A Filipino named Augusto Lakandula (later identified as Reginald Chua), hijacked a Philippine Airlines jet on May 25, 2000. After collecting the passengers' money and other valuables, the suspect, who appeared to be deranged, jumped at 6,000 feet from the plane's rear door using a homemade parachute.
Did he survive his dramatic escape or mysteriously vanished like 'D.B. Cooper', the infamous skyjacker?
On May 25, 2000, Philippine Airlines Flight 812 was en route from Francisco Bangoy International Airport in Davao City to Ninoy Aquino International Airport in Manila, carrying a total of 278 passengers and 13 crew members on board.
A few moments after the plane took off, a man wearing a blue ski mask and swimming goggles declared hijacking. Apparently prepared to commit his crime, the man was also carrying ropes, a gun and a hand grenade.
During the incident, the man fired a shot into the aircraft bulkhead, forcing the pilots to let him into the cockpit. Carlos Neri, the plane's co-pilot, described the hijacker as a deranged man who lamented that his family left him while his wife had an affair with a policeman.
The hijacker, first identified as "Augusto Lakandula" based on his plane ticket, demanded all passengers to place all their belongings in a bag. He also asked the pilot to have the plane diverted and return to Davao, but the pilot refused to do so due to insufficient fuel left.
After his initial request was denied, the hijacker commanded the pilot to descend and depressurize the airplane so the door could be opened, allowing him to escape using a makeshift parachute.
Just before his dramatic exit, the hijacker asked the crew to show him how to jump using a nylon parachute he brought with him. Seeing that the parachute apparently had no rip cord, one of the pilots took a curtain sash and attached it to the homemade parachute.
The hijacker finally jumped at 6,000 feet from the plane's left rear door while the aircraft was circling 13 miles from Manila. A flight attendant pushed him out of the door so he could escape against the strong force of the air. He reportedly separated from the parachute in midair and plunged to his death somewhere on a mountain in Real, Quezon.
With it's rear door still opened, the plane finally landed at Ninoy Aquino International Airport. All passengers and crew members arrived safe.
The body of Augusto Lakandula, later identified by his family members as Reginald Chua, was recovered about 40 miles east of Manila by local officials in Real, Quezon.
The hijacker's body was nearly buried in mud and his belt bag containing cash and jewelry was reportedly missing. The remains of his homemade parachute was recovered about a kilometer away from where his body was found.
In an interview with the Radio Mindanao Network, Chua's brother said that the hijacker actually had dreamed of becoming a skydiver and even collected books on skydiving. Unfortunately, he never had the chance to jump from an actual plane before the fateful event, which may explain why the hijacking ended up an epic failure.


In 1973, America's most famous tennis players, Billie Jean King and Bobby Riggs, competed in a match dubbed the Battle of the Sexes. Spoiler alert: She won, girls rule, boys drool.
Emma Stone and Steve Carell will play King and Riggs in a movie titled, well, "Battle of the Sexes." It's the latest project from "Little Miss Sunshine" and "Ruby Sparks" directors Jonathan Dayton and Valerie Faris. The trailer, which premiered Tuesday, features a lot of "can-a-woman-beat-a-dude?" squabbling. Guess we know the answer to that one.


TS Police Constable Answer Key 2018 Telangana SCT PC Question Papers Set A, B, C & D @ TSLPRB.IN – The Telangana State Level Police Recruitment Board (TSLPRB) invited online applications for the direct recruitment to the posts of Police Constable Civil and/or Equivalent. Candidates appeared in the exams are now looking to download TS Police Constable Answer Key 2018. In the newly formed state, this was a golden opportunity for the interested and eligible persons to join such a prestigious government organization. The online registration process started from 09.06.2018 and the registration was closed on 30.06.2018, wherein a good amount of candidates applied for it. The examination for the post of SCT Police Constable was successfully conducted on 30th September 2018.
A large number of candidates appeared in the written examination for these posts and are now waiting for TS Police Constable Answer Key to be declared. The result in such government exams may take time and it can be little frustrating for the candidates to wait, so TS Police Constable Answer Key 2018 will help the candidate to at least have an idea of their performance in the examination. In case if the candidates after evaluating the performance with Telangana Police Constable Exam Key find that they have not done good enough to qualify the exam, then in such a case they can without wasting their time look or prepare for another job or posts.
Telangana State SCT PC Answer Sheet Set A, B, C & D 2018 – The TS Police Constable examination is conducted by the Telangana State Level Police Recruitment Board. The Board releases the notifications for the recruitment to the posts as per the vacancies available. The Board has this time released total vacancies of 16,925 including that of Firemen, wardens (Male and Female) in Prisons and Correctional Services Department. As the examination has been successfully conducted, the Department will soon update the answer key for all question paper set A, B, C and D. Candidates should be very particular about referring to the correct TS Police Constable answer key 2018 against the Question Paper set.
The TS Police Constable answer sheets help the candidate to analyze their answer and evaluate their chances of selection in the next round of the recruitment as it contains the correct answers to every question asked in the SCT PC examinations, according to the Question Paper Set. The candidate can also raise the objection to the Telangana Constable answer keys 2018 updated by the authority. The authority will check the correctness of the objection and if required update the Civil Police Constable answer key accordingly. This will also ensure that the result declared is error free. For now, continue with the article and download TS Police Constable Answer Key 2018 either official or unofficial, we will provide both.
TSLRPB Civil Police Constable Exam Answer Key 2018 – The Selection Scheme for Civil Police Constable posts is as such that the candidates have to first go through the preliminary written examination. The candidate who qualifies the written test will have to undergo the physical measurement test also known as PMT. Further, the candidates who successfully qualify the physical measurement test will have to go through the physical efficiency test (PET). This shall be then followed by the final written examination for those candidates who are shortlisted after PET. The selection to the posts will be on the merit basis through their marks scored in the final examination.
TSLPRB Police Constable preliminary examination has been recently conducted. TS Police Constable answer key 2018 is generally generated on the official website within 7 days after the examination but may take time accordingly. The exam solution marks its importance for every exam. Candidates who wish to cross-check and calculate their exact marks beforehand can refer to the answer key of their respective set of question paper. There will also be many unofficial answer keys updated by the coaching centers such as Vyoma, Spark Academy, Sakshi education, however, the candidates shall not solely rely on them and wait until the official TS Police Constable answer key 2018 for Set A, B, C & D has been updated by exam conducting authority whiz; TSLPRB.
Telangana Police Constable Answer Key 2018 – The answer sheet is that important document for which every candidate waits impatiently. However, it is to be noted that the TS Police Constable Exam answer key shall not be treated as a final result of the written examination. It must be treated as merely an instrument to analyze the performance, the probable chances of selection to the next round. For more information regarding Telangana Constable answer keys, results and cut-off, the candidates shall stay tuned to our website. We have provided links through which interested candidates can download TSLPRB Police Constable Answer Keys for Set A, B, C & D online in pdf format.
How to Download TS Police Constable Answer Key?
On the homepage, locate "SCT PC Civil and/or Equivalent" Section.
Click on Preliminary Key under SCT PC Civil and/or Equivalent.
Answer Sheet will be displayed on the screen.
Download PDF file and take a printout of the same.
Compare the official answers with your's ones.


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Q: Is it possible to add a header to a mandrill webhook for inbound emails? In the inbound emails second, it said in the docs that when an email is received, it will forward the email through a POST request, so I should have an endpoint that will receive these POST requests. The thing is, we want the POST request sent by mandrill to have specific header values, possibly a token to verify it's from mandrill. Otherwise, I think anyone will be able spam us with fake POST requests. Is it possible to add a custom header?

A: It's not currently possible to add a custom header for POSTs. Mandrill, does, however, offer webhook authentication so you know it's coming from Mandrill and hasn't been modified.

