Samsung Galaxy S4 Mail App not syncing while set to Automatic sync(Push)

It does sync automatically(push) only when I have opened the app. But today, eventhough I had it opened it didn’t.
I would like my phone to sync auto even after I close the app.

I would also download another email app if necessary.

Android version 5.0.1

gr.group theory – Can one reduce to ‘reversing’ the right multiplier finite-state automata of an automatic group to obtain a biautomatic structure?

Let $left( G, A, W, left{ R_{a} right}_{a in A cup { 1 }} right)$ be a group equipped with an automatic structure, where $G$ is the group, $A$ is a finite set of generators of $G$, $W$ is the word-acceptor finite-state automaton, and $R_{a}$ is the right multiplier finite-state automaton for $a in A cup { 1 }$. Recall that $R_{a}$ accepts (up to padding by a symbol which I’ll denote by $p$) a pair of words $(w_1, w_2)$ in $A$ if and only if $w_1$ and $w_2$ are accepted by the word-acceptor and if $w_1a = w_2$ in $G$.

I will view finite-state automata as decorated finite directed graphs (states are nodes, arrows are labelled, an arrow from $s_1$ to $s_2$ with label $l$ means that the state-transition (partial) map takes $(s_1, l)$ to $s_2$, some node is picked out as an initial state, and some set of nodes is picked out as the set of acceptor states).

It seems to me that the following simple construction equips an automatic group with a bi-automatic structure under some simplifyng assumptions.

Construction. Suppose that $1 in A$, and that if $a in A$ then $a^{-1} in A$, suppose that a word $w$ is accepted by $W$ if and only if $w^{-1}$ is accepted by $W$, suppose $R_{a}$ has a single acceptor state for every $a in A$, and suppose that the padding symbol is not used in the label of any arrow of $R_{a}$. Then we can construct $L_{a}$, the required left-multiplier finite-state automaton for a biautomatic structure for $a$, as follows.

  1. Take $R_{a^{-1}}$.

  2. Reverse the direction of all arrows.

  3. For every occurrence of $a in A$ in a label of an arrow, replace it by $a^{-1}$.

  4. Make the single acceptance state of $R_{a^{-1}}$ the initial state of $L_{a}$.

  5. Make the initial state of $R_{a^{-1}}$ the only acceptance state of $L_{a}$.

Since $aw_1 = w_2$ if and only if $w_1^{-1}a^{-1} = w_2^{-1}$, it is clear that $L_{a}$ accepts exactly the required pairs of words.


The following matters need to be addressed to be able to generalise the above construction. Firstly, if $R_{a}$ has more than one acceptor state, then the graph constructed as above does not quite define a finite-state automaton, since a finite-state automaton must have a single initial state. Secondly, we must handle the fact that the padding symbol might be used in some labels of of $R_{a}$. Thirdly, if $W$ accepts a word $w$, it might not accept $w^{-1}$.

I believe that these three matters can be addressed. The possibility of using the identity element of the group gives flexibility in being able to manipulate the finite-state automata of an automatic structure to obtain ones with the required properties.

But whether every automatic group is bi-automatic has been an open question for 20 or 30 years, and the answer is generally expected to be negative. Presumably there is therefore a serious problem somewhere. My question is what that problem is.

To give a hint that these three matters can be addressed, I will address a couple of them below.

Proposition 1. Suppose that the same assumptions as in the Construction above hold, except that a padding symbol may be used in some labels of $R_{a}$. Assume in addition that $W$ has the property that if $w$ is accepted by $W$, then every word obtained by adding $1$‘s at the beginning or end of $w$ is also accepted by $W$. Then we can still construct a left-multiplier finite-state automaton $L_{a}$.

Proof. We can construct $L_{a}$ as follows.

  1. Begin with $R_{a^{-1}}$.

  2. Reverse the direction of all arrows.

  3. For every occurrence of $a in A$ in a label of an arrow, replace it by $a^{-1}$.

  4. For every occurrence of the padding symbol $p$ in a label of an arrow, replace it by $1$.

  5. For every path of arrows in $R_{a^{-1}}$ whose first arrow has the initial state of $R_{a^{-1}}$ as its source, and which has the property that the left (resp. right) component of the label of every arrow in the path is $1$, carry out the following steps in $L_a$.

    a) Add a disjoint copy of this path (i.e. add a new arrow for each arrow in the path, and a new state for each state in the path, with the source (resp. target) of the new arrow being the new state corresponding to the source (resp. target) of the original arrow), and carry out steps 2) and 3) for this copy.

    b) In $L_a$, glue the initial state of $R_{a^{-1}}$ to the new state corresponding to the source of the first arrow of the path in $R_{a^{-1}}$.

    c) In $L_a$, glue the source state in $R_{a^{-1}}$ (viewed as a state of $L_a$) of the last arrow of the path to the corresponding new state in the copied path in $L_a$.

    d) For every occurrence of $1$ in a label of the form $(1,-)$ (resp. $(-,1)$) in the copied path, replace it by $p$.

