Can Slayers pick the Combat Trick talent multiple times?

Question: Can Slayers pick the rogue talent Combat Trick multiple times or not? I've seen long debates about this, but never found a good answer.
Comment: Other than the discussion you linked to, I couldn't find any *other* debates; I would like to read more of the controversy surrounding this if you can provide further links.
Comment: http://paizo.com/threads/rzs2rfrn?Rogue-Talent-AND-Slayer-Talents this is close to same. Mostly there just my party arguing back and forth about it
Answer: No, you cannot take Combat Trick multiple times.
The reason for this has nothing to do with anything in the Slayer class description; it has to do with the Rogue Talent class feature description.

A rogue cannot select an individual talent more than once. Source.

Some Rogue Talents, such as the Terrain Mastery talent, can be taken multiple times because they explicitly say so.

Special: A rogue can take this ability multiple times, each time applying it to a new terrain. Source.

Combat Trick does not contain any permission to take it multiple times, so it cannot be taken multiple times. That applies to anyone who can take Rogue Talents, not just Rogues themselves.
Whether Slayer talents can be used to take the same Rogue Talent multiple times if the Rogue Talent allows it depends on your interpretation of "it grants the slayer a new rogue talent," as discussed in Hey I Can Chan's answer. I would argue that each time a Rogue can pick a new Rogue Talent, even if he repeatedly takes Terrain Mastery, he is gaining new Rogue Talents each time. In my mind it works much like Weapon Focus: you take Terrain Mastery(Forest) as one Talent and later decide you want Terrain Mastery(Plains). They are in that sense separate talents. This is an interpretation, however, and there is no exact rules-as-written answer like there is for Combat Trick.
Comment: My one question is piazo was very specific about the ninja http://paizo.com/pathfinderRPG/prd/ultimateCombat/classes/ninja.html using rogue talent but not the slayers http://paizo.com/pathfinderRPG/prd/advancedClassGuide/classes/slayer.html
Comment: The classes were probably written by different designers. If you think about it, the different wordings for the Ninja and Slayer mean the same thing most of the time. They both have the same general intent. The Ninja wording is better because it is more clear, and that was overlooked with the Slayer. I don't think Paizo is trying to make meaningful design differences by using slightly different wordings that are otherwise so close in meaning.
Comment: I really want to agree with this because, darn it, it makes sense, but I can't get over the fact that the slayer is *not* a rogue, so while a rogue can take Terrain Mastery multiple times, that just doesn't affect the slayer, like, *at all.* It certainly *should*, and it's *dumb* that it doesn't, and maybe even the GM should rule it *does*, but that's not what it *says.* (Totally not *your* fault, obviously. Just let me vent: Argh! *Paaathfinder!*)
Comment: (I wish there were something addressing this in the description of [hybrid classes](http://www.d20pfsrd.com/classes/hybrid-classes).)
Comment: Unless an errata is written, the class works like the text says it does. There have been multiple cases of classes they are JUST like another but were intentionally different. The ACG is full of examples, investigator and alchemists being the one that caused the most discussions on the boards.
Answer: A slayer can pick the slayer talent rogue talent (combat trick) only once
The class feature slayer talent is introduced with the following text:

As a slayer gains experience, he learns a number of talents that aid him and confound his foes. Starting at 2nd level and every 2 levels thereafter, a slayer gains one [n.b. not new] slayer talent. Unless otherwise noted, a slayer cannot select an individual talent more than once.

Emphasis mine. Later, the slayer talent rogue talent says, in part, that

This talent can be selected multiple times; each time, it grants the slayer a new rogue talent.

Thus, while the individual slayer talent rogue talent can be picked more than once, each rogue talent picked must be different from one already possessed or else the newly picked rogue talent isn't a new rogue talent but an old rogue talent picked a second time.
(By the way, were each time, it grants the slayer a new rogue talent omitted, I'd totally agree with this answer, but allowing picking the same rogue talent multiple times ignores that sentence as, if it's read as permitting multiple picks of the same talent, the clause merely repeats the introduction to the class feature slayer talents, which already grants a slayer talent, in essence, saying A slayer gains a slayer talent twice for no reason. I know these folks are paid by the word, but that's why editors are salaried.)
In short, a slayer who picks the slayer talent rogue talent to pick the rogue talent combat trick is barred from picking combat trick a second time.
For comparison, with its full base attack bonus, two good saving throws, sneak attack, studied target, and track, the slayer is already a heady combination of ranger and rogue. Permitting the slayer also to take 10 bonus combat feats over the course of his career means virtually obviating the fighter, which Paizo has gone to great (some would say unreasonable) lengths to keep relevant despite Pathfinder's evolving character class landscape.
Comment: @Cutter8812 Literally *everything* is at the GM's discretion; sometimes, however, it's useful to be able to justify using that discretion in a different fashion. `:-)`
Comment: I still feel like there are good reason why the answer could still being yes or no. Unless there are more points that other people make would it be fair to leave it to dm discretion.
Answer: YES

This talent can be selected multiple times; each time, it grants the slayer a new rogue talent

This is fairly unambiguous.  Note that you gain a new rogue talent, not a 'new rogue talent not currently possessed', and especially note the lack of phrases like "do not stack'.
Compare Spell Focus's similar line:

You can gain this feat multiple times. Its effects do not stack. Each time you take the feat, it applies to a new school of magic.

When you can take something like this multiple times, the rules have to specifically call it out.  When they want to limit how that multiple times works, they have to call that out, too.  No such restriction is placed on the Rogue Talent Slayer Talent, and indeed it seems clear that some of the options (e.g. terrain mastery) are likely intended to be allowed to be taken multiple times-- no other class with access to favored terrains can't get more than one.
From a RAW standpoint, slayers are most indubitably permitted multiple combat tricks, allowing them to attempt to fill their more combat-focused role somewhat more handily.
Comment: To clarify, *new* here means merely *additional* instead of *not already selected* and *additional* can mean *already selected*? Wouldn't saying *it grants the slayer a rogue talent* (without the word *new* at all) then mean the same thing? Further, a wizard can then repeatedly pick the same school of magic for the feat Spell Focus, too, right?
Comment: @HeyICanChan Yes, and yes (but note that the latter case may as well not be allowed; it explicitly doesn't stack, like most such things)
Comment: +1; Combat Trick (Critical Focus) is not the same rogue talent as Combat Trick (Improved Initiative), just as Weapon Focus (Cestus) is not the same feat as Weapon Focus (Longsword). They are distinctly different, and therefore when you take Combat Trick with a different feat, it is indeed a *new talent* even though you may have taken Combat Trick previously with a different combat feat.
Comment: But the slayer being a different class, so wouldn't his list different than the rogue and follow his own rules. Also the the slayers list is much smaller than the rogues list so that's where I got confused because I didn't think he would follow the rogues talent rules because he's picking from a certain pool of them.
Comment: Nowhere in the description of combat trick it says that it can be taken multiple times for different feats, unlike Weapon Focus does. You learn one feat once you learn combat trick, and you are limited to taking it only once as per "Rogue Talents" and "Slayer Talents" abilities descriptions.
Comment: Thank you that was a well put answer, and cleared up a lot of confusion I had about that wording.
Comment: @Dyndrilliac Unlike similar feats, the talent [combat trick](http://www.d20pfsrd.com/classes/core-classes/rogue/rogue-talents/paizo---rogue-talents/combat-trick) **is** limited in the number of times a rogue can select it. Were that *not* the case, the rogue could take multiple times the talent [combat trick](http://www.d20pfsrd.com/classes/core-classes/rogue#TOC-Rogue-Talents) despite the talent being listed but once on [its list](http://www.d20pfsrd.com/classes/core-classes/rogue/rogue-talents).
Answer: Appended feat names are considered different individual feats throughout Pathfinder (this is a deductive argument).  For example, the Investigator talent alchemist discovery (type-appended), and investigator talent rogue talent (type-appended) can be taken multiple times as long as its of a different type.  "This talent can be selected multiple times; each time grants a new ..." alchemist or rogue discovery/talent. Ranger favored terrain [types] can also be selected multiple times as long as it isn't of the same type.  The slayer's ninja trick [type-appended] "can be selected multiple times "  The slayer's rogue trick [type-appended] "can be selected multiple times; each time, it grants the slayer a new rogue talent."
The rule "A rogue cannot select an individual talent more than once." is not abrogated by the talent terrain mastery [type], which can be taken multiple times as long as its a new terrain type.  Likewise, the rogue ninja trick can be taken more than once without violating the individual talent rule.
The second argument is a language argument.  We should expect Combat Trick (agile maneuvers) to be a different feat name, representing a different individual feat, than Combat Trick (alchemical strike) in much the same way as we should expect George Bush Jr. to be a different individual than George Bush Sr.  When we append names to proper nouns, we are differentiating between those nouns.
Hence, I believe we can pick either a Slayer or Rogue 'combat trick (type)' multiple times as long as it is a different type each time.  Likewise, we can pick the Rogue talent 'weapon focus (type)' or favored terrain (type) multiple times.
If however there is a circumstance where appended names represent the same thing, they do not stack (as is stated in the Ranger section of Terrain Master).  If somehow there was 'Jungle-Forest', and we had both Terrain Mastery (Jungle) and Terrain Mastery (Forest), then we could use only one non-stacking Terrain Mastery (type).
Comment: There is no "above" and "below" here, and comments or answers that are here now may not be here in the future. Consequently, if your answer depends on a particular argument, saying "so-and-so's argument above" does not help anyone find the information, and instead you need to include it in the answer directly (in your own words, of course).
Comment: My edit helps but isn't really enough, so please do make your own edit to finish incorporating the substance of Dyndrilliac's argument.
Comment: [James Jacobs disagrees](http://paizo.com/threads/rzs2kkm2?Rogue-Talents#4), and no Paizo-published creature so far as I'm aware has multiple iterations of the rogue talent (combat trick). In fact, most creatures having the rogue talent (combat trick) have the talent listed in stat blocks as *rogue talent (combat trick)*, with the feat in their feats section. Arguing that from a rules-as-written standpoint the designers here are *wrong* is something I totally support, but it may strengthen your position to explain why no published creature has seized the opportunity to take combat trick twice.
Comment: It looks like you've created multiple accounts. You can read how to resolve that [here](http://rpg.stackexchange.com/help/merging-accounts).
Comment: Can you find just *one* Paizo-published example of, like, a ninja or [rogue](http://paizo.com/pathfinderRPG/prd/npcCodex/iconic/merisiel.html) having more than once the ninja or rogue talent (combat talent (`combat feat name`)) or a slayer having more than once the slayer talent (rogue talent (combat trick (`combat feat name`))) . That is, is the argument *This is true because this* can *be read this way, and it's of no consequence if the publisher opts to or not* or something? Like I said, I'm okay with that—that's bold!—, but if that *is* the argument, it should be made clear it is.


