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Home Archive September 2010 Issue Issue Content Bringing Energy Security to East Central Europe: Regional Cooperation Is the Key
Bringing Energy Security to East Central Europe: Regional Cooperation Is the Key
Wednesday, 29 September 2010 00:00 Keith Smith
Much of the blame for the lack of greater Central and Southeastern Europe (CSEE) energy cooperation can be laid at the feet of the large EU member states who seek closer energy ties with Russia, along with the lack of support from the European Commission, which has so far hindered regional cooperation.
Nevertheless, by far the biggest obstacle to collaboration and more effective resistance to Russian pressure is the lack of sufficient reform within the CSEE countries themselves. Their vulnerability to energy coercion and questionable agreements with Russian leaders in large measure stem from the lack of transparency in their own governments. In addition, there are regulatory, licensing, and taxation issues that have to be tackled by each of the region's governments in order to effectively implement the funding programs approved by the European Commission. Investment laws have to be adopted in order to attract foreign energy investors—those who follow the best business practices and who bring the most innovative technology into the country. They need to attract foreign firms who intend to stay involved in the long term, and they must not act as future sales agents for nontransparent firms representing Russian interests.
The situation, however, may now be moving in a more positive direction. The intergovernmental group that was announced at the February 24, 2010 Budapest energy summit could be put together quickly and should include the best energy policy specialists from each of the 11 participating states. The group should be given broad authority by each government to implement the best practices in developing domestic and cross-border projects. Total transparency should be required of the group in order to prevent nontransparent or corrupt local business interests from overriding the need for regional energy security. Openness is also needed in order to combat efforts by supplier nations to subvert the goal of greater regional cooperation. It will be a difficult task to harmonize totally the energy activities of 11 nations, but a high degree of combined action should be possible, particularly if it has strong political support from member governments and it is allowed to reach out to international banks and development institutions for technical and financial advice.
Effective use of this intergovernmental group would likely increase the region's influence within the European Commission's Directorate-General for Energy and with the commissioner for energy. Strong and effective support from member governments could persuade Russia's closest energy backers in the European Union (Germany, France, Italy, and Austria) to pay greater attention to investment opportunities and even to the energy security needs of the region. Most importantly, the kind of cooperation envisioned at Budapest would increase the weight of the CSEE countries on a wide range of energy security issues, including the formation of a common energy market and the enforcement of competition and antitrust laws that are now openly flouted by several of the larger states. An intergovernmental commission, however, cannot resolve many of the most important impediments to regional cooperation. Under international law, two countries cannot link together their energy transmission systems without a formal state-to-state agreement. Several countries that want to build gas interconnectors with EU funding have not yet reached agreements with their neighbors. Even Hungary and Romania, which plan to tie part of their pipeline systems together, do not yet have the required intergovernmental agreement, which has caused some delay in completing the interconnector. This is one area in which a nudge from the Directorate-General for Energy could help move things along.
All EU members are already subject to the rules contained in the Energy Charter Treaty. And yet, several member states have taken no action to force their domestic firms to open their pipelines as "common carriers" so that multiple companies would have access to spare capacity. Doing so would increase competition and efficiency, and would ultimately lower prices for the consumer. Unfortunately, several of the larger EU members, such as France and Germany, have resisted both unbundling and a common carrier system, but this should not prevent the CSEE countries from moving ahead on their own.
A stable regulatory and licensing system that covers pipelines, LNG facilities, and nuclear plants, as well as pricing and environmental concerns, is also a must. The challenge is to devise a regulatory system that encourages, rather than stifles, competition and that is transparent without tying projects up with endless political conflict. The United Kingdom's energy regulatory system is a good model for the new democracies, as is the US Federal Energy Regulatory Commission, although the United States has a separate agency for oil and nuclear plants. Both countries appear to strike a reasonable balance between promoting open energy markets and being able to protect the interests of the consumer. Governments should provide a regulatory framework and not try to "manage" the market. All of the countries must avoid the temptation to regulate consumer prices to the benefit of either producer or consumer groups, rather than have them reflect the real market price of the final product.
Although for political reasons many governments are reluctant to publicize negotiations with non-EU or non-European energy suppliers, the European Union has so far refused to require greater transparency. The CSEE states could take on this challenge. It would prevent the smaller and more economically vulnerable countries from being played off against each other by phantom promises of future riches made by supplier states, such as Russia. The history of Russian negotiations with CSEE states regarding participation in the South Stream project and possible benefits to individual countries is a good example of why greater transparency is needed in bilateral discussion involving energy deals. The shrouded negotiation between Croatia and Russia regarding the Druzhba pipeline and the pipeline from the Krk Island LNG facility is a good case in point. Both pipelines are controlled by a Croatian company that is independent of Hungary's MOL. The unspoken goal of Gazprom appears to be one of shutting off competition to Russian oil and South Stream gas by limiting shipments to Central Europe through Adriatic ports. After Gazprom's takeover of the key NIS oil refinery in Serbia, a Croatia-Russia deal would pose a significant danger to CSEE energy security. Although the Croatians deny that they are "breaking ranks" with their CSEE neighbors on energy security, they have agreed to Russian suggestions that they support the South Stream pipeline in return for possible economic benefits, such as insuring that the pipeline would pass through Croatia to the gas storage terminals in eastern Austria. Even if South Stream never materializes (a good possibility), Russia will still be in a strong position through its control of Croatia's pipelines from the Adriatic.
The Croatians are not the only CSEE country to sign up for South Stream while at the same time declaring their support for Nabucco. Austria, Hungary, and Bulgaria have agreed to participate in both pipelines. There is not sufficient gas demand for both pipelines, and it may appear prudent for governments to hedge their bets on which pipeline (assuming one of them is built) can demonstrate that it will be constructed and be the first to bring new gas supplies. Not only are there supply questions regarding Nabucco, but in addition, Turkey's demands for effective control of the gas and its touchy relationship with Azerbaijan has stalled the project. At the same time, Russia's South Stream project shows even fewer signs that it can supply the necessary gas volumes. South Stream's reported $26-billion construction costs stimulate the question whether the project is only being put forward in order to kill off Nabucco. Only Italy appears to claim that it is a serious proposal to supply the region with new gas supplies. Russia also sees South Stream as a means of putting pressure on Ukraine to turn its pipeline system over to Gazprom.
Recently, Russia and Turkey have floated the idea of Gazprom becoming part of the Nabucco project, with Russian gas from the Blue Stream or a possible new Russia-Turkey pipeline contributing a large share of the gas. The difficulty with such an option is that Russia has never been content to play a minority role in any of its pipeline deals. It should be assumed that even if Gazprom were to take a minority and non-blocking share of Nabucco ownership at the beginning, it would, within a short time, start applying pressure on other partners to sell their shares to Gazprom, until it acquired a majority, or at least a blocking position in the consortium.
A look at Baltic cooperation: slow but steady on regional integration
The "energy island" countries of Estonia, Latvia, and Lithuania are nearly 100 percent dependent on Russia for oil and gas imports, reflecting the legacy energy policies of the former Soviet Union. The three Baltic states have been a prime target for Russian energy supply disruptions for the past 20 years. An oil cutoff was used by Moscow in 1990 in an attempt to stifle the region's new independence movements. In 1992, in a futile effort to keep the Russian officer corps stationed in Estonia and Latvia, energy supplies were again shut off in the middle of an especially cold winter. Oil shipments to Lithuania were disrupted nine times between 1997 and 1999 in an attempt to influence negotiations over ownership of an oil refinery. Russian oil shipments through the Druzhba pipeline have now been permanently stopped to Latvia and Lithuania (both EU member states) as a result of these two countries' unwillingness to sell their pipelines, oil ports, and refineries to Russian companies cheaply (or at no cost). Gazprom has succeeded in gaining control of the natural gas companies of all three Baltic states, along with ownership of power plants in Latvia and Lithuania.
Surprisingly, few Western Europeans are aware of the extensive use by Moscow of energy coercion in the Baltic states, even though more than 25 politically motivated disruptions have occurred in the past 20 years. These tactics by the Kremlin have been widely documented but have attracted little attention in Germany, France, Italy, or Austria—Moscow's closest energy partners in Europe. Requirements imposed on the three Baltic states as a condition for EU membership, such as the premature closing of the nuclear power stations in Lithuania, have made it certain and unavoidable that in the short term the three countries will be more, rather than less, dependent on energy supplies from Russia.
Since the mid-1990s, there have been numerous meetings among officials of the Baltic states in an attempt to adopt a coordinated response to Russian pressure and a strategy for enlisting more support from Western Europe. In addition, a Baltic States Energy Forum is held each year in one of the capitals, with representatives from the United States joining officials of the three governments. The goal of the forum is to draw up realistic proposals for greater regional energy cooperation and diversity of energy import sources.
Although there have been repeated meetings at the governmental and nongovernmental level, progress toward greater regional energy cooperation has been slow. Nevertheless, there has been development toward establishing a common electricity market that would cover the three Baltic states. Lithuania has even introduced an electricity exchange, in which 43 percent of the country's domestic electricity use is now traded. Electricity exchanges in the other two countries would advance considerably the prospects for a real common Baltic market for energy. With the help of Sweden and Finland, construction of a Baltic-wide power grid is under way, linking the three Baltic states' electricity systems with those of Poland and the two Nordic countries. Estlink I, the first electricity inter-connector between Estonia and Finland (350 megawatts) was completed in 2007, and a second line (Estlink II) with a capacity of 650 megawatts will likely be completed in 2013–2014. In times of need, the three Baltic states will in the future be able to draw on excess electricity from Finland, particularly when Finland's two newest nuclear power plants go on stream in the next few years. Another significant development has been the signing in early February 2010 of a deal between Estonia and the Norway-based regional electricity bourse called Nord Pool, with the aim of creating a single Nordic and Baltic owner market by 2013. The latest agreement allows Nord Pool to expand its connections with Latvia and Lithuania. Until February, the Baltic–Nord Pool agreement provided only for electricity ties with Estonia manifested through Estlink I.
The European Union has approved funding of a large-capacity electricity inter-connector between Lithuania and Sweden, and the governments of Latvia and Estonia have agreed to the arrangement. It also appears as if the long talked about "power bridge" between Poland and Lithuania will be constructed in the next five years. All of these interconnectors are designed to work in two directions. Nevertheless, there is some uncertainty whether a new nuclear power station will ever be constructed in Lithuania to replace the two reactors that were shut down, the last at the end of 2009. The Baltic states have been discussing for the past five years the common use of power from a new Lithuanian nuclear reactor. The project's delays, however, only add to the energy insecurity in the region. Russia's announcement that it will construct a nuclear power reactor in the small Russian enclave of Kaliningrad and that it will sell excess electricity to neighboring states appears to be an attempt to weaken support for constructing a new Lithuanian reactor.
Within the Baltic states there is growing interest in importing LNG in order to compete with Russia's present gas monopoly. There is the inevitable competition among the three countries regarding where to establish the LNG gasification plant. Lithuania seems to have taken the lead over Latvia, by securing funding in 2008 from the US Trade and Development Agency for an $800,000 feasibility study for a plant located on the country's Baltic west coast. One Polish energy planner, however, voiced a preference for a plant in Latvia, arguing that an LNG receiving plant located at the port of Ventspils could serve to ship the piped gas more easily to the three Baltic markets.
In any case, with the Baltic economies only slowly emerging from a deep recession, it is difficult to see how a proposed LNG plant could secure the necessary financing in the near term. One Latvian businessman has discussed with Qatar the possibility of that country financing an LNG port with the guarantee that the three countries would take enough gas to make the venture profitable. It would be very difficult, however, to persuade the three governments to agree to this formula—and to overcome the inevitable opposition of Gazprom, which at present controls the gas distribution systems in the region. Surprisingly, the Latvian branch of Russia's Itera is supporting the idea of building an LNG plant in Latvia. Since Itera relies on Gazprom for its product, this may simply be an attempt by Gazprom to control any possible competition or to discredit the deal.
Another plan, with better prospects, is the attempt to reach an agreement among the three countries on the construction underground of additional gas storage facilities. Currently, the largest underground storage facility is in Latvia and is controlled by Gazprom. The firm uses gas imported in the summer to supply the St. Petersburg area during periods of peak demand in the winter. A new and larger Baltic-wide storage facility, including a revised financing formula, would have to be agreed upon by the three governments in order to construct new facilities. This project is already receiving funding for a feasibility study by the European Union, and it might receive construction funding as part of the European Union's next energy facilities funding package. Latvia is the only Baltic state that appears to have enough underground cavern space for constructing new storage facilities, although Lithuania is financing a new geological survey that will examine possible storage sites in that country. In any case, the region is under pressure by the European Union to increase significantly its gas storage capacity; in itself, this is a positive step.
One roadblock that has defied resolution for the past 15 years is the demarcation of the sea boundary between Latvia and Lithuania. An intergovernmental agreement has been signed between the two states and has been approved by the Lithuanian parliament. However, its prospects for passage in the Latvian parliament are not good. This is a serious setback, since there is good reason to believe that profitable quantities of oil and gas lie offshore in almost the entire Baltic Sea basin. The Chevron Company of the United States was ready to begin exploratory drilling in the disputed zone in the late 1990s but pulled up stakes after it became apparent that the two countries were far from reaching a sea boundary agreement. Even the possibility of significant tax revenues has not been enough incentive to break the legal and political logjam.
Meanwhile, Russian companies have already been drilling off Kaliningrad, and while the results have not been made public, it is reasonable to assume that significant amounts of gas and oil will be found there. Polish seismic work in the Baltic basin just west of the Kaliningrad offshore economic zone indicates that there are reasonable prospects for commercial exploitation of some oil and larger quantities of gas. Therefore, the failure of the Latvian government to ratify the most recent treaty is a major disappointment to those hoping for some relief from the tight grip that Gazprom has over the gas markets of Latvia and Lithuania.
Lithuania proposed in November 2009, at the NATO Industrial Planning Committee, the establishment of a NATO Energy Security Center of Excellence, with a possible location at a university in Kaunas, Lithuania. Although Lithuania has been floating the idea for over a year, it has not been able to gather enough support from other member states. It is difficult to ascertain how much opposition is coming from countries like Germany that oppose any NATO role in energy security. In order to move ahead, the proposal would require a strong push from the United States and several other key NATO members.
Keith C. Smith is currently a Senior Associate in the Center for Strategic and International Studies' New European Democracies Project in Washington, D.C. He retired from the US Department of State in 2000. From 1997 to 2000, he was US ambassador to Lithuania
Note: this article was adapted from a much longer report published by CSIS in Washington. The full report can be downloaded here.


