Moving from one job to another isn't easy especially if you're attached to your current workplace. Your final weeks at your job are some of the bitterest you'll experience because you're going to miss some of your co-workers, your routine and being involved with your company.
You must transition to a new job with new responsibilities, compensation, benefits, and co-workers. This is the reason why transitioning from one job to another isn't easy, but there may be a time that you have to switch from one job to another or alternatively, pick up a second job in order to make ends meet. The first 90 days at your new job is called the probationary period. In this blog, we'll share some tips that will help you transition to a new job successfully.
When you change careers, you can experience roller-coaster emotional feelings. With a better job opportunity, you'll be excited at the beginning, but the moment you enter your new workplace you will have to face reality with change and to the possibility of having to jump through hoops and over some obstacles. Some people can quickly fit in while with others it will some time. Whatever the case may be you should always maintain a positive attitude towards your new working environment. As time goes on, you'll become accustomed to your new employer and your new co-workers.
Research shows that it took an average of eight (8) months for unemployed individuals to find a new job. This means that there are some habits or routines that you develop in this period that might affect how you transition into a new job. Breaking these bad habits can be a challenge for some people. A good way to adapt to your new job is to manage your routine carefully, and this will help increase your effectiveness and efficiency.
If you really want to fit in very well with the new company, you may want to participate in the various companies' activities and do all you can to adapt to their overall culture. If you're hired by a company it means they found a lot of value in you and your skills. You can adapt to the company's culture by making office norms your personal habits. For example, A norm of accountability will help make your organization successful.
When you're new to a workplace environment there is a need to focus your attention and observe your co-workers and see how they're running things, this will help you to grasp the behaviors and attitudes of the people that are around you and how to go with the flow. You should take note of the different things that are going on around you and how people react in certain situations. Being self-aware is a trait of any good leader. Being a good leader requires being a good observer. Being a good observer is more than just self-awareness. Observing oneself is essential, but observing others provides one a way of looking at things from another's point of view.
One of the important things to take note is how your boss reacts to situations. There may be problems that you can solve in the company based on this observation. In general, you should take the time to understand how your company functions. Think of ways to improve the current status quo. It can be risky but challenging the status quo can be a great way to get ahead in business.
Wherever you are, you should set goals; make a list of clear goals that you want to achieve, you should have both short-term and long-term goals. Set the goals you want to achieve within the first three (3) months of your new job, plan on how you want to improve yourself. Great goal setting means you can take steps towards improving any aspect of work that's relevant and specific to you, building on professional knowledge, skills, and effective workplace practices.
Relationships are very important in any work environment. The type of people you are connecting with is going to play a great role in your career; relationships can help you excel in your career. With the right co-workers, you can definitely adapt to new working environments quickly. There is a need to accept various invitations from your colleagues as long as it's not going to affect your time management. Try as much as you can to network with people and build relationships with people that have a positive mindset just like you.
These are just a few tips that will help you transition to a new job and if you need any help regarding your job transition either with a severance agreement from your previous employer or/and an employment agreement with your new employer, please give us a call at Massey & Duffy for a free consultation.


text:
Son-Vidéo.com: blog
Home Reviews Review: McIntosh RS200
Review: McIntosh RS200
Sommaire afficher
McIntosh RS200: the brand
McIntosh RS200: packaging & accessories
McIntosh RS200: presentation
McIntosh RS200: WiFi, DLNA, AirPlay, DTS Play-Fi, Bluetooth aptX HD
McIntosh RS200: multi-room, wireless surround sound
McIntosh RS200: configuration
McIntosh RS200: listening impressions
McIntosh RS200: compared to…
McIntosh RS200: conclusion
This week we reviewed the McIntosh RS200 active speaker, an ultra-versatile model that serves as a wireless speaker, a multi-room speaker, a USB DAC and a soundbar. The McIntosh RS200 is compatible with the DTS Play-Fi and AirPlay 2 multi-room technologies, features WiFi (DLNA) and Bluetooth connectivity, an HDMI ARC input, an optical input and a USB port so that it can be used as a DAC with a computer. All of this with McIntosh's iconic design and famous VU meters. Sold for €3 800, will this all-in-one speaker prove to be worthy of the American manufacturer's legendary electronics?
The McIntosh RS200 active speaker is an ultra-versatile model that serves as a wireless speaker, a multi-room speaker, a USB DAC and a soundbar.
Is there really any need to introduce McIntosh? Since its creation in Maryland in 1949 by Frank McIntosh, the brand, whose full appellation is McIntosh Laboratory, has been a frontrunner in the hi-fi domain. Its very first device, the McIntosh 50W-1 power amplifier was a great success, characterized by its ability to deliver an extensive amount of power whilst retaining an extremely low distortion rate. Very quickly, the company branched out and commercialized its first preamplifiers, the EA1 and C8. It was only in the sixties that the brand first integrated its large, blue backlit VU meters housed behind a polished glass front panel. To this day, this distinctive look remains a trademark of McIntosh. 70 years after its creation, McIntosh features an extensive catalog. Its power amplifiers, vacuum tube preamplifiers, DAC amplifiers, stereo amps, A/V preamps, turntables and, more recently, wireless speakers all provide extraordinary performance. No matter the model, musical enjoyment has been at the forefront for over 70 years. The manufacturer already impressed us during our reviews of the McIntosh MA-252 vacuum tube amp and the McIntosh MHA100 headphone amplifier and DAC.
The McIntosh RS200 features 8 drivers and offers 650 watts of total power.
Covered by a white fabric pouch, the McIntosh RS200 wireless speaker comes in a cardboard box and is protected by large blocks of polystyrene. It comes with a user manual, a warranty card, a remote control identical to that included with the McIntosh MTI100 turntable (which we will be reviewing soon), two AAA batteries and foam feet.
The McIntosh RS200 wireless speaker comes with a user manual, a warranty card, a remote control and foam feet.
As soon as your take it out of its box, there's no doubt about the lineage of the McIntosh RS200 wireless speaker. This imposing speaker is 63cm wide and 14cm high, weighing 13.3kg. Its glass chassis follows the brand's esthetic code and features the famous McIntosh VU meters at its center, along with two dials on either side. The first dial is dedicated to volume control and the second is used to select the source. Very conveniently, the name of the selected source is displayed beside the dial and is illuminated by an elegant green light, identical to the one that lights up the name of the brand. This dial can also be used to adjust the backlight of the VU meter so that it isn't too bright when it is being used at night. It can also be turned off completely. Lastly, the top of the McIntosh RS200 features touch controls allowing you to activate the WiFi connection and select one of the four presets that can be configured in the DTS Play-Fi app for iOS and Android. These presets can also be assigned to different online music services that are compatible with the RS200.
The McIntosh RS200 wireless speaker features McIntosh's famous VU meters on its front panel, along with two dials on either side.
When it comes to connections, the McIntosh RS200 is very generous and is fitted with an auxiliary input (3.5mm mini jack), a type B USB port so that the device may be used as a DAC with a computer, an optical digital audio input, and an HDMI ARC input to retrieve sound tracks from a TV (Dolby and DTS) so the RS200 can serve as a soundbar. This HDMI input is compatible with CEC, meaning that the soundbar, TV, Blu-ray player and any device connected to the TV can be controlled using only one remote control. Lastly, the McIntosh RS200 wireless speaker also has a subwoofer output (RCA mono).
The McIntosh RS200 features an auxiliary input (3.5mm mini jack), a type B USB port so that the device may be used as a DAC with a computer, an optical digital audio input, and an HDMI ARC input.
The American brand's DNA can be seen throughout this connected speaker which boasts no less than 8 drivers. Its front panel accommodates two tweeters at the left and right which are oriented towards the listener. These tweeters are 1" titanium dome fluid-filled models and are paired with four 2" midrange drivers featuring inverted titanium domes.
The McIntosh RS200 speaker's left and right channels both feature a 1" titanium dome fluid-filled tweeter that is paired with two 2" midrange drivers featuring inverted titanium domes.
The lows are handled by two elliptical drivers that are 4" wide and 6" long and feature a long throw suspension. They are situated underneath the speaker, hidden behind fabric covers and loaded in a bass-reflex enclosure. A polymer component evenly distributes the air at the front and the back of the speaker.
The two low frequency drivers are situated underneath the speaker and are loaded in a bass-reflex enclosure.
The McIntosh RS200 speaker features considerable amplification, delivering a total power output of 650 watts. From this total power, 400 watts are assigned to the tweeters and midrange drivers. The remaining 250 watts are assigned to the woofers. This configuration ensures a frequency response between 40Hz and 20kHz, as well as excellent dynamics.
The McIntosh RS200 wireless speaker is an all-in-one-solution to get the most out of your music, regardless of the source. Once the speaker is connected to the internet via WiFi, it can play HD tracks up to 24-bit/192kHz shared via the local network thanks to its compatibility with DTS Play-Fi and DLNA. The RS200 also provides access to multiple online music services such as Spotify, Tidal, Qobuz, Napster and Amazon Music. These services can be quickly accessed using the 4 preset buttons on top of the speaker or directly via the remote control.
The top of the McIntosh RS200 features touch controls allowing you to activate the WiFi connection and quickly access online music services.
The McIntosh RS200 also supports the AirPlay 2 and Bluetooth protocols to stream music from a smartphone, tablet or computer with ease. Compatibility with the aptX HD codec ensures high-quality transmission, while aptX Low Latency reduces the latency in the transmission, eliminating any delay between the picture of videos played on a smartphone or tablet and the sound streamed to the speaker via Bluetooth. It goes without saying that the source must also be compatible with aptX Low Latency in order to use this function. If this isn't the case, the McIntosh RS200 may be manually synchronized by the user (Lip sync function). This function is accessible via the remote control.
The McIntosh RS200 speaker's remote control provides user-adjustable sound and image synchronization (Lip sync function).
When connected via WiFi, the McIntosh RS200 can be added to a group of multi-room speakers compatible with the DTS Play-Fi or AirPlay 2 protocols. Therefore, it is possible to stream the same track in every room of the house or listen to a different track in each room. But the true advantage of the multi-room function is that you can create a wireless home theater system by pairing the McIntosh RS200 soundbar with two surround speakers or any Play-Fi compatible receiver. When associated with a UHD 4K television via its HDMI ARC or optical input, the RS200 allows you to fully enjoy your TV programs.
The McIntosh RS200's initial setup begins by connecting the speaker to the WiFi network. To do this, you must press and hold the WiFi button situated on the top of the speaker for 8 seconds. After 8 seconds, an audio signal is emitted and the "Net" indicator above the left dial blinks slowly, indicating that the speaker is emitting its own WiFi signal and that it is possible to connect to it with a mobile device. Theoretically at least… In reality, the indicator was flashing rapidly and we had to repeat the process multiple times so that it would blink correctly and we could carry out the pairing. The rest of the setup, however, is very simple. First you must go into the WiFi settings on your smartphone or tablet to connect to the speaker's WiFi. Once the latter is selected, the speaker automatically connects to the home WiFi network (a Netgear Orbi Satellite in our case). The rest of the procedure is performed in a matter of seconds via the DTS Play-Fi app. The app lets you associate your different online music service accounts, assign the 4 quick access buttons on the top of the speaker and pair the RS200 with other DTS-compatible speakers. This app also lets you play music shared over the local network.
The DTS Play-Fi app lets you configure the McIntosh RS200, associate online music services and play music via DLNA.
We started our review of the McIntosh RS200 speaker by listening to 24-bit/96kHz and 24-bit/192kHz files streamed from Qobuz (Sublime Plus subscription). The McIntosh RS200 displayed exceptional balance with Michael Jackson's Beat It. The first notes of the percussions and electric guitar were very naturally restituted. The drums and bass were lively and very dynamic. The RS200 also surprised us with its ability to produce abyssal and powerful lows. Spatialization-wise, the soundstage was deep and each instrument was perfectly delineated. Naturally, the width of the soundstage wasn't extensive as that of a stereo system comprised of several speakers placed around the room, but it was airy and seemed a lot broader than that of a mono speaker. However, the McIntosh RS200 is very authoritative and it needs to be centered and placed at ear height in order to fully enjoy this spatialization.
The RS200 surprised us with its ability to produce abyssal and powerful lows.
Next, we tested the performance of the McIntosh RS200's optical digital input. For this we connected it to the Pioneer UDP-LX500 UHD 4K Blu-ray player using a Norstone Jura Optic optical cable. After we inserted the jazz album The Famous Sound of the Three Blind Mice Vol.2 and selected a piano piece with substantial dynamic shifts, the McIntosh RS200 revealed its full potential. The highs and mids were reproduced with finesse, while the lows filled the room so that we could almost imagine that the pianist was right in front of us. On a more groovy track, the brass instruments were exhilarating and superbly spatialized. However, to avoid a bass-saturated restitution it is important not to turn the volume all the way up.
The highs and mids were reproduced with finesse, while the lows filled the room.
Finally, we connected the McIntosh RS200 to an OLED 4K TV using an Audioquest Cinnamon HDMI cable in order to receive sound from the Pioneer UDP-LX500, itself connected to the television via HDMI. We were able to analyze the RS200's performance when used as a soundbar. With the blu-ray version of Mad Max Fury Road, the McIntosh RS200 soundbar was very impressive. It delivered dynamic sound with a soundstage that was much more extensive than it was when we had previously listened to music. During the car chase scene through the desert, the spatialization of sound effects was excellent, with machine gun fire that seemed to come from different areas in the room, as well as tangible, believable explosions. Even though we didn't get the opportunity to test it with surround speakers, the RS200 had no trouble plunging us right into the heart of the action. For nightime sessions, the VU meter's backlighting can be reduced or even turned off to eliminate any distractions.
Used as a soundbar, the McIntosh RS200 provides dynamic sound, along with a very extensive soundstage.
KEF LSX Wireless: this is a pair of true speakers capable of producing a much wider soundstage. That said, the LSX are slightly more analytical and aren't as deep. The RS200 is more balanced, has more powerful lows and is overall more pleasant to listen to. The RS200 also has the advantage of being more versatile, in particular thanks to its HDMI ARC input which makes it possible to use the speaker as a soundbar or add surround speakers.
Naim Mu-so 2: here, Naim provides a very similar solution in terms of connectivity and network functionality. It offers a little more detail in the highs, but isn't as impressive in the lows, which are deeper and more present with the RS200. The RS200 also has the advantage of integrating the DTS Play-Fi protocol and can be used as a USB DAC.
The McIntosh RS200 is a complete success, both in sound and appearance. Featuring a wide range of connections, DLNA and multi-room compatibility with support for diverse online services, it is a versatile and effective solution to fully enjoy all of your sources. Its numerous drivers provide a highly balanced and dynamic sound. It is important, however, to not turn the volume all the way up in order to avoid the lows saturating the restitution. The McIntosh RS200 is a true pleasure to listen to and use. We were rapidly captivated by the iconic blue light of the VU meters on the front panel which invited us to enter a world of rich sound. The regular succession of tracks was the only thing to remind us of the passing time… The McIntosh RS200 also revealed itself to be an excellent soundbar for enjoying movies, providing superior immersion and a more extensive soundstage than the TV's speakers. Sold for €3 800, the McIntosh RS200 is currently the most "affordable" solution if you wish to discover and enter the iconic American brand's universe.
The build quality
The integration of DTS Play-Fi and AirPlay 2 technologies
The vitality
The tonal balance
The magical VU meters
What we would have liked:
A wider stereo effect when listening to music
A virtual surround sound mode for movies
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Traductrice et rédactrice avec des goûts très éclectiques en matière de musique et de cinéma. Lorsque je ne suis pas au travail, vous pouvez me retrouver en train de regarder "Lost in Translation" de Sofia Coppola pour la centième fois, ou d'écouter un disque de David Bowie, Kate Bush, Joy Division ou Daft Punk sur ma platine Rega Planar 1. Étant d'origine britannique, je suis également adepte de séries à l'humour absurde comme Monty Python's Flying Circus et The Mighty Boosh !
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Review: Braun LE01 and Braun LE03 wireless speakers
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: creating a plot using ggplot2 I was trying to plot using ggplot but I am a beginner. The data frame (newdata) contains 2002 rows and 3 columns. When I use basic graphics plot function I get a perfect plot like an inverted bell shape but when I use ggplot2 I don't get a similar plot.
(i solved it..my dataframe had character columns,i converted them to numerical values and i plotted it..i works fine now..)
now i have two plots val vs dat1 and val vs dat2 , now i need to combine them using facet wrap...any help would be appreciated..
val = -1000,-999,-998 to 998,999, 1000.
dat1 = 1.2, 3.4, 5.5, 33.3, 55.4, ...
so on. dat2 is similar to y values.
str(newdata)
'data.frame':   2001 obs. of  3 variables:
 $ val  : chr  "-1000" "-999" "-998" "-997" ...
 $ dat1     : num  0.229 0.235 0.247 0.25 0.249 ...
 $ dat2   : num  1.97 1.98 1.98 1.98 1.98 ...

