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Vectrex games that *should* come out...
japjac
Battlezoned
Vectrex games that *should* come out... Jan 2, 2018 16:38:38 GMT -5
Post by japjac on Jan 2, 2018 16:38:38 GMT -5
A Vectrex version of Nintendo's Sheriff would be rather spiffing. I played the original upright cab in Natsuge and I was quite smitten by it.
Post by VectorX on Jan 2, 2018 18:21:56 GMT -5
Bandido looks similar to it.
Jan 2, 2018 16:38:38 GMT -5 japjac said:
I played the original upright cab in Natsuge and I was quite smitten by it.
I can imagine, as Bandido looked cool. I only saw it at one game room ever and wished I had played it later.
EDIT: I just looked up the games and Nintendo licensed Bandido to Exidy after Sheriff came out.
Last Edit: Jan 2, 2018 20:54:40 GMT -5 by VectorX
50tbrd
Causing Major Havoc
Tony Holcomb - Velocity Design Concepts
Vectrex games that *should* come out... Jan 3, 2018 0:00:09 GMT -5
Post by 50tbrd on Jan 3, 2018 0:00:09 GMT -5
Microvision Collection
hcmffm
Wanted: Vectorcade
Post by hcmffm on Jan 3, 2018 5:43:39 GMT -5
Jan 3, 2018 0:00:09 GMT -5 50tbrd said:
I'm not familiar with Microvision Games but assume that these games are quite simple. Luchssoft has ported various LCD games to the Vectrex and I must say that I think that they are a bit too simple (especially for a cartridge release). Though, a (comparatively) simple gameplay with well done graphics and music and some variety (levels, ...) is also fine, Vectrexagon or Big Blue are good examples.
Just my 2 cents...
Last Edit: Jan 3, 2018 5:48:46 GMT -5 by hcmffm
ledzep
Vectrex games that *should* come out... Jan 10, 2018 6:28:32 GMT -5
Post by ledzep on Jan 10, 2018 6:28:32 GMT -5
Not sure if it's been mentioned or not, but Stern's "Moon War" should be on the Vectrex, it is practically a vector game already -
www.youtube.com/watch?v=UcNBkTdEmx8
It also has one of the best arcade flyers ever made -
flyers.arcade-museum.com/flyers_video/stern/44135901.jpg
Vectrex games that *should* come out... Jan 10, 2018 10:07:41 GMT -5
Post by VectorX on Jan 10, 2018 10:07:41 GMT -5
Jan 10, 2018 6:28:32 GMT -5 ledzep said:
Might not be considered work safe
Interesting, I guess that game didn't get anywhere, as I never saw it. Wonder how the gameplay is, rather than it being just for the voice synthesis.
Jan 10, 2018 10:07:41 GMT -5 VectorX said:
Well, depends on where you work, I guess, right?
Ya, I recently went on a binge hunt of every PDF I could find of arcade game manuals for old favorite games that I have or might try to get in the future. You know, like owners manuals for Red Baron, Xevious, Asteroids Deluxe, etc., along with schematics and whatever (lots of good info antelopearcade.com/files/?dir=Arcade%20Manuals ). Moon War is one of the few games that was from that era from a known company that I couldn't even get a whiff of a scanned manual. Odd.
I certainly never saw that game in an arcade and I would have played it seeing as how it seemed like a vector game and had some cool Stern sounds (very Berzerk-sounding). But I just looked at the entry at the KLOV and it shows a different style of game -
www.arcade-museum.com/game_detail.php?game_id=8752
so now I don't know what that video is showing. Another version of that same video labels it as a prototype but the KLOV description makes it sound like it was a regular game.
Oh, wait, here's a video that matches the images from the flyer (the game, not the girl, haahaha) -
www.youtube.com/watch?v=yYb-nlOTmlY
The description says it plays like a combination of Asteroids and Defender but to me it seems more like Star Trek and Defender because of the overhead view and having to give a sh*t about bases being attacked (sort of like Star Raiders, in that sense, too). Even though this is a different game, it still would work on the Vectrex, though I think it's got some oddball controls.
Post by VectorX on Jan 11, 2018 9:32:06 GMT -5
I certainly never saw that game in an arcade and I would have played it seeing as how it seemed like a vector game and had some cool Stern sounds (very Berzerk-sounding).
But I just looked at the entry at the KLOV and it shows a different style of game -
It still could be right; gameplay could've really changed in between the original prototype and its release. Or maybe it had a name change, as there were a lot of games out there with the word "moon" in them. Who knows...
Drat!
^Looks cool either way. And it'd make use of the VecVox if someone were to do whatever the hell game was in the first video you linked to. Hell, maybe TWO games could come out of this!
Vectrex games that *should* come out... May 1, 2018 12:28:37 GMT -5 via mobile
Post by japjac on May 1, 2018 12:28:37 GMT -5
I'm a bit obseesed with Pandora's Palace (pi$$es on DK)
at the moment after having played it in Natsuge for the first time quite recently. Ambitious but Becky's Message was possible after all.
Also Nectaris for the Vectrex would just be insanely mint.
Vectrex games that *should* come out... May 1, 2018 13:56:36 GMT -5
Post by VectorX on May 1, 2018 13:56:36 GMT -5
May 1, 2018 12:28:37 GMT -5 japjac said:
after having played it in Natsuge
Had to look that up to see what that was. Looks like a good Tokyo gaming museum.
Developing "Outlaw Hill" / "Frontier"
Vectrex games that *should* come out... May 11, 2018 6:00:18 GMT -5 TrekMD likes this
Post by cNp on May 11, 2018 6:00:18 GMT -5
Jan 2, 2018 18:21:56 GMT -5 VectorX said:
I missed this discussion but I own a "Taito" Sheriff and love the game... Bandido is the same game as is Western Gun Pt 2. The one player element of my western themed game Frontier will take Sheriff as the staring inspiration
Some details in an interview I just did: www.homebrewheroes.co.uk/homebrew-heroes-interview-vector-republic/
And a short vid of an early build...
youtu.be/TKEFV0NR-M0
Last Edit: May 11, 2018 6:06:22 GMT -5 by cNp
Vectrex games that *should* come out... May 11, 2018 9:06:27 GMT -5 cNp likes this
Post by VectorX on May 11, 2018 9:06:27 GMT -5
I was *thinking* it might be a little difficult to pull off/some compromises might have to be made with a lot of vectors onscreen but it looks like it might not be too much of a problem
Vectrex games that *should* come out... Jan 9, 2019 13:23:26 GMT -5 marblemad and SiLic0ne t0aD like this
Not a *game*, but perhaps an accessory here...
I got Stramash Zone out of my personal box yesterday. I was really blown away by how it looks like you've got a miniature version of Battlezone sitting right in front of you with that overlay, with the corners having the orange/reddish colors on them!
So, not the Battlezone clone that most of us have wanted since the 80s, but I'd say a Stramash Zone Vectrex Wrap would be killer
Vectrex games that *should* come out... Dec 1, 2019 22:28:51 GMT -5
Post by VectorX on Dec 1, 2019 22:28:51 GMT -5
There's already a crapload of shooters on the Vectrex, but hear me out on this one...
When I went to my local gaming expo the other week, on a Pandora's Box I was thumbing through tons of ROMs when I ran across The End, which, even though it wasn't very good, I hadn't played it in years and fired it up. Stern had a nice try in trying to make the Space Invaders/Galaxian formula have a nice twist to it by having blocks on the screen being scooped up by aliens or something and once they spelled out "END" the game was, indeed, over (if you hadn't lost all of your ships before then).
The only problem was, even though the game was divided into waves and one of them had a certain name to it (I forget what it was, but it sounded like there was something different about the round, yet it seemed to be the same as all the others), it just seemed endless, as nothing really changed (even though you kept on *expecting* something to change). I give it a 5 out of 10.
Then I saw on High Score where someone made a homebrew port out of it for the Atari 2600. The enemy sprites are a bit bigger and the aliens will sometimes pause before they dive down for a block, so it looks like they improved it a little.
If they gave you different things to do with every wave (or every few waves) it could actually make a decent game rather than the POS they originally came up with. There's a spaceship at the top of the screen that you can't destroy...unless you have a bonus wave where it's just them darting about or something. Have power-ups or something to make the blocks disappear occasionally so it won't "end" the game as quickly...SOMETHING.
TrekMD
Official OUYA rep
Vectrex games that *should* come out... Dec 2, 2019 7:18:55 GMT -5
Post by TrekMD on Dec 2, 2019 7:18:55 GMT -5
Oh yes, The End. It could work on the Vectrex. There was another shooter I saw recently that could be cone on the Vectrex as well. I'm trying to remember the name of it!
Going to the final frontier...gaming!


