Magenta@48: There is one convex regular 4D polytope built out of dodecahedra. It is the only one with pentagonal faces. There is another convex regular 4D polytope with the same symmetry group; just like the icosahedron has the same symmetry group as the dodecahedron.
The convex regular 4D polytopes are: the 5-cell (4-simplex, hyper-tetrahedron [snrk], self-dual), the 8-cell (tesseract, hypercube), the 16-cell (dual to the tesseract, tetrahedral cells), the 24-cell (sort of like a hyper-octahedron, octahedral cells, self-dual), the 120-cell above, and the 600-cell (sort of like a hyper-icosahedron, tetrahedral cells, dual to the 120-cell).
What's even more interesting are the non-convex regular 4D polytopes -- like star polygons in 2D, and stellated polyhedra in 3D. There are a bunch of them with the same symmetry group as the 120-cell, and some have dodecahedral cells.
The dodecahedron was the basis of a Rubik's Cube-like puzzle, called the Megaminx. There is also a 4-dimensional analogue of the dodecahedron, called the "120-cell" -- and someone just released a program to let you play with the 4D Megaminx.
Sarah@152:
At present, it is believed that RSA is largely secure. That is, so long as you don't do something stupid in terms of picking your modulus or public/private keys (where stupid is defined by certain mathematical tests), the only way to crack your message is by factoring the modulus. Which is believed to be very difficult.
If they do crack one msg, there are two possibilities: First, they did so by obtaining your private key (which means they have the factorization of the modulus). Second, they did so by some side-channel attack (like a keylogger on your machine) which doesn't involve breaking the factorization. In the first case, your future msgs are insecure if you use the same modulus. Using a different modulus might provide better security. In the second case, your computer is insecure, and it doesn't matter what encryption you use, so long as you're using that computer.
Xopher@138:
Okay. So your modulus is N = 15 (which we know, in secret = 3*5).
Your text, which you wish to encrypt, is t = 7.
Pick an exponent e as your private key. It should be relatively prime to the special exponent which gets you back to 1 (in this case, (3-1)*(5-1) = 8).
Let's pick e = 3. Then d, your public key, is a number such that
d * e = 1 (mod 8).
Well, d = 3 works. So your public key is 3.
To encrypt your text, you take
c = t^e (mod N) = 7^3 (mod 15) = 343 (mod 15) = 13.
So your ciphertext is c = 13.
To decrypt, someone else uses your public key, and takes
c^d (mod N) = 13^3 (mod 15) = 2197 (mod 15) = 7.
They now have the plaintext, t, as expected.
Xopher@69: Let me just do a simple example.
Suppose N = 15 (P, Q = 3,5).
Take a number t, not divisible by P or Q. Say, 7.
The t^{(3-1) * (5-1)} = t^8 = 5764801.
t^8 = 384320 * 15 + 1 = 1 (mod 15).
Does this help?
Xopher@22: Let me know if this helps with the math.
First of all, the specific type of public-key encryption Marcus mentioned in the book (RSA) relies on modular arithmetic, also known in grade school as "clock math" ("What's 8 o'clock plus 5 hours?" "1 o'clock." etc.).
In modular arithmetic, you can add, subtract, and multiply just like you would normally -- but at the end you divide by your "modulus" (12, for clocks) and take the remainder.
There is a major theorem regarding modular arithmetic, called "Fermat's Little Theorem", which says that if your modulus is a prime (P), and you take a number not divisible by the modulus and raise it to the appropriate power (P-1), you get 1.
That is, t^{P-1} = 1 (mod P) for t not divisible by P.
This can be generalized to non-prime moduli. Specifically, if your modulus is the product of two primes (N = P * Q), then you have
t^{(P-1) * (Q-1)} = 1 (mod N)
for t not divisible by P or Q.
Now, this also means that if you have two numbers (d and e), whose product is one more than a multiple of the special exponent above, you can get back to your original text (t), i.e.
if d * e = 1 + k * (P-1) * (Q-1)
then
t^{d * e} (mod N) = t^{1} * (t^{(P-1)*(Q-1)})^k (mod N)
= t * 1^k (mod N) = t (mod N).
This turns this whole arithmetic exercise into a good cryptosystem. Why? Because you pick one of the numbers (d). Since you know your two prime factors (P and Q), you can generate the other number (e). Then your public key is (d,N), your private is e, and to encrypt with your private key you raise your message to the e'th power mod N,
c = t^e (mod N)
and then anyone can verify that the message came from you with the public key:
c^d (mod N) = t^{d*e} (mod N) = t (mod N).
Let me know where you got lost, and I can try to elaborate.
In re KnitML:
My prediction is that they will find that there is a formal equivalence between knitting patterns and Turing machines (or possibly Wang tiles), and that therefore showing that a given pattern is physically possible is equivalent to the Halting Problem. (And therefore can only be solved in the general case by an appeal to the gods -- or at least an oracle).
Bear in mind, though, that this prediction is made in total ignorance of knitting mechanics.
Giving blood -- oh yes, I know that fun, fun combination of hypovolemic and psychogenic shock well. And under relatively silly circumstances too: I was doing your basic blood-typing Bio lab, and had to use the finger-punching thing on myself to get the blood. Well, at the time, I was severely phobic of needles (less so now), and I kept jerking away.
I finally managed to break the skin, but couldn't get enough blood out. The professor running the lab was ever-so-helpful, and squeezed my arm to get the blood going to my hand. I got enough blood in the pipette, and about thirty seconds later I started feeling really clammy and nauseated. As my vision tunneled out, I wobbled over to the man and tugged weakly on his sleeve. He realized immediately what was going on, walked me out into the hall and sat me down on a chair with my head between my legs.
Clifton @279:
I haven't been following those particular experiments, but they sound interesting and the results look somewhat more plausible than the various other superconductor/gravity experiments.
Background: There are at least two or three separate theoretical motivations for believing that superconductors should have unusually large interactions with gravitomagnetic fields (not what we normally think of as gravity; electrical field : magnetic field ~:: gravity : gravitomagnetic field).
There have also been a number of very crank-buzzer-inducing claims of gravitational effects associated to elaborate experimental setups involving superconductors (to give an example of elaborateness, I believe one setup required a rotating superconducting disk that was a) accelerating, b) having a capacitor discharge thru it, and c) going thru its transition temperature -- simultaneously).
There have been a few decent experiments with YBCO superconductor -- all of which said "effect too small to show up with this setup", which was more or less expected.
This one looks interesting, possibly plausible, but the theoretical background in the second paper looks iffy to me (but I am not up on my superconductor theory).
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