## group theory – Topswaps Puzzle Termination Proof

Consider the following one-person card game. It is played with 13 cards of
the same suit; each card is considered as having a numerical value: the ace
is 1, the deuce is 2, and so on, where the jack, queen, and king are 11, 12,
and 13, respectively. Before the game begins, the cards are shuffled. After
that the following operation is repeated. The top card of the deck is turned
up. If it is the ace, the game stops. Otherwise, the top n cards, where n is the
value of the top card, are removed from the deck and then returned there
in reverse order. An example of the game’s step is shown below:

5 7 10 K 8 A 3 Q J 4 9 2 6 -> 8 K 10 7 5 A 3 Q J 4 9 2 6

Does the game always stop after a finite number of iterations for every

I’m only an amateur mathematician and this is a bit beyond my ability I think for now.

My work so far is to observe that there is probably nothing special about 13. So a smaller puzzle can be described by the following permutations, where H means “Halt”.

#### N = 4

``````(1, 2, 3, 4) H
(1, 2, 4, 3) H
(1, 3, 2, 4) H
(1, 3, 4, 2) H
(1, 4, 2, 3) H
(1, 4, 3, 2) H
(2, 1, 3, 4) (1, 2, 3, 4) H
(2, 1, 4, 3) (1, 2, 4, 3) H
(2, 3, 1, 4) (3, 2, 1, 4) (1, 2, 3, 4) H
(2, 3, 4, 1) (3, 2, 4, 1) (4, 2, 3, 1) (1, 3, 2, 4) H
(2, 4, 1, 3) (4, 2, 1, 3) (3, 1, 2, 4) (2, 1, 3, 4) (1, 2, 3, 4) H
(2, 4, 3, 1) (4, 2, 3, 1) (1, 3, 2, 4) H
(3, 1, 2, 4) (2, 1, 3, 4) (1, 2, 3, 4) H
(3, 1, 4, 2) (4, 1, 3, 2) (2, 3, 1, 4) (3, 2, 1, 4) (1, 2, 3, 4) H
(3, 2, 1, 4) (1, 2, 3, 4) H
(3, 2, 4, 1) (4, 2, 3, 1) (1, 3, 2, 4) H
(3, 4, 1, 2) (1, 4, 3, 2) H
(3, 4, 2, 1) (2, 4, 3, 1) (4, 2, 3, 1) (1, 3, 2, 4) H
(4, 1, 2, 3) (3, 2, 1, 4) (1, 2, 3, 4) H
(4, 1, 3, 2) (2, 3, 1, 4) (3, 2, 1, 4) (1, 2, 3, 4) H
(4, 2, 1, 3) (3, 1, 2, 4) (2, 1, 3, 4) (1, 2, 3, 4) H
(4, 2, 3, 1) (1, 3, 2, 4) H
(4, 3, 1, 2) (2, 1, 3, 4) (1, 2, 3, 4) H
(4, 3, 2, 1) (1, 2, 3, 4) H
``````

It seems the key is to find a way to bring the 1 into the first position. One idea I had was thinking in terms of a function “distance_to_1()”.

The solutions with just one step have distance_to_1(p(0)) = p(0) -1

E.g. for (1, 2, 3, 4) H, distance_to_1(1) = 0

The same holds for two step versions:

e.g. (2, 1, 3, 4) (1, 2, 3, 4) H

distance_to_one(2) = 1

Then for 3 steps, we have something like distance_to(distance_to_1(p0)) = ??? and some kind of recursive function that eventually is guaranteed to cover all possibilities and thus prove that the solution will work for any size puzzle.. ???

That’s as far as I got going down that road.

Another idea is that there is probably some result from permutations in the group theory sense, although I know very little about this. I think it would be helpful to frame the problem in terms of composition of permutations, but I don’t know how to approach this.

Any help much apprecieated.

## applications – How to purchase premium features of an Android app for a group of people?

For a research project, we are utilizing an app available in the Google Play Store to collect data. However, to do this properly, we need to provide all participants with the paid full version of the app.

One version would be buying Google Play gift codes and telling our participants to use them on the full version.

However, I was wondering if there is any alternative way to directly enable them to download the premium version such as buying the version for the Google Play user with the email address example@gmail.com.