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Trump Teen Rape Allegation Resurfaces, Ronan Farrow Claims National Enquirer Tried to Protect Him in New Book
By Tufayel Ahmed On 10/16/19 at 1:46 PM EDT
U.S. Donald Trump Jeffrey Epstein
Amid behind-the-scenes details of his explosive reporting on Harvey Weinstein's alleged sexual misconduct, Ronan Farrow's new book Catch and Kill: Lies, Spies and a Conspiracy to Protect Predators claims that American Media Inc. (AMI), the former publisher of the National Enquirer, tried to help Donald Trump bury allegations he raped a teenager in 1994.
Farrow claims in the book that AMI CEO David Pecker was in close contact with Trump when the rape allegations were made public in a 2016 lawsuit, and that then-National Enquirer editor Dylan Howard assured Trump's personal lawyer at the time, Michael Cohen, "that they would track down the woman with the rape allegation and see what they could do about her." An AMI spokesperson told Newsweek Tuesday that Farrow's claims are "completely untrue."
AMI has been accused in the past of purchasing potentially damaging stories about Trump in order to keep them from becoming public—a practice known in the industry as "catch and kill." Last December, AMI admitted that in 2016 it made a $150,000 payment "in concert" with Trump's election campaign to former Playboy model Karen McDougal, who allegedly had an affair with Trump a decade earlier. The publisher made the pact with McDougal "in order to ensure that the woman did not publicize damaging allegations about the candidate before the 2016 presidential election," the U.S. Attorney's Office said.
At the crux of the lawsuit filed against Trump in the lead-up to the 2016 presidential election were claims that Trump raped a woman when she was 13 years old in 1994. At the time, Trump's lawyer Alan Garten responded to the lawsuit saying that the allegations are "categorically untrue."
US President Donald Trump leaves after signing an executive order regarding Medicare at Sharon L. Morse Performing Arts Center October 3, 2019, in The Villages, Florida. Brendan Smialowski/Getty
The anonymous plaintiff—identified only as "Katie Johnson" in an initial legal filing that was dismissed in California, and "Jane Doe" in two subsequent legal filings in New York—said that she was raped by Trump during a party hosted by the now-deceased pedophile Jeffrey Epstein at his New York City apartment. In the third and final lawsuit, Doe alleged she had numerous sexual encounters with Trump and Epstein at the latter's parties and said she was also raped by Epstein, as BuzzFeed News reported at the time.
Among the lurid details of the lawsuit, Jane Doe alleged Trump tied her to a bed, "forcibly raped" her and threatened her and her family with physical harm, if not death, if she told anyone about the assault. "I understood that Mr. Trump and Mr. Epstein knew that I was 13 years old," Jane Doe wrote in an affidavit.
Farrow alleges that after the suit was filed in September 2016, Enquirer editor Howard and Trump lawyer Cohen were in contact frequently. (Cohen was sentenced to three years in prison last December on charges including campaign finance violations for his part in hush payments to McDougal and adult film star Stormy Daniels.)
"There was no opportunity to buy this story," Farrow writes, claiming that AMI chief Pecker—a longtime friend of Trump's—only found out about the lawsuit after it was filed.
Before Proposing to Melania, Trump Engaged in 'Wave' of Alleged Touching
Still, Farrow says, Howard, now chief content officer at AMI, tried to use his influence to convince Lisa Bloom, a power attorney who agreed to represent Jane Doe, to drop her client.
In November 2016, just days before the presidential election, Bloom suddenly announced a press conference with Jane Doe had been canceled, saying Doe had become frightened after receiving death threats. Two days later, Doe's lead attorney, Thomas Meager, filed to dismiss the case. Jane Doe has not been heard from since.
Speaking to Newsweek Tuesday, Bloom said that while the Enquirer editor "did tell me he thought Jane Doe lacked credibility ... that wasn't the reason she asked her other attorney to drop her case."
"After we received numerous death threats and my law firm's website and emails were hacked, she did not want to go forward," Bloom added.
Bloom also said that she did not enter any agreements with AMI on Doe's behalf. "I represented Ms. Doe for free and there was never any discussion of money or settlement as I strongly believed her allegations should be made public given that Donald Trump was running for president at that time."
Paul Tweed, an attorney for Howard, told Newsweek: "We have advised Mr. Howard to make no further comment regarding the book at this stage, while all his legal remedies in appropriate jurisdictions are being considered."
An AMI spokesperson said: "Mr. Farrow's narrative is driven by unsubstantiated allegations from questionable sources and while these stories may be dramatic, they are completely untrue."

redpajama_set_name: RedPajamaCommonCrawl


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2022 bmw x5 vs mercedes gle
If you're in the market for a midsize premium SUV that is functional, stylish and available in Vancouver, then the 2022 BMW X5 should be on your list. This SUV offers everything you could want, competing with the Mercedes-Benz GLE. Both premium SUVs are similar in size, performance and features but have some key differences.
BMW vs. Mercedes
The choice between a BMW or Mercedes isn't always easy, but the below comparison will outline how they are the same and how they are different.
Price and Models
The 2022 BMW X5 is priced between $59 400 and $105 900, while the Mercedes GLE is between $55 700 and $80 200. There are four models available for the 2022 BMW X5 in Vancouver, where the first model is enhanced to provide better power.
The Mercedes has two body types: the Standard and the Coupe GLE. This offers slightly fewer choices for buyers compared to BMW models.
A 2022 BMW X5 seats five people, with an optional third row to seat seven people. The standard model has a base 6-cylinder rear-wheel-drive and an all-wheel-drive model. Models are also available with a 6-cylinder and electric motor and battery pack or a V8-powered engine. It's a sports activity vehicle and has an overall length of 493.5cm, a width of 200.34cm and a height of 174.5cm.
The Mercedes is nearly identical in dimensions and seats five people, with an optional third row. The standard base GLE 350 has a 4-cylinder engine, with the coupe GLE 450 having a 6-cylinder engine. If you want a V8-powered engine, you only have the option of the Mercedes GLE 580.
Both vehicles have standard features like an anti-theft system, tire pressure monitoring system, etc. However, the 2022 BMW X5 comes with a standard rear spoiler, rear parking aid and a leather-wrapped steering wheel. It also has digital key access, a wifi hotspot and an MP3 player, along with the standard remote access features, live emergency assistance, etc., that the Mercedes has.
The 2022 BMW X5 starts with the standard turbocharged 3L, 6-cylinder engine, offering 335 hp, as well as 448 newton-metres of torque. You can get up to 600 hp and 749 newton-metres depending on the model. A mild hybrid electrification system of 48-volts powers the electrical system and 8-speed automatic transmission. The tow limit is approximately 3265kg.
The standard Mercedes GLE starts at a 2L, 4-cylinder engine with 255 hp and 370 newton-meters and has a 9-speed automatic transmission. Depending on the model, you can go up to 483 hp and 699 newton-metres, with a towing capacity of around 2492kg if adequately outfitted.
Another benefit that BMW X5 offers that Mercedes doesn't is complimentary scheduled maintenance up to 57 936km or for the first three years. There is no such plan available with Mercedes.
BMW or Mercedes
It's always a big decision when buying a new car, so it's essential to ensure you buy the vehicle that will work for your needs. While both brands offer great vehicles and use the latest technology, the 2022 BMW X5 offers more range in power and customizations, along with scheduled maintenance.


20 Beauty Hacks Every Girl Should Know !
Which girl doesn't want to look beautiful all of the times? It isn't entirely possible to look good all the time as there are many factors that influence this. Yet there are a few handy beauty hacks that would make us look satisfactorily good.
Make any lipstick matte with loose powder. Apply your lipstick and then place a tissue paper on it. Now brush some translucent powder onto the tissue, this powder will seep through the tissue and give your lipstick an authentic matte look.
Also Read:MAKEUP BRUSHES 101 – THE PERFECT GUIDE FOR BEGINNERS!
Avoid mascara smudges by holding up a business card behind your lashes as you apply. Perfect for lower lashes and the perfect beauty hack.
Also Read:GLITTER MAKEUP – SHIMMER AND SHINE IN JUST 4 STEPS !!
Dunk your digits in a bowl of cold water for 1-3minutes to reduce drying time.
Mix a dab of highlighter or moisturizer with your foundation to get a dewy, model-esque glow.
Mix sugar with coconut or olive oil and gently rub over your lips.
If your feet are feeling crack and dry, rub Vaseline over them before bed and put on a pair of socks. You'll wake up to baby soft soles.
Dry your hair with a t-shirt instead of a towel. Why? Well, because the rough fibers of a towel can roughen up your hair and result in frizz. The perfect hack of all beauty hacks!
Make your pony tail look longer and fuller by making two pony tails, one on top of the other. Then, wrap a 1-inch section of hair around both elastics and pin it in place underneath.
Bruises aren't pretty, especially when they occur on your leg before a big event. Next time one pops up, dab a bit of mouthwash on the area in question and it will lose its purple tinge.
If you want to smell good all day long, simply smear some Vaseline over your pulse points – the back of the neck, temples and wrists before applying your signature scent.
Put the tip of the pencil near a flame for about 30 seconds and let it cool for 15 minutes or so. It will now give you a thick and intense line, the way any gel liner would. The perfect beauty hack, right?
Very often, bobby pins slide out of your hair. To stop them from doing so, simply lay a tissue paper on a table, put the pins on it and spray them with any stiff hold hairspray. This will make them stay on your hair longer and stop them from sliding out every other minute.
It's impossible to get your hands on makeup wipes everywhere. Simply replace them with baby wipes. They work the same and baby wipes are usually kinder to your skin.
Braid your hair on both sides really tight and go to bed. When you wake up, untangle them, spray and scrunch it and you'll be left with perfect beach waves without using a curling iron, making this a perfect beauty hack.
Take the curved end of a band-aid and line it a little away from the tip of your nails. Now apply your nail color and wait till it's dry before you peel the band-aid off. Perfect French manicure at home.
While applying your eyeliner, put a spoon beside you so that you would get the perfect winged lashes you always want. It's simple, brush your brows with the spoon immediately after applying the liner, this will set them in the desired shape.
Use pale frosted eye shadow on your lipstick to get fuller lips.
Use a few drops of contact solution to bring your flaky mascara back to life.
Remove armpit stains by using a mixture of peroxide and soap.
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Next article8 Nail Cleansers That Work Like Magic!


The learning of CAD for construction: technical abilities or visual?
Abstract: The increasing demands of the construction industry for individuals with good IT skills add continuous pressures on higher education to improve their methods of teaching. CAD training, as an important part of IT training for construction students, is becoming an essential part of the curriculum in most built environment schools. However, general CAD training is mostly concerned with providing students with technical skills rather than the initial ability of spatial visualisation. Indeed, existing training methods of CAD applications, do not take into consideration students? learning styles, and the differences in their spatial visualisation abilities. Considering that CAD students need to perform various activities within CAD applications to develop an understanding of building concepts and components, their spatial visualisation abilities and their learning style, remain the main barriers. This paper identifies the learning strategies required to assist with the learning of 3D modelling and describes a new approach adopted to examine students' Special Visualisation Skills. The paper also describes innovative e-learning approaches developed to reinforce students' learning of 3D CAD, tracking their progress and highlighting qualitative measures of their effectiveness.
Abstract: Construction projects are complex, fragmented and highly risk business, due to the nature of construction operations. Therefore project managers require more efficient techniques and tools to plan and monitor the construc-tion project. In recent years many research studies have been carried out in order to make construction industry more efficient, profitable and attractive business. The IMCS-CON developed as decision support system for project mangers to assist project-controlling processes using a holistic approach. The IMCS-CON provide a framework to measure, analyse, review, and report performance data and enabling project management team to make corrective decision and keep project on track. The IMCS-CON system was evaluated using a case study of £2.3 million, three-story residential apartment building project in UK. The IMCS-CON system utilises multivariate statistical process control techniques to monitor the construction site variables. The MSPC combines a large number of variables into few independent vari-ables, which then can be monitored and any process deviations from the normal operating conditions can be identified with corrective actions suggested. The IMCS-CON models on-site information as quantitative variables and uses his-torical data and establishes patterns of correlated variables and assists project management in making future decisions. The outputs can also be visualised in multi-dimensional graphs. Statistics of external variables and internal variables influencing construction site operations were identified using a real life case study. The results of modelling the vari-ables and conducting experiments with IMCS-CON are analysed and discussed in this paper.