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redpajama_set_name: RedPajamaCommonCrawl


Doing this will offer your opponent much less time to earn a wise choice. If your opponent gets predictable by repeatedly throwing to the exact same route, think about changing to a defensive back and positioning yourself in which https://prowritingpartner.com/essayshark-com-review/ you think they wish to throw. As a consequence, its easier to stop the counter by playcall frequency. On the reverse side, if you're boxed out with higher strength you've got a better prospect of doing a swim move for passed your man. A normal reaction is to refrain from flying as much as possible. On to the manner Monsta likes to set this up.


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Q: Can quantum black hole be in superposition state? Quantum black hole can have electric charge and angular momentum or spin, I am wondering if a quantum black hole can be described as a probability wavefunction or not? If so can it quantum tunnels too?

A: There is currently no widely-accepted theory of quantum black holes, so any statements about their behavior are highly speculative. I think most physicists assume that a black hole probably does have a wavefunction, but what equation that wavefunction satisfies is not obvious or agreed on. Therefore discussion of scenarios like tunneling is not particularly productive.

redpajama_set_name: RedPajamaStackExchange


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I'm delighted that you are considering joining us. I can assure you that you will be joining a brilliant sixth form college where our students achieve outstanding results. In 2018 ,99.7% of our A Level students passed their exams, with 80% achieving A*- C grades. Our BTEC students did equally well with 100% passing their Diploma course, many with top distinction grades. Our A Level and BTEC results, as measured by ALPs, places the College in the top 10% in the entire country. We are extremely proud of this achievement.
Achieving outstanding results at college is challenging, but we work closely with each and every one of our students to support you in achieving the best possible grades. Our teachers are subject specialists and are passionate about their subjects. They will inspire you and provide the expertise and support you need to achieve and progress to your chosen destination, be that university, an apprenticeship or the world of work. Our excellent support staff also enable you to have a very positive experience at the college.
We are ambitious for you. We want you to leave Lowestoft Sixth Form College proud of what you have contributed and achieved; ready to do more. We will support you when you need it, provide honest feedback about your progress and really care about your goals.
That you want to learn; that you are willing to work hard and invest the time and effort in your future.


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Business, Financial & Legal
Thieves break into the late Steve Jobs' home, steal over $60,000 worth of goods.
Anthony Garreffa
@anthony256
Published Aug 15, 2012 2:18 AM CDT
0 minutes & 47 seconds read time
The late Steve Jobs' home was broken into recently, where thieves stole more than $60,000 worth of "computers and personal items" from the former CEO of Apple, reports the San Jose Mercury News. Jobs' house is on the 2100 block of Waverley Street in Palo Alto, and was broken into on July 17.
It was on August 2, however, that police arrested Kariem McFarlin, a 35-year-old man from Alameda. Fast forward to August 7, McFarlin was arraigned on one count of residential burglary and selling stolen property. McFarlin's bail has been set at a hefty $500,000, with the man potentially facing a sentence as long as seven years and eight months, this includes an extra year for "excessive taking of property" according to Santa Clara County Deputy District Attorney Tom Flattery.
Flattery hasn't yet said if the stolen property included some of Jobs' posthumous possessions or if it instead belonged to other members of his family. Police and DA's office have kept the mouth's sealed on details of the case, but Flattery does hint that McFarlin didn't know that the house he was breaking into belonged to Steve Jobs, the attorney says "the best we can tell is it was totally random".
NEWS SOURCE:electronista.com
Anthony joined the TweakTown team in 2010 and has since reviewed 100s of graphics cards. Anthony is a long time PC enthusiast with a passion of hate for games built around consoles. FPS gaming since the pre-Quake days, where you were insulted if you used a mouse to aim, he has been addicted to gaming and hardware ever since. Working in IT retail for 10 years gave him great experience with custom-built PCs. His addiction to GPU tech is unwavering.
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Q: DSpace with links to social media At the DSpace repositories of Massey University and University of Waikato in New Zealand there are buttons to share the link to a document in the repository with Twitter, Facebook and other social media.
Also have both sites a Window with Twitter tweets.
I hope someone of you wants to explain how both functionalities are added to the  DSpace Mirage 2 theme.

A: We use the addThis service with our DSpace installation: http://www.addthis.com/
The service allows you to customize the share options for your site.
For Mirage2, we added the following javascript to Mirage2/xsl/core/page-structure.xsl to add the service to our site.  The link will contain a unique id for your configuration.

We also added some custom CSS to Mirage2/styles/_style.scss
.container {
  padding-left: 50px;
  padding-right: 50px;
}

You can preview this on our repository site: https://repository.library.georgetown.edu/

A: terrywb's answer covers the share button aspect - that is exactly the service used at the University of Waikato's repository that Peter linked to. We add the JavaScript in navigation.xsl not page-structure.xsl, and we also add

just after the opening tag for the #ds-options div, in line with the AddThis embedding instructions.
As to the tweets on the home page, refer to the Twitter embedded timeline instructions. I believe we add a placeholder div via news-xmlui.xml that then triggers a template in theme.xsl that pulls in the JavaScript and generates the required a element - example below is from the Twitter documentation.

Tweets by @TwitterDev

A: If you want to avoid services like AddThis tracking your users or just want to keep it simple, you can use simple share links. Add to dspace-xmlui-mirage2/src/main/webapp/xsl/aspect/artifactbrowser/item-view.xsl this template:
<xsl:template name="share-section">

<i18n:text>xmlui.mirage2.itemSummaryView.ShareSection</i18n:text>

        <xsl:variable name="url">
            <xsl:value-of select="encoder:encode(confman:getProperty('dspace.baseUrl'))" />
            <xsl:value-of select="encoder:encode('/handle/')" />
            <xsl:value-of select="encoder:encode(substring-after(/mets:METS/@OBJID,'handle/'))" />
        </xsl:variable>

</xsl:template>

Here you can find more share links for other social networks.
Then, you have to call the template from where you want it to appear. For example, we add it to the first column of the item metadata, at the <xsl:template match="dim:dim" mode="itemSummaryView-DIM"> template of the same file.
We got the icons from Expando (another option for sharing with social networks, and it's free software). This CSS is to show the icons:
.social-network-icon {
  background-image: url("../images/social-networks-icons.png");
  background-repeat: no-repeat;
  background-origin: 0;
  display: inline-block;
  height: 16px;
  margin-left: 4px;
  vertical-align: middle;
  width: 16px;
}

.social-network-icon.facebook {
  background-position: -0*16px 0;
}

.social-network-icon.google-plus {
  background-position: -2*16px 0;
}


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Q: Parsing JSON with PHP multiple arrays to MYSQL table stackoverflow newbie here. I've been reading through similar articles trying to piece together the code that I'm after. Fairly new to PHP and JSON, so here goes.
I'm trying to split some JSON data with the end goal of storing the data as two separate columns in a MySQL table. The JSON data looks like the below -
{
"request": {
    "command": "series",
    "series_id": "PET.RBRTE.A"
},
"series": [{
    "series_id": "PET.RBRTE.A",
    "name": "Europe Brent Spot Price FOB, Annual",
    "units": "Dollars per Barrel",
    "f": "A",
    "unitsshort": "$\/bbl",
    "description": "Europe Brent Spot Price FOB",
    "copyright": "Thomson-Reuters",
    "source": "Thomson-Reuters",
    "start": "1987",
    "end": "2015",
    "updated": "2016-05-16T16:56:05-0400",
    "data": [
        ["2015", 52.32],
        ["2014", 98.97],
        ["2013", 108.56],
        ...
    ]
}]

