text:
ZMF.ORG ABOUT US CONTACT US TEACHINGS RESOURCES CONGREGATIONS
Does Israel have the right to the Land of Israel?
This special topic was recorded on March 29, 2008. Our speaker for this session was Elie Nessim.
Cannot play video. Please make sure you have Flash installed.
The text for this session:
Today's question is concerning Israel and the Land. The question is as follows:
Does Israel have the right to the Land of Israel? What is the Biblical perspective and how should believers be responding? Is there a Biblical basis for a two-state solution in Israel?
In reply, let us consider the Biblical evidence. It's as follows: -
GOD's promise to Abraham, His promise to Isaac, His promise to Jacob, His promise to the twelve tribes. This promise is irrevocable, judgement is threatened on usurpers, and strangers are welcomed.
Let's look at these in order. First of all, GOD's promise to Abraham.
In Genesis 12:1 "Now the LORD had said to Abram: "Get out of your country, from your family and from your father's house, to a land that I will show you."" And again, when Abraham arrived at the land – verse 7: "Then the LORD appeared to Abram and said, "To your descendants I will give this land." And there he built an altar to the LORD, who had appeared to him.""
Abraham's title to the land was confirmed. In Genesis 13, verse 14 through 17: "And the LORD said to Abram after Lot had separated from him: "Lift your eyes now and look from the place where you are – northward, southward, eastward, and westward; for all the land which you see I give to you and your descendants forever. And I will make your descendants as the dust of the earth; so that if a man could number the dust of the earth, then your descendants also could be numbered. Arise, walk in the land through its length and its width, for I give it to you.""
Also, Abraham's title to the land was confirmed by Covenant. In Genesis 15, verse 7: GOD said to Abraham – "I am the LORD, who brought you out of Ur of the Chaldeans, to give you this land to inherit it."
And then from verse 18 through 21: "On the same day the LORD made a covenant with Abram, saying: "To your descendants I have given this land, from the river of Egypt to the great river, the River Euphrates – the Kenites, the Kenezzites, and Kadmonites, the Hittites, the Perizzites, the Rephaim, the Amorites, the Canaanites, the Girgashites, and the Jebusites.""
That was GOD's promise to Abraham. This was passed on to Isaac. In Genesis 17, verse 7 and 8, GOD said to Abraham: "And I will establish My covenant between Me and you and your descendants after you in their generations, for an everlasting covenant, to be God to you and your descendants after you. Also I give to you and your descendants after you the land in which you are a stranger, all the land of Canaan, as an everlasting possession; and I will be their God."
There's the land secured to Abraham's children, and also in Genesis 17, verse 20 and 21, the choice was narrowed down to Isaac. "And as for Ishmael, I have heard you. Behold, I have blessed him, and will make him fruitful, and will multiply him exceedingly. He shall beget twelve princes, and I will make him a great nation. But My covenant I will establish with Isaac, whom Sarah shall bear to you at this set time next year."
Not only was the choice narrowed down to Isaac, but it was personally confirmed to Isaac by GOD himself. In Genesis 26, verse 3, the LORD said to Isaac: "Dwell in this land, and I will be with you and bless you; for to you and your descendants I give all these lands, and I will perform the oath which I swore to Abraham your father."
Thirdly, we see GOD's promise to Jacob. In Genesis 28, we read about Jacob fleeing from his brother Esau, and GOD appearing to him. Genesis 28, verse 12 and 13: "Then he dreamed, and behold, a ladder was set up on the earth, and its top reached to heaven; and there the angels of God were ascending and descending on it. And behold, the LORD stood above it and said: "I am the LORD God of Abraham your father and the God of Isaac; the land on which you lie I will give to you and your descendants.""
So the promise was repeated to Jacob and his sons. Later on also in Genesis chapter 35, the LORD again confirmed the promise to Jacob. Verse 12: "The land which I gave Abraham and Isaac I give to you; and to your descendants after you I give this land." Then God went up from him in the place where He talked with him."" So it was divinely confirmed twice to Jacob.
Fourthly, it was then passed on to the twelve tribes when GOD called the people of Israel out of the land of Egypt, in Exodus chapter 6, verse 7 and 8:
"I will take you as My people, and I will be your God. Then you shall know that I am the LORD your God who brings you out from under the burdens of the Egyptians. And I will bring you into the land which I swore to give to Abraham, Isaac, and Jacob; and I will give it to you as a heritage: I am the LORD."
So here is GOD's promise announced to the nation of Israel, and as it is celebrated in the Psalms, it was by an everlasting covenant. Psalm 105 verse 7 through 12:
"He is the LORD our God; His judgments are in all the earth. He remembers His covenant forever. The word which He commanded, for a thousand generations, The covenant which He made with Abraham, And His oath to Isaac, And confirmed it to Jacob for a statute, To Israel as an everlasting covenant, saying "To you I will give the land of Canaan as the allotment of your inheritance, when they were few in number, indeed very few, and strangers in it.""
Fifthly, GOD's promise is irrevocable. Speaking of a time of apostasy, in Leviticus chapter 26, and the judgements that would overtake His people for their apostasy, GOD goes on to say this in Leviticus 26, verse 40:
"But if they confess their iniquity and the iniquity of their fathers, with their unfaithfulness in which they were unfaithful to Me, and that they also have walked contrary to Me, and that I also have walked contrary to them and have brought them into the land of their enemies; if their uncircumcised hearts are humbled, and they accept their guilt – then I will remember My covenant with Jacob, and My covenant with Isaac and My covenant with Abraham I will remember; I will remember the land. The land also shall be left empty by them, and will enjoy its Sabbaths while it lies desolate without them; they will accept their guilt, because they despised My judgments and because their soul abhorred My statutes. Yet for all that, when they are in the land of their enemies, I will not cast them away, nor shall I abhor them, to utterly destroy them and break My covenant with them; for I am the LORD their God. But for their sake I will remember the covenant of their ancestors, whom I brought out of the land of Egypt in the sight of the nations, that I might be their God: I am the LORD."
The question arises, how does GOD make up the deficiency of His people? And here is the answer. In the Song of Moses, Deuteronomy 32, verse 43:
"Rejoice, O Gentiles, with His people; for He will avenge the blood of His servants, and render vengeance to His adversaries; He will provide atonement for His land and His people."
The price that Israel should have paid, GOD promises to pay Himself. "He will provide atonement for His land and His people." The land is secured by GOD Himself.
Sixthly, GOD, in the Prophets denounces those nations that seek to usurp the land, which He had promised His people.
In Jeremiah chapter 12, verses 14 and 15, this is what GOD has to say about those nations that seek to usurp the land He promised His people: "Thus says the LORD: "Against all My evil neighbours who touch the inheritance which I have caused My people Israel to inherit – behold, I will pluck them out of their land and pluck out the house of Judah from among them. Then it shall be, after I have plucked them out, that I will return and have compassion on them and bring them back, everyone to his heritage and everyone to his land."
Evil is threatened against any other nation that seeks to expropriate the land promised to Abraham, Isaac, Jacob, and the twelve tribes of Israel.
But there is lastly, provision made for strangers. In Ezekiel chapter 37, verse 22, the first thing GOD says there is that the nation would be one united nation.
Ezekiel 37, verse 22: "And I will make them one nation in the land, on the mountains of Israel; and one king shall be king over them all; they shall no longer be two nations, nor shall they ever be divided into two kingdoms again." So there's no such thing as a two-state solution.
Also, in Ezekiel chapter 47 and verse 22, the period called the Millennium, when Messiah reigns in Israel, the LORD says in Ezekiel 47, verses 21 – 23 for completeness:
"Thus you shall divide this land among yourselves according to the tribes of Israel. It shall be that you will divide it by lot as an inheritance for yourselves, and for the strangers who dwell among you and who bear children among you. They shall be to you as native-born among the children of Israel; they shall have an inheritance with you among the tribes of Israel. (Not besides them or apart from them). And it shall be that in what ever tribe the stranger dwells, there you shall give him his inheritance," says the Lord GOD.""
By way of conclusion, In view of the above Scriptures there is only the State of Israel occupying the land that was promised to them by GOD.
(Q): In Genesis 15:18, it says to your descendants and it describes from the Nile to the Euphrates. Do you see that as a promise that Israel could have owned all that land? Or is that really only going to be fulfilled in the Messianic Kingdom?
(A): Genesis chapter 15 and verse 18: "To your descendants I have given this land, from the river of Egypt to the great river, the River Euphrates…" As I understand it, they will never fully occupy the land until they are forgiven as a nation. When GOD promised the land to Israel, He didn't give it all to them when they entered the land after the Exodus because of their unbelief. And so the fullness of the promise awaits the time when Israel will be a righteous nation, and will be entitled to possess the whole land. So yes, that means that this promise here will be fully realized only in the Millennium under the Messiah.
(Q): What answer would you give to someone who says there's been so much warfare over the Land of Israel for centuries and why do you think that's been the case?
(A): The reason is because of the disobedience of the nation. GOD promised them that if they observed His statutes, their enemies may come against them one way and flee from them seven ways. The problem was that they did not trust and obey, and the result was what the LORD warned them of – that they would have wars and conflicts. It's all a proof of the consequences of disobedience and of disbelief.
(Q): GOD has placed His presence in the Land of Israel, and His eyes watch over the Land and it's a place that He delights in. In reference to that, is that a special place obviously that GOD holds to? Is that also why there have been problems over this piece of land? Or could there be other reasons?
(A): I believe you have hit upon the right reason why the Land is coveted by others, because Satan is the god of this world and whatever GOD loves, Satan hates. That's why he tempted man to fall – man who is the crown of GOD's creation. So it just shows that this piece of Middle Eastern real estate is very precious to GOD. GOD says 'the eyes of the LORD your God are upon it from one year's end to the other.' And for that reason, the evil one seeks to mar it and to deprive GOD of His pleasure in it.
(Q): You know that Abraham nearly sacrificed His son Isaac on that Mount; the same mount which Messiah died upon. Could we say that this Land has been precious to GOD and maybe goes right back to the point that the Garden of Eden could have been residing within this part of the world or is that too speculative?
(A): That might be somewhat speculative. No doubt the Land was precious to GOD because He is a GOD who knows no time, and so we can say that His love for the Land was a constant, but it may have been enhanced because it became the site not only of Abraham's act of obedience at the Akkedah when he nearly sacrificed Isaac, but because it was the Land where Messiah, GOD's Isaac, gave His Life for the sins of the world.
(Q): We are called to pray for the peace of Jerusalem, yet in some sense we know that it won't occur until Messiah reigns. For someone who wishes to pray for the peace of Jerusalem what's the encouragement at this time to continue to pray that prayer? What's the reality of that prayer that we are hoping for?
(A): The Bible says "Pray for the peace of Jerusalem; may they prosper who love you." There's a promise of blessing to those who pray for the peace of Jerusalem, and although we do not see the outcome of our prayers, yet our prayers contribute much to the comparative peace of Jerusalem. Jerusalem would be worse off without our prayers and so would the world, because the future of the world, the happiness of the world are all dependent on the peace of Jerusalem.
One old Jewish saint said that Jerusalem is like the heart of the world. When the heart beats regularly, the world is in good health. He compared it to a human body. When the heart is sound, the body is sound. When the heart is unsound so is the body. Therefore when we pray for the peace of Jerusalem, we may well by our prayers be not only anticipating the day when perfect peace will remain in Jerusalem, but we are also holding back much that could harm, that could destroy the peace of Jerusalem, and thereby the peace of the world.
We see it in our international situation where the troubles of Israel reverberate throughout the whole world like ripples from a pond. Therefore it is a responsibility as Isaiah put it in chapter 62, and
verse 6 and 7: "I have set watchmen on your walls, O Jerusalem; they shall never hold their peace day or night. You who make mention of the LORD, do not keep silent, and give Him no rest till He establishes and till He makes Jerusalem a praise in the earth."
And the speaker Himself has set the example in verse 1: "For Zion's sake I will not hold My peace, and for Jerusalem's sake I will not rest, until her righteousness goes forth as brightness, and her salvation as a lamp that burns."
Much is involved in our prayers for the peace of Jerusalem.
(Q): There is a lot of obvious political wrangling over the Land of Israel today. What advice would you give to believers that are watching these attempts and political solutions being thrust, sometimes heavily, upon Israel trying to attempt to bring peace and how should believers respond biblically?
(A): My advice to them first of all is that GOD is true to His word, and that what He says will certainly come to pass. My second piece of advice would be to plead GOD's promises to Israel, to plead them before the Throne of Grace, and to ask GOD for the accomplishment of His promise. And thirdly I would advise them to encourage those of our brothers and sisters in the Land itself to stand firm, and to believe GOD despite all appearances to the contrary. One thing we can be sure of, GOD said, "the Land shall not be sold forever, for the Land is Mine." And GOD will make good that precept.
(Q): And the true peace of Jerusalem will come when Messiah reigns?
(A): The true peace of Jerusalem only comes when Messiah returns and reigns. Look at how Psalm 72 puts it concerning the Messianic King.
"He will judge Your people with righteousness, and Your poor with justice, the mountains will bring peace to the people, and the little hills by righteousness. He will bring justice to the poor of the people; He will save the children of the needy, and will break in pieces the oppressor." That's Psalm 72, verses 2 through 4. That is the glorious promise, which we depend upon.
(Q): When we pray for the peace of Jerusalem, is it in effect a prayer directly to the LORD for His coming, because after all is He not the one who brings peace and no one else?
(A): I quite agree. You've 'hit the nail on the head.' When we pray for the peace of Jerusalem, we are also praying at the same time, for the coming of the Prince of Peace who brings peace to Jerusalem. Thank you for that thought.
Back to special topics list.
For more information on Zion Messianic Fellowship
email us at info@tsion.org, or call (604) 800-3663.