redpajama_set_name: RedPajamaStackExchange


Partition on hard drive, SD card, flash drive or external HDD become RAW and cannot access files? Here is an easy and safe solution to help you perform RAW partition recovery and get back lost data quickly.
Why does disk partition become RAW file system?
A partition is a section of hard drive, which can be used to store data. A hard drive can be divided into one or more partitions to organize data. When you partition a hard drive, you can select filesystem for partition such as NTFS, FAT32, exFAT, etc. In Windows, NTFS is the default file system on hard drive, and FAT for removable disks. However, during the process of using storage device, partition may become RAW file system. RAW drive issue can happen to various devices such as hard disk, SD card, external hard drive, USB flash drive, etc.
A RAW drive means the file system of the partition is RAW and it cannot be recognized by operating system. Thus Windows cannot access and mount the partition properly. The RAW partition reports error message when you try to open it. If you check its status in Disk Management, the RAW drive is displayed as RAW, unallocated or uninitialized.
Most often, RAW partition is the result of logical data corruption. All data on the corrupt partition seems to be lost and you won't be able to repair it by assigning a drive letter. Partition can be regarded as a container of data, and the file system is the manager which is responsible for managing data. When the manager is corrupted or missing, operating system get nowhere to know where the partition is located and how files are stored. Thus we cannot access the drive anymore.
Fortunately, RAW partition recovery is available and it can be done many partition recovery applications. If the RAW drive contains important files for you and you can retrieve these files with the help of partition recovery software. When files are recovered successfully, you can format the RAW partition to get it repaired.
When you attempt to open the drive in File Explorer, a message would appear saying "Disk is not formatted. The disk in drive G is not formatted. Do you want to format it now?"
You need to format the disk in drive J: before you can use it. Do you want to format it?
H:\ is not accessible. The parameter is incorrect.
If you run chkdsk for RAW drive, you will receive the error message: The type of the file system is RAW. Chkdsk is not available for RAW drives or Convert is not available for RAW drives, chkdsk reports RAW file system.
Location is not available. G:\ is not accessible. The volume does not contain a recognized file system. Please make sure that all required file system drivers are loaded and that the volume is not corrupted.
Location is not available. H:\ is not accessible. The file or directory is corrupted and unreadable.
If you check properties for the drive, you can see the file system is RAW. In some cases, the free space is showing 0 byte in size.
Note: Do NOT format RAW partition even though Windows asks you to do so. Formatting will cause further damage to RAW drive and adds difficulty for RAW partition recovery.
ŸRead-only recovery process: the entire recovery process is read-only, which does not affect original data or devices.
ŸPartition recovery: it supports recovering deleted partition, restore lost partition, and retrieve files from formatted or corrupted partitions.
File recovery: Get back deleted or formatted files from local disk as well as removable devices.
File systems: support file recovery and partition recovery from NTFS/FAT/exFAT/Ext2/Ext3/Ext4 disks.
Hex editor: advanced users can recover lost files or partition manually via editing hex raw data on disk.
File preview: it allows users to preview various files (photos, MS office files, PDF, video, audio and Text files) to help check if files are correct.
WinPE bootable disk: it supports creating Windows PE bootable disk which can be used to recover data when system cannot boot.
Connect the disk that contains RAW partition to computer and make sure you can see the drive in either File Explorer or Disk Management. Formatting or running chkdsk command for RAW drive is of no help in RAW partition recovery, thus you should ignore error messages that Windows prompts to format it. Download and install DiskGenius on your computer and you can follow steps below to recover RAW drive.
Step 1 Launch DiskGenius and you can see the RAW drive from the main interface.
Step 2 Select the drive from the drive directory on the left pane and click File tab on the right pane, and check files in the RAW partition can be displayed directory.
In some cases, files and folders in RAW drive can be listed by DiskGenius with doing any scanning. If you can see your files, you can copy files to a different drive and get data recovered. If files cannot be displayed in this way, you should move to following steps.
If the RAW drive was NTFS format before it got corrupted, you can also try this: select the RAW drive and select "Load Current Partition Intelligently" option from Tools menu. This feature is designed to the recovery of NTFS partition whose $MFT is damaged not that badly. If this feature works, you can view all files in RAW drive within a couple of seconds. Then you can copy all files you need to a different drive.
Step 3 Right-click on the RAW drive and select "Recover Lost Files" option from context menu. This opens the file recovery window.
Step 4 Select recovery mode Complete Recovery and click Start button to search recoverable files from RAW partition.
If you want to try recovering files deleted before the drive got corrupted, you can select option "Recover Deleted Files". This recovery mode will scan and analyze sectors tagged as Free by system, so that deleted files can be found and recovered.
If the RAW drive is damaged quite badly, for example, you performed chkdsk or formatting on it, the option "Search For Known File Type" should be selected. This option is able to recover files even when file directory of partition is totally destroyed. Thus you can find more recoverable file with this option selected.
Step 5 Preview files during or after scanning to check whether files can be recovered correctly.
DiskGenius supports preview various types of files, including photos, documents (MS Office files, PDF file, TEXT files, etc.), audio and video files. File preview helps to judge if the file is correct and whether it is damaged. Select a file and you can view the file in thumb and hex data in the lower pane; right-click a file and select "Preview" or double-click a file, and you can view the file content in a previewing window which shows file content in its original size.
Step 6 Copy files to a different drive and get files recovered. Select files and folders you want to restore from the RAW partition, and right-click on them to select "Copy To". Then you'll be asked to select a location to save these files. You should select or create a folder on a different disk instead of the RAW drive and save recovered files.
Tip: If you are advanced user, you can try RAW partition recovery manually with the help of Sector Editor. With the help of editing hex data, you can repair damaged MFT or DBR to fix corrupt partition, which enables you to recover files from the disk directly without copying data to a different drive.
RAW file system means the drive is not formatted and there is not correct file system on it. Reasons why a partition becomes can be various. Now we sum up the common reasons that could lead to the issue.
Disk drive cable issue: It rare circumstances, the RAW drive can be caused by loose or broken cables. For example, the USB cable connecting external hard drive and computer is broken or loose; there will be something wrong during the data transfer and then lead to RAW partition.
Improper shutdown: A sudden power cut such as improper system shutdown while hard drive or other storage media is working threats data security. File system is likely to get damaged during the process and hence result in RAW partition.
Error while changing file system: Since different operating system uses different file system types, many users want to change file system format so that the drive can be used under multiple platforms. Some third-party software is able to change file system without formatting drive, but once error occurs during the converting process, the partition may turn to RAW and inaccessible.
Improper operations: Many users do not click the "Safely Remove Hardware and Eject Media" icon before disconnect USB flash drive, SD card or external hard drive. This is not a good habit, as it is one of the major causes of RAW drive.
Files system structural damage: As has mentioned above, RAW drive is the result of corrupt or incorrect file system. Thus if file system structural gets damaged, the drive will turn to RAW.
Virus attack: Virus and malware are widespread throughout the Internet, which accounts for many undesired issues like data loss, system crash, data damage, etc. Virus is destructive software which is able to infect all kinds of storage devices and corrupts file system. The affected drive might be RAW or gets encrypted due to virus. Therefore, users should install anti-virus software and scan computer regularly.
This instruction explains RAW file system and RAW drive and tells how to recover data from RAW partition. There are a number of causes leading to RAW partition, but you can easily carry out RAW partition recovery with the help of partition recovery software. However, it is still recommended to back up data in regular basis to avoid data loss issue. Hope you'll be able to do RAW drive recovery after reading this tutorial. If you feel this article useful, please share it to help more people.


Here we see twigs, red maple leaves, acorns, apples, evergreen clippings, straw and pinecones. It makes for an eye-catching modern take on the cornucopia.
We could see doing something similar on an autumn tabletop, maybe with a series of cheap Rektangel vases at IKEA. Gathering the materials yourself for a display like this could make for a cost-effective centerpiece at the autumn dining table. Make it into a fun activity with family or friends: the outcome will be personalized little specimens of autumn color from your own yard or neighborhood.


HardOCP has a look at the PC Power & Cooling Turbo-Cool 1KW-SR power supply. With the latest in quad cores and DX10 multi-GPU based setups pulling increasing amounts of power the company known for high-quality power supplies steps up to the plate with this behemoth. It'll certainly power any consumer grade machine you throw at it but the [H] has some thoughts on it.


The average person moves every 7 years in their life time the cost of professional fees for these 7 moves is at least equal to their annual salary.
This is not including buy to let properties and holiday homes.
Or the cost of care home fees in later life.
Yet many people believe that these assets can be protected with a simple will equal to a few hours work. Some believe they don't need to make any provision and the Rules of Intestate will be sufficient.
Properly structured Trust and Wills can protect your property for less than a month's salary.
Resulting in the assets going to who you wanted irrespective of events after your death. I.e. your spouse remarrying or children divorcing etc.
Comparing the cost of Trust and other legal services is very confusing, when we researched the cost of these services from banks and other providers; we were unsure exactly what they referred to. Similar to comparing car servicing cost without knowing if you are being offered a basic safety check ,mini service, standard service, or major service with M.o.T.
Many people know what they want to achieve and know there must be a product but become confused by all the different names used by different providers for the same product. We suspect this is to stop you comparing. Also the more complicated they make a subject appear the more they can justify high fees.
Therefore we have listed some of the questions people ask us we hope this helps.
Requirement. You want to put your property and assets in to a Discretionary Trust to take them out of your estate. Suitable in making sure assets pass to your intended beneficiaries, protect against sideways inheritance, care home fees, bankruptcy.
Requirement. you want your property to pass to your children even if your spouse remarries use a Discretionary Trust.
Requirement. You do not want your child to inherit at 18. A Discretionary trust gives the trustees control until any age or event, they could release some assets for education or deposit for a property or any reason set out in your guidance notes.
Requirement you want your property to pass to your children even if your spouse remarries use a Discretionary Trust.
Requirement you want your property to pass to your children even if your spouse remarries but you want your spouse to have the uses of the property for they life time. Lifetime Interest Trust.
Requirement you want your "Buy to Let" property to pass to your children but you want the rental income to go to your mother for her life time, product; Lifetime Interest Trust.
Requirement you are going abroad and want a relative / friend to take care of your assets in the UK including the sale of your property. General Power of attorney or specific Power of attorney with just gives them control over a named item i.e. one property.
Requirement you want a relative / friend to be able to act for you regarding property and financial affairs if you where physical incapable i.e. in hospital. Lasting Power of attorney (Property & Financials) or Lasting Power of attorney (Health & Welfare) for medical treatment etc.
Requirement you want a relative / friend to be able to act for you if you where mentally incapable i.e. dementia or as a result of an accident. Lasting Power of attorney,(Health & Welfare) for medical treatment etc.
Requirement you have nominated guardians for your children, you have also set up adequate life insurance provision, you want a trust that will release part of the funds on a annual basis for their upkeep to the guardians, a provision for education, and a provision for deposit on a property you want to nominate trustees to carry out your wishes at their discretion. Discretionary Trust.
Requirement you have a business partner on your death you want your partner to be able to buy your share of the business ;product Life Insurance paid for by the business but how do make sure that your business partner uses this money to buy out your family product Cross Option Agreement & Business Trust. If you want to make other arrangements regarding your business then Business Trust.
You will have noticed that a Discretionary Trust is mentioned many times one trust can for fill all these functions and many others. We do not give the same trust 3 or 4 different names and advise you that you need one of each (we are not a multi-national bank) A Absolute Trust performs a similar function but there is no flexibility, e.g. if you state your son inherits at 25 that when he inherits irrespective of the fact he is about to get divorced or to be made bankrupt or joined a sect.
*1 £250 deposit Safe Hands Plans.
*2 £99 deposit Golden Charter.
*3).Mirror Wills are where there are two identical wills with just the names changed.
*4) Pair of Wills. This is where there are differences between the Wills. Such as different distribution, choice of executors etc but where both Wills are instructed at the same time.
*5) Property Protection Trust helps protect the property against care home fees. For more information clink link.
*6) Lasting Power of Attorney (LPA) includes a LPA Health & Welfare and a LPA Property & Finance together with the application forms completed. It does not include the Ministry of Justice registration fee of £82 this has been reduced from £110. Registration does not have to be completed immediately. Price only applies where we are also producing the Wills.
*7) Two complete sets of LPAs arranged at the same time.
*8) Funeral Plan purchase through us (Safe Hands direct deal and Golden Charter Base Plan 25% discount).
*9). Price quoted is for an online telephone and virtual adviser appointment, home visits are available at additional cost.
*10). Price applies if purchasing a funeral plan through us.
*11). Does not include severance of title. Not always required depends on type of ownership.


Your family will love this delicious wholesome dish.
Heat butter and fry onion and garlic until fragrant.
Add paprika and flour, and cook for 2 minutes, stirring constantly.
Gradually add milk, whisking until smooth. Reduce heat and cook until sauce thickens.
Add broccoli, kale, macaroni, bacon and cheese. Season with pepper.
Transfer into oven dish. Arrange tomato slices on top, and sprinkle with Parmesan cheese. Bake for 15–20 minutes or until golden.
Serve hot with basil leaves and fresh thyme.
For hot food lovers, try this peri peri chicken recipe.
Try this 20 minutes recipe for your meat-free days.


text:
Audio Stories by APMD
Remember the Mastodon
Staten Island SEEN
Opening the Treasure Box
From Farm to City
Field Notes: Seed Stories and the Power of Plants
Impractical Jokers: Homecoming
We Are Here: Superstorm Sandy and Her Aftermath on Staten Island
Staten Island in Photographs: Images from the History Archives and Library of Staten Island Museum
This online gallery has been created to commemorate the historic event of Superstorm Sandy and the aftermath on Staten Island, as seen through eyes and camera lenses of those whose communities were ravaged by the water and winds on the night of October 27, 2012. Thirty-seven images from thirteen artists are presented here to illustrate the onslaught of the storm and the resulting losses.
Learn More and View Selected Works >>
With Staten Island now the fastest-growing borough in New York City, and the fastest-growing county in New York State, the island's past, as recorded in these pictures, has never been more relevant or timely.

redpajama_set_name: RedPajamaCommonCrawl


text:
Galîna Vasîlivna Berejna (în ) (n. 2 august 1943, Bahmaci, raionul Bahmaci, regiunea Cernigău) este o profesoară de filologie și traducătoare ucraineană din limbile română, bulgară, slovacă, bielorusă și italiană.