This JSON data is available direct via a URL so I'm using $json = file_get_contents and $array = json_decode($json, true) to get the data initially into an array.
Dumping the array var_dump($array) shows it as the following -
Array
(

[request] => Array
    (
        [command] => series
        [series_id] => PET.RBRTE.A
    )

[series] => Array
    (
        [0] => Array
            (
                [series_id] => PET.RBRTE.A
                [name] => Europe Brent Spot Price FOB, Annual
                [units] => Dollars per Barrel
                [f] => A
                [unitsshort] => $/bbl
                [description] => Europe Brent Spot Price FOB
                [copyright] => Thomson-Reuters
                [source] => Thomson-Reuters
                [start] => 1987
                [end] => 2015
                [updated] => 2016-05-16T16:56:05-0400
                [data] => Array
                    (
                        [0] => Array
                            (
                                [0] => 2015
                                [1] => 52.32
                            )

                        [1] => Array
                            (
                                [0] => 2014
                                [1] => 98.97
                            )

                        [2] => Array
                            (
                                [0] => 2013
                                [1] => 108.56
                            )
                        [3] ...
                    )

            )

    )

All good so far. I then wanted to obtain the sub-array data with the data array series, of which there are 29 in this particular JSON dataset. [0] within each sub-array is a date (year) and [1] is a value associated with that year.
The closest I've been able to get to this data is with the code below -
$array_data = implode(",", $array['series'][0]['data'][0]);

Which when using echo '<pre>'; print_r(($array_data)); echo '</pre>'; gives me 2015,52.32, but what I'd like to get is the result for each of the sub-arrays as their own variable that I can then use to insert into MySQL, for example -
2015,52.32
2014,98.97
2013,108.56
...,...

I would then use an insert statement to add the first value into a datetype column and the second value into a decimal column.
I feel I'm going about this the long way round, so if anyone has suggestions on how to improve this to get the end result, I would be most grateful. Thanks for the help.

A: Use a for loop:
$dataArray = $array['series'][0]['data'];
for ($i = 0; $i < count($dataArray); $i++) {
  $implosion = $dataArray[$i]);
}

If you want a dictionary:
$dataArray = $array['series'][0]['data'];
$dict = [];

for ($i = 0; $i < count($dataArray); $i++) {
  $dict[dataArray[$i][0]] = $dataArray[$i][1];
}

Giving:
Array
(
[2015] => 52.32
...
)

EDIT For separate loops:
$dataArray = $array['series'][0]['data'];
$dates = [];
$values = [];

for ($i = 0; $i < count($dataArray); $i++) {
  array_push($dates, $dataArray[$i][0]);
  array_push($values, $dataArray[$i][1]);
}

$dateImplosion = implode(", ", $dates);
echo $dateImplosion;


I am in San Diego for the annual convention of the National Lesbian and Gay Journalists Association. I hadn't planned to come this year but on Thursday afternoon I was informed I was receiving a very important honor from the organization to which I have devoted 15 years of volunteerism. So I popped in my car and drove down on Thurs night and this photo is me getting the Distinguished Service Award at lunch on Friday at which, in my brief remarks, I invoked a little Larry Craig humor and told the audience, "I am gay, I always have been gay..." Har Har.
Anyhow, if there were ever a question of how much Las Vegas casinos lust after the queer dollar, allow me to show you. Harrah's, MGM Mirage, the Las Vegas Convention and Visitors Authority and newcomer to the party Boyd Gaming all had booths in a row at the center of the expo hall. Harrah's hosted a reception Friday evening at which they announced they recently became the first gaming company to receive a 100 percent score from the gay-rights lobby Human Rights Campaign. It's hard to imagine old Bugsy, the original owner of the Flamingo, wishing to brag about such a thing.
I particularly liked the LVCVA's booth, with was staffed by a Vegas showboy and showgirl with whom folks could take pictures on a Polaroid camera that printed the pictures on Polaroid sheets framed by the Only-Vegas logos. I'll scan in mine to show when I get home. But I also thought the very idea that "Zumanity" has condoms printed up was intriguing. Somehow I doubt that the Vegas memorabilia collectors had that in mind when they started hoarding swizzle sticks and place mats. I wonder where else they distribute them!
Finally, this is my favorite -- and believe it or not unposed -- photo: The Vegas "couple" reading a very interesting new book that every one of you ought to buy!
When it first happened, we were sure the remarks would prompt angry protests, demands for discipline, outrage and shock and righteous indignation. Instead, bafflingly, they prompted nothing of the sort.
The Amazing Johnathan is an edgy comic magician over at the Sahara who will be moving shortly to the Riviera. He's been around Las Vegas forever, packs a good, consistent crowd and makes regular appearances on Comedy Central. And on Aug. 9, he came on our weekly celebrity-interview podcast "The Strip" and let loose the n-word five times and the c-word twice.
Our listeners around the world were certainly stunned. They wrote in droves debating whether what was stated was offensive, what he meant, whether his use of the word "honky" was equally alarming, whether we should have bleeped out the words or not.
African-American activists, however, took a startling pass. Las Vegas Review-Journal's Norm Clarke said he didn't even get the courtesy of returned calls or emails from leaders of the National Association of Black Journalists, which led the charge to have Don Imus fired earlier this year and who were coincidentally meeting in Vegas the day he lead his column with Johnathan's remarks. Some of our black listeners told us they wrote to the NAACP, the NABJ and other civil rights groups but similarly heard no answers.
Since that prompted no reaction from African-American forces, we've been pondering why. Is it that Johnathan's not a big enough star for their bother? That would be an odd way to decide what to be offended by -- and give an awful lot of license to an awful lot of people to use the word willy-nilly. And it's also not true: Johnathan is arguable a bigger star than Richards, who had one claim to fame and wasn't even selling out the Laugh Factory when he was heckled into insanity.
Perhaps the NAACP, having "buried" the n-word recently, is over it? (I posit that facetiously, actually.) Or maybe it's that there's no video, a la the Richards or Imus things? That could be, but the odd thing here is that Richards used the words in anger in a momentary loss of control and Imus was attempting, however badly, to be funny. Johnathan essentially defined what he believes a "n*****" to be in a way that offers real insight into his thought process.
No, I don't buy any of that. What this silence reflects is how the liberal elite view Las Vegas. They'd rather react to things that emerge in the coastal media -- Hollywood, D.C. or New York -- even though Vegas is where real Americans come for an awful lot of their entertainment.
In Vegas, we can only conclude, performers are free to say or do whatever they wish because the city and its entertainment offerings are already believed to be crass, low-brow and inconsequential. I can't decide if that's a victory for free speech or a defeat for good taste.
It's a slowish day around here -- no chlorine tankers running amok through the streets of Sin city to speak of like yesterday -- so I thought I'd share this really fun photo. I'm not going to let on where it is because that will be a question I'll ask on "The Strip" podcast in a couple of weeks, but I shot it and I loved it and wanted to share.
So that you folks don't spoil it, I'm not allowing comments for this post.
The current trivia question is: What groundbreaking actress was once married to the owner of the classic Vegas mob hangout Italian restaurant Piero's?
OK - JERRY LEWIS show is online and will hit the RSS feed later today. You can hear it now by clicking here or right-click here to download it and listen to it at your leisure. Also, enjoy the YouTube clip of Lewis in "At War With The Army," the audio of which was used on this week's program.
* Jerry thinks Vegas is going to hell.
* Jerry asserts that Princess Diana might have been a guest on his Labor Day Telethon the weekend she died had she not, uh, died.
* Jerry has another book coming out in December tentatively titled, "All The Things I Didn't Get To Say In The Other Book." No telling the publisher. He says he's writing it himself.
I've been off this soapbox for a while because it was getting redundant and a bit too easy to point out terrible abuses of the already-overused What Happens Here Stays Here -- known here as WHHSH -- tourism slogan.
But my friend Shaun in Phoenix pointed this one out and it simply cannot be left unnoticed. It's too terrible.
Larry McShane is a TV writer for the Associated Press. He clearly was desperate for a way to end his glowing review of a new Sundance Channel series starting Sept. 10 called "Sin City Law," which goes over a four particularly gruesome crimes that happened in Clark County in recent years in two hour-long dissections each. McShane's dazzled by the storytelling, the camera work, the interviews. Fine. I'm not convinced we're going to be as enchanted, but it bears giving a chance if only because the filmmakers have won Peabodys and Oscars.
It's McShane's pathetic ending that just killed Shaun and then, of course, me.
"And it's compelling television, here only through Oct. 1. Be glad that, thanks to "Sin City Law," what happens in Clark County no longer stays in Clark County."
The Strip is LIVE tonight with Jerry Lewis!
Join us live from 7-8 pm PT at LVROCKS.COM as we celebrate our 2nd anniversary by airing the much-ballyhooed chat with legend Jerry Lewis. We'll also be joined live by the Review-Journal's Norm Clarke to answer last week's Trivia Question, discuss the news of the day and much more. And, if you're lucky, I'll have finished a montage of bloopers and other outtakes!
Of course, if you miss the live show, the episode will post on Tuesday, Aug. 30.
Is there a worse idea?
Harrah's and AEG want to put a major sports stadium behind Bally's and Paris. I'll expound on this more on "The Strip" this week and in my Las Vegas Weekly column, but just curious if I'm alone on this one. Notwithstanding the question of whether Vegas wants/needs/should have a major league sports franchise or two, what sort of crack are they on that they think putting a massive center like that in that location is a good idea? And what sort of acquiescent fool legislators would allow it? What, the traffic's not bad enough already?
Am I crazy? Am I missing something?
Loyal listener Jeff Leatherock of Oklahoma City married his longtime girlfriend Lisa on Thursday. Since they knew nobody else in Vegas, Jeff asked me if I wanted to attend in the garden at the Flamingo Las Vegas. I was honored, it was lovely and here are a couple of pictures.
The slide is terrific, although the lines are probably quite long on the weekends, and the scenery is certainly a huge step up from the old Golden Nugget pool area that was actually shuttered for part of the year and turned into a makeshift poker room.
Good news is that, while the pool is intended for hotel guests, by law they must admit you if you say you're going to gamble at one of the three poolside tables. Don't know if that means you can bring along your kids, though. Also, local Nevadans are permitted in for free with a Nevada ID after 6 pm.