PRIVATE_KEY is not importing from .env file when deploying the contract using node.js

Question: I put my private key in .env file and then import the .env file using require("dotenv").config() but still it is not compiling the code and throws an error. I even installed the dotenv using "yarn add dotenv".
In .env file, I put my private key as:
PRIVATE_KEY = "(private-key)";
In const wallet, I call the private key as : process.env.PRIVATE_KEY
<code>const ethers = require("ethers");
const fs = require("fs-extra"); //To import the 'fs' fuctionality of node.js to read the files of abi and bytecode
require("dotenv").config();
async function main() {

  await console.log("Smart contract address : 0x123456....");
  //include the provider and connect the local blockchain to our script
  const provider = new ethers.providers.JsonRpcProvider(
    "RPC-URL"
  );

  const wallet = new ethers.Wallet(process.env.PRIVATE_KEY, provider);
  const abi = fs.readFileSync("./Fundme_sol_Storage.abi", "utf8");

 //Here we've included our abi file and it readFileSync mainly takes two arguments obviously one is path of abi file and another is encoding for the smart contract

  const byteCode = fs.readFileSync("./Fundme_sol_Storage.bin", "utf8");

  //Now we have to use contract factory to deploy the smart contract on our local blockchain

  const contractFactory = new ethers.ContractFactory(abi, byteCode, wallet);
  console.log("Contract is deploying.....");
  const contractDeploy = await contractFactory.deploy();
  await contractDeploy.deployTransaction.wait(1);

  //This is going to give the number....

  const contractRet = await contractDeploy.retrieve();
  console.log(contractRet.toString());

  //This is going to increement the number..

  const contractStore = await contractDeploy.store("42");
  await contractStore.wait(1);

  //Now, we're going to retrieve number..

  const newNum = await contractDeploy.retrieve();
  console.log(`The update Fav number is:${newNum}`);
</code>
This is the error I am getting in terminal while deploying the code.
Answer: Change to PRIVATE_KEY="803b89dh9f89"
sometimes it's okay with PRIVATE_KEY=803b89dh9f89 too
Answer: my .env file :

my deploy.js:

Hope that helps, I'm a beginner
Comment: The next time include code and snippets as text. That way it is much easier if someone wants to use them.


Any alternative to finding singularity of Puma 560 than differentiating the Jacobian and putting the Det=0?

Question: Any alternative to finding singularity of Puma 560 than putting the Jacob Det=0? Also when it says the axes intersect, which axes do they mean? The z?
Comment: There are some good videos demonstrating robot arm singularities such as this https://www.youtube.com/watch?v=lD2HQcxeNoA that should greatly help in understanding what they are and why.
Answer: The determinant of the Jacobian will be zero at singularity.  The $j^{th}$ column of the Jacobian (a 6-vector) is the end-effector Cartesian velocity due to unit-velocity motion of the $j^{th}$ joint.  At singularity these column vectors are not independent, if the $k^{th}$ column is some constant multiple of the $j^{th}$ column then this means that the axes of joints $k$ and $j$ are aligned, that is motion of either joint will cause the same end-effector Cartesian motion.
We can illustrate this using the Robotics Toolbox for MATLAB
<code>>> mdl_puma560   % create a model of a Puma560 robot
>> J = p560.jacob0([0 0 0 0 0 0])  % compute the Jacobian for world coordinate frame

J =

    0.1500   -0.4318   -0.4318         0         0         0
    0.4521    0.0000    0.0000         0         0         0
         0    0.4521    0.0203         0         0         0
         0         0         0         0         0         0
         0   -1.0000   -1.0000         0   -1.0000         0
    1.0000    0.0000    0.0000    1.0000    0.0000    1.0000

>> det(J)

ans =

     0
</code>
which we see is singular
<code>>> jsingu(J)
1 linearly dependent joints:
  q6 depends on: q4
</code>
tells us that joints 4 and 6 are aligned, and we can easily see that columns 4 and 6 of the Jacobian are linearly related, in fact they are identical.
Since two joints control the same Cartesian DoF, and we have 6 joints, then only 5 Cartesian DoF can be controlled independently – the definition of a singularity.
-- Peter Corke, developer of the Robotics Toolbox for MATLAB
Comment: Peter, how do you describe the singularity which occurs when the wrist center is located in the plane defined by the first two axes (the over-the-shoulder singularity)?  Is that an alignment of q2 and q3, or is q1 also involved in the "alignment"?
Comment: @SteveO not sure what you mean by "describe", they are all singularities and the kinematic structure might allow alignment between wrist and base axes, but often robots have physical offsets to prevent that.  The only 2 types of singularity I know of are axis alignment, as described, and maximum reach where a DoF is lost because the arm can stretch no further.
Comment: I have struggled to describe this other singularity.  It is easier to show the wrist singularity because of the actual physical alignment of the axes of q4 and q6.  But when the arm is elbow-down, with the wrist center in the plane of q1 and q2 (it would be on the q1 axes except for the elbow offset), then q1 only rotates the wrist center in place - there is no axis to cause motion horizontally in this configuration.   You described the wrist singularity so well I was hoping you had magic words to describe that arm singularity also!
Comment: No magic, sorry!  It's just instanteous alignment of axes X and Y, they don't have to both be from the wrist.
Comment: BTW the singularity is strictly an arm singularity.  The wrist joints are not involved.   I mention wrist center only because that enables partitioning the inverse kinematics.
Answer: Mathematically you need to set det(J) = 0.  However, you can also visualize "singular planes" of a manipulator.  Check out papers by Stanisic.  Here is one.


How to write this kind of sentence for a distinguished lecture series?

Question: I work in a Chinese university and recently our department has decided to start a Distinguished Lecture Series in Physics. We want to give the invited speaker some kind of document showing that, basically, he has been a speaker at this series. I'm not sure what's the best way to write this sentence. I'm not even sure what this kind of document should be called. A certificate? Some kind of award? Please give me some clue. Thank you very much.
Comment: "Certificate of Appreciation" is a good catch-all choice, if given when the speaker presents or afterwards.  "Certificate of Acknowledgement" would work before or after.
Comment: This seems like a weird thing to give a certificate for.  Presumably the speaker can just download the announcement for the talk if proof of having been a speaker is needed.  Or perhaps sending a thank you letter afterward would work?
Answer: Firstly, on what to call this document, I would suggest "Bonafide Certificate of Acknowledgement" or simply "Certificate of Acknowledgement" as that's what it is exactly, as opposed to being an award of any kind.
Secondly, on what to write in the document, I would suggest something on the lines of the following:
"This is to certify the participation (or) contribution of ____Mr/Ms. XXXXXX____ as ____Subject Matter Expert / Distinguished Speaker____ in Distinguished Lecture Series in Physics held at ____XXXXXX University____ on DD-MMM-YYYY. [[Optional note of thanks]] [[Authorized Signatory]] "
Comment: I would not use "bona fide" (two words, BTW) on a certificate like that.
Comment: @HotLicks: Agreed. Bona fide would be ideal to certify the reality/authenticity of something as the purpose.
Comment: It's best to print it on the thickest paper available, and to put a decorative but dignified BORDER (e.g., scrollwork) around the certificate. Make it about half-page size (about 5x8") and orient it landscape (i.e., long edges are top & bottom).
Comment: Thank you very much for all your kind help! Cannot vote up because it requires 15 reputations. Sorry.
Comment: @Bill As the OP you can tick the answer as correct though (if you wish) and that's worth more than all the up-votes in the world.


How can i disable CC section in Publish a major Version Approval Workflow?

Question: I want to hide these sections , but every person can change the workflow. So how to hide this?
Answer: The Approval Workflow forms are InfoPath forms that can be modified using Infopath Designer.
To modify the form do the following:

Open the SharePoint Designer
Connect to your Site.
Got to Workflow > Click on the Approval Workflow.
In forms Section,> Click on the form to open and edit it with InfoPath.

Now, you can remove the CC field or hide it as mentioned at Hide a control based on values on the form


"baptized into Christ" vs "baptized into the name of Christ"

Question: As I research the use of "baptized into Christ" (Romans 6:3; Gal 3:27), can we legitimately distinguish this from "baptized into the name of Christ" (Matt 28; Acts 2:38; 8:16; 19:16).  One seems to speak to the baptism of the Holy Spirit (into Christ) the other seems to always refer to water baptism (into the name of Christ).  Is there anything there?  I know some theologians see these as identical.  Is there a distinction to be made?
Comment: I'm voting to close this question because this is a theological question suited to CSE
Comment: Welcome to BHSX.  Thanks for your question.  Please remember to take the tour (link below left) to better understand how this site works.
Answer: As an aside, theophilus, before I respond, I thought it'd be good to give you a brief introduction as to what to expect here on the Hermeneutics SE. In theory, Hermeneutics is the main issue that is discussed.  But, in actuality, there is no real definition as to what Hermeneutics actually is. So, in practice, people are more than content to pour out their own theology here (Michael16). But, when others do, the very same people try to bar and forbid them, saying that their question is not "Hermeneutics"-based.
So also, in general topics, which all Christians agree on (Trinity, Virgin Birth, Two natures in Christ, etc.), you'll find some unity here. But, when it comes to the sacraments (Baptism, Lord's Supper), you'll find much disunity. And this disunity shows itself in unneeded downvoting of people's answers. I do not find, for example, Dottard's answer all that convincing. But I acknowledge his right to speak. And I don't find a reason to downvote it.  His answer illustrates the main failing of our Hermeneutics SE. His answer is a blobbing together of both exegesis and dogmatics. These two naturally belong together.  Exegesis and Dogmatics are vital pieces of Hermeneutics.
But it does show how the word, "Hermeneutics" without any boundaries and borders will lead to different conclusions.  I offer the following as an alternate explanation, knowing that, historically, any discussion of the sacraments here is often met with verbal vitriol.

