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# How high to go without tracking?

### Help Support The Rocketry Forum:

#### Cabernut

##### Well-Known Member
I have a 2.6" rocket weighing about 29oz with a 38mm motor mount that I'll be bringing to NYPOWER in a couple weeks. I would like to take advantage of the large field and really put it up there, but I don't have any kind of tracking device currently other than a pair of binoculars...

It will be motor eject only with a Chute Release. I would love to lob it up to 4.7K ft on an I350 but I doubt I'll see it that high.

How high should I / would you go before it gets too hard to track visually?

#### EXPjawa

##### Well-Known Member
I fly there and over in Potter (at an even larger field). My personal feeling is the cut off is around 2500', depending on cloud cover and wind. But I've had plenty of flights that tickled 2000', and even with the CR, it was starting to get difficult to locate. Especially if the MARStians insist on putting the flight line right next to the bean field, even when the wind is in that direction. At that site, is quite possible to drop the rocket within a few hundred feet of your camp and still not find it... So, I've held off on putting up anything that was expected to breach 2500' until I've got tracker of some sort. I'd love to punch a mile, and have a few rockets that would do it if asked, but things can still drift a ways from that height when the chute is still reefed.

#### ActingLikeAKid

##### Well-Known Member
My specs are similar to yours. Having an I in it is why I got an Eggfinder

If you're really keen on doing that height without tracking, some things to mitigate the risk:
-While you watch with binoculars, have a friend with a smartphone (if they do see it on the descent) get a bearing on it with their compass app as it comes down. Then at least you know that you need to head at bearing X, instead of "I think towards that old tree that looks half-dead?"
-Mylar streamer. It's amazing what a bit of flashy plastic does for visibility.
-Buy a security "screamer" alarm and either activate it right before launch or rig it up so the separation makes the screamer open. It's not perfect, but you'll hear it within 200 feet or so.

Good luck!

#### Nick@JET

##### Well-Known Member
Yea, I had the same plan at Potter at LDRS...New York still has my rocket. I even asked for announcer to ask for all eyes since no tracker, but not really their responsibility. I have floaters in my eyes so thought I saw it but followed the floater and never saw the rocket again

Risky plan. Ask someone to borrow an RF tracker.

I agree with above altitude or if your really hell bent on it, screamer, polished rocket, shiny streamers etc

#### mpitfield

##### Moderator
Staff member
Global Mod
Hey Chris,

I just built three Eggfinder Minis and the Eggfinder LCD, you are more than welcome to borrow what you need. I am brutal at visually tracking so I plan to try to put a tracker in anything, it will fit into, that is expected to go over 1,500'. If you want to borrow the gear then we should meet up so I can hand it off to you as you will need to figure out how to integrate it. Nosecone would be ideal if you can do a quick retrofit.

#### Cabernut

##### Well-Known Member
I do have an 8" wide mylar "Happy Birthday" banner... maybe I'll attach that to it as well. If weather isn't all that great I might have to settle for a baby H. However if conditions are good, I know I'm going to get Go Fever.

#### Cabernut

##### Well-Known Member
Hey Chris,

I just built three Eggfinder Minis and the Eggfinder LCD, you are more than welcome to borrow what you need. I am brutal at visually tracking so I plan to try to put a tracker in anything, it will fit into, that is expected to go over 1,500'. If you want to borrow the gear then we should meet up so I can hand it off to you as you will need to figure out how to integrate it. Nosecone would be ideal if you can do a quick retrofit.
Thanks, I'll keep that in mind. I've been looking into getting an Eggfinder but my soldering skills are about caveman level. I'll have to search around for how others have integrated them into nose cones and see if I can swing it.

#### Bat-mite

##### Rocketeer in MD
Besides altitude, it also matters what the surrounding territory is like. Could it come down in a cornfield, wheat field, pond, trees? Sometimes a tracker is the only thing that can get you there even if it lands close by. And remember this important rule: it is already farther away than you thought it was when you saw it come down.

#### ksaves2

Listen to Rick, 2000' to 2500' is a good reference. Does depend upon how large the rocket is. The bigger the rocket, the easier it is to see at altitude. The gist of the matter is one wants to avoid a lengthy "out of sight" time as believe me, the upper
air winds can play havoc with where a rocket finally ends up. I've tracked a lot via APRS on a live map and have seen rockets out of sight and at altitude go in directions 180 degrees to prevailing ground winds. Many a time, I've tracked mine and other
APRS equipped rockets and have yelled at those scanning the skies to turn their eyes 90 to 180 degrees from where they are looking for the expected point in the sky where the main deployment is going to occur. Also nice to be able to see on a map
that the rocket has landed on the field or on the other side of a road whathave you . The other thing is with a fiberglass rocket, if the chute release doesn't deploy the chute properly, the rocket can land pretty close to you and it can remain sight unseen. Without an APRS or NMEA tracker (AKA an EggFinder or Missileworks NMEA trackers) one won't know where to go without a position recorded to go to. Had that happen a few times with glass rockets that totally survived the hard landing in soft dirt and were completely flyable again without any work other than re-prepping. They would have been likely lost without a position to go to.
Yes RDF can give one a bearing and be workable but GPS can give one an edge. Especially if the recovery is ballistic. Since it's high speed, get one or two positions on descent and you'll likely run into the fincan sticking out of the ground or the
splattered remains when you go out there. Again if it's a fiberglass rocket might just have to replace the nosecone and the electronics and good to go again. (Been there and done that too.)

Choose your motor wisely for a visual recovery until you can afford a tracker. Also an audible alarm or as I call them "screamers" can help. Sometimes if the rocket is coming in under drogue or drogueless, you can hear where to turn your eyes to get a visual.

Kurt

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#### thomas

##### Well-Known Member
If you have GSM/GPRS reception at the field you can buy a GPS tracker for less than 10 Dollar on ebay.

#### amarillo_rocket

##### Well-Known Member
If the area is fairly quiet use a 4"-6" wide, 5-6ft streamer. They make a fair amount of noise flapping in the wind while descending. Roll the streamer part way and add a good amount of Johnsons Baby Powder to get a visual reference at apogee. Johnsons does not have corn starch like a lot of baby powder so it doesn't burn. The risk is if the rocket does not head into the prevailing wind and everyone is looking in the wrong direction. I have seen this happen many times.An audible alarm would also help. The obvious best solution is to use a tracker.

#### EXPjawa

##### Well-Known Member
Listen to Rick...

Kurt
Can I quote you on that? Wait, I just did! Sorry, I need to go get lunch...

When I first got the Chute Release, my naive expectation was that I'd never have to worry about tracking stuff again, within reason. But I learned real quick what folks flying traditional DD already knew - that upper altitude winds can push even a drogueless rocket pretty far, and not necessarily in the direction I'd expect. So, getting the rocket to drop (dramatically, with Chute Release popping a few hundred feet up) back into the pad area is a game of judging winds and rod angle. Same as DD, in that regard. And the higher you go, the farther off your judge will probably be. I've watched the couple of flights I've done into the mid-2000' range go pretty far down range on the CR, so that was the basis of my self-imposed cap. I was able to find them each time, but that's as much luck as not, especially as crops grow over the season...

With that in mind, I did wrap bands of aluminum foil tape on my MAC Scorpion to make it more reflective, and I have a roll of holograohic mylar streamer in my tool box with the same intent (though haven't tried it). Reflective streamers only help when the sun is bright, though we fly under overcast skies a lot. And I'm still not certain about combining a chute on CR and a streamer at the same time...

#### blackjack2564

##### Crazy Jim's Gone Banana's
TRF Supporter
True short story expounding the reason to always use a tracker...................

Wanna drag race that? [Gizmo]

"Sure" what motor?"

"OK you wanna borrow one of my trackers?"

Nah they're only going 1500-1800 ft.

"Better to have one & not need it...rather than need it & not have it"

I'll be fine.

3....2....1...launch OOPS looks like one of the Gizmo's rainbowed!

His rocket had the rear button hang on rail, after leaving the rail, rocket does a left turn, flies like a cruise missile in shallow arch and land in trees almost 1/2 mile away. 5hrs that day...3hrs the next it was located.

The end.

PS there are always things that can go wrong, you never think about, when suddenly one will rear it's ugly head & happen. Tracker will find your rocket.

#### boatgeek

##### Well-Known Member
I launched a 2.6" FG rocket to about 5000' on a baby J with a CR earlier this year on a crystal clear day. It was out of sight for a short while but we picked it up again from flashes off the body tube. It turned out I didn't need the tracker that was in it, but took a little more time finding it by walking (snowshoeing) around than was ideal. Funny thing, it was a little further away than I thought.

My general rule of thumb is 2500' for 54mm rockets and 5500' for 4" rockets is reasonably reliably visible as long as there aren't clouds in the way. It helps to have access to some younguns with sharp eyes.

#### dhbarr

##### Amateur Professional
I launched a 2.6" FG rocket to about 5000' on a baby J with a CR earlier this year on a crystal clear day. It was out of sight for a short while but we picked it up again from flashes off the body tube. It turned out I didn't need the tracker that was in it, but took a little more time finding it by walking (snowshoeing) around than was ideal. Funny thing, it was a little further away than I thought.

My general rule of thumb is 2500' for 54mm rockets and 5500' for 4" rockets is reasonably reliably visible as long as there aren't clouds in the way. It helps to have access to some younguns with sharp eyes.
A quarter mile per inch diameter? Seems like a decent rule of thumb.

#### jnelson

##### Well-Known Member
At Black Rock a couple years ago I flew a 54mm x 5ft rocket on a couple research I's, to 3400 ft, I don't recall having issues seeing it. Last year I flew the same rocket on a J420 to about 5000 ft and lost sight of it. So I think my personal limit for that size of rocket is 3500 ft. FWIW

#### Nick@JET

##### Well-Known Member
Hey Chris,

I just built three Eggfinder Minis and the Eggfinder LCD, you are more than welcome to borrow what you need. I am brutal at visually tracking so I plan to try to put a tracker in anything, it will fit into, that is expected to go over 1,500'. If you want to borrow the gear then we should meet up so I can hand it off to you as you will need to figure out how to integrate it. Nosecone would be ideal if you can do a quick retrofit.
And Michael steps up! Gotta love the rocket community.
I hear you about go fever, at a launch I the guy jumping up and down, clapping "press the button, press the button!"

As a backup to CJ, At LDRS 34, they had some issues with equipment after Friday night rain and launched a few rockets on accident and those people weren't even watching. I found a big 6" rocket while I was searching for mine and that guys was still trying to log on to his computer to track. 6" and they never saw it

#### ActingLikeAKid

##### Well-Known Member
Thanks, I'll keep that in mind. I've been looking into getting an Eggfinder but my soldering skills are about caveman level. I'll have to search around for how others have integrated them into nose cones and see if I can swing it.
I built one over the winter and it wasn't bad. My soldering experience before was building something that didn't work. So before the Eggfinder, I ordered a few kits from Amazon to practice on. Two things that helped a lot:
-An illuminated magnifying visor (like $8 from Amazon) -A cheap LED desk light. Depending on what you're flying, you can get a nosecone with a removable electronics sled. They aren't super expensive and would do exactly what you want. #### Oberon ##### Well-Known Member I flew an Estes Partizon to 3600 ft with motor eject and chute release and had no issues seeing it without binoculars the whole way, but the LCO and one other guy spotting lost it. So depends on your eyes. I probably would have lost it another 1000ft up, but if you could get the binoculars on it... Then again, I only felt comfortable launching it because a) it was a nearly dead calm day b) I had seen other rockets that day fly to nearly the same altitude, and had a good idea what to expect and where to look in the sky if I temporarily lost sight, and c) I did have a tracker, though I ended up not needing it at all as the rocket landed about 50 yards away. One thing I did do, and recommend, is use a large, bright streamer for visibility prior to Chute Release activation. It also depends a lot on the field conditions for actually finding the rocket once down. At our desert scrub field, you can usually spot a rocket from a significant distance away. But dense grass or crops or big trees to block your view would make a big difference - really the question is, "how close do I need to get to the rocket to find it?" At one extreme, something like a dry lake bed you just need a general bearing, easily obtained visually. A corn field you need GPS or RDF even if its 200 yards away. #### Banzai88 ##### Lvl 1,Wallet....Destroyed TRF Supporter I think that it all depends on the field, the potential landing areas, the size of the rocket, the type of rocket recovery, and the weather conditions. I fly mostly at Bayboro, which is square miles of crop stubble or beans, slightly interrupted by ditches, and ringed by Mordor (trees). For anything that's single deploy, or single deploy with JLCR, I try to go no higher than 2500 ft. Dual deploy without tracker is about 3K, 4 inch I can see pretty reliably up to about 3500. Anything minimum diameter gets a tracker, always. I've seen quite a few high flights go into the trees, and never return. People get into their cars, drive around the field to the trees, and attempt to walk into Mordor. No tracker often means a sacrifice to the rockets gods has been performed. #### mccordmw ##### Well-Known Member If you think you can get within 100-200 yards of where it will land, and are willing to take the risk with no tracking, how about investing a small amount in a sonic beeper? I toss these into my rockets to help find them when they land in tall grasses. Sure, my Eggfinder will get me almost right on top of my rocket, but the cheap added insurance is worth it. These come in a 2-pack for$15. Pull-pin activation; perfect for chutes. Loud as HELL. About a 2" diameter by 1/3" thick disc.