Among the dialog lines in the drama series The Moon Embracing the Sun, the most impressive line which chosen by Han Ga In is the line of Lee Hwon (Kim Soo Hyun), attracting everyone's attention.
On February 18th, 2012 in the MBC Theme Park at Yangju city, Han Ga In gave an interview for MBC official website. In the interview, when asked about the most impressive line, Han Ga In said that the line of "you want me to forget you, sorry, I want to forget but can't forget" during the childhood period of Lee Hwon was making her getting goose bumps when watching. As for the lines of adult Lee Hwon, because the wonderful lines are too many, so every time she can only sighs. But if really have to select one line, then Han Ga In selected "don't come near, but also don't stay away" as her favorite. Han Ga In continued to say that the impressive lines of Hwon are too many, so it's difficult to choose just one.
In addition, on the day's interview, in regard to all heroes including the minister is attracted by the charm of Wol, Ga In said that when she saw the scene, she could not help but laughed, and expressed deep gratitude to everybody.


text:
\section{Introduction}
\label{sec:introduction}
Young's lattice is the set $\Lambda$ of integer partitions, partially ordered by containment of Young diagrams.
It has a unique minimal element $\emptyset$, the trivial partition of $0$.
Its Hasse diagram is known as Young's graph.
For each $\lambda\in \Lambda$, let $f_\lambda$ denote the number of saturated chains from $\emptyset$ to  $\lambda$.
This number $f_\lambda$ is also the number of standard tableaux of shape $\lambda$, and the dimension of the irreducible representation of the symmetric group associated with $\lambda$.
It can be computed by the hook-length formula of Frame, Robinson and Thrall \cite[Theorem~1]{frame1954hook}.

Let $\Lambda^{\mathrm{odd}}$ denote the subset of partitions $\lambda\in \Lambda$ for which $f_\lambda$ is odd.
The partitions in $\Lambda^{\mathrm{odd}}$ will be referred to as \emph{odd partitions}.
Macdonald \cite{macdonald1971degrees} has shown that the number of odd partitions of $n$ is $2^{\alpha(n)}$, where,  if $n$ has binary expansion $n = a_0 + 2a_1 + 2^2a_2 + 2^3a_3 +\dotsb$, with $a_i\in \{0, 1\}$, then
\begin{displaymath}
  \alpha(n) = a_1 + 2a_2 + 3a_3 + \dotsb.
\end{displaymath}
In this article, we show (Theorems~\ref{theorem:unique-parent} and~\ref{theorem:children}) that the subgraph induced in Young's graph by $\Lambda^{\mathrm{odd}}$ is an incomplete binary tree (Figure~\ref{fig:macdonald-tree}).
We call this tree the \emph{Macdonald tree}.
For each $\lambda \in \Lambda^{\mathrm{odd}}$, we determine the number of branches of $\lambda$ in this tree.
This tree has self-similarities at all scales (Lemma~\ref{lemma:fractal}), only two infinite rays (Theorem~\ref{theorem:ray}) and a simple recursive description (Section~\ref{sec:recursive}).
Hook-shaped partitions form an order ideal of Young's lattice.
When each partition $\lambda$ in the Hasse diagram of this ideal is replaced by $f_\lambda$, Pascal's triangle is obtained.
The intersection of this ideal with $\Lambda^{\mathrm{odd}}$ is the tree of odd binomial coefficients in Pascal's triangle (see Section~\ref{sec:pascal}).
\begin{figure}
  \centering
  \includegraphics[scale=0.67]{young_lattice.pdf}

  \caption{The Macdonald tree (blue edges) in Young's graph}

  \label{fig:macdonald-tree}
\end{figure}

{\bf Note added in proof:} While this paper was being reviewed, its results were applied by Gianelli, Kleshchev, Navarro and Tiep \cite{gagp} to construct an explicit McKay correspondence for symmetric groups.

\section{Preliminary Results}
\label{sec:preliminary-results}
Our first lemma is based on the theory of cores and quotients of partitions.
An exposition of this theory, including definitions concerning hooks and their anatomy, can be found in \cite[Section~2.7]{2009JamesKerber}.
For a partition $\lambda$ and an integer $p>1$, let $\core\lambda p$ denote the $p$-core of $\lambda$ and $\quot\lambda p$ denote the $p$-quotient.
By definition, $\core\lambda p$ is a partition with no hook-length divisible by $p$, and $\quo_p(\lambda)$ is a family $(\mu^0,\dotsc,\mu^{p-1})$ of partitions such that $|\lambda| = |\core\lambda p| + p(|\mu^0| + \dotsb + |\mu^{p-1}|)$.
\begin{lemma}
  \label{lemma:unique_hook}
  Suppose that $2^k \leq n <2^{k+1}$, and $\lambda$ is a partition of $n$.
  Then $\lambda$ is odd if and only if $\lambda$ has a unique $2^k$-hook, and $\core\lambda{2^k}$ is odd.
  Moreover, for each odd partition $\mu$ of $n-2^k$, there are $2^k$ odd partitions $\lambda$ of $n$ with $\core\lambda{2^k}=\mu$.
\end{lemma}
This lemma follows from the discussion in \cite[Section~6]{olsson}.
A self-contained proof is given in Section~\ref{sec:append-result-from} of this article.

\pagenumbering{arabic}
\addtocounter{page}{1}
\markboth{\SMALL ARVIND AYYER, AMRITANSHU PRASAD, AND STEVEN SPALLONE}{\SMALL
ODD PARTITIONS IN YOUNG'S LATTICE}

If $\lambda$ covers $\mu$ in Young's lattice, we write $\lambda\in \mu^+$, or $\mu\in \lambda^-$.
\begin{lemma}[\sc Main Lemma]
  \label{lemma:hook-removal}
  Suppose that $\lambda$ and $\mu$ are odd partitions with $\mu\in \lambda^-$.
  Assume that $2^k\leq |\mu|$, and $|\lambda|<2^{k+1}$.
  \begin{enumerate}
  \item We have $\core\mu{2^k} \in \core\lambda{2^k}^-$.
  \item
    Let $r$ be the rim of the unique $2^k$-hook of $\lambda$, and $s$ be the rim of the unique $2^k$-hook of $\mu$ (see Lemma~\ref{lemma:unique_hook}).
    Let $c$ be the unique cell of $\lambda$ that is not in $\mu$,
    and $c'$ be the unique cell of $\core\lambda{2^k}$ which is not in $\core\mu{2^k}$, then exactly one of the following holds:
    \begin{enumerate}
    \item $c$ has no neighbour in $r$, $r = s$, and $c' = c$,
    \item $c = \hand(r) = \hand(s)^E$, $c' = \foot(r)^W = \foot(s)$.
    \item $c = \foot(r) = \foot(s)^S$, $c' = \hand(r)^N = \hand(s)$.
    \item $\{c^N, c^W\}\subset r\cap s$, $c'=c^{NW}$.
    \end{enumerate}
    In \emph{(b)-(d)}, $s$ is obtained from $r$ by removing $c$ and adding $c'$.
  \end{enumerate}
\end{lemma}
Here $c^N$, $c^S$, $c^E$, and $c^W$ denote the cells directly to the north, south, east, and west of $c$.
Also, $\hand(r)$ denotes the node $c$ of $r$ for which neither $c^N$ nor $c^E$ is in $r$.
Similarly, $\foot(r)$ denotes the node of $r$ for which neither $c^W$ nor $c^S$ is in $r$.

Before proving Lemma~\ref{lemma:hook-removal}, we formulate two simpler lemmas which will be used in its proof.
\begin{lemma}
  \label{lemma:large-hook-lemma}
  Let $\lambda$ be any partition and $c$ be a cell of $\lambda$ such that its hook $h(c)$ satisfies $|h(c)|\geq |\lambda|/2$.
  Then $c$ lies in the first row or in the first column of $\lambda$.
\end{lemma}
\begin{proof}
  If not, consider the cell $c^{NW}$ which lies to the north-west of $c$.
  The hooks $h(c^{NW})$ and $h(c)$ are disjoint.
  Also, $|h(c^{NW})| \geq |h(c)| + 2$.
  So $|\lambda|\geq |h(c)| + |h(c^{NW})|\geq 2|h(c)|+2\geq |\lambda|+2$, a contradiction.
\end{proof}
\begin{lemma}
  \label{lemma:two-big-hooks}
  If $c_1$ and $c_2$ are two cells in the Young diagram of a partition $\lambda$ such that $|h(c_1)|\geq |\lambda|/2$ and $|h(c_2)|> |\lambda|/2$, then $c_1$ and $c_2$ lie either in the same row or in the same column.
\end{lemma}
\begin{proof}
  If either $c_1$ or $c_2$ is the cell $(1,1)$, then the result follows from Lemma~\ref{lemma:large-hook-lemma}.
  Otherwise, if $c_1$ and $c_2$ do not lie in the same row or column, then the hooks $h(c_1)$ and $h(c_2)$ share at most one cell.
  Also, the cell $(1,1)$ is in neither hook.
  So we have:
  \begin{displaymath}
    |h(c_1)|+|h(c_2)| - 1\leq |\lambda|-1,
  \end{displaymath}
  contradicting
  \begin{displaymath}
    |h(c_1)|+|h(c_2)| > |\lambda|
  \end{displaymath}
  from the hypotheses.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lemma:hook-removal}]
  Suppose $c$ does not have a neighbour in $r$.
  Then $c$ is a removable cell of $\lambda \setminus r = \core\lambda{2^k}$, and $r$ is a $2^k$-rim hook of $\mu = \lambda \setminus c$.
  So $\core\mu{2^k} = \mu \setminus r = (\lambda \setminus r) \setminus c\in \core\lambda{2^k}^-$, giving the first part of the lemma, and Case~(a) of the second part.

  Now suppose that $c$ has a neighbour in $r$.
  Since $c$ is removable from $\lambda$, $c^E$ and $c^S$ cannot be in $r$.
  But if $c^W$ or $c^N$ is in $r$, then $c$ must also be in $r$, because $r$ is a removable rim hook.

  Let $x$ (respectively $y$) denote the node of $\lambda$ for which $h(x)=r$ (respectively $h(y)=s$).
  We may rule out $x=(1,1)$, because then the longest hook of $\mu$ would be of strictly smaller length.

  Suppose $c^W\in r$, but $c^N\notin r$.
  Then $c = \hand(r)$.
  Note that the hook-lengths of $\mu$ are the same as the hook-lengths of $\lambda$ except in the row and column of $c$, where they decrease by one.
  Since $\lambda$ has no other $2^k$-hook, $y$ must lie either in the row of $c$ or in the column of $c$.
  But $y$ cannot lie in the column of $c$, for then it would not lie in the same row or column as $x$, contradicting Lemma~\ref{lemma:two-big-hooks}.
  So $y$ must lie in the row of $c$, which is also the row of $x$.
  This would imply that $\hand(s) = c^W$, and so $\foot(r)^W = \foot(s)$.
  Then $c' = \foot(r)^W$.
  We have $\core\mu{2^k} = \core\lambda{2^k} \setminus c'$, giving the first part of the lemma and Case~(b) of the second part.
  The case where $c^N\in r$, but $c^W$ not in $r$ can be dealt with similarly, and leads to Case~(c).