ggplot(data = newdata) +
  geom_point(mapping = aes(x = val, y = dat1))

updated..dput(data)
since the data is very huge im adding a part of it.
structure(list(**val** = c("-1000", "-999", "-998", "-997",
"-996", "-995", "-994", "-993", "-992", "-991"), dat1 = c(0.229377104377104,
0.23526936026936, 0.246843434343434, 0.250210437710438, 0.248526936026936,
0.252314814814815, 0.226641414141414, 0.230218855218855, 0.223484848484848,
0.236952861952862),  dat2 = c(1.97385049452862,
1.97675496296296, 1.97780065909091, 1.97756823063973, 1.97745205218855,
1.98053092087542, 1.98291262079125, 1.98401634175084, 1.98796655597643,
1.98639806102694)), row.names = c(NA, 10L), class = "data.frame")

A: Try:
ggplot(data = newdata, aes(x = val, y = dat1)) + geom_line()

A: To facet, you need data in long format (see this paper for a description of "tidy" data).  Basically, you need one column that holds values of either dat1 or dat2, and another column that distinguishes between rows for dat1 and rows for dat2.
Here's how to accomplish this for your dataset:
library(dplyr)
library(tidyr)
library(ggplot2)

# Convert from wide to long.  Convert strings to numbers.
data.long = newdata %>%
  gather(dat.type, dat.value, -val) %>%
  mutate(val = as.numeric(val))

# Plot and facet.
ggplot(data.long, aes(x = val, y = dat.value)) +
  geom_line() +
  facet_wrap(~ dat.type)

redpajama_set_name: RedPajamaStackExchange


There's a debate in great length going on at The Washington Post's On Parenting column. Stacey Garfinkle posed the question to the floor.
"Home-schooling has its faults. You don't spend as much time with kids your own age, no prom, and no highly competitive sports (community league just isn't the same).
Montessori is based on the fundamental principles that children ENJOY learning and do so naturally and that children learn best when they physically interact with what their learning. To that end, Montessori uses materials for all its lessons that facilitate the learning of concepts.


Constant innovation for a better experience to consumers is what all Grover Zampa Vineyards has aimed and achieved, says Kapil Grover, chairman, Grover Zampa Vineyards.
Grover Zampa Vineyards is a Nashik-based winery which crossed their 25 glorious years this November. On certain occasion, they launched Auriga – India's first-ever sparkling wine with edible gold flakes. According to owners, such an idea symbolizes embodiment of grace and artistry. The wine has a lemon-flavored shade with aroma of citrus and floral fragrance giving sure-fire refreshment with traces of dried apricot.
Grover Zampa Vineyards is a joint venture of Grover Vineyards Ltd and Vallee De Vin Pvt Ltd. The company marking their name on top in Karnataka and Maharashtra carry out the wine making done by a team of local oenologists under the direction of International French Wine Consultant Michel Rolland and winemaker Mathias Pellisard. Being one of the largest wine exporters from India, the company distributes their wines to over 26 countries and has won 92 awards since 2013. The Vineyards was also featured on many international platforms for red, white, rose and sparkling wines they produce in India.


text:
International Scientific Committee
Please note that fee has to be transferred under the registrant's name.
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A copy of the receipt of the bank remittance should be sent to the secretariat (idcam.info@gmail.com).
A direct bank transfer should be made to the account below.
During the payment of the participation fee; name and surname of the participant and a definition as "Dentistry Faculty Additive Manufacturing Congress participation fee" must be written in the explanation part of the receipt. Bank receipts must be sent to the congress e-mail (idcam.info@gmail.com). Participants requesting an invoice are required to indicate their request in this e-mail.
Bank Name: Ziraat Bankası
Account Name: Ç.Ü. Diş Hekimliği Fakültesi Döner Sermayesi
TRY IBAN: TR06 0001 0016 9025 5940 2050 12
Cancellations received after May 01, 2023 will not be refunded. All cancellations must be made by sending an e-mail to idcam.info@gmail.com. In order to get the payment back in case of cancellation of congress participation, it is necessary to apply to the revolving fund unit with a petition.
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Whew, okay, hopefully now fewer of these will clutter my posts. Wanting to do other kinds of shots besides subjects that stay perfectly still. 🙂 In fairness, if you think you would like to see more of these (flower shots in general), let me know in the comments, it's always up to you guys. And with that, I'm out, have a great night!


text:
package io.mycat.manager.response;

import java.nio.ByteBuffer;

import io.mycat.backend.mysql.PacketUtil;
import io.mycat.config.Fields;
import io.mycat.manager.ManagerConnection;
import io.mycat.net.mysql.EOFPacket;
import io.mycat.net.mysql.FieldPacket;
import io.mycat.net.mysql.ResultSetHeaderPacket;
import io.mycat.net.mysql.RowDataPacket;

/**
 * @author mycat
 */
public final class SelectVersionComment {

    private static final byte[] VERSION_COMMENT = "MyCat Server (monitor)".getBytes();
    private static final int FIELD_COUNT = 1;
    private static final ResultSetHeaderPacket header = PacketUtil.getHeader(FIELD_COUNT);
    private static final FieldPacket[] fields = new FieldPacket[FIELD_COUNT];
    private static final EOFPacket eof = new EOFPacket();
    static {
        int i = 0;
        byte packetId = 0;
        header.packetId = ++packetId;

        fields[i] = PacketUtil.getField("@@VERSION_COMMENT", Fields.FIELD_TYPE_VAR_STRING);
        fields[i++].packetId = ++packetId;

        eof.packetId = ++packetId;
    }

    public static void execute(ManagerConnection c) {
        ByteBuffer buffer = c.allocate();

        // write header
        buffer = header.write(buffer, c,true);

        // write fields
        for (FieldPacket field : fields) {
            buffer = field.write(buffer, c,true);
        }

        // write eof
        buffer = eof.write(buffer, c,true);

        // write rows
        byte packetId = eof.packetId;
        RowDataPacket row = new RowDataPacket(FIELD_COUNT);
        row.add(VERSION_COMMENT);
        row.packetId = ++packetId;
        buffer = row.write(buffer, c,true);

        // write last eof
        EOFPacket lastEof = new EOFPacket();
        lastEof.packetId = ++packetId;
        buffer = lastEof.write(buffer, c,true);

