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March 2

Find the missing perspective

This should be a fairly straightforward problem in 3D representation of 4D objects, but I'm having no luck in wrapping my head around this. At first I thought I could treat each as a cube within a 4-cube (since there should be eight of those), with each cell being adjacent to a corresponding cell of the cube, but I can't get the seven below to line up right, so I now believe that I'm thinking about the problem wrong (i.e., I shouldn't think of it as an unfolded tesseract but as a 2x2x2x2 something being projected in some other way). Any thoughts or advice would be greatly appreciated. I've exhausted all that I can really think of to solve this sort of problem, since it doesn't seem to fold the right way - assuming that's the right way to think about this.

Given the seven 3D projections of a hyper-object, provide the eighth:

(Each █ is a cell, with the left side being one layer and the right being the other layer of the 2x2x2 slice.)

_	_		█	█
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_	_		█	_
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_	_		█	█
_	_		█	_

█	█		_	_
_	█		_	█

_	_		_	█
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_	█		_	_
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Spability951 (talk) 02:39, 2 March 2015 (UTC)[reply]

Since there are only 4 dimensions to squish, if they're listing 8 different projections, they must be including projecting onto opposite faces as different projections. The projections onto opposite faces would be reflections of one another, and indeed 1 is a reflection of 5, 2 is a reflection of 7, and 3 is a reflection of 6. So that leaves 4. I don't see a pattern to how they're doing the reflections, so maybe any reflection of 4 will do?80.109.80.31 (talk) 02:13, 3 March 2015 (UTC)[reply]

Y'know, I think you're right. So these must be projections of the 4 choose 3 dimensions. Question is: do they "line up" properly to mean that actually would work? I've thought about it and I'm not really sure which pairs belong to XYZ, XYW, YZW or XZW yet.Spability951 (talk) 17:19, 3 March 2015 (UTC)[reply]

Finding a proper primitive polynomial

I'm interested in the Massey-Omura cryptosystem at the moment and I need a primitive polynomial for $GF(2^{256})$ because I want to use a 256-bit long key. Is there a list I can look at or a webpage with an appropriate calculator our there for me to obtain such a polynomial? — Melab±1 ☎ 06:54, 2 March 2015 (UTC)[reply]

(As I mentioned last time, you could use nimber arithmetic; then you don't need a primitive polynomial.) -- BenRG (talk) 07:44, 2 March 2015 (UTC)[reply]

It's unclear to me, though, if the polynomial mask is applied after every multiplication/squaring or not. Also, the concept of nimbers isn't made clear enough in the article for me. — Melab±1 ☎ 20:04, 2 March 2015 (UTC)[reply]

I'm not sure what you mean by "polynomial mask". If you're talking about reducing modulo the polynomial in the usual GF representation, then it doesn't matter, correctness-wise, when you do it. It's like reducing modulo a prime when doing integer arithmetic modulo that prime. The difference between integer arithmetic and polynomial-over-Z₂ arithmetic is just that in the latter case there are no carries/borrows. In particular, addition is bitwise xor (and so is subtraction). Recent x86 processors have carryless multiply instructions to make Galois field arithmetic more efficient.

To multiply 2ⁿ-bit nimbers (n>0) you can split them into 2ⁿ⁻¹-bit upper and lower halves and recursively multiply and add those, similar to Karatsuba multiplication. There's an implementation in C++ here. It's surprisingly inefficient, though. I don't know whether that's fundamental to nimber arithmetic or whether there just hasn't been any research into more efficient algorithms. -- BenRG (talk) 22:48, 4 March 2015 (UTC)[reply]

This page has an example primitive polynomial for

\mathrm {GF} (2^{256})

. --173.49.77.202 (talk) 02:35, 3 March 2015 (UTC)[reply]

Integral

About this integral $\int _{a}^{b}\!f(x)g(x)\,dx/\int _{a}^{b}\!g(x)\,dx$

Can be equal

$\int _{a}^{b}\!f(x)\,dx$

Of course the two expressions can be equal, but they usually aren't. -- Meni Rosenfeld (talk) 17:24, 2 March 2015 (UTC)[reply]

Just to clarify for OP, the expressions are equal if g(x)=k, for any non-zero k, or if f(x)=k. OP might like to refresh their memory on the product rule and integration by parts. SemanticMantis (talk) 18:13, 3 March 2015 (UTC)[reply]

Consider an extremely simple example:

f(x)=1,g(x)=x

\int _{a}^{b}\!f(x)\,dx=b-a

\int _{a}^{b}\!g(x)\,dx=\int _{a}^{b}\!f(x)g(x)\,dx=(b^{2}-a^{2})/2

\int _{a}^{b}\!f(x)g(x)\,dx/\int _{a}^{b}\!g(x)\,dx=1

—Tamfang (talk) 03:18, 3 March 2015 (UTC)[reply]