Ps: i still wished they did not make him jump from the roof for an easy way out. I'd rather he had some after surgery complications hua3. I really dont support suicide eventhough its in a drama. But it was somewhat staying true to its original source.
My only disappointment is no kiss. I was still holding out for it since I saw the spoiler video of the BBC version.
I suppose they did his death jump in a very classy way.
He did not feel pain when he clenched his fist around the file clip but remembers feeling NY's heart beat under his hands. So NY is more real to him than 2018.
You can see he jumped in a very confident manner and seemed to be embracing the possibility of going back to 1988. I thought the drama would end at this point but PD nim and writer nim gave us an epilogue of sorts to see how the team were doing after the warehouse attack and to highlight reasons why TJ would decide to stay.
I think TJ got his happy ending.
Excuse me for my nagging but I just don't understand how a person can survive a jump from a rooftop.
LOL... we could always assume that the voices now on the television and radio could be the divine one calling him to redemption, not 2018.
If you think about it, he was run over by a car (twice!) , shot in the head and survived a fall off a bridge. Superman he is!
I think I will have to watch the episode again with subs but today it was easier to see where it was headed.
Just now, aisling said:Excuse me for my nagging but I just don't understand how a person can survive a jump from a rooftop.
Actually I don't think he can survive logically speaking. So that's why I believe he died. To make myself feel better , I think of 1988 as an Alternate reality for TJ. But the call from KMS messed with my mind .
I couldn't get to watch the show live, but reading the comments I actually quite like the ending (though I wished there was a kiss or even a scene with TJ&NY in a belated cinema date lol).
I read though that in the BBC version, there was a sequel. Did the K version speed through the trilogy or was it just the first season?
Cause I want more the group dynamic was amazing.
All I can say is I'm glad that they did the BBC ending. I cheated yesterday and looked it up on a website and that ending seemed very well received even with viewers still trying to interpret their take on the ending.
Trawling internet for any updates..
From the tweets, it looks like a lot of speculation about Season 2.
KMS is from 2018 trying to bring him back so his spirit is in 1988. He's in a coma again.
@Dramanoona @aisling yeah, he's superman in 1988 so we know his body is not physically in 1988.
Seriously this drama is open to interpretation.
Ok... I read more meaning in the lyrics now.
I need the subs asap, and several hours to digest/think what happened.
Wow the final was great - even so I got what I feared I still love it.
And yes - we did not get a kiss - sob - but to see TJ happy and singing and not looking ill anymore - that was even better for me. So cute when he started to sing along in the car he looked so surprised about himself.
I read somewhere (but can´t find it now) that in the original the final scene was planned as an open ending the moment when he jumped from the roof. So it would have been up to the imagine of the viewer what will happen next. Oh gosh - i would have freaked out like that.
I wonder - if this is a world somehow created between life and death or even for the near death and death people - died the villian and followed TJ? Its really open for season 2. This Show was amazing - not one minute was draggy or boring. I will wait for the subs now.
Whole cast was amazing !
Jung Kyung-Ho had one more great performance.
His role as Paksa in" Cruel city" is still the best of him for me but he is such a great actor he never disappoints you.
When I saw Go Ah-sung in " Heart through the grapevine" i knew she is more than a beautiful face.
I never liked second season in dramas and I hope they will leave us with this open ending.
Han Tae Joo (Jung Kyung Ho) manages to go back to the past by jumping from the roof top.
He manages to save his teammates, and arrest Ahn Mi Sik (Choi Jin Ho).
He decides to stay in year 1988 along with them.
At the end of the episode, Han Tae Joo receives a mysterious call from Kim Hyun Seok?
The best drama in 2018!! You guys have worked hard in such a hot weather. ㅠㅠ They didn't reveal why Han Tae Joo's father died.. And the criminal is alive too, let's go with Season 2!!
I had goosebumps at the end.... Season 2, let's go!!!!!!!!!
What am I going to do on Saturdays and Sundays now? ㅠㅠㅠㅠ The casts and staff have worked hard.
I had this thought that Park Sung Woong only suits evil roles. But he fits comical characters well too. ㅠㅠ Tsundere. Life on Mars' actors and staff, thank you for making the best drama this summer. ♥Season 2!!!!
It seems like a happy ending, and it also seemed like a sad ending..
Saying that you are alive wherever you can smile.. It's good, but dangerous too.. At the end, it feels a bit empty.. but the epilogue turned everything around.. It would be good to catch the criminal Kim Hyun Seok in season 2.. and differentiate reality and dreams properly..
Please be happy, Han Tae Joo.. Whether you go back or not... Just be happy wherever you are... and I'll be waiting for season 2.
I enjoyed watching. It seems short. What a pity.
They opened up the possibility for season 2 and ended it. I enjoyed it. Members of Detective team 3 have good chemistry. I feel so empty now that it ended..
Seems like there's season 2 after watching the epilogue!! Season 2 please!!!
YAY for ending on a high note. Well deserved. Be happy always Tae-joo and Na-young.
5.8% and 6.37%! Very well desereved!
Eng subs are out. It was a surprisingly feel good episode. And theres build up for taejoo's decision, which i appreciate.


text:
Home News SPIRITS OF FIRE Releasing Debut Album in February
SPIRITS OF FIRE Releasing Debut Album in February
After an intense brainstorming session, a dream team of renowned metal musicians including Tim "Ripper" Owens (ex-Judas Priest, Iced Earth), Chris Caffery (Savatage, Trans-Siberian Orchestra), Steve DiGiorgio (Testament, Death) and Mark Zonder (ex-Fates Warning, Warlord) was born and put together under the direction of renowned LA-based producer Roy Z (Bruce Dickinson, Halford, Tribe Of Gypsies).
Called SPIRITS OF FIRE (after the title of one of their songs), the band has completed work on their debut album which is set for release on February 22, 2019, via Frontiers Music Srl.
The music can best be described as "classic Priest meets Savatage".
Recently, SPIRITS OF FIRE released their first video and single, Light Speed Marching.
Guitarist Chris Caffery on his influences for the record and working with Roy Z: "My influences for writing on this record were the bands that we were and are still in! I wanted to write songs that Priest, Savatage, Testament, and Fates Warning fans would like. But, I wanted them to have a personality of their own as well. To give it a vibe that was SPIRITS OF FIRE! I grew up listening to Judas Priest and spent 30-plus years working with Jon Oliva and Paul O'Neill with Savatage and TSO. Many years playing Criss' music. These influences for me came naturally! Especially because I'm a fan of all of our bands! This was the first time that I ever worked with Roy and it was so much fun. I have been a huge fan of his work for a long time. He's written music, played on, and produced some of my favorite metal albums ever. He had a difficult job bringing this all together and I can't think of anyone that would have been a better producer and co-worker for this album!"
Vocalist Tim "Ripper" Owens says, "I always just sing to the song, whatever I feel it should be. It's like a character thing! I like layers! This music is really perfect for me and in my natural wheel house. I like to sing all over the place and change it up and this all happens on this album. This is just the start of an amazing band, I mean sky is the limit and I actually can't wait to record the next album, because this one is so great and perfect."
Drummer Mark Zonder adds about his playing and writing process: "I am the kind of player that plays what I think fits best with the music. With SPIRITS OF FIRE, the music was more aggressive and hard hitting which I really liked. Again, I tried to be very mindful of the song and the vocals, but tried to add parts that were interesting and creative for the listener."
SPIRITS OF FIRE Tracklisting
Light Speed Marching
Temple of the Soul
All Comes Together
Spirits of Fire
It's Everywhere
Dispensable
Meet Your End
Never to Return
Alone in the Darkness
The album will be available on CD, red colour vinyl, standard black vinyl, collector's box (CD + t-shirt, size L), and digital formats.
Pre-order "SPIRITS OF FIRE" on CD/LP/digital and stream the singles here.
North American fans can get LIMITED EDITION, exclusive red colour vinyl and collector's box (feat. CD + t-shirt – Size L) HERE. These are very limited and this is the only retailer in North America carrying these items.
European fans can get the same items HERE.
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What Is Sme Finance And How Can It Help You?
Small and medium enterprises or SMEs hold a special place in the economy of developing countries as it is critical for the development and growth of the society and economy. The role of SMEs in emerging markets acquires immense importance due to its ability to assist low income groups in generating income and in creating jobs.
Governments in the developing countries are eager to support this sector of business that has contributed to the development of a strong private sector and has also fostered social stability and economic development. Like any other business, the growth of SMEs is dependent on proper access to the financial services. That banks attach special importance to SMEs would be evident from special financing schemes exclusively for them and is termed as SME finance.
The SME sector as a whole is considered important for many economies. In the UK, half of the national income and half of the employment are generated by SMEs, thus making it extremely important for the country's economy. The size of SMEs often works to its advantage as its structure allows a lot of flexibility that also facilitates quick innovation. It is also easy and quick for SMEs to adapt new technologies. These abilities of SMEs make it attractive to other big companies that lose no time in acquiring the smaller entities to exploit the full potential of its innovative capabilities. SMEs thus add more value to the economy by allowing scale up of innovations.
Despite huge potential in creating new businesses and adding more value to it through innovation and better use of technology, SMEs still find financing a difficult proposition. It is common to hear that SMEs are deprived of proper financing opportunities that can boost their creative capabilities and growth. SMEs do receive finance but not in the way they want. This creates a gap in financing.
Investor funds are mostly directed to big companies and government projects as investors feel that SMEs are less attractive. In addition, there is a general perception that it could be risky to invest in SMEs due to uncertainties that are quite high. From the investor's perspective, return on investment from SMEs does not make investors happy and some SMEs have limited internal controls as they are out of the purview of stock exchanges. Many SMEs can even get a waiver on annual accounts auditing in the U.K.
Funds for SMEs are usually available from banks and other modes of financing. Banks finance SMEs against collateral securities like land and machinery and can even arrange for overdraft facilities. Leasing assets instead of investing in it is another financing option that is available. Other methods of financing can include trade credit from suppliers and other creditors, invoice factoring and discounting as well as depending on friends and family for financial help line whenever it is needed.
The best thing about SME financing is that although it is not always easy, it seldom becomes show stopper.