In my notes for this theological issue, I have the following:
<code> Jesus is the Mediator between the Triune God and man.

a) Baptism seals to the recipient the reestablished union between him and the Triune God.
    Cf Mt 28:19; Nu 6:22-27. (the name of God is in both, but what are you doing with them?  you are putting God's name on them.  When you speak the benediction you are actually blessing people, you are putting God's name on them.  The same blessing we recieve in Baptism)

b) This must be clearly expressed when administering Baptism.
    -1) On the importance.
        cf Ga 3:26,27; Jn 1:12; Ro 8:16. (Gal 3:  υἱός emphasizes the heir, the firstborn, while τέκνα emphasized the children)

    -2) No other words may be found so well adapted to this purpose as the words of Jesus.
        cf Mt 28:19.
        cf The Apostolic Creed; all ancient "regulae fidei".

c) Since there is no adoption to sonship by the Triune God except through the mediatorship of Christ: the sacrament may be briefly called a baptism in the name of Christ Jesus.
    Cf Ac 2:38; 8:16; 10:48; Ro 6:3; Ga 3:27,
    cf. Didache, VII, 1.3: . . . βαπτίσατε εἰς τὸ ὄνομα τοῦ πατρὸς καὶ τοῦ υἱοῦ καὶ τοῦ ἁγίου πνευμ́ατος. . . ἔκχεον εἰς τὴν κεφαλὴν τρὶς (?) εἰς ὄνομα πατρὸς καὶ ὑιοῦ καὶ ἁγιοῦ πνεύματος  [AF-T, V, did 7:1-3]
    Coll. Didache IX, 5 μηδεὶς δὲ φαγέτω μηδὲ πιέτω ἀπὸ τῆς εὐχαριστίας ὑμων, ἀλλ' οἱ βαπτισθεντες εἰς ὄνομα κυρίω·  καὶ γὰρ περὶ τούτου εἴρηκεν ὁ κύριος·  Μὴ δῶτε τὸ ἅγιον τοῖς κυσί.  [AF-T, V, did 9:5]
</code>
Jesus is the mediator between God and man (1 Tim. 2:5; Hebrews).  Point C, in my notes puts it well then: "Since there is no adoption to sonship by the Triune God except through the mediatorship of Christ: the sacrament may be briefly called a baptism in the name of Christ Jesus".
As a result "into Christ" and "into the name of Christ" are the same.
From the passages you list then, we see what baptism is and does.  Baptism is a good work that God performs in us, in which water is applied to the person being baptized in the name of the Triune God (Matt 28).  What the person receives through this is forgiveness (1 Peter 3; Acts 2; etc.) and freedom from death (Rom. 6).
Comment: Apologies, friends.  I am new to this board.  I thought the meaning of the phrase 'baptized into Christ" vs baptized into the name of Christ was a hermeneutic issue.  Again, apologies if this violated the intended course of discussion.  Feel free to close the board.
Comment: @theophilus There is no need to apologize at all.  Your question was just fine. My comments were for your sake, but directed towards others. You have done nothing out of place. It grieves me when I see newer people to SE who ask honest questions that have a right to be asked, and then have people vote to close down the question for no other reason than that they don't like the question. Ask away.
Answer: What escapes the notice of many people as they study the NT is that "Christ" is not identical to "Jesus." Just as "Jacob" referred to an individual man, it also came to be a synonym for the People, Israel. The NT authors all refer to "Christ" as a corporate Body, as well being individual members of this Body, those in the New Covenant are also Branches of God's True Vine, Subjects of the Son of David, God's Sheep under the care of God's Good Shepherd, etc.
I think we can say this:

baptizing "in the name of [the LORD] Jesus, the Christ" refers to the authority of the baptism: Jesus

"baptized into Christ" refers to what one is joining, and that is to become part of the New Covenant People of God: Christ corporate
Comment: It is your assertion that "into Christ" suggests corporate membership appears to ignore the fact that in each case, an explanatory clause follows which suggests otherwise.
Comment: I understand your suggestion.  However, have you seen Rom 6:3 and Gal 3:27?
Comment: Do you see "in the name of" in either passage? What is your concern?
Comment: Your first point is  somewhat correct because Christ took first shape in Jesus when he was 30 years old, after he received Holy Spirit anointing. Your second point that Christ is the body of believers in him is a little bit dodgy. Christ means Messiah, and only applies to him. Although believers in Christ are Christians. But that is different.
Answer: First, the only places in the NT that discuss being "baptized into Christ" are the two listed by the OP:

Rom 6:3 - Or are you unaware that as many as have been baptized into Christ Jesus have been baptized into His death?
Gal 3:27 - For as many of you as were baptized into Christ have put on Christ.

Thus, according to these two references, being "baptized into Christ" means two things:

being baptized into His death, ie, accepting Jesus death as our vicarious death, both for the expiation of sin and death to the old way of life.
putting on Christ, ie, living a life dedicated to imitating the life of Christ, ie, the new life in Christ.

By contrast, "baptized in/into the Name of Christ/Jesus/Lord" is much more frequent, eg, Matt 28:19, Acts 2:38, 8:12, 16, 10:48, 19:5, 22:16, etc.  That is, when a person is baptized into Christ, one takes the name of "Christian" and thus leads a life dedicated to imitating Christ.  The above references spell this out in more detail.  Being baptized into the Name of Christ means:

becoming a "Christian" in name
publicly showing one's repentance (Acts 2:38)
becoming a disciple/student/follower of Christ (Matt 28:19)
initiation into a life of Christian service empowered by the Holy Spirit (Acts 19:1-6)
starting a new life of forgiveness and one where one's sins are "washed away" (Acts 22;16)

In summary, I see little difference between the NT expressions of being "baptized into Christ" and being "baptized in the Name of Christ".
Answer: It may help to understand exactly what is meant by the term "eis to onoma" – "into the name of." In the classical style of the first century language, the phrase "eis to onoma" – "into the name of" was often used as a legal term. This expression is found among ancient legal documents that recorded the transfer of property. If one purchased a section of land or a dwelling for example, a title transfer would be drawn up to show that this property was now eis to onoma – in the name of - the new owner. When Jesus commanded his disciples to preach the gospel to every creature and "baptize them into the name of the Father, Son, and the Holy Spirit," what he is commanding them to do is to baptize them into the possession of God. Christians are the objects of a property transfer – "out of the kingdom of darkness and into his marvelous light." Baptism then is a property transfer. This same language is used in Acts 2:38 Peter commanded those present to be baptized "en to enomati – into the name of – thus, into the possession of Jesus Christ. This was for a two-fold purpose:

For the removal of sin – Spiritual circumcision, Colossians 2:9-13.
To receive the gift of the Holy Spirit. He is the seal of the transfer of property, Ephesians 4:30, 1 Corinthians 6:19-20. He is the seal of ownership to show that we have been bought with a price.
Answer: There is power in the name of Jesus. You must be baptised "in" the name of God - Jesus to be saved.
1 Corinthians 1:13-15

13 Is Christ divided? was Paul crucified for you? or were ye baptized in the name of Paul? 14 I thank God that I baptized none of you, but Crispus and Gaius; 15 Lest any should say that I had baptized in mine own name.

Acts 4:10-12

10 Be it known unto you all, and to all the people of Israel, that by the name of Jesus Christ of Nazareth, whom ye crucified, whom God raised from the dead, even by him doth this man stand here before you whole. 11 This is the stone which was set at nought of you builders, which is become the head of the corner. 12 Neither is there salvation in any other: for there is none other name under heaven given among men, whereby we must be saved.
Answer: Baptism in the name of Christ means being baptized confessing faith that Jesus is the Messiah. Baptism into Christ is baptism as the means, condition, instrument of incorporation into Christ. Matthew 28:19 is as BAGD says baptism into the possession of the Son.


OpenGL benchmark for Linux?

Question: I'm looking for an OpenGL benchmark tool for Linux. Something sort of similar to 3DMark on Windows. Is there anything at all? I tried Phoronix, was not impressed (nothing seemed to run at all).
Answer: You might want to try this benchmark. There are a few more on the site if this is not what you are looking for (downloadable by mirror torrent interestingly enough).

OpenGL benchmark for Linux?