Biografie
A absolvit Facultatea de Filologie a Universității Naționale "Taras Șevcenko" din Kiev. Predă ca profesoră de limba ucraineană la Institutul Politehnic din Kiev.

A tradus în limba ucraineană din operele literare ale unor scriitori români (Mihail Sadoveanu, Ioan Slavici, Titus Popovici, Petre Sălcudeanu, I. Rațiu, Octav Pancu-Iași, I. Ciobanu, N. Vieru, G. Gheorghiu, Sergiu Celac), italieni (Luigi Pirandello), slovaci (Helena Križanová-Brindzová), bieloruși (Ianka Brîl) sau bulgari.

Este membru al Uniunii Naționale a Scriitorilor din Ucraina din anul 1987.

A fost activistă și director adjunct la Asociația prietenilor orașului Cernigău.

Traduceri
 Mihail Sadoveanu, Ostriv kvitiv (Nada Florilor), Veselka, Kiev, 1980.
 Helena Križanová-Brindzová, Поштарик : повість, Veselka, Kiev, 1986. - 279 p.

Legături externe
 http://nspu.org.ua/dovidnyk/%D0%91
 Галина Бережна на кафедрі української мови, літератури та культури КПІ

Nașteri în 1943
Traducători ucraineni
Ucraineni în viață

redpajama_set_name: RedPajamaWikipedia


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Home Blogs Tap On the first day of 2018, The New York Times reported on a technological breakthrough
On the first day of 2018, The New York Times reported on a technological breakthrough
by Harold Meyerson
Google Street View's images of America's neighborhoods, a Stanford University study concludes, can now be interpreted by artificial intelligence to predict a neighborhood's—or a street's or a block's—politics.
"Image recognition technology, much of it developed by major technology companies, has improved greatly in recent years," the Times reported, noting that the primary data on which AI drew its conclusions were the cars parked on the street. "The Stanford project gives a glimpse at the potential. By pulling the vehicles' makes, models, and years from the images, and then linking that information with other data sources, the project was able to predict factors like pollution and voting patterns at the neighborhood level."
The story didn't say if a name has been given to this project, but I know what its name was in 1980: Michael Berman. In that year, the most prominent California Democrat in the House of Representatives, and the most brilliant legislative strategist of House liberals, San Francisco's Phil Burton, was the Democrats' choice to head up the state's decennial redistricting. California was growing by leaps and bounds, and Burton used the opportunity not only to give Democrats the edge in most of the state's newly added districts, but to redistrict several right-wing Republicans—most notably, former John Birch Society Western Regional Director John Rousselot—into districts they could not win. Rousselot and several other Republicans lost their re-election bids in 1982.
Burton was an acknowledged genius at redistricting; he later called his 1980 line-drawing "my contribution to modern art." (In fairness, I should note that he also steered to passage expansions of welfare, protections for mine workers, and the creation of Redwoods National Park; led the anti-Vietnam War Democrats in the House, engineered the abolition of the House Un-American Activities Committee, and ended the practice of assigning committee chairs strictly by seniority, which had the effect of dethroning the party's remaining old guard Dixiecrats.)
What was Burton's secret? How did he redistrict so masterfully, in a time when computers couldn't yet spit out the data routinely used today to draw the lines? The answer is Michael Berman.
A onetime Burton aide who later became one of California's most successful political consultants, in 1980, Berman got in his car and drove all over the state, assessing a neighborhood's socioeconomic status by noting which cars were parked on which streets. His methods were essentially those of Google Street Views as interpreted not by artificial intelligence but his own. (As early as Henry Waxman's first campaign for the California State Assembly, in 1968, Berman was experimenting with an embryonic version of micro-targeting voters, decades before it became common practice.) In 1990 and 2000, Berman was to play a similar role working with his brother Howard, a San Fernando Valley congressman who succeeded Burton (who died in 1983) as California Democrats' capo de redistricting. But with each succeeding decade, the capacity to redistrict using computerized data grew substantially stronger. And today, in our age of digital marvels and artificial intelligence, tech has finally caught up with Michael Berman c. 1980, tooling down the street, noting all the cars.
Blog: ON TAP
Harold Meyerson
Harold Meyerson is editor at large of The American Prospect.
Read more by Harold Meyerson


Do you adore elegance and style? Let your beloved American Bulldog take joy in wearing such collars on! Purchase this high quality leather dog collar with studs from FDT Artisan . This accessory is designed for those who just love elegance and style. In this exquisite collar, your four-legged friend will have more stylish, but attractive look. Your attractive four-legged friend will get the best treatment when wearing this gear. Full grain natural leather is gentle and totally eco-friendly.
Brass plated hardware will stand out even the hardest American Bulldog tension. Incredible embellishments are brass plated and have impressive shining. Accentuate your dog's individuality, purchase this Artisan gorgeous genuine leather collar!
Full grain genuine leather was chosen to provide total comfort and safety while wearing. The material is perfectly oiled that keeps it from cracking. Among the best features of genuine leather are perfect adjustability, wear-resistance and durability. The layer of leather is quite thick that prevents the tool cracking. It helps you to walk even big and strong dog safely. Rounded and smooth edges of the collar are non-cutting and don't rub the American Bulldog's skin. Besides, genuine leather doesn't consist of any toxic elements being totally eco-friendly.
Speaking about the studs, they are stylish! Catchy studs are placed along the leather strap. These adornments are brass plated and, therefore shine with stunning glittering. The sophisticated design of this tool is matchless and unique. Due to these decorations, this genuine leather dog collar has a spark. That means your American Bulldog will have his individual and exquisite style .
As for the hardware of this decorated leather supply, it is brass plated. It matches perfectly with the embellishments. This covering provides the best protection for the details as it prevents rust. The set of fittings consists of a classic elegant buckle and reliable D-ring for attaching a leash. These fittings are easy to use and reliable in service they provide. Add more zest to your American Bulldog's outlook, purchase this attractive leather collar!


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\section{Introduction}

Let $\mathfrak{g}$ be a Lie algebra over an algebraically closed field
$\Bbbk$ of characteristic zero. For $f \in \mathfrak{g}^{*}$, we denote by
$\mathfrak{g}^{f}=\{ X\in \mathfrak{g} ; f([X,Y])= 0$ for all $Y \in \mathfrak{g} \}$,
the annihilator of $f$ for the coadjoint representation of $\mathfrak{g}$. The
\highlight{index} of $\mathfrak{g}$, denoted by $\chi ( \mathfrak{g})$, is defined
to be
$$
\chi (\mathfrak{g})=\min_{f\in \mathfrak{g}^{*}} \dim \mathfrak{g}^{f}.
$$
It is well-known that if $\mathfrak{g}$ is an algebraic Lie algebra and $G$ denote
its algebraic adjoint group, then $\chi (\mathfrak{g})$ is the transcendence degree
of the field of $G$-invariant rational functions on $\mathfrak{g}^{*}$.

The index of a semisimple Lie algebra $\mathfrak{g}$ is equal to the rank of $\mathfrak{g}$.
This can be obtained easily from the isomorphism between $\mathfrak{g}$ and $\mathfrak{g}^{*}$
via the Killing form. There has been quite a lot of recent work on the determination
of the index of certain subalgebras of a semisimple Lie algebra:
parabolic subalgebras and related subalgebras
(\cite{DK}, \cite{Panyseaweed}, \cite{TYseaweed}, \cite{Mor2}),
centralizers of elements and related subalgebras
(\cite{Pany}, \cite{Char}, \cite{Yaki}, \cite{Mor1}).

Let $\mathfrak{g}$ be a semisimple Lie algebra, $\mathfrak{p}$ a parabolic subalgebra of
$\mathfrak{g}$ and $\mathfrak{u}$ (resp. $\mathfrak{l}$) the nilpotent radical
(resp. a Levi factor) of $\mathfrak{p}$.
In \cite[Corollary 1.5 (i)]{Pany}, Panyushev showed that
\begin{equation}\label{oldupperbound}
\chi (\mathfrak{p})+\chi (\mathfrak{u}) \leq \dim \mathfrak{l}.
\end{equation}
He then suggested \cite[Remark (ii) of Section 6]{Pany} that
\begin{equation}\label{indexsumformula}
\chi (\mathfrak{p})+\chi (\mathfrak{u}) \geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g}.
\end{equation}
For example, it is well-known that if $\mathfrak{b}$ is a Borel subalgebra of $\mathfrak{g}$
and $\mathfrak{n}$ is its nilpotent radical, then
$\chi (\mathfrak{b})+\chi (\mathfrak{n})=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$ (see for example
\cite{TYstable}, \cite[Chapter 40]{TYbook}). It is therefore also interesting to characterise
parabolic subalgebras where equality holds in \Eqref{indexsumformula}. Indeed, in
\cite{Rais}, Ra\"\i s looked for examples of direct sum decompositions
$\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{n}$ verifying the ``index
additivity condition'', namely $\mathfrak{m}$ and $\mathfrak{n}$ are Lie subalgebras of
$\mathfrak{g}$ and
$$
\chi (\mathfrak{g})=\chi (\mathfrak{m}) + \chi (\mathfrak{n}).
$$

If $\mathfrak{u}_{-}$ denotes the nilpotent radical of the opposite parabolic subalgebra
$\mathfrak{p}_{-}$ of $\mathfrak{p}$, then $\mathfrak{g}=\mathfrak{p}\oplus \mathfrak{u}_{-}$
and the Lie algebras $\mathfrak{u}$ and $\mathfrak{u}_{-}$ are isomorphic. Thus
parabolic subalgebras such that equality holds  in \Eqref{indexsumformula}
would provide examples of direct sum decompositions verifying the index
additivity condition.

Using the formula, conjectured in \cite{TYseaweed} and proved in
\cite{MFJ,Jindex}, for the index of parabolic subalgebras,
we obtain a formula for the sum $\chi (\mathfrak{p}) + \chi (\mathfrak{u})$.
By a careful analysis of root systems, we prove the inequality \Eqref{indexsumformula} and give
a necessary and sufficient condition of equality to hold in \Eqref{indexsumformula} (See Theorem
\ref{Main}).

To describe the index of a parabolic subalgebra, and the index of its nilpotent radical
we need to recall Kostant's cascade construction of pairwise strongly
orthogonal roots (\cite{Jan}, \cite{Jos}, \cite{TYbook}).