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Children's workforce training strategy 2018 to 2021
Published: 2020-01-09 Updated: 2020-01-09 Subject: Children, young people and families
This strategy provides the framework and direction for developing a Children's Workforce in Bath and North East Somerset to ensure it is capable and confident to achieve the best possible outcomes for all children and young people in the Local Authority. It aims to ensure that the workforce is inducted, supported and appropriately trained to perform to the highest standard; a workforce which parents, children and young people engage, trust and respect.
Children's workforce training strategy
Copyright © 2020 Bath & North East Somerset Council


text:
Home » When Will Cloud ERP Take Off?
When Will Cloud ERP Take Off?
Pat Garrehy, President & CEO, Rootstock
Nonetheless, today on-premise ERP remains king of the installed base, even though many of these systems are 10, 20 and 30 years old - ancient in the world of software applications. So, with the knowledge of the benefits of cloud ERP, which include reduced IT operational costs by outsourcing hardware and software maintenance and support to the cloud provider among others, what is causing manufacturing cloud ERP to presently trail in the market?
One major obstacle is that the larger ERP vendors, especially those investing big marketing dollars, are on-premise providers who continue to focus on their on-premise systems, even if they could offer a cloud ERP solution. The second obstacle is that end users with on-premise systems are not being quick to replace them. Either they are trying to wring a little more ROI out of their former investment or, remembering the implementation frustrations of their present suite, don't want to change regardless of cloud ERP benefits. However, industry analysts report that an increasing number of companies in the market for a replacement ERP will consider a cloud solution. Two years ago, it was estimated that 25 percent would consider a cloud solution; today, that estimate has grown to 40 percent.
Who Is Implementing Cloud ERP Now?
Cloud ERP is presently being embraced by three different types of organizations. The first are new, emerging businesses that have never had an on-premise system and want to immediately harness the benefits of cloud ERP. Astrum Solar, a full-service residential solar provider headquartered in Annapolis Junction, Md., is such a company.
"Astrum Solar's rapid expansion facilitated our requirement for a solution that could interweave the ever-changing environment of the solar industry with the rapid development in our company," said Vadim Polikov, president and chief executive officer of Astrum Solar. "We chose [cloud ERP] "¦ because [it will] enable our organization to sustain our flexibility and commitment to the customer as we continue to grow our business and spread solar power to roofs across the country."
The second customer group is those smaller plants of larger corporations that never successfully implemented the larger on-premise solution installed at corporate.
EBARA International, a Japanese world leader in the engineering and production of specialized cryogenic liquefied gas pumps and turbine expanders, is typical. Frank Lowery, director of IT at EBARA International, has a highly functional on-premise system but knows that the cloud is the wave of the future and EBARA can invest in expanded capability at a reduced cost with it.
"EBARA needed to move away from the closed and rigid processes of legacy ERP to the very flexible and open workflows possible [with cloud ERP]" said Lowery. "[We needed a] cloud platform that [could] support our complex, project-oriented, engineer-to-order manufacturing environment today."
The third group comprises companies that are experiencing the end of life of their on-premise systems, such as Oclaro (formerly Xtellus), a San Jose, Calif.-based supplier of scalable Wavelength Selective Switches (WSS) for reconfigurable optical networks. Oclaro grew rapidly and could no longer afford to depend on disparate systems, one for accounting and the other for manufacturing.
Given Oclaro's expansion into a global multi-facility manufacturing and engineering operation, it required an integrated manufacturing, planning, sales, accounting, purchasing, inventory and shop floor control ERP system with standard cost and project cost capabilities. Oclaro also wanted a contemporary, SaaS ERP system which insured that could minimize investment in an IT infrastructure.
Many On Premise ERP Users Staying Put - For Now
There are a number of ERP systems developed 20 and 30 years ago that are still effectively running today. In addition, there are ERP vendors that are continuing to profitably reap software maintenance fees from these systems, keeping a small support development crew to maintain these systems, at least for a few more years.
As a result, when considering replacing on-premise ERP with cloud ERP, one must take the longer view. Massive investments similar to the Y2K challenge that spurred spending on ERP systems in the late '90s will not happen with cloud ERP. Even though there are many benefits from a cloud ERP solution, the eventual replacement of on-premise systems will take time.
While the adoption rate for cloud ERP solutions increases year by year and while new system buys are predominantly falling to cloud ERP vendors, the replacement of on-premise ERP (especially manufacturing cloud ERP) has not yet reached the tipping point and will not do so for several years. Once present on-premise ERP software maintenance practitioners either retire or leave for better job opportunities, the momentum will switch to cloud ERP. As a result, within 10 years, the majority of ERP investments will be in in the cloud.
Source: Rootstock
Keywords: enterprise resource planning, legacy systems, supply chain management IT, supply chain solutions, ERP in manufacturing SaaS ERP, ERP in the cloud
Technology Cloud & On-Demand Systems ERP & Enterprise Systems High-Tech/Electronics
KEYWORDS Cloud, SaaS & On-Demand Systems Enterprise Resource Planning ERP & Enterprise Systems ERP in manufacturing SaaS ERP ERP in the cloud High-Tech/Electronics legacy systems supply chain management IT supply chain solutions Technology
Taking S&OP Out of ERP - Turning to Cloud-based Technology for Sales and Operations Planning
Top 5 Reasons S&OP Belongs in the Cloud: Integrating Cloud S&OP With ERP vs. Customizing ERP for S&OP


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Q: RANK vs DENSE_RANK for postgreSQL queries I just started learning SQL about a month ago and have been practicing writing queries from the NorthWind sample database on postreSQL in pgAdmin4. After posting about this specific query I am trying to solve the other day, this is what I have come up with:
-- Return Month, Product name, SalesForMonth, SalesForMonthRanking for nth best selling product
-- for each month in 1997

This is the query I wrote in order to solve said prompt:
SELECT *
FROM (
    SELECT
        CAST( EXTRACT( MONTH FROM o.orderdate) AS integer) AS month,
        p.productname,
        ROUND (CAST (SUM (od.quantity * od.unitprice) AS numeric),2) salesmonth,
        RANK () OVER (PARTITION BY EXTRACT( MONTH FROM o.orderdate)
        ORDER BY SUM (od.quantity * od.unitprice) DESC) monthrank
    FROM order_details od
    INNER JOIN products p ON od.productid = p.productid
    INNER JOIN orders o ON od.orderid = o.orderid
    WHERE o.orderdate >= '1997-01-01' AND o.orderdate <= '1997-12-31'
    GROUP BY EXTRACT( MONTH FROM o.orderdate), p.productname
) subq
WHERE monthrank = n
ORDER BY month, monthrank, productname

While the result is correct, I noticed that when finding the 27th highest selling product for each month there is no result for month = 6 because there are 2 products tied for the 26th highest selling product. My question is: is this supposed to happen or should I change my RANK statements to DENSE_RANK statements to combat this issue. Also if there is any advice anyone wanted to give me on the actual structure of my query that would be greatly appreciated.

redpajama_set_name: RedPajamaStackExchange


Rusty Griswold takes his wife and their two sons on a family vacation to the coolest theme park in America, "Walley World". This road trip is a chance to rekindle his marriage and build lasting relationships with his sons. Needless to say things don't go according to plan!


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Q: why can't I take input while debugging in VS code? #include<stdio.h>
int main()
{
   int a,b,c;
   scanf("%d %d",&a,&b);
   c=a+b;
    printf("%d\n",c);

return 0;
}

this is my simple program. my terminal does not take input while I am in debugging mode. here I am attaching my launch.json setting.
I can take input in terminal when I am not in debugging mode. This problem accours when I am debugging a program. I have set my "externalConsole": true, but its not working afterall.


text:
November 27, 2014 December 15, 2014 Alicia Sakal
Mike Luzi: 30 Years of Humanitarian Service to the Town of Sherman
Contributed Photo
This original article ran in the October 8th Edition of the Citizen News, a local newspaper serving Sherman and New Fairfield, Connecticut.
By Alicia Sakal
In our small New England town of Sherman, many local people know Mike Luzi as "one of the nicest guys ever, who runs the Sherman IGA". For longtime residents, if they really think hard enough, they may actually be able to count the years he first opened the store. As for newer residents, the total count happens to be 30 years of amazing service and dedication to the Town of Sherman.
What is common knowledge in our community is that Mike provides residents with a convenient market and deli with "something for everyone", his selection is expansive, and he'll special order anything for you. He also has the friendliest of employees on earth that will help you out. As for the countless fundraising initiatives in Sherman, plenty of people know that Mike provides food and beverages at cost and is always their go-to person.
Mike directly touches the lives of families and residents in need, especially during economic downturns, and he is synonymous with local charity. Mike also provides much needed food and organizes food drives on their behalf. In addition, as if this weren't enough on his list of good deeds, Mike makes donations to local nonprofit charities that help to further support many noble causes in the community. No wonder why the Office of the First Selectman declared this Saturday, Oct. 11 as the official "Mike Luzi Appreciation Day".
When asked how he picked Sherman as his new store location 30 years ago, Mike replied, "I was already running another grocery store in Winsted, Connecticut at the time and just happened to read in the newspaper about a grocery store that was for sale. The next thing I know, I found myself jumping in the car and heading to Sherman." The rest is history, as the saying goes, and Mike immediately fell in love with the store, the building, and the Town of Sherman.
He quickly became the new, proud owner of the Sherman IGA. "Family and friends helped me to clean, paint, and stock the store," he said. Literally, three weeks after receiving the keys to the storefront, the Sherman IGA doors re-opened for business. Mike recalls opening day vividly and knew instantly that he made the right decision. "Much to the relief of the town, they welcomed me with open arms.
Mike commutes from Bantam where, even during inclement weather, like with Hurricane Sandy and other major storms over the past years in which the town lost power, Mike was there with his doors open for business. "I mostly open the store myself. I leave home at 4:30 A.M. and I usually arrive at the store around 5:00 A.M." The store's official hours are 6:00 A.M. to 8:00 P.M. seven days a week, year-round. The one-and-only day the Sherman IGA is not open is on Christmas Day because Mike keeps a promise that he made to his Mom decades ago.
Mike takes pride in his early morning ritual which consists of meeting people for deliveries, lining up the newspapers, getting the bakery going to make the bread, and making the coffee for his customers. A "Day in the life of Mike" starts with the first shift of customers around 6:00 A.M. "I must say, the Sherman IGA is the preferred meeting place for the majority of the local contractors like the town's construction workers, painters, and floor sanding guys. They come in for coffee and talk about the day and weather. They even come here to network with each other and have formed a group," said Mike energetically.
The second shift of shoppers Mike sees are the moms and dads that suddenly realize they ran out of sandwich meats for their children's school lunches, and they come into the store in a panic. The commuters also come in at this time to buy newspapers before heading out to the train station. "It is routine to see mothers doing their shopping while the kids are in school. Then there's the after 5:00 wave of parents with their children that need to shop for sporting events, or people just stopping in to shop after working all day," said Mike.
"The Sherman IGA is a hometown market, a meeting spot, and is the center of the town. People, my customers, come to me with all kinds of issues and I just want to be there for them. It can be anything: a loss in the family, children having problems, someone needs food. I also love helping all of the nonprofits like the Boy Scouts and the Girl Scouts, and I just love to do good things for the community. The people feel like the Sherman IGA is their very own private store."
As for Mike's other charitable causes, aside from the 12 listed in the Mike Luzi Appreciation Day Proclamation, he donates to other nonprofits outside of the Town of Sherman. For well over 30 years, he has supported the Special Olympics. His store is also a major contributor to the Connecticut Food Bank. Last year, this nonprofit picked up 6,000 pounds of food just from the Sherman IGA to support their cause for feeding one out of every seven people who go to bed hungry in this state. "They couldn't believe it and how beyond generous our small town community is," Mike declared with pride. Once a month, and with the help of many others, he also helps feed 40 to 50 homeless people in New Milford.
As for the challenges Mike faces as a small town business owner, there are plenty that worry him. He said, "I've had some great days and good years, but right now, this location struggles. It's a sign of the times when you see all of the small businesses around us that closed. Factor in the added costs of running a local business, like increased taxes, it's not easy but I love it because I am an integral part of this community."
When asked what more we can do as a community to give back to him, Mike confided, "I need your support for my business and your loyalty. Please come into my store and shop whenever you need something. Support your local businesses in town because once the small shops are gone, they are gone forever. The big stores are gobbling us up. In order to be successful, we all have to work together."
Mike also discussed how important it is in real estate to have a local grocery store in town. He thinks that if the Sherman IGA didn't exist, then some potential buyers may choose other small towns. "Sherman is a sought after town to live in. Realtors showing homes for sale always stop here. In order to make and keep Sherman as a desirable New England town, my store and supporting it is very important in making the decision to move here."
For all of these reasons, and then some, the Office of the First Selectman is honoring Mike and recognizing him for his years of service and dedication to our community. Whenever you have a minute Saturday, October 11th try to make a special trip to the Sherman IGA to say "thank you" to Mike for everything he does for the people of Sherman. You never know when you or a neighbor may one day need his help.
Business owner. Local humanitarian. Person with a heart of solid gold. Now, this is an exceptional human being and Sherman is beyond lucky to have Mike Luzi. Thank you, Mike, for embracing our Town.
mike luzi
sherman iga
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Rebecca Black says:
Lovely article about a lovely person.