  Finally, suppose $c^W$ and $c^N$ are both in $r$.
  Then replacing $c$ by $c^{NW}$ in $r$ results in a $2^k$-hook of $\mu$; this must be $s$, giving the first part of the lemma and Case~(d) of the second part.
\end{proof}

By Lemma~\ref{lemma:unique_hook}, the odd partitions of $2^k$ are precisely the hook-shaped ones.
In general, if $2^k\leq n<2^{k+1}$, the function $\mathrm{core}_{2^k}:\lambda\mapsto \core\lambda{2^k}$ maps odd partitions of $n$ onto odd partitions of $n-2^k$.
\begin{prop}
  \label{prop:coring-tree}
  Let $\lambda$ be an odd partition of $n$.
  \begin{enumerate}
  \item If $2^k< n< 2^{k+1}$, then the map
    \begin{displaymath}
      \mathrm{core}_{2^k}:\lambda^-\cap \Lambda^{\mathrm{odd}}_{n-1}\to \core\lambda{2^k}^-\cap\Lambda^{\mathrm{odd}}_{n-2^k-1}
    \end{displaymath}
    is injective.
  \item If $2^k-1<n< 2^{k+1}-1$, then the map
    \begin{displaymath}
      \mathrm{core}_{2^k}:\lambda^+\cap \Lambda^{\mathrm{odd}}_{n+1}\to \core\lambda{2^k}^+\cap\Lambda^{\mathrm{odd}}_{n-2^k+1}
    \end{displaymath}
    is injective.
  \end{enumerate}
\end{prop}
\begin{proof}
  Suppose that $2^k<n< 2^{k+1}$.
  Let $\mu$ and $\nu$ be distinct elements of $\lambda^-\cap \Lambda^{\mathrm{odd}}_{n-1}$.
  Let $c$ and $d$ be the cells of $\lambda$ that are not in $\mu$ and not in $\nu$ respectively, and $c'$ and $d'$ be the cells of $\core\lambda{2^k}$ that are not in $\core\mu{2^k}$ and $\core\nu{2^k}$ respectively.
  Since $\mu\neq \nu$, $c\neq d$.
  We need to show that $c'\neq d'$.

  Consider first the case where $c$ has no neighbour in $r$.
  Then, if $d$ also has no neighbour in $r$, Lemma~\ref{lemma:hook-removal} tells us that $c'=c$ and $d'=d$, so $c'\neq d'$.
  On the other hand, when $d$ satisfies one of cases (b), (c), and (d) of Lemma~\ref{lemma:hook-removal}, then $d$ lies in $r$, and $d'$ has a neighbour in $r$.
  Since $c'=c$ does not have a neighbour in $r$, $c'\neq d'$.

  Now suppose $c=\hand(r)$.
  Then $c'=\foot(r)^W$.
  Since $d\neq c$, $d$ corresponds to one of the cases (a), (c), and (d) in Lemma~\ref{lemma:hook-removal}.
  But none of these can give rise to $c'$, so $d'\neq c'$.
  The remaining cases are similar.
  This concludes the proof of Part (1) of the proposition.
  The proof of Part (2) is similar, but the roles played by $\lambda$ and $\mu$ in Lemma~\ref{lemma:hook-removal} are interchanged.
\end{proof}

\section{Tree structure}
\label{sec:tree-structure}
\begin{theorem}
  [\sc Unique Parent Theorem]
  \label{theorem:unique-parent}
  For every odd partition $\lambda$ with $|\lambda|>1$, there exists a unique odd partition $\mu\in \lambda^-$.
In other words, the subgraph induced in Young's graph by $\Lambda^{\mathrm{odd}}$ is a rooted tree.
\end{theorem}
\begin{proof}
  Let $n = |\lambda|$, and let $\nu(n)$ denote the sum of the binary digits of $n$.
  We proceed by induction on $\nu(n)$.
  If $\nu(n)=1$, then $n=2^k$.
  In this case, $\lambda$ is a hook, say $(r, 1^{2^k-r})$, and $f_\lambda = \binom{2^k-1}{r-1}$ (see Section~\ref{sec:pascal}).
  Recall the well-known result (see e.g., \cite[Exercise~1.14]{ec1}) that a binomial coefficient $\binom nm$ is odd if and only if the sets of place values where $1$ occurs in the binary expansions of $m$ and $n-m$ are disjoint (in other words, there are no carries when $m$ and $n-m$ are added in binary).
  The set $\lambda^-$ consists of the partitions $(r-1, 1^{2^k-r})$ and $(r, 1^{2^k-r-1})$ whose dimensions are $\binom{2^k-2}{r-2}$ and $\binom{2^k-2}{r-1}$.
  The former is odd when $r$ is even, and the latter is odd when $r$ is odd.
  In any case, one of them is odd, and the other is even, proving the theorem for $\nu(n) =1$.

  If $\nu(n)>1$, then $2^k<n<2^{k+1}$ for some $k\geq 1$.
  Since $\nu(n-2^k)<\nu(n)$, by induction, we may assume that $|\core\lambda{2^k}^-\cap \Lambda_{n-2^k-1}^{\mathrm{odd}}| = 1$.
  Proposition~\ref{prop:coring-tree} implies that
  \begin{displaymath}
    |\lambda^-\cap \Lambda_{n-1}^{\mathrm{odd}}| \leq 1.
  \end{displaymath}
  By Pieri's rule, we have
  \begin{displaymath}
    f_\lambda = \sum_{\mu\in\lambda^-} f_\mu.
  \end{displaymath}
  So if $f_\lambda$ is odd, then $f_\mu$ is odd for at least one $\mu\in \lambda^-$. In other words,
  \begin{displaymath}
    |\lambda^-\cap \Lambda_{n-1}^{\mathrm{odd}}| \geq 1,
  \end{displaymath}
  hence equality holds.
\end{proof}
\begin{prop}
  \label{prop:counting-by-core}
  Suppose $2^k-1<n<2^{k+1}-1$ for some positive integer $k$.
  Then, for any odd partition $\lambda$ of $n$,
  \begin{displaymath}
    |\lambda^+\cap \Lambda^{\mathrm{odd}}_{n+1}| = |\core\lambda{2^k}^+\cap \Lambda^{\mathrm{odd}}_{n-2^k+1}|.
  \end{displaymath}
\end{prop}
\begin{proof}
  We have
  \begin{align*}
    2^{\alpha(n+1)} & = |\Lambda_{n+1}^{\mathrm{odd}}|\\
    & = \sum_{\lambda\in \Lambda_n^\mathrm{odd}} |\lambda^+\cap \Lambda_{n+1}^{\mathrm{odd}}| & \text{(by Theorem~\ref{theorem:unique-parent})}\\
    & = \sum_{\mu\in \Lambda_{n-2^k}^{\mathrm{odd}}}\sum_{\core\lambda{2^k} = \mu} |\lambda^+\cap \Lambda_{n+1}^{\mathrm{odd}}| \\
    & \leq \sum_{\mu\in \Lambda_{n-2^k}^{\mathrm{odd}}}\sum_{\core\lambda{2^k} = \mu} |\mu^+\cap \Lambda_{n-2^k+1}^{\mathrm{odd}}| & \text{(by Proposition~\ref{prop:coring-tree})}\\
    & = 2^k \sum_{\mu\in \Lambda_{n-2^k}^{\mathrm{odd}}}|\mu^+\cap \Lambda_{n-2^k+1}^{\mathrm{odd}}| & \text{(by Lemma~\ref{lemma:unique_hook})}\\
    & = 2^k |\Lambda^{\mathrm{odd}}_{n-2^k-1}| & \text{(by Theorem~\ref{theorem:unique-parent})}\\
    & = 2^k\times 2^{\alpha(n-2^k+1)}\\
    & = 2^{\alpha(n+1)}.
  \end{align*}
  Since the first and last terms are equal, equality holds at each step, and the proposition follows.
\end{proof}
Let $v_2(n)$ denote the $2$-adic valuation of $n$.
\begin{theorem}
  \label{theorem:children}
  Let $\lambda$ be an odd partition of $n$.
  If $n$ is even, then there exists a unique odd partition $\mu\in \lambda^+$.
  If $n$ is odd and $v_2(n+1) = v$, then
  \begin{displaymath}
    |\lambda^+\cap \Lambda_{n+1}^{\mathrm{odd}}| =
    \begin{cases}
      2, & \text{if $\core\lambda{2^v}$ is a hook,}\\
      0, & \text{otherwise}.
    \end{cases}
  \end{displaymath}
  In particular, the induced subgraph of Young's graph consisting of partitions in $\Lambda^{\mathrm{odd}}$ is an incomplete binary tree.
\end{theorem}
\begin{proof}
  When $n$ is even, the theorem is proved by induction on $\nu(n)$.
  If $\nu(n) = 0$, then $n=0$, and the result is obviously true.

  If $n$ is even and $\nu(n)>0$, then $2^k-1<n<2^{k+1}-1$ for some $k>0$.
  Proposition~\ref{prop:counting-by-core} reduces the result to $\core\lambda{2^k}$ of size $n-2^k$.
  Since $\nu(n-2^k)<\nu(n)$, the theorem follows by induction.

  When $n$ is odd, the theorem is proved by induction on $\nu(n+1)$.
  If $\nu(n+1) = 1$ then $n=2^v-1$, and $\core\lambda{2^v} = \lambda$.
  If $\lambda$ is not a hook, then no element of $\lambda^+$ is a hook.
  If $\lambda$ is a hook, then $\lambda^+$ has two hooks.
  Since a partition of $2^v$ is odd if and only if it is a hook, the theorem holds for $\nu(n+1)=1$.

  If $n$ is odd and $\nu(n+1)>1$, then $2^k-1<n<2^{k+1}-1$ for some $k>0$.
  Proposition~\ref{prop:counting-by-core} then reduces the result to $\core\lambda{2^k}$, a partition of $n-2^k$, and, as before, the theorem follows by induction on $\nu(n+1)$.
\end{proof}

The subtrees consisting of the first $2^k-1$ rows, for each $k$, are repeated infinitely many times in the Macdonald tree.
Given a partition $\lambda\in \Lambda^{\mathrm{odd}}$, define $\lambda^{+[0, k]}$ to be the induced subtree rooted at $\lambda$ consisting of nodes of $\Lambda^{\mathrm{odd}}$ which descend from $\lambda$ (more precisely, nodes which are greater than or equal to $\lambda$ in the containment order) and whose ranks lie between $|\lambda|$ and $|\lambda|+k$.
\begin{theorem}
  [\sc Self-similarities of the Macdonald tree]
  \label{lemma:fractal}
  Let $n$ be a positive integer such that $v_2(n)\geq v$.
  Let $\lambda$ be an odd  partition of $n$.
  Then
  \begin{displaymath}
    \mathrm{core}_{2^v}: \lambda^{+[0,2^v-1]}\to \emptyset^{+[0,2^v-1]}
  \end{displaymath}
  is an isomorphism of trees.
\end{theorem}
\begin{proof}
  Suppose that $2^k\leq n<2^{k+1}$ for some $k\geq v$.
  Then, by Proposition~\ref{prop:coring-tree}, the map $\mu\mapsto \core\mu{2^k}$ gives rise to an isomorphism $\lambda^{+[0,2^v-1]}\to \core\lambda{2^k}^{+[0,2^v-1]}$.
  Repeating this operation until $k$ is reduced to $0$, and noting that $\mathrm{core}_{2^k}\circ \mathrm{core}_{2^l} = \mathrm{core}_{2^k}$ for all $k\leq l$, we obtain the desired result.
\end{proof}

