I found some older posts dealing with this(5 yrs old or more), but I am confused on the process or if this information is out of date, hence my new question. I am on sql 2016 SP2-CU11 enterprise.
I need to enable service broker on a database that is in an availability group. Its 1.8TB in size, so I really don’t want to take it totally out of the AG and reseed it all over again with a fresh backup if I can avoid it. Some older posts say you have to completely remove and add the db back in the AG, others reference just running some code to create endpoints, and routes which I have never run across before.
Is the process below the only way to enable service broker on a DB in an AG where it is not enabled?
remove the secondary copy of the DB, then the primary copy from the AG.
Run below code to activate service broker:
ALTER DATABASE XXXXXX SET ENABLE_BROKER;
GO
Take a full backup, restore it to the secondary with no recovery.
join the DB to the AG on the secondary.
Background
The scenario I am struggling with is the following: I have a standalone Windows 10 Enterprise system (not connected to any network / AD) which I want to configure with local group policies. The group policies should be different for different user groups. (User Groups: custom_admin, custom_user1, custom_user2)
Issue
What I have done so far is to create local GPOs for the machine via gpedit
and GPOs for the “Admin” users via mmc
(as shown here).
Now I want to also configure different GPOs for the custom_user1 and custom_user2 user groups, so that all users that will be created and assigned to these groups are restricted by these policies.
So far I have not been able to find a solution to this, but I suppose this has to be possible somehow. Is there any way to achieve this?
While you shouldn’t have to play in a game that you don’t enjoy, I think it would be a good idea to step back and try to understand why this is happening.
You’ve been running this group for a long time. Now once someone else is GMing, they are basically running wild, right? And picking on you? Maybe it’s just random, but maybe there’s a reason for it. I am reminded of a GM who posted a question on this site about how his characters wanted an “evil campaign” and were now running around doing whatever they wanted instead of playing “realistic evil,” and he came to understand that it was probably because they really just wanted freedom of action after being controlled tightly by the GM for so long. I’ve seen this a number of times over the years when a very controlling GM has the tables turned on them.
So:
This tells me that possibly, they feel like you have been an overly controlling GM/deity for some time and they are enjoying both the freedom of being able to have their characters do what they want (even if there are consequences) and hazing you in some kind of retribution. Naturally you are still trying to be the force of order and control even as a PC, given how used you are to being in charge, and are doing your part in creating an adversarial relationship with the other PCs. However, this then continues their motivation to resist your influence, and now that you’re not in charge, you can’t just stop them with your fiat.
I’m not saying they should be bullying you, but this is an opportunity to learn. Both about being a player from the player’s perspective (a mortal who frequently has a lot less control over their circumstances than they’d like) and about the GM from the player’s perspective (you need them to have a game, but they don’t see themselves as an equal and their actions affect you greatly).
You could just quit. But then what? Take GMing back over? Well, maybe they “got it out of their systems” but maybe they are just going to take this as “oh sure, she can dish it out but not take it” and act up even more.
You could steer into it and use it as a learning experience. The first two lessons to learn are:
I suspect you are entirely too invested in both of these things. Give them up, and focus instead on your relationships with the other people at the table and all of you having fun, NOT at the expense of each other.
You can start playing a hothead character. Why not? That’s not “you” but that’s why we call it role-playing. Next time an NPC y’all are questioning gives you lip, punch him in the throat! That’ll surprise the others. Let yourself off the leash and have fun. Have your character drink too much and dance on the bar. Have fun with the other players to be accepted.
Instead of focusing on the plot (which clearly neither the PCs or the DM care about) – focus on how you can help the other PCs achieve their goals. Be their wingman in the bar. Assist the higher skill characters on skill attempts. Some of the problem is that they don’t believe you are “on their side” – and they’re right. You’re still trying to control them, and that is just about as rude and inconsiderate as them calling you a hothead. Ideally your SO would have clued you in to this, but maybe he agrees and thinks openly discussing it’ll “get him in trouble” with you. (Also be sensitive to that if the GM sides with his SO too much it erodes player confidence in his judgement, so he may be taking a light hand to not fall into that trap.)
As you do this you will learn a lot from the player perspective about how you have to work with other players and how the GM sets a lot of the boundaries for you, consciously and unconsciously, that affect your fun. This will make you a better GM when you resume the screen and if you can show yourself a friend and ally of the other PCs and not just “their boss”, they should be more willing to settle down a little when it happens.
Again, I’m not saying they’re right to bully, and you can always just quit, but I strongly suspect that there’s something to be learned from this and if you can just reorient your thinking to have fun as a player in this game being a bit more chaotic and supporting the other PCs you will come out of it much better.
(As a side note this isn’t rally that much of a RPG-specific dynamic, and can be found in workplace relationships and friend-group relationships – developing the skill to understand why you’re not being accepted and what to do about it is super valuable and it’s actually great to be able to work it out in the safer confines of a RPG.)
Like the question says, I was the sole admin of the Messenger chat. We had the setting enabled where new people had to be approved to join by an admin. Since I left, nobody can approve me to join.
Is there any way an existing member can become an admin? Or is there any way around this so that I can get back in?
In his Notes on Group Theory, 2019 edition (http://pdvpmtasgaon.edu.in/uploads/dptmaths/AnotesofGroupTheoryByMarkReeder.pdf p. 83 and ff.)