text:
How COVID reshaped Seattle, season by season
Each season of the pandemic altered the look and feel of the city. With a trusty Prius and cameras, disaster researchers mapped the morphing disruption.
Jane C. Hu
A team of public health researchers and engineers is capturing how COVID-19 policies and the behavior of Seattle residents have changed the look and feel of the city. (Kiliii Yuyan/High Country News)
Just before 7 on a cool, misty Seattle morning, Jacqueline Peltier stands alone on the University of Washington campus. Nearby, squirrels and rabbits frolic in the morning dew. Peltier, part of a National Science Foundation-funded research team, will spend the next hour securing a 360-degree camera to the roof of a rental Toyota Prius Prime, ensuring it's level and synced with its smartphone controller. In the past, this setup traveled to Puerto Rico, capturing the aftermath of Hurricane Maria. Today, Peltier is crisscrossing Seattle on a 100-mile route to document a different emergency: the COVID-19 pandemic.
The idea arose in the first days of COVID-19's spread across Seattle. Joe Wartman, Peltier's colleague, an environmental engineer who studies natural disasters, was walking home from work one day when he noticed the empty streets, abandoned businesses and quiet sidewalks. "It struck me how much Seattle looks like a disaster zone," he said.
Unlike earthquakes or hurricanes, however, the pandemic has been raging for months. Since March, public health recommendations have tightened and loosened, businesses have closed and reopened, and travel has ebbed and flowed. Peltier and Wartman, along with a team of public health researchers and engineers, are capturing how COVID-19 policies and the behavior of Seattle residents have changed the look and feel of the city. "We haven't had a pandemic at this scale in modern history," said study co-investigator Nicole Errett, an assistant professor of public health at the University of Washington. "This data provides the novel opportunity to really see, as we change policy, what happens in the short term."
Peltier has retraced the same path through Seattle roughly every three weeks. The route was carefully mapped out using neighborhood census data to examine how socioeconomic status influences business recovery and neighborhood activity. "We really want to understand what role this pandemic is going to play in either bringing us closer together economically, or pushing us further apart," said Wartman.
Each drive generates tens of thousands of images, and it will take the researchers years to pore over the details. In addition to clues about economic recovery, they're hoping to analyze people's behavior in response to public health measures, everything from compliance with mask mandates to average group sizes in city parks. Peltier has already noticed trends in the various phases.
In May, near the pandemic's start, Seattle was a ghost town. "I had free rein of the streets," she said. The researchers looked to landmarks like Seattle's Harborview Medical Center to observe changes, said Errett. Initially, the area outside the hospital was empty, save a row of triage tents set up to accommodate the potential surge of COVID patients. At the time, local businesses were closed, elective surgeries were canceled, and people generally avoided hospitals unless they had COVID-19 or an acute emergency, so it's no surprise car and pedestrian traffic around the hospital was practically nonexistent. People were clearly abiding by stay-at-home orders.
Over the summer, as restrictions loosened and weather improved, triage tents were taken down and signs of life emerged. In June, pedestrian and car traffic reappeared, and by July, nurses and doctors took breaks outside the hospital, standing alongside people at a nearby bus stop. "You're seeing people return to spaces around health care facilities, so one hypothesis is that they're more comfortable, or less scared, than they were at the beginning of the pandemic," said Errett. Peltier noticed that local businesses were reopening and often leaving their doors open — either as an invitation to customers or a means to improve ventilation, or perhaps both.
But over the fall, Peltier noted a new phase of the pandemic: the failure of small businesses. While Seattle retail behemoth Amazon and food delivery apps like DoorDash have seen profits soar during the pandemic, local stores and restaurants don't appear to be doing as well. Signs reminding customers to mask up or stay distanced have been replaced by closing notices, or letters of thanks to loyal customers. In some cases, storefronts are empty and dark. "Maybe they've just put things in storage," Peltier said. "But maybe they're not going to open again."
In mid-November, amid a nationwide surge in COVID-19 cases, Washington Gov. Jay Inslee announced a new set of restrictions similar to the stay-at-home order at the beginning of the pandemic. The economic and social impacts remain to be seen, but the data Peltier is capturing will likely tell us more — and may even help in future disaster preparation. "I would be surprised if this is the last pandemic I see in our lifetime," said Errett. "I want my community to know more, so that next time we're advancing our response, it's based on what we learned in this pandemic."
This story was first published at High Country News on Dec. 9, 2020.
2020 was the year of empathy fatigue
We were too ready to abandon our older selves.
Katie Wilson
/ December 22
Inside Crosscut
The top stories readers helped us tell in 2020
Some of the most important stories we published this year were the ones you asked us to tell.
Mohammed Kloub
WA youth in the state's care endure rain, cold for holiday visits
The state's inpatient psychiatric facility for youth requires outdoor visits during the pandemic. But that means shivering in the elements for many families.
Emily McCarty
Jane C. Hu is a freelance science writer. Her work has appeared in Slate, Nautilus and Pacific Standard. You can find her at janehu.net, or on Twitter as @jane_c_hu.
The congresswoman from Washington state details a harrowing day at the Capitol and how the House will move forward on the pandemic, infrastructure and climate.