  6. Make the single acceptance state of $R_{a^{-1}}$ the initial state of $L_{a}$.

  7. Make the initial state of $R_{a^{-1}}$ the only acceptance state of $L_{a}$.

Given $(w_1, w_2)$ for which $w_1$ (resp. $w_2$) must be padded with $n$ copies of $p$ to make it the same length as $w_2$ (resp. $w_1$), then $(w_1, w_2)$ is accepted by $R_{a^{-1}}$ if and only if the pair $(w_1′, w_2)$ (resp. $(w_1, w_2′)$) also is accepted by $R_{a^{-1}}$, where $w_1’$ (resp. $w_2’$) is obtained by adding $n$ copies of $1$ to the beginning of it (here we appeal to our assumption that $w_1’$ (resp. $w_2’$) is accepted by $W$). In turn, $(w_1′, w_2)$ (resp. $(w_1, w_2′)$) is accepted by $R_{a^{-1}}$ if and only if $(w_1^{-1}, w_2^{-1})$ is accepted by $L_{a}$. Note that this argument goes through whether or not $w_1$ (resp. $w_2$) begins with a series of $1$‘s; this is the reason for making a copy of the reversed path before re-labelling it in step 5) above, as opposed to simply re-labelling it.


Proposition 2. Suppose that the same assumptions as in the Construction above hold, except that, for any $a in A$, a padding symbol may be used in some labels of $R_{a}$, and $R_a$ may have more than one acceptance state. Assume in addition that $W$ has the property that if $w$ is accepted by $W$, then every word obtained by adding $1$‘s at the beginning or end of $w$ is also accepted by $W$. Then we can replace the automatic structure by one in which $R_a$ is replaced by a finite-state automaton $R’_a$ with a single acceptance state.

Proof. We carry out the following steps to construct $R’_a$.

  1. Add a new state, which I’ll denote by $T$, to $R_a$.

  2. Carry out the following pair of steps for every acceptor state $S$ of $R_a$.

    a) Add an additional three states $S’$, $S’_l$, and $S’_r$ to $R’_a$. Add three arrows from $S’$ to $T$, one each labelled by $(1,1)$, $(p,1)$, and $(1,p)$. Add an arrow from $S’_l$ to $T$ labelled by $(p,1)$. Add an arrow from $S’_r$ to $T$, labelled by $(1,p)$.

    b) Duplicate every arrow in $R_a$ with target $S$ whose label does not make use of the padding symbol, and make each duplicate have target $S’$ in $R’_a$. Duplicate every arrow in $R_a$ with target $S$ whose label is of the form $left( p, a’ right)$ (resp. $left( a’, p right)$ for some $a’ in A$, and make each duplicate have target $S’_l$ (resp. $S’_r$) in $R’_a$.

  3. Make $T$ the (only) acceptance state of $R’_a$.

We also replace the word-acceptor $W$ by one, which I’ll denote by $W’$, which is the same as $W$ except that we add a new state $T$, add an arrow labelled by one from every acceptance state of $W$ to $T$, and make $T$ the (only) acceptance state of $W’$.

The effect of the replacement of $W$ by $W’$ is that every accepted word finishes with at least one $1$. We make use of this when modifying $R_a$ to $R’_a$.


At this point, we have reduced the general case to being able to replace an automatic structure by one in which the word-acceptor $W$ has the following properties.

  1. If $w$ is accepted by $W$, then $w^{-1}$ is accepted by $W$.

  2. If $w$ is accepted by $W$, then any word obtained from $w$ by adding a series of $1$‘s at the beginning and/or end is also accepted by $W$.

It is easy to modify $W$ itself to a finite-state automaton with these properties, but one has to modify the right multiplier automata thereafter as well, which is more subtle, but can I believe be done.