Let us fix a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ and a
Borel subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ containing $\mathfrak{h}$. Denote
by $R$, $R^{+}$ and $\Pi=\{ \alpha_{1},\dots ,\alpha_{\ell} \}$
respectively the set of roots, positive roots and simple roots with
respect to $\mathfrak{h}$ and $\mathfrak{b}$. For any $\alpha \in R$, let $\mathfrak{g}_{\alpha}$
be the root subspace associated to $\alpha$. Choose $X_{\alpha}$ such that
$\alpha ([X_{\alpha},X_{-\alpha}])=2$. We shall write
$\alpha^{\vee}=[X_{\alpha}, X_{-\alpha}]\in \mathfrak{h}$,
and for $\lambda \in \mathfrak{h}^{*}$, $\langle \lambda, \alpha^{\vee} \rangle =\lambda
(\alpha^{\vee})$. For $S\subset \Pi$, we denote by
$R_{S}=R\cap \mathbb{Z}S$, $R_{S}^{+}=R_{S} \cap R^{+}$. If $S$ is connected,
then we shall denote by $\varepsilon_{S}$ the highest root of $R_{S}$.

Let $S\subset \Pi$. We define $\mathcal{K}(S)$ inductively as follows:
\begin{List}{(iii)}
\item[a)] $\mathcal{K}(\emptyset)=\emptyset$.
\item[b)] If $S_{1},\dots ,S_{r}$ are the connected components of $S$, then
$
\mathcal{K}(S)=\mathcal{K}(S_{1})\cup \cdots \cup \mathcal{K}(S_{r}).
$
\item[c)] If $S$ is connected, then
$
\mathcal{K}(S)=\{ S \} \cup \mathcal{K}(\widehat{S})
$
where $\widehat{S}=
\{ \alpha \in S ; \langle \alpha , \varepsilon_{S}^{\vee} \rangle=0 \}$.
\end{List}

It is well-known that (see for example \cite[Chapter 40]{TYbook}) elements of $\mathcal{K}(S)$
are connected subsets of $S$. Moreover, if we denote by
$\mathcal{R}(S)=\{ \varepsilon_{K} ; K\in \mathcal{K}(S) \}$, then $\mathcal{R}(S)$
is a maximal set of pairwise strongly orthogonal roots in $R_{S}$.

Let us also recall the following properties of $\mathcal{K}(S)$:

\begin{lemma}\label{lemma}
Let $S$ be a subset of $\Pi$, $K,K'\in \mathcal{K}(S)$ and set
$$
\begin{array}{rl}
\Gamma^{K}& =\{ \alpha \in R_{K} ; \langle \alpha ,\varepsilon_{K}^{\vee} \rangle > 0 \}\\
& =\{ \alpha= \sum_{\beta\in K} n_{\beta}\beta \in R_{K}^{+} ; n_{\beta} > 0 \hbox{ for some }
\beta\in K\setminus \widehat{K} \}.
\end{array}
$$
\begin{List}{(iii)}
\item[i)] We have either $K\subset K'$ or $K'\subset K$ or $K$ and $K'$ are connected components
of $K\cup K'$.
\item[ii)] $\Gamma^{K} = R_{K}^{+} \setminus \{ \beta \in R_{K}^{+} ;
\langle \beta , \varepsilon_{K}^{\vee} \rangle =0 \}$. In particular,
$R_{K}^{+}$ is the disjoint union of the $\Gamma^{K}$'s, $K\in \mathcal{K}(S)$.
\item[iii)] $\sum_{\alpha \in \Gamma^{K}} \mathfrak{g}_{\alpha}$
is a Heisenberg Lie algebra whose centre is
$\mathfrak{g}_{\varepsilon_{K}}$. Thus if $\alpha ,\beta \in \Gamma^{K}$ verify
$\alpha + \beta \in R$, then $\alpha +\beta =\varepsilon_{K}$.
\item[iv)] Suppose that $\alpha \in \Gamma^{K}$ and $\beta \in \Gamma^{K'}$ verify
$\alpha + \beta \in R$, then either $K\subset K'$ and $\alpha + \beta \in \Gamma^{K'}$
or $K'\subset K$ and $\alpha + \beta \in \Gamma^{K}$.
\end{List}
\end{lemma}

\goodbreak
\begin{state}{Table}\label{table}
The cardinality of $\mathcal{K}(\Pi )$ is listed below for an irreducible root system $R$:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\hline
Type & $A_{\ell}, \ell \geq 1$ &
$B_{\ell}, C_{\ell}, \ell \geq 2$ &
$D_{\ell} , \ell \geq 4$ & $E_{6}$ & $E_{7}$ & $E_{8}$ & $F_{4}$
& $G_{2}$ \\ \hline
$\sharp \mathcal{K}(\Pi )$ &
\strut\rule[-2.8ex]{0ex}{7ex}
$\left[\displaystyle\frac{\ell+1}{2}\right]$ & $\ell$ & $2 \left[
\displaystyle\frac{\ell}{2}\right]$& $4$ & $7$ & $8$ & $4$ & $2$ \\ \hline
\end{tabular}
\end{center}
where for any $x\in\mathbb{Q}$, $[x]$ is the unique integer such that
$[x]\leq x<[x]+1$.
\end{state}

\begin{examples}\label{examples}
Let $R$ be an irreducible root system. We shall use the numbering of simple roots in
\cite[Chapter 18]{TYbook}. Set $k=\sharp \mathcal{K}(\Pi )$.

1) Let $R$ be of type $A_{\ell}$. Then
$\mathcal{K}(\Pi )=\bigcup_{i=1}^{k}\{ K_{i}\}$ where
$K_{i}=\{ \alpha_{i},\dots ,\alpha_{\ell +1-i} \}$. For $1\leq i\leq k$,
$$
\Gamma^{K_{i}} = \{ \alpha_{i}+ \cdots + \alpha_{i+r}\ , \
\alpha_{\ell+1-i-r} + \cdots + \alpha_{\ell+1-i}
; 0\leq r \leq \ell -2i \}  \cup \{ \varepsilon_{K_{i}} \}.
$$

2) Let $R$ be of type $D_{2n+1}$. Then $k=2n$,
$$
\mathcal{K}(\Pi )=\bigcup_{i=1}^{n} \{ K_{i}\} \cup \{ L_{i}\}
$$
where
$K_{i}=\{ \alpha_{2i-1}, \dots ,\alpha_{2n+1} \}$
and
$L_{i}=\{ \alpha_{2i-1} \}$. For $1\leq i\leq n$,
$$
\Gamma^{K_{i}} = \bigg\{ \sum_{j=2i-1}^{\ell} m_{j}\alpha_{j} ; m_{2i}\neq 0\bigg\} \ , \
\Gamma^{L_{i}} = \{ \alpha_{2i-1} \}.
$$

3) Let $R$ be of type $E_{6}$. Then
$$
\mathcal{K}(\Pi )=\{ \Pi \} \cup
\{ \{ \alpha_{1},\alpha_{3},\alpha_{4},\alpha_{5},\alpha_{6} \} \}
\cup
\{ \{ \alpha_{3} , \alpha_{4},\alpha_{5} \} \}
\cup
\{ \{ \alpha_{4} \} \} .
$$
\end{examples}

\section{Main result}\label{main}

Recall that for any subset $S\in\Pi$,
$$
\mathfrak{p}_{S}=\mathfrak{h}\oplus \bigoplus_{\alpha\in R_{S}\cup R^{+}} \mathfrak{g}_{\alpha}
$$
is a (standard) parabolic subalgebra of $\mathfrak{g}$.
Any parabolic subalgebra of $\mathfrak{g}$ is conjugated to a standard
parabolic subalgebra. The Lie subalgebra
$$
\mathfrak{u}_{S}=\bigoplus_{\alpha\in R^{+}\setminus R_{S}} \mathfrak{g}_{\alpha}
$$
is the nilpotent radical of $\mathfrak{p}_{S}$.

For $S\subset \Pi$,
denote by
$V_{S}$ the vector subspace of $\mathfrak{h}^{*}$ spanned by the elements
of $\mathcal{R}(S)$ and $\mathcal{R}(\Pi )$. Set
$$
\begin{array}{c}
\mathcal{E}_{S}=\{ K \in \mathcal{K}(\Pi ) ; X_{\varepsilon_{K}} \in \mathfrak{u}_{S} \}
=\{ K \in \mathcal{K}(\Pi ) ; \varepsilon_{K} \not\in R_{S}\} \\
\hbox{ and } \
\mathcal{Q}_{S}=\bigg( \bigcup_{K\in \mathcal{E}_{S}} \Gamma^{K}\bigg) \cap R_{S}^{+}.
\end{array}
$$
The subset $\mathcal{E}_{S}$ has the following simple characterisation.

\begin{lemma}\label{TSlemma}
Let $T_{S}$ be the union of  $K\in \mathcal{K}(\Pi )$ verifying $K\subset S$. Then
$\mathcal{E}_{S}=\mathcal{K}(\Pi ) \setminus \mathcal{K}(T_{S})$.
\end{lemma}
\begin{proof}
This is straightforward.
\end{proof}

Our main result is the following theorem.

\begin{theorem}\label{Main}
Let $S\subset \Pi$. Then
\begin{equation}\label{newupperbound}
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})
=\mathop{\mathrm{rk}}\nolimits \mathfrak{g} +  \sharp \mathcal{K}(S) - \sharp \mathcal{K}(T_{S}) +
2 (\sharp \mathcal{K}(\Pi )-\dim V_{S} ) + \sharp \mathcal{Q}_{S}.
\end{equation}
We have $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})\geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g}$,
and equality holds
if and only if the following conditions are satisfied:
\begin{List}{(iii)}
\item[i)] $\sharp ( \mathcal{K}(S)\cup \mathcal{K}(\Pi ) )= \dim V_{S}$.
\item[ii)] For any connected component $S'$ of $S$,
we have either $S'\in \mathcal{K}(\Pi)$ or $\sharp (S'\setminus T_{S})=1$.
\end{List}
\end{theorem}
\begin{proof}
The formula for the sum $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})$
is a direct consequence of the formula of the index of parabolic subalgebras
conjectured in \cite{TYseaweed} and proved
in \cite{MFJ,Jindex} :
\begin{equation}\label{paraformula}
\chi (\mathfrak{p}_{S}) = \mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp \mathcal{K}(\Pi) +
\sharp \mathcal{K}(S) - 2 \dim V_{S} \ ,
\end{equation}
and the formula for the index of $\mathfrak{u}_{S}$
(see for example \cite[Chapter 40]{TYbook}), which, in view of Lemma \ref{lemma},
can be expressed in the following way:
\begin{equation}\label{radformule}
\chi (\mathfrak{u}_{S})=
\sharp \mathcal{E}_{S} + \sum_{K \in \mathcal{E}_{S}} \sharp \Gamma^{K} - \dim \mathfrak{u}_{S}
= \sharp \mathcal{E}_{S} +  \sharp \mathcal{Q}_{S}
\end{equation}

To prove the rest of the theorem, we may clearly
assume that $\mathfrak{g}$ is simple. Observe that
\begin{equation}\label{S-T}
\sharp \mathcal{K}(S) - \sharp \mathcal{K}(T_{S}) \geq 0.
\end{equation}
\vskip0.5em