I planned only to do some fast AH canceling and reposting and ended up doing 6 games with my all mighty warlock mate.
We lost 4 games due to bad preparation, mainly on my side. We had a break of playing together for maybe two weeks and I was out of practice. It was pity but reloging into my paladin after doing Ulduar 25 raid last week I had the interface set up to raids. Therefore, the group pictures where not shown into arena. Somehow the checkbox "do not show group in raids" considers arena matches also as raids.
In one match the opposing team made lot of pressure onto my warlock. I thought he was dead. Irritated by the missing frame of him. But he was not. The moment I realized (couple of seconds) he was dead.
Maybe the problem is also that we are both have to press button to speak via Team speak. Therefore, communication is limited.
Another mistake witch is very bad within an arena math is the targeting of my team members with the F1 F2 F3 … keys. If you have a teammate with a pet, you get the pet into target if you press the F2 button again. Damn.
Its an extremly sad moment when you realize that you are not healing your mate but the pet instead.
After 2-3 games we where getting better. At least I felt much more comfortable with my key bindings and situational awareness. We one 2 games out of those. Mainly because my warlock mate is very good with fearing other players and self heals. In many matches he is doing more heals than the opposing healer. Funny.
It was great to be back into arena. Even if I have not thought about it, I missed it.
We need more practice and we will further improve.
Double check gear after using outfitter, should have 450 resilience, 2000 spell power, 18K mana, 19K live.
After 3 days without playing. It is Wednesday again, this means today start the next run for points in arena. On Sunday evening I joined my first 3v3 team. And holy good we sucked. We started with the setup Dot-Warlock, Destruction-Warlock an myself as holy paladin. Is still have no clues to separate one WL from the other but probably some have advantages above the other. The plan was, that the Destruction-WL would be able to kill one opponent player faster than they would be able to kill us.
The main changes for my side, 2v2 as my experience so far, I am know more often either main target or crowd controlled. This way, I saw our side more often loosing than that we ware able to pull out a victory. So we switched the Dest.-WL to a Warrior with mortal strike. This way the match last a little bit longer, but I was still not able to give a victory through good moves.
I have to improve, but also we have to find a valid strategy as a team to work with.
It was hard for me to figure out if I was now main target or should be close to my mates to heal them. In 2v2 you usually have at least some time to get close and make the heal. In 3v3 everything is much more urgent. And it is also very hard to run between my two mates to heal them if they are separated.
One match we played against two healers, and where not able to down any player. And after a couple of minutes the opposing ret paladin was able to down me. Sure somehow my mistake, but after 7 minutes you should also expect that the first opposing player is down.
One game was great because it was that close we have been in 3v2 lead. But shortly after the kill lost our first WL; me and the opposing healer dying at the same time. In the end it was our destro. WL against the opposing DK and he managed to pull out our win.
This is still my aim to archive, if left in 2v2 match alone against opposing damage dealer, I want to will pull out the win, man was it close last time. I have to keep myself under control and do the right things.
Oh I totally forget: We managed to win 10 straight games in 2v2 in the region of 870 team rating. We now have a rating from 1030 and looking forward to improve in this area.
Funny side note is that my teammate didn't notice that he still was not using the best version of his spell. He only lately hit 80 and they did not update automatically. Strange.
So thumbs up, hope for much fun today.
So finally I decide to write something to the world outside. So this is my first post. Free welcome to make comments.
It took me around 4-6 hours of preparation. I changed some of my emblems into epic PVP equip. I was finally lucky to at least get gloves and trousers from Archavon. I had some 1K winter emblems and some 60K honor to spent.
With all sockets used for resilience and proper enchanting I could come up to 710 resilience. I have no idea if this is good or not, but at least it gives me the feeling of preparation.
The rest of the equip is some 213/200 ilvl stuff.
So now it is up to me to find another team member for arena. I don't like the thought of running with a high equipped and experienced char, since I was quit sure that I don't even had a clue how to perform well. And man was I right, I had absolutely no clue. Therefore, I want to share my thoughts from yesterday.
I teamed up with a warlock who is an alt of an holy paladin. I meet him some weeks before, where I could join his 5v5. At that time I learned being focus without any resilience is a bad situation.
This warlock has some 1300 spell power which leads to the idea that there will be allot of room for improvement. But he has experience and he is a dot warlock. After our first 2 games which we lost of course I was introduced into the concept of run and hide. Since most games would with our team look like this if opposing team has one healer and one dd. They will try to eliminate the WL. But the damage they do is in the team rating around 800 not that much. It can be handled very easily. Thinks get only tricky if they get me into crowd control chains and do there burst.
So being out of range for there fears, dots, sheep's was really very helpful.
After I got myself out of the line more and more, we start winning games. Now it comes to the point of communication and executions.
Hitting the right buttons on the right target is tough?
You may ask why? Let me explain the problem. I was always a fan of getting my heal target by pressing the F1 or F2. The bad thing here is if you press F2 you may end up with the pet in target. This at least cost us 3 games.
How do I plan to improve: There are several blogs speaking about gaming devices. I will buy a gamers mouse. I will try to figure out how to use the best key binding. I will build suitable macros for modifiers. I expect all those changes will take me a lot of time. But for now I decided that I will go into that direction.
Did I have fun. Yes due to the skill of my warlock it was great. He had lots of suggestions on special situation and was very positive at the moment I made something right. It is always a very good attitude if you don't blame a beginner, but instead help to improve and to compliment on the progression.
– 30seconds after poping of wings no pala bubble possible.
– If your close weaver is under main attack at the beginning of the match, give him hand of protection (no melee damage anymore), if this runs out give him another hand which transfers damage from him to you.
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Lawrence, Massachusetts residents have finally returned to their homes several months after a disastrous series of explosions, only to be faced with mounting utility bills.
Paul Craney, one of the thousands of residents who were affected by the Sept. 13 tragedy, was greeted with an invoice from Columbia Gas stating he owed $820.92 for his gas bill. Residents have recently been receiving their first gas bill since the explosive disaster and have been caught off guard at the high cost.
"We were hoping that people would get some communication beforehand about what they were expected of in their bills," Lawrence Mayor Dan Rivera said.
Columbia Gas tweeted on Wednesday that Merrimack Valley customers who had their gas shut off as a precaution are receiving bills, and that anyone with concerns should contact the utility's customer service line at 1-866-388-3239.
In response to the bills, Rivera set up an email address, mygasbilliswrong@cityoflawrence.com, that will also review residents' concerns.
"We'll try to get to the bottom of it, too," Rivera said. "The reality is, is that the last thing Columbia Gas should be worried about is making money off the people that they hurt the most."
The Sept. 13 natural gas disaster killed a teenager, injured more than 20 people and damaged more than 100 structures. Authorities say over-pressurization during a routine pipeline replacement prompted the natural gas explosions in Andover, North Andover and Lawrence.
Columbia Gas did not comment on the new utility bills.


September North American Freight Number | Dalko Resources Inc.
The value of overall U.S. trade with its NAFTA partners, Canada and Mexico, rose 5.0 percent from September 2012 to September 2013 according to North American Freight numbers released by the Bureau of Transportation Statistics (BTS).
Pipelines were reported as having the largest year-to-year growth, 14.2 percent above last September's value of freight carried by pipelines. Trucks are the most heavily utilized mode for moving freight between the U.S. and its NAFTA partners, carrying three-fifths of U.S.-NAFTA trade. Freight moved by trucks in September increased 5.4 percent from last year, and rail rose 8.1 percent. However, both freight carried by vessel and air declined from last year, 4.3 percent and 0.1 percent respectively.
In September 2013 trucks carried 59.4 percent of U.S.-NAFTA trade followed by rail at 16.1 percent, vessels at 8.6 percent, pipelines at 7.2 percent and air at 3.9 percent.
Increasing 33.0 percent, U.S.-Canada trade by vessel had the largest percentage increase of any mode from September 2012 to September 2013. Pipeline trade came in second in September, increasing 15.2 percent from last year. Freight moved by rail between the U.S. and Canada only increased 3.0 percent from last year.
In September 54.2 percent of freight carried between the U.S. and Canada was moved by trucks, 16.8 percent by rail, 12.3 percent by pipeline, 5.8 percent by vessel and 4.5 percent by air.
From September 2012 to September 2013, U.S.-Mexico trade by rail had the largest percentage increase of any mode, increasing 16.4 percent. According to the BTS this was mainly due to a 20 percent increase in vehicles and vehicle parts moving by rail. Freight moved by vessel between the U.S. and Mexico decreased 18.2 percent from September 2012.
Trucks carried 66.0 percent of U.S.-Mexico trade in September, followed by rail at 15.1 percent, vessel at 12.3 percent, air at 3.1 percent and pipelines at 0.7 percent.


Umang Singhar is Member of the Madhya Pradesh Legislative Assembly and ready for Assembly Election 2018 Madhya Pradesh. He completed his BA and thereafter he became a politician.
Warning: Content on this wiki page completly user generated. If any information found incorrect about Umang Singhar, you can suggest edit through comment/opinion to improve this page.


text:
Q: Javascript swapable calculated field I am working on a "calculator" (game) where you click on Form Input buttons and the numbers automatically Add together in the "Total" Input. However - I need to have Three (or more) "Total" Windows. And I would like to swap which "Total Window" you are using via a radio button.
I have it working with one "Total Window" - but I am unsure how to swap the Field that is being used in my calculations.
Probably something simple - but I am new to javascript and the syntax...