        // post write
        c.write(buffer);
    }

}

redpajama_set_name: RedPajamaGithub


Have you checked for colon cancer?
Constipation — irregular bowel movements or difficulty emptying the bowel is a common problem that bothers 20% of people. When constipation bowel movements less than three times a week. A condition is considered chronic if it persists for two weeks or longer.
Constipation is quite a common phenomenon, however, it should not be ignored. In each case it is necessary to establish the cause of constipation, especially if symptoms persist after a few weeks. It is important to consult with your doctor to make sure that the constipation is not caused by some serious disease.
A sedentary lifestyle , constipation may be caused by a lack of physical activity, resulting in impaired normal bowel movements.
Nutrition – diet low in fiber is one in which the little legumes, vegetables and fruits. Eating large amount of fiber helps to regulate intestinal microflora and to establish a regular chair.
Constipation can cause processed foods – fried, cakes, candies, pastries, pasta and bread. Cheese also is often a cause of constipation. If you suspect that your bowel problems causes a specific product or products, start a food journal to determine the cause of constipation.
Dehydration, often resulting from insufficient water intake or excessive drinking coffee and tea can interfere with bowel movements. To dehydration and can cause some medications, such as diuretics for high blood pressure.
As you know, constipation is caused by some medications – Allergy medications (diphenhydramine, loratadine, Fexofenadine, etc.) and opiate drugs (hydrocodone, oxycodone, morphine, etc.) from chronic pain.
Dietary supplements – calcium and iron – can also contribute to constipation. Calcium in a daily dose of 1000 mg per day can cause constipation, if with it not to take adequate amounts of magnesium. Similarly, constipation can cause iron supplements, which can easily be mistaken for iron deficiency anaemia (must be diagnosed by a doctor). However, if you take iron with vitamin C, it can promote the absorption of iron and prevent constipation.
Irritable bowel syndrome (IBS) – IBS is sometimes manifested by diarrhea, and constipation. Some patients with IBS constipation alternates with diarrhea. For those who have when IBS is dominated by constipation, doctors often prescribe medication if diet and lifestyle changes are not enough.
Excessive bacterial growth in the small intestine (SIBO) – according to some, the excessive growth of bacteria that produce methane gas, can increase the risk of constipation due to methane gas slows intestinal transit time.
Hypothyroidism – underactive thyroid, which affects up to 10% of the inhabitants of the Earth. In this disease slows down the transit of food through the digestive tract, causing constipation.
Frequent intake of antibiotics – disruption of the microbiome of the intestine can be caused by antibiotics. In some cases, it causes diarrhea. This will disturb the beneficial microflora, which can lead to chronic constipation come SIBO, as mentioned above.
Stress everyone reacts to stress differently. Someone stress may lead to constipation. To preserve the mental and physical health it is important to find a healthy way to reduce internal stress. As I said, the constipation arises due to the containment in the literal sense of the word. So to establish the regularity of the chair will help the reduction of stress, biofeedback and restoration of normal mode of work and rest.
Chronic diseases, such as Parkinson's disease, colon cancer, spinal cord injury, and sometimes even stroke, can lead to chronic constipation.
Most doctors recommend that people over the age of 50 to get checked for colon cancer. This can be done via colonoscopy, sigmoidoscopy, or stool test for blood or DNA of cancer cells.
For those who have a family history of colon cancer may need to be screened at age 40 or even earlier. Consult your doctor and check health recommendations based on your age and family risk factors.
If the main or reversible cause of constipation is not identified, doctors often prescribe docusate sodium (Colace), polyethylene glycol (PEG or miralax), lactulose (enulose) or sorbitol.
Chronic constipation prescribed with other drugs, including linaclotide (lenses, konstella), placental (trumans) and lubiprostone: (amitiza).
If the cause of constipation is opiatnye the drug (constipation – a typical side effect for this class of drugs), pharmaceutical companies have developed drugs (methylnaltrexone) to eliminate this side effect.
Many people want to avoid taking prescription drugs because of the side effects that they cause. Listed below are natural approaches to the treatment of constipation, favoured by many patients who want to alleviate the symptoms of constipation and help the body to establish a regular chair. The decisive factors underlying the normal functioning of the intestines, diet and lifestyle. With them always and should start. However, sometimes this is not enough.
Exercise – to stimulate regular bowel movements it is important to possible, active lifestyle. Doctors recommend that most people, moderate physical activity the total duration at least 150 minutes a week. It promotes healthy and regular bowel movement. Simple daily half-hour walk can be very beneficial to health. Some good help and yoga.
A lot of fiber in the diet is very important. Doctors advise to eat at least 25 grams of fiber a day, you need to eat 5-9 servings of fruits and vegetables. For example, in one avocado contains 13 grams of fiber in a medium Apple – 4 grams, and a banana – 3 grams. Berries are not only rich in fiber but also help to protect the body from colon cancer, according to research by 2016, published in Molecules. Chia seeds are also a good product — just one teaspoon of Chia seed contains almost 6 grams of fiber. In a Cup of black beans as much as 15 grams of fiber.
Herbal products will help to normalize the frequency of bowel movements. Plant food has several other useful properties, including enhances the absorption of vitamins A, K and E as well as vitamin C and other phytonutrients rich in antioxidants.
Study of 2015 has shown that fiber is effective in mild and moderate constipation and constipation caused by IBS. A diet with a high content of plant foods is associated with a lower risk of developing colon cancer and has a beneficial effect on intestinal microflora. A 2012 study published in World Journal of Gastroenterology, showed that fiber helps to increase the frequency of bowel movements in adults.
Products with probiotics– probiotics are beneficial bacteria. They can be very effective for restoring balance to the microbiome of the intestine, especially if this balance was disturbed as a result of taking antibiotics or the chronic use of drugs that reduce the acidity.
Restore populations of healthy bacteria (such as lactobacilli, bifidobacteria) and other important and useful microorganisms are important for reinforcing the General health and treatment of constipation. The study of 2017, published in Advances in Nutrition, showed a tendency to decrease populations of these vital bacteria in chronic constipation.
Consumption of fermented milk and fermented foods and drinks such as sauerkraut, miso soup, tempeh, yogurt, and Kombucha (Kombucha) can also be useful to restore this natural balance. Held in Iran in 2016, the study showed that the condition of constipated pregnant women has improved in the daily consumption of yogurt. Many also take food supplements with probiotics. We will talk about them later.
Magnesium – magnesium is a ubiquitous trace element that is involved in more than 350 biochemical reactions throughout the body. Magnesium deficiency is among the most common deficiencies of nutrients and can manifest in the form of headaches, palpitations, muscle cramps and even constipation.
Therefore, supplementation with magnesium can be effective for those who suffer from constipation. If the background intake of magnesium is observed laxative effect, the dosage of magnesium should be reduced. Patients with advanced kidney disease need to consult with your doctor. In a study of 2017, published in Magnesium Research, assessed the condition of patients undergoing open-heart surgery. Magnesium helped prevent not only constipation, but also arrhythmia, especially atrial fibrillation.
In a study published in 2014 in Clinical Gastroenterology and Hepatology, attended by the French, suffering from constipation. The researchers found that the daily consumption of magnesium-rich mineral water symptoms constipation decreased significantly.
Vitamin C – commonly found in citrus fruits and pepper. Vitamin C is important for preventing scurvy. Helps vitamin C and constipation.
The husk of psyllium is the psyllium husk can be very effective in chronic constipation. It can also help to control blood sugar in diabetes. A study of 2018, published in Complementary Therapies in Medicine, showed that when taking psyllium suffering from constipation patients with type II diabetes not only reduced weight, but also saw improvements in glucose control. A study of 2016, published in Appetite showed that supplementation shells of psyllium has contributed to the feeling of saturation, which allowed the subjects do not experience feelings of hunger in between meals.
Prebiotics – prebiotics are nutrients and/or foods consumed by the beneficial bacteria of the intestine. Introducing in the diet of prebiotics, such as inulin, can provide adequate nutrition of beneficial intestinal bacteria that help restore harmonious balance in the gastrointestinal tract.
The prebiotics include foods such as apples, asparagus, banana, barley, chicory root, dandelion greens, flax seed and garlic. Read more about prebiotics.
A study of 2019 have demonstrated the benefits of optimizing reception diabetikov in the treatment of constipation. A similar study of 2017, published in the American Journal of Medical Sciences, showed that the prebiotic separately and in combination with probiotics can improve stool frequency and reduce symptoms of stress and bloating.
Probiotics – for many years many of my patients notice that when taking probiotic they had constipation. The authors of the study in 2017, published in the Archives of Gerontology and Geriatrics, compared the probiotic Supplement with a pill placebo and found that probiotics reduced the constipation in 10-40 % of patients.
Another study 2017 with the participation of suffering from constipation Asian children showed that probiotics helped to increase the frequency of chair. A study of 2018 has demonstrated the effectiveness of probiotics in the treatment of constipation in patients with Parkinson's disease.
Cascara Sagrada (Frangula) buckthorn is a medicinal plant that have been used for centuries as a remedy for chronic constipation. Name of the plant translates as "sacred bark". This plant was an important healing tool among indigenous peoples. Scientists discovered that the active component of the plant is anthraquinone. This substance has helpful for digestive properties.
Triphala– triphala is often used as a remedy for constipation, but I was unable to find studies supporting its effectiveness that's when the constipation. But I found a study showing the positive impact of triphala on the microbiome of the intestine, which may explain its effectiveness in the treatment of constipation.
Aloe Vera– a cactus of the aloe have been used for centuries for a variety of purposes in Asia and Mexico. There are many reports of its effectiveness in the treatment of chronic constipation. Studies, proving the benefit of aloe were conducted in 1974 (and the most recent in 2008). Aloe can be taken in form of juice or nutritional supplements.
Senna is used for thousands of years. The efficacy of Senna was clearly demonstrated in a study of 2017 with the participation of suffering from constipation of children. Another study in 2018 showed that Senna not only effective but also safe. It can be taken in form of tea or supplements.
Rhubarb – the Chinese almost 3000 years use rhubarb for medical purposes, in particular as a laxative. Modern science supports ancient practices of the use of this plant. The 2018 study has shown that rhubarb may be helpful to those people who have constipation while in the hospital. Rhubarb can be eaten as they are, or take as a dietary Supplement.
It may be useful to ancient Ayurvedic medicine a glass of warm milk with melted ghee. I recommend adding Ki in coffee or tea — this drink is called armor-piercing coffee or tea.
The powder of licorice root can also be useful if you dilute it in warm water.
Prunes and prune juice is a known folk remedy for constipation, which many find effective. In the course of the study in 2011 revealed that the constipation prunes more effective than psyllium.
You can also stimulate bowel movements, regularly drinking a Cup of warm herbal tea, like black tea, chamomile tea, green tea, tea with ginger, peppermint or extract of Senna.
Paré P, Fedorak RN. Systematic review of stimulant and nonstimulant laxatives for the treatment of functional constipation. Can J Gastroenterol Hepatol. 2014;28(10):549-57.
Molecules. 2016 Jan 30;21(2):169. doi: 10.3390/molecules21020169.
Aliment Pharmacol Ther. 2015 Jan;41(12):1256-70. doi: 10.1111/apt.13167. Epub Apr 22 2015.
Cancer Causes Control. 1997 Jul;8(4):575-90.
World J Gastroenterol. 2012 Dec 28;18(48):7378-83. doi: 10.3748/wjg.v18.i48.7378.
Clinical Gastroenterology and Hepatology. 2014 Aug;12(8):1280-7. doi: 10.1016/j.cgh.2013.12.005. Epub 2013 Dec 14.
Complementary Therapies in Medicine. 2018 Oct;40:1-7. doi: 10.1016/j.ctim.2018.07.004. Epub 2018 Jul 10.
Appetite. 2016 Oct 1;105:27-36. doi: 10.1016/j.appet.2016.04.041. Epub May 7 2016.
Am J Med Sci. 2017 Mar;353(3):282-292. doi: 10.1016/j.amjms.2016.09.014. Epub Oct 4 2016.
Arch Gerontol Geriatr. 2017 Jul;71:142-149. doi: 10.1016/j.archger.2017.04.004. Epub Apr 14 2017.