1) Let $S_{1},\dots ,S_{r}$ be the connected components of $S$.
For each $i$, there is a unique $K_{i}\in \mathcal{K}(\Pi )$ (see Lemma \ref{lemma})
such that $\varepsilon_{S_{i}} \in \Gamma^{K_{i}}$.
If $K_{i}=S_{i}$, then $S_{i}$ is a connected component of $T_{S}$.
Otherwise $K_{i} \in \mathcal{E}_{S}$, and we have
$$
\varepsilon_{S_{i}} \in \Gamma^{K_{i}} \cap \Gamma^{S_{i}} \subset \mathcal{Q}_{S}.
$$

2) It follows from Point 1) that $\mathcal{Q}_{S}=\emptyset$ if and only if
$\mathcal{K}(S)\subset \mathcal{K}(\Pi )$ (or equivalently $S=T_{S}$).
\vskip0.5em

3) Note that the connected components of $T_{S}$ are the connected components of
$T_{S}\cap S_{i}$, it follows again from Point 1) that:
$$
\begin{array}{rcl}
\sharp \mathcal{K}(S)-\sharp \mathcal{K}(T_{S})
& = & \displaystyle\sum_{i=1}^{r} (\sharp \mathcal{K}(S_{i}) -\sharp \mathcal{K}(T_{S}\cap S_{i}) )\\
& = & \displaystyle\sum_{K_{i} \in \mathcal{E}_{S}}
( \sharp \mathcal{K}(S_{i}) - \sharp \mathcal{K}(T_{S}\cap S_{i}) ).
\end{array}
$$

4) From Table \ref{table}, we have $\dim V_{S} = \sharp \mathcal{K}(\Pi )=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$
in the cases where $\mathfrak{g}$ is of type
$B_{\ell}$, $C_{\ell}$, $D_{2n}$, $E_{7}$, $E_{8}$, $F_{4}$ et $G_{2}$.
The inequality follows immediately from \Eqref{newupperbound} and \Eqref{S-T}, and
the condition for equality follows from Point 2).
\vskip0.5em

5) \underbar{Type $A_{\ell}$}.

For any $i$ verifying $S_{i}\neq K_{i}$,  by Point 1), Lemma \ref{lemma}
and Examples \ref{examples}, half of $\Gamma^{S_{i}}\setminus \{  \varepsilon_{S_{i}} \}$ belongs
to $\mathcal{Q}_{S}$. Since $\sharp (\Gamma^{S_{i}})=2\sharp (S_{i})-1$ (Examples \ref{examples}),
such an $S_{i}$ contributes $\sharp (S_{i})$ elements of $\mathcal{Q}_{S}$.

Again, since we are in type $A$, $\mathcal{K}(\Pi )$ is totally ordered by inclusion. It
follows that $T_{S}$ is connected. Without loss of generality, we may assume that
$T_{S}\subset S_{1}$.

Suppose that $S_{1}= T_{S}$. Then from the previous discussion, we deduce that
$$
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) \geq
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp \mathcal{K}(S\setminus S_{1})
+ 2 (\sharp \mathcal{K}(\Pi )- \dim V_{S}) + \sharp (S\setminus S_{1}).
$$
But our hypothesis implies that
\begin{equation}\label{eqnA}
\dim V_{S} \leq \sharp \mathcal{K}(\Pi ) + \sharp \mathcal{K}(S \setminus S_{1})
=\sharp (\mathcal{K}(\Pi )\cup \mathcal{K}(S) ),
\end{equation}
so
$$
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})
\geq
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp (S \setminus S_{1} ) -  \sharp \mathcal{K}(S\setminus S_{1}).
$$
Hence $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) \geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g}$.
For equality to hold, we must have equality in \Eqref{eqnA} and
$$
\sharp (S \setminus S_{1} )=\sharp \mathcal{K}(S\setminus S_{1}).
$$
This latter is only possible if $\sharp (S_{i})=1$ for $i\geq 2$, so we have conditions
(i) and (ii). Conversely, suppose that conditions (i) and (ii) are verified, then
$\sharp (S_{i})=1$ for $i\geq 2$. Consequently $\mathcal{Q}_{S}=S\setminus S_{i}$ by Point 1)
and the definition of $\mathcal{Q}_{S}$.

Suppose that $S_{1}\supsetneq T_{S}$ (this includes the case $T_{S}=\emptyset$). Then
$\mathcal{K}(S)\cap \mathcal{K}(\Pi )=\emptyset$.
Thus
\begin{equation}\label{eqnA2}
\dim V_{S} \leq \sharp \mathcal{K}(\Pi ) + \sharp \mathcal{K}(S)
=\sharp ( \mathcal{K}(\Pi )\cup \mathcal{K}(S) ).
\end{equation}
We deduce from Point 1) and
the remark in the first paragraph of Point 5) that
$$
\begin{array}{rl}
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) & \geq
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp \mathcal{K}(S) - \sharp \mathcal{K}(T_{S})
+ 2 (\sharp \mathcal{K}(\Pi )- \dim V_{S}) + \sharp S \\
& \geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp (S) - \sharp \mathcal{K}(S) - \sharp \mathcal{K}(T_{S}) .
\end{array}
$$
Hence
$$
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})
\geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g} +  \displaystyle \sum_{i=1}^{r} \left(
\sharp (S_{i}) - \left[ \frac{\sharp (S_{i}) +1}{2} \right] \right) -
\displaystyle\left[ \frac{\sharp (T_{S}) +1}{2} \right] .
$$
Since $T_{S}\subsetneq S_{1}$, we deduce from Table \ref{table} that
$$
\sharp (S_{1}) - \displaystyle\left[ \frac{\sharp (S_{1}) +1}{2} \right]  -
\displaystyle\left[ \frac{\sharp (T_{S}) +1}{2} \right] \geq 0.
$$
So we have our inequality $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) \geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g}$.

Now for the equality $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) =\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$
to hold, we must have equality in \Eqref{eqnA2} and
$\sharp \mathcal{Q}_{S}=\sharp (S)$,
$$
\sharp (S_{1}) - \left[ \frac{\sharp (S_{1}) +1}{2} \right] -
\displaystyle\left[ \frac{\sharp (T_{S}) +1}{2} \right]=0
\hbox{ and }
\sharp (S_{i}) - \left[ \frac{\sharp (S_{i}) +1}{2} \right]=0
$$
for $i\geq 2$. This implies that $\sharp (S_{1})=\sharp (T_{S})+1$, and $\sharp (S_{i})=1$ for
$i\geq 2$. So we have conditions (i) and (ii). Conversely,
if conditions (i) and (ii) are verified,
then $\sharp (S_{i})=1$ for $i\geq 2$, and $\sharp (S_{1})=\sharp (T_{S})+1$. In view of the above,
to show that $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) =\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$, it suffices to check
that $\sharp (\mathcal{Q}_{S}\cap R_{S_{1}}^{+})=\sharp (S_{1})$, which is a straightforward
verification.
\vskip0.5em

6) \underbar{Type $D_{2n+1}$}.

In this case, $\sharp \mathcal{K}(\Pi )=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}-1$. Let us use the numbering of
simple roots in \cite[Chapter 18]{TYbook}. We check easily that
$\alpha_{1},\dots ,\alpha_{\ell-2}\in V_{\Pi}$.

If $\dim V_{S}=\sharp \mathcal{K}(\Pi )$, then the inequality follows from
\Eqref{newupperbound} and \Eqref{S-T}, and the condition for equality follows from Point 2).

Suppose now that $\dim V_{S} = \mathop{\mathrm{rk}}\nolimits \mathfrak{g}$ and
$\alpha_{\ell-1}\in S$ (the case $\alpha_{\ell}\in S$ being analogue).
Then
\begin{equation}\label{eqnD1}
\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})
= \mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp \mathcal{K}(S) - \sharp \mathcal{K}(T) -
2 + \sharp (\mathcal{Q}_{S}),
\end{equation}
and the connected component $S_{1}$ of $S$ containing $\alpha_{\ell-1}$ is not in
$\mathcal{K}(\Pi )$ (for otherwise, we would have
$\dim V_{S} = \sharp \mathcal{K}(\Pi )$).

By Point 1), $\varepsilon_{S_{1}} \in \mathcal{Q}_{S}$. By examining
the possibilities for $S_{1}$ and $K_{1}$ (Examples \ref{examples}), we verify that
\begin{equation}\label{eqnD2}
\sharp \mathcal{K}(S_{1}) - \sharp \mathcal{K}(T_{S}\cap S_{1}) +
\sharp (\Gamma^{K_{1}} \cap \Gamma^{S_{1}}) \geq 2
\end{equation}
with equality if and only if
$S_{1}$ is of type $A_{1}$ or $A_{2}$. We have therefore obtained the inequality.

In fact, we showed in the previous paragraph that already we have
$$
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} + \sharp \mathcal{K}(S_{1}) - \sharp \mathcal{K}(T\cap S_{1}) - 2 +
\sharp (\Gamma^{K_{1}} \cap \Gamma^{S_{1}}) \geq \mathop{\mathrm{rk}}\nolimits \mathfrak{g}.
$$
So if $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) =\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$,
then from \Eqref{eqnD1} and the above inequality,
we must have $\mathcal{K}(S\setminus S_{1}) \subset \mathcal{K}(\Pi )$,
and also equality in \Eqref{eqnD2}. Hence conditions (i) and (ii).
Conversely, suppose that conditions (i) and (ii) are verified, then
the fact that $\alpha_{1},\dots ,\alpha_{\ell-2}\in V_{\Pi}$ implies that
$\mathcal{K}(S\setminus S_{1}) \subset \mathcal{K}(\Pi )$ and
$\sharp \mathcal{K}(S_{1})=1$. Hence $S_{1}$ is of type $A_{1}$, $A_{2}$.
It is then easy to check that
$\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S}) =\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$.
\vskip0.5em

7) \underbar{Type $E_{6}$}.