<FORM NAME="Calc">

<input type="radio" name="BetSelector" value="InputWin"   onclick="BetWIN(this.form)" checked="checked" /> WIN
<INPUT TYPE="text"   NAME="InputWIN" Size="16" value="0" />
<input type="radio" name="BetSelector" value="InputPLACE" onclick="BetPLACE(this.form)" /> PLACE
<INPUT TYPE="text" NAME="InputPLACE" Size="16" value="0" />
<input type="radio" name="BetSelector" value="InputSHOW"  onclick="BetSHOW(this.form)" /> SHOW
<INPUT TYPE="text"  NAME="InputSHOW" Size="16" value="0" />

        <INPUT TYPE="button" NAME="b" VALUE="1"     OnClick="a_plus_b(this.form)" class="ChipB">
        <INPUT TYPE="button" NAME="c" VALUE="5"     OnCLick="a_plus_c(this.form)" class="ChipC">
        <INPUT TYPE="button" NAME="d" VALUE="25"    OnClick="a_plus_d(this.form)" class="ChipD">
        <INPUT TYPE="button" NAME="e" VALUE="100"   OnClick="a_plus_e(this.form)" class="ChipE">
        <INPUT TYPE="button" NAME="f" VALUE="500"   OnCLick="a_plus_f(this.form)" class="ChipF">
        <INPUT TYPE="button" NAME="g" VALUE="1000"  OnClick="a_plus_g(this.form)" class="ChipG">
        <INPUT TYPE="button" NAME="h" VALUE="5000"  OnClick="a_plus_h(this.form)" class="ChipH">

</FORM>

I started a jsfiddle: http://jsfiddle.net/bigfatdesigns/eSAPk/

A: Side note: if you use jQuery it will ease the interaction between fields
To specify which total to use you can simply add the following lines to your a_plus_X functions to specify which total to use.
Code using jQuery calls
var TotalField = "#total_" + $(':radio[name="BetSelector"]:checked').val();

$(TotalField).val() = calculation();

redpajama_set_name: RedPajamaStackExchange


B&W ws to check the use of simple past & past perfect.
This downloadable worksheet is recommended for high school students and elementary school students at Elementary (A1), Pre-intermediate (A2) and Intermediate (B1) level. It was designed for strengthening your groups' Spelling, Reading and Writing skills.


text:
\section{Introduction}

Calculations of higher-order corrections are very important for precision
tests of the Standard Model in present and future high-energy-physics
experiments.
The complexity of such calculations and the final results strongly increases
with the number of quantum loops considered, and one rapidly reaches the
limits of the present realm of possibility when one attempts to exactly
account for all mass scales of a given problem, already at the two-loop level.
To simplify the calculations and also the final expressions, various types of
expansions were proposed during the last couple of years
(see Ref.~\cite{AS1,AS2,DaTa} and references cited therein).
These approaches usually provide a possibility to reduce the problem of
evaluating complicated Feynman integrals to the calculation of tadpoles
$T_{0,m}(\alpha_1,\alpha_2)$
and loops with massless propagators $L_q(\alpha_1,\alpha_2)$,
defined in Eqs.~(\ref{eq:one}) and (\ref{eq:two}), respectively, which have
simple representations.

As a first step, let us to introduce some definitions to be used below.
All the calculations are performed in Euclidean momentum space of dimension
$D=4-2\epsilon$ using dimensional regularisation with 't~Hooft mass scale
$\mu$.
For the Euclidean integral measure, we use the short-hand notation
$Dk\equiv {\mu}^{4-D}d^Dk/{\pi}^{D/2}$.
The dotted and solid lines of any diagram correspond to massless and massive
Euclidean propagators, represented graphically as
\begin{equation}
\frac{1}{(k^2)^\alpha}=\hspace{3mm}
\mbox{{
\begin{picture}(30,10)(5,4)
\Gluon(0,5)(30,5){1}{5}
\Vertex(0,5){1}
\Vertex(30,5){1}
\Text(15,11)[b]{$\scriptstyle\alpha$}
\end{picture}
}}\hspace{1mm}
,\qquad \frac{1}{(k^2+m^2)^\alpha}=\hspace{3mm}
\mbox{{
\begin{picture}(30,10)(5,4)
\Line(0,5)(30,5)
\Vertex(0,5){1}
\Vertex(30,5){1}
\Text(15,11)[b]{$\scriptstyle\alpha$}
\Text(15,3)[t]{$\scriptstyle m^2$}
\end{picture}
}}\hspace{1mm},
\end{equation}
where $\alpha$ and $m$ denote the index and mass of the considered line,
respectively.
Unless stated otherwise, all solid lines have the same mass $m$.
Lines with index 1 and mass $m$ are not marked.

Let us now discuss the rules used in our calculation.
Firstly, massive tadpoles $ T_{0,m}(\alpha_1,\alpha_2)$ are integrated
according to the identity
\begin{eqnarray}
T_{0,m}(\alpha_1,\alpha_2)&\equiv&
\int\frac{Dk}{(k^2)^{\alpha_1}{(k^2+m^2)}^{\alpha_2}}\equiv\hspace{3mm}
\mbox{{
\begin{picture}(50,30)(0,13)
\GlueArc(20,15)(15,90,270){1}{6}
\CArc(20,15)(15,-90,90)
\Vertex(20,30){2}
\Vertex(20,0){2}
\Line(20,0)(17,-5)
\Line(20,0)(23,-5)
\Text(3,18)[r]{$\scriptstyle\alpha_1$}
\Text(38,18)[l]{$\scriptstyle\alpha_2$}
\end{picture}
}}
\nonumber\\
&=&\frac{\displaystyle R(\alpha_1,\alpha_2)}
        {\displaystyle (m^2)^{\alpha_1+\alpha_2-D/2}},
\label{eq:one}
\end{eqnarray}
where
\begin{equation}
R(\alpha_1,\alpha_2)=
\frac{\Gamma(D/2-\alpha_1)\Gamma(\alpha_1 +\alpha_2 -D/2)}{\Gamma(\alpha_2)
\Gamma(D/2)}
\end{equation}
and $\Gamma$ is Euler's Gamma function.
Secondly, massless loops $ L_q(\alpha_1,\alpha_2)$ with the external
momentum $q$ are integrated according to the identity
\begin{eqnarray}
L_q(\alpha_1,\alpha_2)
&\equiv&\int\frac{Dk}{(k^2)^{\alpha_1}{(k-q)}^{\alpha_2}}
\equiv\hspace{3mm}
\mbox{{
\begin{picture}(60,30)(0,13)
\GlueArc(30,0)(30,25,150){1}{6}
\GlueArc(30,30)(30,-25,-150){1}{6}
\Line(5,15)(-5,15)
\Line(55,15)(65,15)
\Vertex(5,15){2}
\Vertex(55,15){2}
\Text(30,37)[b]{$\scriptstyle\alpha_1$}
\Text(0,12)[t]{$q$}
\Text(30,-5)[t]{$\scriptstyle\alpha_2$}
\end{picture}
}}
\nonumber\\
&=&A(\alpha_1,\alpha_2)\hspace{3mm}
\mbox{{
\begin{picture}(70,30)(0,4)
\Gluon(5,5)(65,5){1}{6}
\Vertex(5,5){2}
\Vertex(65,5){2}
\Line(5,5)(-5,5)
\Line(65,5)(75,5)
\Text(33,10)[b]{$\scriptstyle\alpha_1+\alpha_2-D/2$}
\Text(-3,-5)[b]{$q$}
\end{picture}
}}\hspace{3mm},
\label{eq:two}
\end{eqnarray}
where
\begin{equation}
A(\alpha_1,\alpha_2)=
\frac{a(\alpha_1) a(\alpha_2)}{a(\alpha_1 +\alpha_2 -D/2)},\qquad
a(\alpha)=\frac{\Gamma(D/2-\alpha)}{\Gamma(\alpha)}.
\end{equation}

Recently, two examples of four-loop tadpoles were calculated in
Ref.~\cite{CKMS}.
Certain $O(\epsilon^0)$ parts of these results, denoted as $N_{10}$ and
$N_{20}$, were presented there only numerically.\footnote{%
During the preparation of this article, we were informed by K.G. Chetyrkin
about a paper \cite{Schroder} which also contains analytic results for
$N_{10}$ and $N_{20}$.
Our results are in full agreement with those of Ref.~\cite{Schroder}.}
Also the $O(\epsilon)$ terms beyond $N_{10}$ and $N_{20}$ are of current
interest in the literature \cite{dec} and so far only known numerically.
In conjunction with the results of Ref.~\cite{KKOV06}, they allow us to derive
analytic results for several four-loop master integrals which are
indispensable for modern calculations at this level of accuracy~\cite{KnKo06}.

The subject of this letter is to advertise a powerful technique to evaluate
such tadpoles analytically.
The simplifications encountered in the examples at hand suggest that this
technique might also be useful for more complicated diagrams.
Therefore, we wish to introduce it to the interested reader.

The content of this letter is as follows.
Section~\ref{sec:two} explains the calculational technique in general.
Sections~\ref{sec:three} and \ref{sec:four} describe its application to the
two integrals $N_{10}$ and $N_{20}$ of Ref.~\cite{CKMS}, respectively, also
including their $O(\epsilon)$ terms.
A summary is given in Section~\ref{sec:five}.

