Complexity Theory – Post Correspondence Problem: Finding the Total Number of Solutions

To let $ A = {a, b } $ be an alphabet.

$ P = A ^ * times A ^ * $

An instance of the PCP is a non-empty list $ D = (d_1, d_2, …, d_n) in P ^ n $ of couples of words.

To a PCP instance $ D in P ^ n $ an index sequence $ (i_1, i_2, …, i_m) in {1, …, n } ^ m, m in mathbb {N ^ +}, $ is a solution for D if it has the following property:

to the $ (t, b) = d_ {i_1} diamond d_ {i_2} diamond … diamond d_ {i_m}: t = b $

d) Now: $ d_1 = (ab, a), d_2 = (ab, ba), d_3 = (ba, b). $ Explain why this PCP instance has no solution.

My Answer: The first part of the pair always has two letters, while the second part has only one letter except a couple. This means that the first part always gets bigger than the second, unless you just use it $ d_2 $ since both the first and the second part each have two letters. But you can not do it $ d_2 $ since the initial letters do not match, you would already have to start with $ d_1 $ or $ d_3 $ which means that the first part becomes much larger than the second part. So you would never come to a solution.

Is there a more formal way to answer the question?

e) What are the possibilities for the total number of solutions for a random PCP instance?

My Answer: I've tried to think about the different number of letters every Domino has for each part. But it does not seem right.

Can you give me an indication of how to find out if and how many solutions exist?

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Invisibility by correspondence

How you do that indicates (page 34) that a magician could become invisible with a personnel 2 effect. I would like to know if it is possible (and better if referenced in any source) to use it Correspondence to somehow fold the space around the magician (instead of bending the rays of light as you would with strength) to become invisible.

Undecidability – How is the postal correspondence problem undecidable?

Mail correspondence problems can be solved with a brute-force approach.

Brute force can only be used if the number of dominoes we use is limited. For example, if we choose to use each Domino at most once, the number of possibilities is limited and we can use a brute-force algorithm.

With the PCP we can use any Domino as often as you like, so the number of possible configurations is infinite.

Proof of undecidability:
We will prove this by reducing the "Turing Machine Acceptance Problem" to an instance of PCP. We already know the acceptance problem of Turing Machine: $$ L ^ A = { | M : is : a : TM : and : M : accepts : w } $$
is undecidable. So if our case is decidable by PCP, then the acceptance problem of Turing Machine will be crucial. A contradiction.
The reduction involves creating dominoes for all possible configurations of a TM as they are executed on the input string. A full discount can be found here. Let me know in the comments if you need clarification on the reduction.

Lying Groups – Mark manifolds as incidence correspondence

To let $ G $ be a reductive group, $ B $ a Borel and $ P_j $ the maximum parabolics, indicated by the vertices $ j $ the dynkin diagram. Then $ B = bigcap_j P_j $So the flagpole $ G / B $ injected in $ prod_j G / P_j $,

In the classic types, this is a useful concrete description of $ G / B $: Like $ A_ {n-1} $, $ GL_n / P_j $ is the Grassmannian of $ j $Aircraft in $ n $Space, so a point of $ prod_j GL_n / P_j $ is a sequence of subspaces $ (V_1, V_2, cdots, V_ {n-1}) $With $ dim V_j = j $, The subspace $ G / B $ is the flag $ V_1 subset V_2 subset cdots subset V_ {n-1} $,

Similar to types $ B $ and $ C $, the $ G / P_j $ are isotropic flags, and we get the standard descriptions of the type $ B $/$ C $ Filler manifold as complete isotropic flags.

Like $ D_n $, the $ n-2 $ Vertices on the long leg of the graph correspond to isotropic subspaces of dimensions $ 1 leq j leq n-2 $and the two additional vertices correspond to the two isotropic types $ n $-plane: We get that $ G / B $ is flags $ V_1 subset V_2 subset cdots subset V_ {n-2} subset V_n ^ +, V_n ^ – $, from where $ V_j $ is isotropic $ j $ Plane and $ V_n ^ + $ and $ V_n ^ – $ are from the $ + $ Type and the $ – $ Art. This is not the default description for complete isotropic flags, but it's easy to convert them to one: Given $ (V_n ^ +, V_n ^ -) $we can relax $ V_ {n-1} $ as $ V_n ^ + cap V_n ^ – $ and given $ (V_ {n-1}, V_n ^ +) $we can relax $ V_n ^ – $ like the other isotropic $ n $Level between $ V_ {n-1} $ and $ V_ {n-1} ^ { perp} $, Note that the condition $ V_ {n-2} subseteq V_n ^ +, V_n ^ – $ along with the types of $ V_n ^ { pm} $ actually forces $ dim (V_n ^ + cap V_n ^ -) = n-1 $,

I find that in each of these cases, I only have to impose conditions on couples $ (V_i, V_j) $ where is there a border $ (i, j) $ in the dynkin diagram. This leads me to my question:

For two dynkin corners $ a $, $ b $, To let $ pi_ {ab} $ be the projection of $ prod G / P_j $ on to $ G / P_a times G / P_b $, Is it still true with the exceptional types? $ G / B = bigcap pi_ {ab} ^ {- 1} left ( pi_ {ab} (G / B) right) subset prod G / P_j $ where the intersection runs over edges $ (a, b) $ the dynkin diagram? Is this question useful and is the answer for Kac-Moody groups yes?

Motivation: I've committed myself to giving a course on Bruhat Commands and Schuberty things in the fall, and I'm thinking of telling the Combiner about flag variety in other types, not the entire construction of reductive / Kac-Moody To enter groups.

Algebraic Groups – L Packages in the Local Langlands Correspondence: Why Finite Quantities?

To let $ G $ be a connected, reductive group over a local field $ k $, and let $ ^ LG $ Be the Langlands double group. As Borel has pointed out in his Corvallis article, the general local correspondence of Langlands (1) should classify the classes of irreducible representations of $ G (k) $ in finite sets, called L-packets, and (2) a bijection between the L-packets and the equivalence classes of allowed homomorphisms of the Weil-Deligne group $ W_k # $ in $ ^ LG $,

When $ G = operator name {GL} _n $the L packages are just singleton sets. I believe that only the local Langlands guesses $ operatorname {GL} _2 $ were proven at the time when Borel's article was written. As far as I know, there were no elaborate examples of L packages with more than one element.

Why did Borel and others expect the L packages to be finite in the 1970s? Why do we still expect this today?

Tls Certificate Correspondence – Information Security Stack Exchange

I'm not very good at these details, just know a few basics.

In that case, I really recommend that you read how SSL / TLS works.

Is there a way to decrypt tls in wireshark if the encryption method is diffie hellman? I know that using a web browser is easy as I have a session key there. But in general, how can I extract session keys if I only have bare TCPs? #

The goal of TLS is to protect the transport from snooping. This means that if used properly, there will be no information in the TCP connection that will allow decryption. Only the client and server have the information they need to encrypt and decrypt the data.

If I have multiple clients connecting to the server, how does the server distinguish the different certificates and select the correct certificate to decrypt the message?

The server does not use certificates to decrypt the traffic. The certificate is used only for authentication to prevent man-in-the-middle attacks. In the case of an RSA key exchange, the server certificate is additionally used in the key exchange.

It is common for a server to have only a single certificate. If the server has multiple certificates for different domains, the client must specify which domain within the SNI extension of the ClientHello (first message of the TLS handshake) to access.

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