redpajama_set_name: RedPajamaCommonCrawl


This makes so much sense it's scary. A $200 computer, by Google, on sale NOW at WalMart that leverages the power of the Internet with browser-based and open source software (thus no "Microsoft tax").
I might just be geeky enough to go out and buy one. We'll see. Meanwhile, click here for details of the Everex gPC as covered by John Biggs in the NYTimes.
http://www.combridges.com/wp-content/uploads/2017/06/ComBridges-Logo-300x60-300x60.jpg 0 0 JonLeland http://www.combridges.com/wp-content/uploads/2017/06/ComBridges-Logo-300x60-300x60.jpg JonLeland2007-11-02 12:05:002007-11-02 12:05:00Revolutionary gPC is a $200 Web-facing Computer with no "Microsoft tax"
We are way too early in this to be talking about blossoming, but the sprouts or seeds that are the revolution of the advertising business going totally digital, totally accountable, totally targeted and totally interactive are HUGE!
I cover the basics in my internet marketing ebook, and today's New York Times covered how Microsoft is trying to catch up with Google in this arena. This is a good article packed with perspective.
Hey, isn't this the guy who runs a software company that sells more software (and software upgrades) than any other company on earth?
In other words, free ad-supported digital media will rule, and digital advertising will blossom into channels of humongous proportions… proportions that make Google as it exists today seem small.
Bottom line, they all agree that it's ALL going digital and this will change everything. Personally, I'm interested in building an advertising "on-ramp" for small businesses… A web application platform that would guide small business users so that they could leverage the advantages of the new media & advertising world. This is seriously needed by millions of small companies.
Like I've been saying for years, we need communication bridges spanning the grand canyon gap between people and technology… Anyone want to play?
I don't think there are two better visionaries on the planet (of course, I'm partial to Steve Jobs), and while their joint appearance with Wall Street Journal's Walt Mossberg was covered in the news, I just discovered the complete coverage including transcripts and video clips (an edit of video excerpts is included below). Good stuff!
There are many ways to look at the launch of Google's new Apps platform, but for me, it looks like a pretty cool and very low cost ($50/year/user) way for a small business to do a web site without traditional hosting, and with a bunch of neat features.
Because it supports direct domain (as well as sub-domain) hosting and includes integrated e-mail as well as a bunch of features (from Google Page Creator for creating web pages to calendars and other forms of collaboration), it's quite powerful. More info here specifically for small businesses.
I think the biggest challenge will be having small businesses understand the scope of what's being offered.
On the other hand, the bigger business story for larger enterprises and the broader IT market is that Google is Challenging Microsoft (NYTimes) with a service that offers email and more for $50/user vs $225/user annually using Office and Exchange.


text:
How does a Laser Pointer Work? The Power Rating and Green Laser Attention
Home » Laser Pointers » How does a Laser Pointer Work? The Power Rating and Green Laser Attention
Ever wondered how does a laser pointer work? Albeit its compact size, there are a lot of things going on behind the scenes to produce the tiny laser beam from your pointer.
If you're curious about what goes on inside your trusty pocket laser pointer, or just lasers in general, keep on reading!
Cheap green laser pointers may not filter out any unconverted infrared laser light. If this invisible infrared laser light shines in your eye, you won't be able to see it, so you don't blink or move, causing permanent damage to your eyes.
What is a laser pointer?
When was the laser pointer invented?
How does a laser pointer work?
Laser pointer's power ratings
Why you don't want to use a cheap green laser pointer
How to safely use your pointer
The word "LASER" itself is an acronym of Light Amplification by Stimulated Emission of Radiation.
Only a handful of modern lasers use light amplification but the name stuck even for lasers with one color or wavelength (monochromatic laser).
In a nutshell, a laser pointer is a device typically used to highlight something important. It works by powering a laser diode with an energy source which will then emit a coherent low-powered beam of visible light.
The uses for a laser pointer vary. It can be used during a presentation, to point out stars, for science experiments, and more. Laser pointers are available in various colors, sizes, and prices as well.
For instance, the pocket laser pointer that we're familiar with emits a red laser and is quite affordable. On the other hand, laser pointers used for stargazing are usually green and have higher wattage so we can use it during the night.
Other laser colors include blue, violet, and yellow.
Affordable, pocket-sized laser pointers that we know today can be dated back to the 1980s. However, the concept of the laser can be dated back to the late 1950s.
Researchers at Bell Labs produced a theory to create lasers in 1958. Two years later, Hughes Aircraft built the very first laser, using rubies, light, and mirrors to produce the beam.
Then, in 1962, the first semiconductor laser was created. Also known as diode lasers, this technology made laser pointers more affordable and accessible to the mass market.
Laser pointers have three important components: lasing medium, energy source, and a resonator.
The lasing medium refers to a material that can be powered by an energy source, which can be in the form of light or electricity. Light mediums can be in the form of gas, liquid, or minerals like the rubies used in the first laser pointer.
Once the laser medium is powered or pumped, it will release the energy in the form of monochromatic radiation. The function of the resonator then comes to play. It holds the energy and builds it up before it releases the energy.
The resonator is made up of two mirrors located at each end of the lasing medium. One would reflect the light back to the medium and the other would divide it between the medium and the chamber's exit.
The resonator is also responsible for aligning the light in one direction, which causes coherency of the laser beam. The alignment of the light is done through the bouncing of the light that goes back and forth between the two mirrors.
Semiconductor lasers or diode lasers produce light when electricity passes through them in a single direction. Laser diodes usually contain a circuit board, optics, and a case.
The semiconductor is usually made from compounds like aluminum, arsenic, phosphorus, gallium, or indium. Traces of tantalum and gold can also be found.
Refer to the video below for more information on how the laser works.
How Lasers Work - A Complete Guide
Read more: How Does Mosquito Laser Work?
As a safety measure, laser pointers are classified into four main classes and subclasses based on their wavelength and maximum output power.
The classes are categorized based on how damaging the laser can be to the naked eye, with 1 being no danger to 4 being severely dangerous for the eyes and skin.
The following is the complete list of the laser classifications:
Safety Level
Class 1 Completely safe
Class 1M Safe without optics or viewing aids
Class 2 Visible laser; safe when accidentally exposed
Class 2M Visible laser; safe when accidentally exposed without optics or viewing aids
Class 3R Unsafe; low risk
Class 3B Dangerous but safe to view the reflection of radiation on surfaces
Class 4 Dangerous; unsafe to view the reflection of radiation on surfaces; fire risk
Some low-quality laser pointers can be wrongly classified. They could possess higher power than they're supposed to, which can be dangerous if wrongly used.
The National Institute of Standards and Technology conducted tests on laser pointers labeled as Class IIIa or 3R in 2013.
The results showed that about half of the laser pointers emitted power at twice the Class limit. Hence, instead of Class IIIa, the correct classification should be Class IIIb, which is more dangerous than Class IIIa.
The highest measured power output was 66.5 milliwatts; more than 10 times the limit.
Green laser light is generated from an infrared laser beam, which should be confined within the laser housing. However, more than 75% of the laser pointers tested were found to emit infrared light in more than the limit.
If you're hesitant on which type of laser pointer to choose, it's best to avoid cheap green lasers. Green lasers often use an invisible infrared laser that is then converted into visible green light by clever optical materials.
This can be further explained scientifically through the concept of infrared diodes.
Infrared diode lasers are normally cheaper, which led to the creation of infrared-pumped green, violet, and blue diode laser pointers with double frequency. With infrared (IR) diode laser modules of up to 1000 mW, those laser pointers may have visible power up to 300 mW.
Filtering invisible infrared light in the laser beam is highly challenging. Doing so can contribute to a higher heat level that would be hard to distribute within a pocket-sized laser pointer.
The infrared then is left as particles in cheaper high-power laser pointers because of the difficulty to filter the infrared. This invisible infrared component in these cheap green laser pointers can be a source of an extra potential hazard when pointed at nearby objects and people.
Now you have already known the answer to the question "how does a laser pointer work?". In addition, when using a laser pointer, no matter how low or high the power of the light is, you should always maintain the safety protocols.
Here are some important things to always keep in mind when using a laser pointer:
Never direct your laser pointer towards someone's eyes
Never point the laser beam directly towards an animal
Never look directly into the laser beam
Never aim your laser pointer towards any kinds of vehicles
Take note of reflective surfaces and avoid them
Avoid prolonged contact between the laser and the skin
Wear safety goggles and gloves when using high-powered lasers
Categories Laser Pointers Post navigation
[Fun Tricks] 13 Cool Things To Do With Your Laser Pointers
How To Make Slides Transition Automatically In Google Slides?
[Review] Doosl Wireless Presenter With A Green Laser Pointer
[Review] KNORVAY N76RBK Wireless Presenter With A…
[Review] Dinofire Wireless Presenter With A Long…
Green vs. Red Laser: What are the differences?
How Does Mosquito Laser Work? Here's How To…
Do Laser Beams Work Underwater? Yes, But What Are Use Cases?
How Exactly Do Laser Tag Guns Work? Here Are Simple…
How To Fix A Broken Laser Pointer
Can You Use A Laser Pointer On LED/LCD TV Screens?
[Review] Red Star Tec PR-819 Laser Pointer