By a ray in the Macdonald tree, we mean a sequence $\{\lambda^{(n)}\}_{n=0}^\infty$ of odd partitions such that $\lambda^{(0)}=\emptyset$ and $\lambda^{(n+1)}\in \lambda^{(n)+}$.
\begin{theorem}
  \label{theorem:ray}
  The only rays in the Macdonald tree are $\{(n)\}_{n=1}^\infty$ and $\{(1^n)\}_{n=1}^\infty$.
\end{theorem}
\begin{proof}
  The only hooks in $\Lambda^{\mathrm{odd}}_{2^k+1}$ are $(2^k+1)$ and $(1^{2^k+1})$.
  So if $\lambda\in \Lambda^{\mathrm{odd}}_{2^k}$ is different from $(2^k)$ or $(1^{2^k})$, then $\lambda^+\cap \Lambda^{\mathrm{odd}}_{2^k+1}$ has no hooks in it.
  It follows that none of the partitions of $2^{k+1}-1$ in $\lambda^{+[0, 2^k-1]}$ are hooks, and so, by Theorem~\ref{theorem:children}, have no children.
  Thus $\lambda$ cannot be contained in any ray.
  Thus each ray in the Macdonald tree must pass through the points $(2^k)$ or $(1^{2^k})$ for each $k$, and so must be either $\{(n)\}_{n=1}^\infty$ or $\{(1^n)\}_{n=1}^\infty$, as claimed.
\end{proof}

\section{Recursive description of the Macdonald tree}
\label{sec:recursive}
\begin{figure}[h]
  \centering
  \includegraphics[scale = 0.62]{macdonald_tree.pdf}
  \caption{Hooks (joined by red and green edges) in the Macdonald tree}
  \label{fig:macdonald_tree}
\end{figure}
The Macdonald tree, when viewed as an abstract rooted binary tree, has a very short recursive description.
Let $T_k = \emptyset^{+[0, 2^k-1]}$, the subtree formed by nodes which are at distance at most $2^k-1$ from the root.
The recursive description allows for the construction of $T_{k+1}$ from $T_k$.

As a first step, construct a tree $\tilde T_k$ by adding one node $*$ to the root of $T_k$ (thus $\tilde T_k$ may be viewed as a rooted tree with root $*$).
Note that, for $k\geq 2$, $T_k$ has $2^{\binom k2}$ partitions of $2^k-1$, of which $2^{k-1}$ are hooks.
Among these hooks, there are two partitions $\lambda$ with $f_\lambda = 1$ (we will call these one-dimensional partitions).

\begin{quote}
  For $k\geq 2$, the tree $T_{k+1}$ is obtained from $T_k$ by attaching two branches to each partition of $2^k-1$ in $T_k$ that is a hook, each of these branches being isomorphic to $\tilde T_k$.
\end{quote}

Thus the nodes of $T_{k+1}$ can be partitioned into $2^k + 1$ subsets, in such a way that the induced subgraph on each of these subsets is isomorphic to $T_k$.
In order to be able to extend the recursive process further, we need to mark the hooks and one-dimensional partitions of $2^{k+1}-1$ in $T_{k+1}$.
To mark the hooks, choose one branch descending from each of the one-dimensional partitions in $T_k$.
This branch is, by construction, isomorphic to a copy of $T_k$.
The partitions in this branch which correspond to hooks of $T_k$ are the partitions of $2^{k+1}-1$ in $T_{k+1}$ which are hooks.
To mark partitions of $2^{k+1}-1$ which are one-dimensional, simply choose one of the one-dimensional partitions of $2^k-1$ in each branch.

These markings, of course, are only defined up to an automorphism of $T_{k+1}$ (which is not a problem, because we are only describing the Macdonald tree as an abstract binary tree).
The Macdonald tree up to partitions of $16$ is shown in Figure~\ref{fig:macdonald_tree}.
The green and red edges are the ones which join hooks.
Among these, the red edges join one dimensional partitions.
All remaining edges are coloured blue.
The horizontal lines mark powers of two.

\section{Hooks and Pascal's triangle}
\label{sec:pascal}
Among all partitions, consider the hooks, which are partitions of the form $(n_1+1, 1^{n_2})$ for nonnegative integers $n_1$ and $n_2$.
Pascal's triangle may be regarded as the Hasse diagram of the poset of pairs of non-negative integers, with $(n_1,n_2)\leq (m_1,m_2)$ if and only if $n_1\leq m_1$ and $n_2\leq m_2$.
The unique minimal element of this poset is $(0, 0)$, and it is graded with $(n_1,n_2)$ having rank $n_1 + n_2$.
In this poset, $(n_1, n_2)$ is covered by two elements, $(n_1+1, n_2)$ and $(n_1, n_2+1)$.
The number of saturated chains from $(0, 0)$ to $(n_1,n_2)$ is the binomial coefficient $\binom{n_1 + n_2}{n_1}$.
The subgraph of Pascal's triangle consisting of pairs $(n_1, n_2)$, where $\binom{n_1+n_2}{n_1}$ is odd  (see Figure~\ref{fig:sierpinski}) is closely related to the Sierpi\'nski triangle \cite{reiter1993determining}.
\begin{figure}[h]
  \centering
  \includegraphics[scale = 0.5]{pt.pdf}
  \caption{Odd binomial coefficients (joined by green and red edges) in Pascal's triangle up to $n=32$}
  \label{fig:sierpinski}
\end{figure}

The map $(n_1,n_2)\mapsto (n_1+1, 1^{n_2})$ is an embedding of Pascal's triangle into Young's lattice (with a shift of one in rank).
The image, consisting of all the hooks in Young's lattice, taken together with the empty partition $\emptyset$, is an order ideal in $\Lambda$, which we will denote by $P$.
Thus if $\lambda\in \Lambda$ is a hook, then every saturated chain from $\emptyset$ to $\lambda$ in $\Lambda$ is also a saturated chain in $P$.
This gives an amusing proof of the well-known formula:
\begin{displaymath}
  f_{(n_1+1, 1^{n_2})} = \binom{n_1+n_2}{n_1}.
\end{displaymath}
It follows that the subgraph induced in Young's graph by the set of odd-dimensional hooks is isomorphic to the graph of odd binomial coefficients in Pascal's triangle.
These are the green and red edges in Figures~\ref{fig:macdonald_tree} and ~\ref{fig:sierpinski}.

\section{Proof of Lemma~\ref{lemma:unique_hook}}
\label{sec:append-result-from}
Our proof of Lemma~\ref{lemma:unique_hook} relies on the following result of Frame, Robinson, and Thrall \cite[Lemma~2]{frame1954hook}.
\begin{lemma}
  \label{sec:macd-lemma-1}
  There exists a bijection from the set of cells in $\quot\lambda p$ onto the set of cells in $\lambda$ whose hook-lengths are divisible by $p$
  under which a cell of hook-length $h$ in $\quot\lambda p$ maps to a cell of hook-length $p h$ in $\lambda$.
\end{lemma}
  For a partition $\lambda$ of $n$, let $\alpha = \core\lambda 2$ and $(\mu^0,\mu^1) = \quot\lambda 2$.
  Let $a = |\alpha|$, $m_0 = |\mu^0|$, and $m_1 = |\mu^1|$ (so $n = a + 2m_0 + 2m_1$).
\begin{lemma}
  \label{sec:recurse-f}
  The partition $\lambda$ is odd if and only if $a\leq 1$ (so $\alpha$ is $\emptyset$ or $(1)$), the sets of place values where $1$ appears in the binary expansions of $a$, $2m_0$ and $2m_1$ are disjoint, and $\mu^0$ and $\mu^1$ are odd.
\end{lemma}
\begin{proof}
  By the hook-length formula \cite[Theorem~1]{frame1954hook}, we have
  \begin{equation}
    \label{eq:3}
    v_2(f_\lambda) = v_2(n!) - v_2(H_\lambda),
  \end{equation}
  where $H_\lambda$ is the product of all the hook-lengths of $\lambda$.
  By a well-known theorem of Legendre, $v_2(n!) = n - \nu(n)$, where $\nu(n)$ is the number of times $1$ occurs in the binary expansion of $n$.
  Moreover, by Lemma~\ref{sec:macd-lemma-1}, $v_2(H_\lambda) = m_0 + m_1 + v_2(H_{\mu^0}) + v_2(H_{\mu^1})$.
  Using these facts, the identity \eqref{eq:3} can be rewritten as
  \begin{displaymath}
    v_2(f_\lambda) = [a - \nu(a)] + [\nu(a) + \nu(2m_0) + \nu(2m_1) - \nu(n)] +  v_2(f_{\mu^0}) + v_2(f_{\mu^1}).
  \end{displaymath}
  The right-hand side is a sum of four nonnegative parts.
  It  is zero if and only if each part is zero.
  This happens only under the conditions of the lemma.
\end{proof}
\begin{proof}
  [Proof of Lemma~\ref{lemma:unique_hook}]
  Suppose that $\lambda$ is an odd partition of $n$ with $n>1$.
  Choose $k$ such that $2^k\leq n<2^{k+1}$.
  By Lemma~\ref{sec:recurse-f}, both $f_{\mu^0}$ and $f_{\mu^1}$ are odd, and exactly one of $m_0$ and $m_1$ satisfies $2^{k-1}\leq m_i<2^k$.
  By induction, this $\mu^i$ has a unique $2^{k-1}$-hook, while the other has none.
  By Lemma~\ref{sec:macd-lemma-1}, $\lambda$ has a unique $2^k$-hook.

  Moreover, by \cite [Theorem~4]{frame1954hook}, we have
  \begin{align}
    \label{eq:1}
    \core{\core\lambda{2^k}}{2} & = \alpha,\\
    \label{eq:2}
    \quot{\core\lambda{2^k}}{2} & = (\core{\mu^0}{2^{k-1}}, \core{\mu^1}{2^{k-1}}).
  \end{align}
  If $2^{k-1}\leq m_i< 2^k$, then $f_{\core{\mu^i}{2^{k-1}}}$ is odd by induction.
  Otherwise, $m_i<2^{k-1}$, so again $\core{\mu^i}{2^{k-1}} = \mu^i$, so $\core{\mu^i}{2^{k-1}}$ is odd.
  Thus $\core\lambda{2^k}$ is odd by Lemma~\ref{sec:recurse-f}.