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#### blackbrandt

##### That Darn College Student
I think that it all depends on the field, the potential landing areas, the size of the rocket, the type of rocket recovery, and the weather conditions. I fly mostly at Bayboro, which is square miles of crop stubble or beans, slightly interrupted by ditches, and ringed by Mordor (trees).

For anything that's single deploy, or single deploy with JLCR, I try to go no higher than 2500 ft. Dual deploy without tracker is about 3K, 4 inch I can see pretty reliably up to about 3500. Anything minimum diameter gets a tracker, always.

I've seen quite a few high flights go into the trees, and never return. People get into their cars, drive around the field to the trees, and attempt to walk into Mordor. No tracker often means a sacrifice to the rockets gods has been performed.
I've flown an accidental single deploy at Bayboro to 6300, and man was I glad I had a tracker in there. Even though you have a line on the rocket, with the amount of side-stepping you have to do to get across the ditches, it's almost impossible to trace a line like that at Bayboro. With a tracker, it's easy peezy.

My opinion on tracking: If you have a tracker, put it in there (and put the biggest battery you can fit in there with it). I will NOT be flying again without a tracker in any of my flights. I look at it as insurance. Of course, if I'm flying below about 1500 feet, I might be lazy and go without a tracker. But if I don't have a tracker in the rocket, you can bet that there is marking chalk, a bright red streamer, and a pull-pin alarm in the rocket.

#### Barkley

##### Well-Known Member
I have a 2.6" rocket weighing about 29oz with a 38mm motor mount that I'll be bringing to NYPOWER in a couple weeks. I would like to take advantage of the large field and really put it up there, but I don't have any kind of tracking device currently other than a pair of binoculars...

It will be motor eject only with a Chute Release. I would love to lob it up to 4.7K ft on an I350 but I doubt I'll see it that high.

How high should I / would you go before it gets too hard to track visually?
My L2 bird is almost 8 feet and 3 inches in diameter, and at 4200 it passed through the only cloud in the sky, a microscopic thing. The only reason why I found it was another rocket had tipped over the same way about 2 hours earlier, and I was told where that one was found. Otherwise I literally never would have found it. My bird is much bigger than yours, and had two big ejection charges to listen to.

Bought a BigRedBee, and now I'm much happier. My advice to you: Borrow Michael's tracker. Please.

#### grouch

##### Well-Known Member
Funny, the first time I flew my tracker in anger was at NYPOWER. It's a great field but the year I went the ground was really soft and the grass was high. I mean like waist high to the left of the flight line due to all the recent rains. I was certainly glad I was tracking each flight.

#### Steve Shannon

##### Well-Known Member
TRF Supporter
As others have said, the size and color of the rocket, the terrain, and the clarity of the sky determine whether you need a tracker. We routinely launch rockets over a mile in altitude without trackers and are able to view them the entire flight. Then we walk our line and recover them.
We've also lost sight of rockets that didn't fly that high and wished we had a tracker.
If I were flying that high at a place where I was unfamiliar with the terrain and didn't have a feel for where things land I'd probably want some kind of help, either spotters or a tracker.

Steve Shannon

#### rharshberger

##### Well-Known Member
I launched a 2.6" FG rocket to about 5000' on a baby J with a CR earlier this year on a crystal clear day. It was out of sight for a short while but we picked it up again from flashes off the body tube. It turned out I didn't need the tracker that was in it, but took a little more time finding it by walking (snowshoeing) around than was ideal. Funny thing, it was a little further away than I thought.

My general rule of thumb is 2500' for 54mm rockets and 5500' for 4" rockets is reasonably reliably visible as long as there aren't clouds in the way. It helps to have access to some younguns with sharp eyes.
At that same site I launched my L3 to 7400' , it drifted over a hill, thank goodness for my Eggtimer TRS and John with his binocs between the two we pretty much walked up to it. The rocket was in sight from apogee till it disappeared over the low hills due north of the racetrack. Now I use trackers on every flight they can fit in.

#### DavidMcCann

##### Well-Known Member
Track everything. It'll land in sight and 30 yards away

#### tomsteve

##### Well-Known Member
How high to go without tracking?
i can get about 14,200' and recover easily-with an A8 launchin at the summit of pikes peak.

#### Q-Aero

##### Well-Known Member
Find someone who go to NYPOWER with a Comm-Spec system, then you can just purchase the transmiter and if you lost the rocket the member just select your frequency and he will find your rocket. \$ 50 and you run from 3 to 6 week on a battery depending if you get a 1 mile or 3 mile range . No need to charge LiPo battery betwen flight. Later you can purchase the Receiver, everything integrated, no hassle

http://www.com-spec.com/rcplane/index.html

#### mrwalsh85

##### Well-Known Member
Surprised no one from MARS has chimed in... MARS should have trackers available for loan (I think for club members, anyway)... Just make sure you get it back in one piece! I would contact one of the BOG members to coordinate with them on that regard.

mike


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# 100 chairs in a circle problem solving with Angular.js and visualization

Here is the problem statement:

Take a second to imagine that you are in a room with 100 chairs arranged in a circle. These chairs are numbered sequentially from One to One Hundred. At some point in time, the person in chair #1 will be told to leave the room. The person in chair #2 will be skipped, and the person in chair #3 will be told to leave. Next to go is person in chair #6. In other words, 1 person will be skipped initially, and then 2, 3, 4.. and so on. This pattern of skipping will keep going around the circle until there is only one person remaining- The survivor! Note that each chair is removed when the person leaves the room.

Please review my attempt below considering the following points:

1. Correctness (I am getting the winner as # 31)
2. Any chances for optimization

var myApp = angular.module('myApp',[]);

myApp.controller('MyCtrl', [ '$scope', '$interval', '$log', function($scope, $interval,$log){

$scope.init = function(){$scope.people = [];
$scope.logs = []; // get the 100 people/chair for(var i=1; i<=100; i++){$scope.people.push({ index: i });
}
};

$scope.startFiring = function(){ var fireIndex = 0, // for removing the person at this index skipCount = 0; // for incremental counting if($scope.people.length >= 1){ // if start button clicked again
$scope.gotWinner = false; // hide the winner$interval.cancel(fireInterval); // stop the interval
$scope.init(); } var fireInterval =$interval(function(){

$scope.disableStartButton = true; if ($scope.people.length > 1) { // if more than 1 people
$scope.logs.push({text: 'Firing #' +$scope.people[fireIndex].index}); // show the log

$scope.people.splice(fireIndex, 1); // fire! // increment skipCount += 1; fireIndex += skipCount; //if fireIndex becomes greater than$scope.people.length, then set fireIndex to remainder from fireIndex / $scope.people.length if(fireIndex >$scope.people.length){
fireIndex %= $scope.people.length; } } else {$interval.cancel(fireInterval); // stop the interval
$scope.gotWinner = true; // show the winner$scope.disableStartButton = false;
}
}, 500); // every 1/2 sec.

};

$scope.init(); // intialize the chair/people when the view is loaded or start is clicked again }]); body{ font-family: "HelveticaNeue-Light", "Helvetica Neue Light", "Helvetica Neue", Helvetica, Arial, "Lucida Grande", sans-serif; } span{ display: inline-block; padding: 5px; margin: 4px; border-radius: 3px; background: #000000; color: #FFFFFF; } <!DOCTYPE html>   <meta charset="utf-8"> <title>JS Bin</title>

 {{person.index}}
 <button ng-click="startFiring()" ng-disabled="disableStartButton">Start!</button>
And the winner is.... # {{people[0].index}}!

Logs:

{{log.text}}

    • The firing sequence you have described (1, 3, 6, 10, 15, …) is different from the typical "Josephus Problem". Feb 14, 2016 at 5:05 • @200_success I have edited the title. However, the problem statement remains the same. Feb 14, 2016 at 7:27 • 31 is correct. The fiddle could be improved codewise but is correct. You can check it by using a smaller number for initialization (instead of 100 set 10. Feb 15, 2016 at 9:54 ## 1 Answer Mathematical % simplification Remember that $$a\ \%\ b == a \ \ if\ b > a$$ I don't know the math, but Python confirms: >>> import random >>> def random_ints(): for _ in range(1000): yield random.randint(1, 100) >>> all(a % b == a for a in random_ints() for b in random_ints() if b > a) True  So:  fireIndex += skipCount; //if fireIndex becomes greater than$scope.people.length, then set fireIndex to remainder from fireIndex / $scope.people.length if(fireIndex >$scope.people.length){
fireIndex %= $scope.people.length; }  can be simplified to fireIndex = (fireIndex + skipCount) %$scope.people.length

  if ($scope.people.length > 1) { // if more than 1 people  It is already obvious from the code that you are checking if there are more than 1 people, the comment just doubles the effort required to change the code. The below comment is very confusing instead: if($scope.people.length >= 1){ // if start button clicked again

• Thank you! What do you think about the correctness? Is 31 correct? Feb 14, 2016 at 18:22

format: plain


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# MathJax v2.1 now available

After a successful beta testing, we’re happy to officially release MathJax v2.1.

Version 2.1 is available on the CDN at

http://cdn.mathjax.org/mathjax/2.1-latest/MathJax.js

and on 12pm PDT, Tuesday, Oct 23, 2012 the files at the

http://cdn.mathjax.org/mathjax/latest/MathJax.js

address will be switched over the v2.1, but it may take some time for them to propagate out to the distributed cloud servers.

During the time that the files are making their way out to the CDN’s servers, there may be a mixture of files in a browser cache, and so users may need to clear their cache and restart their browser in order to get a consistent version of the files.

If you are a page author and concerned about this, you can change (temporarily) to the mathjax/2.1-latest URL instead of mathjax/latest since that is a new address that will not have any cached older versions to worry about. You can switch back to mathjax/latest after a day or two when the new version has migrated to all the locations in the cloud.

See http://docs.mathjax.org/en/latest/whats-new-2.1.html for details about the changes in v2.1, and some caveats about the effect of these changes on existing sites.

We anticipate a smooth upgrade from v2.0 to v2.1, but as always, let us know on the bug tracker if you experience problems with this new version of MathJax.  Thank you for your continued support.

The MathJax Team.