text:
var c = require('net').createConnection({host : 'PUBLIC_IP_OF_CLIENT_A', port : PUBLIC_PORT_OF_CLIENT_A},function () {
  console.log('> (clientB) connected to clientA!');

  c.on('data', function (data) {
    console.log(data.toString());
  });
});

redpajama_set_name: RedPajamaGithub


Monthly variations of fungal biomass and assemblages on leaves exposed to the air and submerged, of sporulation and specific richness in the water, and of cel- lulolytic activity in the leaves were investigated in a warm temporary river. The bio- mass of terrestrial fungi was relatively low in the dry air and cellulolytic activity also was low . During submersion, biomass and activity of the aquatic hyphomycetes were greater. The fungi were most active in spring when the water was flowing and when breakdown was fast. Complete exclusion of one fungal species on one leaf species, presumably by competition, was observed twice. From June onwards, though fungal biomass remained high, fungal diversity and activity decreased, and further decay was presumably due to other microorganisms. No major difference to the same processes in temperate permanent streams appeared except that initiation of the sequence was the onset of the aquatic phase and not deciduous leaf fall.


This weekend we took LM bowling for the first time. I loved her tiny bowling shoes and she loved picking out which color ball to throw down the lane.
In the end, even though I got the only strike of the game (pat on back), I got beat out by a two-year-old! I'm serious. I'm not even going to show you the picture Husband took of the scoreboard (although I will tell you that he only beat her by 4 points, ha ha).
Cute Lark was our sleepy cheerleader on the sidelines - she didn't even know we left the house.
Random fact about me: I was in bowling club in high school and our team was called the Ditsy Chicks (no comment). It was a lot of fun, although I think I was the worst one on the team - it always takes me a game or two to get warmed up before I'm any good (this time was no different).
We ended our game with an impromptu dance party, then commiserated/laughed about our mediocre bowling skills over yummy cupcakes. How was your weekend?
I'm terrible at bowling, yet I enjoy it too. Haven't been in a long while but it's always good times.
shop update and happy easter!


text:
\section{Introduction}
\label{sec:intro}

 The Tutte polynomial is a two variable polynomial which can be
 defined for  a graph $G$ or, more generally, a matroid $M$. The Tutte
 polynomial  has many interesting combinatorial interpretations
 when evaluated at different points ($x$, $y$) and along several
 algebraic
 curves. For example, for a graph $G$, the Tutte polynomial along the
 line $y=0$ is the chromatic polynomial, after a suitable change of
 variable and multiplication by an easy term. Similarly, we can
 get the flow polynomial of a graph and the all terminal reliability
 of a network and the partition function of the $Q$-state Potts
 model. When considering  a GF($q$)-representable matroid, the Tutte
 polynomial gives us the weight enumerator of  linear codes over
 GF($q$) associated to $M$. All the necessary background on the Tutte
 polynomial is contained in Section~\ref{sec:prelim}.

 It is well-known~\cite{Bjorner} that the Tutte polynomial of a matroid $M$
 has an expansion
\[ T_M(x,y) = \sum_{i,j} t_{ij}x^iy^j,\]
in which each coefficient $t_{ij}$ is non-negative. Consequently, for
$m\geq 0$ and for any $b$, if $(x,y)$ lies on the portion of the line
$y=mx+b$ lying in the positive quadrant, then $T_M$ increases as $x$
increases. The simplicity of the behaviour of $T$ along lines with
positive gradient suggests the study of the behaviour of $T_M$ along
lines with negative gradient in the positive quadrant. Merino and
Welsh~\cite{MrWl} were the first to consider this and were particularly
interested in resolving the question of whether the Tutte polynomial
is convex along the portion of the line $x+y=2$ lying in the positive
quadrant. They made the following intriguing conjecture.

\begin{conjecture}\label{old}
Let $G$ be a 2-connected graph with no loops. Then
\begin{equation}
\max\{T_G(2,0),T_G(0,2)\} \geq T_G(1,1). \label{eq:MW}
\end{equation}
\end{conjecture}
Any graph with at least one loop and at least one isthmus fails to
satisfy~\eqref{eq:MW}, so~\eqref{eq:MW} cannot hold for all
graphs. The main reason for the particular interest in the points
$(2,0)$, $(0,2)$ and $(1,1)$ is that in a connected graph $G$,
$T_G(2,0)$, $T_G(0,2)$ and $T_G(1,1)$ give the number of acyclic
orientations, totally cyclic orientations and spanning trees in
$G$. Definitions of acyclic and totally cyclic orientations are
contained in Section~\ref{sec:prelim}.

A related question is to determine whether any loopless 2-connected
graph $G$ satisfies the apparently stronger requirement
\[ T_G(2,0)T_G(0,2) \geq (T_G(1,1))^2.\]
Relatively little progress has been made to resolve these
questions. However, Jackson in~\cite{Jackson} has shown, with a
clever argument, that for any connected matroid $M$,
\[T_G(3,0)T_G(0,3) \geq (T_G(1,1))^2.\]

In this paper we make three contributions. First, in Section 4, we
show that the Tutte polynomial $T$ of a coloopless paving matroid
satisfies the inequality
\begin{equation}
tT(x_1,y_1)+(1-t)T(x_2,y_2) \geq T(tx_1+(1-t)x_2,ty_1+(1-t)y_2), \label{eq:convex}
\end{equation}
where $0\leq t \leq 1$ and $x_1,x_2,y_1,y_2$ are non-negative and
satisfy $x_1+y_1=x_2+y_2$. That is, $T$ is convex along the portions
of the line segments $x+y=p$ lying in the positive quadrant.
A paving matroid is one in which all circuits have size at least
$r(M)$. Interest in them stems from a conjecture in~\cite{MyNwWlWh}
which says that asymptotically almost every matroids is paving.
The special case of~\eqref{eq:convex}, obtained by setting
$x_1=y_2=2$, $x_2=y_1=0$ and $t=1/2$, establishes \eqref{eq:MW} for
the class of paving matroids. Therefore if the above conjecture is true
then we have established~\eqref{eq:MW} for, asymptotically, almost all
coloopless matroids.