text:
LA STAGE Times Archive / September 30, 2010
Fresh From the Fringe…and Other Plays About America's Wars
Last July I expressed skepticism — on a Hollywood Fringe panel and later in writing on this site — about the euphoric hopes that the Hollywood Fringe would somehow transform L.A. theater. That particular edition of LA Stage Watch drew a lot of comments — most of them still hopeful after all those words.
One of the responders, Gregory Crafts, asked whether I had seen a Fringe show called The Birthday Boys. He maintained that "it was better than anything I've seen at one of the 'top tier' LA companies. I put it right up there with some Broadway shows I've seen. I would be interested to hear what you thought."
Unfortunately, The Birthday Boys was not among the nine Fringe shows I saw in June. And I was hardly alone. Even the helpful cooperative effort by the LA Weekly and Backstage to review as many Fringe shows as possible managed to miss The Birthday Boys — although at least it was acknowledged on a few web sites, including this one.
But the play went on to win the Hollywood Fringe First award as the Fringe's outstanding world premiere. Thanks in part to that honor, the production was invited to participate in an extension of some of the Fringe shows at Theatre Asylum.
The Birthday Boys. Photo by Steve Kozak.
So I finally got to The Birthday Boys last weekend, three months after the Fringe. And I'm glad I did. While it was hardly "better than anything I've seen at one of the 'top tier' LA companies," it was indeed among the top two of the 11 Fringe shows I've now seen (I also saw another of the shows extended at Theatre Asylum extensions since the Fringe).
The Birthday Boys details the ordeal of three Americans soldiers who are kidnapped and held hostage in Iraq. At first that set-up seems almost too easy for playwright Aaron Kozak (although not for the characters or the actors). Almost any writer could create something interesting out of such a naturally tense situation.
Kozak makes the three young actors remain in blindfolds, with hands and feet bound, throughout most of the play. The hard-working cast succeeds at transcending such restrictions and helps Kozak create somewhat involving characters. But the biggest challenge facing Kozak was to figure out how to end it.
He came up with a truly remarkable ending — something that I didn't anticipate but that, in retrospect, seems at least somewhat plausible. Of course I can't discuss it — the play has another weekend in its extension, and you really don't want to know about it before you enter the theater. Even after the production closes, no one should publicly discuss the ending. It could easily be spoiled for those who see the next production. But it's a fascinating topic for face-to-face discussion among friends who have seen the play together.
One topic for that private discussion — does the ending make the play feel a little too much like a stunt? Perhaps, but that doesn't necessarily invalidate it. At least Kozak appears to be in complete control of the ride that he takes his audience on.
The Birthday Boys, Theatre Asylum, 6320 Santa Monica Blvd., Hollywood. Tonight-Sun, 8 pm. 800-838-3006. www.brownpapertickets.com/event/123759
That sense of artistic control was missing from another recent play that refers to America's recent wars, Titus Redux at the Kirk Douglas. This multimedia adaptation of Shakespeare's Titus Andronicus was fascinating, but its transposition to the current Iraq-Afghanistan era felt sloppy and confused.
After my initial remarks about it in LA Stage Watch, its creator and director, John Farmanesh-Bocca, invited me to see it again. He wrote to me that it had been rushed into production without workshops or sufficient time. Even some of the favorable reviews indicated a degree of misunderstanding about what went on in the play, he wrote. "We knew we had missed them."
Jack Stehlin as Titus and John Famanesh-Bocca as Aaron, his neighbor who isn't really a Jihadist. Photo by Ed Krieger.
Farmanesh-Bocca wants future audiences to understand that just about everything that happens in the production, after Titus returns home to D.C., is presented from the perspective of his post-traumatic-disordered brain. No, his wife doesn't really collude in their daughter's rape and mutilation, nor do their sons. The friendly neighbor (played by Farmanesh-Bocca) is not a sinister Jihadist waiting to pounce — that's all in Titus's head. "Giving the audience a simply harrowing experience was not the main goal," wrote Farmanesh-Bocca. That goal was "to make them feel some of the heart-wrenching disorientation and confusion that PTSD sufferers face."
From my own first viewing, it was clear that Titus was somewhat delusional — but it wasn't clear to what extent. I saw the production again on its closing night — and it made a lot more sense for me, but then I had the advantage of having read a personal explanation from the director in my e-mail. I'm not sure if others who saw it for the first time at that closing performance necessarily understood it better than did the audience on opening night. Farmanesh-Bocca wrote that he hopes to "repair that airplane" before it flies again.
It's good to see playwrights think imaginatively about how to dramatize America's 21st-century wars. But by far the most affecting such plays to be seen in L.A. in recent weeks were the revivals of Tom Burmester's and Los Angeles Theatre Ensemble's Wounded and Nation of Two (the latter is the new title for the former Survived), at the Powerhouse. (A third play in LATE's War Cycle trilogy is still waiting in the wings — it was workshopped along with the recent revivals of the earlier plays but has yet to officially premiere.)
These now-closed productions offer none of the high concepts at work in The Birthday Boys and Titus Redux — no surprise endings or Shakespearean plots. In Wounded and Nation of Two, Burmester employs old-fashioned realism. Many theater critics tend to disparage realism because it isn't sufficiently different from the methods available in other media, but it worked like gangbusters in these two plays and in this small theater. These plays also could move audiences at a somewhat larger, actor-paying venue. I hope that midsize and even larger theaters are lining up to take a crack at them.
PianoFight Secures Its Position in LA Theater
Oliver Mayer Heats Up LA with Two New Projects
Don Shirley
Don Shirley writes about theater for LA Observed. He is the former longtime theater writer for the Los Angeles Times, LA Stage Times and other publications.
Nina Harada heads up San Francisco's PianoFight's expansion to LA with The Position. READ MORE


text:
\section{Introduction}

When calculating certain functions of quantum systems, in many instances
the running time of classical computers increases exponentially with
the number of elementary parts that compose those systems. This issue
is a hurdle to current research in many areas of science. But it is
also a motive for the quest towards the construction of a large scale
quantum computer \cite{Feynman82,Loyd,Ito}. For now, we have to resort
to several other alternative techniques, with which one can extract
approximate information about quantum systems using only the available
classical computing power.

Among those methods, some famous examples are: stochastic Monte Carlo
simulations \cite{Krauth,Binder,Kroese,Lutsyshyn}, mean field approximations
\cite{Kadanoff,Liu_mft,Sen,Yamamoto}, density functional theory \cite{Capelle,Ullrich,Jones,Kanjouri_DFT},
renormalization group \cite{Fisher_RG,Kadanoff_RG,Alvarez,Osborne_RG},
and matrix product states and projected entangled pair states \cite{Orus,Cirac,Troyer,Karrasch}.
On the other hand, recently several authors have shown that some general
patterns of the many-body behavior, usually accessed in the thermodynamical
limit, may be disclosed by analyzing systems with a moderate number
of particles \cite{Campbell0,Monroe,Giorgi,Campbell1,Kwek,Liu,Sen_SwNqb,Campbell2}.
In such kind of investigation, it is desirable to use a system as
large as practically possible. And for that purpose we would like
to optimize the implementation of basic and frequently used functions
in order to reduce the computation time as much as possible.

In the quantum mechanics of composite systems, one ubiquitous function
is the partial trace (PTr) \cite{Nielsen=000026Chuang,Preskill,Paris,Wilde}.
The PTr function has a unique place, for instance, for the computation
of reduced density matrices and related functions. In spite of the
physical interpretation of the partial trace not being a trivial matter
\cite{Garola,Fortin}, it's mathematical and operational meaning is
well established \cite{Nielsen=000026Chuang,Preskill,Paris,Wilde}.
Besides, the PTr appears very frequently, for instance, in the context
of correlation quantifiers (mutual information \cite{Winter,Maziero_MI},
quantum entanglement \cite{Bennett_EE,Okano,Fritzche,Loss}, quantum
discord \cite{Maziero_RevD,Vedral_rd,Maziero_Opt}, etc), in the generation
of random density matrices \cite{Maziero_rrho,Maziero_Fort}, and
in investigations regarding phase transitions \cite{Sacramento,Zeng}.
It is also a fundamental ingredient in the quantum marginal and extension
problems \cite{Bellomo2,Christandl,Modi,Viola1,Viola2,Chen1}, for
the strong subadditivity property of von Neumann entropy and related
results \cite{Bathia,Petz_SSAeq,Lieb_ESSA}, and in the theories of
quantum measurement and decoherence \cite{Schlosshauer,Zurek_Dar,Maziero_Zimmer,Fritzsche2}.

Our aim here is to examine in details the partial trace function with
special focus on its numerical calculation. The remainder of the article
is structured as follows. In Sec. \ref{PTr_def} we present two definitions
for the PTr involving bipartitions of a system, verify their equivalence,
and discuss the uniqueness of the partial trace function. In Sec.
\ref{num_bip} we address the numerical calculation of the partial
trace, firstly via its direct implementation (Sec. \ref{sec:1st})
and afterwards using two levels of optimization which are obtained
simply by avoiding making null multiplications and sums (Secs. \ref{sec:2nd}
and \ref{sec:3rd}). In Sec. \ref{PTr_mult}, the calculation of the
PTr for the general case of multipartite systems is regarded; and
Fortran code produced to implement the PTr numerically is described.
We consider the computation of the partial trace via Bloch's parametrization
with generalized Gell Mann's matrices in Sec. \ref{sec:su}. A brief
summary of the article is included in Sec. \ref{sec:cc}.

\section{Partial traces for bi-partitions}

\label{PTr_def}

Let $O$ be a linear operator defined in the Hilbert space $\mathcal{H}$,
that is to say $O:\mathcal{H}\rightarrow\mathcal{H}$. The space composed
by these operators is denoted by $\mathcal{L}(\mathcal{H})$. As a
prelude, let us recall that the trace of $O$ is a map $\mathrm{Tr}:\mathcal{L}(\mathcal{H})\rightarrow\mathbb{C}$
defined as the sum of the diagonal elements of $O$ when it is represented
in a certain basis $|\psi_{j}\rangle\in\mathcal{H}$, i.e., $\mathrm{Tr}(O)=\sum_{j=1}^{d}\langle\psi_{j}|O|\psi_{j}\rangle,$
with $d$ being the dimension of $\mathcal{H}$ \cite{Arfken}.

By its turn, in the quantum mechanics of composite systems with Hilbert
space $\mathcal{H}=\mathcal{H}_{a}\otimes\mathcal{H}_{b}$, the \emph{partial
trace function}, taken over sub-system $b$, can be defined as \cite{Watrous}
\begin{equation}
\mathrm{Tr}_{b}(O)=\sum_{j=1}^{d_{b}}(\mathbb{I}_{a}\otimes\langle b_{j}|)O(\mathbb{I}_{a}\otimes|b_{j}\rangle),\label{eq:ptr1}
\end{equation}
with
\begin{equation}
|b_{j}\rangle=[b_{j1}\mbox{ }b_{j2}\mbox{ }\cdots\mbox{ }b_{jd_{b}}]^{t}
\end{equation}
being any orthonormal basis for $\mathcal{H}_{b}$, $\langle b_{j}|=|b_{j}\rangle^{\dagger}$,
$d_{b}=\dim\mathcal{H}_{b}$, and $\mathbb{I}_{b}$ is the identity
operator in $\mathcal{H}_{b}$ ($X^{t}$ denotes the transpose of
$X$ and $X^{\dagger}$ stands for its conjugate transpose). So the
partial trace is a map
\begin{equation}
\mathrm{Tr}_{b}:\mathcal{L}(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H}_{a});
\end{equation}
and the analogous definition follows for $\mathrm{Tr}_{a}:\mathcal{L}(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H}_{b})$.