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# Killing vectors - Schwarzschild metric

Given the Schwarzschild metric, $$\mathrm{d}s^{2}=-\left(1-\frac{R_s}{r}\right)\mathrm{d}t^{2}+\left(1-\frac{R_s}{r}\right)^{-1}\mathrm{d}r^{2}+r^{2}\mathrm{d}\theta^{2}+ r^2 \sin^{2}\theta\mathrm{d}\phi^{2},$$ I'm asked to show that $$K^{\mu}=\left(1,0,0,0\right),\; R^{\mu}=(0,0,0,1)$$ are Killing vectors, i.e. they satisfy the Killing equation $$\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0.$$

First of all, using the metric I lower the indices: $$K_\mu=\left(-\left(1-\frac{R_s}{r}\right),0,0,0\right),\; R_\mu=\left(0,0,0,r^2\sin^2\theta\right).$$ Then, for $$R_\mu$$ I tried to compute $$\nabla_\mu R_\nu\equiv\partial_\mu R_\nu - \Gamma^\lambda_{\mu\nu} R_\lambda=\partial_\mu R_\nu - \Gamma^\phi_{\mu\nu} R_\phi;$$ so $$\nabla_r R_\phi=r\sin^2\theta, \; \nabla_\theta R_\phi=r^2\sin\theta\cos\theta.$$ But now it's not clear for me what should I do next.

• What happened to the Christoffel symbols? I think you're missing some terms in your expressions for the covariant derivatives – N. Steinle Nov 24 '18 at 17:54
• @N. Steinle, I have an implicit sum over $\lambda=t,r,\theta,\phi$. Then I realize that the only non-vanishing Christoffel symbol is that one whit $\lambda=\phi$. – Stig Nov 24 '18 at 18:42

Your equation for $$R_\mu$$ reads $$\nabla_\mu R_\nu + \nabla_\nu R_\mu =\left(\partial_\mu R_\nu -\Gamma^\lambda_{\mu\nu} R_\lambda\right) + \left(\partial_\nu R_\mu - \Gamma^\sigma_{\nu\mu}R_\sigma\right)=\partial_\mu R_\nu + \partial_\nu R_\mu -2\Gamma^\phi_{\mu\nu} R_\phi.$$ If you use $$\mu=r$$ and $$\mu=\theta$$ (the only non-vanishing terms), you will find that Killing equation for $$R_\mu$$ it's satisfied.

$$\def\bg{{\mathbf g}}$$ It depends on how Killing vector has been defined. If definition was "a vector field obeying Killing equation", then you are only left to pursue the computation and I can add nothing. But there is an alternative definition, which I prefer:

A Killing field is one whose flow is an isometry for all values of parameter.

This requires to know what is the flow of a vector field and what means "isometry" in a Riemannian manifold. I shortly summarize.

The flow of a vector field $$X$$ is simply the ensemble of its integral curves, each parametrized by a real variable which I will name $$u$$. For each value of $$u$$ the flow of $$X$$ defines a mapping $$\mu$$ of the manifold into itself.

An isometry of a Riemannian manifold is a (differentiable) mapping leaving invariant the metric tensor: $$\mu^*\bg = \bg$$.

Then it can be shown that $$X$$ is a Killing field for the metric $$\bg$$ iff it satisfies the Killing equation (where covariant derivative is defined by Levi-Civita connection of $$\bg$$).

In your problem the flow of $$K$$ is obviously the mapping $$\mu_u: (t,r,\theta,\phi) \mapsto (t+u,r,\theta,\phi)$$ i.e. a time translation by $$u$$. Since $$\bg$$ (i.e. $$ds^2$$) doesn't depend on $$t$$, $$\mu_u$$ is an isometry for all $$u$$.

Analogously, the flow of $$R$$ is $$\nu_u: (t,r,\theta,\phi) \mapsto (t,r,\theta,\phi+u)$$ i.e. a rotation of angle $$u$$. And $$\bg$$ doesn't even depend on $$\phi$$, so $$\nu_u$$ is an isometry as well.


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# Swift Program to Find the Intersection of Two Sets

SwiftServer Side ProgrammingProgramming

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13 Lectures 2 hours

#### How to create Tic-Tac-Toe Game for iPhone using Swift in Xcode - iOS App

7 Lectures 1 hours

This tutorial will discuss how to write swift program to find the intersection of two sets.

Set is a primary collection type in Swift. It is an unordered collection which stores unique values of same data type. You are not allowed to store different type of values in the same set. A set can be mutable or immutable.

Intersection is one of the set operation which is used to create a set of elements that are common in both the sets. For example, we have two sets: set X and set Y, now the intersection of both the sets contains only those elements that are common in both the sets.

To find the intersection of two sets Swift provide in-built function named as intersection(). This function returns a new set which contain the intersection of given two sets.

Below is a demonstration of the same −

Suppose our given input is −

Set 1 = [34, 5, 78, 9, 10]
Set 2 = [34, 45, 5, 10, 100]

The desired output would be −

Intersection Set = [34, 5, 10]

## Syntax

Following is the syntax −

Set1.intersection(Set2)

## Algorithm

Following is the algorithm −

• Step 1 − Declare and initialise two sets with values.

• Step 2 − Finding the intersection of two sets using intersection() function and store the result into a variable −

var intersectionSet = mySet1.intersection(mySet2)

For sorted set −

var intersectionSet = mySet1.intersection(mySet2).sorted()

• Step 3 − Print the output

## Example 1

The following program shows how to find the intersection of two sets.

import Foundation
import Glibc

var mySet1 : Set = ["apple", "mango", "jackfruit"]
var mySet2 : Set = ["jackfruit", "kiwi", "mango", "dragon fruit"]

// Finding intersection of two sets
var intersectionSet = mySet1.intersection(mySet2)

print("Set 1:", mySet1)
print("Set 2:", mySet2)
print("Intersection Set:", intersectionSet)

## Output

Set 1: ["mango", "apple", "jackfruit"]
Set 2: ["jackfruit", "dragon fruit", "kiwi", "mango"]
Intersection Set: ["mango", "jackfruit"]

Here, in the above code, we have two sets: Set 1 and Set 2 of String type. Now we find their intersection using intersection() function. This function will return a set which contain the intersection of both the sets −

var intersectionSet = mySet1.intersection(mySet2)

Hence the result is ["mango", "jackfruit"].

## Example 2

The following program shows how to find the intersection of two sets.

import Foundation
import Glibc

var mySet1 : Set = [3, 45, 67, 2, 4]
var mySet2 : Set = [34, 5, 65, 2, 3]

// Finding intersection of two sets
// Here the result will in sorted form
var intersectionSet = mySet1.intersection(mySet2).sorted()

print("Set 1:", mySet1)
print("Set 2:", mySet2)
print("Intersection Set:", intersectionSet)

## Output

Set 1: [4, 3, 45, 67, 2]
Set 2: [2, 3, 5, 34, 65]
Intersection Set: [2, 3]

Here, in the above code, we have two sets: Set 1 and Set 2 of integer type. Now we find their intersection using intersection() function. This function will return a set which contain only the common elements present in both the sets that is [2, 3]. Here the result is sorted in ascending order because we use sorted() function with intersection() function using dot operator −

var intersectionSet = mySet1.intersection(mySet2).sorted()

Updated on 12-Oct-2022 07:18:07


text:
# [R] a nlm() question

Jiang Peng jp021 at sjtu.edu.cn
Thu Oct 30 09:49:00 CET 2008

  Dear R listers,
I have a very annoying problem using nlm().
I want to find the minimizer of my target function, if written in
\LaTeX is
f(\mu1,\mu2,\sigma1,\sigma2) = \sum_i^n( w_ig_t(z_i) ), where
g_t(z) is a pdf of  bivariate normal distribution and z_i is my samples.
I cannot get the estimation result generated by nlm(), and I got
the following errors

"
Error in nlm(foo, theta.start, hessian = TRUE) :
non-finite value supplied by 'nlm'
"

from time to time( If I ran the code second time , it does not have
the error .) I am new to R so I don't know if there is any debug
method of R code. i hope somebody can help me out .

------------------------
Jiang Peng, Ph.D. Candidate
Department of mathematics &
Antai college of Economics and management
Shanghai Jiao Tong University
Address: Room 127, #3 Building, Xuhui Campus
E-mail: jp021 at sjtu.edu.cn
jp021 at 126.com
MSN: jp021 at hotmail.com
iChat: mathfrog at mac.com
Mobile: +86-138 168 780 95


text:
# Can I add a table through PGF \nextgroupplot functionality?

I have made a minimal working example which shows how far I am right now:

\RequirePackage{luatex85}
\documentclass[tikz]{standalone}
% Default preamble
\usepackage{pgfplots}
\usepgfplotslibrary{groupplots}
\usepgfplotslibrary{polar}
\usepgfplotslibrary{smithchart}
\usepgfplotslibrary{statistics}
\usepgfplotslibrary{dateplot}
% Custom preamble from global variable:
\usepackage{pgfplotstable}
\begin{document}
\begin{tikzpicture}
\begin{groupplot}[group style={group size={2 by 2}}, height={4cm}, width={4cm}]
\nextgroupplot[title={1}]
coordinates {
(0,0)
(1,1)
(2,2)
}
;

\nextgroupplot[title={2}]
%                        \pgfplotstabletypeset[
%                       col sep=&,  % specify the column separation character
%                       row sep=\\, % specify the row separation character
%                       columns/E/.style={string type} % specify the type of data in the designated column
%                        ]{
%                           A & B & C & D & E \\
%                           1 & 10 & 0.1 & 1000 & 2011-01-01 \\
%                           2 & 20 & 0.2 & 2000 & 2012-02-02 \\
%                           3 & 30 & 0.3 & 3000 & 2013-03-03 \\
%                           4 & 40 & 0.4 & 4000 & 2014-04-04 \\
%                           5 & 50 & 0.5 & 5000 & 2015-05-05 \\
%                        }

\nextgroupplot[title={3}]
coordinates {
(0,2)
(1,1)
(2,1)
}
;
\nextgroupplot[title={4}]
coordinates {
(0,2)
(1,1)
(1,0)
}
;
\end{groupplot}
\end{tikzpicture}
\end{document}

Running this code produces the following figure:

Instead of the empty figure to the top right, ("2"), I want to insert a table. This table has already been defined in my code, but commented (with %) - if I uncomment and run my LaTeX code again the result becomes:

Would anyone know how to stay in this workflow with \nextgroupplot, and be able to resolve this problem, such that I can insert a table instead of a figure?

Kind regards

• Does the table datas correspond to the plots? – cis Sep 21 '20 at 15:58

You could remove the empty axis with

\nextgroupplot[hide axis]

and then add the table in a node after \end{groupplot}. The axes are named group cNrM, where N and M are column and row numbers respectively, so you can use for example

\node [below right, xshift=-5mm] at (group c2r1.north west) { <table code> };

Whether you're using \pgfplotstabletypeset or a tabular directly doesn't matter, but you can't have a table environment in the node.

\documentclass[tikz]{standalone}
% Default preamble
\usepackage{pgfplots}
\usepgfplotslibrary{groupplots}
\usepgfplotslibrary{polar}
\usepgfplotslibrary{smithchart}
\usepgfplotslibrary{statistics}
\usepgfplotslibrary{dateplot}
% Custom preamble from global variable:
\usepackage{pgfplotstable}
\begin{document}
\begin{tikzpicture}
\begin{groupplot}[group style={group size={2 by 2}}, height={4cm}, width={4cm}]
\nextgroupplot[title={1}]
coordinates {
(0,0)
(1,1)
(2,2)
}
;

\nextgroupplot[hide axis]

\nextgroupplot[title={3}]
coordinates {
(0,2)
(1,1)
(2,1)
}
;
\nextgroupplot[title={4}]
coordinates {
(0,2)
(1,1)
(1,0)
}
;
\end{groupplot}

\node [below right, xshift=-5mm] at (group c2r1.north west) {
\pgfplotstabletypeset[
col sep=&,  % specify the column separation character
row sep=\\, % specify the row separation character
columns/E/.style={string type} % specify the type of data in the designated column
]{
A & B & C & D & E \\
1 & 10 & 0.1 & 1000 & 2011-01-01 \\
2 & 20 & 0.2 & 2000 & 2012-02-02 \\
3 & 30 & 0.3 & 3000 & 2013-03-03 \\
4 & 40 & 0.4 & 4000 & 2014-04-04 \\
5 & 50 & 0.5 & 5000 & 2015-05-05 \\
}};

\end{tikzpicture}
\end{document}

• Thanks! That is what I wanted, in a much more robust way in what I figured out at the end. Could you show me how I would go about adding a table like this using the "normal" table enviroment in LaTex? For an example one generated by tablesgenerator.com/latex_tables# When I try I get an error about a missing item Kind regards – CuriosLatex Sep 19 '20 at 20:48
• @CuriosLatex You can't add a table environment in that context, but just remove/comment \begin{table} and \end{table}. Do you want a caption? – Torbjørn T. Sep 19 '20 at 20:54
• Oh you are right, my bad. I got it to work by removing the lines you mentioned and only keeping the tabular. Afterwards I discovered I had forgot to escape a \$ correctly. I have updated my question with the final result for my MWE, thanks once again! – CuriosLatex Sep 19 '20 at 20:59
• @CuriosLatex You actually tried editing my answer, but anyways, I'll just add a note saying that if you have a tabular instead of a \pgfplotstabletypeset that works the same. – Torbjørn T. Sep 19 '20 at 21:04
• Sorry I suppose I am bit tired. I spent a good few hours trying to figure this out, so I am pretty happy that you took time to write me a detailed answer Once again thanks! – CuriosLatex Sep 19 '20 at 21:05

Sometimes a table next to a group of plots has something to do with these plots.