Second, in Section 5, we prove that \eqref{eq:MW} holds for some
smaller classes of matroids and graphs that are not paving
matroids. Finally, in Section 3, we prove that if the ground set of
$M$ contains two disjoint bases then $T_M(0,2a) \geq T_M(a,a)$ and
dually if the ground set of $M$ is the union of two bases then
$T_M(2a,0)\geq T_M(a,a)$. These results cannot be obtained with the
methods used by Jackson in~\cite{Jackson}.

We conclude with a brief discussion of the natural question of for
which matroids is $T_M$ a convex function in the positive quadrant?

\section{Preliminaries}
\label{sec:prelim}

 We assume that the reader has some familiarity with matroid and graph
 theory. For matroid theory we follow Oxley's book \cite{Oxley} and
 for graph
 theory we follow Diestel's book \cite{Diestel}.

 The Tutte polynomial is a matroid invariant over the ring
 $\mathbb{Z}[x,y]$. Further details of many of the concepts
  treated here can be found in Welsh \cite{Welsh} and Oxley and
 Brylawski \cite{BrOx}.

 Some of the richness of the Tutte polynomial is due to its numerous
 equivalent definitions. One of the simplest
 definitions, which is often the easiest way to prove properties of
 the Tutte polynomial, uses the notion of rank.

 If $M=(E,r)$ is a  matroid, where $r$ is the rank-function of $M$,
 and $A\subseteq E$, we denote $r(E)-r(A)$ by $z(A)$  and $|A|-r(A)$ by
 $n(A)$.

\begin{definition}\label{definition}
 The Tutte polynomial of $M$, $T_{M}(x, y)$, is defined as follows:
 \begin{equation}\label{eq:expansion}
     T_{M}(x, y) = \sum_{A\subseteq E} (x-1)^{z(A)}(y-1)^{n(A)}\;.
 \end{equation}
\end{definition}

 Almost immediately  we see that $T_{M}(1,1)$ equals the number of
 bases of $M$ and $T_{M}(2,2)$ equals $2^{|E|}$.
 Recall that if $M=(E,r)$ is a matroid, then $M^{*}=(E,r^{*})$ is its
 dual matroid, where $r^{*}(A)=|A|-r(E)+r(E\setminus A)$. Because
 $z_{M^{*}}(A)=n_{M}(E\setminus A)$ and $n_{M^{*}}(A)=z_{M}(E\setminus
 A)$ it follows that $T_{M}(x,y)=T_{M^{*}}(y,x)$.

 For a graphic matroid $M(G)$, the evaluations of the Tutte
 polynomial at $(2,0)$ and $(0,2)$ equal
 the number of acyclic orientations and the number of totally cyclic
 orientations of $G$, respectively. An acyclic orientation of a graph
 $G$ is an orientation where there are no directed cycles. A totally
 cyclic orientation is an orientation where every edge is in a
 directed cycle. See~\cite{BrOx} for a proof of this result. In this
 situation we let $\alpha(G)$ and $\alpha^{*}(G)$ denote $T_{G}(2,0)$ and
 $T_{G}(0,2)$ respectively. If $G$ is connected, the
 number of spanning trees of $G$ is the evaluation of the Tutte
 polynomial at $(1,1)$ and this quantity is denoted by  $\tau(G)$.

 The Tutte polynomial may be also defined by a linear recursion relation
 given by deleting and contracting elements that are neither loops nor
 isthmuses.

\begin{definition}\label{def:recursive}
If $M$ is a matroid, and $e$ is an element  that is neither an isthmus
nor a loop, then
\begin{equation}\label{eq:deletion-contraction}
   T_{M}(x,y) = T_{M\setminus e}(x, y) + T_{M/e}(x, y).
\end{equation}  \label{end_recursion}
 If there is no such element $e$, then  $T_{M}(x, y) = x^{i}y^{j}$
  where $i$ and $j$ are the number of isthmuses and loops of $M$ respectively.
\end{definition}

 The proof that Definition~\ref{definition} and~\ref{def:recursive} are
 equivalent can be found in~\cite{BrOx}. We still require another
 (equivalent) definition of the Tutte polynomial but first we
 introduce the relevant notions.

 Let us fix an ordering $\prec$ on the elements of $M$, say
 $E=\{e_1,\ldots,e_m\}$, where $e_i\prec e_j$ if $i<j$. Given a fixed
 basis $S$, an element $e$ is called \emph{internally active} if $e\in S$
 and it is the smallest edge with respect to $\prec$ in the only
 cocircuit disjoint from $S\setminus
 \{e\}$. Dually, an element $f$ is \emph{externally active} if
 $f\not\in S$ and
 it is the smallest element in the only circuit contained in $S\cup
 \{f\}$.  We
 define $t_{ij}$ to be the number of bases with $i$
 internally activity elements and $j$ externally activity
 elements.  In \cite{Tutte} Tutte defined $T_{M}$ using these
 concepts. A proof of the equivalence with Definition~\ref{definition}
 can be found in~\cite{Bjorner}.

 \begin{definition} \label{def:activities}
 If $M=(E,r)$ is a matroid with a total order on its ground set, then
\begin{equation}\label{tutte_activities}
   T_{M}(x,y)=\sum_{i,j}t_{ij}x^iy^j\;.
\end{equation}
In particular, the coefficients $t_{ij}$ are independent of the total order
used on the ground set.
\end{definition}

 By an inductive argument using equation
 (\ref{eq:deletion-contraction}), it can
 be proved that $t_{10} = t_{01}$ when $E(M)\geq 2$. This is one of
 a  number of identities known to hold for the coefficients $t_{ij}$.
 For a complete characterization of all the
 affine linear relations that hold among the coefficients  $t_{ij}$
 see Theorem 6.2.13 in~\cite{BrOx}. From there we extract the
 relations that we need.

 \begin{thm}\label{relations}
  If  a rank-$r$ matroid $M$ with $m$ elements has neither loops nor
  isthmuses, then
     \begin{enumerate}
     \item $t_{ij}=0$, whenever $i>r$ or $j>m-r$;
     \item $t_{r0}=1$ and $t_{0, m-r}=1$;\label{max_rank_nullity}
     \item  $t_{rj}=0$ for all $j>0$ and $t_{i,m-r}=0$ for all $i>0$.
     \end{enumerate}
 \end{thm}
  The previous result follows easily from
  Definition~\ref{def:activities}. In~\cite{BrOx} the statement is for
  simple matroids (geometries) but
  it is easy to extend it to matroids with parallel elements.

\section{Some inequalities for the Tutte polynomial}
\label{sec:ineq}
 From the  results in the previous section it is easy to prove the
 following result stated in~\cite{MrIbRd}.
\begin{thm}\label{basic_2_2}
 If a matroid $M$ has neither loops nor  isthmuses, then
 \begin{equation*}
\max\{T_{M}(4,0),T_{M}(0,4)\}\geq T_{M}(2,2).
\end{equation*}
\end{thm}
\begin{proof}
 Let $r$ be the rank  and $m$ the number of elements of $M$.
\begin{align*}
  \max\{T_{M}(4,0),T_{M}(0,4)\}&\geq \max\{4^{r},4^{m-r}\}\\
                               &=\max\{2^{2r},2^{2(m-r)}\}\\
                               &\geq 2^{m}=T_{M}(2,2)\;,
\end{align*}
 where the first inequality follows from equation~(\ref{tutte_activities})
 combined with~\ref{max_rank_nullity} in Theorem~\ref{relations}.
\end{proof}

Note that, for a matroid $M=(E,r)$ with dual
$M^{*}=(E,r^{*})$, the following inequalities are equivalent for any
$A\subseteq E$.
\begin{eqnarray}
   |A| &\leq& |E|-2(r(E)-r(A)),\label{2-bases} \\
  |E\setminus A| &\leq& 2r^{*}(E\setminus A)\ \text{and}\label{dual_2-bases}\\
   z(A)+n(A) &\leq&  |E|-r.
\end{eqnarray}

 We now restrict attention to matroids $M$ in which all
 subsets $A$ of the ground set $E$ satisfy the (equivalent)
 inequalities above.  By a classical result of
 J.~Edmonds~\cite{Edmonds2}, these
 are the matroids that contain two disjoint bases; by duality, these are
 the matroids $M$ whose ground set is the union of two bases of
 $M^*$.