It is worthwhile observing here that the definition above is equivalent
to \emph{another definition} which appears frequently in the literature
\cite{Nielsen=000026Chuang}:
\begin{equation}
\mathrm{Tr}_{b}(|a\rangle\langle a'|\otimes|b\rangle\langle b'|)=|a\rangle\langle a'|\otimes\mathrm{Tr}_{b}(|b\rangle\langle b'|).\label{eq:ptr_def2}
\end{equation}
In the last equation $|a\rangle,|a'\rangle\in\mathcal{H}_{a}$ and
$|b\rangle,|b'\rangle\in\mathcal{H}_{b}$ are generic vectors in the
corresponding Hilbert spaces. In order to verify this assertion, let
us use two basis $|a_{j}\rangle\in\mathcal{H}_{a}$ and $|b_{k}\rangle\in\mathcal{H}_{b}$
and the related completeness relations to write
\begin{equation}
O=(\mathbb{I}_{a}\otimes\mathbb{I}_{b})O(\mathbb{I}_{a}\otimes\mathbb{I}_{b})=\sum_{j,l=1}^{d_{a}}\sum_{k,m=1}^{d_{b}}(\langle a_{j}|\otimes\langle b_{k}|O|a_{l}\rangle\otimes|b_{m}\rangle)|a_{j}\rangle\langle a_{l}|\otimes|b_{k}\rangle\langle b_{m}|.\label{eq:Orep}
\end{equation}
The linearity of the partial trace and $\mathrm{Tr}_{b}(|b_{k}\rangle\langle b_{m}|)=\sum_{l=1}^{d_{b}}\langle b_{l}|b_{k}\rangle\langle b_{m}|b_{l}\rangle=\delta_{lk}\delta_{ml}$
(we applied the base independence of the trace function) lead to
\begin{equation}
\mathrm{Tr}_{b}(O)=\sum_{j,l=1}^{d_{a}}\sum_{k=1}^{d_{b}}|a_{j}\rangle(\langle a_{j}|\otimes\langle b_{k}|)O(|a_{l}\rangle\otimes|b_{k}\rangle)\langle a_{l}|=\sum_{k=1}^{d_{b}}(\sum_{j=1}^{d_{a}}|a_{j}\rangle\langle a_{j}|)\otimes\langle b_{k}|O(\sum_{l=1}^{d_{a}}|a_{l}\rangle\langle a_{l}|)\otimes|b_{k}\rangle,\label{eq:ptr_def2c}
\end{equation}
which is equivalent to the definition in Eq. (\ref{eq:ptr1}). To
obtain the last equality in Eq. (\ref{eq:ptr_def2c}), we verified
that
\begin{equation}
(|a\rangle\otimes|b\rangle)\langle a'|=|a\rangle\langle a'|\otimes|b\rangle
\end{equation}
for any vectors $|a\rangle,|a'\rangle\in\mathcal{H}_{a}$ and $|b\rangle\in\mathcal{H}_{b}$.

Another important fact about the the partial trace function $\mathrm{Tr}_{b}(O)$
is that it is the only function $f:\mathcal{L}(\mathcal{H}_{a}\otimes\mathcal{H}_{b})\rightarrow\mathcal{L}(\mathcal{H}_{a})$
such that $\mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O)=\mathrm{Tr}_{a}(Af(O))$,
for generic linear operators $A\in\mathcal{L}(\mathcal{H}_{a})$ and
$O\in\mathcal{L}(\mathcal{H}_{a}\otimes\mathcal{H}_{b})$. To prove
this assertion, let us start \emph{assuming} that $f(O)=\mathrm{Tr}_{b}(O)$.
Then, using $O$ as written in Eq. (\ref{eq:Orep}) leads to
\begin{equation}
\mathrm{Tr}_{b}(O)=\sum_{k,l,m,n}(\langle a_{k}|\otimes\langle b_{l}|O|a_{m}\rangle\otimes|b_{n}\rangle)|a_{k}\rangle\langle a_{m}|\otimes\delta_{nl}=\sum_{k,l,m}(\langle a_{k}|\otimes\langle b_{l}|O|a_{m}\rangle\otimes|b_{l}\rangle)|a_{k}\rangle\langle a_{m}|.
\end{equation}
Now, utilizing the eigen-decomposition $A=\sum_{j}a_{j}|a_{j}\rangle\langle a_{j}|$
we shall have
\begin{eqnarray}
\mathrm{Tr}_{a}(Af(O)) & = & \mathrm{Tr}_{a}(A\mathrm{Tr}_{b}(O))\\
 & = & \mathrm{Tr}_{a}(\sum_{j}a_{j}|a_{j}\rangle\langle a_{j}|\sum_{k,l,m}\langle a_{k}|\otimes\langle b_{l}|O|a_{m}\rangle\otimes|b_{l}\rangle|a_{k}\rangle\langle a_{m}|)\\
 & = & \sum_{j}a_{j}\sum_{k,l,m}(\langle a_{k}|\otimes\langle b_{l}|O|a_{m}\rangle\otimes|b_{l}\rangle)\underset{=\delta_{jm}\delta_{jk}}{\underbrace{\mathrm{Tr}_{a}(|a_{j}\rangle\langle a_{j}|a_{k}\rangle\langle a_{m}|)}}\\
 & = & \sum_{j,l}\langle a_{j}|\otimes\langle b_{l}|a_{j}\mathbb{I}_{a}\otimes\mathbb{I}_{b}O|a_{j}\rangle\otimes|b_{l}\rangle=\sum_{j,l}\langle a_{j}|\otimes\langle b_{l}|a_{j}\sum_{k}|a_{k}\rangle\langle a_{k}|\otimes\mathbb{I}_{b}O|a_{j}\rangle\otimes|b_{l}\rangle\\
 & = & \sum_{j,l}\langle a_{j}|\otimes\langle b_{l}|\sum_{k}a_{k}|a_{k}\rangle\langle a_{k}|\otimes\mathbb{I}_{b}O|a_{j}\rangle\otimes|b_{l}\rangle=\sum_{j,l}\langle a_{j}|\otimes\langle b_{l}|A\otimes\mathbb{I}_{b}O|a_{j}\rangle\otimes|b_{l}\rangle\\
 & = & \mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O).
\end{eqnarray}

To complete the proof we \emph{assume} that $\mathrm{Tr}_{a}(Af(O))=\mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O)$
and use a basis of linear operators $\Upsilon_{j}\in\mathcal{L}(\mathcal{H}_{a})$
to write \cite{Paris}
\begin{equation}
f(O)=\sum_{j=1}^{d_{a}^{2}}\mathrm{Tr}_{a}(\Upsilon_{j}^{\dagger}f(O))\Upsilon_{j}=\sum_{j=1}^{d_{a}^{2}}\mathrm{Tr}_{ab}(\Upsilon_{j}^{\dagger}\otimes\mathbb{I}_{b}O)\Upsilon_{j}=\sum_{j=1}^{d_{a}^{2}}\mathrm{Tr}_{a}(\Upsilon_{j}^{\dagger}\mathrm{Tr}_{b}(O))\Upsilon_{j}.
\end{equation}
Above, the second equality is obtained applying $\mathrm{Tr}_{a}(Af(O))=\mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O)$
and the last equivalence follows from $\mathrm{Tr}_{a}(A\mathrm{Tr}_{b}(O))=\mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O)$,
with $A=\Upsilon_{j}^{\dagger}$. Hence the uniqueness of the decomposition
of an element of a Hilbert space in a given basis \cite{Arfken} implies
in the uniqueness of the partial trace function, i.e., $f(O)\equiv\mathrm{Tr}_{b}(O)$.
Summing up, we proved that
\begin{equation}
f(O)=\mathrm{Tr}_{b}(O)\Longleftrightarrow\mathrm{Tr}_{ab}(A\otimes\mathbb{I}_{b}O)=\mathrm{Tr}_{a}(Af(O)).
\end{equation}

\section{Numerical computation of the partial trace}

\label{num_bip}

In this section we analyze the numerical calculation of the partial
trace for bi-partite systems by first considering the direct implementation
of its definition and afterwards optimizing it by identifying and
avoiding doing null multiplications and sums.

\subsection{Direct implementation}

\label{sec:1st}

Let us analyze the number of basic operations, \emph{scalar multiplications}
(mops) and \emph{scalar sums} (sops), needed to compute the partial
trace directly as given in Eq. (\ref{eq:ptr1}). Considering that
the tensor product of two matrices of dimensions $m\mathrm{x}n$ and
$o\mathrm{x}p$ requires $monp$ mops and zero sops and that the multiplication
of two matrices of dimensions $m\mathrm{x}n$ and $n\mathrm{x}o$
requires $mon$ mops and $mo(n-1)$ sops, we arrive at the numbers
of basic operations shown in Table \ref{table:bo_ptr1}.
\begin{table}[H]
\begin{center}%
\begin{tabular}{|c|c|c|}
\hline
Operation & No. of mops & No. of sops\tabularnewline
\hline
\hline
$\mathbb{I}_{a}\otimes\langle b_{j}|$ & $d_{a}^{2}d_{b}$ & $0$\tabularnewline
\hline
$\mathbb{I}_{a}\otimes|b_{j}\rangle$ & $d_{a}^{2}d_{b}$ & $0$\tabularnewline
\hline
$(\mathbb{I}_{a}\otimes\langle b_{j}|)O$ & $d_{a}^{3}d_{b}^{2}$ & $d_{a}^{2}d_{b}(d_{a}d_{b}-1)$\tabularnewline
\hline
$(\mathbb{I}_{a}\otimes\langle b_{j}|O)(\mathbb{I}_{a}\otimes|b_{j}\rangle)$ & $d_{a}^{3}d_{b}$ & $d_{a}^{2}(d_{a}d_{b}-1)$\tabularnewline
\hline
\end{tabular}\end{center}

\caption{Number of basic operations taken by each one of the steps needed to
compute the partial trace when implemented numerically directly from
its definition in Eq. (\ref{eq:ptr1}).}

\label{table:bo_ptr1}
\end{table}
So, to calculate Eq. (\ref{eq:ptr1}) numerically we would make use
of a total of $d_{a}^{2}d_{b}^{2}(2+d_{a}(d_{b}+1))$ mops and $d_{a}^{2}d_{b}(d_{a}d_{b}-1)(d_{b}+1)$
sops. If $d_{b}\gg1$ then
\begin{equation}
\mbox{mops}=\mbox{sops}\approx d_{a}^{3}d_{b}^{3}=d^{3}
\end{equation}
would be needed, where $d=\dim\mathcal{H}$. As the complexity for
the multiplication is, in the ``worst case'', the square of that for
the addition, then this ops is about $\mathcal{O}(d^{6})$.