In this version the plots are as big as the table.

To place the plots and the table by a loop (where the table was setted to the number 0)

\pgfplotsforeachungrouped \n in {1,0,2,3}{%%
\ifnum\n=0 \nextgroupplot[title=The Table,hide axis]
\else
\edef\tmp{
\noexpand\nextgroupplot[title=Title\n, ylabel=y\n, xlabel=x]
\noexpand\addplot+[black, mark options={color\n}, thick] table[x=X, y=Y\n]{\noexpand\data};
}\tmp
\fi}%%

has the advantage, that you can control where the table (plots) shall appear: You can choose {1,0,2,3} or {1,2,3,0} or...

With \def\ShowHelps{0}% 1 "yes" 0 "no"

\documentclass[a4paper]{article}
\usepackage[margin=2cm, showframe=false]{geometry}
\usepackage{calc}
\usepackage{colortbl}
\usepackage{pgfplotstable}
\usepackage{pgfplots}
\usepgfplotslibrary{groupplots}

\def\ShowHelps{1}% 1 "yes"   0 "no"

X       Y1     Y2    Y3       Date
1      10     0.1   100      2011-01-01
2      20     0.2   200      2012-02-02
3      30     0.3   300      2013-03-03
4      40     0.4   400      2014-04-04
5      50     0.5   500      2015-05-05
}\data

\foreach \n/\mycolor in {1/red, 2/blue!66, 3/purple}{
\xglobal\colorlet{color\n}{\mycolor}          }

\pgfplotsinvokeforeach{1,...,3}{
\pgfplotstableset{every col no #1/.style={
}

\newsavebox{\mytable}
\sbox{\mytable}{%
\pgfplotstabletypeset[
columns/Date/.style={string type}
]{\data}}
\pgfmathsetlengthmacro\tablewidth{width("\usebox{\mytable}")}
%\pgfmathsetlengthmacro\tableheight{height("\usebox{\mytable}")}% not good
\newlength{\tabheight}
\settototalheight{\tabheight}{\usebox{\mytable}} % good
\pgfmathsetlengthmacro\tableheight{\the\tabheight}

\ifnum\ShowHelps=1
\tikzset{Help/.style={draw=#1,
nodes={scale=0.6, fill=yellow, draw=#1, text=black, font=\tiny, midway, above, align=left,} },
Help/.default=red,
}\else
\tikzset{Help/.style={opacity=0, overlay},
}\fi

\begin{document}
\begin{tikzpicture}[]
\begin{groupplot}[group style={
group size={2 by 2},
horizontal sep=44pt, vertical sep=45pt,
},
scale only axis,    % 1/2
width=\tablewidth, height=\tableheight, % 2/2
]
\pgfplotsforeachungrouped \n in {1,0,2,3}{%%
\ifnum\n=0 \nextgroupplot[title=The Table,hide axis]
\else
\edef\tmp{
\noexpand\nextgroupplot[title=Title\n, ylabel=y\n, xlabel=x]
\noexpand\addplot+[black, mark options={color\n}, thick] table[x=X, y=Y\n]{\noexpand\data};
}\tmp
\fi}%%

% Help 1/2
\xdef\verticalsep{\pgfkeysvalueof{/pgfplots/group/vertical sep}}
\xdef\horizontalsep{\pgfkeysvalueof{/pgfplots/group/horizontal sep}}
\end{groupplot}
% Help 2/2
\node[draw=gray, anchor=north west, inner sep=0pt] (X) at (group c2r1.north west) {\usebox{\mytable}};
\draw[Help, -|] (X.north west) -- +(\tablewidth,0) node[Help, near start]{\tablewidth \\ tablewidth};
\draw[Help, -|] (X.north west) -- +(0,-\tableheight) node[Help, rotate=90]{\tableheight \\ tableheight};
\draw[Help,-|] (group c2r1.south west) -- +(0,-\verticalsep) node[Help, rotate=90]{\verticalsep \\ vertical sep};
\draw[Help,-|] (group c2r1.south west) -- +(-\horizontalsep,0)  node[Help]{\horizontalsep \\ horizontal sep};
\draw[Help,dashed] (group c2r1.north west) -- +(-\tablewidth-\horizontalsep,0);
\draw[Help,dashed] (group c2r1.south west) -- +(-\tablewidth-\horizontalsep,0);
\draw[Help,dashed] (group c2r2.south west) -- +(0,\tableheight) ;
\draw[Help,dashed] (group c2r2.south east) -- +(0,2*\tableheight+\verticalsep) ;
\end{tikzpicture}
\end{document}


text:
# Quantum Machine Learning for Data Classification

• Physics 14, 79

Quantum machine-learning techniques speed up the task of classifying data delivered by a small network of quantum sensors.

Quantum technologies are growing at great pace, driven by the interplay between academia, government, startups, and large companies such as Google, IBM, Microsoft, and Amazon. Quantum communication systems, including quantum cryptographic networks, have already been deployed on large scales; quantum metrology schemes, such as atomic clocks, offer the state of the art for high-precision measurements; and quantum computers are entering an early industrial era. Yet the “quantum jungle” of available devices and protocols remains hard to navigate, and researchers still need to work on identifying the most promising paths to quantum technologies that can be societally useful. Now, a collaboration between two teams at the University of Arizona—led by Zheshen Zhang and by Quntao Zhuang, respectively—shows that quantum entanglement can provide a quantifiable advantage to data classification, a task relevant for imaging and navigation [1]. Using a machine-learning algorithm, the team classifies data from a network of entangled sensors (Fig. 1). By comparing their scheme with one using classical data processing, they show that entanglement can boost both the accuracy and the speed of classification. The work paves the way for a broad range of quantum-enhanced classification methods that could be enabled by near-future quantum technologies.

There are good reasons to be optimistic about the future of quantum computers. But building a universal, fault-tolerant quantum computer—one that can correct errors deriving from imperfections in its operations or from environmental disturbances—is a daunting task. Fault-tolerant quantum computers that can be scaled up to solve meaningful problems are a decade away (plus or minus infinity, given the incertitude inherent in such predictions). In the meantime, however, the field of quantum simulation is forging ahead. Devices known as variational quantum eigensolvers are showing promise for solving hard problems in quantum chemistry [2, 3] and for performing data-classification tasks [4]. The power of these quantum devices can be further boosted by incorporating techniques borrowed from the emerging field of machine learning. But can this potential be realized with the quantum hardware that will be available in the near term, that is, noisy, intermediate-scale quantum (NISQ) devices [5]? And can quantum machine-learning algorithms provide a true advantage over the powerful, classical machine-learning methods that are already available?

The work of Zhang, Zhuang, and co-workers explores a route for quantum-enhanced data processing that derives from the marriage of quantum machine learning with the most established quantum technologies: quantum sensing and metrology. A well-known example of such technologies is the quantum logic clock first developed by Nobel laureate David Wineland and colleagues [6]. Using methods of quantum computation, including the manipulation of entanglement, this type of clock achieves unprecedented precision in the measurement of time— it is, by many criteria, the most accurate measurement instrument ever constructed by humans. Related techniques have advanced measurements of other quantities (electric and magnetic fields, mass, acceleration, and more), approaching precision levels close to the ultimate limits set by the laws of quantum mechanics.

Combining techniques from the mature field of quantum metrology with techniques from quantum computing is thus especially promising. Recent theoretical and experimental work has shown the potential of using quantum machine-learning techniques to directly process “quantum data” acquired by quantum measurement and sensing devices, optimizing tasks such as measurement, discrimination, and data classification. Such direct processing has advantages compared with processing classical data using quantum techniques, which requires an interface called a quantum random access memory (qRAM), which loads classical data into a quantum computer [7]. This technology, however, is still in its infancy.

The work of Zhang’s and Zhuang’s teams focuses on a particularly intriguing case—a network of quantum sensors. The use of quantum information processing techniques to combine and analyze the quantum outputs of multiple sensors holds tremendous promise for realizing a quantum advantage. The potential gain stems from a fundamental feature of quantum metrology: The advantage provided by the coherent processing of sensor data scales as the square root of the dimension of the so-called Hilbert space that represents the quantum states sensed by the network. Since the dimension of that Hilbert space scales exponentially with the number of analyzed states, the quantum advantage for a quantum sensor network scales exponentially with the number of sensors.

The teams perform the first experimental study that applies quantum machine learning to quantum sensor networks, providing a compelling indication of a quantum advantage. The authors construct a sensor network that generates entangled states and encodes such states in radio-frequency signals. They then design a quantum protocol to classify these signals based on certain features of their amplitude-phase relationship. Such a protocol, dubbed supervised learning, assisted by an entangled quantum network uses techniques of variational optimization (analogous to the training of a deep neural network) to identify an optimal quantum measurement for analyzing the states and classifying the quantum data.

The authors investigate a framework that may be generalized to data-classification problems in many other physical domains. They analyze an experimental configuration in which a classical signal generates a quantum state in the quantum sensors. They then compare the optimal signal-discriminating measurements for two cases. In the first, the states generated by the quantum sensors are measured individually, and the results of the measurements are processed classically. In the second, the states generated by the sensors are processed by a quantum network that performs entangling operations between the states generated by different sensors. In both cases, the measurements are optimized by variational techniques to minimize the classification error using the established support vector machines (SVM) method. SVM classifies data on the quantum states by finding optimal “hyperplanes” that separate the states in a vector space. The researchers find that the quantum SVM provides a significant reduction in the error rate of classification compared with the error rate achieved with a classical SVM.

This first experimental demonstration is performed with just three quantum sensors—a tiny “quantum step” for the tiny “quantum feet” offered by current technologies. But the method can readily be extended to higher numbers of sensors and, given the exponential scaling, holds the promise of dramatically enhancing the power of large sensor networks. Importantly, such enhancement can be obtained with quantum information processors containing hundreds or thousands of quantum logic gates—those that might be available in the near term.

Zhang, Zhuang, and co-workers chose to explore the quantum jungle by following a path that runs at the border between quantum sensing and quantum machine learning. The beautiful “quantum fruits” they discovered along the way show that this path deserves further exploration.

## References

1. Y. Xia et al., “Quantum-enhanced data classification with a variational entangled sensor network,” Phys. Rev. X 11, 021047 (2021).
2. Y. Cao et al., “Quantum chemistry in the age of quantum computing,” Chem. Rev. 119, 10856 (2019).
3. S. McArdle et al., “Quantum computational chemistry,” arXiv:1808.10402.
4. V. Havlíček et al., “Supervised learning with quantum-enhanced feature spaces,” Nature 567, 209 (2019).
5. J. Preskill, “Quantum Computing in the NISQ era and beyond,” Quantum 2, 79 (2018).
6. C. W. Chou et al., “Frequency comparison of two high-accuracy
${\text{Al}}^{+}$

optical clocks,” Phys. Rev. Lett. 104, 070802 (2010).

