight

What if you curve the pizza as you perform a straight cut? Do pizzas count as two-manifold surfaces?

No, they are flat and round (or rectangles if you prefer rectangled pizza)

I feel like I remember a problem like this from combinatorics.

The first values seem to be

[eqn]f(0)=1[/eqn]
[eqn]f(1)=2[/eqn]
[eqn]f(2)=4[/eqn]
[eqn]f(3)=7[/eqn]
[eqn]f(4)=11[/eqn]
[eqn]f(5)=16[/eqn]

And then based on the construction, I'm guessing you could make a recursive sequence to fit that. Am I on the right track?

Yes.

The recursive sequence should be pretty easy if you got that far, and there's also a closed form expression for [math]f(n)[/math]

You can already get 8 pieces with 3 cuts.

see
It's flat (2D)

a sphere slicing version would be interesting though

f(n+1)=f(n)+n

This, but with a +1

A simple geometric argument shows that

[eqn]f(n)=f(n-1)+n[/eqn]

And since [math]f(0)=1[/math], we must have

[eqn]f(n)=1+\frac{n(n+1)}{2}=\frac{n^2+n+2}{2}[/eqn]

Nice problem OP

Explain the simple arguement

Correct

To get the maximum amount of pieces, each cut must intersect with each previous cuts, but not at a position where two other cuts already intersect (i.e. no 3 cuts through the same point)

To get the max number of pieces, you want to make the cuts in such a way that no more than two cuts intersect at a given point (because otherwise you could make an extra piece by expanding this into a triangle) and so that every pair of cuts intersect each other exactly once (otherwise, you can create another piece by making them intersect).

To get [math]f(n)[/math] pieces, start with [math]f(n-1)[/math] pieces and then make an extra cut. If you do it in the way described above, you'll always get [math]n[/math] new pieces.

If the pizza cutter has zero thickness then the amount of slices it can cut is infinite. Though, it may be a bit messy if you start splitting Pizza Atoms.

>splitting Pizza Atoms.

how many quarks has a pizza

The amount of single quarks it contains.

Quarks can be split into their constitute parts. There is always something smaller even if it can't be detected. The answer literally is "infinite".

Back to /pol/

Quality post, deserving of a board for Science and Math

don't you mean 2^{n}, and also this will fail at n = 5

Spherical pizza? Has science gone too far?!?

It's the next step in the evolution of the perfect pizza

-1/12

americans would buy this

wrong :P

slices = 1+(N!/(2!(N-R)!))+(N!/(2!(N-R)!))

get on my level.

I fucked that up, replace the first R with two, then the second two, and the second R with four.

When you cut the k'th time you have k-1 cuts already. The maximal number of cuts the k'th cut can cross is then k-1, every time you cross a cut you slice a piece into two pieces. Since you also split the piece you start on you add k new pieces. This obviously doesn't work if you cross at the intersection of two other cuts, but that wouldn't maximize the number so we discard the situation.

>assuming all pieces need to be the same sizes and shapes

With this method I can get 7 slices with just 3 cuts on a 2D pizza, higher than the expected 6

so shit just keeps getting smaller and smaller infinitely rlly makes u think

what if the pizza is a k dimensional hypersphere

What formula are referring to?
If you're using this one, the result is 7 for n = 3

Define a straight cut in k dimensions

for n>0 the right answer is:

n!+1

but i just tested it for n

A straight cut in 2D would be a line:
[math]L = p + r_1v_1[/math] where [math]r_1 \in \mathbb{R}[/math] and [math]p,v_1 \in \mathbb{R}^2[/math]

A straight cut in 3D would be a plane:
[math]P = p + r_1v_1 + r_2v_2[/math] where [math]r_1,r_2 \in \mathbb{R}[/math] and [math]p,v_1,v_2 \in \mathbb{R}^3[/math] and [math]v_1, v_2[/math] are linearly independent

Generalising for k dimensions:
[math]P_k = p + r_1v_1 + r_2v_2 + ... + r_{k-1}v_{k-1}[/math] where [math]r_1,r_2,...,r_{k-1} \in \mathbb{R}[/math] and [math]p,v_1,v_2,...,v_{k-1} \in \mathbb{R}^k[/math] and any two vectors of [math]v_1, v_2,...,r_{k-1}[/math] are linearly independent

2 cuts and 4 pieces sombrero

the answer is 8.

>straight cuts through
...through *what*, faggot? The centre?

A hyperplane. The set of points satisfying some particular (nontrivial) linear equation

In k dimensions, clearly you can double the number of pieces each time for the first k cuts.

7

Literally just copied the answer from three blue one brown like a retard.

V+E-F=2