  For the converse, suppose that $\lambda$ has a unique $2^k$-hook, and that $\core\lambda{2^k}$ is odd.
  The first condition implies that $|\core\lambda{2^k}| = n - 2^k$.
  Let $m'_i = |\core{\mu^i}{2^{k-1}}|$.
  We have
  \begin{displaymath}
    n - 2^k = a + 2m'_1 + 2m'_2,
  \end{displaymath}
  and the place values of the $1$s in the binary expansion of the summands on the right-hand side are disjoint.
  Since $2^{k-1}\leq m_i$ for exactly one $i$, \eqref{eq:2} implies that $m'_i = m_i$ for exactly one $i$ (say, $i=0$), and thus, for the other value of $i$ (say $i=1$), $m'_i = m_i - 2^{k-1}$.
  Thus $\mu^0$ has a unique $2^k$-hook and (by Lemma~\ref{sec:recurse-f} applied to $\core\lambda{2^k}$) $f_{\core{\mu^0}{2^{k-1}}}$ is odd.
  Thus, by induction on $k$, $f_{\mu^0}$ is odd.
  Since $m_1<2^{k-1}$, $\mu^1 = \core{\mu^1}{2^{k-1}}$, so (again by Lemma~\ref{sec:recurse-f} applied to $\core\lambda{2^k}$) $f_{\mu^1}$ is odd.
  Finally, the application of Lemma~\ref{sec:recurse-f} to $\lambda$ shows that $f_\lambda$ is odd.

  Since the odd partitions of $n$ have a unique $2^k$-hook, the partition $(1)$ occurs in their $2^k$-quotients once, and the partition $\emptyset$ occurs $2^k-1$ times.
  Thus there are $2^k$ possibilities for the $2^k$-quotient of such a partition once its $2^k$-core is fixed.
  Since a partition is determined by its core and quotient, the second assertion of the lemma follows.
\end{proof}

\section{Concluding remarks}
\label{sec:concluding-remarks}
In this article, we have described how Macdonald's enumerative results on odd partitions are reflected in Young's lattice.
The enumerative result in Macdonald's paper \cite{macdonald1971degrees} is a simple special case of his more general result on the enumeration of partitions $\lambda$ for which $f_\lambda$ is not divisible by a prime number $p$.
It would be interesting to see how these more general enumerative results are reflected in Young's lattice.
However, this can not be achieved by using only the methods here.
For instance, Lemma~\ref{lemma:unique_hook} does not hold for $p>2$.
Also, the partitions $\lambda$ with $f_\lambda$ coprime to $3$ do not form a tree (the partitions $(2)$ and $(1,1)$ both cover $(2,1)$; all three partitions have dimension coprime to $3$).

Another promising direction of generalization is to replace Young's lattice by an arbitrary $1$-differential poset.
Besides Young's lattice, the best-known example of a $1$-differential poset is the Young--Fibonacci lattice (see \cite[Section~5]{sta-diffpo}), denoted $Z(1)$.
For each $x\in Z(1)$, let $f_x$ denote the number of saturated chains in $[\hat 0, x]$.
Using the construction of $Z(1)$ by reflection extension \cite[Section~2.2]{roby-thesis}, it is easy to prove the following.
\begin{theorem}

  The subgraph induced in the Hasse diagram of $Z(1)$ by the set of elements $x\in Z(1)$ for which $f_x$ is odd is a binary tree where every element of even rank has one branch and every element of odd rank has two branches.

\end{theorem}
The following analogue of Macdonald's enumerative result is an immediate corollary.
\begin{theorem}

  The number of elements $x$ in $Z(1)$ of rank $n$ with $f_x$ odd is $2^{\lfloor n/2\rfloor}$.

\end{theorem}
For the Fibonacci $r$-differential poset with $r>1$, the subgraph induced in its Hasse diagram by the subset of elements $x$ with an odd number of saturated
chains in $[0, x]$ is a rooted tree if and only if $r$ is even. In this tree
every node has $r$ branches.

\subsection*{Acknowledgements}
This research was driven by computer exploration using the open-source
mathematical software \texttt{Sage}~\cite{sage} and its algebraic
combinatorics features developed by the \texttt{Sage-Combinat}
community~\cite{Sage-Combinat}.
A. A. was supported in part by a UGC Centre for Advanced Study grant.
The authors thank an anonymous referee for some helpful comments.

redpajama_set_name: RedPajamaArXiv


Google, Mirrors and Dying Birds: Is Ivanpah Evil?
Thursday marked the grand opening of Ivanpah - the first-of-it's-kind large scale solar thermal plant. But was it worth the cost and is this really where our future is headed?
Could Micro-Grids Be The Answer to Renewable Energy Challenges?
There is a quiet, growing trend of micro-grids that can provide redundancy for the grid we rely on. Is this solution a win-win for everyone?
We live in the age of startups, fledgling companies in search for a repeatable and scalable business model. Startups are subject to high rates of failure, and venture capitalists assume significant risk when investing in these companies. The solar industry, an industry that may be considered akin to a startup relative to other established fossil-fuel energy sources, is equally subject to failure... So why are some politicians so eager to point this out?


"80th Anniversary Commemorative Jubilee Edition" Swiss-Made Mechanical Key Winding Movement. Extra Large Satin-finished Stainless Steel case with exposition back to view movement. Limited Series of 87 pieces each individually numbered. Diameter 58mm.
Movement: Swiss-Made Mechanical/Manual Hand Winding. Specially modified original One Hundred Fifty year old key wind pocket watch movement, each carefully dissembled and re-assembled with the latest modern production methods and procedures (winding key included). 11 Jewels - Hand Adjusted Regulation - Beautiful nickel finished bridge movement with blued screws.


text:
Taking the money out of marijuana, and how to feel like you've won the lottery
Health·Second Opinion
Second Opinion is a vital dose of the week's news in health and medicine from reporter Kelly Crowe and CBC Health.
Plus, could an old drug defeat new superbugs?
Kelly Crowe · CBC News · Posted: Mar 19, 2017 8:00 AM ET | Last Updated: March 19, 2017
Forget about buying a lottery ticket. New research suggests you can feel just as good by getting a good night's sleep. (Gabriel Petrescu/Shutterstock)
This week, research shows a good night's sleep can feel like hitting the jackpot, and a Canadian researcher may have found a new weapon in the fight against superbugs. But first, as the government gets set to legalize marijuana, one drug policy expert argues public health and profits don't mix.
Get the full newsletter in your inbox every Friday morning. Subscribe to Second Opinion.
Taking the big bucks out of bud
With Prime Minister Justin Trudeau promising to introduce marijuana legislation by the summer, and growers of medical marijuana grappling with pesticide contamination and other quality control issues, we have a second opinion about how to legalize recreational cannabis without affecting public health.
Drug policy researcher Michael DeVillaer says the government should set up a non-profit institution to manage marijuana in Canada. (CBC)
It's easy, says one McMaster University researcher.
No marketing, no advertising, no Big Cannabis. Instead, Michael DeVillaer says the supply and sale of cannabis could be managed by a non-profit institution with a mandate to protect public health and still allow legal recreational use.
"If we were to do this really cautiously, we could do it with a near neutral impact on public health in Canada. But if we don't, if we go down the same path as alcohol and tobacco, I really think 10, 20 years from now people could be looking back on this and saying, 'What were they thinking?'" said DeVillaer, who has released a report detailing his plan for a non-for-profit cannabis authority.
Listen to our full interview here:
Could an old drug defeat deadly superbugs?
There's an old drug at the back of the shelf that just might help win the war against superbugs. But right now no doctor would think of using it that way.
"It would never enter your head," said Dr. Michael Gardam, a Toronto infectious disease specialist.
The drug is pentamidine and has been used since the 1930s for parasite infections caused by insect bites. It also treats a rare form of pneumonia, although doctors avoid it because of toxic side effects. Now, a Canadian scientist has discovered it could be used to kill antibiotic resistant bacteria.
Eric Brown, a biochemistry professor at McMaster, tested it against those deadly bacteria at the top of the WHO's list of priority pathogens published a few weeks ago. And it worked, at least in a petri dish, and in mice, by breaking down the tough outer membrane allowing an ordinary antibiotic to penetrate and kill the pathogen.
Could a long-shelved drug be the secret to killing these deadly superbugs? (Kateryna Kon/Shutterstock)
But how does a discovery like that go from the lab to the doctor's office?
"If there was a protocol specifying dose and drug combinations doctors would start using it," said Gardam. But who will do those clinical trials for an old drug long off-patent with no real potential to generate profits? Brown agrees it's a problem.
"Given that pentamidine is an off-patent drug this sort of development research is unlikely to be done by a company," he told us. "So it falls into the realm of drug development research paid for with [public] health research dollars."
Ontario scientist thinks old drugs could become our new antibiotics
Meanwhile pentamidine sits on the shelf, already approved for human use and available to any doctor who dares to use it against a superbug infection without having an established protocol to follow.
"It would be a pretty gutsy move to prescribe," Brown said.
And so far no one has announced a plan to follow up on Brown's discovery with clinical tests.
Want to win the lottery? Hit the sleep jackpot instead
A good night's sleep feels wonderful, but can it compare to winning the lottery?
Yes. That's what researchers from the University of Warwick report in this month's journal SLEEP. They asked more than 30,000 U.K. residents about their sleep. After four years, those who reported an increased quantity of sleep or a decreased use of sleep medications had better scores on tests for health and well-being. But the biggest change was in those who reported an improvement in the quality of their sleep.
Researchers say their results show that public health messages should be broadened to focus on improving quality of sleep. (Getty Images/Image Source)
The changes in health and well-being were comparable to those seen in people who completed an eight-week program of mindfulness-based cognitive therapy, or even those who won up to a $300,000 prize in the lottery.
More than a quarter of Canadians get less than 7 hours of sleep
The researchers say their results show that public health messages should be broadened to focus on improving quality of sleep as well as reducing use of sleep medication, on top of the current emphasis on getting enough sleep.
Would you like these stories and others delivered to your inbox every Friday morning? Subscribe to Second Opinion.