Derivation of the Klein-Gordon equation

Question: My question arises from studying section 8.1.3 of Sakurai. I am confused on the way Klein–Gordon equation is created. In Sakurai, the book said that it was derived by taking another time derivative to the Schrodinger equation to obtain a covariant derivative form (as explained in equation 8.5 and 8.6 of Sakurai). This derivation looks to me as stating implicitly that Schrödinger equation works in the relativistic case. However, we knew that the Schrödinger equation is not compatible with special relativity. Even though Klein-Gordon did explain some experiments (as stated in 8.1.4), I don't understand how this derivation is legal, and I am wondering whether the strong limitation of Klein–Gordon equation is related to this ill-defined presumption.
Comment: While Greiner's series are wonderful and he definitely knows his stuff, the whole conception of RQM is indefensible. One cannot have RQM without being able to create and destroy particles (this is also true of statistical thermodynamics) and so only in QFT may we have a tolerable union of SR and QM.
Comment: To people with the same question, there is an alternative source that may be useful: Chapter 1 and 2 of "Relativistic Quantum Mechanics" by Greiner. I found it in this post: [link](https://physics.stackexchange.com/questions/212536/what-do-the-density-function-and-the-current-vector-in-the-continuity-eq-form-o)
Answer: No, the issue with the Klein–Gordon equation does not come from this derivation, and the Klein–Gordon equation actually predates the Schrödinger equation (to the best of my knowledge, Schrödinger got it first, but discarded it due to the issues it predicts).
The difficulty with the KGE has to do with it being relativistic. When you bring QM and special relativity together, you get to the possibility of creating and annihilating particles, but the KGE itself can't predict this. If you understand the wavefunction as we usually do in regular QM, it describes the probability associated with a single particle, and you end up with difficulties. The solution is to perform something called second quantization: we start treating the wavefunction itself as an operator that can create and annihilate particles. This is the basis of quantum field theory, which does employ the Klein–Gordon equation and obtains correct results. The Higgs boson, for example, is described by the Klein–Gordon equation.
Notice also that you can't really derive the Klein–Gordon equation, or the Schrödinger equation: you can only guess it and then test it experimentally. Sakurai's argument is indented to illuminate why one would expect the KGE to solve some issues with Schrödinger's equation, but it shouldn't be seen as a proof of the KGE. The proof of the pudding is in the eating, and the proof of a physical law is in an experiment. Schrödinger guessed his equation by drawing in inspiration from other bits of physics, but at the end of the day that is at most a clever guess. What's truly remarkable is that the equation actually is useful to describe the world.
Bonus: but how do QM and SR predict the creation and annihilation of particles?
I'll give a sketch argument, that is purposed to give intuition, but is also not quite a proof.
Suppose you put a particle of mass $m$ in a small box of size $L$, as we often do in QM. Now, since we know the particle is inside this box, Heisenberg's uncertainty principle tell us that
$$L \Delta p \geq \frac{\hbar}{2},$$
and hence the particle's momentum must be such that
$$\Delta p \geq \frac{\hbar}{2 L}.$$
Now, if we make $L$ small enough, it is reasonable to assume the particle is in the ultrarelativistic limit, and hence it has $E \approx pc$. We then find that the particle has
$$\Delta E \geq \frac{\hbar c}{2 L}.$$
Notice that the uncertainty in $E$ grows as $L$ diminishes. If we make the box small enough, we can get $\Delta \geq 2 mc^2$ when
$$L \leq \frac{\hbar}{4 mc} \sim \lambda_{\text{Compton}}.$$
Hence, when the box is smaller than the particle's Compton wavelength $\lambda_{\text{Compton}} = \frac{h}{mc}$, the energy uncertainty is about enough to create a particle-antiparticle pair, at which point it is no longer reasonable to assume we are dealing with a fixed number of particles. Hence, when dealing with relativistic quantum mechanics, we expect to need to deal with creation and annihilation of particles, and the Klein–Gordon equation is not capable of doing this without quantum field theory.
Comment: Thank you for your picture on particle and anti-particles. I do have a further question on this topic. Sakurai also introduced the idea of anti-particles by turning the KGE into two SE like equations. Is this the way we look for anti particles for spinless particles? If possible, could you please give a bit more intuition on why this does not work for particles with spins? Thank you so much!
Comment: @H-JJeffWang I'm not really familiar with this viewpoint. In field theory, while the KGE is there and it is important, the solution techniques are vastly different and we hardly ever deal directly with solutions to the KGE. I suggest posting this as a new question =)
Comment: @H-JJeffWang I skimmed over Sakurai's discussion and I'm not really familiar with this point of view (it does resemble a lot of what I've seen of relativistic quantum mechanics with the Dirac equation, but I'm not used to thinking like that in field theory). Nevertheless, there are analogous discussions for spin-1/2 fields, for example. I haven't checked, but I find it likely that Sakurai will do that in the section about the Dirac equation
Comment: Got it, thank you so much!
Answer: Firstly, in general the Klein-Gordon equation can't be derived from non-relativistic QM. If it could, it would not be relativistic, as you stated. It is never possible to derive a "bigger" theory from a "smaller" one, at least not without some further assumptions (for example it is possible to derive the laws of relativistic dynamics from classical mechanics if you further assume that special relativity is correct and especially the Lorentz transformations work the way they do).
Secondly, all this derivation assumes is "$E = i \hbar \partial_t$", it doesn't require Schrödinger's equation. For Schrödinger's equation, $\hat H$ has a certain form which is not assumed in Sakurai's derivation of the Klein-Gordon equation. In fact, he later uses $H^2 = p^2 + m^2$ which is not true in non-relativistic dynamics.
So, in other words, he says "$E^2 | \psi \rangle = - \partial_t^2 | \psi \rangle \Rightarrow (\square + m^2) | \psi \rangle = 0$, since $E^2 = p^2 + m^2 = - \nabla^2 + m^2$.
Comment: The wilder guess is that we took the square. The (at first glance) easier ansatz would be $i \partial_t | \psi \rangle = \sqrt{- \nabla^2 + m^2} | \psi \rangle$. But that equation has a lot of problems, for example it is not relativistically covariant (since the square root is defined by it's Taylor series and therefore the RHS contains second, third, forth etc. derivatives in x while the LHS only contains a first derivative in t). All those problems vanish if you "take the square" of that equation, however other problems arise, namely the probability density is no longer positive definite.
Comment: Thank you so much! Can I understand the motivation of this "derivation" as we made a wild guess that we can replace the Hamiltonian in the RHS of Schrodinger equation with barely the "energy operator" to fit it into SR?
Comment: No problem. Yes, you can understand the "derivation" like that. But also in non-relativistic QM the Hamilton operator normally is the "energy operator" (If the Lagrangian doesn't depend on the velocities, so in most practical cases without a magnetic field).

Derivation of the Klein-Gordon equation


CATiledLayer changing images with different sizes when zooming (Map related :)

Question: I have an UIScrollView with a UIView which in turn contains a CATiledLayer which displays a map.
I have to start with an overview where one can zoom in VERY deep. So I made 4 different images (actually PDFs) with 4 different sizes of the map (the largest has something like 5700 x 6700px)
I add the smallest (the overview) to the CATiledLayer and then swap the images on certain zoom levels.
So far, this kinda works.
The problem is, that when I display the largest map (closest zoom), the tiles get VERY big and it takes ages to draw them (and depending on the levels of detail/bias, the app produces a memory warning an crashes).
I tried to start with the largest map and set the zoom-level to the minimum when the map first shows, but then it takes ages to draw the overview because the tiles become very small (when I don't set it the minimum-zoom, everything works smooth).
I tried to adjust the tile size when I switch the maps - but apparently, this doesn't work.
Any idea how I can solve this? ;)
Thanks for any help...
Edit:
I'm trying now to define two separate CATiledLayers and then swap them when a certain zoom-level is reached - but that makes the app crash?
<code>[myContentView.layer replaceSublayer: tiledLayer with: tiledLayerBig];
</code>
oh my... :)
Edit2:
Ok - I got it working - more or less. The only thing that happens now is, that the CATiledLayer behaves strange when I add the largest image. It works fine for a while, but when a certain amount of tiles is drawn, I get a memory warning and the app crashes eventually.
I actually would expect that the invisible tiles would get disposed when they leave the viewport. This doesn't seem to happen, so eventually, all the tiles will be drawn which seems to be too much for the iPhone-memory... :(
Any ideas, how to solve this?
Answer: It sounds like the tiles that are no longer being used are not being released from memory.  What does your custom override's
<code>- (void)drawRect:(CGRect)rect method
</code>
look like?


Centos OS command to unlock domain account

Question: I have CentoOS servers that are on my MS domain.  I had another admin login to one with the wrong password 3 times.  We have unlocked him on the domain but he is still locked out on that server.  What is the command to unlock him on that server because he cannot log back into that particular server? Of course, no domain account shows under that directory only local accounts. This is entry under ssh of course if that changes anything.
Answer: The answer to my own question is
sudo pam_tally2 --file /var/log/tallylog --user [username]  --reset

Answer: Maybe you just need to set him a new password?
Otherwise this might work:
passwd -u username


Average value of creation and annihilation operators - are these two expressions the same?

Question: Suppose we want to study many-fermions quantum mechanics on a lattice, so we start with a finite-dimensional Hilbert space $\mathcal{H}$ and go to its Fock space $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $H$ is the Hamiltonian of the one-particle system in $\mathcal{H}$, let $d\Gamma(H)$ be the Hamiltonian on $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $A$ is an operator on $\mathcal{F}_{\text{fer}}(\mathcal{H})$, then the basic hypothesis of the theory is that the average value of $A$ is given by:
$$\langle A \rangle_{\beta,\mu} := \frac{1}{Z}_{\beta,\mu}\text{Tr} \bigg{(}e^{-\beta (d\Gamma(H)-\mu N)}A\bigg{)} \tag{1}\label{1}$$
where $\beta > 0$ is the inverse temperature, $\mu$ is the chemical potential, $N$ the number operator and $Z$ the partition function.
The above is saying that the equilibrium state is described by the operator $e^{-\beta(d\Gamma(H)-\mu N)}$.
Now, set $K:= d\Gamma(H)-\mu N$. On page 20 of this book, the following formula is displayed:
$$(-1)^{p}\langle \Omega, \mathbb{T}\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})\Omega \rangle = \frac{\int \prod_{l=1}^{p}\bar{\psi}_{x_{l},\sigma_{l}}e^{\mathcal{A}(\bar{\psi},\psi)}\prod_{x,\sigma}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}}{\int e^{\mathcal{A}(\bar{\psi},\psi)}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}} \tag{2}\label{2}$$
where $\mathbb{T}$ denotes the time-ordering operator and $\bar{\varphi}$ on the left hand side can be either a creation operator $\varphi^{\dagger}(x_{l},\sigma_{l})$ or annihilation operator $\varphi(x_{l},\sigma_{l})$, while the $\bar{\psi}$ stands for the associated Grassmann variable. From the text, $\Omega$ is a eigenstate of $K$, but it is not clear if it is the ground state of $H$ as well. Thus, the right hand side of (\ref{2}) is just a path integral representation.
I want to reconcile formulas (\ref{1}) and (\ref{2}). I recognize the right hand side of (\ref{2}) as the mean value of the product $\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})$ written in terms of a path integral. But then, the right hand side of (\ref{2}) is:
$$\langle \prod_{l=1}^{p}\bar{\varphi(x_{l},\sigma_{l})}\rangle_{\beta,\mu} \tag{3}\label{3}$$
How can (\ref{3}) be also the average with respect to the eigenstate $\Omega$? The expression on (\ref{3}) carries a sum over all basis states, how can it be that it is written just with respect to $\Omega$ as well? What is wrong with my reasoning?
Answer: You can reconcile the formulas by specifying the boundary conditions of your path integral. For (2), the time boundaries (at $0,\beta$) are assumed to be in state $|\Omega\rangle$ but to retrieve (1), you only assume periodic boundary solutions. Things usually get clearer once you've actually computed a concrete case (which is doable when your hamiltonian is quadratic since you only have gaussian integrals to calculate).
Check out QFT in a Nutshell by A. Zee for a good introduction (chapter V.2) on the matter.
Hope this helps.


How can I own 100% of a district?