7. I. Chiorescu et al., “Magnetic strong coupling in a spin-photon system and transition to classical regime,” Phys. Rev. B 82, 024413 (2010); D. I. Schuster et al., “High-cooperativity coupling of electron-spin ensembles to superconducting cavities,” Phys. Rev. Lett. 105, 140501 (2010); Y. Kubo et al., “Strong coupling of a spin ensemble to a superconducting resonator,” 105, 140502 (2010); H. Wu et al., “Storage of multiple coherent microwave excitations in an electron spin ensemble,” 105, 140503 (2010).


text:
# Search for direct pair production of supersymmetric top quarks decaying to all-hadronic final states in pp collisions at √s = 8 TeV

CMS Collaboration; Khachatryan, V.; Sirunyan, A. M.; Tumasyan, A.; Adam, W.; Asilar, E.; Bergauer, T.; Brandstetter, J.; Brondolin, E.; Dragicevic, M.; Erö, J.; Flechl, M.; Friedl, M.; Frühwirth, R.; Ghete, V. M.; Hartl, C.; Hörmann, N.; Hrubec, J.; Jeitler, M.; Knünz, V.; König, A.; ... mehr

Abstract (englisch):
Results are reported from a search for the pair production of top squarks, the supersymmetric partners of top quarks, in final states with jets and missing transverse momentum. The data sample used in this search was collected by the CMS detector and corresponds to an integrated luminosity of 18.9 fb-1 of proton-proton collisions at a centre-of-mass energy of 8 TeV produced by the LHC. The search features novel background suppression and prediction methods, including a dedicated top quark pair reconstruction algorithm. The data are found to be in agreement with the predicted backgrounds. Exclusion limits are set in simplified supersymmetry models with the top squark decaying to jets and an undetected neutralino, either through a top quark or through a bottom quark and chargino. Models with the top squark decaying via a top quark are excluded for top squark masses up to 755 GeV in the case of neutralino masses below 200 GeV . For decays via a chargino, top squark masses up to 620 GeV are excluded, depending on the masses of the chargino and neutralino.

 Zugehörige Institution(en) am KIT Institut für Experimentelle Kernphysik (IEKP) Publikationstyp Zeitschriftenaufsatz Jahr 2016 Sprache Englisch Identifikator ISSN: 1434-6044KITopen ID: 1000059583 Erschienen in The European physical journal / C Band 76 Heft 8 Seiten 460 Bemerkung zur Veröffentlichung Gefördert durch SCOAP3
KIT – Die Forschungsuniversität in der Helmholtz-Gemeinschaft KITopen Landing Page


text:
# Hexagonal Numbers

Probability Level 2

In the diagram above, we see a series of triangles and hexagons. How many dots are contained in the $20^\text{th}$ hexagon?

×


text:
# Lane-line Detector#

## Overview top#

The lane-line detector tool can used to prepare lane-line data for the segmentation and lane-line sensors. Depending on the desired result, it can use data from map annotations and/or road intensity maps.

If your annotation data is precise and aligns with the environment well, it's suggested to use it as the only input for this tool. On the other hand, if environment and annotations are slightly misaligned, you might want to try using the optional feature that will attempt to detect lines on intensity maps and correct the annotations based on this. Refer to the main options section for more details.

The main output from this tool is a set of meshes matching lane-lines on the road. These meshes are not visible for normal cameras, but will be used by the segmentation sensor to differentiate between road and lane-lines. If Generate Line Sensor Data in main options is enabled, output will also include correction data for the lane-line sensor. You can find more details about this on lane-line sensor page.

If line visibility for segmentation camera is unnecessary or undesired, you can safely ignore this tool. Comparison of segmentation image with and without lane-lines is shown below.

## Accessing the Detector top#

To open the lane-line detector window, open a Simulator project in Unity editor, then select Simulator/Detect Lane-lines on the menu bar.

## Main Options top#

There are two main options that must be set before any other, more detailed, settings are available.

The Line Source dropdown describes which data should be used to generate lines that will be visible by segmentation sensor. Three options are available: - HD Map - only annotation data will be used and all generated lines will follow lines from the high-definition (HD) map. - Intensity Map - only road intensity maps will be used. This method is based on image processing and results may vary depending on the quality of the intensity maps. - Corrected HD Map - annotation data will be used, but the tool will attempt to correct any misaligned lines based on road intensity maps.

The Generate Line Sensor Data checkbox determines whether the annotations should be corrected for the lane-line sensor. Correction is based on road intensity maps, similar to the Corrected HD Map option in the Line Source setting. If this option is disabled, raw annotation data will be used instead. Note that this setting is separate to Line Source, and both sensors can use differently processed data.

After generation is done, a new object named LineData_generated will be created on the scene. When it is selected, the scene view will show outlines of the generated lines. If you're not satisfied with the result, try using a different mode, or changing the advanced settings.

## Line-detection Settings top#

This section is only visible if any of the main options was configured to use line-detection based on intensity maps.

### Grouping Settings top#

Parameter Name Description
Line Distance Threshold The maximum distance between line segments that can be classified as a single lane-line when classifying lines detected on intensity maps.
Line Angle Threshold The maximum angle between line segments that can be classified as a single lane-line when classifying lines detected on intensity maps.
Max Line Segment Length The maximum length of a single detected line segment. Longer lines will be split into multiple segments.
Worst Fit Threshold The maximum valid distance between a line center and the segments creating it. This is used to filter out large clusters of spread-parallel line-segments.
Min Width Threshold The maximum distance between segments that should be considered parts of a single line. This is used to filter out large clusters of spread-parallel line-segments.
Joint Line Threshold The minimum viable width of a lane-line. This is used to filter out linear artifacts from detected line-segments.

### Postprocessing Settings top#

Parameter Name Description
World Space Snap Distance The maximum distance between detected lines that could be considered parts of a single curve. Lines below this threshold will have their ends snap together to create a better curve approximation.
World Dotted Line Distance The maximum distance between detected lines that could be considered separate parts of a single dotted line.
World Space Snap Angle The maximum angle between detected lines that could be considered parts of a single curve (solid or dotted).

## Line Mesh Generation Settings top#

This section is only visible if Line Source, in the main options, was configured to use map annotations.

Parameter Name Description
Snap Lane Ends If this option is enabled, end vertices from subsequent lanes will be snapped together, within the Snap Threshold, to remove holes.
Snap Threshold The maximum distance (in meters) within which lane ends can be snapped together. This is only available if Snap Lane Ends is enabled.
Line UV Unit The distance, in meters, defining the size of a single UV coordinates-tile for lane-lines. Increasing this value will stretch the line texture over a longer distance.
Line Width The width, in meters, of a single lane-line. Double lines will be scaled proportionally.
Line Bump The vertical distance, in meters, between road and lane-lines. This elevates lines slightly to avoid clipping.


text:
# Draw reversed path in TikZ

Is there a way to draw a TikZ path reversed?

I am trying to create some code that generates tessellating figures (to be used in How do we Draw a Bird in LaTeX).

Here is what I got:

\documentclass{minimal}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\def\x{2}
\def\y{4}
\def\mypath{(0,0) -- (1,1) -- (\x,0) -- (\x,\y)}
\draw \mypath;
\draw[shift={(\x,\y)}, rotate=180, blue] \mypath;
\end{tikzpicture}
\end{document}

It generates this: But what I wanted is this:. So it is obviously not correct to rotate the path. I need to break it into two an draw it reversed.

-
If you just put the URL of a question in your post, it'll be converted to the question title automatically, as can be seen in my edit. – doncherry Mar 17 '12 at 22:55
The shape is already defined in the shape library with the name signal. Check the manual page 439. – percusse Mar 17 '12 at 23:43
@percusse: I am trying to generate tessellating figures. I do not want to draw this specific shape. – Hans-Peter E. Kristiansen Mar 17 '12 at 23:58
I don't see how "drawing it reversed" (whatever that actually means) gives you the result you want. You seem to want to get a path that is reflected in x everywhere, and reflected in y half of the time. Perhaps you should draw a different part of the path and reflect that (since you want the reflection over the x axis everywhere, I suggest drawing either the left or right hand side and mirroring that). Perhaps I am just misunderstanding you, it's not terribly clear. – Roelof Spijker Mar 18 '12 at 0:35
Yes, but they are not symmetric (a slight smell of group theory, yuck!). In other words they don't fit together to form a shape that you want. – percusse Mar 18 '12 at 0:43

It's not possible with the definition of \mypath

\documentclass{minimal}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
\def\x{2}
\def\y{4}
\def\mypath{(0,0) -- (1,1) -- (\x,0) -- (\x,\y)}
\draw[red] \mypath;
\end{tikzpicture}
\hspace{1cm}
\begin{tikzpicture}
\def\x{2}
\def\y{4}
\def\mypath{(0,0) -- (1,1) -- (\x,0) -- (\x,\y)}
\draw[rotate=180, blue] \mypath;
\end{tikzpicture}
\end{document}

There is no shift possible to complete correctly your shape

but with

\documentclass{minimal}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
\def\x{2}
\def\y{4}
\def\mypath{((1,1) -- (\x,0) -- (\x,\y)--++(-1,1)}
\draw[red] \mypath;
\draw[blue,x=-0.5*\x cm,xshift=\x cm]  \mypath;
\end{tikzpicture}

\end{document}

-

A solution to what you are trying to attempt is as follows :

1. draw path forward
2. shift
3. draw path backwards
4. close path

To make this easier, don't specify the path by actual points, but rather by displacements. The code is

\documentclass[border=5pt]{standalone}

\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}

\coordinate (start) at (0,0);
\def\shape{{(1,1)},{(1,-1)}}
\coordinate (shift) at (0,4);

\foreach \point [count=\n] in \shape {
\node[coordinate] (d-\n) at \point {};
}
\pgfmathtruncatemacro{\size}{\n}

\draw (start)
\foreach \i in {1,...,\size} {-- ++(d-\i)} %draw forward
-- ++(shift) %shift
\foreach \i in {\size,...,1} {-- ++($-1*(d-\i)$)} % draw backward
-- cycle; %close

\end{tikzpicture}
\end{document}

The result is

-


text:
Proof of convexity with logs

I'm trying to prove the following result. Let $$p$$ be an input probability distribution and $$Q$$ be a transition matrix. $$q = Qp$$ is a valid output probability distribution. The components of $$p$$ and $$q$$ are given as $$p_i$$ and $$q_i$$ respectively.

For any $$p, p'$$, define

$$J(p, p') = -D(p||p') + \sum_ip_i\left(\sum_{j}Q_{ij}\log(Q_{ij})\right) - \sum_{j}q_j\log q'_j,$$

where $$q'_j = \sum_k Q_{jk}p'_k$$ and $$D(p||p') = \sum_i p_i\log \frac{p_i}{p'_i}$$, the divergence.

The claims are the following

1) $$J$$ is concave in both $$p$$ and $$p'$$. I can see this for $$p$$ quite easily and numerically, it seems to be true for $$p'$$ but how can I prove it? Somehow, I need to combine the concave term $$-D(p||p')$$ and the convex $$-\log$$ term together to make a concave function but I don't know how.

2) $$J(p, p') \leq I$$, where $$I$$ is the mutual information between probability distributions $$p$$ and $$q$$, maximized over all possible input distributions $$p$$. It is given by

$$I = \max_{p}\left[\sum_ip_i\left(\sum_{j}Q_{ij}\log(Q_{ij})\right) - \sum_{j}q_j\log q_j\right]$$

Again, I have verified this numerically over random choices of $$Q$$ but have no proof.

EDIT: For the concavity problem, it follows from what I posted before but is easier to analyze. We require the function

$$\sum_i p(i)\log p'(i) - \sum_j q_j\log q'(j)$$

is concave in $$p'$$ for any choice of stochastic matrix $$Q$$ that gives us $$q_k = \sum_j Q_{kl}p_l$$ and $$q'_k = \sum_j Q_{kl}p'_l$$.