Here, we have $\sharp \mathcal{K}(\Pi )=4$ and $\alpha_{2},\alpha_{4} \in V_{\Pi}$.
Let $S_{1}$ be a connected component of  $S$ such that
$\dim V_{S_{1}} > 4$. Under these conditions, the possibilities are:
$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
S_{1} & \dim V_{S_{1}} & T_{S}\cap S_{1} &
\sharp \mathcal{K}(S_{1}) & \sharp \mathcal{K}(T_{S}\cap S_{1}) & K_{1} &
\sharp (\Gamma^{K_{1}}\cap \Gamma^{S_{1}}) \\
\hline
A_{1} & 5 & \emptyset & 1 & 0 & A_{3}\, \hbox{or}\, A_{5} &  1 \\
\hline
A_{2} & 5 & \emptyset & 1 & 0 & A_{5} & 2 \\
\hline
A_{2} & 5 & A_{1} & 1 & 1 & A_{3} & 2 \\
\hline
A_{3} & 6 & A_{1} & 2 & 1 & A_{5} & 3 \\
\hline
A_{3} & 5 & A_{1} & 2 & 1 & E_{6} & 3 \\
\hline
A_{4} & 6 & A_{1} & 2 & 1 & E_{6} & 4 \\
\hline
A_{4} & 6 & A_{3} & 2 & 2 & A_{5} & 4 \\
\hline
D_{4} & 5 & A_{3} & 4 & 2 & E_{6} & 4 \\
\hline
D_{5} & 6 & A_{3} & 4 & 2 & E_{6} & 9 \\
\hline
\end{array}
$$
Thus
$$
\sharp \mathcal{K}(S_{1})-\sharp \mathcal{K}(T_{S}\cap S_{1})
+ \sharp (\Gamma^{S_{1}} \cap \Gamma^{K_{1}} )
\geq 2(\dim V_{S}-\sharp \mathcal{K}(\Pi )).
$$
A direct verification gives the result. Note that as in the case of type $A_{\ell}$,
$\mathcal{K}(\Pi )$ is totally ordered by inclusion, so $T_{S}$ is connected.
\end{proof}

\begin{remarks}
1. Theorem \ref{Main} says that if $\mathcal{K}(S) \subset \mathcal{K}(\Pi )$ or equivalently,
$S'\in \mathcal{K}(\Pi )$ for any connected component $S'$ of $S$,
then $\chi (\mathfrak{p}_{S})+\chi (\mathfrak{u}_{S})=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$.

2. When $\mathfrak{g}$ is of type $B_{\ell}$, $C_{\ell}$, $D_{2n}$, $E_{7}$, $E_{8}$,
$F_{4}$ or $G_{2}$, we have $\sharp \mathcal{K}(\Pi )=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$. In these cases,
the condition (i) in Theorem \ref{Main}
is equivalent to $\mathcal{K}(S)\subset \mathcal{K}(\Pi )$, and consequently,
condition (ii) is automatically satisfied.
\end{remarks}

\begin{example}
Let us consider the case of minimal parabolic subalgebras. So
$S=\{ \alpha \}$ and \Eqref{newupperbound} is an equality. It follows that
$$
\chi (\mathfrak{p}) + \chi (\mathfrak{u} )=\left\{
\begin{array}{ll}
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} & \hbox{if } \{ \alpha \} \in \mathcal{K}(\Pi ), \\
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} & \hbox{if } \{ \alpha \} \not \in \mathcal{K}(\Pi ) \hbox{ and }
\dim V_{S}=\sharp \mathcal{K}(\Pi )+1, \\
\mathop{\mathrm{rk}}\nolimits \mathfrak{g} + 2 & \hbox{if } \{ \alpha \} \not \in \mathcal{K}(\Pi ) \hbox{ and }
\dim V_{S}=\sharp \mathcal{K}(\Pi ).
\end{array}\right.
$$
Thus the minimal parabolic subalgebras $\mathfrak{p}_{S}$
verifying $\chi (\mathfrak{p}_{S}) + \chi (\mathfrak{u}_{S}) = \mathop{\mathrm{rk}}\nolimits \mathfrak{g}$
are (in the simple roots numbering of \cite[Chapter 18]{TYbook}):
$$
\begin{array}{|c|c|c|}
\hline
\hbox{Type} & \mathcal{K}(S) \not\subset \mathcal{K}(\Pi ) & \mathcal{K}(S)\subset \mathcal{K}(\Pi) \\
\hline
A_{\ell} & \hbox{\vrule height20pt width0pt depth12pt} i\neq {\displaystyle\frac{\ell+1}{2}} &
\hbox{\vrule height20pt width0pt depth12pt} i= {\displaystyle\frac{\ell+1}{2}} \\
\hline
B_{\ell} & \hbox{none} & i \hbox{ odd} \\
\hline
C_{\ell} & \hbox{none} & i=\ell \\
\hline
D_{2n+1} & i=2n, 2n+1 & i < 2n \hbox{ odd} \\
\hline
D_{2n} & \hbox{none} & i \hbox{ odd or } i=2n \\
\hline
E_{6} & i\neq 2,4 &  i=4 \\
\hline
E_{7} & \hbox{none} & i=2,3,5,7 \\
\hline
E_{8} & \hbox{none} & i=2,3,5,7 \\
\hline
F_{4} & \hbox{none} & i=2 \\
\hline
G_{2} & \hbox{none} & i=1  \\
\hline
\end{array}
$$
\end{example}

\begin{example}
In the other extremity, it is easy to check that maximal parabolic subalgebras of
$\mathfrak{g}=\mathrm{sl}_{\ell+1}$ verifying
$\chi (\mathfrak{p})+\chi (\mathfrak{u})=\mathop{\mathrm{rk}}\nolimits \mathfrak{g}$
are exactly the ones associated to simple roots at the extremities
of the Dynkin diagram.
\end{example}

redpajama_set_name: RedPajamaArXiv


North Korean leader Kim Jung Un in an undated photo released by KCNA.
WASHINGTON, D.C. — Less than a week after North Korea's unsuccessful missile test, the Hermit Kingdom conducted another ballistic missile test.
A look at North Korea's nuclear facilities and capabilities.
On Wednesday at 5:00 p.m. CDT near the northwestern city of Kusŏng, US Strategic Command detected a failed launch of a presumed Musadan intermediate-range ballistic missile.
The Musadan missile is speculated to have a range of approximately 1,500 to 2,400 miles, capable of targeting military installations in Guam and Japan, based on estimates from the Missile Defense Project at the Center for Strategic and International Studies.
So far this year, North Korea has conducted eight Musudan missiles tests. All launches except the sixth one on June 22 were considered to be failures.
"This year, Pyongyang successfully conducted two nuclear tests, an intercontinental ballistic missile test, breakthrough successes with its road-mobile intermediate-range missile and submarine-launched ballistic missile, a new solid-fuel rocket engine, and an improved liquid-fuel ICBM engine," Bruce Klingner, senior research fellow of Northeast Asia at the Heritage Foundation, told Business Insider in a previous interview.
"Based on these successes, as well as North Korean-released information, experts reassessed several missile programs as being more capable and closer to deployment," Klingner added.
Meanwhile, US Secretary of Defense Ash Carter will host South Korea's Minister of National Defense at the Pentagon for a joint press conference on Thursday.


StaticWolf had the most liked content!
I like the profession idea.
Hey there! Welcome aboard I'm StaticWolf (a.k.a Kate), the one-woman-welcoming-club of this website. Enjoy your stay here!
My night was a little shaky. I went to an open house with my mom at this awesome school where my best friend goes, and I got to see all the classrooms and talk to all my bf's friends. Everyone was so nice and it was all going great, but in the car on the way home my mother informed me that she found out from a teacher that the school is doing a new curriculum that is severely against our religion. :[ I was excited because I don't want to be home schooled again next year. I'd go to public school but the one in our area has some serious issues with drugs and bullying.
I want to see MSI in April @Orlando, but I thnk my ma is scared to let me go. Advice for anything I could say that could convince her it's safe?
I don't think she would want to, but my 16 yr old sister might want to come!
Hellooooo guys! I've been great. Today's my best friend's birthday, so we're gonna party till we're purple.
Awesome, thanks Yeap, I just figured I would start at their fansite, then work outward. Their topics have to be approved before they can be posted, so it's gonna be a little while before anyone sees it.
Thanks MajorSunshine Yeah, I understand your point about only having 12 people, but the only reason there's so few is that I'm not sure I'll be able to find enough Directioners. I'm trying to find their home planet on the internet ;P EDIT: I just found it. I have officially joined the One Direction Fansite.
Hey guys, I've been thinking about doing something for a while, and I would need a dozen of you to help me. It's still a very primitive idea, but I'm working on evolving it into its final form. Basically I have this theory concerning the MCRmy and the One Directioners. I've noticed quite a lot of similarities between us and them, and I had an idea to do this sociology study on it. You know, because random sociology studies are so common. *sarcasm* So I figured out that I'm going to organize a survey designed specifically for the MCRmy and the 1D-ers. I need 12 MCRmy soldiers and 12 Directioners. I don't need anyone to commit for sure, I just need to know if anybody's actually open to it. If no one is, it's a stupid thesis and I'll pull the plug. Thanks all!


Did y'all know that I do branding sessions? Standard headshots are great (I do those, too) but I swear it's more fun to break out of the box and get some creative photography done that captures who you are and the personality behind your business! That's what these fun sessions with my friend Amanda have been all about. The two of us go way back, in fact, I'm not even going to go there because it makes me feel OLD, but there's something special about brainstorming and shooting with a lifelong friend. She's been a super successful entrepreneur with her Isagenix business. Quick shout out – you can see the products she offers here and reach her on her cell at (704) 649-3011. She's in killer shape, is ridiculously funny, and is SO.MUCH.FUN to photograph. Not to mention she has the cutest little ones too who we've been able to pull into some of our shots.
Here are just a handful of the fun images we've taken to showcase her business and lifestyle. If you're looking for an Atlanta branding or lifestyle photographer please reach out, these are some of my favorite sessions to do. I love hearing all about your story and then telling it through pictures!
Hi, I'm Ashley! I'm an Atlanta branding and commercial photographer as well as a family, newborn, and senior photographer. Feel free to poke around my site and check out my work. If you're interested in a session please contact me here! I'd love to chat and come up with the perfect plan to build your brand by telling your story visually.


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#DXCLUSIVE: The Game Says Michael Jackson Wanted Him & 50 Cent To Work Together For His Album
December 17, 2019 | 2:36 PM
by Jeremy Hecht
Los Angeles, CA – The Game recently stopped by the HipHopDX offices in Hollywood to speak about the journey from his debut album The Documentary to his most recent release Born 2 Rap.
During the conversation, Game was asked about a moment in his career involving Michael Jackson and his former collaborator turned enemy, 50 Cent.
After Michael Jackson passed away, Game penned a tribute song to the late King of Pop called "Better on the Other Side" featuring Chris Brown. The two honored MJ's legacy and talked about his importance to the world of music. On the song Game raps:
"First thing I did when I heard, was call Puff/ 'Cause him and Mike tried to stop the beef between us
Who was us? Me and 50, that beef is dead/ Him and Mike Jackson gonna take us to the ledge."
The West Coast rapper detailed the story of Michael Jackson calling his phone to try and end the beef between him and 50 Cent. Game described the situation as "weird as fuck." He continued by saying "this ain't how I wanna meet Michael Jackson."
Before the two artists could speak however, Game was asked to hold on the line for approximately ten minutes. "N**a you called me!" he said jokingly. "But it's Michael Jackson so I'm holding the fuck on."
Once they were finally connected on the line, Game told DX that the Michael Jackson voice we've become used to hearing so much about was not present on their phone call.
"You know that Michael Jackson voice that you hear all the time? That n***a wasn't talking like that." He said while laughing. "It was low-key deep. You could still tell it was Michael Jackson."
Game describes how Michael Jackson told him how much he enjoyed The Documentary, in particular, his 50 Cent assisted track "How We Do."
Perhaps the most shocking part of the conversation came when Game revealed that Michael Jackson rapped back some of his lyrics bar-for-bar.
"He quoted how I did the countdown… In 'No More Fun And Games.'" Michael asked Game if he had really only been rapping for the amount of time that he stated on that song, which was 1 year, 1 month, 17 days, 13 hours, 28 minutes.
Jackson's main intention with the phone call was actually two-fold. First, he hoped to moderate a conversation between The Game and 50 Cent, who were beefing at the time. Secondly, he was hoping the three of them could collaborate for a song on his upcoming album.
"For your album?" Game recalls asking MJ. "Yeah, I want you and 50 on my album!" Jackson replied.
While Game told Jackson he would think about the potential collab, he told DX that he had no intention of working with 50.
"I actually hated 50 and wanted to kill him at that time. And he probably had that same mutual feeling for me." He said in a serious tone. "That's where Mike lost me. In my head, I was not open to doing that." He continued.
Game expressed how he felt like 50 Cent was trying to destroy his career and was not ready to put aside their differences, even for an opportunity to work with the "king of pop."
"I felt like 50 tried to destroy me and I wasn't about to be "Ja Rule'd" he said jokingly. "And I hate that I got to call it that because Ja is a cool dude" he clarified.
Game closes the clip by confirming that the phone conversation was the last time he ever spoke to Michael Jackson.
Watch the entire clip above and be sure to stay tuned for the full interview dropping soon.