\section{Technique}
\label{sec:two}

The core of the technique is a master formula to represent a loop with two
massive propagators as an integral whose integrand contains a new propagator
with a mass that depends on the variable of integration.
This formula has the following form:
\begin{eqnarray}
\mbox{{
\begin{picture}(60,30)(0,13)
\Curve{(5,15)(30,25)(55,15)}
\Curve{(5,15)(30,5)(55,15)}
\Vertex(5,15){2}
\Vertex(55,15){2}
\Line(5,15)(-5,15)
\Line(55,15)(65,15)
\Text(30,27)[b]{$\scriptstyle\alpha_1,\,m_1$}
\Text(30,3)[t]{$\scriptstyle\alpha_2,\,m_2$}
\Text(0,12)[t]{$q$}
\end{picture}
}}\hspace{3mm}&=&
   \frac{\Gamma(\alpha_1+\alpha_2-D/2)}{\Gamma(\alpha_1)\,\Gamma(\alpha_2)}
\nonumber\\
&&{}\times
\int_0^1 \frac{ds}{(1-s)^{\alpha_1+1-D/2}s^{\alpha_2+1-D/2}}\hspace{3mm}
\mbox{{
\begin{picture}(70,30)(0,4)
\Line(5,5)(65,5)
\Vertex(5,5){2}
\Vertex(65,5){2}
\Line(5,5)(-5,5)
\Line(65,5)(75,5)
\Text(33,10)[b]{$\scriptstyle\alpha_1+\alpha_2-D/2$}
\Text(33,3)[t]{$\scriptstyle \frac{m_1^2}{1-s}+\frac{m_2^2}{s}$}
\Text(-3,-5)[b]{$q$}
\end{picture}
}}
\hspace{3mm}.
\label{eq:master}
\end{eqnarray}
It was introduced in Euclidean and Minkowski spaces in Refs.~\cite{FKVpl} and
\cite{FKVnp}, respectively.
Here, we are working in Euclidean space and will, thus, follow
Ref.~\cite{FKVpl}.
Some recent applications in Minkowski space may be found in Ref.~\cite{KKOV}.
Special cases of Eq.~(\ref{eq:master}) were considered in
Ref.~\cite{DEM}, where a differential-equation method
\cite{DEM,DEMRem} was introduced.
The latter only requires that quite simple diagrams are calculated directly.
The results for more complicated diagrams may be reconstructed by
integrating the results for simpler diagrams over external parameters.
In many cases, the results for complicated diagrams may be obtained by
integrating the one-loop tadpoles of Eq.~(\ref{eq:one}).
The method is now very popular for the calculation of complicated Feynman
integrals; for recent articles, see Ref.~\cite{papers} and references cited
therein.

Here, we shall follow a similar strategy.
Applying Eq.~(\ref{eq:master}), we shall represent the results of
Ref.~\cite{CKMS} as integrals over one-loop tadpoles, which in turn contain
propagators with masses that depend on the variable of integration.
The case of $N_{10}$ will be considered in detail.

We would like to note that Eq.~(\ref{eq:master}) can be applied successfully
to diagrams containing one-loop self-energy sub-diagrams.
Such sub-diagrams are frequently generated by the application of the
integration-by-parts technique \cite{byparts}.
Such an application \cite{DEM,DEMRem} leads to a differential equation for the
original diagram (in exceptional cases, the equation degenerates to an
algebraic relation) with an inhomogeneous term that depends on less
complicated diagrams usually containing one-loop self-energy sub-diagrams
(see, for example, Refs.~\cite{FKVpl,FKVnp}).

\section{First example}
\label{sec:three}

As the first example, we consider the case of $N_{10}$ in Ref.~\cite{CKMS},
which may be represented graphically as\footnote{The graphic representations
of $N_{10}$ and $N_{20}$ are adopted from Ref.~\cite{CKMS}.
A propagator with a point is equal to one with the index 2.}
\begin{equation}
m^2
\raisebox{-0.9cm}{{
\includegraphics[height=2.0cm,bb=71 115 697 720]{Topo406112With3Dots}
}}
+m^{2(1-\epsilon)}\Gamma(\epsilon)
\raisebox{-0.9cm}{{
\includegraphics[height=2.0cm,bb=71 114 697 720]{Topo30503With2Dots}
}}
= \frac{\Gamma^4(1+\epsilon)}{(m^2)^{4\epsilon}}
N_{10}(\epsilon),
\label{NM1}
\end{equation}
where $N_{10}(\epsilon)=[N_{10}+M_{10}\epsilon+O(\epsilon^2)]/(1-\epsilon)$.
For $\epsilon=0$, we recover $N_{10}=N_{10}(0)$.
Applying Eq.~(\ref{eq:two}) to the massless loop in the left diagram and
Eq.~(\ref{eq:master}) with $\alpha_1=1$ and $\alpha_2=2$ to the interior
massive loops and to the loops with exterior propagators, we easily obtain the
following representation for $N_{10}(\epsilon)$:
\begin{equation}
N_{10}(\epsilon) ={m^2\over\epsilon}\left[\frac{1}{m^{2\epsilon}}
J_1(0)-\frac{\Gamma^2(1-\epsilon)}{\Gamma(1-2\epsilon)}J_1(1)\right],
\end{equation}
where
\begin{equation}
J_{1}(a)=
\frac{(1-\epsilon)(m^2)^{4\epsilon}}{\Gamma(1+\epsilon)}
\int_0^1 \frac{ds_1}{s_1 \eta_1^{\epsilon}}
\int_0^1 \frac{ds_2}{s_2 \eta_2^{\epsilon}}
T_{0,M_1,M_2}(1+a\epsilon,1+\epsilon,1+\epsilon),
\end{equation}
$M^2_i=m^2/\eta_i$, $\eta_i=s_i(1-s_i)$,
and $T_{M_1,M_2,M_3}(\alpha_1,\alpha_2,\alpha_3)$,
where the index $\alpha_i$ belongs to the mass $M_i$, is the one-loop tadpole
involving three massive propagators,
\begin{equation}
T_{M_1,M_2,M_3}(\alpha_1,\alpha_2,\alpha_3)=
\hspace{3mm}
\trg{\alpha_1}{\alpha_2}{\alpha_3}
\hspace{3mm}.
\vspace{5mm}
\end{equation}
Expanding the propagator $(k^2+M^2_2)^{-\alpha_2}$ in
$T_{0,M_1,M_2}(\alpha_3,\alpha_1,\alpha_2)$ as
\begin{equation}
\frac{1}{(k^2+M^2_2)^{\alpha_2}}=\sum_{n=0}^{\infty}
\frac{\Gamma(n+\alpha_2)}{n!\Gamma(\alpha_2)} \,
\frac{\mu_{12}^n}{(k^2+M^2_1)^{n+1+\epsilon}},
\label{exp}
\end{equation}
where $\mu_{ij}=M^2_i-M^2_j$,
and using Eq.~(\ref{eq:one}) for the tadpole
$T_{0,M_1}(\alpha_1,\alpha_2)$, we obtain
\begin{equation}
T_{0,M_1,M_2}(\alpha_3,\alpha_1,\alpha_2) =
\frac{R(\alpha_3,\alpha_1+\alpha_2)}{(M_1^2)^{\overline{\alpha} -D/2}}
\,{}_2F_1(\overline{\alpha} -D/2,\alpha_2;\alpha_1+\alpha_2;\overline{x}),
\end{equation}
where $\overline{\alpha}=\alpha_1+\alpha_2+\alpha_3$, $\overline{x}=1-x$,
$x=M^2_2/M^2_1=\eta_1/\eta_2$, and ${}_2F_1$ denotes a hyper-geometric
function  \cite{Gradshteyn}.
Setting $\alpha_i=1+a_i\epsilon$ and expanding the hyper-geometric function, we
have
\begin{eqnarray}
\lefteqn{
T_{0,M_1,M_2}(1+a_3\epsilon,1+a_1\epsilon,1+a_2\epsilon)
=\frac{1}{(M_1^2)^{1+(\overline{a}+1)\epsilon}}\,
\frac{\Gamma(1+\epsilon)}{\overline{x}(1-\epsilon)}
\left\{-\ln x
\vphantom{\frac{(a_2+a_3+1)^2}{6}} \right.}
\nonumber\\
&&{}+ \epsilon \left[\frac{a_2+a_3+1}{2} \ln^2 x
-(a_1+a_2) \mathop{\mathrm{Li}}\nolimits_2(\overline{x}) \right]
+ \epsilon^2 \left[ \frac{(a_2+a_3+1)^2}{6} \ln^3 x
\right.
\nonumber\\
&&{}-\Bigl((a_3+1)(\overline{a}+1)-1\Bigr)\zeta(2)\ln x
+a_1(a_3+1)\ln x\mathop{\mathrm{Li}}\nolimits_2(\overline{x})
\nonumber\\
&&{}+\left.\left.\vphantom{\frac{(a_2+a_3+1)^2}{6}}
(a_1+a_2)^2\mathop{\mathrm{Li}}\nolimits_3(\overline{x})
-\Bigl(a_2(\overline{a}+1)-a_1(a_3+1)\Bigr)\mathop{\mathrm{S}}\nolimits_{1,2}(\overline{x})
\right]\right\}
+ O(\epsilon^3),
\end{eqnarray}
where $\overline{a}= a_1+a_2+a_3$ and
\begin{eqnarray}
\mathop{\mathrm{S}}\nolimits_{n,m}(x)&=&\frac{(-1)^{n+m-1}}{(n-1)!m!} \int_0^1 \frac{dy}{y}
\ln^{n-1}(y) \ln^{m}(1-xy),
\nonumber\\
\mathop{\mathrm{Li}}\nolimits_n(x)&=&\mathop{\mathrm{S}}\nolimits_{n-1,1}(x)
\end{eqnarray}
denote the generalised and ordinary poly-logarithms (see, for example,
Ref.~\cite{DD}), respectively.
Thus, we obtain
\begin{eqnarray}
N_{10}(\epsilon)&=&
\int_0^1 \frac{ds_1}{s_1\eta_1^{\epsilon}}
\int_0^1 \frac{ds_2}{s_2\eta_2^{\epsilon}} \,
\frac{\eta_1}{\overline{x}}
\left\{-\frac{1}{2}\ln^2 x
+\ln x \ln \eta_1
\right.
\nonumber\\
&&{}+\epsilon \left[\frac{19}{6} \ln^3 x
+4 \zeta(2)\ln x
-\ln x\mathop{\mathrm{Li}}\nolimits_2(\overline{x})
- \left(3 \ln^2 x - 2 \mathop{\mathrm{Li}}\nolimits_2(\overline{x})\right) \ln \eta_1
\right.
\nonumber\\
&&{}+\left.\left.
\frac{7}{2} \ln x  \ln^2 \eta_1
\right]\right\}
+ O(\epsilon^2).
\end{eqnarray}
Exploiting the $s_1 \leftrightarrow s_2$ symmetry,
we find
\begin{eqnarray}
N_{10}(\epsilon)&=& \frac{1}{8}
\int_0^1 ds_1\int_0^1
 \frac{ds_2}{\eta_2 - \eta_1}
\left\{\ln x \ln (\eta_1\eta_2)
+\epsilon \left[ \frac{61}{12} \ln^3 x
\right.\right.
\nonumber\\
&&{}+8 \zeta(2) \ln x
+\left( \mathop{\mathrm{Li}}\nolimits_2(\overline{x})-\mathop{\mathrm{Li}}\nolimits_2\left(-\frac{\overline{x}}{x}\right)\right)
\ln (\eta_1\eta_2)
\nonumber\\
&&{}+\left.\left.
\frac{3}{4}\ln x\ln^2 (\eta_1 \eta_2)
\right]\right\}
+ O(\epsilon^2).
\end{eqnarray}