• What are $\lambda$ and $\lambda’$? – David M. Mar 8 at 13:29
• @DavidM. sorry, my mistake with the notation. It was p and p' – user1936752 Mar 8 at 13:32

First, for convenience rewrite $$J(p,p')$$ as \begin{align} J(p,p') &= -D(p\Vert p')+\sum_{i}p_{i}\sum_{j}Q_{ij}\log\frac{Q_{ij}}{q_{j}}+\sum_{j}q_{j}\log\frac{q_{j}}{q'_{j}} \\ &= I({p})-\left[D(p\Vert p')-D(q\Vert q')\right]\\ & = I(p) - \Delta D(p\Vert p') \end{align} where I have used your definition of $$I(p)$$, and defined \begin{align} \Delta D(p\Vert p')= D(p\Vert p')-D(q\Vert q') \end{align}

$$\Delta D(p\Vert p')$$ reflects the "contraction" of Kullback-Leibler divergence between $$p$$ and $$p'$$ under the stochastic matrix $$Q$$. $$\Delta D$$ is convex in the first argument. To see why, consider any two distributions $$p$$ and $$\bar{p}$$, and define the convex mixture $$p^\alpha = (1-\alpha) p + \alpha \bar{p}$$. We will show convexity by demonstrating that the second derivative with respect to $$\alpha$$ is non-negative.

First, compute the first derivative w.r.t. $$\alpha$$ as \begin{align*} & {\textstyle \frac{d}{d\alpha}}\Delta D(p^{\alpha}\Vert q) ={\textstyle \frac{d}{d\alpha}}\left[\sum_{i}p_{i}^{\alpha}\log p_{i}^{\alpha}-\sum_{i}p_{i}^{\alpha}\sum_{j}Q_{ij}\log q_{j}^{\alpha}+\sum_{i}p_{i}^{\alpha}\sum_{j}Q_{ij}\log\frac{q_{j}'}{p'_{i}}\right]\\ & =\sum_{i}(\bar{p}_{i}-p_{i})\log p_{i}^{\alpha}-\sum_{i}(\bar{p}_{i}-p_{i})\sum_{j}Q_{ij}\log q_{j}^{\alpha}+\sum_{i}(\bar{p}_{i}-p_{i})\sum_{j}Q_{ij}\log\frac{q_{j}'}{p'_{i}} \end{align*} Then, we compute the second derivative at $$\alpha=0$$ as \begin{align} {\textstyle \frac{d^{2}}{d\alpha^{2}}}\Delta D(p^{\alpha}\Vert q)\vert_{\alpha=0}=\sum_{i}\frac{\left(\bar{p}_{i}-p_{i}\right)^{2}}{p_{i}}-\sum_{j}\frac{\left(\bar{q}_{j}-q_{j}\right)^{2}}{q_{j}} \end{align} $$\sum_{i}\frac{\left(\bar{p}_{i}-p_{i}\right)^{2}}{p_{i}}$$ is the so-called $$\chi^2$$ divergence from $$p$$ to $$\bar{p}$$, and $$\sum_{j}\frac{\left(\bar{q}_{j}-q_{j}\right)^{2}}{q_{j}}$$ is the same once $$Q$$ is applied to $$p$$ and $$\bar{p}$$. Note that $$\chi^2$$ divergence is a special case of a $$f$$-divergence, and therefore obeys a data-processing inequality (see e.g. Liese and Vajda, IEEE Trans on Info Theory, 2006, Thm. 14). In particular, that means that $${\textstyle \frac{d^{2}}{d\alpha^{2}}}\Delta D(p^{\alpha}\Vert q)\vert_{\alpha=0} \ge 0$$.

At the same time, $$\Delta D(p\Vert p')$$ is not convex in the second argument. Consider $$p = (0.5,0.5,0)$$, $$q=(0.5,0.25,0.25)$$, and $$Q = \left( \begin{smallmatrix} 0.95 & 0.05 \\ 1 & 0 \\ 0 & 1\end{smallmatrix} \right)$$. Here is a plot of $$\Delta D$$ where the first argument is $$p$$ and the second argument is a convex mixture of between $$p$$ and $$q$$, $$\alpha p + (1-\alpha) q$$, for different $$\alpha$$:

(see code at https://pastebin.com/q8XLnGK8)

By visual inspection, it can be verified $$\Delta D(p\Vert p')$$ is neither convex nor concave in the second argument.

(1) $$I({p})$$ is known to be concave in $$p$$ (Theorem 2.7.4 in Cover and Thomas, 2006). As we've shown $$\Delta D(p\Vert p')$$ is convex in $$p$$, so $$-\Delta D(p\Vert p')$$ is concave in $$p$$. Since the sum of concave functions is concave, $$J(p,p') = I(p) - \Delta D(p\Vert p')$$ is concave in $$p$$.
At the same time, as a function of the second argument $$p'$$, $$J(p,p') = \mathrm{const} - \Delta D(p\Vert p')$$, and we've shown above that $$\Delta D(p\Vert p')$$ is neither convex nor concave in the second argument. Thus, $$J(p,p')$$ is not concave in the second argument.
(2) $$\Delta D(p\Vert p') \ge 0$$ by the data processing inequality for KL divergence (Csiszar and Körner, 2011, Lemma 3.11). That means that \begin{align} J(p,p') = I(p) - \Delta D(p\Vert p') \le I(p) , \end{align} from which $$J(p,p') \le \max_s I(s)$$ follows immediately.
• Great answer but regarding the convexity proof, I still have a question. You've used $\alpha > 1$ in your plot but the second argument of $\Delta D(p, p')$ is still a probability distribution so you'd have to restrict $\alpha$ between 0 and 1, no? – user1936752 Mar 10 at 19:35
• Sorry for the confusion -- the plot was not labelled correctly, but the x-axis did actually span $\alpha \in [0,1]$. I also fixed a typo in the specification of $Q$ and added a link to pastebin. – Artemy Mar 11 at 0:50
• Regarding the counterexample: I first tried to show convexity in the second argument, and computed some of the second derivatives of $\Delta D(p\Vert p')$ with respect to $p_i'$. I got something like $p_i/({p_i'}^2)-\sum_{j}q_{j}Q_{ij}Q_{ij}/({q_j'}^{2})$. – Artemy Mar 11 at 18:42
• I then tried to "make these negative" by coming up with a matrix $Q$, $p$, and $p'$ such that there is some $i,j$ such that $Q_{ij}$ is large, $p_i$ is smaller than $p_i'$ and $q_j$ is larger than $q_j'$. This is pretty hand-wavy and not at all rigorous, but it led me to values that are close to the ones shown in the plot. – Artemy Mar 11 at 18:42


text:
# Jump it up Solution | CodeChef [Easy]

Problem

Three dinos go on a camping trip. There they decide to play a game. They mark distances of 1 km each on the ground up to 100 km i.e. 1, 2, 3…100. They can sit only on these 100 spots, with each one at a different spot. In one move, one of the outer dinos can hop into a spot between the other two. Help them maximise their duration of play.

#### Input

Three numbers(integers), l, m and n (0 < l < m < n < 100), the starting spots of the dinosaurs.

#### Output

Output the largest number of jumps the dinosaurs can make in all.

#### Constraints

• 0 < l < m < n <  100

#### Sample Input

5 6 11

#### Sample Output

4

#### Solution

Greedy Approach: Choose the maximum distance between the dinosaurs among 1st & 2nd and 2nd & 3rd.

Code Implementation

//
//  main.cpp
//  Jump it up
//
//  Created by Himanshu on 16/04/22.
//

#include <iostream>
using namespace std;

int main() {
int a, b, c;
cin>>a>>b>>c;

if ((b-a)>(c-b)) {
cout<<(b-a-1)<<endl;
} else {
cout<<(c-b-1)<<endl;
}

return 0;
}

Time Complexity: O(1)


text:
Volume 352 - XXVII International Workshop on Deep-Inelastic Scattering and Related Subjects (DIS2019) - Working Group 2: Low-x and Diffraction
Transverse Spin Asymmetries in the $p^{\uparrow}p \to p\pi^{0} X$ Process at STAR
C. Dilks* on behalf of the STAR collaboration
*corresponding author
Full text: pdf
Pre-published on: 2019 July 19
Published on: 2019 October 04
Abstract
A significant sample of $p^{\uparrow}p\to p\pi^0 X$ events has been observed at STAR
in $\sqrt{s}=200$ GeV transversely polarized $pp$ collisions, where an isolated
$\pi^0$ is detected in the forward pseudorapidity range $2.65<\eta<3.9$ along
with the forward-going proton $p$, which scatters with a near-beam forward
pseudorapidity into Roman Pot detectors. The sum of the $\pi^0$ and the
scattered proton energies is consistent with the incident proton energy of 100
GeV, indicating that no further particles are produced in this direction. It is
postulated that the forward incident proton may have fluctuated into a $p+\pi^0$
system, with an angular momentum correlated with the initial proton spin. The
backward-going proton interacts with the $p+\pi^0$ system, which then separates
such that the $\pi^0$ has a transverse momentum of ${\sim}2$ GeV/$c$ and the
proton has a transverse momentum of ${\sim}0.2$ GeV/$c$, while the backward
proton shatters into the remaining particles $X$. Correlations between the
$\pi^0$ and scattered proton will be presented, along with single-spin
asymmetries which depend on the azimuthal angles of both the pion and the
proton. This is the first time that spin asymmetries have been explored for
this process, and a model to explain their azimuthal dependence is needed.
DOI: https://doi.org/10.22323/1.352.0050
Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.


text:
## AI 生成解读视频

AI抽取解析论文重点内容自动生成视频

AI解析本论文相关学术脉络

# Transverse spin diffusion in ferromagnets

Physical Review B, no. 9 (2009)

We discuss the dissipative diffusion-type term of the form $\mathbf{m}\times\nabla^2\partial_t\mathbf{m}$ in the phenomenological Landau-Lifshitz equation of ferromagnetic precession, which describes enhanced Gilbert damping of finite-momentum spin waves. This term arises physically from itinerant-electron spin flows through a perturbed f...更多

0


text:
# Diagnosing the Steric effect in sea surface height

Changes in steric sea level are caused when changes in the density of the water column imply an expansion or contraction of the column. It is essentially produced through surface heating/cooling and to a lesser extent through non-linear effects of the equation of state (cabbeling, thermobaricity...). Non-Boussinesq models contain all ocean effects within the ocean acting on the sea level. In particular, they include the steric effect. In contrast, Boussinesq models, such as NEMO, conserve volume, rather than mass, and so do not properly represent expansion or contraction. The steric effect is therefore not explicitely represented. This approximation does not represent a serious error with respect to the flow field calculated by the model [Greatbatch, 1994], but extra attention is required when investigating sea level, as steric changes are an important contribution to local changes in sea level on seasonal and climatic time scales. This is especially true for investigation into sea level rise due to global warming.

Fortunately, the steric contribution to the sea level consists of a spatially uniform component that can be diagnosed by considering the mass budget of the world ocean [Greatbatch, 1994]. In order to better understand how global mean sea level evolves and thus how the steric sea level can be diagnosed, we compare, in the following, the non-Boussinesq and Boussinesq cases.

Let denote the total mass of liquid seawater ( ), the total volume of seawater ( ), the total surface of the ocean ( ), the global mean seawater (in situ) density ( ), and the global mean sea level ( ).

A non-Boussinesq fluid conserves mass. It satisfies the following relations:

 (11.3)

Temporal changes in total mass is obtained from the density conservation equation :

 (11.4)

where is the in situ density, and emp the surface mass exchanges with the other media of the Earth system (atmosphere, sea-ice, land). Its global averaged leads to the total mass change

 (11.5)

where is the net mass flux through the ocean surface. Bringing (11.5) and the time derivative of (11.3) together leads to the evolution equation of the mean sea level

 (11.6)

The first term in equation (11.6) alters sea level by adding or subtracting mass from the ocean. The second term arises from temporal changes in the global mean density; from steric effects.

In a Boussinesq fluid, is replaced by in all the equation except when appears multiplied by the gravity ( in the hydrostatic balance of the primitive Equations). In particular, the mass conservation equation, (11.4), degenerates into the incompressibility equation:

 (11.7)

and the global average of this equation now gives the temporal change of the total volume,

 (11.8)

Only the volume is conserved, not mass, or, more precisely, the mass which is conserved is the Boussinesq mass, . The total volume (or equivalently the global mean sea level) is altered only by net volume fluxes across the ocean surface, not by changes in mean mass of the ocean: the steric effect is missing in a Boussinesq fluid.