Ryan Christian found CrossFit in 2009 and currently has his CrossFit Level 1, CrossFit Kids, CrossFit Football, and Krav Maga Level 3 certification and holds a Level 2 Krav Maga certification. Ryan has competed in the 2009 Lumberjack 20 competition and 2010 Fight Gone Bad. Coaching has always been a passion of Ryan's since he can remember. He loves to teach and help others reach as high as they dare strive for. He enjoys working with people and seeing the happiness that comes with reaching their goals. Be it from weight loss, to building muscle or walking around with a stronger sense of safety and self-confidence they all add up as a win.


Complete TV Series MacGyver by . Seasons included: 1 in high quality and low size. Click Scroll to download to select the season.
McGuire-an agent who is able to perform any particularly dangerous and serious assignment. He is a real superhero, as he has all the data to take on complex missions, has extraordinary abilities and extensive scientific knowledge.
Charismatic, smart, optimistic, strong and self-confident hero of the series "McGuire" season 2 can throw everything and rush to solve the most dangerous problems that can cost him his life. Epic skirmishes, clever spy operations and other important events of this genre – something that will make the audience fully believe in what is happening and feel sympathy for the crazy and positive guy. McGuire has a loyal assistant – Jack Dalton, who is always there and supports the agent in difficult moments. Self-confidence, good team, fighting spirit – that will help the young man to cope with all difficulties. Insidious and incredibly dangerous violators of world peace may regret that they contacted the invincible team: Angus, Jack, Patricia, Riley and Wilt. Now people can sleep peacefully, because the guard of their lives are real professionals who are capable of much.


Fred Wagner focuses on environmental and natural-resources issues associated with major infrastructure, mining, and energy project development. Fred manages and defends environmental reviews performed under the National Environmental Policy Act (NEPA) or equivalent state statutes. He works with public agencies and private developers to secure permits and approvals from federal and state regulators under Section 404 of the Clean Water Act (CWA), the Endangered Species Act (ESA), and the National Historic Preservation Act (NHPA). Fred understands the full range of issues surrounding U.S. Department of Transportation (USDOT) surface transportation programs, including grant management, procurement, suspension and debarment, and safety regulations.
During his career, Fred has handled a wide variety of environmental litigation in federal trial and appellate courts across the country, from citizen suits to government enforcement actions and Administration Procedure Act (APA) challenges.
Fred was appointed chief counsel of the U.S. Federal Highway Administration (FHWA) during the Obama administration. He managed all legal matters involving the $40 billion Federal-Aid Highway program, including environmental and natural resources issues for highway and multimodal transportation projects. Among other high-profile projects, he oversaw the agency's defense of the following: New York's Tappan Zee Bridge, San Francisco's Presidio Parkway, Chicago's Elgin-O'Hare Expressway, Kentucky and Indiana's Ohio River Bridges, North Carolina's Bonner Bridge, Alabama's Birmingham Northern Beltline, Wisconsin's Zoo Interchange, and Washington's State Road 520 Bridge. He represented the FHWA on the government-wide Transportation Rapid Response Team, a multi-agency task force focused on improving project delivery and environmental review reforms.
Fred began his career as a trial attorney in the Environment Division of the U.S. Department of Justice. He also served as a special assistant U.S. attorney in the Misdemeanor Trial Section of the U.S. Attorney's Office for the District of Columbia. Prior to joining Venable, he spent more than 20 years in private practice at a national law firm focusing on environmental and natural resources issues.
Co-authored a Supreme Court amicus brief on behalf of a coalition of seven major real estate associations in the controversial Rapanos case, involving the definition of "waters of the United States"


text:
\section{Introduction}
\vspace{-1ex}
\subsection{Problem statement}
A graph $G=(V,E)$ is said to be $k$-colorable for some positive
integer $k$ if there exists a $k$-coloring $\chi: V \rightarrow
\{1,2,\dots,k\}$ such that for all edges $(u,v) \in E$, $\chi(u) \neq
\chi(v)$. For $k \ge 3$, finding a $k$-coloring of a $k$-colorable
graph is a classic NP-hard problem.  The problem of coloring a graph
with the fewest number of colors has been extensively studied.  In
this paper, our focus is on hardness results for the following
maximization version of graph coloring: Given a $k$-colorable graph
(for some fixed constant $k \ge 3$), find a $k$-coloring that
maximizes the fraction of properly colored edge. (We say an edge is
properly colored under a coloring if its endpoints receive distinct
colors.) Note that for $k=2$ the problem is trivial --- one can find a
proper $2$-coloring in polynomial time when the graph is bipartite
($2$-colorable).

We will call this problem \maxcolor{k}. The problem is equivalent to
partitioning the vertices into $k$ parts so that a maximum number of
edges are cut. This problem is more popularly referred to as
\maxcut{k} in the literature; however, in the \maxcut{k} problem the
input is an arbitrary graph that need not be $k$-colorable. To
highlight this difference that our focus is on the case when the input
graph is $k$-colorable, we use \maxcolor{k} to refer to this
variant. We stress that we will use this convention throughout the
paper: \maxcolor{k} {\bf always} {\em refers to the ``perfect
  completeness'' case, when the input graph is
  $k$-colorable.}\footnote{While a little non-standard, this makes our
  terminology more crisp, as we can avoid repeating the fact that the
  hardness holds for $k$-colorable graphs in our statements.} Since
our focus is on hardness results, we note that this restriction only
makes our results stronger.

A factor $\alpha=\alpha_k$ approximation algorithm for \maxcolor{k} is
an efficient algorithm that given as input a $k$-colorable graph
outputs a $k$-coloring that properly colors at least a fraction
$\alpha$ of the edges. We say that \maxcolor{k} is NP-hard to
approximate within a factor $\beta$ if no factor $\beta$ approximation
algorithm exists for the problem unless ${\rm P} = {\rm NP}$. The goal
is to determine the approximation threshold of \maxcolor{k}: the
largest $\alpha$ as a function of $k$ for which a factor $\alpha$
approximation algorithm for \maxcolor{k} exists.
\vspace{-1ex}
\subsection{Previous results}
The algorithm which simply picks a random $k$-coloring, without even
looking at the graph, properly colors an expected fraction $1-1/k$ of
edges. Frieze and Jerrum~\cite{FJ97} used semidefinite programming to
give a polynomial time factor $1-1/k + 2 \ln k/k^2$ approximation
algorithm for \maxcut{k}, which in particular means the algorithm will
color at least this fraction of edges in a $k$-colorable graph. This
remains the best known approximation guarantee for \maxcolor{k}\ to
date. Khot, Kindler, Mossel, and O'Donnell~\cite{KKMO} showed that
obtaining an approximation factor of $1-1/k + 2 \ln
k/k^2 + \Omega(\ln\ln k/k^2)$ for \maxcut{k}\ is Unique Games-hard,
thus showing that the Frieze-Jerrum algorithm is essentially the best
possible. However, due to the ``imperfect completeness'' inherent to
the Unique Games conjecture, this hardness result does {\em not} hold
for \maxcolor{k} when the input is required to be $k$-colorable.

For \maxcolor{k}, the best hardness known prior to our work was a
factor $1-\Theta(1/k^2)$. This is obtained by combining an
inapproximability result for \maxcolor{3} due to
Petrank~\cite{petrank} with a reduction from Papadimitriou and
Yannakakis~\cite{PY91}. It is a natural question whether is an
efficient algorithm that could properly color a fraction $1-
1/k^{1+\varepsilon}$ of edges given a $k$-colorable graph for some absolute
constant $\varepsilon > 0$. The existing hardness results do not rule out the
possibility of such an algorithm.

For \maxcut{k}, a better hardness factor was shown by Kann, Khanna, Lagergren, and Panconesi~\cite{KKLP} ---
for some absolute constants $\beta > \alpha > 0$, they
showed that it is NP-hard to distinguish graphs that have a $k$-cut in
which a fraction $(1-\alpha/k)$ of the edges cross the cut from graphs whose
Max $k$-cut value is at most a fraction $(1-\beta/k)$ of edges.
Since {\sf MaxCut}\ is easy when the graph is $2$-colorable, this reduction
does not yield any hardness for \maxcolor{k}.
\vspace{-1ex}
\subsection{Our results}
Petrank~\cite{petrank} showed the existence of a $\gamma_0 > 0$ such that
it is NP-hard to find a $3$-coloring that properly colors more than a
fraction $(1-\gamma_0)$ of the edges of a $3$-colorable graph. The
value of $\gamma_0$ in \cite{petrank} was left unspecified and would
be very small if calculated. The reduction in \cite{petrank} was rather
complicated, involving expander graphs and starting from the weak
hardness bounds for bounded occurrence satisfiability. We prove that
the NP-hardness holds with $\gamma_0 = \frac{1}{33}$. In
other words, it is NP-hard to obtain an approximation ratio bigger
than $\frac{32}{33}$ for \maxcolor{3}. The
reduction is from the constraint satisfaction problem corresponding to
the adaptive $3$-query PCP with perfect completeness from \cite{GLST}.

By a reduction from \maxcolor{3}, we prove that for every $k \ge 3$,
the \maxcolor{k} is NP-hard to approximate within a factor greater
than $\approx 1-\frac{1}{33 k}$
(Theorem~\ref{thm:kcolor-NP-hard}).  This identifies the correct
asymptotic dependence on $k$ of the best possible approximation factor
for \maxcolor{k}. The reduction is similar to the one in \cite{KKLP},
though some crucial changes have to be made in the construction and
some new difficulties overcome in the soundness analysis when reducing
from \maxcolor{3} instead of {\sf MaxCut}.

In the quest for pinning down the {\em exact} approximability of
\maxcolor{k}, we prove the following {\em conditional}
result. Assuming the so-called $2$-to-$1$ conjecture, it is hard to
approximate \maxcolor{k} within a factor $1-\frac{1}{k} + O\left(\frac{\ln
  k}{k^2}\right)$. In other words, the Frieze-Jerrum algorithm is optimal up to
lower order terms in the approximation ratio {\em even for instances of
\maxcut{k} where the graph is $k$-colorable.} %

Unlike the Unique Games Conjecture (UGC), the $2$-to-$1$ conjecture
allows perfect completeness, i.e., the hardness holds even for
instances where an assignment satisfying {\em all} constraints
exists. The $2$-to-$1$ conjecture was used by Dinur, Mossel, and
Regev~\cite{DMR} to prove that for every constant $c$, it is NP-hard
to color a $4$-colorable graph with $c$ colors. We analyze a similar
reduction for the $k$-coloring case when the objective is to maximize
the fraction of edges that are properly colored by a $k$-coloring. Our
analysis uses some of the machinery developed in \cite{DMR}, which in
turn extends the invariance principle of \cite{MOO}. The hardness
factor we obtain depends on the spectral gap of a certain $k^2 \times
k^2$ stochastic matrix.

\begin{remark}
In general it is far from clear which Unique
Games-hardness results can be extended to hold with perfect
completeness by assuming, say, the $2$-to-$1$ (or some related)
conjecture. In this vein, we also mention the result of O'Donnell and
Wu~\cite{odonnell-wu} who showed a tight hardness for approximating
satisfiable constraint satisfaction problems on $3$ Boolean variables
assuming the $d$-to-$1$ conjecture for any fixed $d$. While the UGC
assumption has led to a nearly complete understanding of the
approximability of constraint satisfaction problems~\cite{Raghavendra08}, the
approximability of {\em satisfiable} constraint satisfaction problems
remains a mystery to understand in any generality.
\end{remark}

\begin{remark}
\label{rem:wt-unwt}
  It has been shown by Crescenzi, Silvestri and Trevisan \cite{CST01}
  that any hardness result for weighted instances of \maxcut{k}
  carries over to  unweighted instances {assuming} the
  total edge weight is polynomially bounded. In fact, their reduction
  preserves $k$-colorability, so an inapproximability result for the weighted
  \maxcolor{k} problem also holds for the unweighted version. Therefore
  all our hardness results hold for the unweighted \maxcolor{k} problem.
\end{remark}