We first concentrate on the $O(\epsilon^0)$ term $N_{10}$.
Using the standard replacement $s_i=(1+\xi_i)/2$, we obtain
\begin{eqnarray}
N_{10} &=& \frac{1}{2}\int_0^1d\xi_1
\int_0^1 \frac{d\xi_2}{\xi_1^2-\xi_2^2}
\left(\ln^2\frac{1-\xi_1^2}{4} - \ln^2\frac{1-\xi_2^2}{4}\right)
\\
&=&\frac{1}{2}\int_0^1d\xi_1
\int_0^1 \frac{d\xi_2}{\xi_1^2-\xi_2^2}
\left\{\left[\ln^2 (1-\xi_1^2) -2\ln 4 \ln (1-\xi_1^2)\right]
- [\xi_1\leftrightarrow\xi_2]\right\}.
\nonumber
\end{eqnarray}
Exploiting the $\xi_1 \leftrightarrow \xi_2$  symmetry, we find
\begin{eqnarray}
N_{10} &=& \int_0^1 d\xi_1
\left[\ln^2 (1-\xi_1^2) -2\ln 4 \ln (1-\xi_1^2) \right]
\int_0^1 \frac{d\xi_2}{\xi_1^2-\xi_2^2}
\nonumber\\
&=& - \frac{1}{2}\int_0^1\frac{d\xi_1}{\xi_1}
\ln\frac{1-\xi_1}{1+\xi_1}
\left[\ln^2 (1-\xi_1^2) -2\ln 4 \ln (1-\xi_1^2) \right].
\label{Sec3.1}
\end{eqnarray}
Evaluating the r.h.s.\ of Eq.~(\ref{Sec3.1}), we obtain
\begin{eqnarray}
N_{10} &=&
\frac{17\pi^4}{720}
+ \frac{7}{2}\zeta(3) \ln 2 - 4  \mathop{\mathrm{S}}\nolimits_{1,3}(-1)
\nonumber\\
&=&
\frac{49\pi^4}{720} -  \frac{1}{2} b_4,
\label{res1}
\end{eqnarray}
where
\begin{equation}
b_4 =-\frac{1}{3}(\pi^2-\ln^2 2)\ln^2 2+8\mathop{\mathrm{Li}}\nolimits_4\left(\frac{1}{2}\right).
\label{b4}
\end{equation}
Equation~(\ref{res1}) coincides with the result obtained in
Ref.~\cite{Schroder}.

The evaluation of the $O(\epsilon)$ term $M_{10}$ is rather tedious and cannot
be described in this brief communication.
Here, we merely list the result, which reads
\begin{equation}
M_{10} = -\frac{149\pi^4}{180} \ln 2+ \frac{279}{8} \zeta(5)  - 7 b_5,
\label{M10}
\end{equation}
where
\begin{equation}
b_5 =\frac{1}{45}(5\pi^2-3\ln^2 2)\ln^3 2+8\mathop{\mathrm{Li}}\nolimits_5\left(\frac{1}{2}\right).
\label{b5}
\end{equation}
We note in passing that also the other master integrals presented in
Ref.~\cite{Schroder} can be written in a more compact form if the combinations
$b_4$ and $b_5$ of Eqs.~(\ref{b4}) and (\ref{b5}), respectively, are
introduced.

\section{Second example}
\label{sec:four}

Let us consider the second example of Ref.~\cite{CKMS}, which may be
graphically represented as
\begin{equation}
m^2
\raisebox{-0.9cm}{{
\includegraphics[height=2.0cm,bb=71 114 697 720]{Topo406111With3Dots}
}}
=\frac{\Gamma^4(1+\epsilon)}{(m^2)^{4\epsilon}}
N_{20}(\epsilon),
\label{NM2}
\end{equation}
where $N_{20}(\epsilon)=[N_{20}+M_{20}\epsilon+O(\epsilon^2)]/(1-\epsilon)$.
For $\epsilon=0$, we recover $N_{20}=N_{20}(0)$.
Applying Eq.~(\ref{eq:master}) with $\alpha_1=1$ and $\alpha_2=2$ to the two
interior massive loops and to the loops involving exterior propagators, we
easily obtain the following representation for $N_{20}(\epsilon)$:
\begin{equation}
N_{20}(\epsilon)=
\frac{(1-\epsilon)(m^2)^{4\epsilon}}{\Gamma(1+\epsilon)}
\int_0^1\frac{ds_1}{s_1\eta_1^{\epsilon}}
\int_0^1 \frac{ds_2}{s_2\eta_2^{\epsilon}}
\int_0^1 \frac{ds_3}{s_3\eta_3^{\epsilon}}
T_{M_1,M_2,M_3}(1+\epsilon,1+\epsilon,1+\epsilon).
\label{N20ep}
\end{equation}
Expanding the propagators $(k^2+M^2_2)^{-\alpha_2}$ and
$(k^2+M^2_3)^{-\alpha_3}$ in\break
$T_{M_1,M_2,M_3}(\alpha_1,\alpha_2,\alpha_3)$ as
in Eq.~(\ref{exp}) and using Eq.~(\ref{eq:one}) for the tadpole
$T_{0,M_1}(0,\alpha_2)$, we obtain
\begin{eqnarray}
T_{M_1,M_2,M_3}(\alpha_1,\alpha_2,\alpha_3) &=&
\sum_{n=0}^{\infty}
\frac{\Gamma(n+\alpha_2)}{n!\Gamma(\alpha_2)}
\sum_{l=0}^{\infty}
\frac{\Gamma(l+\alpha_3)}{l!\Gamma(\alpha_3)}\,
\frac{\Gamma(l+n+\overline{\alpha}-D/2)}{\Gamma(l+n+\overline{\alpha})}
\nonumber\\
&&{}\times
\frac{\mu_{12}^n\mu_{13}^l}{(M^2_1)^{n+l+\overline{\alpha}-D/2}}
\nonumber\\
&=&
\frac{R(0,\overline{\alpha})}{(M_1^2)^{\overline{\alpha} -D/2}}
F_1(\overline{\alpha} -D/2,\alpha_2,\alpha_3;\overline{\alpha};\overline{x},
\mu_{13}/M^2_1),
\nonumber
\end{eqnarray}
where $F_1$ denotes an Appel hyper-geometric function \cite{Gradshteyn}.
Expanding the hyper-geometric function, we have
\begin{eqnarray}
\lefteqn{
T_{M_1,M_2,M_3}(1+a_1\epsilon,1+a_2\epsilon,1+a_3\epsilon)
=\frac{1}{(M_1^2)^{(\overline{a}+1)\epsilon}}\,
\frac{\Gamma(1+\epsilon)}{\mu_{32}(1-\epsilon)}}
\nonumber\\
&&{}\times\left\{ \frac{M_2^2}{\mu_{12}}\ln\frac{M_2^2}{M_1^2}
+ \epsilon \left[
- \frac{a_2+1}{2}\, \frac{M_2^2}{\mu_{12}}\ln^2\frac{M_2^2}{M_1^2}
+\left(\overline{a} \frac{M_1^2}{\mu_{12}} -a_1-a_3 \right)
\mathop{\mathrm{Li}}\nolimits_2\left( \frac{\mu_{12}}{M_1^2}\right)
\right.\right.
\nonumber\\
&&{}+\left.\left.
a_3\frac{M_1^2}{\mu_{12}}\mathop{\mathrm{Li}}\nolimits_2\left(\frac{\mu_{13}}{M_1^2}\right)
+a_3\frac{M_2^2}{\mu_{12}} \left(\mathop{\mathrm{Li}}\nolimits_2\left( \frac{\mu_{21}}{M_2^2}\right)-
\mathop{\mathrm{Li}}\nolimits_2\left( \frac{\mu_{23}}{M_2^2}\right)\right)\right]
+ O(\epsilon^2)\right\}
\nonumber\\
&&{}+(2 \leftrightarrow 3).
\label{compli}
\end{eqnarray}

As in the previous section, we describe the evaluation of the $O(\epsilon^0)$
term in some detail, while we merely list the final result for the
$O(\epsilon)$ term for reasons of space.
At $O(\epsilon^0)$, the right-hand side of the Eq.~(\ref{N20ep}) contains the
tadpole $T_{M_1,M_2,M_3}(1,1,1)$, which can be calculated in an essentially
simpler way.
Indeed, using
\begin{equation}
\frac{1}{(k^2+M^2_1)(k^2+M^2_2)}=\frac{1}{\mu_{21}}
\left(\frac{1}{k^2+M^2_1}-\frac{1}{k^2+M^2_2}\right)
\end{equation}
and Eq.~(\ref{eq:one}), we find
\begin{eqnarray}
T_{M_1,M_2}(1,1)&=&\frac{\Gamma(\epsilon-1)}{\mu_{21}}
\left[{(M^2_1)}^{1-\epsilon}-{(M^2_2)}^{1-\epsilon}\right]
\nonumber\\
&=&\frac{1}{\mu^{2\epsilon}}\, \frac{\Gamma(1+\epsilon)}{1-\epsilon}
\left(\frac{1}{\epsilon}+\frac{M^2_1}{\mu_{21}}\ln\frac{M^2_1}{\mu^2}\right)
+(M_1\leftrightarrow M_2)
\nonumber\\
&&{}+O(\epsilon),
\label{Sec4.0}\\
T_{M_1,M_2,M_3}(1,1,1)&=&\frac{1}{\mu_{32}}
\left[T_{M_1,M_2}(1,1)-T_{M_1,M_3}(1,1)\right]
\nonumber\\
&=&\frac{M^2_1}{\mu_{12}\mu_{13}}\ln\frac{M^2_1}{\mu^2}
+(M_1\leftrightarrow M_2)+(M_1\leftrightarrow M_3)
\nonumber\\
&&{}+O(\epsilon).
\label{Sec4.1}
\end{eqnarray}
Notice that the right-hand side of Eq.~(\ref{Sec4.1}) is $m^2$ and $\mu^2$
independent.
Moreover, it coincides with Eqs.~(\ref{N20ep}) and (\ref{compli}) at
$O(\epsilon^0)$.
Every term in Eqs.~(\ref{Sec4.0}) and (\ref{Sec4.1}) contains only one
logarithm depending on the variable $\eta_i$.
Thus, the integrals $\int^1_0 ds_i$ can be evaluated at the end of the
calculation (see Eq.~(\ref{Sec3.1}), for a similar procedure).
After some algebra, we obtain
\begin{equation}
N_{20}=\frac{9}{2}\zeta(3),
\end{equation}
which agrees with Ref.~\cite{Schroder}.