Nevertheless, following [Greatbatch, 1994], the steric effect on the volume can be diagnosed by considering the mass budget of the ocean. The apparent changes in , mass of the ocean, which are not induced by surface mass flux must be compensated by a spatially uniform change in the mean sea level due to expansion/contraction of the ocean [Greatbatch, 1994]. In others words, the Boussinesq mass, , can be related to , the total mass of the ocean seen by the Boussinesq model, via the steric contribution to the sea level, , a spatially uniform variable, as follows:

 (11.9)

Any change in which cannot be explained by the net mass flux through the ocean surface is converted into a mean change in sea level. Introducing the total density anomaly, , where is the density anomaly used in NEMO (cf. §5.8.1) in (11.9) leads to a very simple form for the steric height:

 (11.10)

The above formulation of the steric height of a Boussinesq ocean requires four remarks. First, one can be tempted to define as the initial value of , set , so that the initial steric height is zero. We do not recommend that. Indeed, in this case depends on the initial state of the ocean. Since has a direct effect on the dynamics of the ocean (it appears in the pressure gradient term of the momentum equation) it is definitively not a good idea when inter-comparing experiments. We better recommend to fixe once for all to . This value is a sensible choice for the reference density used in a Boussinesq ocean climate model since, with the exception of only a small percentage of the ocean, density in the World Ocean varies by no more than 2 from this value (Gill [1982], page 47).

Second, we have assumed here that the total ocean surface, , does not change when the sea level is changing as it is the case in all global ocean GCMs (wetting and drying of grid point is not allowed).

Third, the discretisation of (11.10) depends on the type of free surface which is considered. In the non linear free surface case, key_ vvl defined, it is given by

 (11.11)

whereas in the linear free surface, the volume above the z=0 surface must be explicitly taken into account to better approximate the total ocean mass and thus the steric sea level:

 (11.12)

The fourth and last remark concerns the effective sea level and the presence of sea-ice. In the real ocean, sea ice (and snow above it) depresses the liquid seawater through its mass loading. This depression is a result of the mass of sea ice/snow system acting on the liquid ocean. There is, however, no dynamical effect associated with these depressions in the liquid ocean sea level, so that there are no associated ocean currents. Hence, the dynamically relevant sea level is the effective sea level, the sea level as if sea ice (and snow) were converted to liquid seawater [Campin et al., 2008]. However, in the current version of NEMO the sea-ice is levitating above the ocean without mass exchanges between ice and ocean. Therefore the model effective sea level is always given by , whether or not there is sea ice present.

In AR5 outputs, the thermosteric sea level is demanded. It is steric sea level due to changes in ocean density arising just from changes in temperature. It is given by:

 (11.13)

where and are the initial salinity and pressure, respectively.

Both steric and thermosteric sea level are computed in diaar5.F90 which needs the key_ diaar5 defined to be called.

Gurvan Madec and the NEMO Team
NEMO European Consortium2017-02-17


text:
# Math Help - Simple problem but no clue

1. ## Simple problem but no clue

For finding the shape of the hyperbola, trace the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

I have no clue what to do???

2. Originally Posted by TomJerry
For finding the shape of the hyperbola, trace the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

I have no clue what to do???
I suggest you Google hyperbola if you don't know how to draw a graph of this. Things to keep in mind include diagonal asymptotes and x-intercepts.

3. Originally Posted by TomJerry
For finding the shape of the hyperbola, trace the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

whereby

$c^2 = a^2 + b^2$ .

In your hyperbola, (h,k) = (0,0), so we have a start to this: The center is at the origin.

Note that if

$\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$

then the transverse axis would be vertical. However, the way we have it, with the x-term positive and the y-term negative, the traverse axis is horizontal


text:
# leetcode #62 in cpp

42人阅读 评论(0)

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Solution:

Suppose if w are currently in  a cell, we can only move right or down. Thus the number of paths from this cell to the destination is the sum the number of paths to destination from the cell right next to it,  and the number of paths to the destination from the cell blow it .

We use DP to solve this.

dp[i][j] = number of paths of the position(i+1, j + 1) to the destination

= dp[i+1][j] + d[[i][j+1]

The initialization involves the bottom row and rightmost column, both initialized to 1 since they can move in only 1 direction.

Code:

class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m, vector<int>(n,0));//dp[i][j] = number of paths to destination
dp[m-1][n-1] = 1;//destination to 1
for(int i = 0; i < m && n-1 >=0; i ++){// right border to 1
dp[i][n-1] = 1;
}
for(int i = 0; i < n && m-1 >= 0; i ++){//bottom border to 1
dp[m-1][i] = 1;
}
for(int i = m-2; i >=0; i --){
for(int j = n-2; j>=0; j --){
dp[i][j]+= dp[i][j+1] + dp[i+1][j];//current path number = num of its righr cell paths + num of its down cell paths to destination
}
}
return dp[0][0];
}
};

0
0

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text:
## November 16, 2006

### Segal on QFT

I spent a delightful afternoon, yesterday, discussing quantum field theory, the renormalization group, and such matters with Graeme Segal. Earlier, he gave a nice talk in the Geometry and String Theory seminar on his approach to QFT.

As you might expect, his formulation involves a Riemannian $d$-manifold, $X$, which is a cobordism between $d-1$ manifolds, $Y_1$ and $Y_2$. From that data, one imagines that a QFT is supposed to manufacture an isomorphism

(1)
$U_X:\, \mathcal{H}_{Y_1}\to \mathcal{H}_{Y_2}$

For a free, massive scalar field, $\mathcal{H}_{Y}= L^2_q\left(\Omega^0(Y)\right)$ the space of “square-integrable” functions on the space of smooth 0-forms on $Y$. I put a subscript “$q$” to remind us that “integration” over the topological vector space, $V= \Omega^0(Y)$, depends on a choice 1 of quadratic form $q\in\mathrm{Sym}^2 V^*$. This quadratic form, in turn, depends on the germ of a metric, $g$, on $X$, restricted to $Y$. $L^2_q\left(\Omega^0(Y)\right)$ is the quantization of the infinite-dimensional symplectic vector space, $\Sigma_Y = \Omega^0(Y)\oplus \Omega^{d-1}(Y)$ of “Cauchy data”, $(\phi,(*d\phi)|_Y)\in \Sigma_Y$.

Schematically, for $\Psi\in L^2_q\left(\Omega^0(Y_1)\right)$, $(U_X \Psi)(f_2) = \int_\substack{\phi:X\to \mathbb{R}\\ \phi_{Y_2}= f_2} e^{-S(\phi)} \Psi(\phi|_{Y_1}) \mathcal{D}\phi$

Graeme’s story is in Euclidean signature, but we can continue to complex metrics on $X$, so long as the quadratic form in $S(\phi)$ has positive definite imaginary part. The Minkowski theory arises (for $X=Y\times I$) at the boundary of this region in the space of complex metrics. There are several subtleties

• $L^2_q(V)$ is isomorphic to $L^2_{\tilde q}(V)$ if and only if $\tilde q q^{-1} -1$ is trace-class.
• Even then, the isomorphism is only projective. On one side, $\left\Vert e^{-\frac{1}{4}q}\right\Vert =1$; on the other, $\left\Vert e^{-\frac{1}{4}\tilde{q}}\right\Vert =1$. The above isomorphism gives the latter, only up to a factor of ${\det\left(\tfrac{\tilde{q}q^{-1}}{2\pi}\right)}^{-1/2}$
• To fix this, we set $\left\Vert e^{-\frac{1}{4}q}\right\Vert = {\text{det}_\zeta\left(\tfrac{q}{2\pi}\right)}^{-1/2}$ because the $\zeta$-function regulated determinant has the desired property $\det(\tilde{q}q^{-1}) \text{det}_\zeta(q) = \text{det}_\zeta(\tilde{q})$

All of the above is relatively standard stuff. Segal wants to go further, and study the notion of locality in this framework. To this end, he chops $Y^{d-1}$ into two pieces, $U_1$ and $U_2$, glued together on their common boundary, $Z^{d-2}$.

Naïvely, one is tempted to write

(2)
$\mathcal{H}_Y = \mathcal{H}_{U_1}\otimes \mathcal{H}_{U_2}$

with $\mathcal{H}_{U_i}= L^2(\Omega^0(U_i))$. But this is false2.

Let $\mathcal{G}_Y=\text{Heis}(\Sigma_Y)$ be the Heisenberg group, which is the central extension of $\Sigma_Y=V\oplus V^*$, $0 \to U(1) \to \text{Heis}(\Sigma_Y) \stackrel{\alpha}{\to} \Sigma_Y\to 0$ where $\alpha(g_\xi) = \xi$ and the group multiplication law is $g_{\xi_1}g_{\xi_2} = e^{i\omega(\xi_1,\xi_2)} g_{\xi_1+\xi_2}$ and $\omega(\cdot,\cdot)$ is the symplectic form on $\Sigma_Y$.

$\mathcal{H}_Y= L^2\left(\Omega^0(Y)\right)$ is an irreducible representation of $\mathcal{G}_Y$. It is highly reducible, as a representation of $\mathcal{G}_{U_1}\subset \mathcal{G}_Y$. Still the Reeh-Schlieder theorem tells us that $e^{-\frac{1}{4}q}$ is a cyclic vector for $\mathcal{G}_{U_1}$. In fact, $\mathcal{H}_Y$ is an irreducible representation of $\mathcal{G}_{U_1}\times \mathcal{G}_{U_2}\subset \mathcal{G}_Y$, but is not decomposable into (sums of) tensor products of irreps of $\mathcal{G}_{U_1}$ and $\mathcal{G}_{U_2}$. (Infinite dimensional groups are funny, that way…)

Still, there is something that makes sense, and replaces (2). It’s called the Connes Fusion Tensor Product. This was the punchline of the whole lecture and, alas, it went by too fast for me to give a coherent rendition of it. The idea is to let $V_1\cup V_2$ be a tubular neighbourhood of $Z$ (where $V_i\subset U_i$), and then let $\mathcal{H}_{L}= L^2\left(\Omega^0(U_1\cup V_2)\right)$ and $\mathcal{H}_{R}= L^2\left(\Omega^0(U_2\cup V_1)\right)$ be “thickened” versions of the Hilbert spaces we wrote before. The Connes Tensor product $\mathcal{H}_{L} \underset{\mathcal{G}_{V_1}\times \mathcal{G}_{V_2}}{*} \mathcal{H}_{R}$ can be completed to a Hilbert space isomorphic to $\mathcal{H}_Y$.

1 That is, we know how to do Gaussian integrals of the form $\int e^{- \frac{1}{2}q(v)} P(v) dv$ for some polynomial function, $P\in \mathrm{Sym}^n V^*$.

2 A pet peeve of mine is people who begin a discussion of the blackhole information paradox by drawing a spacelike slice ($Y$), dividing it into regions inside and outside the horizon ($U_1$, $U_2$), and writing $\mathcal{H}_Y = \mathcal{H}_{\text{in}}\otimes \mathcal{H}_{\text{out}}$.

Posted by distler at November 16, 2006 4:37 PM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1041

### Re: Segal on QFT

Re your pet peeve:

As we know by now, the cosmological constant is positive. Now, you turn the holographic crank and conclude (à la Banks) that the relevant Hilbert spaces are finite dimensional and all these subtleties go away! ;-)

Posted by: Robert on November 17, 2006 1:35 AM | Permalink | Reply to this

### À bas l’analyse fonctionnelle!

Now, you turn the holographic crank and conclude (à la Banks) that the relevant Hilbert spaces are finite dimensional and all these subtleties go away! ;-)

Well, yes, that would save me the trouble of having to really understand those accursed Type III factors, which are the source of all these difficulties!

Posted by: Jacques Distler on November 17, 2006 1:49 AM | Permalink | PGP Sig | Reply to this

### Re: Segal on QFT

I went to a great talk by Segal in Oz a few years ago, on Langlands. By the way, Lambda is ZERO. Come on, string theorists don’t want a lambda. Get rid of it.