\section{Unconditional Hardness Results for \maxcolor{k}}
We will first prove a hardness result for \maxcolor{3}, and then
reduce this problem to \maxcolor{k}.
\subsection{Inapproximability result for \maxcolor{3}}
\label{sec:3color-hard}
Petrank~\cite{petrank} showed that \maxcolor{3} is NP-hard to
approximate within a factor of $(1-\gamma_0)$ for some constant
$\gamma_0 > 0$. This constant $\gamma_0$ is presumably very small,
since the reduction starts from bounded occurrence satisfiability (for
which only weak inapproximability results are known) and uses expander
graphs. We prove a much better inapproximability factor below, via a
simpler proof.

\begin{theorem}[\maxcolor{3} Hardness]
\label{thm:3color}
  The \maxcolor{3} problem is NP-hard to approximate within a factor of
  $\frac{32}{33}+\varepsilon$ for any constant $\varepsilon > 0$.
\end{theorem}
\newcommand{{\cal V}}{{\cal V}}
\newcommand{{\cal X}}{{\cal X}}
\newcommand{{\cal Y}}{{\cal Y}}
\newcommand{{\cal Z}}{{\cal Z}}
\begin{proof}
  For the proof of this theorem, we will use reduce from a hard to
  approximate constraint satisfaction problem (CSP) underlying the
  adaptive 3-query PCP given in \cite{GLST}. This PCP has perfect
  completeness and soundness $1/2+\varepsilon$ for any desired constant
  $\varepsilon$ (which is the best possible for $3$-query PCPs).

  We first state the properties of the CSP. An instance of the CSP
  will have variables partitioned into three parts ${\cal X},{\cal Y}$ and
  ${\cal Z}$. Each constraint will be of the form $(x_i \vee (Y_j = z_k))
  \wedge (\overline{x_i} \vee (Y_j = z_l))$, where $x_i \in {\cal X}$,
  $z_k,z_l \in {\cal Z}$ are variables (unnegated) and $Y_j$ is a literal
  ($Y_j \in \{y_j, \overline{y_j}\}$ for some variable $y_j \in
  {\cal Y}$).

For {\sc Yes} instances of the
  CSP, there will be a Boolean assignment that satisfies {\bf all} the
  constraints. For {\sc No} instances, every assignment to the
  variables will satisfy at most a fraction $(1/2+\varepsilon)$ of the
  constraints.

\begin{remark}
  We remark the condition that the instance is tripartite, and that the
  variables in ${\cal Z}$ never appear negated are not explicit in
  \cite{GLST}. But these can be ensured by an easy modification to the
  PCP construction in \cite{GLST}. The PCP in \cite{GLST} has a
  bipartite structure: the proof is partitioned into two parts called
  the $A$-tables and $B$-tables, and each test consists of probing one
  bit $A(f)$ from an $A$ table and 3 bits $B(g),B(g_1),B(g_2)$ from
  the $B$ table, and checking $(A(f) \vee (B(g) = B(g_1)) \wedge
  (\overline{A(f)} \vee (B(g) = B(g_2))$. Further these tables are
  {\em folded} which is a technical condition that corresponds to the
  occurrence of negations in the CSP world. If the queries at
  locations $g_1$ and $g_2$ are made in a parallel $C$-table, and even
  if the $C$-table is not folded (though the $A$ and $B$ tables need
  to be folded), one can verify that the analysis of the PCP
  construction still goes through. This then translates to a CSP with
  the properties claimed above.
\end{remark}

\iffalse
This test has perfect
  completeness and soundness $\frac{1}{2}+\varepsilon$. Moreover we can
  assume exactly half of the literals will be negations, since the
  proof is assumed to be folded.
The proof has $n$ bits,
  $\{x_1,\ldots,x_n\}$ and each query is of the form $(x_i \vee (X_j =
  x_k)) \wedge (\overline{x_i} \vee (X_j = x_l))$, where $X_j$ is a
  literal ($X_j \in \{x_j, \overline{x_j}\}$). This test has perfect
  completeness and soundness $\frac{1}{2}+\varepsilon$. Moreover we can
  assume exactly half of the literals will be negations, since the
  proof is assumed to be folded.
\fi

\begin{figure}[ht]
  \centering
  \begin{minipage}[b]{0.44\linewidth}
    \centering
    \includegraphics[scale=1]{gadget3.eps}
    \caption{Global gadget for truth value assignments. Blocks $X_i$,
      $Y_j$ and $Z_l$ are replicated for all vertices in
      $\mathcal{X}$, $\mathcal{Y}$ and $\mathcal{Z}$. Edge weights are
      shown next to each edge.}
    \label{fig:gadget-global}
  \end{minipage}
  \hspace{0.05\linewidth}
  \begin{minipage}[b]{0.47\linewidth}
    \centering
    \includegraphics[scale=1]{localgadget.eps}
    \caption{Local gadget for each constraint of the form $(x_i \vee
      Y_j = z_k) \wedge (\overline{x_i} \vee Y_j = z_l)$. All edges
      have unit weight. Labels $A,A',B,B'$ refer to the local nodes in
      each gadget.}
    \label{fig:gadget-local}
  \end{minipage}
\end{figure}

Let ${\cal I}$ be an instance of such a CSP with $m$ constraints of
the above form on variables ${\cal V} = {\cal X} \cup {\cal Y} \cup {\cal Z}$. Let
${\cal X}=\{x_1,x_2,\dots,x_{n_1}\}$, ${\cal Y} = \{y_1,y_2,\dots,y_{n_2}\}$ and
${\cal Z}=\{z_1,z_2,\dots,z_{n_3}\}$. From the instance ${\cal I}$ we
create a graph $G$ for the \maxcolor{3} problem as follows.  There is
a node $x_i$ for each variable $x_i \in {\cal X}$, a node $z_l$ for each
$z_l \in {\cal Z}$, and a pair of nodes $\{y_j,\overline{y_j}\}$ for the
two literals corresponding to each $y_j \in {\cal Y}$. There are also three
global nodes $\{R,T,F\}$ representing boolean values which are
connected in a triangle with edge weights $m/2$ (see
Fig. \ref{fig:gadget-global}).

For each constraint of the CSP, we place the local gadget specific to
that constraint shown in Figure \ref{fig:gadget-local}. Note that
there are 10 edges of unit weight in this gadget. The nodes
$y_j,\overline{y_j}$ are connected to node $R$ by a triangle whose
edge weights equal $w_j =
\frac{\Delta(y_j)+\Delta(\overline{y_j})}{2}$. Here $\Delta(X)$
denotes the total number of edges going from node $X$ \emph{into} all
the local gadgets. The nodes $x_i$ and $z_l$ connected to $R$ with an
edge of weight $\Delta(x_i)/2$ and $\Delta(z_l)/2$ respectively. The proofs of the following lemmas appear in Appendix~\ref{app:pf-3col}.
\begin{lemma}[Completeness]
\label{lem:3col-compl}
  Given an assignment of variables $\sigma: {\cal V} \rightarrow \{0,1\}$ which
  satisfies at least $c$ of the constraints, we can construct a
  $3$-coloring of $G$ with at most $m-c$ improperly colored edges
  (each of weight 1).
\end{lemma}
\begin{lemma}[Soundness]
\label{lem:3col-soundness}
  Given a 3-coloring of $G$, $\chi$, such that the total weight of
  edges that are not properly colored by $\chi$ is at most $\tau <
  m/2$, we can construct an assignment $\sigma' : {\cal V} \to\{0,1\}$ to
  the variables of the CSP instance that satisfies at least $m-\tau$
  constraints.
\end{lemma}

Returning to the proof of Theorem~\ref{thm:3color}, the total weight
of edges in $G$ is
\[ 10 m + \frac{3 m}{2} + \underbrace{\sum_{i=1}^{n_1}
  \frac{\Delta(x_i)}{2}}_{m} + \sum_{j=1}^{n_2} 3 w_j +
\underbrace{\sum_{l=1}^{n_3} \frac{\Delta(z_l)}{2}}_{m} = \frac{27}{2}
m + \frac{3}{2} \underbrace{\sum_{j=1}^{n_2} (\Delta(y_i) +
  \Delta(\overline{y_j}))}_{2m} = \frac{33}{2} m \ . \]

  By the completeness lemma, {\sc Yes} instances of the CSP are mapped
  to graphs $G$ that are $3$-colorable. By the soundness lemma, {\sc
    No} instances of the CSP are mapped to graphs $G$ such that every
  $3$-coloring miscolors at least a fraction $\frac{(1/2-\varepsilon)}{33/2}
  = \frac{1-2\varepsilon}{33}$ of the total weight of edges. Since $\varepsilon > 0$ is an arbitrary constant, the proof of Theorem~\ref{thm:3color} is complete.\footnote{Our reduction produced a graph with edge weights, but by Remark~\ref{rem:wt-unwt}, the same inapproximability factor holds for unweighted graphs as well.}
\end{proof}
\subsection{\maxcolor{k} Hardness}
\begin{theorem}
\label{thm:kcolor-NP-hard}
  For every integer $k \ge 3$

  and every $\varepsilon > 0$, it is NP-hard to
  approximate \maxcolor{k} within a factor of $1 - \frac{1}{33 (k + c_k)+c_k} +
  \varepsilon$ where $c_k = k \mod 3 \le 2$.
\end{theorem}

\begin{proof}
  We will reduce \maxcolor{3} to \maxcolor{k} and then apply
  Theorem~\ref{thm:3color}. Throughout the proof, we will assume $k$
  is divisible by $3$. At the end, we will cover the remaining cases
  also. The reduction is inspired by the reduction from {\sf MaxCut}\ to
  \maxcut{k} given by Kann {\it et al.}~\cite{KKLP} (see
  Remark~\ref{rem:KKLP}). Some modifications to the reduction are
  needed when we reduce from \maxcolor{3}, and the analysis has to
  handle some new difficulties. The details of the reduction and its
  analysis follow.

  Let $G=(V,E)$ be an instance of \maxcolor{3}. By
  Theorem~\ref{thm:3color}, it is NP-hard to tell if $G$ is
  $3$-colorable or every $3$-colors miscolors a fraction $\frac{1}{33}
  -\varepsilon$ of edges. We will construct a graph $H$ such that $H$ is
  $k$-colorable when $G$ is $3$-colorable, and a $k$-coloring which
  miscolors at most a fraction $\mu$ of the total weight of edges of
  $H$ implies a $3$-coloring of $G$ with at most a fraction $\mu k$ of
  miscolored edges. Combined with Theorem~\ref{thm:3color}, this gives
  us the claimed hardness of \maxcolor{k}.