  As every term $(x-1)^{z(A)}(y-1)^{n(A)}$ in $T_{M}$ has
  $x^{z(A)}y^{n(A)}$ as its monomial of maximum degree, the following
  theorem  follows directly from the set of inequalities above.

\begin{thm}\label{thm:max_degree}
   If a matroid $M$ contains two  disjoint bases,  then $t_{ij}=0$, for
  all $i$ and $j$ such that
  $i+j> m-r$. Dually, if its ground set is the union of two bases, then
  $t_{ij}=0$, for all $i$ and   $j$ such that $i+j>r$.
\end{thm}

 Now, it is easy to prove an infinite set of inequalities for the
 Tutte polynomial of a matroid that contains two disjoint bases or
 whose ground set is the union of two bases. This theorem was stated
 in~\cite{MrIbRd}.

\begin{thm}\label{main}
 If a matroid $M$ contains two disjoint bases, then
 \begin{equation}
   \label{eq:main}
   T_{M}(0,2a)\geq T_{M}(a,a)\;,
 \end{equation}
   for all $a\geq 2$. Dually, if its ground set is the union of two
   bases, then
\begin{equation}
   \label{eq:main2}
    T_{M}(2a,0)\geq T_{M}(a,a),
 \end{equation}
  for all $a\geq 2$.
 \end{thm}
 \begin{proof}
    Let us  consider just the case when $M$ has two disjoint bases: the
   other case follows from duality. In this situation $m-r\geq r$. From
   the proof of Theorem~\ref{basic_2_2} and
   equation~(\ref{tutte_activities}) we  have
   $4^{m-r}\geq T_{M}(2,2)=\sum_{i,j}t_{ij}2^{i+j}$. Multiplying this
   inequality by $(a/2)^{m-r}$ we get
\[
  (2a)^{m-r}\geq \sum_{i,j}t_{ij} \left( \frac{a}{2}\right)^{m-r} 2^{i+j}
            \geq \sum_{i,j}t_{ij} \left( \frac{a}{2}\right)^{i+j}2^{i+j}
               = \sum_{i,j}t_{ij} a^{i+j}.
\]
 The second inequality follows from Theorem~\ref{thm:max_degree}. Thus
\[T_{M}(0,2a)\geq (2a)^{m-r} \geq  \sum_{i,j}t_{ij}a^{i+j} = T_{M}(a,a). \]
\end{proof}

 We can sum up the previous result by saying that if $M$ contains two
 disjoint bases or  its ground set is the union of two bases then
  \begin{equation}\label{main_inequality}
    \max\{T_{M}(2a,0),T_{M}(0,2a)\}\geq T_{M}(a,a),
  \end{equation}
 for $a\geq 2$.
 Some classes of matroids which contain two disjoint bases or whose
 ground set is the union of two bases are mentioned in the following

 \begin{cor}\label{list_matroids}
 For a matroid $M$, we have that $T_{M}$ satisfies (\ref{main_inequality}),
 for all $a\geq 2$ whenever $M$ is  one of the following:
  \begin{itemize}
  \item an identically self-dual matroid $M$,
  \item a rank-$r$ projective geometry over GF($q$) or its
  dual, for $r\geq 2$.
   \end{itemize}
\end{cor}
  \begin{proof}
    A matroid $M=(E,r)$ is  identically self-dual  if $M=M^{*}$,
     so,  $B$ is a basis of $M$ if and only if $E-B$ is a basis of
     $M$.

    The matroid PG($r$, $q$) contains the graphic matroid $W_{r+1}$,
    the $r+1$-wheel, as a submatroid for
    $r\geq 3$, see~\cite{Oxley}. The latter  contains two disjoint bases. Thus,
    PG($r$, $q$) contains two disjoint bases. A projective plane of
    order $m\geq 4$ contains $U_{2,4}\oplus_2 U_{2,4}$ as a
    submatroid. Again, the latter  contains two
    disjoint bases. Thus, such a projective plane
    contains two disjoint bases. The only projective plane of order 3
    is the Fano matroid which clearly contains two disjoint bases.
  \end{proof}

There are more classes of matroids that can be added to the previous
list, for instance, coloopless paving matroids. However, in the next
section we will prove a much stronger result for them. The graphic
matroids corresponding to the families of graphs in our next result
may also be added to the list.

  \begin{cor}\label{list_graphs}
    For a graph $G$, $T_{G}$ satisfies
    (\ref{main_inequality}),
    for all $a\geq 2$ whenever $G$ is  one of the following:
    \begin{itemize}
    \item a  4-edge-connected graph,
    \item a  2-connected threshold graph,
    \item a complete bipartite  graph,
    \item a series-parallel graph,
    \item a  3-regular graph,
    \item a bipartite planar graph,
    \item a Laman graph,
    \item a triangulation,
    \item  the wheel graph $W_n$, for $n\geq 2$,
    \item  the square lattice $L_n$, for $n\geq 2$,
    \item  the $n$-cycle $n\geq 2$,
    \item  a tree with  $n$ edges, for $n\geq 1$.
    \end{itemize}
    \end{cor}
  \begin{proof}
   By the classical result in \cite{Tut61} every 4-edge-connected
  graph has two edge-disjoint
  spanning trees. It is easy to see that 2-connected threshold and
  wheel graphs have two edge-disjoint spanning trees. Using the
  expression for computing the arboricity of a graph given in
  \cite{NW64} we get that series-parallel,
  3-regular, bipartite planar, and Laman graphs all have arboricity
  two,  which is equivalent to having two spanning trees that cover all the
  edges of the graph. Triangulations are geometric duals of 3-regular
  planar graphs, so they have two edge-disjoint spanning trees.

   It is easy to see that each of $K_{2,m}$ for $m\geq 2$, $K_{3,3}$, the
   square lattice $L_n$ for $n\geq 2$, the $n$-cycle for $n\geq 2$, and a
   tree have two spanning trees which cover all the edges in the
   graph. With the exception of the case $n=m=3$, if both $n$ and $m$
   are at least 3, then $K_{n,m}$ always has two edge-disjoint
   spanning trees.
\end{proof}

\section{Paving matroids}\label{sec:paving}

 A paving matroid $M=(E,r)$ is a matroid whose circuits all have size
 at least $r$. Paving matroids are closed under minors and the set of
 excluded minors
 for the class consists of the matroid $U_{2,2}\oplus U_{0,1}$, see for
 example~\cite{IbMrNb}.  The interest about paving matroids goes back
 to 1976 when Dominic Welsh
 ask if  most matroids are paving, see~\cite{Oxley}. More recently,
 the authors
 in~\cite{MyNwWlWh} pose as a conjecture that asymptotically almost
 every matroid is paving.

 First, we prove that most paving matroids either contain two disjoint
 bases or  their ground set is the union of two bases. Consequently
 paving matroids fall within the class of matroids considered in the
 previous section.

 \begin{thm}\label{paving_DB_or_UI}
  Let $M=(E,r)$ be a rank-$r$ paving matroid with $n$ elements,
  \begin{itemize}
  \item if $2r>n$, then $E$ is the union of two bases,
  \item if $2r\leq n$ and $M$ is coloopless, then $M$ contains two
  disjoint bases.
  \end{itemize}
 \end{thm}
 \begin{proof}
  In the first case, take $B_1$ to be a basis of $M$, then
  $I_2=E\setminus B_1$ has size $n-r<r$, so it is independent and we
  can extend it to a basis $B_2$. Thus $E=B_1\cup B_2$.

 In the second case, if $M$ has a circuit $C$ of size $r+1$, then
 $C'=E\setminus C$ has size $n-r-1\geq r-1$. Let  $I$ be a set
 of size $r-1$ contained in $C'$. As $I$ is independent and $C$ is
 spanning, there exists $a\in C\setminus I$ such that $I\cup\{a\}$ is
 a basis. But $C\setminus\{a\}$ is also a basis. Thus, we have two
 disjoint bases.

 Let $M$ be a coloopless paving matroid with no circuits of size $r+1$
 and suppose that $2r\leq n$. Let  $B$ be a basis of $M$. Then
 either $E\setminus B$ contains a basis, in which case we have
 finished the
 proof, or $r(E\setminus B)=r-1$. In the latter case, let $H$ be  the
 hyperplane
 defined as the closure of $E\setminus B$, and
 $I=E\setminus  H\subseteq B$. The set $I$ has size $p+1$ with $p\geq
 1$ as $M$ is  coloopless.