Question: In Saints Row: The Third I own 64% of the 'Deckers' district,  but I don't know how to own the remaining 36%.
Answer: You gain percentage towards district control from a number of things:

Removing gang operations
Owning buildings
Owning shops
Completing side missions in that district

When you hover over a side mission, gang operation, or building/shop, it will indicate how much more control you'll have by completing that activity.
Some activities won't show up until you've finished certain plot missions or earlier levels of that same activity.  I think the cribs you unlock at certain points in the plot may also contribute to this percentage, so you'll only be able to get 100% after a certain point in the game.
Each of these also has a small radius that they turn "purple" on your phone's map, so you can tell if you're missing something in an area by checking to see what parts of a district aren't purple.  Gang operations don't show up on the map until you've been close to them, so if there's an area that isn't purple in a district and there's not an icon on your map inside of it, go drive around that area and look for a gang operation.
If you manage to get 100% control over a district, you'll get a call from the relevant homie for that district as congratulations, and enemy gang members will no longer spawn there.  Owning buildings and shops contributes to your earned income, which rolls over every 15 (real) minutes or so.
Comment: Quote:
Some activities won't show up until you've finished certain plot missions or earlier levels of that same activity. I think the cribs you unlock at certain points in the plot may also contribute to this percentage, so you'll only be able to get 100% after a certain point in the game.

Thank you, now I understand why not all of the area's turn purple.
Comment: Note also that the map screen has counters for all of these things, so you can see if the things you're missing are gang operations, cribs, etc.
Comment: Also, as an addendum, some of the gang territories are not at ground level, so you may need some for of flight to get to them!
Answer: You have to do all the side missions, gang ops, own all the shops and buildings, and you have to have done all the activities in the area. All that will make you own all of that territory.
NOTE: The hardest territory to get is the territory that belongs to the morning star which is the biggest territory and most of the activities there are rated hard. You also have to look out for when you are taking over the luchadores territory because they have some of the most advanced weaponry and they have brutes.  When you have taken over a territory at 100 percent you will get a phone call and no enemy gang members will spawn and when you have taken over all of the territories you will have no enemy gangs!
Answer: The island cannot be fully 100% after a certain point because of the zombies so get Arapice Island 100% cleared first.
Comment: OP is asking about districts, not the entire island. Thus, this doesn't really answer the question.
Comment: @Vemonus; Arapice Island is the small industrial island that makes up a part of the Deckers' district. It's where the plane crashes and creates the zombies

How can I own 100% of a district?


Get all authors with at least one post of 'custom post type'

Question: How can i achive that?
I found count_user_posts and get_users, but how would I merge them into the desired result? Do I really have to use a WP DB Query?
Answer: For now post_type support is not currently available for count_user_posts(),
so put the below code in theme's functions.php file.
function count_user_posts_by_type( $userid, $post_type = 'post' ) {
    global $wpdb;
    $where = get_posts_by_author_sql( $post_type, true, $userid );
    $count = $wpdb->get_var( "SELECT COUNT(*) FROM $wpdb->posts $where" );
    return apply_filters( 'get_usernumposts', $count, $userid );
}

Now this function would take user id and post_type as parameter post_type='post' being the default and gets the count of post for that user for the concerned post_type
You can verify it by echoing the below where you need to check the post count by the user.
$user_posts = count_user_posts_by_type( 1, 'page' );
echo $user_posts; // This would return the post count for user id 1 and post_type page.

You can do the same for the custom post type. Now you need to get the user ids and pass through this function and display the name of the authors having 1 or more posts
or
You can add the below code where you want to display the list of users with one or more post
$blogusers = get_users();
foreach ( $blogusers as $user ) {
    if ( count_user_posts_by_type( $user->ID, 'post') ) {
         echo $user->user_nicename . '
';
    }
}

The above code lists all the users with one or more posts published of post_type post.
Answer: Very old query but still there are no direct methods to get all authors with at least one post of 'custom post type'. WP_User_Query also does not support the post type. Using $wpdb will solve the issue.
Code:
$post_type = 'gutentor';/*Your Post type*/
global $wpdb;

$all_authors = $wpdb->get_results(
    "
    select
A.*, COUNT(*) as post_count
from
$wpdb->users A
inner join $wpdb->posts B
on A.ID = B.post_author
WHERE ( ( B.post_type = '$post_type' AND ( B.post_status = 'publish' OR B.post_status = 'private' ) ) )
GROUP BY A.ID
ORDER BY post_count DESC"
);

if ( $all_authors ) {
    foreach ( $all_authors as $author) {
        echo $author->display_name;
    }
}

You just need to assign $post_type with your post type.


Testing software: fake vs stub

Question: There are quite a few written about stub vs mocks, but I can't see the real difference between fake and stub. Can anyone put some light on it?
Comment: possible duplicate of [What's the difference between faking, mocking, and stubbing?](http://stackoverflow.com/questions/346372/whats-the-difference-between-faking-mocking-and-stubbing)
Answer: I assume you are referring to the terminology as introduced by Meszaros. Martin Fowler does also mentions them regularly. I think he explains the difference pretty well in that article.
Nevertheless, I'll try again in my own words :)
A Fake is closer to a real-world implementation than a stub. Stubs contain basically hard-coded responses to an expected request; they are commonly used in unit tests, but they are incapable of handling input other than what was pre-programmed.
Fakes have a more real implementation, like some kind of state that may be kept for example. They can be useful for system tests as well as for unit testing purposes, but they aren't intended for production use because of some limitation or quality requirement.
Answer: A fake has the same behavior as the thing that it replaces.
A stub has a "fixed" set of "canned" responses that are specific to your test(s).
A mock has a set of expectations about calls that are made. If these expectations are not met, the test fails.
All of these are similar in that they replace production collaborators that code under test uses.
Comment: the term "keep it simple" does not apply here. I mean, if you can't find a simple, concrete example to explain differences, then it's just phrasing. IMO All these definitions are the same unless we can demonstrate it it with a simple example. I mean, far from help people, this confuses people a lot.
Answer: To paraphrase Roy Osherove in his book The Art of Unit Testing (second edition):
A Fake is any object made to imitate another object.  Fakes can be used either as stubs or mocks.
A Stub is a fake that is provided to the class you are testing to satisfy its requirements, but is otherwise ignored in the unit test.
A Mock is a fake that is provided to the class you are testing, and will be inspected as part of the unit test to verify functionality.
For example, the MyClass class you are testing may utilize both a local logger and a third-party web service as part of its operation.  You would create a FakeLogger and a FakeWebService, but how they are used determines whether they are stubs or mocks.
The FakeLogger might be used as a stub: it is provided to MyClass and pretends to be a logger, but actually ignores all input and is otherwise just there to get MyClass to operate normally.  You don't actually check FakeLogger in your unit tests, and as far as you're concerned it's there to make the compiler shut up.
The FakeWebService might be used as a mock: you provide it to MyClass, and in one of your units tests you call MyClass.Foo() which is supposed to call the third party web service.  To verify that this happened, you now check your FakeWebService to see if it recorded the call that it was supposed to receive.
Note that either of these could be reversed and depend on what it is you're testing in a particular unit test. If your unit test is testing the content of what is being logged then you could make a FakeLogger that dutifully records everything it's told so you can interrogate it during the unit test; this is now a mock.  In the same test you might not care about when the third-party web service is called; your FakeWebService is now a stub.  How you fill in the functions of your fake thus depends on whether it needs to be used as a stub or a mock or both.
In summary (direct quote from the book):

A fake is a generic term that can be used to describe either a stub or a mock object because they both look like the real object. . . . The basic difference is that stubs can't fail tests.  Mocks can.

All the rest is implementation details.
Comment: While I personally think this definition of stub makes more sense, it seems to me that isn't consistent with the definition used by many others (which maybe all come from Martin Fowler). =/
Answer: These might help

http://xunitpatterns.com/Mocks,%20Fakes,%20Stubs%20and%20Dummies.html
http://hamletdarcy.blogspot.com/2007/10/mocks-and-stubs-arent-spies.html
Comment: Sorry to nitpick, but it's best practice to summarize relevant parts of links in case they go dead.  An answer should be more than links.
Comment: That first link shows just how complex this topic is, and how no two people seem to share the same definitions.  As an example the answer to the StackOverflow question [What's the difference between faking, mocking, and stubbing?](http://stackoverflow.com/questions/346372/whats-the-difference-between-faking-mocking-and-stubbing) seems to reverse the definitions of Stub and Fake compared to Thorarin's answer to this question.

Testing software: fake vs stub


Does entropy change with Reynolds number?

Question: It seems like, as a flow becomes more turbulent, there should be more possible states in the fluid. Shouldn't that correspond to an increase in entropy? Steam tables seem to only depend on temperature and pressure and not Reynolds number.
Comment: "Steam tables seem to only depend on temperature and pressure and not Reynolds number".  That statement is true, but I must ask what you would want steam tables to show.
Comment: The rate of entropy generation is higher in turbulent flow because viscous dissipation is more intense in turbulent flow.
Comment: @DavidWhite I can imagine flows with the same temperture and pressure but different Reynolds numbers. It confused me that the specific entropy doesn't change with Reynolds number according to steam tables. I thought the more turbulent flow should have a higher specific entropy, which seems to contradict the information in steam tables.
Comment: I've looked this up but I can't find a known relationship between entropy and Reynolds number
Comment: @kilojoules, steam tables are normally used to evaluate the enthalpy of steam entering and exiting a turbine or heat exchanger, in order to do an energy balance.  Reynolds number is not related to energy balances, so this redundant information will not be found in a steam table.
Answer: In Thermodynamics, entropy is introduced as an equilibrium  state function, i.e. as a quantity depending on the thermodynamic state only. The basic request is that a macroscopic system, whose microscopic state depends on a huge number of degrees of freedom, can be fully characterized by only a few physical quantities like temperature and pressure. Such a huge reduction of the number of effective parameters required to describe the system can only be achieved under special conditions: the presence of the so-called thermodynamic equilibrium states.
At thermodynamic equilibrium, no macroscopic motion is present. Therefore, no velocity field, no Navier-Stokes equation and no Reynolds number. This is the reason one cannot find any reference to Reynolds number in steam tables.
In many cases, it is possible to introduce a local thermodynamic description of a macroscopically moving fluid, by introducing a field of thermodynamic variables describing the thermodynamic behavior of a fluid element small enough to be uniform, but large enough to contain still a large number of molecules. Within such local thermodynamic equilibrium (LTE) approach, turbulent fluxes can be  characterized by a local entropy production term.
What about turbulent flux and the increase of the number of possible states? This has to do with a statistical mechanics description. However  most of what we learn  is connected with statistical mechanics at equilibrium. Only at thermodynamic equilibrium, we can interpret entropy as a measurement of the number of microstates compatible with a fixed volume and energy. However, all the possible mechanical states of the same energy are counted  and no dependence on the kind of flux is allowed.


iCloud and iPhone Calendar have different timezones

Question: The updates for my events in iCloud and iPhone differ by a day due to the difference in timezone. My iPhone's timezone is set to Singapore time. How do I set my iCloud timezone to Singapore time as well? iCloud only has a few timezones for us to choose from.
Answer: Even though the timezone was set correctly on a Mac, an iPhone and in iCloud settings, syncing across devices has always been off by a couple of hours for me.
Turns out, the time zone, though specified, was disabled for Calendar.
To fix this, click the gear icon in the upper right corner and choose Preferences:

Go to Advanced and make sure the first checkbox is ticked.