## distributed systems – How does organizing several processs on a group makes processes fault tolerant?-process resilience

Has it do with the fact that if f+1 processes are there and f processes fail, still 1 will be able to perform the operation? Or why?

I am studying process resilience.(Process resilience=processes can be made fault tolerant by organizing several identical processes in a group).

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## MySQL Group Replication Query – Database Administrators Stack Exchange

In case of MySQL Group Replication how will the application connects to DB? For 3 node group replication, do we need to configure all the 3 DB nodes hostnames in the connection sting to connect to DB?

Are MySQL Group Replication and InnoDB cluster available in Community edition?

Regards,
varun

## nt.number theory – Galois group of zeta function of hyperelliptic curve

Let $$f in mathbb F_q(T)$$ be monic, squarefree.

Can we say anything on the Galois group of $$Z_f$$, the zeta function of the hyperelliptic curve $$y^2=f$$, directly in terms of $$f$$ (coefficients or Galois group for example) ? I am especially interested in conditions implying maximality (or not !) of the Galois group of $$Z_f$$.

## The definition of a rational representation of a linear algebraic group

I’m reading Concini-Procesi’s “the invariant theory of matrices”. My question regards their setup in the introduction, and its relation/translation into the language of modern algebraic geometry.

The setup

Here is the setup they use in $$S$$2.3:

Let $$F$$ be an algebraically closed field. We view $$GL_n(F)$$ as the affine subvariety of $$F^{n^2+1} = M_n(F)times k$$ defined by the pairs $$(A,u)$$ satisfying $$det(A)u = 1$$. Thus its coordinate ring is $$F(x_{i,j})(det^{-1})$$ (polynomial ring in $$n^2$$ coordinates with the determinant inverted).

They define a linear algebraic group $$G$$ over $$F$$ to be a subgroup of $$GL_n(F)$$ cut out by polynomial equations. It’s coordinate ring is defined to be the restriction of the coordinate ring of $$GL_n(F)$$ to the subvariety $$G$$, denoted $$A(G)$$. Thus $$A(G)$$ is the quotient of $$F(x_{i,j})(det^{-1})$$ by some ideal.

They define an $$N$$-dimensional rational representation of $$G$$ to be a homomorphism
$$rho : Glongrightarrow GL_N(F)$$
such that the matrix elements $$rho(g)_{h,k}$$ for $$h,k = 1,ldots,N$$ belong to $$A(G)$$.

My question

Here, I assume they mean to take $$rho(g)_{h,k}$$ as functions on $$G$$? Is this the same as saying that $$rho$$ defines a morphism of affine group schemes?

Wikipedia defines a representation to be rational if $$rho$$ defines a rational map of algebraic groups $$Grightarrow GL_N$$, so e.g. the morphism needs only be defined on a dense open.

Are these two definitions equivalent?

## How to check current logged in user in SharePoint exist in specific group?

I want to check current logged in user in SharePoint exist in specific group using C# (.Net)

``````public static bool IsCurrentUserMemberOfGroup(string groupName)
{
/* check current user exist in give group name */
/*return true/false; */
}
``````

## sql server – Error: While creating clusterless Availbility Group

I was trying to create Clusterless AlwaysON for SQL Server 2019 and getting the below error.

Error Message

Msg 47106, Level 16, State 3, Line 11
Cannot join availability group ‘100’. Download configuration timeout. Please check primary configuration, network connectivity and firewall setup, then retry the operation.
Msg 41158, Level 16, State 3, Line 11
Failed to join local availability replica to availability group ‘100’. The operation encountered SQL Server error 47106 and has been rolled back. Check the SQL Server error log for more details. When the cause of the error has been resolved, retry the ALTER AVAILABILITY GROUP JOIN command.
Msg 15151, Level 16, State 1, Line 15

I made the both SQL server are running in same windows account and has full sysadmin privileges. All firewall settings are off and also able ping and connect SQL1 and SQL2

## query performance – is HAVING clause before GROUP BY clause processed differently than after?

Is there any difference how query is processed (translated to extended relational algebra), if HAVING condition is written before GROUP BY (compared to after)?
In logical query plan it should be initially upper in the tree (after GROUP BY) ? As far as I understand the topic(having selection should execute later in any case)? Is that correct?

I was translating some sql query to logical query plan (extended relational algebra) and in the query HAVING was before GROUP BY, which I thought might be incorrect, but on google I found it might be possible in some implementations.