automation – How to build an Automatic Warehous Model?

I have been using Anylogic for few months and I’m interested in finding out how I can model this kind of  industrial application:

I find out on the case studies of the AnyLogic web site this interesting conference:
It would have been great if the model was accessible.

Can you give me some advice in order to obtain a flexible model that allow me to specify just the number of cells that compose this automatic warehouse (in x, y and z direction)?
Furthermore if a pallet has been dropped in a specific cell, the shuttle (carrier/platform) needs to know it and consequently change its route.

Thank you so much.

Best regards

Fedora 33 how to disable automatic updates

I recently updated to fedora 33 and now the system has a very very ugly behavior:

From time to time I am not able to switch the system off. Instead it is doing “something during the shutdown”. By restarting, it boots to an update mode and shutdowns again. I have no control about all that ugly things.

I want to do updates as needed with the good old “sudo dnf update” if I like and not to get any broken package as someone else means. Especially the multiple boot/reboot/shutdown feels like windows.

Q: How to disable automatic updates and get back update control. 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.

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)

Yes, under the conditions you described, you are eligible for revalidation after a brief trip from the US to Canada or Mexico. Here are the conditions for automatic revalidation listed at The last paragraph is about your case, revalidation after changing status to H-1B/H-2B/L/J:

A person is eligible for automatic visa revalidation provided the following conditions are met:

  • The underlying authorization for the current status continues to be valid for the Form I-129 for non-immigrant workers and Form I-20 for students in F status.

  • The person’s absence from the U.S. was 30 days or less.

  • The person did not visit any countries other than Mexico or Canada in that period.
    Travelers who are on a F visa or J visa status are allowed to visit
    adjacent islands to the U.S.(i.e., the Caribbean).

  • The person does not have a pending or rejected application for a new
    visa. Since it is not possible to renew a non-immigrant visa in the
    U.S., a person on a non-immigrant visa may travel to a nearby country
    to apply for a new visa.

  • The person is not a citizen of one of the countries designated by the
    U.S., as a state sponsor of terrorism.

If you changed your status to H-1B/H-2B/L/J and want to re-validate
with a valid or expired visa of a different classification, you can do
so provided you meet the requirements under 8 CFR 214.1 (a)(3)(b) (sic) and
22 CFR 41.112(d), you may apply for admission to resume your
non-immigrant status.

The regulations it mentions say essentially the same rules: 8 CFR 214.1 (a)(3) and (b) and 22 CFR 41.112(d)

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());




iPhone sometimes dims display quickly (a few seconds) even when Automatic Display and Brightness is turned off

My iPhone is over 90% charged – and in fact was connected to power. But it was dimming even just a few seconds after accepting a call. I could not look at another app on the phone since the display kept turning off (more than once). The phone call proceeded otherwise normally.

I have looked at this other question that talks about five minutes . That is short but nothing at all like about five seconds: How to stop iPhone from automatically dimming the brightness?

This does not happen often but when it does it seems to only get resolved by entering into the Settings app and looking around (not necessarily even doing any changes).

What might be happening here – and is there a workaround/fix? I have an iPhone6S and am on iOS 14.2