Click Save and enjoy your timezone.
Answer: When you log in to iCloud you should be able to select Singapore's timezone.

Click your name in top right corner
Click Time Zone to open Time Zone modal window
Try to aim for Singapore as showed in the pic below. It doesn't have to be precise. It's good enough to hit timezone and any city in it.
If you didn't manage to click Singapore precisely and other city got selected just click the name of the city that appears on the button below the map and select Singapore from the dropdown.

Good luck as it's not that easy to click your timezone! Using the Zoom feature (Preferences -> Accessiblity) might help.


Difference between Multistage Sampling and Stratified Random Sampling?

Question: I know the question is a very elementary one, but I simply cannot understand the difference other than the fact that an SRS is a form of Multi-Stage Sampling. Can anyone provide a simple example(s) to help me understand the critical difference between these two sampling designs?
Answer: Multi-Stage Sampling:
Population: USA elementary school students
First stage sampling: 10 States from total of 50 States.
Second Stage: 20 Counties from total XX counties in selected XXXXX state in the first stage.
Third stage: 50 elementary schools from total yy elementary schools in the selected ZZZ county in the second stage
Fourth stage: 10 students from each selected school.
Finally, I got 10 * 50 * 20 * 10 = 100,000 students.
Stratified sampling:
Population: USA elementary school students
Geographically stratify USA into ten regions
<code>Region I: Connecticut, Maine, Massachusetts, New Hampshire, Rhode Island, Vermont
Region II: New Jersey, New York, Puerto Rico, US Virgin Islands
Region III: Delaware, District of Columbia, Maryland, Pennsylvania, Virginia, West Virginia
Region IV: Alabama, Florida, Georgia, Kentucky, Mississippi, North Carolina, South Carolina, Tennessee
Region V: Illinois, Indiana, Michigan, Minnesota, Ohio, Wisconsin
Region VI: Arkansas, Louisiana, New Mexico, Oklahoma, Texas
Region VII: Iowa, Kansas, Missouri, Nebraska
Region VIII: Colorado, Montana, North Dakota, South Dakota, Utah, Wyoming
Region IX: Arizona, California, Hawaii, Nevada, American Samoa, Guam, Northern Mariana Islands
Region X: Alaska, Idaho, Oregon, Washington
</code>
Randomly select 10,000 students from elementary schools from each region. I got a sample with 100,000 students.


How can someone hack our bank account?

Question: Can someone hack and withdraw money from my bank account if they have my full name, bank name, last four digit bank account number and my phone number. Yes or No, Please explain why?
Comment: "hack" is a rather meaningless word. There are a number of things that only you should be able to do with your bank account, such as view the balance, spend the money, and terminate the account. It should be obvious that these are not equal. Even so, no bank should allow anyone to even view your balance even if they had your **full** account number - your employer might very well have that.
Comment: Your bank account isn't going to be hacked with only the last 4 digits of your account.  However, if the bank's records are successfully hacked into then everyone's  accounts will be at risk.
Comment: Account numbers are *not* private information; they are printed on every check you write. (They may be considered personal *protected* information depending on jurisdiction, but there is no assumption that you and only you know what your account number is for authentication purposes.)
Answer:
My full name on bank, my last four digit account number and they also know my phone number that link to my bank account. So, Will they able to hacked my account through this information.

Maybe. It's hard to say how secure your bank is, but this is the sort of information that might be enough to take over your bank account, depending on the bank. I wouldn't risk it, personally. It's not clear to me why you want to do this, but one possible reason is that a stranger has asked you to send them money, and now they are asking to see the transaction.
If that's the case, then you may already be the victim of a scam. There are numerous scams which involve unknown people contacting victims and asking them to send money through all sorts of different channels. I'm not specifically familiar with scams involving Google Pay, but it would not surprise me if this were the case. In general, unsolicited requests for money are a red flag regardless of how the money is transmitted and regardless of who or what the scammer claims to be or represent.
On the other hand, if you initiated this transaction, in exchange for some real good or service, it may be legitimate, depending on the circumstances. You should still make sure that the agreement is in writing, that this person is an identifiable individual or works for an identifiable company, and that you have the practical ability to initiate a chargeback or pursue some other form of redress in the event that they fail to produce the good or service that you purchased. Unless, of course, you've already received the good or service and just need to pay for it.
As for your immediate problem, I would ask the recipient whether they received a receipt from Google (they should have). If they did not, then you should inform Google that you sent the money to the wrong person, and ask them to reverse the transaction. However, I would suggest telling the recipient that you are going to do this. If they suddenly change their mind and assert that they did receive the cash, that's another red flag.
If you believe that you have been scammed, you should cease all contact with the recipient, notify Google, and hopefully they can pull your funds back. You should do this as soon as is practical, if indeed it is a scam.
Answer: No. If you are in the US, then that information is printed on every check you write.
Answer: "Hacking" these days means either:

They guessed your password or tricked you into telling them
They tricked the bank into thinking the hacker was you (e.g. "forgot password)
They got into the bank's database and stole a large number of account info, and yours happened to be one of them

They can do these anyway, and knowing your full name or a few digits of the account number does not make it much easier or harder. As an exercise, consider:

Everyone knows Jeff Bezos's full name
Everyone knows he has money
There are only so many banks (and he probably has accounts with most of them)
There are only 10,000 possible last four numbers so it's not that hard to guess, especially if many people are trying

So if these thing allowed you to hack someone's bank account, how come the evil hackers didn't hack Bezos?
Answer: I'm not going to claim that there aren't places in the world where this information may be sufficient.
However, here in the SEPA area (≈ Euro zone), all this information including the full account number (which actually includes bank and branch as well) are usually handed out on pretty much every business letter.

Full account number incl. bank and branch is required to pay money into this account. Account owner is not really checked any more since the account numbers contain check digits.
Depending on the legal form, a business may have to state full name of owner in their official letters
Their web page (here in Germany, this varies substantially across countries) has to state a responsible person and a fast way of reaching this person. That is often the same business owner.
For obvious reasons, businesses want to spread their phone number.
A small business opening a bank account may find that the bank wants to use the phone number that said business uses for official communication (as part of their IDing the account holder [which is part of anti-money-laundering requirements]).

In order to even only see the account, other authentication factors are used. Nowadays typically 2 other factors such as a password and a TAN/OTP.

How can someone hack our bank account?


Sources of cardano rest api

Question: While testing some  tangocrypto api , it is asking for "collection_id" which I don't know what is it.
So are there any Rest API providers other than (tangocrypto.com, blockFrost.com and opencnft.io)?
Answer: Firstly, here's the documentation for the Tangocrypto API:
https://www.tangocrypto.com/api-reference/#/operations/nft-collection
The collection_id here is just the ID of an NFT Collection that you can create on the platform. (You receive it in the response when you create one via the "Create NFT collection" endpoint or list your collections via "List NFT collections".)
If you're looking to find assets on the blockchain, e.g. via policy id you can look at this endpoint:
https://www.tangocrypto.com/api-reference/#/operations/get-asset-by-policy
Secondly, here's another alternative you could look at, if needed: https://dandelion.link/
Comment: but , How can I fetch all data of NFT's like buy Price , sold price . rarity, meta data through wallet address in one api ?
Comment: The metadata is there, but the others are marketplace specific. Haven't used it but looks like the one you mentioned, opencnft.io, has aggregated lots of data, eg floor prices, traits for rarity etc. You can try it out in their docs: https://api.opencnft.io/1/#/Policy/PublicController_getPolicyFloorPrice. To get them "by address" you'd have to use any of the other APIs to fetch assets sitting at it and then get the NFT data for each.
Comment: I am doing the same process as you mentioned above. I want to fetch NFT data through users wallet address, so for that I am using (cardano-mainnet.tangocrypto.com /{app_id}/v1/addresses/{address}/assets) and this gives me policy_id and asset name which I am using in (api.opencnft.io/1/asset{asset}, where asset is "policy_id + asset name") which gives me some data i want and another api (api.opencnft.io/1/asset{asset}/tx) also gives some useful data. Problem is that all these Processes are taking time
Comment: Ok, I don't know of any API that can get you the data faster unfortunately. The fastest way is probably to run your own node, query the chain locally und monitor the marketplace contracts yourself. I don't have experience setting that up though.

Sources of cardano rest api


What are the top schools for online part-time Masters in data science/analytics programs? Is it much better to do the program in person?

Question: I am looking to get my Masters in Science in Data Science or Analytics. I will continue working during the program so it needs to be part-time. I would prefer to do an online course, but if in-person is much better, I could be open to it. So far, I found Georgia Tech and Johns Hopkins for online, part-time programs. For in-person, I could attend Georgetown part-time. Any insight you can provide regarding which of these programs would be best, if there is another school that could be better for my situation, or if any school is essentially the same would be very helpful! Thanks!
Answer: Michigan University has a nice and good dual program that you get your Master in Data Science and also a MBA.
I'd take a look at MIT's Master of Business Analytics as well.
If you want to get a job, I don't think a Master is necessary. This Forbes article explains that pretty well https://www.forbes.com/sites/metabrown/2016/07/29/4-reasons-not-to-get-that-masters-in-data-science/#160f195a40c0


How do you have a parented object stay in one place while the parent moves

Question: I am trying to make it look like a sword is being pushed into the ground but i don't know how to keep the sword in the ground while the parent moves.
Answer: i would use a "child of" constraint for that.