Lack of space prevents us from going into details with our derivation of the
$O(\epsilon)$ term $M_{20}$, and we merely list our result,
\begin{equation}
M_{20} = -\frac{4\pi^4}{15} - 27 \zeta(3) + 3 b_4,
\end{equation}
where $b_4$ is defined in Eq.~(\ref{b4}).

\section{Conclusion}
\label{sec:five}

We demonstrated the usefulness of a simple technique for the analytic
evaluation of four-loop tadpoles, which allows one to obtain the results by
integrating one-loop tadpoles with masses that depend on integration
variables.
By means of this method, we calculated two Feynman diagrams which were
presented in Ref.~\cite{CKMS} in numerical form.
Our results are in full agreement with the analytic results obtained just
recently in Ref.~\cite{Schroder}.

We note that the integral $N_{10}$ was found in Ref.~\cite{Schroder} with the
aid of the PSLQ program \cite{PSLQ}, which is able to reconstruct the
rational-number coefficients multiplying a given set of transcendental numbers
from a high-precision numerical result, i.e.\ the algebraic structure of the
result must be known a priori.
Such structures are often more complicated, and they are in general not known
from other considerations (see, for example, Ref.~\cite{FKalK}), so that the
PSLQ program cannot be applied.
The method presented here does not suffer from such a limitation and should
be applicable also for more complicated tadpoles
depending on one \cite{Schroder} or two \cite{difmass} nonzero masses.
Moreover, it can be applied in combination with the PSLQ program.
In fact, by evaluating some part of a complicated Feynman diagram by means of
this method, the underlying set of transcendental numbers may emerge and can
then be injected into the reconstruction of the full analytic result with the
help of the PSLQ program.

\vspace{0.5cm}

\noindent{\bf Acknowledgements}

We are grateful to K.G. Chetyrkin for useful communications and for bringing
Ref.~\cite{Schroder} to our attention, and to O.L. Veretin for his interest in
this work, for his helpful comments, for checking some of our formulae, and
for technical advice related to the PSLQ program \cite{PSLQ}, which helped us
in checking the $O(\epsilon)$ terms of the four-loop tadpoles $N_{10}$ and
$N_{20}$.
A.V.K. was supported in part by the RFBR Foundation under Grant No.\
05-02-17645-a and the Heisenberg-Landau-Programm.
This work was supported in part by BMBF Grant No.\ 05 HT4GUA/4 and HGF Grant
No.\ NG-VH-008.

redpajama_set_name: RedPajamaArXiv


text:
Q: After you create a class, how do you add information from a csv to a python object, without using a module I am struggling to understand classes/objects and using a csv with them.
I have a CSV with 26 rows, 1 being the header, the other containing rows of info. Small example below
id,food,food_print,cal1,cal2,expi1999,expi2000,expi2001
1,bun,bun_bun,45.3434,199.32323,23.3333,45.4444,33.33333
2,burger,burger_bun,45.342343,200.34243,34.3333,0,9
3,pickle,pickle_seed,67.345454,34.3434,34,56,33
4,chicken,chicken_egg,44.34343,43.343343,43,434,34343

I have my class as follows:
class City(object):
    def __init__(self, food = 'n/a', foodprint = 'n/a', cal1 = -999, cal2 = -999,
    expi1999 = -999, expi2000 = -999, expi2001 = -999)
        self.food = food
        self.foodprint = foodprint
        self.cal1 = cal1
        self.cal2 = cal2
        self.expi1999 = expi1999
        self.expi2000 = expi2000
        self.expi2001 = expi2001

meals = []

foodfile = open('Food.csv', 'rt')
headers = foodfile.readline().strip().split(',')
headers = headers.split(',')

for line in foodfile:
    foodfields = foodfile.readline().strip().split(',')

How do I write in the rows from my food csv into an object to be referenced in the class?

A: Assuming all colums are filled in every row:
try:
for line in foodfile:
    foodfields = foodfile.readline().strip().split(',')
    meals.append(City(foodfields[1],foodfields[2],foodfields[3],foodfields[4],foodfields[5],foodfields[6],foodfields[7]))

A: true tone to avoid dynamic classes.  Because your code become unpredictable in future changes. Use dataclass instead.  Also this behaviour let you refer to this "data model" as annotation during data exchange with any callable.
from dataclasses import dataclass
@dataclass
class City:
    food: str = 'n/a'
    foodprint: str = 'n/a'
    cal1: int = -999
    cal2: int = -999
    expi1999: int = -999
    expi2000: int = -999
    expi2001: int = -999

But if you want to now how to do it - setattr function exists in global namespace.
class DynamicContainer:
    @classmethod
    def create(cls, headers: list[str], data: list[str]):
        obj = cls()
        for header, value in zip(headers, data):
            setattr(obj, header, value)
        return obj

headers = 'id,food,food_print,cal1,cal2,expi1999,expi2000,expi2001'.split(',')
data = '1,bun,bun_bun,45.3434,199.32323,23.3333,45.4444,33.33333'.split(',')
cont = DynamicContainer.create(headers, data)

print(cont.id, cont.cal1, cont.expi2000)

A: Here is how I would do it:
import csv

class City(object):
    def __init__(self, food=None, foodprint=None, cal1=None, cal2=None,  expi1999=None, expi2000=None, expi2001=None):
        self.food = food
        self.foodprint = foodprint
        self.cal1 = float(cal1)
        self.cal2 = float(cal2)
        self.expi1999 = float(expi1999)
        self.expi2000 = float(expi2000)
        self.expi2001 = float(expi2001)

    def __repr__(self):
        return (
            f"{self.__class__.__name__}("
            f"food={self.food!r}"
            f", "
            f"food_print={self.foodprint!r}"
            f", "
            f"cal1={self.cal1!r}"
            f", "
            f"cal2={self.cal2!r}"
            f", "
            f"expi1999={self.expi1999!r}"
            f", "
            f"expi2000={self.expi2000!r}"
            f", "
            f"expi2001={self.expi2001!r}"
            f")"
        )

with open("Food.csv") as stream:
    next(stream)  # Skip the header row
    reader = csv.reader(stream)
    meals = [City(*row[1:]) for row in reader]

Here is what rows looks like:
[City(food='bun', food_print='bun_bun', cal1=45.3434, cal2=199.32323, expi1999=23.3333, expi2000=45.4444, expi2001=33.33333),
 City(food='burger', food_print='burger_bun', cal1=45.342343, cal2=200.34243, expi1999=34.3333, expi2000=0.0, expi2001=9.0),
 City(food='pickle', food_print='pickle_seed', cal1=67.345454, cal2=34.3434, expi1999=34.0, expi2000=56.0, expi2001=33.0),
 City(food='chicken', food_print='chicken_egg', cal1=44.34343, cal2=43.343343, expi1999=43.0, expi2000=434.0, expi2001=34343.0)]

Notes

*

*The repr function is to show the contents of the class object, you can ignore it

*I use the csv module to ease parsing the CSV data

*I also converted the values into floating point numbers.


The 2019 Saloon Stock Car World Championship will be held at the Racewall Cowdenbeath over the weekend of the 17th and 18th August, just one of a a number of great events lined up for season 2019. Whilst full details of the event will be confirmed in due course, we have provided some general information below to help long-distance visitors plan their visit to the Racewall.
The Racewall's postcode is KY4 9QQ.
By Road: Cowdenbeath lies just off the A92 between Dunfermline and Kirkcaldy. The A92 links Dunfermline and Dundee and ajoins the M90 at the Dunfermline end. The M90 links Edinburgh and Perth via the Forth Road Bridge (A90). On entering Cowdenbeath carry-on down Bridge Street to the round-about and take the second exit to the High Street. About 100m up the High Street, before the steel Railway Bridge overhead, there is a left-hand turn that is sign-posted 'Central Park'. Follow this road under the railway arch and you have arrived!
Parking is limited in the free public car park directly in front of the stadium, but there are two other free public car parks if you follow the one-way road round to the right and up to Stenhouse Street - you can't miss them! On street parking is also available, but be careful to adhere to parking restrictions and please do not block the access of our neighbours.
Rail: Cowdenbeath train station is a stop on the Fife Circle line that is serviced by Scotrail. The station is a hop-skip-and-a-jump from the stadium. Timetables and connections can be found here.
Bus: Cowdenbeath bus stations are serviced by Stagecoach. Timetables and connections can be found here. The X54 is an express service that links Edinburgh, Dunfermline, Cowdenbeath, Glenrothes and Dundee.
Park and Ride: Combine your private car and bus travel! Making use of the local park and ride facility makes sense when the journey time between it and Cowdenbeath is only 10 mins. . The no. 19/19A Stagecoach bus passes through Cowdenbeath frequently (every 10-30 mins depending on time of day). The timetable can be found here. Alternatively, park and ride facilities are available at a number of train stations on the Fife Circle line including Dunfermline Queen Margaret.
There are a range of local accommodation options to suit all budgets. You can use the Welcome to Fife website to search for options. Dunfermline, Kirkcaldy and Glenrothes are all just 10 minutes drive from the stadium and as well as a range of local hotels and B&B options there are also a number of nationwide chain hotels in the local vicinity including Premier Inn, Travelodge, Holiday Inn Express and JD Weatherspoons. Please note that hotels in Edinburgh and South Fife tend to book up early and charge higher rates during the Edinburgh Festivals (approx. 4-28 August), and during this time you may wish to consider more local options or towns to the North of Cowdenbeath.
The Racewall has two covered grandstands providing seating for more than 1600 spectators. A disabled spectator viewing area is located in the East (New) Grandstand. There is no additional charge for these grandstands. There is open terracing around the rest of the raceway providing a mix of concrete standing and grassed areas. Spectators are welcome to bring their own chairs for use around the terracing.
There are catering outlets located around the stadium serving hot and cold drinks and fast-food style items. A Raceview licenced bar is located in the East (New) Grandstand, access to which is from the pit side of the building. Spectators are not permitted to bring into the stadium any glass containers.
Your safety is important to us. Please follow the direction of our dedicated security personnel at all times. There is no smoking (including the use of e-cigarettes) in both of our Grandstands. You must also remain seated when racing is in progress – there is no standing at the front of the grandstand area. Around the terracing you must stay behind the blue perimeter barrier at all times.