Posted by: Kea on November 17, 2006 2:29 AM | Permalink | Reply to this

### Re: Segal on QFT

Hi Jacques,

Sorry to post a somewhat off-topic question, but you guys here probably have the answer I have been seeking for some time now:

Question:

What is “elliptic cohomology” exactly? And why is it considered so central to current theory physics research?

Thanks,
Brian

Posted by: Peter on November 17, 2006 4:12 AM | Permalink | Reply to this

### Elliptic Cohomology

What is “elliptic cohomology” exactly? And why is it considered so central to current theory physics research?

Who said it’s “central” to current theoretical physics research? Is that some weird “New Yorker map of the world” view of theoretical physics research?

Elliptic Cohomology is the biggest and baddest of the Generalized cohomology theories. A generalized cohomology theory satisfies all of the Eilenberg-Steenrod axioms, except for the dimension axiom: $\mathrm{H}^n(\text{point})= 0,\quad\forall n\gt 0$

Some generalized cohomology theories have a multiplicative structure, so that $\mathrm{H}^\bullet(X)$ forms a ring. Elliptic cohomology is one of those.

One characterization of elliptic cohomology is in terms of the formal group law satisfied by $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)$ where ${\mathbb{C}P}^\infty=BU(1)$ is the classifying space of $U(1)$ bundles. $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)$ is always a formal power series $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)= H^\bullet(\text{point})[[t]]$ with a single generator, $t$, in degree 2 and coefficients in the ring $H^\bullet(\text{point})$.

For ordinary cohomology, this satisfies the formal group law $F(x,y) = x+y$ (multiplication is the cup product). For K-theory, the formal group law $F(x,y) = x+y + x y$ (multiplication is the tensor product of line bundles).

For elliptic cohomology, the formal group law is defined by an elliptic curve.

I think Urs had a nice description of all of this here.

Posted by: Jacques Distler on November 17, 2006 8:25 AM | Permalink | PGP Sig | Reply to this
Read the post Categorical Trace and Sections of 2-Transport
Weblog: The n-Category Café
Excerpt: A general concept of extended QFT and its relation to the Kapranov-Ganter 2-character.
Tracked: November 17, 2006 10:43 AM

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text:
# Longest Common Subsequence

#### Algorithms Dynamic Programming (DP)

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Reading time: 30 minutes | Coding time: 10 minutes

A subsequence is any string formed by any collection of characters of the string based on their indices, like ogs is a subsequence of the string opengenus.We have presented an efficient way to find the longest common subsequence of two strings using dynamic programming.

Examples

str1 = ashutosh
str2 = amitesh
The largest common subsequence is "atsh" as it is present in both the strings.

str1 = opengenus
str2 = engineers
The largest common subsequence can either be "engns" or "enges" as the length of both the subsequences are equal i.e. 5.

# Approach

We have presented two approaches to find the longest common subsequences:-

• Naive Recursive approach in O(2n) time-complexity.
• Dynamic Programming approach in O(n * m) time-complexity.

# Recursive Approach

The naive solution for this problem is to generate all subsequences of both given sequences and find the longest matching subsequence. This solution is exponential in term of time complexity.Time complexity of the above naive recursive approach is O(2^n) in worst case and worst case happens when all characters of both the strings mismatch i.e., length of LCS is 0.

The pseudo-code for the recursive implementation is :-

function LCS(S, T)
if S is empty or T is empty then
return empty string

if first char of S == first char of T then
return (first char of S) + LCS(S - first char, T - first char)

otherwise // first chars are different
return longer of LCS(S - first char, T) and LCS(S, T - first char)

# Dynamic Programming Approach

In dynamic programming approach we store the values of longest common subsequence in a two dimentional array which reduces the time complexity to O(n * m) where n and m are the lengths of the strings.

Let the input sequences be X and Y of lengths m and n respectively. And let dp[n][m] be the length of LCS of the two sequences X and Y.

We iterate through a two dimentional loops of lengths n and m and
use the following algorithm to update the table dp[][]:-

• If any of the loop variable i or j is 0 , then dp[i][j] = 0.
• if X[i-1] = Y[j-1] ,i.e., when the characters at ith and jth index matches, dp[i][j] = 1 + dp[i-1][j-1].
• Otherwise, store the maximum value we get after considering either the charater X[i] or the character Y[j],i.e.,dp[i][j] = max(dp[i][j-1],dp[i-1][j]).

To retrieve the subsequence, follow a bottom-up approach.When the length of subsequence considering the Y[j] is greater than that considering X[i],decrement j, increment i when the length of subsequence considering the Y[j] is less than that considering X[i].Whenever the the length of subsequence considering the Y[j] is equal to that considering X[i],include the character in the answer.

For example:-

Consider the strings aabbc and abacc.Here length of both strings is 5.When we iterate through two nested loops where 0<=i<=5 and 0<=j<=5,When i=0 or j=0 ,dp[0][0..5] = 0 and d[0..5][0] = 0.When i=1 and j=1 ,since X[i-1] = Y[j-1] = 'a',therefore dp[1][1] = 1 + dp[0][0] = 1.

Since X[i-1] != X[j-1] for i=1 and j=2, therefore dp[1][2] = max(dp[0][2],dp[1][1]) = 1.

Similarly filling the table, the final table dp[][] would look like:-

a b a c c
0 0 0 0 0 0
a 0 1 1 1 1 1
a 0 1 1 2 2 2
b 0 1 2 2 2 2
b 0 1 2 2 2 2
c 0 1 2 2 3 3

To retrieve the longest common substring, let us create a character array ans[] of length ind equal to the length of longest common subseuence i.e. dp[5][5].So in our example ind=3.Let us initialize two variables i=5 and j=5(since length of both strings is 5).Let the string aabbc be a[] and abacc be b[].
Since a[i-1]=b[j-1],i.e, a[4]=b[4]='c', ans[ind-1]=a[i-1].So ans[2]=c and i,j and ind all are decremented.Now i=4,j=4,ind=2.

Now dp[i-1][j] = dp[3][4] = 2 and dp[i][j-1]=dp[4][3] = 2.Since dp[3][4] = dp[4][3],therefore, only j is decremented.Now i = 4,j = 3,ind=2.

Now dp[i-1][j] = dp[3][3] = 2 and dp[i][j-1] = dp[4][2] = 2.Since dp[3][3] = dp[4][2],therefore, only j is decremented.Now i = 4,j=2,ind=2.

Since a[i-1] = b[j-1],i.e, a[3] = b[1] = 'b', ans[ind-1] = a[i-1].So ans[1]=b and i,j and ind all are decremented.Now i=3,j=1,ind=1.

Now dp[i-1][j] = dp[2][1] = 1 and dp[i][j-1] = dp[3][0] = 0.Since dp[2][1]>dp[3][0],therefore, only i is decremented.Now i=2,j=1,ind=1.

Since a[i-1] = b[j-1],i.e, a[1] = b[0] = 'a', ans[ind-1]=a[i-1].So ans[0] = a and i,j and ind all are decremented.Now i = 2,j = 0,ind = 0.Since j=0,therefore the algorithm will stop.So, the longest common subsequence would be ans[]=abc.

Pseudo code of the above implementation:-

bottom-up-lcs() {
lcs = new array [0..m, 0..n]
for (i = 0 to m) lcs[i,0] = 0 // basis cases
for (j = 0 to n) lcs[0,j] = 0
for (i = 1 to m) { // fill rest of table
for (j = 1 to n) {
if (x[i] == y[j]) // take x[i] (= y[j]) for LCS
lcs[i,j] = lcs[i-1, j-1] + 1
else
lcs[i,j] = max(lcs[i-1, j], lcs[i, j-1])
}
}
return lcs[m, n] // final lcs length
}

Following is the cpp implementation of the above approach :-

#include<bits/stdc++.h>
using namespace std;
void lcs(string a,string b,int n,int m)
{
int dp[n+1][m+1];
// Building the table in bottom-up manner
for(i=0;i<n+1;i++)
{
for(int j=0j<m+1;j++)
{
if(i==0 || j==0)
dp[i][j]=0;
else if(a[i-1]==b[j-1])
dp[i][j]=1+dp[i-1][j-1];
else
{
dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
}
}
}

int ind=dp[n][m]; // length of longest common subsequence
char ans[ind+1]; // longest common subsequence
int i=n,j=m;
ans[ind]='\0';
while(i>0 && j>0)
{
if(a[i-1]==b[j-1])
{
ans[ind-1]=a[i-1];
i--;j--;ind--;
}
else if(dp[i-1][j]>dp[i][j-1])
i--;
else
{
j--;
}
}
cout << ans;
}
int main()
{
string a,b;
cin >> a >> b;
cout << "The largest common subsequence is " ;
lcs(a,b,a.length(),b.length());
}
/*
Sample Input:-
aabbc
abacc
Sample Output:-
The largest common subsequence is abc
*/

# Time Complexity

• Worst case time complexity: O(n*m)
• Average case time complexity: O(n*m)
• Best case time complexity: O(n*m)
• Space complexity: O(n*m)

Since we are using two for loops for both the strings ,therefore the time complexity of finding the longest common subsequence using dynamic programming approach is O(n * m) where n and m are the lengths of the strings.Since this implemetation involves only n rows and m columns for building dp[][],therefore, the space complexity would be O(n * m).

With this, you have complete idea of effectively solving the Longest Common Subsequence problem using Dynamic Programming.

#### Ashutosh Singh

Intern at OpenGenus | Pursuing B. Tech in Computer Science at Indian Institute of Information Technology (IIIT) Sricity

Improved & Reviewed by:


text:
# Link .lib file Windows Octave

I am trying to compile some code using Octave and I need to link some .lib files. I can't seem to get the link to work properly and when I compile I am getting errors about references to functions that don't exist. This is what lead me to believe that my links are not working.

this is the command I am running:

mex -v someFile.cpp -IC:\path\to\includes -LC:\path\to\libs -lmylib.lib

I am running Octave 3.2.4 in Windows.

Thanks.

-

mex -v -IC:\path\to\includes -LC:\path\to\libs someFile.cpp -lmylib.lib


text:
# $\sum \sqrt{a_n b_n}$ converges when $\sum a_n$ and $\sum b_n$ converge

I am to show that $\sum \sqrt{a_n b_n}$ converges when $\sum a_n$ and $\sum b_n$ converge (here the series are assumed to have non-negative terms). I am unsure how to approach this problem; since I don't know what the series would converge to, I tried using Cauchy's criterion. Hence my goal was to bound

$$\sum_{i=n+1}^{n+k} \sqrt{a_i b_i}$$

for some $n$ large enough and any $k \geq 1.$ I tried to express this in terms of the $a_i$ and $b_i$ separately (to use convergence of the series $\sum a_n$ and $\sum b_n$) by writing the above expression as

$$\frac{1}{2} \left( \sum_{i=n+1}^{n+k} (\sqrt{a_i}+\sqrt{b_i})^2 - \sum_{i=n+1}^{n+k} a_i - \sum_{i=n+1}^{n+k} b_i \right),$$

but I'm not sure if this really helps.

Thanks for any help.

• See this. – David Mitra Feb 26 '14 at 19:45
• If the terms are non-negative, the sequence of partial sums is monotonically nondecreasing. Thus it is convergent if and only if it is bounded. – Daniel Fischer Feb 26 '14 at 19:46
• @DavidMitra So I can bound each term by a term of a converging series... so simple! – user131708 Feb 26 '14 at 19:51
• @DanielFischer Does this help at all, if we don't think about using the AM-GM inequality? Or did you have something else in mind? – user131708 Feb 26 '14 at 19:53
• @user131708 The next step would be AM-GM. The point is that you don't need to bother with the Cauchy criterion and $\sum\limits_{i=n+1}^{n+k}$ once you know that $\sqrt{a_nb_n} \leqslant \frac12(a_n+b_n)$ and the sum of the larger terms is finite. – Daniel Fischer Feb 26 '14 at 20:00

$$ab\le\frac12(a^2+b^2)$$