Let $K'_{k/3}$ denote the complete graph with loops on $k/3$
vertices. Let $G'$ be the tensor product graph between $K_{k/3}$ and
$G$, $G'=K'_{k/3}\otimes G$ as defined by
Weichsel~\cite{weichsel62}.
Identify each node in $G'$ with $(u,i),
u\in V(G), i\in \{1,2,\dots,k/3\}$. The edges of $G'$ are $((u,i),(v,i'))$ for $(u,v) \in E$ and any $i,i' \in \{1,\dots,k/3\}$. Next we make $3$ copies of $G'$, and identify
the nodes with $(u,i,j), (u,i)\in V(G'), j\in \{1,2,3\}$, then put edges
between all nodes of the form $(u,i,j)$ and $(u,i',j')$ if either
$i\neq i'$ or $j\neq j'$ with weight $\frac{2}{3} d_u$, where $d_u$ is
degree of node $u$. The total weight of edges in this new construction $H$ equals
\[ \sum_{u\in V} \left({k\choose 2} \frac{2}{3} d_u + \frac{3}{2}
  \left(\frac{k}{3}\right)^2 d_u \right) \le k^2 m \ . \]
\iffalse
\begin{figure}[htb]
\centering
\includegraphics[width=0.7\textwidth]{maxkcut.eps}
\caption{Tensor product, $K'_{k/3} \otimes G$.}
\label{fig:maxkcut-constr}
\end{figure}
\fi

\begin{lemma}
  If $G$ is $3$-colorable, then $H$ is $k$-colorable.
\end{lemma}
\begin{proof}
Let $\chi_G:V(G)\to \{1,2,3\}$ be a 3-coloring of $G$. Consider the
following coloring function for $H$, $\chi_H:V(H)\to \{1,2,\dots,k\}$. For node
$(u,i,j)$, let $\chi_H((u,i,j)) = \pi^{j}(\chi_G(u)) + 3 (i-1)$. Here
$\pi$ is the permutation $\left(\begin{array}{ccc} 1 & 2 & 3
  \\ 2&3&1 \end{array}\right)$, and $\pi^j(x) =
\underbrace{\pi(\ldots(\pi(}_{\text{j times}} x)))$. Equivalently
$\pi(x) = x\mod 3 + 1$.

Consider edges of the form $\{(u,i,j),(v,i',j)\}$. If $i\neq i'$, then
colors of the endpoints are different. Else we have
$\chi((u,i,j))-\chi((v,i,j)) \equiv \chi(u)-\chi(v) \not\equiv 0 \mod
3$. For edges of the form $\{(u,i,j),(u,i',j')\}$, if $i\neq i'$,
clearly edge is satisfied. When $i=i',j\neq j'$, $\chi((u,i,j))-\chi((u,i,j'))
\equiv \pi^{j}(u) - \pi^{j'}(u) \equiv j-j' \not\equiv 0 \mod 3$.
\end{proof}

\begin{lemma}
  If $H$ has a $k$-coloring that properly colors a set of edges with
  at least a fraction $(1-\mu)$ of the total weight, then $G$
  has a $3$-coloring which colors at least a fraction $(1-\mu k)$ of
  its edges properly.
\end{lemma}
\begin{proof}
Let $\chi_H$ be the coloring of $H$, $\mathsf{Sugg}^j_u = \{\chi_H((u,i,j)) \mid 1 \le i\le k/3\}$ and $\mathsf{Sugg}_u = \bigcup_j \mathsf{Sugg}^j_u$. Denote the total weight of uncut
edges in this solution as
\begin{equation}
C^{total} = \sum_{u \in V(G)} \frac{2}{3} d_u C^{within}_u +
C^{between},
\end{equation}
where $C^{within}_u$ and $C^{between}$ denotes the number of
improperly colored edges within the copies of node $u$ and between
copies of different vertices $u,v \in V(G)$ respectively. We have the
following relations:
\begin{equation}
\label{eq:c-between}
  \begin{array}{rclr}
  C^{between} &=& \sum_{j=1}^3 \sum_{uv \in E(G)} \sum_{1\le i\le
    i'\le k/3} 1_{\chi_H((u,i,j))=\chi_H((v,i',j))} \\ &\ge&
  \sum_{j=1}^3 \sum_{uv \in E(G)} |\mathsf{Sugg}^j_u \cap \mathsf{Sugg}^j_v|
  \end{array}
\end{equation}

\begin{equation}
  \begin{array}{rclr}
    C^{within}_u &=& \sum_{c \in \mathsf{Sugg}_u} {|\chi_H^{-1}(c) \cap B_u|
      \choose 2} & \mbox{($B_u=\{(u,i,j)|\forall i,j\}$)} \\ &=&
    \sum_{c \in \mathsf{Sugg}_u} \frac{|B_{u,c}|^2}{2} -
    \frac{k}{2} & \mbox{($B_{u,c} = B_u \cap \chi_H^{-1}(c)$)}\\
    &\ge& \frac{1}{2 |\mathsf{Sugg}_u|} \left(\sum_{c \in \mathsf{Sugg}_u}
    |B_{u,c}|\right)^2  - \frac{k}{2} & \mbox{(Cauchy-Schwarz)}\\
    &\ge& \frac{k}{2}\left(\frac{k}{|\mathsf{Sugg}_u|}-1\right)
      \ge \frac{k}{2} \frac{|\overline{\mathsf{Sugg}_u}|}{|\mathsf{Sugg}_u|}
      \ge \frac{|\overline{\mathsf{Sugg}_u}|}{2}
  \end{array}
\end{equation}

Now we will find a (random) $3$-coloring $\chi_G$ for $G$.  Pick $c$ from $\{1,2,\dots,k\}$
uniformly at random.  If $c \notin \mathsf{Sugg}_u$, select $\chi_G(u)$
uniformly at random from $\{1,2,3\}$. If $c \in \mathsf{Sugg}_u$, set
$\chi_G(u) = j$ if $j$ is the smallest index for which $c \in
\mathsf{Sugg}^j(u)$.  With this coloring $\chi_G(u)$, the probability that an
edge $(u,v) \in E(G)$ will be improperly colored is:
\begin{eqnarray*}
\prob{}{\chi_G(u) = \chi_G(v)}
& \le & \sum_{j=1}^3 \prob{c}{c \in \mathsf{Sugg}^j_u \cap \mathsf{Sugg}^j_v} + \frac{1}{3} \prob{c}{c \in \overline{\mathsf{Sugg}_u},c\in \mathsf{Sugg}_v} \\
& & \quad + ~~
\frac{1}{3} \prob{c}{c \in \mathsf{Sugg}_u,c\in \overline{\mathsf{Sugg}_v}}
+ \frac{1}{3} \prob{c}{c \in \overline{\mathsf{Sugg}_u},c\in \overline{\mathsf{Sugg}_v}}\\
&\le& \sum_{j=1}^3 \frac{|\mathsf{Sugg}^j_u\cap \mathsf{Sugg}^j_v|}{k} +
\frac{|\overline{\mathsf{Sugg}_u}|}{3k} + \frac{|\overline{\mathsf{Sugg}_v}|}{3k}
\end{eqnarray*}
\noindent We can thus bound the expected number of
miscolored edges in the coloring $\chi_G$ as follows.
\begin{eqnarray*}
\mathbb{E} \Biggl[ \sum_{(u,v) \in E(G)} 1_{\chi_G(u)=\chi_G(v)}\Biggr] &\le& \sum_{uv \in E} \Biggl[  \biggl( \sum_{j=1}^3 \frac{|\mathsf{Sugg}^j_u\cap
      \mathsf{Sugg}^j_v|}{k} \biggr)+ \frac{|\overline{\mathsf{Sugg}_u}|}{3k} +
    \frac{|\overline{\mathsf{Sugg}_v}|}{3k} \Biggr] \\
&\le& \frac{1}{k} \Bigl( C^{between} +
\sum_{u\in V(G)} \frac{d_u}{3} |\overline{\mathsf{Sugg}_u}|\Bigr) \quad \mbox{(using (\ref{eq:c-between}))} \\
&\le& \frac{1}{k} \Bigl(C^{between} +
      \sum_{u\in V(G)} \frac{2 d_u}{3}C^{within}_u  \Bigr) = \frac{C^{total}}{k}
\end{eqnarray*}
This implies that there exists a $3$-coloring of $G$ for which the
number of improperly colored edges in $G$ is at most
$\frac{C^{total}}{k}$. Therefore if $H$ has a $k$-coloring which
improperly colors at most a total weight $\mu k^2 m$ of edges, then
there is a $3$-coloring of $G$ which colors improperly at most a fraction
$\frac{\mu k^2 m}{k m} = \mu k$ of its edges.
\end{proof}

This completes the proof of Theorem~\ref{thm:kcolor-NP-hard} when $k$ is divisible by $3$. The other cases are easily handled by adding $k \mod 3$ extra nodes connected to all vertices by edges of suitable weight. See Appendix~\ref{app:k-mod-3} for details.
\end{proof}
\begin{remark}[Comparison to \cite{KKLP}]
\label{rem:KKLP}
The reduction of Kann {\it et al}~\cite{KKLP} converts
  an instance $G$ of {\sf MaxCut}\ to the instance $G' = K'_{k/2} \otimes G$
  of \maxcut{k}. Edge weights are picked so that the optimal $k$-cut
  of $G'$ will give a set $S_u$ of $k/2$ different colors to all
  vertices in each $k/2$ clique $(u,i)$, $1 \le i \le k/2$. This
  enables converting a $k$-cut of $G'$ into a cut of $G$ based on
  whether a random color falls in $S_u$ or not. In the $3$-coloring
  case, we make $3$ copies of $G'$ in an attempt to enforce three ``translates''
  of $S_u$, and use those to define a $3$-coloring from a
  $k$-coloring. But we cannot ensure that each $k$-clique is properly
  colored, so these translates might overlap and a more careful
  soundness analysis is needed.
\end{remark}
\section{Conditional Hardness Results for \maxcolor{k}}
We will first review the (exact) \dtoone{2} Conjecture, and then
construct a noise operator, which allows us to preserve
$k$-colorability. Then we will bound the stability of coloring
functions with respect to this noise operator. In the last section, we
will give a PCP verifier which concludes the hardness result.
\vspace{-1ex}
\subsection{Preliminaries}
We begin by reviewing some definitions
and \dtoone{d} conjecture.
\begin{definition}
  An instance of a bipartite Label Cover problem represented as
  $\mathcal{L} = (U, V, E, W, R_U, R_V, \Pi)$ consists of a weighted
  bipartite graph over node sets $U$ and $V$ with edges $e=(u,v) \in
  E$ of non-negative real weight $w_e \in W$. $R_U$ and $R_V$ are
  integers with $1\le R_U\le R_V$. $\Pi$ is a collection of projection
  functions for each edge:
  $\Pi=\{\pi_{vu}:\intset{R_V}\rightarrow\intset{R_U}\big|u\in U, v\in
  V\}$. A labeling $\ell$ is a mapping $\ell:U\rightarrow
  \intset{R_U}$, $\ell:V\rightarrow \intset{R_V}$. An edge $e=(u,v)$ is
  satisfied by labeling $\ell$ if $\pi_e(\ell(v))=\ell(u)$. We define the
  value of a labeling as sum of weights of edges satisfied by this
  labeling normalized by the total weight. $\mathsf{Opt}(\mathcal{L})$ is the
  maximum value over any labeling.
\end{definition}
\begin{definition}
  A projection $\pi:\intset{R_V}\rightarrow \intset{R_U}$ is called
  \dtoone{d} if for each $i \in \intset{R_U}$, $|\pi^{-1}(i)|\le
  d$. It is called {\em exactly} \dtoone{d} if $|\pi^{-1}(i)|= d$ for
  each $i \in \{1,2,\dots,R_U\}$.
\end{definition}
\begin{definition}
  A bipartite Label-Cover instance $\mathcal{L}$ is called $d$-to-$1$
  Label-Cover if all projection functions, $\pi \in \Pi$ are
  $d$-to-$1$.
\end{definition}
\begin{conjecture}[\dtoone{d} Conjecture \cite{Khot02}]
  For any $\gamma > 0$, there exists a \dtoone{d} Label-Cover instance
  $\mathcal{L}$ with $R_V=R(\gamma)$ and $R_U \le d R_V$ many labels
  such that it is NP-hard to decide between two cases,
  $\mathsf{Opt}(\mathcal{L}) = 1$ or $\mathsf{Opt}(\mathcal{L})\ge \gamma$.  Note that
  although the original conjecture involves \dtoone{d} projection
  functions, we will assume that it also holds for \emph{exactly}
  \dtoone{d} functions (so $R_U = d R_V$), which is the case in
  \cite{DMR}.
\label{conj:dtoone}
\end{conjecture}
Using the reductions from \cite{DMR}, it is possible to show that the
above conjecture still holds given that the graph $(U\cup V, E)$ is
left-regular and unweighted, i.e., $w_e = 1$ for all $e\in E$.
\subsection{Noise  Operators}
For a positive integer $M$, we will denote by $[M]$ the set
$\{0,1,\dots,M-1\}$.  We will identify elements of $[M^2]$ with $[M]
\times [M]$ in the obvious way, with the pair $(a,b) \in [M]^2$
corresponding $a + M b \in [M^2]$.
\begin{definition}
  \label{def:markovop}
  A Markov operator $T$ is a linear operator which maps probability
  measures to other probability measures. In a finite discrete
  setting, it is defined by a stochastic matrix whose $(x,y)$'th entry
  $T(x \rightarrow y)$ is the probability of transitioning from $x$ to
  $y$. Such an operator is called symmetric if $T(x \rightarrow y) =
  T(y \rightarrow x) = T(\xiffy{x}{y})$.
\end{definition}
\begin{definition}
\label{def:beckner}
  Given $\rho\in [-1,1]$, the Beckner noise operator, $T_{\rho}$ on
  $[q]$ is defined by as $T_\rho(x\rightarrow x) = \frac{1}{q} +
  \left(1-\frac{1}{q}\right) \rho$ and $T_\rho(x\rightarrow y) =
  \frac{1}{q}(1-\rho)$ for any $x\neq y$.
\end{definition}
\begin{observation}
  All eigenvalues of the operator $T_{\rho}$ are given by $1 =
  \lambda_0(T_\rho)\ge \lambda_1(T_\rho) = \ldots =
  \lambda_{q-1}(T_\rho) = \rho$. Any orthonormal basis
  $\alpha_0,\alpha_1,\ldots,\alpha_{q-1}$ with $\alpha_0$ being
  constant vector, is also a basis for $T_{\rho}$.
\end{observation}
\begin{lemma}
  \label{lemma:T}
For an integer $q \ge 6$, there exists a symmetric Markov operator $T$ on $[q]^2$ whose diagonal entries are all $0$ and with
 eigenvalues  $1=\lambda_0\ge
  \lambda_1\ge \ldots\ge \lambda_{q^2-1}$ such that the {\em spectral radius}
  $\rho(T) = \max\{|\lambda_1|,|\lambda_{q^2-1}|\}$ is at most
  $\frac{4}{q-1}$
\end{lemma}
\begin{proof}
  Consider the symmetric Markov operator $T$ on $[q]^2$ such that, for
  $x=(x_1,x_2), y=(y_1,y_2)\in [q]^2$,
  \[T(\xiffy{x}{y}) =
  \begin{cases}
    \alpha &\mbox{if $\{x_1,x_2\}\cap\{y_1,y_2\}=\emptyset$
      and $x_1\neq x_2, y_1\neq y_2$,}\\
    \beta &\mbox{if $x_1\not\in\{y_1,y_2\}$
      and $x_1= x_2, y_1\neq y_2$,}\\
    \beta &\mbox{if $y_1\not\in\{x_1,x_2\}$
      and $x_1\neq x_2, y_1= y_2$,}\\