text:
https://lonzowire.usatoday.com/2017/11/07/lonzo-ball-lamelo-ball-lavar-ball-lakers-luke-walton/
LaMelo Ball still gets on Lonzo Ball's nerves
Despite the media circus surrounding him, Lonzo Ball has kept an even keel both on and off the court with the Los Angeles Lakers. The one person who can still agitate the NBA's most talked-about rookie? His 16-year-old brother, LaMelo Ball.
"My little brother for sure," Ball told NBA.com's David Aldridge. "The youngest one. They take pride in it."
Melo's antics aside, Zo has managed to keep his composure, in part—perhaps surprisingly so—because of lessons learned from his outspoken father, LaVar Ball.
"I was just taught, the team goes how the point guard goes," Ball said. "If I'm mad or showing my frustration, the whole team's gonna be like that, techs, and people are going to go down. So I just try to keep the even keel. That's why I don't get too high or too low. I've been playing like that my whole life. It's just natural for me."
The 20-year-old said he's "still working on" commanding the locker room—one of the focal points for improvement during his freshman year at UCLA under coach Steve Alford—because "playing with grown men, you can't always be quiet. You have to get into them sometimes, especially in the locker room."
"A lot of the credit goes to him, honestly," Lakers coach Luke Walton said. "I don't know if his upbringing has prepared him for it, the insanity that comes with everyone always watching and judging everything that comes with being a star. But he just has this ability where it seems that it doesn't really get to him. He's calm. He doesn't seem like stress and pressure really affect him that much."
Walton was also impressed with how Ball has dealt with all the hoopla around and about him for someone his age.
"Obviously a lot of people have a lot of different opinions, but he was raised properly," Walton added. "He handles himself, he handles situations, he handles his teammates with dignity and class."


Truly a one-of-a-kind home hidden in the heart of Florence, convenient to all amenities. This custom-built executive estate has been completely renovated by Master Builder Marvin Tisdale. Come see the new gourmet kitchen and remodeled downstairs master suite, each having their own fireplace. Huge bonus room above the 3-car garage is a perfect game room retreat. Large swimming pool with a private gazebo and amazing brick work surrounding the entire house. Call to schedule your private tour today!


text:
Q: How can you find and delete a section of an array with a parameter of a string? How can you find a section of a JS array and delete it? What I mean is a function that can find a string, looks at an array, sees if one of the "sections" are exactly like the string, and deletes that "section". Here's what I have:
var array = ['a1','f4'];
function findAndDelete(string){
    delete array.find('a1');
}

When I try it, nothing happens. It seems that there's a syntax error. How do I fix this?

A: var array = ['a1','f4'];
function findAndDelete(string){
    var index = array.indexOf(string);
    if (index > -1) {
      array.splice(index, 1);
    }
}

Here we can use splice() method of array to delete specific element from it.

A: This should work
var array = ['a1','f4'];
var index = array.indexOf('a1');

if (index > -1) {
   array.splice(index, 1);
}

splice will actually return a new array containing elements which were removed.

A: Use this:
function findAndDelete(array, string){
  if(array.indexOf(string) !== -1) array.splice(array.indexOf(string), 1);
}

var array = ['a1', 'f4'];
findAndDelete(array, 'a1');

JSON.stringify(array); // ['f4'];

a couple of things are happening here:
array.indexOf finds the numeric index of the value (in this case 0 is the index because array[0] == 'a1')
(If the value is not found the index returned will be -1)
array.splice uses the index to delete the item located at that index, the second numer specifies how many elements to delete from there.

A: There are two possible solutions to this problem, depending on what you want to get out of the function. The first solution will work well if there is only ever one element in the array that matches the string, of it you only want to delete the first element that matches. The second solution will delete all instances of the string passed into the function.
Delete the First Element to Match the String Passed In
The solution that I found to delete one element that matches your passed in string is to use the array.indexOf() method, test the value from that method and, if the index returns a value other than -1, then to pass the index to splice and delete the element from the array. My code is as follows:
var array = ['a1','f4'];

function findAndDelete(arg){

    var testString = arg;
    var index = array.indexOf(testString); //cache index

    if(index !== -1){
        array.splice(index,1);
    }
}//end findAndDelete()

Once a string is passed into the function, it is stored in the testString variable. Then, we call array.indexOf(testString) and cache the value returned by the function in the index variable. This value is cached to optimize the script and to keep the code DRY (Don't Repeat Yourself). The value of index is then tested to see if it is anything other than -1. The reason index is tested against -1 is because, if the string passed into indexOf() does not exist, the function will return -1. So, if index is not -1, we know that the string passed in exists in the array. The index of that value is then passed to splice as the first argument, which tells where to begin the splice in the array. The second argument passed into splice is the number of element to delete from the array. Since we are only deleting one element of the array, we always pass 1.
You could also create a function that loops through the array with a for-loop and checks every element against the testString. This will work just as well, but is going to be a less efficient way of solving your problem if there will only ever be one element in the array that could match the string you are passing into the function.
Deleting All Instances of the String Passed In
However, the solution I listed above will not work if there are two identical string in the array. For example, if array=['a1','f4','a1'], the function will only delete the first element that it finds. So, after running the function with an argument of 'a1' the array will be: array=['f4','a1']. If you want to be able to delete all elements of an array that match the string you have input, the for-loop solution will work best. The following code would work if you want to delete all elements of an array that match your input:
var array = ['a1','f4', 'a1'];

function findAndDelete(arg){

    var testString = arg;

    for(var i = 0; i< array.length; i++){
        if(testString === array[i]){
            array.splice(i,1);
        }

    }//end for-loop

}//end findAndDelete()

Additionally, if the script is running on IE 8 or older, the indexOf() method will not work. You will need to use the for-loop solution or look for another way to solve your problem.

A: You are using find() incorrectly. The function you want is indexOf().
IndexOf() returns the index of the value you give it. So you just need to replace
delete array.find('a1');

with
delete array[array.indexOf('a1')];

Note that this only removes the first instance of it.


text:
Q: ICS Emulator will not create Calendar Account Please help.  I absolutely need to be able to test the Calendar.contract provider in my app using the SDK emulator.  Purchasing a $400 hardware device just to test is not a real option.
I have seen the posts that remind us that we need to target the Google API platform, which we have done.  Next, I have seen a work around where it is suggested that we create a google account as an exchange account by specifying the m.google.com server and clicking "Accept all SSL".  Unfortunately, that generates the error that:
You have typed an incorrect server address or the server requires a protocol version that Email doesn't support.
We created a basic Google account.  However, the emulator still says we must create at least one calendar account and forces us into creating the Echange Account.  According to the emulator, the account we did create as a Google account has no calendars associated with it, even though I can go to Google Calendar, access that account, and add items to a calendar.
Can anyone suggest a solution?
Thanks,


Welcome to Countryside Landscaping LLC, where our main priority is our customer's satisfaction! Countryside operates out of northern New Jersey by owner Jay Meerendonk, who has been working in the landscaping industry for over ten years. In those ten years he has obtained his Associate's Degree in Landscape Design and Management as well as attended many continuing education classes and seminars. He orchestrates a highly qualified team that can handle any and every aspect of lawn maintenance and landscape construction for residential and commercial service which includes but is not limited to: spring and fall cleanups, retaining walls, paver walkways and patios. Countryside Landscaping operates year round by providing snow plowing for private driveways and public parking lots during the snowy winter months. If you are interested in lawn maintenance, landscape construction, or even snow plowing please give us a call or contact us by clicking here!
Content copyright . Countryside Landscaping LLC. All rights reserved.


The video of a cabbie confronting the driver of a stolen SUV in the world's worst hit and run accident is a big hit at parties and smartphones everywhere. Now, courtesy of Chicago Dispatcher, we can see the hit and run and its setup via the taxi's dashboard camera.
The video clocks in at nearly nine minutes and 30 seconds and the dashboard camera doesn't provide the vantage point of the video that went viral. But there are some notable moments in the video starting at the one second mark when you can see the camera shake from the SUV rear-ending the taxi.
At the 55 second mark you can see the SUV make a sharp left and leave the scene of the accident, followed by the cabbie in hot pursuit, to quote Roscoe P. Coltrane.
The cabbie follows the SUV tightly down Southport when the driver of the stolen vehicle turns west on Belmont. Note at the 5:19 mark as the cabbie waits for some pedestrians to cross before finally cutting off the SUV on Belmont. The rest, as they say, is history. The dashboard camera records the growing congestion on Belmont Avenue's eastbound lane, a sign of the cabbie confronting the driver of the SUV. At the 8:55 mark the cab is hit again, followed two seconds later by the SUV slamming into the dealer vehicle on the left.
As you can see, the cabbie took a hell of a chance by chasing down the driver of the stolen SUV. It was a dumb move in retrospect but we can see why he did it. And the driver of the SUV did eventually wind up getting arrested.


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