PayPal automatic billing

PayPal provides its own recurring billing facility for customers, I am curious to know if there is any tool that syncs the billing profiles … | Read the rest of https://www.webhostingtalk.com/showthread.php?t=1832745&goto=newpost

sharepoint online – Automatic folder creation when a Teams user creates a site (warning: rank noob poster)

Warning again: This is my first attempt to use SharePoint beyond the web UI. I laid out a folder hierarchy to store per-project files and later got a surprise requirement that any user in the organization has to be able to create a new project, and therefore a new instance, preferably from Teams. I’m trying to salvage the design work and make it usable in a way acceptable to the organization.

I will probably misuse SharePoint terms and reserved words, and welcome corrections. Here’s my best attempt to describe the ideal workflow:

  1. A user creates a new team in Teams from a template I have created. This much is working – the creation of the team results in a team site in SharePoint, and two channels in the template do create channels in the team, and parallel libraries in SharePoint.

  2. We want automatic creation of folders in each of the channel libraries. The names of these folders are consistent and could be hard-coded.

Step two seems out of reach. I appreciate any replies.

usability – How to let user know that an automatic process in a mobile app can still be triggered from a different screen?

I’m working on a mobile app with Bluetooth Low Energy, that can trigger some hardware by either holding the phone near it (primary action) or trigger the hardware manually by looking for nearby devices and showing them in a list (secondary action).

My 2 screens basically looks like this:

enter image description here

The main tab (Scan) is a simple animation showing the user to hold his phone near the device to trigger it. It will show another animation if it detects a device nearby a phone and sends a signal to trigger it.

The secondary tab (Manual) will show all nearby hardware that can be triggered manually (if applicable). The user can click on the circle to trigger the device.

Now, even if the user is in the “manual” subtab, he can still trigger a device by holding his phone next to the hardware device. But it might look that the user HAS to be in the “SCAN” tab for that to work. This might lead to a case where the user accidentally triggers because he thought that it would not work.

How do i make clear that the SCAN is still active, even if in the other tab. By extension, the trigger can even work if the user is in the application settings, or has the app running in the background.

canada – Automatic Visa Re validation (US)

So i am currently on H1B visa. Planning to move to Canada soon, i would like to complete my obligation to finish the soft landing for the Canadian PR application.I want to transition to US to Canada very smoothly and would very much like to comeback to take care of final formalities in USA. Can i come back into the USA without any Visa Stamping in the US Consulate in Canada, if i come back in 30 days or less. I may be will have 4 months left on my H1B and would like to use the Automatic Visa Re validation to enter back into the USA without hassle.

Thanks
Bharat C P

UPDATE

I WAS ABLE TO COME BACK INTO USA WITHOUT MUCH HASSLE. I DID LEAVE THE COUNTRY WITHOUT A H1B VISA STAMPING STILL ON MY OLD F1 STAMPING. THE AGENT JUST ASKED FOR MY OLD PASSPORT AND VISA.

Automatic date in google sheet data entry. I can apply the scenario I have found to 6 windows in the same worksheet

I have prepared such a scenario, but I want to use it in 6 pages like the question asked above. Thanks in advance for your help how to change my script.

function onEdit() {

var s = SpreadsheetApp.getActiveSheet();

if( s.getName() == “Sheet1” ) { //checks that we’re on the correct sheet

var r = s.getActiveCell();

if( r.getColumn() == 1 ) { //checks the column

var nextCell = r.offset(0, 1);

if( nextCell.getValue() === ” ) //is empty?

nextCell.setValue(new Date());

}

}

}

setting automatic charge levels in ubuntu

I’m on linux, and I’d like to make my computer start charging when I am at 30% level and stop charging when I’m at 80%,because it’s mostly plugged in and I want to increase my battery life. Is this possible to do with some setting, etc.?

windows – The Python.exe process reads recent automatic destinations files and loads filezilla dll. Is that malware activities?

I run a simple python program for testing the WIN32 SHFileOperation api. However, from FileSpy, the python process reads a lot automaticDestinations-ms files in the recent directory. Besides, the process also loads a FileZilla Shell extension dll.

Is that a malware activities or just side effects of SHFileOperation api calling that brings thing from windows explorer.

Thank you!

PYTHON reads automaticDestinations-ms

enter image description here

Plugin or standalone script for wordpress automatic generate passwords for admin users with 24h regenerate

I have clean wordpress installation with one admin user. I want some script or plugin with autogenerate passwords for WordPress’s admin and sent automatically to email notification. Is possible?
Example :
i have user’s admin with password ‘test’ – i want generate random passwords for this admin once at 24h and send automatically thsi fact to email.