    0 & \mbox{else,}
  \end{cases}\] where $\alpha = \frac{1}{(q-1)(q-3)}$ and
  $\beta = \frac{1}{(q-1)(q-2)}$. It is clear that $T$ is symmetric and
  doubly stochastic.

  To bound the spectral radius of $T$, we will bound the second
  largest eigenvalue $\lambda_1(T^2)$ of $T^2$. Notice that $T^2$ is
  also a symmetric Markov operator. Moreover $\lambda_i(T^2) =
  \lambda^2_i(T)$, therefore $\lambda_1(T^2) \ge
  \max(\lambda^2_1(T),\lambda^2_{q^2-1}(T)) \ge \rho(T)^2$.

  Notice that $T^2(\xiffy{x}{y})>0$ for all pairs $x,y\in[q]^2$.
  Consider the variational
  characterization of $1 - \lambda_1(T^2)$ \cite{Sinclair92}:
  \[\min_\psi \frac{\sum_{x,y}
    (\psi(x) - \psi(y))^2 \pi(x) T^2(\xiffy{x}{y})}{\sum_{x,y}
    (\psi(x)-\psi(y))^2 \pi(x)\pi(y)} \ge \min_\psi \min_{x,y}
  \frac{\pi(x) (\psi(x) - \psi(y))^2
    T^2(\xiffy{x}{y})}{(\psi(x)-\psi(y))^2 \pi(x)\pi(y)} = \min_{x,y}
  q^2 T^2(\xiffy{x}{y})\]

  For any two pairs $(x_1,x_2),(y_1,y_2)\in[q]^2$, let $l =
  |[q]\setminus \{x_1,x_2,y_1,y_2\}|$. Then we have
  \begin{eqnarray*}
    T^2(\xiffy{(x_1,x_2)}{(y_1,y_2)}) &=&
  \begin{cases} l (l-1)\beta^2\ge (q-2)(q-3)\beta^2 &
    \mbox{if $x_1=x_2$ and $y_1=y_2$,}\\
    l (l-1) \alpha \beta\ge (q-3)(q-4)\alpha\beta &
    \mbox{if $x_1\neq x_2$ and $y_1=y_2$,}\\
    l (l-1) \alpha \beta\ge (q-3)(q-4)\alpha\beta &
    \mbox{if $x_1=x_2$ and $y_1\neq y_2$,}\\
    l (l-1) \alpha^2 + l \beta^2 \ge (q-4)(q-5)\alpha^2 + (q-4)\beta^2 &
    \mbox{if $x_1\neq x_2$ and $y_1\neq y_2$.}
  \end{cases}\\
  &\ge& \frac{(q-5) (q-4)}{(q-3)^2 (q-2) (q-1)}
  \end{eqnarray*}
  So $\rho(T)\le \sqrt{\lambda_1(T^2)} \le \sqrt{1-\frac{(q-5) (q-4)
      q^2}{(q-3)^2 (q-2) (q-1)}} \le \frac{3}{q} +
      \frac{8}{q^2}\le\frac{4}{q-1}$ for $q\ge 6$.
\end{proof}
\subsection{$q$-ary Functions, Influences, Noise stability}
We define inner product on space of functions from $[q]^N$ to
$\mathbb{R}$ as $\langle f,g \rangle = \expct{x\sim[q]^N}{f(x)
  g(x)}$. Here $x \sim \mathcal{D}$ denotes sampling from distribution
$\mathcal{D}$ and $\mathcal{D} = [q]^N$ denotes the uniform
distribution on $[q]^N$.

Given a symmetric Markov operator $T$ and
$x=(x_1,\ldots,x_N)\in[q]^N$, let $T^{\otimes N} x$ denote the product
distribution on $[q]^N$ whose \texsup{$i$}{th} entry $y_i$ is
distributed according to $T(\xiffy{x_i}{y_i})$. Therefore $T^{\otimes
  N} f (x) = \expct{y \sim T^{\otimes N} x} { f(y)}$.

\begin{definition}
  Let $\alpha_0,\alpha_1,\ldots,\alpha_{q-1}$ be an orthonormal basis
  of $\mathbb{R}^q$ such that $\alpha_0$ is all constant vector. For
  $x\in[q]^N$, we define $\alpha_x \in \mathbb{R}^{q^N}$ as
  \[ \alpha_x = \alpha_{x_1}\otimes \ldots \otimes \alpha_{x_N}.\]
\end{definition}

\begin{definition}[Fourier coefficients]
  For a function $f:[q]^N\to\mathbb{R}$, define $\hat{f}(\alpha_x) =
  \langle f, \alpha_x \rangle$.
\end{definition}
\begin{definition}
  Let $f:[q]^N\rightarrow \mathbb{R}$ be a function. The
  \emph{influence} of \texsup{i}{th} variable on $f$, $\ensuremath{\mathsf{Inf}}_i(f)$ is
  defined by
  \[ \ensuremath{\mathsf{Inf}}_i(f) =
      \expct{}
            {\var{}{f(x)|x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_N}}\]
  where $x_1,\ldots,x_N$ are uniformly distributed.
  Equivalently, $\ensuremath{\mathsf{Inf}}_i (f) = \sum_{x:x_i\neq 0} \hat{f}^2(\alpha_x)$.
\end{definition}
\begin{definition}
  Let $f:[q]^N\rightarrow \mathbb{R}$ be a function. The
  \emph{low-level influence} of \texsup{i}{th} variable of $f$ is
  defined by
  \[ \ensuremath{\mathsf{Inf}}_i^{\le t}(f) = \sum_{x:x_i\neq 0,\ |x|\le t} \hat{f}^2(\alpha_x).\]
\end{definition}
\begin{observation}
  For any function $f$, $\sum_i \ensuremath{\mathsf{Inf}}_i^{\le t}(f) = \sum_{x:|x|\le t}
  \hat{f}^2(\alpha_x) |x|\le t \sum_x \hat{f}^2(\alpha_x) = t
  \|f\|^2_2$. If $f:[q]^N\to[0,1]$, then $\|f\|^2_2 \le 1$, so $\sum_i
  \ensuremath{\mathsf{Inf}}_i^{\le t}(f) \le t$.
\end{observation}
\begin{definition}[Noise stability]
  Let $f$ be a function from $[q]^N$ to $\mathbb{R}$, and let $-1\le
  \rho\le 1$. Define the \emph{noise stability} of $f$ at $\rho$ as
  \[ \mathbb{S}_{\rho}(f) = \langle f, T^{\otimes n}_\rho f\rangle =
  \sum_x \rho^{|x|} \hat{f}^2_i(\alpha_x)\] where $T_{\rho}$ is the
  Beckner operator as in Definition~\ref{def:beckner}.
\end{definition}
A natural way to think about a $q$-coloring function is as a
collection of $q$-indicator variables summing to $1$ at every point.
To make this formal:
\begin{definition}
  Define the unit $q$-simplex as $\Delta_q = \{(x_1,\ldots,x_{q}) \in
  \mathbb{R}^q \mid \sum x_i = 1, x_i \ge 0\}$.
\end{definition}
\begin{observation}\label{obs:infbndsimplex}
  For positive integers $Q,q$ and any function
  $f=(f_1,\ldots,f_q):[Q]^N\to \Delta_q$, $\sum_i \ensuremath{\mathsf{Inf}}^{\le t}_i(f) =
  \sum_i \sum_j \ensuremath{\mathsf{Inf}}^{\le t}_i (f_j) \le t \sum_j \|f_j\|^2 \le t$.
\end{observation}
We want to prove a lower bound on the stability of $q$-ary functions
with noise operators $T$. The following proposition is generalization
of Proposition 11.4 in \cite{KKMO} to general symmetric Markov
operators $T$ with small spectral radii. Its proof appears in Appendix~\ref{app:noise-stab}.
\begin{proposition}
\label{prop:stab}
For integers $Q,q \ge 3$, and a symmetric Markov operator $T$ on $[Q]$
with spectral radius $\rho(T)\le \frac{c}{q-1}$, for some $c>0$, there
is a small enough $\delta=\delta(q)>0$ and $t=t(q)>0$ such that for
any function $f=(f_1,\ldots,f_q):[Q]^N\rightarrow \Delta_q$ with
$\ensuremath{\mathsf{Inf}}_i^{\le t} (f)\le \delta$, for all $i$, satisfies
  \[ \sum_{j=1}^q \langle f_j, T^{\otimes N} f_j\rangle \ge 1/q - 2 c \ln q/q^2 - C
  \ln\ln q /q^2\] for some universal constant $C<\infty$.
\end{proposition}
\begin{definition}[Moving between domains]
  For any $x=(x_1,\ldots,x_{2 N}) \in [q]^{2 N}$, denote $\overline{x}
  \in [q^2]^N$ as
\[ \overline{x} = ((x_1, x_2),\ldots,(x_{2 N-1},x_{2 N})) \ . \]
  Similarly for $y=(y_1,\ldots,y_{N}) \in [q^2]^{N}$, denote
  $\underline{y} \in [q]^{2 N}$ as
  \[ \underline{y} = (y_{1,1}, y_{1,2},\ldots,y_{N,1},y_{N,2}),\]
  where $y_i = y_{i,1} + y_{i,2} q$ such that $y_{i,1},y_{i,2}\in
  [q]$.  For a function $f$ on $[q]^{2 N}$, define $\overline{f}$ on
  $[q^2]^N$ as $\overline{f}(y) = f(\underline{y})$.