To help us make the The Arc Chapter Web Service website a positive place for everyone, we've been using the Web Content Accessibility Guidelines (WCAG) 2.0. These guidelines explain how to make web content more accessible for people with disabilities, and user friendly for everyone.
The guidelines have three levels of accessibility (A, AA and AAA). We've chosen Level AA as the target for the The Arc Chapter Web Service website.
We're working hard to achieve our goal of Level AA accessibility, but we realise there are some areas that still need improving. The following information explains what we're doing to make that happen.
Yes, all error messages provide suggestions to the end user.
Yes, all donation forms and payment forms can be reviewed before final submission.
Yes, all navigation systems appear in consistent locations throughout the site.
Yes. Our website identifies consistently navigation, sub-navigation and footer navigation items.
Yes, all of the pages use adaptable layouts.
Yes, you can see the focus ring on all elements as you tab through the page.
Yes, all headings are descriptive.
Yes, all color combinations meet the AA requirements.
Yes, we have this functionality on all pages on the site.
Yes, all errors are identified and described in the user text.
Yes, all labels provide instructions when a user action is required.
Yes, content changes on focus.
We do not have any automatic changes of context on the site.
Yes each page has the default language set for screen reader capabilities.
Yes, all links are descriptive, we do not use links that use words like "Click here" but rather we use descriptive titles and alt tags.
Yes, using the tab key the focus can be removed from any form field.
All content can be tabbed through in the proper order using the tab key.
Yes, all tab indexes follow a logical path preserving meaning and operability.
Yes, all pages have a title that describes topic and purpose.
Yes, you can tab through all content on the site and can be used as a method of bypassing and or skipping content.
Yes, all language settings have been identified for screen reader purposes.
Yes, audio controls exist on all pages where audio is embedded. Audio can also be stopped by screen readers and as a browser control.
Yes, we use words to convey the importance or severity of content on the site as well as colors.
Any place where there is a video or a form there are clear instructions on how to play and submit the form on the website.
Yes, our content displays correctly in all screen readers programmatically.
Yes, we meet this criteria. All information is structured and relationships between them are conveyed correctly in the markup language.


It is rare to see a home without a dishwasher these days. We love our dishwashers especially if we have a big family or lots of guests visiting. If you entertain in your home, you really depend on it. Three meals a day with a lot of people in and out of your home means a lot of dishes.
Obviously, you want to extend the life of your appliance as long as you can so here are tips for routine maintenance to extend the life of your dishwasher.
Routinely check some of the key components of your dishwasher. This can save you money in repair and replacement costs. Rust, mold and mildew can develop due to the extreme temperatures, constant moisture and moving parts.
Break out your manual and see the timeframes to change your filters and other items the manufacturer recommends doing.
Make sure you rinse off any chunks of food before you place the plate or utensil in the dishwasher. Your dishwasher has a recirculating pump that could spray the food onto your dishes.
Run your dishwasher regularly to get rid of excess. Running your dishwasher frequently will keep debris from settling in the bottom of the appliance.
Make sure you are using the proper detergent and dosage for your specific appliance. Your manufacturer will be able to give you the specifics.
Use an acid based cleaner once or twice a month which will keep it smelling like new. This will help you remove any mold or mildew that might be growing in this damp space. You could even try using bleach in the dishwasher to sanitize it.
Most dishwashers have a rinse aid dispenser. Use one to create a barrier to help water "sheet" off your dishes. It leaves them spot free.
The drain is at the very bottom of your dishwasher. This is a common place where food debris builds up which will cause a clog or mildew and mold to grow and get on your clean dishes. This will also allow the dirty water to exit.
If your appliance is not working at peak performance, a repair might be required or at least a service check. Make sure to call an experienced quality dishwasher specialist who will restore it to top condition.
Gulf Coast Appliance Repair is the local appliance repair company that you can count on. For 10+ years, our family owned business has been serving customers in Pinellas County. We specialize in repairing appliances including refrigerators, washing machines, dryers, stoves, dishwashers and much more. Call our appliance and repair service at 727-510-3524 email us to get the job done or to schedule your appointment.
Dryer Good Health – Why Clean Lint Filter?


text:
Intermediate Listening Lesson 84 – Medical Missionary
https://learnenglishfunway.com/wp-content/uploads/2022/02/084-Medical-Missionary.mp3 Pin
During the reign of Queen Victoria (1837-1901), British people traveled around the whole world.
They charted the seas, mapped out distant countries and studied plants, animals and people.
They also claimed many lands for England.
This kind of international travel was made easier by improved transportation and communication.
New inventions such as steamships, trains,
telegraphs and telephones made long distances seem smaller.
Of course, people had different reasons for going to distant lands.
Some were businessmen who saw economic opportunities overseas.
Soldiers wanted fame and a chance to enlarge the British Empire.
Big-game hunters wanted to be the first to shoot strange animals and bring back trophies to England.
Scientists intended to study unknown animals and plants.
Missionaries planned to be the first to introduce Christianity to faraway people.
In 1836 a young Scotsman called David Livingstone began to study medicine in Glasgow.
Livingstone intended to become a medical missionary.
This means that he would be a doctor, as well as a preacher and teacher.
Livingstone (1813-1873) came from a poor family.
From an early age, he had worked 14 hours a day in a clothing factory for very little pay.
But he was determined to learn.
He took his books with him to the factory and read as he worked.
Then, after work, he would go to his teacher to learn more.
Livingstone's goal was to teach faraway people about Jesus.
However, unlike some missionaries, he was also interested in science, geography and exploring.
He had planned to go to China in 1839,
but because of the Opium Wars no missionaries were being sent there.
Instead, he asked to go to South Africa.
Europeans had traveled around the coasts of Africa for hundreds of years.
But very few white people had traveled inland.
A missionary named Robert Moffatt who had begun a mission at Kuruman in the interior inspired Livingstone.
Livingstone arrived in Kuruman in 1841.
This was the farthest outpost of white settlement, and no one seemed to want to go further inland.
Livingstone felt that the missionaries should go to the Africans,
rather than waiting for the Africans to come to them.
With a fellow missionary he set out.
When they came to an African tribe, they would talk to the chief and ask permission to preach to his people.
Livingstone would also set up a tent and treat the people who had diseases.
After a while, he would move on to the next tribe.
Once Livingstone learned the Bantu language he would talk to many Africans.
But sometimes he needed interpreters.
There were many diseases, including malaria and sleeping sickness.
Livingstone suffered much of his life from river fever.
He was also so weak that he rode on the back of an ox.
Livingstone wanted to stop the slave trade.
At this time, the slave trade was the most profitable business in Africa.
Livingstone hoped that if other kinds of trade were developed,
then slavery could be abolished.
In order to open up trade, he wanted to find an easy route into the center of Africa.
Livingstone kept going further into the interior.
He was probably the first European to cross the Kalahari Desert before reaching Lake Ngami in present-day Botswana.
Not long after, he traveled further inland.
He explored the sources of the Zambezi
and Kasai rivers and eventually reached the west coast of Africa at Luanda, Angola.
Livingstone was being criticized for neglecting missionary work in order to explore.
Livingstone replied that he was opening up the continent for missionaries.
Meanwhile, he was becoming famous as a great explorer.
The British government commissioned him to explore the Zambezi River.
They hoped that ships could sail up the river into the interior.
Unfortunately, the Zambezi had too many rapids.
However, Livingstone did find a route up the Shire River to Lake Nyassa.
He continued to struggle against the slave trade, which was now being taken over by Arabs.
Livingstone died in Africa in 1873.
He was the first white man to explore Botswana, Zimbabwe, Zambia, Malawi and surrounding areas.
He was not only a great explorer, but also a fine doctor and a good missionary.
Nowadays, the countries that Livingstone visited are nearly all Christian, just as he had hoped they would be.
Lesson 85: Gambling
Mariam on 10 Best Books To Learn English Grammar – PDF Free Download
Danil on 10 Best Books for Learning English Writing
Masud on 100 Easy English Listening Lessons for Beginners
D.Keshvari on 10 Best Books for Learning English Vocabulary


text:
Home AM-NC MIL-OSI USA: Feinstein, Padilla Announce Historic $4.2 Billion Investment to Upgrade California's...
MIL-OSI USA: Feinstein, Padilla Announce Historic $4.2 Billion Investment to Upgrade California's Aging Bridges
US Senate News:
Source: United States Senator for California – Dianne Feinstein
Washington — Senators Dianne Feinstein and Alex Padilla (both D-Calif.) today announced that California will receive $4.2 billion over five years to help California address the needs of highway bridges across the state.
This funding comes from the Bipartisan Infrastructure Law – which both Padilla and Feinstein voted to pass last year – and is part of the largest single national investment in bridges in American history. According to the U.S. Department of Transportation, California has 1,536 bridges rated in poor condition.
"California has more than 1,500 bridges that are in desperate need of repair," Senator Feinstein said. "This funding will go a long way to increase the safety, stability and lifespan of our state's bridges, especially in the face of earthquakes. This funding will also generate good-paying jobs in communities throughout California."
"The Bipartisan Infrastructure Law is a game-changing investment in our economy," Senator Padilla said. "Bridges are vital arteries for our economic activity, helping commuters get to work and keeping goods moving efficiently through the supply chain. With the Bipartisan Infrastructure Law providing billions of dollars over the next half-decade, California will be able to renovate and upgrade bridges across the state. Investing in our bridges will create good-paying jobs, improve safety for motorists, and keep our economy moving efficiently."
This funding will be distributed through Federal Highway Administration's Bridge Formula Program, which the Biden administration launched earlier today. The program represents the largest investment ever made in fixing bridges, dedicating $26.5 billion to states, Washington, D.C., and Puerto Rico over five years and $825 million for tribal transportation facilities.
Nationwide, the Bridge Formula Program is expected to help repair approximately 15,000 bridges. In addition to providing funds to states to replace, rehabilitate, preserve, protect, and construct highway bridges, the Bridge Formula Program has dedicated funding for Tribal transportation facility bridges as well as "off-system" bridges, locally-owned facilities which are those not on the federal-aid highway system.
Previous articleMIL-OSI USA: Congressman Troy Carter Delivers $202.7 million for Louisiana's Bridges from Infrastructure Law
MIL-OSI USA: Congressman Troy Carter Delivers $202.7 million for Louisiana's Bridges from Infrastructure Law
MIL-OSI Economics: So you got new gear for the holidays. Now what?