 We show that in this case $M$ also has two disjoint bases. Let
 $I'=I\setminus\{a\}$, for some $a\in I$. Then, $I'$ is a
 non-empty independent set of size $p$ with the property that for any
 circuit $C$ of size $r$ contained in $H$, $I'\cup C$ contains a basis
 of $M$. Thus, there is a basis $B_1$ of $M$ of the form $I'\cup A_1$
 for some subset $A_1$ of $H$ of size $r-p$.  Now, let
 $B_2=\{a\}\cup A_2$ for some $A_2\subseteq H\setminus A_1$ of size
 $r-1$. This is possible as $|H\setminus A_1|=(n-p-1)-(r-p)
 =(n-r)-1\geq r-1$. Thus, $B_1$ and $B_2$ are disjoint bases of $M$.
 \end{proof}

The main goal of this section is prove that for any coloopless paving matroid
\begin{equation}
tT(x_1,y_1)+(1-t)T(x_2,y_2) \geq T(tx_1+(1-t)x_2,ty_1+(1-t)y_2),
\end{equation}
whenever $0\leq t \leq 1$ and $x_1,x_2,y_1,y_2$ are non-negative and
satisfy $x_1+y_1=x_2+y_2$.  Notice
that this inequality is a much stronger statement than~(\ref{eq:main2})
as it says
that $T$ is a convex function along the portions of the line $x+y=p$
lying in the positive quadrant, rather than merely saying that the
value of $T$ at one of the endpoints of the line segment is greater
than the value of $T$ at its midpoint.

 Our main tools for establishing the convexity of $T$ are the
 following easy results.
 \begin{lemma}\label{convexity_duality}
  Let $M$ be a matroid. Either, both $T_M(x,y)$ and $T_{M^{*}}(x,y)$
  are convex along the portion of the line $x+y=p$ lying in the
  positive quadrant or neither is.
 \end{lemma}
  \begin{proof}
   This follows directly from the equality $T_{M}(x,y)=T_{M^{*}}(y,x)$.
  \end{proof}

 \begin{lemma}\label{convexity_contraction_deletion}
 Let $M$ be a matroid and $e$ in $M$ be neither a loop nor a coloop. If
 $T_{M\setminus e}$  and $T_{M/e}$ are both convex along the portion
 of the line $x+y=p$ lying in the positive quadrant, then $T_M$ is
 also convex on the same
 domain.
 \end{lemma}
  \begin{proof}
    This follows directly from the deletion-contraction formula
    (\ref{eq:deletion-contraction}) and that the  sum of convex
    functions is  also a convex function.
  \end{proof}

  The following three results deal with the convexity of $T$ for
  some coloopless paving matroids. We use these
 cases as bases for an inductive argument later on.

 \begin{lemma}\label{parallel_convex}
 If M is isomorphic to  the paving  matroid $U_{1,k+1}\oplus U_{0,l}$,
 where $l\geq 0$  and $k\geq 1$, then  $T_M$ is
 convex along the portion of the line $x+y=p$ lying in the positive quadrant.
 \end{lemma}
 \begin{proof}
  We have
    \[ T_M(x,y) = y^l(y^k+\cdots +y+x) = py^l  +\sum_{m=l+2}^{l+k} y^m.\]
  Since $y^m$ is convex for all $m\geq 0$ in the given region and the sum
  of convex
  functions is convex, the result follows.
 \end{proof}

 \begin{lemma}\label{uniform_convex}
  The Tutte polynomial $T_M$ is a convex function in the positive
  quadrant when $M$ is a
  uniform matroid. In particular, $T_M$ is convex along the portion of
  the line $x+y=p$ lying in the positive quadrant
 \end{lemma}
 \begin{proof}
   The Tutte polynomial of a uniform matroid can be computed easily
   using (\ref{eq:expansion}).
   \begin{equation*}
          T_{U_{r,n}} (x,y)= \sum_{i=0}^{r-1}\binom ni(x-1)^{r-i}+
                           \binom nr+
                           \sum_{i=r+1}^{n}\binom ni(y-1)^{i-r}.
    \end{equation*}
  This can be expanded into the following expression, which may also
  be established directly using~(\ref{tutte_activities}).
   \begin{equation*}
          T_{U_{r,n}} (x,y)= \sum_{j=1}^{n-r}\binom{n-j-1}{r-1}y^{j}+
                           \sum_{i=1}^{r}\binom{n-i-1}{n-r-1}x^{i},
    \end{equation*}
  when $0<r<n$, while $T_{U_{n,n}}(x,y)=x^{n}$ and $T_{U_{0,n}}(x,y)=y^{n}$.

 As each term is a convex function we get the result.
 \end{proof}

 \begin{thm}\label{rank_2_convexity}
  If $M$ is a rank-2 loopless and coloopless matroid, then $T_M$ is
  convex along the portion of the line $x+y=p$ lying in the positive quadrant.
 \end{thm}
 \begin{proof}
   If $M$ is isomorphic to the uniform matroid $U_{2,n}$, the result
   follows from applying the previous lemma. Otherwise, $M$ is
   isomorphic to a
   matroid with parallel elements whose simplification is isomorphic
   to $U_{2,n}$.

   If $n\geq 3$ or there is a parallel class of size at
   least 3, we can choose an element $e$ in a non-trivial parallel
   class of $M$ such that $M\setminus e$ does not have a coloop. In
   this case $M/e$ is isomorphic to $U_{1,k+1}\oplus U_{0,l}$,
   where $l\geq 1$  and $k\geq 1$ and $M\setminus e$ is a rank-2
   loopless and coloopless matroid. The result follows from
   Lemma~\ref{parallel_convex}, induction and
   Lemma~\ref{convexity_contraction_deletion}.

   In the last case, the simplification of $M$ is isomorphic to
   $U_{2,2}$ and every element is in a parallel class of size 2. Then
   $M$ is isomorphic to $U_{1,2}\oplus U_{1,2}$. Then, $T_M=(x+y)^2$
   which is convex (in fact is constant) along $x+y=p$ for $p>0$ and
   $0\leq y\leq p$.
 \end{proof}

   In order to establish or main result, we need the following
   structural result about coloopless paving matroids.

   \begin{lemma}\label{lemma_2-stretching}
    Let $M$ be a rank-r coloopless paving matroid. If for every element $e$
    of $M$, $M\setminus e$ has a coloop, then one of the following
    three cases happens.
    \begin{enumerate}
    \item $M$ is isomorphic to $U_{r,r+1}$.\label{2-stretching_i}
    \item $M$ is the 2-stretching of a uniform matroid $N$ and
       $N$ is isomorphic to  $U_{s,s+1}$ or
       $U_{s,s+2}$, for some $s\geq 1$.\label{2-stretching_ii}
    \item $M$ is isomorphic to $U_{1,2}\oplus U_{1,2}$.\label{2-stretching_iii}
    \end{enumerate}
   \end{lemma}
   \begin{proof}
    If $e$ is such that $M\setminus e$ has a coloop $f$, then
    $\{e,f\}$ are in either a  series or form a parallel class. If there is a
    parallel class in a  paving matroid, its rank is either 1 or
    2. Thus, if $\{e,f\}$ are in a parallel class, $M$ is isomorphic
    to  $U_{1,2}\oplus U_{1,2}$ or $U_{1,2}$.

    We can assume that $M$ contains no non-trivial parallel
    classes. Hence every element belongs to a series class of size at
    least two. Suppose that there is a series class containing at
    least three elements $e,f,g$. In
    this case, $M\setminus e$ will have at least 2 coloops. But as $M$ is
    paving all its minors are also paving. Thus, $M\setminus e$, being
    a paving matroid with at least 2 coloops, cannot
    have circuits  and $M\setminus e$ is isomorphic to $U_{r,r}$. In
    this case, we conclude that $M$ is isomorphic to  $U_{r,r+1}$.

    To finish, we suppose that every element in $M$  is in a series
    class of size 2.  In this case, $M$ is the 2-stretching of a
    rank-$s$ matroid $N$ with $m$ elements and $s\geq 1$. $N$ is
    paving because it is a minor of $M$ and it must have circuits as $M$
    is coloopless.

    If the minimal size of a circuit in  $N$ has size $s$, $M$ has a
    circuit of size $2s$.  But the rank of $M$ is $s+m$ as it is the
    2-stretching of $N$. Then $2s\geq s+m$ and $s=m$. In this case,
    $N$ would be isomorphic to $U_{s,s}$ and we arrive at a
    contradiction. Thus, $N$ does not have circuits of size $s$.