You can keyframe the influence (should be 1 until the sword is pushed in the ground) and 0 one frame after that.
you can keyframe the objects position + rotation at the last frame with influence 1 and at the frame + 1 after that.
result:
Answer: Is parenting it obligatory?
There's an alternate option of parenting which is the constraint: child of. You can turn off location to make it so you can move the parent without moving the child, or use keyframes to turn off it's infuence.
You can also use the limit location constraint, by using the minimum and maximum coordinates as the coordinate of where you want it  to be, to lock it into place regardless of the parent.


How can I Copy and Paste the form to another object?

Question:
Hello, I´m new to blender and want to find out if I can copy the form of the good eye and paste it on to the bad eye.
Comment: Not sure what you mean, but you can use SHIFT-D to duplicate the good eye, then delete the bad eye. Or simply add a UV Sphere as a whole new eye, eyes are (almost) perfect spheres.
Answer: If at any moment you want to change the mesh of an object for another one (here for example you want to switch from the cube to the cone), select the object, go in Object Data and in the top dropdown menu (Browse Mesh Data) select the mesh you want:

Or select the object, shift select the one that you want to copy the mesh from, and press CtrlL > Link Object Data.


Always blowing breaker

Question: We live in a 37 yr. old house that is total electric. When I try using a electric grill and a 1000 watt microwave at the same time it blows the breaker. How can we fix this problem?We live in Missouri if it matters
Comment: Which breaker ?
Answer: The reason you're tripping the breaker is that the peak current draw with both items running exceeds the safe level for that wiring.
The answer is to plug one of them into an outlet that's connected to a different breaker.  (or use them sequentially :-( ).
The answer is most specifically NOT to replace the breaker with a 20 or 30-amp unit unless you've had an electrician verify the existing wiring is rated for the larger current load.
Answer: You are overloading the circuit, plain and simple.  Don't do that.
Modern electrical code requires that kitchen outlets be served by two circuits.  Shut off that breaker, and plug a night-light or outlet tester into every outlet in your kitchen. If any outlets still have power, that means you have the modern standard of more than one circuit in your kitchen.  That's good.  Take the time to identify which are on which circuit (turn off one circuit's breaker, and mark all the outlets now dead).  Armed with that knowledge, you can take care to plug each of those two appliances into a different circuit.
People didn't always own the cavalcade of countertop kitchen gadgets we now do (George Foreman grills, juicers, wafflers, etc. etc.)
If you are stuck with one circuit, either use heavy appliances one at a time, or install additional circuits into the kitchen.  The latter can either be fairly easy or fairly hard depending on the physical construction of the home and space available in the service panel.

Always blowing breaker


Frozen poses after rendering animation

Question: I have rigged character and I want to make an animation. I set few kayframes with different poses. It shows up correctly on the timeline. And when I move the timeline character moves the way I desired. And here is the problem. When I render the animation [button render animation], all I get is frozen (almost) character without any movement. Almost, because bones with wiggle bone addon move on their own, the same with particle hair. To summarize, after rendering is done I have frozen character with jiggling body and dynamic hair - the movement of the bones I set on the timeline is non-existant.
I use blender 2.83 and eevee. The problem appears both when I use png format as well as any movie format. To answer questions, I don't think it's the problem with output directory, because blender renders animation - even with frozen bone movement. Also, I set start and end frame correctly - be it 10, 15, 20 frames. Kayframes are visible on the timeline - made with both auto keying and insert kayframe: LocRot. Rigged characters move when I move the timeline, it's only problem when I try to render it as animation. When I render it's as animation all I get is bunch of frames with characters frozen in one pose, but when I render random frame as image - I get correct pose that matches with the timeline.
Could it be possible that I have bugged pose mode? I made a test - I added sphere to the scene and in object mode I set movement of it (up and down). And it rendered it as animation correctly. Sphere moved up and down. It seems that animation works with everything (moving objects, simulating dynamic hair) - except the bones.
Answer: Ok. I found the reason of my problem. It was wiggle bone addon. It seems that you have to bake animation of wiggle bones and then turn it off, otherwise you will get frozen animation of any other bones.

Frozen poses after rendering animation


eth_call to call a function in my solidity contract fails with "execution reverted"

Question: UPDATE: I figured it out though is doesn't entirely make sense.
When abi-encoding an "address" as an argument, you use "address" in the hash of the function signature, however you abi-encode the actual argument as if it were uint256, padding the address out with 0's. Or at least you need to do so when using curl POST to an RPC client (geth in my case)
It's odd because an address is really uint160 (and i've seen other libraries treat it as such) but that did't work for me...
If this makes no sense, there's more details below...
I have a deployed contact, built from solidity and deployed. For reasons too detailed to go into, I am using curl POST to call functions within the contract via an instance of geth.
This means abi-encoding the function call. I am able to do this successfully for several functions in the contract, but when there is an address as an argument, it fails immediately with a "execution reverted".
I've tried lots of things, but always fails. Here's what I thought should work.
the function in solidity is:
function mintWithTokenID(address to, uint256 theTokenID) public virtual { ...

which is compiled into
{
      "inputs": [
        {
          "internalType": "address",
          "name": "to",
          "type": "address"
        },
        {
          "internalType": "uint256",
          "name": "theTokenID",
          "type": "uint256"
        }
      ],
      "name": "mintWithTokenID",
      "outputs": [],
      "stateMutability": "nonpayable",
      "type": "function"
    },

So here's what I thought would work, but doesn't...
I want to do this (note that actual addresses are fake here...)
mintWithTokenID('0x7b59C1447800000000c4D059ef3AD600000194F',7)
Running the function from truffle works fine, but I need to do the curl POST method.
So... here's the abi encoding:
web3.utils.sha3("mintWithTokenID(address,uint256)")
0xb9ae3621e688e2432b598dd4b1f33a482b375a14224e2ec6aa300bd16598e84a
so we need 0xb9ae3621 as the start, then the arguments are
7b59C1447800000000c4D059ef3AD600000194F
and
0000000000000000000000000000000000000000000000000000000000000007
resulting in complete abi data:
0xb9ae36217E98D8b036B6A2Ef80770c87145Bd1609D1a25d70000000000000000000000000000000000000000000000000000000000000007
and my curl statement is:
curl localhost:8545 -X POST --data '{"jsonrpc":"2.0","method":"eth_call","params":[{"from": "0x7b59C1447800000000c4D059ef3AD600000194F", "to": "0x7b500000008E6242a525c4D059ef3AD6511bd194F", "data":"0xb9ae36217E98D8b036B6A2Ef80770c87145Bd1609D1a25d70000000000000000000000000000000000000000000000000000000000000007"},"latest"],"id":1} -H "Content-Type: application/json"
but when run, it immediately returns:
{"jsonrpc":"2.0","id":1,"error":{"code":-32000,"message":"execution reverted"}}
Any ideas?   I've tried using uint160 instead of address, but documentation says address should be used in the function signature.
Thank you in advance!
Answer: How parameters in a call are encoded is defined in the Contract ABI Specification.
All fixed size types are first padded to 32 bytes. In you case you have:

address: be6fee3756f7be3a0cd492059341cb5b77dd81f9 => 000000000000000000000000be6fee3756f7be3a0cd492059341cb5b77dd81f9
unit256: 7 => 0000000000000000000000000000000000000000000000000000000000000007

If the call selector is 0xb9ae3621, then the call will be:

0xb9ae3621000000000000000000000000be6fee3756f7be3a0cd492059341cb5b77dd81f90000000000000000000000000000000000000000000000000000000000000007

One of the reasons to pad everything to 32 bytes is because the decoding code is much simpler and shorter. Decoding is done by the contract so shorter means less gas and cheaper transaction.


Git: fast diff viewing before commit

Question: Before I commit I run git status and get output like this:
#   modified:   TemplateDemo/src/com/example/templatedemo/Item.java
#   modified:   TemplateLib/res/layout/layout_list.xml
#   modified:   TemplateLib/res/layout/list_header.xml
#   modified:   TemplateLib/res/values/strings.xml
#   modified:   TemplateLib/src/com/saturn/templatelibrary/ISwappableAdapter.java
#   modified:   TemplateLib/src/com/saturn/templatelibrary/ItemAnimator.java
#   modified:   TemplateLib/src/com/saturn/templatelibrary/SwappableAdapter.java
#   modified:   TemplateLib/src/com/saturn/templatelibrary/TemplateListActivity.java
#   modified:   TemplateLib/src/com/saturn/templatelibrary/TemplateListFragment.java

Now I want to view diff for each file. Now I do it like this: I manually select each filename from the command this output and run meld <filename> for each file in the list. Is there a faster way?
Answer: Just running git diff should do exactly what you want.
If you want a more graphical view use git difftool.
Answer: There are many third party programs that make Git much simpler to use, and also show you diffs of each file before you stage them, and even allow you to stage chucks at a time.
SourceTree is my GoTo program for Git.
Answer: Git gui is my weapon of choice for this. You can easily scan through each file and see what's removed, and what is added.
Answer: I manually select each filename... Is there a faster way?

Yes. If you prefer meld for viewing diffs you can run meld . from the same directory where you run git status. Then you can select any filename in meld GUI to view diff or view diffs for all files at once with every diff in separate tab.

Git: fast diff viewing before commit


Prove that $\pi$ has $2$ digits that repeat infinite number of times

Question: How can I prove that $\pi$ has $2$ digits repeating an infinite number of times in its decimal expansion? Proving that $1$ digit repeats an infinite number of times is easy – $\pi$ is irrational, and by the definition of irrational number, a digit must be repeated an infinite number of times?
Answer: HINT: Assume only one digit appears infinitely many times, then after finitely many decimal places we will have only the repeating digit appearing. This means that $\pi$ is periodic (has a repeating decimal pattern), hence it's rational, which is a contradiction.
Answer: Well, if only one digit appears an infinite number of times, then $\pi$ is rational.