\subsection{(Not) Using the zeros of $\mathbb{I}_{a}$}

\label{sec:2nd}

Now, instead of simply sending the matrices to a subroutine that computes
tensor products, let's observe that, for $d_{a}\gg1$, most matrix
elements of the identity operator are null. Thus, using the notation:
\begin{equation}
[\alpha,\beta:\gamma]=[O_{\alpha,\beta}\mbox{ }O_{\alpha,\beta+1}\mbox{ }\cdots\mbox{ }O_{\alpha,\gamma-1}\mbox{ }O_{\alpha,\gamma}],
\end{equation}
and with $|0\rangle$ being the null vector in $\mathcal{H}_{b}$,
$p=d_{a}$, and $q=d_{b}$, follows that each term $(\mathbb{I}_{a}\otimes\langle b_{j}|)O(\mathbb{I}_{a}\otimes|b_{j}\rangle)$
in Eq. (\ref{eq:ptr1}) is equal to:
\begin{eqnarray}
 &  & \mathbb{I}_{a}\otimes\langle b_{j}|\begin{bmatrix}[1,(1-1)q+1:q] & [1,(2-1)q+1:2q] & \cdots & [1,(p-1)q+1:pq]\\{}
[2,(1-1)q+1:q] & [2,(2-1)q+1:2q] & \cdots & [2,(p-1)q+1:pq]\\
\vdots & \vdots & \ddots & \vdots\\{}
[pq,(1-1)q+1:q] & [pq,(2-1)q+1:2q] & \cdots & [pq,(p-1)q+1:pq]
\end{bmatrix}\begin{bmatrix}|b_{j}\rangle & |0\rangle & \cdots & |0\rangle\\
|0\rangle & |b_{j}\rangle & \cdots & |0\rangle\\
\vdots & \vdots & \ddots & \vdots\\
|0\rangle & |0\rangle & \cdots & |b_{j}\rangle
\end{bmatrix}
\end{eqnarray}
\begin{equation}
=\begin{bmatrix}\langle b_{j}| & \langle0| & \cdots & \langle0|\\
\langle0| & \langle b_{j}| & \cdots & \langle0|\\
\vdots & \vdots & \ddots & \vdots\\
\langle0| & \langle0| & \cdots & \langle b_{j}|
\end{bmatrix}\begin{bmatrix}[1,(1-1)q+1:q]|b_{j}\rangle & [1,(2-1)q+1:2q]|b_{j}\rangle & \cdots & [1,(p-1)q+1:pq]|b_{j}\rangle\\{}
[2,(1-1)q+1:q]|b_{j}\rangle & [2,(2-1)q+1:2q]|b_{j}\rangle & \cdots & [2,(p-1)q+1:pq]|b_{j}\rangle\\
\vdots & \vdots & \ddots & \vdots\\{}
[pq,(1-1)q+1:q]|b_{j}\rangle & [pq,(2-1)q+1:2q]|b_{j}\rangle & \cdots & [pq,(p-1)q+1:pq]|b_{j}\rangle
\end{bmatrix}
\end{equation}
\begin{equation}
=\begin{bmatrix}{\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(1-1)q+\alpha,(1-1)q+1:q]|b_{j}\rangle & \cdots & {\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(1-1)q+\alpha,(p-1)q+1:pq]|b_{j}\rangle\\
{\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(2-1)q+\alpha,(1-1)q+1:q]|b_{j}\rangle & \cdots & {\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(2-1)q+\alpha,(p-1)q+1:pq]|b_{j}\rangle\\
\vdots & \ddots & \vdots\\
{\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(p-1)q+\alpha,(1-1)q+1:q]|b_{j}\rangle & \cdots & {\displaystyle \sum_{\alpha=1}^{q}}b_{j\alpha}^{*}[(p-1)q+\alpha,(p-1)q+1:pq]|b_{j}\rangle
\end{bmatrix}
\end{equation}
\begin{equation}
=\begin{bmatrix}{\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(1-1)q+\alpha,(1-1)q+\beta}b_{j\beta} & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(1-1)q+\alpha,(2-1)q+\beta}b_{j\beta} & \cdots & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(1-1)q+\alpha,(p-1)q+\beta}b_{j\beta}\\
{\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(2-1)q+\alpha,(1-1)q+\beta}b_{j\beta} & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(2-1)q+\alpha,(2-1)q+\beta}b_{j\beta} & \cdots & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(2-1)q+\alpha,(p-1)q+\beta}b_{j\beta}\\
\vdots & \vdots & \ddots & \vdots\\
{\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(p-1)q+\alpha,(1-1)q+\beta}b_{j\beta} & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(p-1)q+\alpha,(2-1)q+\beta}b_{j\beta} & \cdots & {\displaystyle \sum_{\alpha,\beta=1}^{q}}b_{j\alpha}^{*}O_{(p-1)q+\alpha,(p-1)q+\beta}b_{j\beta}
\end{bmatrix}.
\end{equation}

So, a generic matrix element of $O^{a}=\mathrm{Tr}_{b}(O)$ shall
take the form:
\begin{equation}
O_{kl}^{a}=\sum_{\alpha,\beta=1}^{d_{b}}O_{(k-1)d_{b}+\alpha,(l-1)d_{b}+\beta}\sum_{j=1}^{d_{b}}b_{j\alpha}^{*}b_{j\beta}.\label{eq:Oa_kl}
\end{equation}
To compute each one of the $d_{a}^{2}$ matrix elements above we need
to do $d_{b}^{2}(d_{b}+1)$ mops and $d_{b}(d_{b}^{2}-1)$ sops. Then,
on the total $d_{a}^{2}d_{b}^{2}(d_{b}+1)$ mops and $d_{a}^{2}d_{b}(d_{b}^{2}-1)$
sops will be necessary. For $d_{b}\gg1$ follows that
\begin{equation}
\mbox{mops}=\mbox{sops}\approx d_{a}^{2}d_{b}^{3}.
\end{equation}
We notice thus a decreasing by a multiplicative factor $d_{a}$ in
ops with relation to the previous direct implementation.

\subsection{(Not) Using the zeros of the computational basis}

\label{sec:3rd}

Now let us recall and verify that \emph{the partial trace is base
independent} and use this fact to diminish considerably the number
of basic operations required for its computation. We regard the following
arbitrary basis for $\mathcal{H}_{b}$: $|j\rangle=\sum_{k=1}^{d_{b}}c_{jk}|b_{k}\rangle$,
with $c_{jk}=\langle j|b_{k}\rangle$. Taking the partial trace in
the basis $|j\rangle$,
\begin{eqnarray}
\mathrm{Tr}_{b}(O) & = & \sum_{j=1}^{d_{b}}(\mathbb{I}_{a}\otimes\langle j|)O(\mathbb{I}_{a}\otimes|j\rangle)=\sum_{j=1}^{d_{b}}(\mathbb{I}_{a}\otimes\sum_{k=1}^{d_{b}}c_{jk}^{*}\langle b_{k}|)O(\mathbb{I}_{a}\otimes\sum_{l=1}^{d_{b}}c_{jl}|b_{l}\rangle)\\
 & = & \sum_{j,k,l=1}^{d_{b}}c_{jk}^{*}c_{jl}(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{l}\rangle)=\sum_{j,k,l=1}^{d_{b}}\langle j|b_{k}\rangle^{*}\langle j|b_{l}\rangle(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{l}\rangle)\\
 & = & \sum_{j,k,l=1}^{d_{b}}\langle b_{k}|j\rangle\langle j|b_{l}\rangle(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{l}\rangle)=\sum_{k,l=1}^{d_{b}}\langle b_{k}|b_{l}\rangle(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{l}\rangle)\\
 & = & \sum_{k,l=1}^{d_{b}}\delta_{kl}(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{l}\rangle)=\sum_{k=1}^{d_{b}}(\mathbb{I}_{a}\otimes\langle b_{k}|)O(\mathbb{I}_{a}\otimes|b_{k}\rangle),
\end{eqnarray}
is then seem to be equivalent to compute the partial trace with $|b_{j}\rangle$.

Hence we shall use the \emph{computational basis}
\begin{equation}
|j\rangle=[\delta_{j1}\mbox{ }\delta_{j2}\mbox{ }\cdots\mbox{ }\delta_{jd_{b}}]^{t},
\end{equation}
(with $\delta_{jk}$ being the Kronecker's delta function) to take
partial traces, avoiding multiplying its $d_{b}-1$ null elements
(for each $|j\rangle$). This is done simply by replacing $|b_{j}\rangle$
by $|j\rangle$ in Eq. (\ref{eq:Oa_kl}) to get\footnote{Another, simpler way to get this result is by applying the definition
in Eq. (\ref{eq:ptr_def2}) to $O$ represented as in Eq. (\ref{eq:Orep}),
but with $|b_{j}\rangle$ being the computational basis $|j\rangle$.
So, utilizing $O^{a}=\mathrm{Tr}_{b}(O)=\sum_{k,l}\sum_{j}\langle k|\otimes\langle j|O|l\rangle\otimes|j\rangle|k\rangle\langle l|=\sum_{k,l}O_{k,l}^{a}|k\rangle\langle l|$
and the fact that only the $((l-1)d_{b}+j)$-th element of $|l\rangle\otimes|j\rangle$
is non-null, we obtain Eq. (\ref{eq:ptre}).}
\begin{eqnarray}
O_{kl}^{a} & = & \sum_{j,\alpha,\beta=1}^{d_{b}}\delta_{j\alpha}\delta_{j\beta}O_{(k-1)d_{b}+\alpha,(l-1)d_{b}+\beta}\\
 & = & \sum_{j=1}^{d_{b}}O_{(k-1)d_{b}+j,(l-1)d_{b}+j}.\label{eq:ptre}
\end{eqnarray}

Therefore, in this last implementation of the partial trace operation,
we need to perform ``only'' $d_{a}^{2}(d_{b}-1)$ sops (and no mops).
Or, for $d_{b}\gg1$
\begin{equation}
\mbox{mops}=0\mbox{ and }\mbox{sops}\approx d_{a}^{2}d_{b}.
\end{equation}
We think this is the most optimized way to calculate partial traces
for bipartite systems. In the case of Hermitian reduced matrices (in
particular for density matrices) $O_{lk}^{a}=(O_{kl}^{a})^{*}$, and
the number of basic operations needed to compute the partial trace
can be reduced yet by $2^{-1}d_{a}(d_{a}-1)(d_{b}-1)$ sops (which
is almost half of the total when $d_{a},d_{b}\gg1$). For the sake
of illustration, it is shown in Fig. \ref{fig:tptr} the time taken
to compute the partial trace via these three methods as a function
of system $a$ dimension $d_{a}$.

\begin{figure}[t]
\begin{centering}
\includegraphics[scale=0.5]{tptr}
\par\end{centering}

\caption{(color online) Time taken to compute the partial trace using Eqs.
(\ref{eq:ptr1}), (\ref{eq:Oa_kl}), and (\ref{eq:ptre}) (from up
down) as a function of system $a$ dimension. In the inset is shown
the log-log plot of the same data. We set $d_{b}=d_{a}$ and used
the maximally mixed global state in all cases. The calculations were
performed using the GNU Fortran Compiler version 5.0.0 in a MacBook
Air Processor 1.3 GHz Intel Core i5, with a 4 GB 1600 MHz DDR3 Memory.
If, for $d_{a},d_{b}\gg1$ and $d_{a}=d_{b}$, we have $t\propto d_{a}^{p}d_{b}^{q}$,
then $\log t\propto(p+q)\log d_{a}$. So, the increasing rate of $\log t$
with the number of quantum bits \cite{Nielsen=000026Chuang} constituting
$a$ in the optimized implementation of the partial trace is one fourth
of that for the direct calculation.}

\label{fig:tptr}
\end{figure}

\section{Partial trace for multi-partitions}

\label{PTr_mult}

Let us consider a multipartite system $\mathcal{H}=\bigotimes_{s}\mathcal{H}_{s}$
and a linear operator $O\in\mathcal{L}(\mathcal{H})$. For two arbitrary
sub-systems $s'$ and $s''$, with $s''>s'$, and bases $|s_{j}'\rangle\in\mathcal{H}_{s'}$
and $|s_{j}''\rangle\in\mathcal{H}_{s''}$, one can verify that
\begin{equation}
(\mathbb{I}_{s'}\otimes\mathbb{I}_{(s'+1)\cdots(s''-1)}\otimes|s_{k}''\rangle)(|s_{j}'\rangle\otimes\mathbb{I}_{(s'+1)\cdots(s''-1)})=|s_{j}'\rangle\otimes\mathbb{I}_{(s'+1)\cdots(s''-1)}\otimes|s_{k}''\rangle.
\end{equation}
We then use this relation to see that the equality
\begin{equation}
\mathrm{Tr}_{s's''}(O)=\mathrm{Tr}_{s'}(\mathrm{Tr}_{s''}(O))
\end{equation}
holds for all $s'$ and $s''$. Therefore the partial trace taken
over any set of sub-systems of $\mathcal{H}$ can be implemented sequentially
via partial tracing over single partitions; and we can do that using
the following two main procedures. In one of these procedures we split
$\mathcal{H}$ in two parts $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$
and trace out $a$ or $b$. And in the other one we divide $\mathcal{H}$
in three parties $\mathcal{H}_{a}\otimes\mathcal{H}_{b}\otimes\mathcal{H}_{c}$
and trace over the inner party $b$. As the first procedure was addressed
in the previous section, we shall regard the details of the last one
in this section.