\end{definition}
\noindent
The relationship between influences of variables for functions $f$ and
$\overline{f}$ are given by the following claim (Claim 2.7 in
\cite{DMR}).
\begin{claim}\label{claim:infrel}
  For any function $f:[q]^{2N}\rightarrow \mathbb{R}$,
  $i\in\intset{N}$ and any $t\ge 1$,
  $\ensuremath{\mathsf{Inf}}_i^{\le t}(\overline{f}) \le \ensuremath{\mathsf{Inf}}_{2i-1}^{\le 2t}(f) +
  \ensuremath{\mathsf{Inf}}_{2i}^{\le 2t} (f)$.
\end{claim}

\vspace{-1ex}
\subsection{PCP Verifier for \maxcolor{k}}
This verifier uses ideas similar to the \maxcut{k} verifier given in
\cite{KKMO} and the $4$-coloring hardness reduction in \cite{DMR}. Let $\mathcal{L} = (U,V,E,R,2 R,\Pi)$ be a \dtoone{2}
bipartite, unweighted and left regular Label-Cover instance as in
Conjecture \ref{conj:dtoone}. Assume the proof is given as the Long
Code over $[k]^{2 R}$ of the label of every vertex $v \in V$. Below
for a permutation $\sigma$ on $\intset{n}$ and a vector $x \in \mathbb{R}^n$, $x\circ \sigma$ denotes $(x_{\sigma(1)},x_{\sigma(2)},\cdots,x_{\sigma(n)})$. For a function $f$ on $\mathbb{R}^n$, $f \circ \sigma$ is defined as $f \circ \sigma(x) = f(x \circ \sigma)$.

\begin{itemize}
\vspace{-1ex}
\itemsep=0ex
\item Pick $u$ uniformly at random from $U$, $u\sim U$.
\item Pick $v,v'$ uniformly at random from $u$'s neighbors. Let $\pi,
  \pi'$ be the associated projection functions, $\chi_v, \chi_{v'}$ be
  the (supposed) Long Codes for the labels of $v, v'$ respectively.
\item Let $T$ be the Markov operator on $[k]^2$ given in Lemma
  \ref{lemma:T}. Pick $x \sim [k^2]^R$ and $y \sim T^{\otimes R}
  x$. Let $\sigma_v,\sigma_{v'}$ be two permutations of $\intset{2 R}$ such
  that $\pi(\sigma^{-1}_v(2i-1)) = \pi(\sigma^{-1}_v(2i)) =
  \pi'(\sigma_{v'}^{-1}(2i-1)) = \pi'(\sigma_{v'}^{-1}(2i))$ (both $\pi$ and
  $\pi'$ are exactly \dtoone{2}, so such permutations exist).
\item Accept iff $\chi_v\ \circ\ \sigma_v (\underline{x}) $ and
  $\chi_{v'} \ \circ\ \sigma_{v'}(\underline{y})$ are different.
\end{itemize}
The proofs of the following two lemmas appear in Appendix~\ref{app:kcol-PCP}.
\begin{lemma}[Completeness]
\label{lem:kcol-compl}
  If the original \dtoone{2} Label-Cover instance $\mathcal{L}$ has a
  labeling which satisfies all constraints, then there is a proof
  which makes the above verifier always accept.
\end{lemma}
\begin{lemma}[Soundness]
\label{lem:kcol-soundness}
  There is a constant $C$ such that, if the above verifier passes with
  probability exceeding $1-1/k+O(\ln k/k^2)$, then there
  is a labeling of $\mathcal{L}$ which satisfies $\gamma'=\gamma'(k)$
  fraction of the constraints independent of label set size $R$.
\end{lemma}
Note that our PCP verifier makes ``$k$-coloring'' tests. By the standard conversion from PCP verifiers to CSP hardness, and Remark~\ref{rem:wt-unwt} about conversion to unweighted graphs with the same inapproximability factor, we conclude the main result of this section by combining Lemmas \ref{lem:kcol-compl} and \ref{lem:kcol-soundness}.
\begin{theorem}
  For any constant $k \ge 3$, assuming \dtoone{2} Conjecture, it is
  NP-hard to approximate \maxcolor{k} within a factor of $1-1/k+O(\ln
  k/k^2)$.
\end{theorem}

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


text:
Reform of court system
[b]The judicial system of Belarus is changing, with general and economic courts uniting under a uniform system of general jurisdiction courts from 1st January 2014. The Supreme Court is to head the new system of economic courts from the regions and Minsk, which will continue to function under their current specialisations. The President signed the 2nd Edict and the Decree formalising the reform at the end of November. [/b]
The judicial system of Belarus is changing, with general and economic courts uniting under a uniform system of general jurisdiction courts from 1st January 2014. The Supreme Court is to head the new system of economic courts from the regions and Minsk, which will continue to function under their current specialisations. The President signed the 2nd Edict and the Decree formalising the reform at the end of November.
The latest reform of the law-enforcement sphere involves advocacy bodies and the powers of the Ministry of Internal Affairs and the Prosecutor's Office. Last year, the Investigative Committee launched, with success; it discussed the need for an improved court system, including the operation of judicial bodies, at its recent November session. The President of Belarus emphasised that reform has a clear goal; it is not change for its own sake or for the purpose of copying the ideas of neighbouring countries.
Naturally, the well-being of society and individuals relies largely on the state of affairs in this sensitive sphere. Many significant questions are solved in court, deciding the destinies of people, the financial health of enterprises and mediating in cases of dispute. Though noticeable improvements have appeared in the work of courts in recent years, a number of problems remain. Judges' workloads have decreased and new methods are improving the quality, efficiency and availability of justice but we lack uniformity of application by general and current economic courts. Moreover, it can be difficult to identify the jurisdiction of specific disputes. Multiplicity of processes and differences in the application of justice complicate matters: general courts apply a classical form of law while economic courts use arbitration. The majority of Post-Soviet countries have already included economic courts in the general judicial system.
The improvement of the judicial system in Belarus was discussed several years ago but found form in 2013, with the creation of an interdepartmental working group considering how best to unify general and economic courts. The issue is being analysed from the point of view of judges at all levels, led by the Deputy Head of the Presidential Administration, Valery Mitskevich. He notes that transformations will not require additional staff or budgetary expenditure. He tells us, "We can embody the best and most progressive achievements of general and economic courts, accelerating the use of information technologies into our activity and taking a unified judicial system to a totally new, higher level of functioning."
The Chairman of the Supreme Court, Mr. Sukalo, notes that an overwhelming majority of countries, including post-Soviet states and those of the Customs Union, are developing specialised courts within a uniform judicial system — with one superior court co-ordinating and directing judicial policy and judiciary practice. He said, "Such is the construction of judicial systems in most European states: a unified system of courts of general jurisdiction and economic courts, with one supreme court at the head, one plenum and one presidium. This, certainly, would form a uniform, clear, distinct judiciary, with uniform, classical forms of justice and uniform status for all judicial authorities." He believes that change would be supported by the whole judiciary.
Organisational, material and personnel approaches are also under review. Until recently, economic courts solved these issues independently, while general courts were supported by the Ministry of Justice at regional and district level. Now, the judicial administration will be unified and all corresponding powers concerning courts of general jurisdiction transferred from the Ministry of Justice to the Supreme Court. The latter will be completely responsible for all questions concerning the functioning of the system, from delivery of justice to the material and technical support of courts. Chairs of regional courts will have additional duties connected with such matters as repair of premises and disposition of funds, solving financial and personnel questions more effectively.
Another innovation involves bodies of enforcement being subordinate to government executive bodies. From 1st January 2014, enforcement will be co-ordinated by the Ministry of Justice, served by central administrative boards of justice of regional executive committees. Local authorities will oversee regional governing bodies of enforcement and regional departments of enforcement.
Court — one for all
From 1st July 2014, another stage of reform will begin, with the Military Collegium of the Supreme Court and military courts abolished. Their powers will pass to regional, district and city courts of general jurisdiction, bringing a uniform approach for civilians and the military. This will exclude the possibility of departmental influence, ensuring independence and impartiality of judges.
The issue of military courts arose during discussion of judicial reform, with the President clear in his opinion, saying, "I think it best to act in the same way with military courts as we do with economic courts. Why do we have special structures for those wearing shoulder markings? A major should not try a major. All should be in one set of hands: one court for all citizens."
Our past left a legacy of specialised military courts, which have already been abolished in most former Soviet states. Common practise has also been to reduce armies and the number of military crimes. Belarus is now following suit, with its 30 military judges receiving civil status and working in regional and district courts. The President emphasises the rationality of the move, explaining, "Our country is compact, so having several structures in the same direction is not rational. We can govern this system from one centre, and will soon see cohesion, acting in one direction. There will be no disagreements in decision-making."
According to Mr. Sukalo, reform of the judicial system will be smooth and won't affect the process of justice.
By Lilia Khlystun


text:
The 1907 All-Ireland Senior Hurling Championship Final was the 20th All-Ireland Final and the culmination of the 1907 All-Ireland Senior Hurling Championship, an inter-county hurling tournament for the top teams in Ireland. The match was held at Fraher Field, Dungarvan, on 21 June 1908, between Kilkenny, represented by a club side from Mooncoin, and Cork, represented by a club side from Dungourney. The Munster champions lost to their Leinster opponents on a score line of 3-12 to 4-8.

The game was regarded as the best All-Ireland decider up to that point.

Match details

1
All-Ireland Senior Hurling Championship Finals
Cork county hurling team matches
Kilkenny GAA matches
June 1908 sports events

redpajama_set_name: RedPajamaWikipedia


On Monday, June 26, 2017, at 9:51 a.m., a resident on the 34000 block of Eucalyptus Terrace called 9-1-1 to report a verified alarm activation and in-progress burglary at his residence. The caller had received the activation while he was out on a bike ride with his children. He advised dispatchers that he had a live video feed of a male trying to gain entry into his residence through a back door.
Several officers and a sergeant arrived on scene within in minutes of the call being received. Shortly after arriving on scene, one officer advised that she observed a male exit the rear of the residence and flee towards the front of the house. A perimeter was quickly established and additional patrol units and detectives responded. Fremont Fire was called for UAS (drone) support and a CHP helicopter responded from Napa.
Upon the scene being locked down, residents in the immediate area were notified and officers began a yard to yard search. Officers at the home of the reporting party found the rear glass sliding door shattered and confirmed a burglary had been committed.
At 11:18 a.m., during the yard to yard search, officers found the suspect hiding in a backyard on the 34200 block of Tupelo St. Demetrius Wells, a 19 year old Newark resident, was taken into custody and arrested for burglary. Arraignment is scheduled for June 28, 2017 at the Fremont Hall of Justice.
The loss from the burglary was recovered and returned to the victim.