Suppose that only $d$ appears an infinite number of times.
Split $\pi$ into the following parts:

$A=$ the integer part, i.e., $\lfloor{\pi}\rfloor$
$B=$ the fraction part's until only $d$ appears

Let $N$ denote the number of decimal digits in $B$, then:
$$\pi=\frac{9A\cdot10^{N}+9B+d}{9\cdot10^{N}}$$
Hence $\pi$ is rational.

For example, suppose $\pi=3.14159\overline{2}$, with $d=2$:

$A=3$
$B=14159$
$N=5$

Then:
$$\pi=\frac{9\cdot3\cdot10^{5}+9\cdot14159+2}{9\cdot10^{5}}=\frac{2827433}{900000}$$
Hence $\pi$ is rational.
Comment: S. Peter: Please see a more detailed explanation + an example in the updated answer.


add pagination control for asp.net mvc

Question: I am new to asp.net MVC. I am trying to understand how to add pagination for a list of items using MVC without code for a view. I am using nunit testing to check whether pager option is tested or not
Answer: There is no built-in pagination for a vanilla MVC project -- you'll need to write one yourself or find something off the shelf.
See PagedList, which is also available on NuGet.

add pagination control for asp.net mvc


How do I cover up the white spots on my cast iron stove?

Question: My friends brought me a paper bag full of peaches. I set the bag on my fisher cast iron stove, and I forgot about them. Lastnight, I picked up the bag to find the peach juice has bled through the paper bag and ate the black finish on the stove. Now i have two large white spots on the stove. I have tried to rub olive oil on the spots. They are still there. How do i get the black finish back on the stove? Thank you! -Annie
Comment: @WayfaringStranger iron is actually metallic gray/silver. The black is the cooked on oil (seasoning) in the case of a pan. In this case, probably stove paint.
Comment: Have you tried firing up the stove? That should burn off any organic matter left by the peaches and olive oil, and iron is naturally blackish.
Comment: As far as I can tell from research (don't personally have any experience with a cast iron stove) the finish is [stove paint](http://www.amazon.com/dp/B000Y4MSPC/) [link for example only, I can't vouch for that product]. If the finish is actually damaged, you'd repaint it to repair—the folks at [diy.se] can probably offer better help than us on how to prep metal for painting... (If you'd like your question moved, flag it and pick moderator attention, and explain you'd like it moved in the text box.)
Comment: (There appears to also be something called "stove black" which is another possible finish. Supposed to have a dull sheen instead of matte; probably similar to an iron pan...)
Comment: If you can't find stove paint, it's possible that radiator paint would work, if it's rated for steam radiators (and not just water radiators, which can't get above 100°C)
Answer: Either something has accumulated on the stove or something was taken off the stove by the acids and sugars in the peaches. If you can't clean the spots, then you're looking at the latter situation.
Your stove has some type of finish on it. As Fisher stoves seem to be out of production, official information seems to be sparse. You could apply some type of high-temperature paint, of which there are many different color depths and finishes. Most products I found in 5 minutes of googling seemed to recommended doing the whole stove at once. This could be good for sprucing up the whole stove or could be more work, time, waiting and fumes than you're interested in.
You could also go for a polish or blackening agent, which works more like seasoning a cast iron pan – an organic, lipid wax or oil is applied gradually in layers and the heating and cooling of the stove transforms it into a polymer, which gives it that dully shiny hardened finish. If I had to guess, I would guess that an old school stove is more likely to have a seasoning based finish than a pain.
There are a lot of websites dedicated to wood burning stoves, such as Hearth.com and woodheat.org. With their help and a little research, you could probably identify the exact model of Fisher you have and the year or era it was made; or maybe someone can find a manufacturer recommended product directly.


Application to save copies of reduced resolution photos on iPhone?

Question: I am looking for an application that will save a copy of a photo on my camera at a smaller resolution. Can't find anything...thanks!
Comment: So you basically want all your photos to be resized and old versions removed?
Comment: Old versions of photos are acceptable to keep. Looking for something that will take the huge file size that Camera.app produces into something more manageable to upload. My end product here is a form that technicians will upload ~15 images to from the field. Obviously, you can't have 1.5-2.0mb images like that.
Comment: Take a look at my answer below. I'm certain that it is exactly what you are looking for, only... **FREE**. That might change your mind.
Comment: I'm having trouble understanding whether you are looking for an iOS *camera* app, or an iOS image manipulation app (that reduces file size smoothly and easily)?
Answer: Try this: Reduce - Batch Resize Images and Photos for iPhone & iPad.
However, if you just want to reduce the photo uploading traffic, the iOS built-in Photo.app allows you to share the photos with size-reduced version.
Answer: I have used "MultiLoader" to do something similar (ie, upload 15-20 pictures for a document); but, it is only effective if you have a WiFi connection.  It creates a mini web server on the iPhone, zips all the pictures together at full or 1/2 or 1/4 resolution, and lets you connect to download them.
https://itunes.apple.com/us/app/wifi-photo-multiloader/id448314967
Answer: I have used ImageOtim with some success.  Having said that it does not resize or reduce the resolution, it just compresses the images better, which might be enough for you or not -- in general I find that having better compressed larger images is easier for me than having not compressed small low res images.
Answer: I have used an app called Resize Photo from the App Store.
It's Free, and that made up all the difference for me!
When using it, you may even save in batch with all the photos the same resolution. This was a phone-saver when I found myself with over 600 photos and only 16 GB.
Hope this helps.
Note: I am not in any way affiliated with the developer of this application, and am only stating my personal favorite application for resizing photos.
Answer: Not free - Photoshop
you can do all of this quickly by doing it once while recording your actions, and exporting your actions as a droplet. A Droplet is essentially a self contained program that you drag image(s) to and it runs on them all.
Actions I'd suggest: Resize, Save For Web, Save As, Close Window, Do Not Save.
I've done this in the past for hundreds of images at a time when converting them for web display- essentially you're doing the same thing, but with some personalized tweaks for your phone.
If you have Photoshop, give it a try... if not, don't buy it just for this one feature!


What's the difference between traveling salesman and chinese traveling?

Question: What's the difference between travelling salesman problem (TSP) and Chinese postman problem (CPP)?
For me, both wants go to a destination, and then back.
Comment: http://wiki.answers.com/Q/What_are_the_differences_between_chinese_postman_problem_and_travelling_salesman_problem
Comment: this is a good article: http://faculty.bracu.ac.bd/~rouf/course/summer06/cse426/TSP_CPP.pdf
Answer: Graphs are made of edges and vertices. CPP requires a visit to all edges. TSP requires a visit to all vertices.
Comment: @Alan: A Hamiltonian path is one that visits all *vertices*. Eulerian visits all edges. I'm not sure what your equivalence is (since you seem to give two definitions for CPP), but this is the correct difference between CPP and TSP.
Comment: wait, I think it's the oppostive o.O
I know that chinese = hamilton and postman = euler.
Comment: I learned that the easiest way to solve chinese thing is with hamilton, and the other with euler.
Answer: The travelling salesman problem is about going to each city once and taking the shortest route.
The Chinese postman problem is about getting a path from each city to each other city.
E.g., with points A, B, C, and D, the travelling salesman could go A-B-C-D-A, but the Chinese postman would go need a route that had A-B and A-C and A-D, etc.
The travelling salesman route does not have a direct between each point (in the above example there is no A-C link).
Each city is a vertex and each inter-city link is an edge. So, I think I'm just restating Xodarap's answer.
Answer: Keeping it ultrasimple:
The travelling salesman problem is about visiting each city exactly once while returning to the original city (thus walking along a Hamiltonian cycle) and also taking the shortest route among all possible routes that fulfill this criterium (if such a route exists). Finding such a cycle, perforce finding the possibly unique optimal cycle with the shortest distance, is "hard".
The Chinese postman problem or route inspection problem is about visiting each road between cities at least once while returning to the original city and taking the shortest route among all possible routes that fulfill this criterium (if such a route exists). A solution that takes each road exactly once is automatically optimal and called an Eulerian cycle. Finding such a cycle is "feasible".
Comment: A better algorithm to generate _approximately optimal_ Hamiltonian cycles: [Computer Scientists Break Traveling Salesperson Record](https://www.quantamagazine.org/computer-scientists-break-traveling-salesperson-record-20201008/)
Answer:
Chinese postman

Travelling salesman

From a brief read of the two articles (and I never took a course in graph theory, so I could be talking through my hat), it appears that the "CPP" involves visiting all edges, and the "TSP" involves visiting all nodes.
Answer: The key difference between the two is:
The travelling salesman problem can not visit a node more than once. The path produced will consist of all different nodes/cities.
The Chinese postman/route inspection problem can have duplicate nodes in the path produced (but not duplicate edges). I.e., nodes can be visited more than once as long as you take a different route out than you took in.
Answer: I think it's just another variation of the pathing problem presented in computer science college courses.
The Chinese traveling salesman problem (C-TSP) is a typical symmetric TSP problem. Its simple description is: Given a list of 31 Chinese capital cities and their pairwise distances, the task is to find the shortest possible tour that visits each city exactly once. The C-TSP is a mediumscale TSP problem, which has (31−1)! / 2 = 1.326 *1032 possible routes.
Comment: I think chinese postman is actually different from TSP in that it requires all edges be visted, not that all vertices be visited. (A la the difference between Hamiltonian and Eulerian paths.)
Answer: Travelling salesman problem:
Given cities and distance between cities, find the shortest distance tour such that it visits each city exactly one. Visualizing this as a graph and a cost or weight associated with each edge, find the cheapest or least weight tour (Hamiltonian path) such that each vertices or node is visited exactly once. We can think of this as finding all possible Hamiltonian path and then find the best among them. Finding all possible routes is an optimization problem and in NP-complete means no polynomial time solution exists for this problem
Chinese postman problem:
Contrary to travelling salesman problem, CPP requires to find a least cost or minimum weight tour through the graph such that each edge is visited at least once. The problem has a polynomial solution and the optimal solution requires to find a Euler tour through the graph if the graph is Eulerian.
Else modify the graph to make it Eulerian and define a Euler tour. A special instance of the Chinese postman problem is where we don't need to travel all the edges of the graph, but only some of them (required edges). This variation is called the Rural postman problem and is NP-complete. In other words, given a graph, find a least cost/ minimum weight tour such that all the required edges are covered at least once, may be using the non-required edges.