Moving from one place to another can be very exciting and may also involve several tasks that even the most essential of them all is merely forgotten. Pennsylvania is a great place to relocate to but this may soon turn around if you find yourself ill-prepared. Getting valid Pennsylvania cheap car insurance should come as a top agenda on your to-do list before even moving to the state.
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In Pennsylvania the first rule you should consider most important is that all motorists regardless of age, race, religion or affiliation, must show financial responsibility anytime they are driving so that in the event of a vehicle accident, they are able to pay for the damages and costs that come as a result of the medical expenses, property repair and more. The government of Pennsylvania has gone further and set out the minimum liability car insurance that one is supposed to have at any one time they are driving. The lowest is the liability insurance limit for damage or destruction of property due to an accident you caused and stands at $5,000. The next is the minimum limit for causing bodily injury or death to one person in a car accident in which you are responsible and stands at $15,000 for every accident you cause. The last is the highest limit which is $30,000 for causing the death of more than one person or causing physical injury to them for every accident you cause.
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If you are bewildered about the placement of furniture and the orientation of the total design layout in your home, apps readily available on the Internet can effectively help you in the chore. On the flip side, you may also think about Commercial Design. You could either think about a design for your home all on your own or seek help from a professional and knowledgeable Singapore homeInterior Design Expert.
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You're able to still have a minimalist workspace within your minimalist home. It is a central tenet of a family and we value what a comfortable home space can do for your family. It is one location that everyone wants to return to.


text:
'She Put Her Dreams Aside for Me.' No. 1 NBA Draft Pick Zion Williamson Gets Emotional Talking About His Mother
NBA Prospect Zion Williamson is introduced before the start of the 2019 NBA Draft at the Barclays Center on June 20, 2019 in the Brooklyn borough of New York City.
Sarah Stier—Getty Images
Basketball phenom Zion Williamson was the No. 1 pick at the NBA's draft on Thursday night and while he was the star of the show, there was one person he wanted to put in the spotlight—his mom.
Williamson, who was selected by the New Orleans Pelicans, shook hands with NBA Commissioner Adam Silver and after flashing his new jersey, walked off stage for his first official ESPN interview as an NBA player. The 18-year old Duke star chatted with ESPN's Maria Taylor and the interview quickly got emotional.
"You have been waiting for this moment since you were 5," Taylor said. "How would you describe being the overall No. 1 pick in this draft?"
"I didn't think I'd be in this position," he said. "My mom sacrificed a lot for me. I wouldn't be here without my mom, she did everything for me, I just want to thank her," he said tearfully, as his mother Sharonda Sampson stood beside him looking proud. Taylor then reminded viewers at home that Sharonda had coached her son until he was 14, just four short years ago. Her son clearly appreciates her efforts, saying, "She put her dreams aside for mine. She always looked out for the family first before herself."
The interview warmed a lot of hearts as people who watched it quickly took to the internet to share it. Seems that Zion has been an NBA player for less than one day and is already a hit with the fans:


I have been exploring the area with my camera for the last 15 years and find that Speyside provides a seemingly endless source of inspiration from a rich variety of subjects. I greatly appreciate the wonderful landscape, from its panoramic impact to the significance of the tiny details encountered while walking with a camera. Seasonal changes and the sometimes breathtaking effect of our variable weather conditions also fascinate me. This often pulls me back to revisit favourite places or to look at the world about me with a more abstract perspective. The shapes, colours, patterns and textures of the natural and man made world have inspired a range of more unusual images featured in the "Art of Nature" section.
This fascination with a more abstract approach led to me developing a series of images I call "PhotoMandalas", where small details are repeated and symmetrically juxtaposed, creating startling new kaleidoscopic patterns.
Above all, I hope that my collection of images helps to celebrate the diverse beauty which can be found in the obvious and also in the less obvious.


WASHINGTON (AP) — U.S. employers have more job openings than at any other time in nearly five years. That's in part because they seem in no hurry to fill them.
And it helps explain why the job market remains tight and unemployment high. Even as openings have surged 11 percent in the past year, the number of people hired each month has declined.
Why so many openings yet so few hires?
Economists point to several factors: Some unemployed workers lack the skills employers want. Some companies may not be offering enough pay. And staffing firms and employment experts say that in a still-fragile economy, many businesses seem hesitant to commit to new hires. They appear to be holding out for the perfect candidate.
"We're living in a fear-based environment right now," says Kim First, CEO of The Agency Worldwide, a recruiting firm for pharmaceutical and biotech companies.
Those who do have jobs these days are unlikely to lose them. Layoffs have sunk to a pre-recession level.
But First says that companies feel they can't afford to take a risk by hiring someone who doesn't appear to be an ideal fit for the job they've advertised.
The Labor Department said Tuesday that the number of job openings rose 8.7 percent in February from January to a seasonally adjusted 3.93 million. That was the most since May 2008.
At the same time, companies hired a seasonally adjusted 4.4 million people, just 2.8 percent more than in January. And hiring remained lower than it was a year ago, when it was nearly 4.5 million.
The figures suggest that the Great Recession may have transformed the job market in ways that economists still don't fully understand. Normally, more openings lead, over time, to stronger hiring and steadily lower unemployment.
Yet in May 2008, when job openings were as numerous as they are now, the unemployment rate was 5.4 percent. Now, it's 7.6 percent.
And in 2007, before the recession began, employers were hiring an average of 5.2 million people a month — 15 percent more than in February this year.
The Labor Department's Job Openings and Labor Turnover survey, or JOLTS, reveals the total number of people hired and laid off each month. It differs from the department's jobs report, which provides each month's net job gain or loss. But by quantifying total hiring and layoffs, the JOLTS can paint a fuller picture of what employers are doing.
From November through February, employers added a net average of about 220,000 jobs a month. The JOLTS report shows that a big reason for those gains was that layoffs fell. Companies cut 1.5 million jobs in January — the fewest since the JOLTS data was first compiled in December 2000.
The number of people quitting is still low compared with pre-recession levels, though it's risen in recent months. A low number of people quitting reduces opportunities for those out of work. About 2.3 million people quit in February, 7 percent higher than a year ago but below the average of nearly 2.9 million that were quitting each month when the recession began in December 2007.
Jason Faberman, an economist at the Federal Reserve Bank of Chicago, likens the job market to a game of musical chairs: If no one gets up, there isn't any room for anyone else to sit.
In March, U.S. employers added a net 88,000 jobs, the fewest in nine months and less than half the pace of the previous six months. At the same time, the increase in openings suggests that net job gains may strengthen in coming months.
Federal Reserve Chairman Ben Bernanke has said that total hiring, as gauged by the JOLTS report, is something he and other Fed officials track in assessing the job market. The Fed has said it plans to keep short-term interest rates at record lows at least until unemployment falls to 6.5 percent.
Steven Davis, an economics professor at the University of Chicago, says his research shows that companies aren't filling jobs as fast as they did in the past. Jobs now remain unfilled for an average of 25 days. That's up from about 16 days in mid-2003, when the job market was recovering from the previous recession.
Davis cites several reasons for the change. Companies may be uncertain about the outlook for the economy. And if the outlook worsens after a company posts a job opening, they may not fill it.
With unemployment high, the cost of missing a good hire doesn't seem as high, Davis says. Managers may figure they can always find someone just as qualified later.
And a bigger proportion of job openings in the United States is in health and education. Those jobs take longer to fill, partly because of higher skill and education requirements. Lower-skilled jobs in areas such as construction, which are typically filled more quickly, have declined.
Some economists also say that there may be a mismatch between the skills the unemployed have and what employers are seeking. Some manufacturers, for example, have reported difficulty in finding higher-skilled machinists, according to the Fed's Beige Book, which provides anecdotal information on economic conditions across the country.
All of which could make it harder to reduce unemployment. Howes says a "normal" level of U.S. unemployment may now be around 7 percent, rather than 5 percent to 6 percent.
Other economists doubt that a skills mismatch is playing a significant role. Elise Gould, an economist at the liberal Economic Policy Institute, says there are more unemployed people than jobs in nearly every industry. That suggests that a broad slowdown in the economy is to blame for still-high unemployment, rather than a shortage of qualified workers in certain industries.
Some companies may also have slowed hiring after steep government spending cuts began taking effect March 1. Those cuts are expected to shave about a half-point from economic growth this year.
That dynamic won't change, Cappelli says, until the economy grows fast enough that people feel comfortable about quitting to find jobs elsewhere.


MacGregor, part of Cargotec, has recently opened a new facility in Arendal, Norway, which houses a training academy for customers. It specialises in advanced simulation training and has a purpose-built virtual reality (VR) showroom.
The academy provides a risk-free environment where the users learn how to make real-time, complex manoeuvres safer and more efficient. It also showcases simulations of MacGregor products in action and demonstrates technical challenges.
"Virtual reality technology has improved dramatically in recent years," says Geir Roland, Director Advanced Offshore Solutions, Global Lifecycle Support at MacGregor. "We can pass these advances on to our customers at our new facility and through portable training programmes".
"The simulation/training software we use has been developed by our experts and is based on their expertise and experience in the field. We believe this is unique on the market," he notes.
VR training is particularly good for customers who have MacGregor equipment on board their vessels. "Customers can offer their crew fully-immersive training programmes, which are so much better than previous offerings," says Jan Finckenhagen, Training Manager, Advanced Offshore Solutions. "This will reduce the likelihood of causing injury to personnel or damage to equipment because they have already tried and tested it. Our aim is to help customers use their equipment safely and efficiently."
The VR showroom is divided into two zones comprising an authentic operating chair for offshore crane simulations and a zone where participants can walk around the simulated ship familiarising themselves with the safe operation of the equipment.
VR headsets offer extremely realistic 3D visualisation, allowing users to view very small details of an operation, as well as the wider picture. They are linked to large, wall-mounted screens to display exactly what the user sees. This is very useful for the instructor monitoring and guiding the training sessions and improves the effectiveness of the training process.
"When you wear VR headsets linked to a computer running MacGregor's VR software, it is just like being on a vessel or offshore installation. As you move around and turn your head you see exactly the same things that you would see if you were on board. It is a very convincing experience," explains Mr Finckenhagen.
"You can also explore restricted, dangerous areas that you would not normally be allowed to enter. This provides otherwise unobtainable perspectives on specific operations, which can prove very useful in the real world."
When customers approach MacGregor early in the design phase of their projects, simulation services can be really beneficial. They can explore and test a product's capabilities before it is a reality.


Charles Sustains Mallett was born in Kessinland, Suffolk, England, on 1 November 1866. His parents were William Mallett and Elizabeth nee Sustains. His wife's name was Susan Josephine (nee Hendren); they had two sons and three daughters. He died at Gunnamatta Bay NSW on 27 June 1949. A requiem mass was held at the Church of St Joseph, Camperdown, before his burial at Waverley Cemetery.
Charles Mallett arrived from England at an early age. He became a butcher with an interest in promoting the industry. He was involved in licensing metropolitan butchers' shops and was on the committee deciding the site of the Central Meat Markets in 1914. He was the chairman of the Master Butchers' Cooperatives Ltd from about 1905 to 1915, president of the Master Butchers Association of NSW, and a committee member of the NSW Retail Butchers Association.
In October 1928 Charles Mallett, small goods merchant, leased Manning Chambers Room A10 at 449A Pitt Street, Sydney. In 1937 Mallett was present and photographed at the first annual Commonwealth Meat Trade Conference held in Sydney. He was the president of the Camperdown Cricket Club, 1905–30, and several athletic clubs. He lived in Camperdown and Glebe before retiring to Gloucester House, Lugarno Avenue, Gunnamatta Bay. Mallet Street in Camperdown is named after him.
Mallett was the President of the Camperdown Cricket Club 1905-30.
Charles Mallett was Alderman of Camperdown from 1902-08 and its Mayor in 1906-07. Mallett as Camperdown Mayor in 1905-6 had to deal with the parlous financial state of Camperdown Council, with its roads neglected and municipal services limited.
In 1906 amalgamation with Newtown Council was proposed, but as the proposal did not proceed as the required two thirds majority was not met. In 1909, Mallett worked for the amalgamation of Camperdown borough with Sydney City Council.
Mallett was Alderman of Sydney City Council for Camperdown from 8 January 1909 to 30 November 1924. He was a member of the Health and By-Laws Committee, 1909-19, and its Vice-Chairman in 1917. He belonged to the Works Committee, 1910-24 and served as its Chairman in 1920. He was a member of the Electric Lighting Committee in 1911, the Finance Committee, 1920-24, and the Electricity Supply Committee in 1923. Mallett was a Labor Party supporter who changed allegiances in 1912.