Let $O\in\mathcal{L}(\mathcal{H}_{a}\otimes\mathcal{H}_{b}\otimes\mathcal{H}_{c})$
and let us consider the partial trace
\begin{equation}
O^{ac}=\mathrm{Tr}_{b}(O)=\sum_{j=1}^{d_{b}}(\mathbb{I}_{a}\otimes\langle b_{j}|\otimes\mathbb{I}_{c})O(\mathbb{I}_{a}\otimes|b_{j}\rangle\otimes\mathbb{I}_{c}).
\end{equation}
In a direct implementation of this equation, $d_{a}^{2}d_{b}^{2}d_{c}^{2}(d_{a}d_{c}(d_{b}+1)+2)$
mops and $d_{a}^{2}d_{b}d_{c}^{2}(d_{b}+1)(d_{a}d_{b}d_{c}-1)$ sops
would be used. For $d_{b}\gg1$,
\begin{equation}
\mbox{mops}=\mbox{sops}\approx d_{a}^{3}d_{b}^{3}d_{c}^{3}.
\end{equation}

With the aim of reaching an optimized computation of the partial trace,
we start considering
\begin{eqnarray}
O^{ac} & = & \sum_{j,m=1}^{d_{a}}\sum_{k,n=1}^{d_{b}}\sum_{l,o=1}^{d_{c}}(\langle j|\otimes\langle k|\otimes\langle l|)O(|m\rangle\otimes|n\rangle\otimes|o\rangle)|j\rangle\langle m|\otimes\mathrm{Tr}_{b}(|k\rangle\langle n|)\otimes|l\rangle\langle o|\\
 & = & \sum_{j,m=1}^{d_{a}}\sum_{l,o=1}^{d_{c}}\left(\sum_{k=1}^{d_{b}}(\langle j|\otimes\langle k|\otimes\langle l|)O(|m\rangle\otimes|k\rangle\otimes|o\rangle)\right)|j\rangle\langle m|\otimes|l\rangle\langle o|\label{eq:ptr_m1}\\
 & = & \sum_{j,m=1}^{d_{a}}\sum_{l,o=1}^{d_{c}}(\langle j|\otimes\langle l|)O^{ac}(|m\rangle\otimes|o\rangle)|j\rangle\langle m|\otimes|l\rangle\langle o|.\label{eq:ptr_m2}
\end{eqnarray}
In this article, if not stated otherwise, \emph{we assume that the
matrix representation of the considered operators in the corresponding
(global) computational basis is given}. Next we notice e.g. that only
the element $\alpha:=(m-1)d_{b}d_{c}+(n-1)d_{c}+o$ of $|m\rangle\otimes|n\rangle\otimes|o\rangle$
is non-null; thus $O|m\rangle\otimes|n\rangle\otimes|o\rangle$ is
equal to the $\alpha$-th column vector of $O$. Thus, Eqs. (\ref{eq:ptr_m1})
and (\ref{eq:ptr_m2}) and these results can be used to write the
following relation between matrix elements (in the corresponding global
computational basis):
\begin{equation}
O_{(j-1)d_{c}+l,(m-1)d_{c}+o}^{ac}=\sum_{k=1}^{d_{b}}O_{(j-1)d_{b}d_{c}+(k-1)d_{c}+l,(m-1)d_{b}d_{c}+(k-1)d_{c}+o}.
\end{equation}
In terms of ops, in this implementation we shall utilize $\mbox{mops}=0$
and $\mbox{sops}=d_{a}^{2}d_{c}^{2}(d_{b}-1)$, which for $d_{b}\gg1$
is
\begin{equation}
\mbox{mops}=0\mbox{ and }\mbox{sops}\approx d_{a}^{2}d_{b}d_{c}^{2}.
\end{equation}
As in the case of bipartite systems, here also we can utilize the
hermiticity of the reduced matrix to diminish the number of basic
operations by $2^{-1}d_{a}d_{c}(d_{a}d_{c}-1)(d_{b}-1)$ sops.

Fortran code to perform all numerical calculations associated with
this article, and several others, can be accessed in \href{https://github.com/jonasmaziero/LibForQ.git}{https://github.com/jonasmaziero/LibForQ.git}.
In particular, we provide the subroutine, \texttt{partial\_trace(rho,
d, di, nss, ssys, dr, rhor)}, which returns the reduced matrix \texttt{rhor}
once provided \texttt{dr} (its dimension), \texttt{rho }(the matrix
representation in the computational basis of the regarded linear operator
in $\mathcal{H}_{1}\otimes\mathcal{H}_{2}\otimes\cdots\otimes\mathcal{H}_{n}$),
\texttt{d} (the dimension of rho), \texttt{nss}$=n$ (the number of
sub-systems), \texttt{di }(a $n$-dimensional integer vector whose
components specify the dimensions of the sub-systems), and \texttt{ssys
}(a $n$-dimensional integer vector whose null components specify
the sub-systems to be traced over; the other components must be made
equal to one). In the \emph{Hermitian case}, just change the subroutine's
name to \texttt{partial\_trace\_he}.

Let's exemplify the application of what was discussed in this section
by considering the thermal ground state \cite{Nielsen_EntPT}, $\rho=\exp(-\beta H)/\mathrm{Tr}(\exp(-\beta H))$
with $\beta\rightarrow\infty$, of a line of qubits with Ising interaction
between nearest-neighbors: $H=-\frac{J}{2}\sum_{j=1}^{n-1}\sigma_{j}^{z}\sigma_{j+1}^{z}-h\sum_{j=1}^{n}\sigma_{j}^{x},$
where $\sigma_{j}^{x(z)}$ are the Pauli operators in the state space
of the $j$-th spin, $h$ is the so called transverse magnetic field,
and we set the exchange interaction strength to unit ($J=1$). Here
we want to compute the nonlocal quantum coherence \cite{Maziero_QCS}
of the edge spins: $C_{nl}(\rho_{1n})=C(\rho_{1n})-[C(\rho_{1})+C(\rho_{n})],$
where the $l_{1}$-norm quantum coherence is given by \cite{Plenio_QQC}:
$C(\rho)=\sum_{j\ne k}\langle j|\rho|k\rangle,$ with $|j\rangle$
being the standard-computational basis in the regarded Hilbert space.
When performing this kind of calculation, we shall need the reduced
states: $\rho_{1n}=\mathrm{Tr}_{2\cdots(n-1)}(\rho_{12\cdots(n-1)n})\mbox{, }\rho_{1}=\mathrm{Tr}_{n}(\rho_{1n})\mbox{, and }\rho_{n}=\mathrm{Tr}_{1}(\rho_{1n}).$
The results for the quantum coherence and a comparison between the
time taken by the optimized and direct implementations of the partial
trace function are shown in Fig. \ref{fig:inner}.

\begin{figure}
\begin{centering}
\includegraphics[scale=0.65]{coh}
\par\end{centering}

\caption{(color online) Non-local quantum coherence (NLQC) of the edge qubits
of an Ising chain with $n$ spins as a function of the reciprocal
of the transverse magnetic field. In the bottom inset is shown the
first derivative of the NLQC, whose minimum is seem to go slowly towards
the point of quantum phase transition ($h=1/2$). In the upper inset
is shown, in logarithmic scale, the difference between the time taken,
to do the calculations for a fixed value of $n$, by the direct and
optimized implementations of the partial trace function. We notice
thus a exponential increase of this time difference with $n$. }

\label{fig:inner}
\end{figure}

\section{``Partial traces'' via Bloch's parametrization}

\label{sec:su}

As the most frequent application of the partial trace function is
to compute reduced density matrices (also called partial states or
quantum marginals), let us consider yet another approach we may apply
to perform that task and analyze its computational complexity. From
the defining properties of a density matrix (positiveness $\rho\ge0$
and unit trace $\mathrm{Tr}(\rho)=1$), follows that it can be written
in terms of $\mathbb{I}_{b}$ and of the orthonormal-traceless-hermitian
generators $\Gamma_{j}$ of the special unitary group, as shown below.
For a bipartite system $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$, the
reduced state of sub-system $b$ can be written as \cite{Petruccione_rrho,Bertlmann}:
\begin{equation}
\rho^{b}=\frac{\mathbb{I}_{b}}{d_{b}}+\sum_{j=1}^{d_{b}^{2}-1}\frac{\gamma_{j}}{2}\Gamma_{j},\label{eq:BlochP}
\end{equation}
where $\mathrm{Tr}_{b}(\Gamma_{j}^{\dagger}\Gamma_{k})=2\delta_{jk}$
and $\gamma_{j}=\mathrm{Tr}_{b}(\Gamma_{j}\rho^{b})$. If the density
matrix $\rho$ of the whole system is known, and we want to compute
$\rho^{b}$, then
\begin{equation}
\gamma_{j}=\mathrm{Tr}_{b}(\Gamma_{j}\rho^{b})=\mathrm{Tr}_{ab}(\mathbb{I}_{a}\otimes\Gamma_{j}\rho)=\sum_{k=1}^{d_{a}d_{b}}(\mathbb{I}_{a}\otimes\Gamma_{j}\rho)_{k,k}\label{eq:BlochVC}
\end{equation}
will yield the $d_{b}^{2}-1$ real components of the Bloch vector
$\vec{\gamma}=[\gamma_{1}\mbox{ }\gamma_{2}\mbox{ }\cdots\mbox{ }\gamma_{d_{b}^{2}-1}]^{t}$.

\subsection{Direct implementation}

Let us start by the most straightforward, unoptimized, implementation
of the partial trace via Bloch parametrization. Counting the basic
operations need to compute each $\gamma_{j}$, we note that for the
tensor product $\mathbb{I}_{a}\otimes\Gamma_{j}$, $\mbox{mops}=d_{a}^{2}d_{b}^{2}$,
for the matrix multiplication $(\mathbb{I}_{a}\otimes\Gamma_{j})\rho$,
$\mbox{mops}=d_{a}^{3}d_{b}^{3}$ and $\mbox{sops}=d_{a}^{2}d_{b}^{2}(d_{a}d_{b}-1)$,
and for the trace, $\mbox{sops}=d_{a}d_{b}-1$. Then, for the $d_{b}^{2}-1$
components $\gamma_{j}$ we shall need $\mbox{mops}=(d_{b}^{2}-1)d_{a}^{2}d_{b}^{2}(d_{a}d_{b}+1)$
and $\mbox{sops}=(d_{b}^{2}-1)(d_{a}^{2}d_{b}^{2}+1)(d_{a}d_{b}-1)$.
After knowing $\vec{\gamma}$, $d_{b}^{4}$ mops and $d_{b}^{2}(d_{b}^{2}-2)+d_{b}$
sops are required to compute $\rho^{b}$. Thus, on total we have to
do $\mbox{mops}=d_{b}^{2}(d_{b}^{2}+d_{a}^{2}(d_{b}^{2}-1)(d_{a}d_{b}+1))$
and $\mbox{sops}=(d_{b}^{2}-1)(d_{a}^{2}d_{b}^{2}+1)(d_{a}d_{b}-1)+d_{b}^{2}(d_{b}^{2}-2)+d_{b}$,
which for $d_{b}\gg1$ become
\begin{equation}
\mbox{mops}=\mbox{sops}\approx d_{a}^{3}d_{b}^{5}.
\end{equation}