The murder of a senior Tajik security official on July 21, 2012, has triggered deadly clashes in Tajikistan's Gorno-Badakhshan province (GBAO) between government troops and the supporters of a local strongman whom the authorities blame for the assassination. The clashes have left the region's residents cut off from the outside world.
Since July 24, Internet, mobile, and landline connections to GBAO have been cut. As a result, Tajikistanis outside the province have had little reliable information about the situation and the extent of human suffering in GBAO.
Badakhshani expatriates have responded to the blackout by pooling what information on the conflict becomes available in public Facebook groups and calling on telecommunications companies to restore connections to Khorog (aka Khorugh), the provincial capital.
One Facebook group, Stop the Military Action in Khorog, compile reports on the clashes in GBAO. Group members share the latest news on the crisis and discuss media reports. Many Tajik users have replaced their Facebook profile pictures with the 'Stop the Killing' sign. Overall, the conflict has dominated discussion in major Tajik groups on Facebook over the last week.


Weight loss is the goal of every adult as well as a teenager in recent times. We often feel shameful when we receive comments such as fatty and overweight. We then join the gym and do regular workouts and pushups just to reduce the extra calories. Instead of that, we should try something new which can really help. We are talking about Keto Insta Cleanse Toned formula. It is a nice weight loss supplement made from natural and organic items. It may give you a perfect figure within 3 to 4 weeks.
As we discussed above Keto Insta Cleanse Diet supplement contains all natural ingredients. It contains Vitamin B2, Acetyl-L-Carnitine, Niacin, Black Pepper Extract, Green tea extract, Linoleic acid powder, vitamin B1, capsicum extract and extract of GarciniaCambogia.
These ingredients are first checked in the labs and then used. Because of natural composition, this supplement does not give side effects and allergies in the body. It is free from any artificial preservative and substance. Furthermore, the formula does not contain gluten, colors, and flavors. Many topnotch doctors suggest taking this supplement for weight loss.
With regular use of Keto Insta Cleanse supplement, life sounds simply amazing.
How Keto Insta Cleanse Toned formula is unique from other products?
Rich quality natural ingredients are what make this product a safe one. The natural ingredients are harmless to the body. It may give you faster results without side effects. Moreover, it works effectively in the body to burn extra fats.
If you gain control over the hunger, it is then possible to reduce the obesity. With regular use of Keto Insta Cleanse supplement, you may control your appetite. It will help in reducing the food cravings which you often get at late nights.
Foods with carbs are the main ways by which fats enter the body. This weight formula may burn the extra fats of the body in stomach, hips, booty, thighs, and neck. It will further block the route for those fats to accumulate again in the body. This will give you a slimmer and toned figure.
By burning more fats instead of carbs, Keto Insta Cleanse formula provides more energy in the body. You may feel energetic even after doing various physical tasks. This supplement may remove the drained feeling and tiredness from the body.
LDL is the bad cholesterol and so, it is necessary to reduce LDL. This weight loss supplement may help in lowering LDL in the body. It will further maintain the proper level of blood pressure.
By sending more blood and oxygen to the brain, Keto Insta Cleanse may improve the brain functions. It will increase the mental focus and make you more attentive in the work. It will also make your memory sharper and better.
In addition to this, this weight loss formula may improve the immune and digestive system.
You should consume only 2 capsules of Keto Insta Cleanse supplement each day. Consume the capsules before 30 minutes of the meals. You should do some exercises, yoga, and workouts for remaining energetic and active throughout the day.
You can also do some physical activities such as playing outdoor sports, swimming, walking, jogging, cycling and others.
The supplement contains natural and organic ingredients.
It burns fats instead of crabs to produce more energy.
It works faster than others.
It will keep you energetic for the full day.
You can get it only on the official website of the company.
It is often out of stock.
You may get side effects such as vomiting or headaches if you take more than 2 capsules in one day.
You cannot use it with other weight loss supplements and capsules.
Most of the customers gave positive comments after using Keto Insta Cleanse formula. Many women got a slimmer body with this weight loss supplement. They state that they lose the extra fats of the body within a few weeks.
Some people got better concentration on their work with this formula. The customers also feel energetic each day after taking these capsules.
It is advisable to order Keto Insta Cleanse supplement from the official site only. You will get 100% original product from the official site. You can first place the order of a monthly pack of 60 capsules to see the results. The product takes 4 to 6 business days to reach your place.


text:
Reviews/Features
Hooked: Dirty Honey's John Notto on Led Zeppelin's
"Whole Lotta Love" -- 11/30/21
Justin Smolian: Rolling The Dice with Dirty Honey 10/19/21
Mammoth WVH, Dirty Honey Ready to Prove the Heart of Rock & Roll Is Still Beating on 2022 Tour - 11-14-21
Dirty Honey Let the Music Stand on its own
Dirty Honey are Putting the Roll Back in Rock
Meet the next generation of rock players putting their own spin on a retro sound - 7/9/21
John Notto on "Have Guitar, Will Travel" Podcast - 7-21
Cleveland Scene Talks with Marc LaBelle about The Black Crowes Tour and his Classic Rock Influences - 7-26-21
Six things you need to know about Dirty Honey - 6/15/21
Dirty Honey Continues its Success...
with a Gritty, Self-titled LP Debut - 4/13/21
Dirty Honey Explore the Darker Side of the Golden State on 'California Dreamin'' • Track will appear on the Los Angeles outfit's upcoming album - 3/5/21
How the classic-rockers became the first unsigned band to reach Number 1 on Billboard's Mainstream Rock Chart - 2/27/21
"Why Would We Ever Sign With A Label?"
"Somebody has to be on mushrooms, right?... - 4/9/21
The Next Big Thing: An Exclusive Interview With DIRTY HONEY - 4/12/21
Dirty Honey were riding high when the pandemic shut things down...here's what they did next
- 4/13/21
The LA rockers are about to drop their debut LP
Screamer Magazine: Dirty Honey, Ready To Sting Twice - 4/12/21
Dirty Honey Build On Indie Success
With Release of Self-Titled Debut Album -- 4/29/21


text:
How a chance encounter with the Mahatma changed one Canadian's life
A Canadian law professor reminisces on his meeting with Gandhi.
Teenaz Javat special to · CBC News · Posted: Jan 29, 2009 4:50 PM ET | Last Updated: January 29, 2009
From his small apartment in a seniors building in Mississauga, Ont., a retired Canadian law professor reminisces about how a chance meeting with Mohandas Gandhi all those years ago changed the course of his life.
At the time, Fali Balsara was a young police officer recruited by the British under the Defence of India Act.
Fali Balsara, then 82, at the Zoroastrian cultural centre in Toronto in July 2001. (Courtesy Fali Balsara)
There were many perks that came with a government job like that. But one of the discouraging elements was the fact that if recruited under this act, leaving the service of the Raj could entail imprisonment.
A uniformed position in India in those days came with considerable clout, but unconditional loyalty was expected in return.
This was August 1942, when the young and dapper sub-inspector was sent off on his first assignment.
Arrest Gandhi
At that point, Gandhi and members of his Indian National Congress had just launched the Quit India movement in Bombay (now Mumbai) to try to move the British colonial powers out of India.
Already a beloved political and spiritual leader, Gandhi's non-violent approach of passive resistance had taken the country by storm. But this was still in the midst of the Second World War and the authorities were not about to have any challenge to their order.
As a result, Bombay's deputy commissioner of police, H.E. Butler, dispatched Balsara and his team to arrest Gandhi.
"I was dumbstruck," Balsara, now 90, recalls. "We used to call him Bapu, meaning father. However, it was part of my job so I complied."
Balsara and his team were to escort Gandhi in a special train to Poona, a quiet city 192 kilometres south of Mumbai, where he was to be held at the Aga Khan palace.
That train journey changed the course of Balsara's life.
'Humble to a fault'
"Bapu spoke to us," reminisces Balsara, "His calm demeanour amid so much uproar over his arrest completely took us by surprise. He was humble to a fault and respected all who served him."
Balsara recalls that, by the time the train reached Poona, he and his police guards were thoroughly mesmerized by Gandhi's personality and his unshakable belief in India's right to its own destiny.
In those few hours it dawned on him that the path of non-violent resistance that Gandhi had chosen to guide India to freedom was so completely at odds with what he was doing as a police officer.
'We were so influenced by Gandhi's charisma and his captivating toothless smile that on reaching Poona we decided to quit our jobs, even though we knew quite well that we would be imprisoned," said Balsara.
That is exactly what happened: all six police officers became prisoners of the Raj.
Just over five years after this incident, on Jan. 30, 1948, Gandhi was killed, felled by an assassin's bullet in New Delhi while on his way to prayers. He was 78 years old and had lived just long enough to see India's independence from Britain the year before.
Since, 1934, there had been six attempts on his life.
A lawyer, politician and, many say, saint, he is now officially remembered as the Father of the Nation.
Change of direction
They went on to spend 14 days in close proximity to Gandhi as the British considered them mutineers.
"The days I spent in jail with Gandhi, I remember vividly to this day," Balsara says. "He convinced me to become a teacher, as according to him it was the noblest of all professions."
Back in Mumbai, Balsara's mother was worried that her son, then just 23, was now a prisoner of the Raj and in so much trouble with his employers.
However, her brother, who worked for the governor of Burma, was able to pull some strings and appealed to the governor of Bombay, Lawrence Roger Lumley, to pardon Balsara for his youthful enthusiasms.
On being pardoned, Balsara was discharged from service and went back to school where he would earn a doctorate in social science and then emigrate to Canada in 1964, first to teach at the University of British Columbia and then in Toronto.
Balsara taught at several universities and colleges in Canada before retiring as a professor of law and economics from York University.
The lessons he learned from Gandhi, Balsara says, have stayed with him his entire life. "He taught me that one's worth as an individual lies only in your action."


All the best to everyone I know for this coming year. I hope you all celebrate safely and have fun! 2010 was insane. 2011 is going to be crazier, but better! Trust me!
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