Mark Reeder gives a proof of the non-existence of simple groups of order 720.
Here : (No simple group of order 720 ) I asked for a clarification about a statement at the beginning of the proof and I got it. (Thanks again to David A. Craven.)
I have another problem, perhaps more serious, at the end of the proof. M. Reeder proves that if $G$ is a simple group of order 720, if $s$ is an involution (element of order 2) of $G$, then $C_{G}(s)$ is a Sylow 2-subgroup of $G$ and that if $s$ and $t$ are two differents involutions of $G$, then $C_{G}(s)$ and $C_{G}(t)$ are different. So far, so good. A little further, M. Reeder says : “Since $s$ is contained in just one Sylow 2-subgroup, namely $C_{G}(s)$ (…)”. I don’t understand how that can be deduced from what precedes.
I must say that I am skeptical about the possibility of proving this in the frame of M. Reeder’s proof, for the following reason. It seems that, after proving that (for a simple group $G$ of order 720), $G$ has exactly ten Sylow 3-subgroups, that these Sylow 3-subgroups are non-cyclic and intersect pairwise trivially, M. Reeder derives a contradiction from only the following properties :
s1 : $G$ is a group (I don’t say “simple group”) of order 720 ;
s2 : $G$ has exactly ten Sylow 3-subgroups ;
s3 : these Sylow 3-subgroups are non-cyclic ;
s4 : these Sylow 3-subgroups intersect pairwise trivially ;
s5 : $G$ has no element of order 6 ;
s6 : every involution of $G$ normalizes at least two Sylow 3-subgroups of $G$ ;
s7 : $G$ has several Sylow 2-subgroups.
(M. Reeder uses another property to prove that some given subgroups of $G$ are isomorphic to $Q_{8}$, but he notes that this fact plays no real role.)
Now, properties s1 to s7 are not contradictory, because the group $M_{10}$ possesses them. (By the way, properties s1 to s6 are not independent : s3 can be deduced from s1, s2 and s5, using the N/C lemma.)
My question is : do you see how the statement “$s$ is contained in just one Sylow 2-subgroup, namely $C_{G}(s)$” can be proved in the frame of Mark Reeder’s proof ? Thanks in advance.
Note : I wrote up Mr. Reeder’s proof by breaking it down into simple elements and indicating, for each element, the minimum hypotheses. If anyone is interested, I can post this work here. I can also post a proof that $M_{10}$ possesses properties s1 to s7.
I have string column that looks usually approximately like this:
https://mapy.cz/zakladni?x=16.3360208&y=49.6718038&z=8&source=firm&id=13123554
I would like to group data by source which is part of the string – four letters behind “source=” (in the case above: firm
) and then simply count them. Is there a way to achieve this directly in SQL code?
It is well known that there are plenty possible applications of ‘fast arithmetic’ (that is, 1. having an algorithm at hand that actually computes in…, and 2. the running time of that algorithm is known and not too bad) in the degree zero Picard group (the ‘Jacobian’) of an integral, smooth and projective curve $C$ with small genus (for instance elliptic curves and hyperelliptic curves of genus 2) over a field $k$. For instance, there are cryptosystems whose safety relies on the discrete logarithm problem in the Jacobian of elliptic curves and thus it is crucial to know how fast the arithmetic in that group can be carried out.
I wonder whether there are applications of ‘fast arithmetic’ in the degree zero Picard group of more general projective curves $C$ over $k$. Especially, I am curious about the case when $C$ is at least one of the following: reducible, singular or has large genus.
For instance, in the paper of Michael Stoll and Peter Bruin the authors use the Mordell-Weil sieve to prove the non-existence of rational points on smooth projective curves $C$ over $mathbb{Q}$ of genus $g geq 2$. They use the embedding of the rational points on $C$ into the Mordell-Weil group (degree zero Picard group) and local data of reductions at primes $p$ of $mathbb{Q}$. And there is a speed up in their computation if they do not restrict themselves to primes $p$ of good reduction, but also consider $p$ with bad reduction. In that case the authors provide a variant of the Cantor algorithm to compute in the Jacobian in the case that the reduction has genus $2$. Here a generalization to higher genus may be useful to also gather information from bad primes whose corresponding reduction has large(er) genus.
I am interested if there are further applications of ‘fast arithmetic’ in the Jacobian/degree zero Picard group in the same manner as above or even in a complete different style (for instance, it might be valuable to know a lower bound for the running time of such an algorithm to provide statements about how hard a problem related to the Jacobian is).
Many thanks in advance for your help!
How do you tell if a group of items is stacked horizontally or vertically given only the items x and y positions?
Vertical Group:
Horizontal Group:
• item 1 • item 2 • item 3
The items in the vertical stack can be moved left or right but each item is further down from the last.
The items in a horizontal stack can be higher up or lower down than the first item.
What I’m doing now to check for a vertical stack is to check if the second item Y position is greater than the first item Y position. But if it is a horizontal group of items and the items are stair-cased down from top to bottom then it would fail.
if (secondItemY>topY) {
return true;
}
I almost had it figured out
I currently have a report produces a matrix, displayed per user group. (1 user group per page, with 35 user groups/pages)
I want to highlight the row in pink in this condition:
=IIf(Fields!CountFailures.Value>=1,”Pink”,”White”)
However, it is not showing any colour. Please help! (I am using PowerBI Report Builder… should be same as SSRS)