\subsection{(Not) Using the zeros of \emph{$\mathbb{I}_{a}$}}

Now let's use the zeros of\emph{ $\mathbb{I}_{a}$} to diminish the
number of ops needed to obtain $\gamma_{j}$. We begin by writing
\begin{equation}
(\mathbb{I}_{a}\otimes\Gamma_{j})\rho=\begin{bmatrix}\Gamma_{j} & 0_{d_{b}\mathrm{x}d_{b}} & \cdots & 0_{d_{b}\mathrm{x}d_{b}}\\
0_{d_{b}\mathrm{x}d_{b}} & \Gamma_{j} & \cdots & 0_{d_{b}\mathrm{x}d_{b}}\\
\vdots & \vdots & \ddots & \vdots\\
0_{d_{b}\mathrm{x}d_{b}} & 0_{d_{b}\mathrm{x}d_{b}} & \cdots & \Gamma_{j}
\end{bmatrix}\begin{bmatrix}\rho^{(1)} & [?] & \cdots & [?]\\{}
[?] & \rho^{(2)} & \cdots & [?]\\
\vdots & \vdots & \ddots & \vdots\\{}
[?] & [?] & \cdots & \rho^{(d_{a})}
\end{bmatrix}=\begin{bmatrix}\Gamma_{j}\rho^{(1)} & \Gamma_{j}[?] & \cdots & \Gamma_{j}[?]\\
\Gamma_{j}[?] & \Gamma_{j}\rho^{(2)} & \cdots & \Gamma_{j}[?]\\
\vdots & \vdots & \ddots & \vdots\\
\Gamma_{j}[?] & \Gamma_{j}[?] & \cdots & \Gamma_{j}\rho^{(d_{a})}
\end{bmatrix},
\end{equation}
where $0_{d_{b}\mathrm{x}d_{b}}$ is the $d_{b}\mathrm{x}d_{b}$ null
matrix and $[?]$ is used to denote those $d_{b}\mathrm{x}d_{b}$
sub-blocks of $\rho$ that we do not need when computing $\gamma_{j}$.
From the last equation we see that
\begin{eqnarray}
\gamma_{j} & = & \sum_{\beta=1}^{d_{a}}\mathrm{Tr}(\Gamma_{j}\rho^{(\beta)})=\sum_{\beta=1}^{d_{a}}\sum_{\alpha=1}^{d_{b}}(\Gamma_{j}\rho^{(\beta)})_{\alpha,\alpha}=\sum_{\beta=1}^{d_{a}}\sum_{\alpha=1}^{d_{b}}\sum_{k=1}^{d_{b}}(\Gamma_{j})_{\alpha,k}\rho_{k,\alpha}^{(\beta)}\\
 & = & \sum_{\alpha=1}^{d_{b}}\sum_{k=1}^{d_{b}}(\Gamma_{j})_{\alpha,k}\sum_{\beta=1}^{d_{a}}\rho_{(\beta-1)d_{b}+k,(\beta-1)d_{b}+\alpha},\label{eq:BV_GM}
\end{eqnarray}
which is valid for any choice of the generators $\Gamma_{j}$. If
computed as shown in the last equation, each $\gamma_{j}$ requires
$\mbox{sops}=d_{b}^{2}(d_{a}-1)$ and $\mbox{mops}=d_{b}^{2}$, and
we need to compute $d_{b}^{2}-1$ of them. Thus, when accounted for
also the ops necessary to compute Eq. (\ref{eq:BlochP}) given $\vec{\gamma}$,
we arrive at a total of $2d_{b}^{2}(d_{b}^{2}-1)$ mops and $d_{b}^{2}((d_{b}^{2}-1)d_{a}-1)+d_{b}$
sops needed to compute $\rho^{b}$. For $d_{b}\gg1$:
\begin{equation}
\mbox{mops}\approx2d_{b}^{4}\mbox{ and }\mbox{sops}\approx d_{a}d_{b}^{4}.
\end{equation}

\subsection{(Not) Using the zeros of $\Gamma_{j}$ (and of $\mathbb{I}_{b}$)}

These numbers can be reduced even more if we\emph{ }use the zeros
of the generators $\Gamma_{j}$. To do that, a particular basis $\Gamma_{j}$
has to be chosen, and we shall pick here the generalized Gell Mann's
matrices \cite{Bertlmann}:

\begin{eqnarray}
 &  & \Gamma_{j}^{(1)}=\sqrt{\frac{2}{j(j+1)}}\left(\sum_{k=1}^{j}|k\rangle\langle k|-j|j+1\rangle\langle j+1|\right),\mbox{ for }j=1,\cdots,d_{b}-1,\label{eq:SU1}\\
 &  & \Gamma_{(k,l)}^{(2)}=|k\rangle\langle l|+|l\rangle\langle k|,\mbox{ for }1\leq k<l\leq d_{b},\label{eq:SU2}\\
 &  & \Gamma_{(k,l)}^{(3)}=-i(|k\rangle\langle l|-|l\rangle\langle k|),\mbox{ for }1\leq k<l\leq d_{b}.\label{eq:SU3}
\end{eqnarray}

From Eq. (\ref{eq:BV_GM}), when computing the components of the Bloch
vector, we see that for the generators corresponding to the diagonal
group in Eq. (\ref{eq:SU1}):
\begin{equation}
\gamma_{j}^{(1)}=\sqrt{\frac{2}{j(j+1)}}\sum_{\alpha=1}^{j+1}(-j)^{\delta_{\alpha,j+1}}\sum_{\beta=1}^{d_{a}}\rho_{(\beta-1)d_{b}+\alpha,(\beta-1)d_{b}+\alpha}.
\end{equation}
Then $6(d_{b}-1)$ mops and $1+(d_{a}-1)(\sum_{j=1}^{d_{b}}j)=1+(d_{a}-1)2^{-1}d_{b}(d_{b}+1)$
sops are used for this first group. Related to the generators belonging
to the symmetric and anti-symmetric groups in Eqs. (\ref{eq:SU2})
and (\ref{eq:SU3}), respectively, from Eq. (\ref{eq:BV_GM}) we get
\begin{eqnarray}
\gamma_{(k,l)}^{(2)} & = & \sum_{\beta=1}^{d_{a}}\left(\rho_{(\beta-1)d_{b}+l,(\beta-1)d_{b}+k}+\rho_{(\beta-1)d_{b}+k,(\beta-1)d_{b}+l}\right)=2\sum_{\beta=1}^{d_{a}}\mathrm{Re}(\rho_{(\beta-1)d_{b}+l,(\beta-1)d_{b}+k}),\label{eq:g2}\\
\gamma_{(k,l)}^{(3)} & = & -i\sum_{\beta=1}^{d_{a}}\left(\rho_{(\beta-1)d_{b}+l,(\beta-1)d_{b}+k}-\rho_{(\beta-1)d_{b}+k,(\beta-1)d_{b}+l}\right)=2\sum_{\beta=1}^{d_{a}}\mathrm{Im}(\rho_{(\beta-1)d_{b}+l,(\beta-1)d_{b}+k}).\label{eq:g3}
\end{eqnarray}
These two groups, formed by $d_{b}(d_{b}-1)/2$ elements each, entail
in $\mbox{mops}=2$ and $\mbox{sops}=2(d_{a}-1)$; but, as we will
see below, we do not need them to compute the reduced state $\rho^{b}$.

In the sequence, we shall rewrite the partial state splitting its
diagonal $\Delta$ and non-diagonal $\Theta$ parts, i.e.,
\begin{equation}
\rho^{b}=\Delta+\Theta.
\end{equation}
We will look first at the diagonal elements of $\rho^{b}$:
\begin{eqnarray}
\Delta & = & \frac{\mathbb{I}_{b}}{d_{b}}+\sum_{j=1}^{d_{b}-1}\frac{\gamma_{j}^{(1)}}{2}\Gamma_{j}^{(1)}=\frac{\sum_{k=1}^{d_{b}}|k\rangle\langle k|}{d_{b}}+\sum_{j=1}^{d_{b}-1}\frac{\gamma_{j}^{(1)}}{2}\sqrt{\frac{2}{j(j+1)}}\left(\sum_{k=1}^{j}|k\rangle\langle k|-j|j+1\rangle\langle j+1|\right).
\end{eqnarray}
If, for $j=1,\cdots,d_{b}-1$, we define the function $\xi_{j}=\gamma_{j}^{(1)}/\sqrt{2(j(j+1))}$
and set $\xi_{0}=d_{b}^{-1}$ and $q=d_{b}$, then
\begin{eqnarray}
\Delta & = & \mathrm{diag}(\xi_{0},\xi_{0},\xi_{0},\xi_{0},\xi_{0},\cdots,\xi_{0})+\mathrm{diag}(\xi_{1},-\xi_{1},0,0,0,\cdots,0)+\mathrm{diag}(\xi_{2},\xi_{2},-2\xi_{2},0,0,\cdots,0)\\
 &  & +\mathrm{diag}(\xi_{3},\xi_{3},\xi_{3},-3\xi_{3},0,\cdots,0)+\cdots+\mathrm{diag}(\xi_{q-3},\cdots,\xi_{q-3},-(q-3)\xi_{q-3},0,0)\\
 &  & +\mathrm{diag}(\xi_{q-2},\xi_{q-2},\cdots,\xi_{q-2},-(q-2)\xi_{q-2},0)+\mathrm{diag}(\xi_{q-1},\xi_{q-1},\xi_{q-1},\xi_{q-1},\cdots,\xi_{q-1},-(q-1)\xi_{q-1}).
\end{eqnarray}
After some analysis, we see that
\begin{eqnarray}
\Delta_{1,1} & =\sum_{k=0}^{d_{b}-1}\xi_{k} & \mbox{ and }\Delta_{j,j}=\Delta_{j-1,j-1}+(j-2)\xi_{j-2}-j\xi_{j-1}\mbox{ for }j=2,\cdots,d_{b}.
\end{eqnarray}
If we know the $\gamma_{j}^{(1)}$'s, for the $\xi_{j}$'s $\mbox{mops}=4(d_{b}-1)+1$
and $\mbox{sops}=d_{b}-1$ and for the $\Delta_{j,j}$'s $2\mbox{mops}=\mbox{sops}=4(d_{b}-1)$.
Then for $\Delta$ we shall need $6(d_{b}-1)+1$ mops and $5(d_{b}-1)$
sops.

Let us now consider the off-diagonal elements of $\rho^{b}$:
\begin{eqnarray}
\Theta & = & \sum_{1\le k<l\le d_{b}}\frac{1}{2}\left(\gamma_{(k,l)}^{(2)}\Gamma_{(k,l)}^{(2)}+\gamma_{(k,l)}^{(3)}\Gamma_{(k,l)}^{(3)}\right)=\sum_{1\le k<l\le d_{b}}\frac{1}{2}\left((\gamma_{(k,l)}^{(2)}-i\gamma_{(k,l)}^{(3)})|k\rangle\langle l|+(\gamma_{(k,l)}^{(2)}+i\gamma_{(k,l)}^{(3)})|l\rangle\langle k|\right).
\end{eqnarray}
Using Eqs. (\ref{eq:g2}) and (\ref{eq:g3}) we arrive at
\[
\Theta_{l,k}=\sum_{\beta=1}^{d_{a}}\rho_{(\beta-1)d_{b}+l,(\beta-1)d_{b}+k}.
\]
Then, as $\Theta_{k,l}=\Theta_{l,k}^{*}$, we shall need $\mbox{sops}=2^{-1}d_{b}(d_{b}-1)(d_{a}-1)$
to calculate $\Theta$. So, after knowing $\vec{\gamma}$, $6(d_{b}-1)+1$
mops and $(d_{b}-1)(2^{-1}d_{b}(d_{a}-1)+5)$ sops are used to compute
$\rho^{b}$. Therefore, counting the ops need to compute the $\gamma_{j}^{(1)}$'s,
on total $\mbox{mops}=12(d_{b}-1)+1$ and $\mbox{sops}=d_{b}^{2}(d_{a}-1)+5(d_{b}-1)+1$
shall be used to get $\rho^{b}$ via Bloch parametrization. And for
$d_{a},d_{b}\gg1$
\begin{equation}
\mbox{mops}\approx12d_{b}\mbox{ and }\mbox{sops}\approx d_{a}d_{b}^{2}.
\end{equation}
As the ``worst case'' complexity for the multiplication is equal to
the square of the complexity for the addition, we see that this number
of operations is comparable with the optimized one obtained in Sec.
\ref{sec:3rd}, for $d_{a},d_{b}\gg1$. Here is another, more dramatic
example of the reduction in the number of elementary operations needed
to compute the partial trace ($\mathcal{O}(d_{a}^{6}d_{b}^{10})\rightarrow\mathcal{O}(d_{a}d_{b}^{2})$),
which is obtained simply by performing some analytical developments.
We also provide the Fortran code to compute partial states of bipartite
systems via Bloch's parametrization using this last procedure, i.e.,
with generalized Gell Mann's matrices.

\section{Concluding remarks}

\label{sec:cc}

In this article, a thorough discussion about the partial trace function
was made. We gave special attention to its numerical implementation.
It is a good programming practice trying to avoid making the computer
perform calculations which make no difference to the final result.
We showed here that by following this truism we can decrease considerably
the number of elementary operations needed to compute the partial
trace, or reduced density matrix, which is an extremely important
and common procedure in the quantum mechanics of composite (and open)
systems. We provided and described Fortran code for computing the
partial trace over any set of parties of a discrete multipartite system.
At last, we analyzed the calculation of partial states via the Bloch's
parametrization with generalized Gell Mann's matrices. We believe
this text will be of pedagogical and practical value for the physics
and quantum information science communities. As the partial trace
function is highly adaptable for parallel calculations, we think this
is a natural theme for future investigations. Extending our approach
to continuous variable systems is an interesting research topic. It
would also be fruitful producing translations of our code to other
open source programming languages such as e.g. Maxima and Octave,
as was already done for \href{https://bitbucket.org/huberfe/quantum-code/src/03994b9fc2a1251b82996cea14d65c5e4e2eb216/qgeo.py?fileviewer=file-view-default\#qgeo.py-1456}{Python}.
\begin{acknowledgments}
This work was supported by the Brazilian funding agencies: Conselho Nacional de Desenvolvimento Cient\'ifico e Tecnol\'ogico (CNPq), processes 441875/2014-9 and 303496/2014-2, Instituto Nacional de Ci\^encia e Tecnologia de Informa\c{c}\~ao Qu\^antica (INCT-IQ), process 2008/57856-6, and Coordena\c{c}\~ao de Desenvolvimento de Pessoal de N\'{i}vel Superior (CAPES), process 6531/2014-08. I gratefully acknowledge the hospitality of the Physics Institute and Laser Spectroscopy Group at the Universidad de la Rep\'{u}blica, Uruguay. I also thank Felix Huber for pointing out an error in one of the partial traces subroutines of a previous version of LibForQ.\end{acknowledgments}

redpajama_set_name: RedPajamaArXiv


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Why Choose Hershey, PA?
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Clinical Pastoral Education Residency (Level II)
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The Clinical Pastoral Education Residency at Penn State Health Milton S. Hershey Medical Center is a 54-week, Association for Clinical Pastoral Education-accredited program that offers continuing specialty training in clinical pastoral care. Five full-time positions are available.
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The Clinical Pastoral Education Residency is designed for individuals who are pursuing a specialized vocation in professional pastoral care (e.g., chaplaincy, congregational care). These residents have successfully completed a master's degree at an accredited seminary and a clinical pastoral education internship. Under the direction of highly trained clinical pastoral education supervisors, the residency attends to residents' development of advanced pastoral care-giving skills, interdisciplinary and teamwork functioning and crisis ministry competency.
The program begins in mid-August and concludes Aug. 31 of the following year, and has a stipend and benefits package.
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Applicants must submit a completed application along with a $10 deposit made payable to Penn State Health Milton S. Hershey Medical Center. The application form is available online.
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Ronald Slye: How, how has this experience changed you?
I think I stated it before. This capacity building, the capacity building work, the outreach work has become – it is, it is not in the library's mandate as a matter of fact but it has become a very important part of our work.
So, yes it feels good that you've really contributed to, you know, training people, to building their capacity and to giving them some autonomy in your field. So, this now tells me that maybe I should continue. Maybe I should do that, maybe continue in Rwanda, ma-, b- but also elsewhere. And it tells me that it is very important.
We've been, we've been so told that library was, you know, not a good job. I remember my daughter telling me that, "Mom, why are you a librarian? Why couldn't you be a lawyer or, or a doctor? Why are you just a librarian?" Because when she said her Mom was a librarian, her friends made fun of her, you see.
But now, this has given even myself another dimension of the work that I'm doing and that I should do it even better and even more.
RS: If somebody, if there was a new tribunal established today and there was somebody who was being asked to be the chief librarian but they weren't sure if they wanted to do it, what would you say to them?
RS: So a, a new tribunal like the ICTR is created . . .
RS: . . . and there is a library that's being created . . .
RS: . . . and they, there is, somebody has come to you . . .
RS: . . . who has been asked to be the librarian of that new tribunal but that person isn't sure if they want to take the job.
RS: What would you say to them?
That they should not consider the role as just sitting in an office and catalogue books and order books. They should actually see the role as ag-, agents of change also and that actually being a librarian of an International Criminal Court is also, is, is also being contributing to the peace process in the world.
So to me it is a very big challenge and I'm honored that I'm doing it. And the-, and they should see a more – they should, they should look at the bigger picture because if you give a wrong information, even if you sit in your office and you give the wrong information to a people who needs information to s-, to make their case, you could screw it for them.
And if you give the right information, the success of the case is your success also, so they should see their role not as a librarian but as a part of, a part, a major part of the process. It, it, it takes some good decision-making and some judgment to contribute to a process like this.
RS: Is there anything else that you, we haven't touched upon that you think would be important for people to know?
I think it is the same thing, the same realization that I did myself – that being a librarian is a big, is a big job. So I think if you want to promote peace and promote reconciliation, you should start with the prevention. And I think that the prevention starts with empowering people, giving them the information they need to know in order to make some decisions.
In Rwanda for instance, the literacy rate is very low. There are no libraries. People were sent to kill and people – most of the people who were killing were killing because they couldn't judge themselves that this is g-, wrong or right, but because they were receiving orders from people who they consider role models.
If they had access to information, they could be role models to themselves and know that this is right or this is wrong. So the message that I'm sending is that – empower people with information, promote libraries wherever there is a need and I think we start there. Peace starts there.


Poor connectivity was always the price a business paid for operating in the middle of nowhere. Locate your office somewhere like central London instead, and you can tap into a multitude of high capacity networks and peering points.
But as nationwide speeds and services have improved, you hear less about the provincial startup forever stuck on dialup, or the big organisation in the countryside waiting months to upgrade to a reliable Ethernet-grade service. By contrast, you increasingly hear about the agonies that city-based businesses suffer getting their fibre services connected.
In a densely-populated city, properties not already served by a fibre connection will be very close to the network infrastructure that enables it. In many cases, the gap could be as little as a few metres. Hypothetically, the technical process of linking up those two points is quick and simple. In practice, it is fraught with legal and logistical problems.
The last word you want to hear when ordering fibre for your business is 'wayleave'. A wayleave is a legal requirement giving permission from the owner of property/land to have building works undertaken by Openreach, such as digging into the ground where fibre cables are buried. Without obtaining the necessary wayleave/s, no work can be conducted.
Is frequently needed to make new fibre connections happen.
Can take months, especially if there are disputes about access rights, compensation etc.
Does not solve the entire issue of obtaining a fibre connection; there is still the process of agreeing a timeline for conducting the works which may be particularly problematic if road/pavement closures are necessary.
Rather than plod along on with slow connectivity for an indeterminate period, or even risk not having fibre installed in time for a move to new premises, organisations have the option of receiving fibre-air services without the need for physical fibre – cutting install times from up to 6 months down to less than 10 days.
The fixed wireless infrastructure needed for Luminet's Fibre-Air normally needs no wayleaves or planning permission ( it is a de-minimis installation on the roof), and delivers 10Mbs to 1Gbps symmetrical clear channel services. Its standard SLA of 99.95% puts it on a par with fibre-based Ethernet services, and can be used for dedicated internet access or for multi-site networking (virtual or physical) using MPLS/VPLS.
Services like Fibre-Air don't just provide a stop-gap measure while you wait for fibre wayleave issues to be overcome – businesses understandably adopt it on a permanent basis. Why? As well as the SLA-assured performance and reliability, customers benefit from the added resilience that comes from truly diverse connectivity. By operating both a fibre-based connection and a wireless alternative like Fibre-Air, continuous 100% uptime for cloud is here..
Organisations don't have to put up with the wayleave delays and costs that often beset traditional fibre rollouts. All that pain can be avoided, and additional resilience achieved, using services such as Fibre-Air.


text:
Q: Remove last character from wxPython wxTextCtrl? I am trying to remove the very last character from a TextCtrl object in wxPython. I'm using wxPython 2.8.12 and Python 2.7.2.
My code is as follows:
def btnBkClicked(self, e):
    self.txtItem.Remove(self, (self.txtItem.GetLastPosition()[-1]), (self.txtItem.GetLastPosition()))

However, that doesn't work :( What do I need to change/do?

A: If you need to remove the very last character from the string, try
self.txtItem.SetValue(self.txtItem.GetValue()[:-1])

This code gets current text from TextCtrl and sets and sets its value to this text up to the last symbol.
Regarding your input, TextCtrl.Remove needs two parameters: from and to, which are integer numbers giving the first and the last positions to be removed. As GetLastPosition returns the number of characters in the control, your code should be modified as
self.txtItem.Remove(self.txtItem.GetLastPosition()-1, self.txtItem.GetLastPosition())


I am super excited. I have been busy creating some amazing Alphabet themed printables for preschool Kids. It's been a great and fun project and brought me right back to my Preschool teaching days. I really miss it and am grateful to be able to share these free printable worksheets here on The Inspiration Edit.
I've made 26 sets, one set per Alphabet letter and today we begin with the letter A. This A is for Acorn preschool printable has 4 fantastic worksheets. The activities involve, colouring or coloring (if your in America), counting and matching.
These sheets have fun ideas which will help with basic literacy and basic maths and they will introduce kids to each letter of the alphabet. These sheets can be used as letter of the week or maybe to help a child struggling. You may even use it to help teach phonics or a child with learning difficulties.
I will be sharing one letter every few weeks for the next year and am excited to have you download these and use them in your classroom or at home. Education should be free for all and I hope these help little ones in developing literacy and Alphabet skills.
Today we are sharing the first colouring page from the A is for Acorn activity pack.
We printed out the Acorn page and my daughter painted the Acorn using her new paint sticks. It's simple and easy to do and Sylvia had fun painting this sheet. I'm sure your kids will too.
We used Little Brian Mini Paint sticks for painting and colouring our Acorn worksheets. They are really good to use and there is no need for brushes or water. You can paint or crayon or use pencils, the choice is yours. Enjoy!
Download Your Free A is for Acorn Worksheets Here!


text:
Q: Google Search Appliance newwavesearchables UDP port 2058 Anyone have any idea what services newwavesearchables rmi UDP port 2058 does?
i cant find any information on it and our GSA is creating an excessive on our firewall. I cant find any options set on the GSA web interface that would utilise this service. Is anyone able to advise if this is normal as i cant find any information about this port in the GSA documentation.
Thank you

A: Never heard of that in my 8 years of work with the GSA.