    Hence all the circuits of $N$ have size $s+1$ and
    $N$ is uniform. Then, there is a circuit  in $M$ of size $2s+2\geq
    s+m$, and $s+2\geq m\geq s+1$. Thus, $N$ is isomorphic to
    $U_{s,s+1}$ or $U_{s,s+2}$.
   \end{proof}

    \begin{lemma}\label{lemma_2-stretching_convexity}
    Let $M$ be a rank-r coloopless paving matroid. If for every element $e$
    of $M$, $M\setminus e$ has a coloop, then $T_M$ is
    convex along the portion of the line $x+y=p$ lying in the positive quadrant.
   \end{lemma}
   \begin{proof}
    We analyse the cases for $M$ given in the previous lemma. If $M$
    is isomorphic to $U_{r,r+1}$, the result follows from
    Lemma~\ref{uniform_convex}. If $M$ is isomorphic to $U_{1,2}\oplus
    U_{1,2}$ or  $U_{1,2}$, the corresponding Tutte polynomials are
    $(x+y)^2$ and $x+y$, which in both cases are convex.

    If $M$ is the 2-stretching of $U_{s,s+1}$, then $M$ is isomorphic
    to $U_{r, r+1}$ and the result follows from
    Lemma~\ref{uniform_convex}. If $M$ is the 2-stretching of
    $U_{s,s+2}$, then $M^*$ is the 2-thickening of $U_{2,n}$ which
    is a rank-2 matroid and the result follows from
    Theorem~\ref{rank_2_convexity} and
    Lemma~\ref{convexity_duality}.
   \end{proof}

 Finally, we arrive at the main result of this section.

 \begin{thm}\label{convexity_Tutte}
  If $M$ is a  coloopless paving matroid, then $T_M$ is
  convex along the portion of the line $x+y=p$ lying in the positive quadrant.
 \end{thm}
  \begin{proof}
   If $M$ has a loop, then $M$ has rank 1 and it is isomorphic to
   $U_{1,k+1}\oplus U_{0,l}$ with $l,k\geq 1$ and the result follows
   from Lemma~\ref{parallel_convex}.

   Otherwise, every element of $M$ is neither a loop nor a coloop. If there
   is an element $e$ such that $M\setminus e$ has no coloop, then both
   $M/e$ and $M\setminus e$ are coloopless paving matroids and the
   result follows from Lemma~\ref{convexity_contraction_deletion}.

   So, we can assume that  for all $e$, $M\setminus e$ has a coloop. Then
   the result follows from Lemma~\ref{lemma_2-stretching_convexity}.
   \end{proof}

 Hence, subject to an affirmative answer to Welsh's
problem mentioned earlier, we have proved Conjecture~\ref{old} and
Theorem~\ref{convexity_Tutte} for asymptotically almost all matroids.

 Paving matroids are not closed under duality but using
 Lemma~\ref{convexity_duality} we obtain the convexity
 of the Tutte polynomial for a bigger class of matroids.

  \begin{cor}
  If $M$ or $M^{*}$ is a  coloopless paving matroid, then $T_M$ is
  convex  along the portion of the line $x+y=p$ lying in the positive
  quadrant.
 \end{cor}

  By Theorem~\ref{paving_DB_or_UI}, the class of matroids $M$ such
  that either $M$ or $M^{*}$ is a coloopless paving matroid is contained in
  the class of matroids that contains two disjoint bases or whose
  ground set is the union of two bases.   Thus, we have a strengthening
  of Theorem~\ref{main}.

 \begin{cor}\label{strengthening_main}
  If $M$ or $M^{*}$ is a  coloopless paving matroid, then $T_M$ satisfies
  inequality (\ref{main_inequality})  for $a\geq 0$.
 \end{cor}

\section{The Merino-Welsh  conjecture}
\label{sec:conjecture}

In this section we return to the original Merino--Welsh conjecture
(Conjecture~\ref{old}) and establish that its conclusion holds for
some fairly
specific classes of graphs and matroids. Recall that the conclusion of
the conjecture is certainly not true for all graphs. Taking any graph
and adding a loop and a bridge results in a graph that does not
satisfy~\eqref{eq:MW}. However, the condition on the connectivity may
not be the most natural because if $G$ consists of 2 cycles of length
2 sharing a common vertex, then the graphic matroid $M(G)$ satisfies
(\ref{eq:main}) for all $a\geq 0$. So (\ref{eq:MW}) is satisfied by some
graphs that are not 2-connected.

\subsection{Wheels and whirls}\label{wheels}
 In this subsection we consider  wheels, a well-known class of
 self-dual planar graphs, and whirls, a related class of matroids which
 are also self-dual. The  wheel
 graph  $W_n$ has $n+1$ vertices and $2n$ edges. The  vertices
 $\{1,\cdots, n\}$ form an $n$-cycle while the vertex $0$ is adjacent
 to every vertex in this cycle. The whirl $W^{n}$ is the matroid with
 ground set $E(W^{n})=E(W_{n})$, while the set of bases of $W^{n}$
 consists of the edge set in the $n$-cycle of $W_n$ together with all
 edge sets of spanning trees of $W_n$, see~\cite{Oxley}.

 It is well-known that $\tau(W_n)=L_{2n}-2$, for $n\geq 1$, where
 $L_k$ is the $k$th-Lucas number which is defined recursively by
 $L_1=1$, $L_2=3$ and $L_k=L_{k-1}+L_{k-2}$ for $k\geq 3$. This result
 was proved by Sedl\'a\v cek \cite{Sed} and also by Myers
 \cite{My}. Using the analogy of
 Binet's Fibonacci formula for Lucas numbers we get
$$
   \tau(W_n)=\left( \frac{3+\sqrt{5}}{2}\right)^{n}
             +\left(\frac{3-\sqrt{5}}{2}\right)^{n}-2.
$$
The same formula can be obtained directly by
using equation~(\ref{eq:deletion-contraction}) for $T_{W_n}(1,1)$ and
then solving
the corresponding recurrence relation.

The chromatic polynomial of $W_n$ is known, see \cite{Biggs}, and is
equal to $\chi_{W_n}(x)$ = $x(x-2)^n+(-1)^n x(x-2)$.  Now, applying the
famous result of R. Stanley \cite{Stanley}
 that relates the number of acyclic orientations and the chromatic polynomial,
 namely $\alpha(G)=|\chi_{G}(-1)|$, we get  $\alpha(W_n)= 3^{n}-3$.
 These results together yield the following

\begin{thm}
   For all $n\geq 2$, $\alpha(W_{n})\geq \tau(W_{n})$ and $M(W_{n})$
   satisfies Conjecture~\ref{old}.
\end{thm}

The Tutte polynomials of  $W^{n}$ and  $M(W_{n})$ are related by the
 equality,  $T_{W^{n}}(x,y)=T_{W_{n}}(x,y)-xy+x+y$. Thus, obtain the
 following result

\begin{thm}
   For all $n\geq 2$, $T_{W^{n}}(2,0)\geq T_{W^{n}}(1,1)$ and $W^{n}$
   satisfies equation~(\ref{eq:MW}).
\end{thm}

 \subsection{3-regular graphs with girth at least 5}\label{3-regular}

 For  3-regular graphs  with
 girth at least 5 a
 lower  bound for the number of  acyclic orientations,
\[
  \alpha(G)\geq (2^{3/8} 3^{3/8} 4^{1/8})^n\;,
\]
is given in \cite{KhSc}, where $n$ is the number of vertices of
$G$. On the other hand, the following upper
bound for the number of spanning trees in a 3-regular graph $G$ is
 given in \cite{ChYao}.
\[ \tau(G)\leq
        \frac{2\beta}{3 n}
        e^{\frac{12}{\sqrt{\pi}}
             \left( \frac{1}{\beta}\right)^\frac{5}{2}}
        \left( \frac{4}{\sqrt{3}}\right)^n\;,
\]
 where $\beta=\lceil \ln(n)/\ln(9/8)\rceil$. From the formulae we
 obtain the following

\begin{thm} If $G$ is a 3-regular  graph of girth at least 5, we have
 $\tau(G)< \alpha(G)$ and $M(G)$  satisfies Conjecture~\ref{old}.
\end{thm}

 \subsection{Complete graphs}
\label{complete_bipartite}
 It is natural to check if Conjecture~\ref{old} is  true for complete
 graphs and complete bipartite graphs.

 A classical result of Cayley~\cite{Biggs} states that $\tau(K_n)=n^{n-2}$.
 For $K_3$ we have
 $\alpha(K_3)=6>3=\tau(K_3)$, thus $K_3$ satisfies
 Conjecture~\ref{old}.

 We use the following lemma which has an easy proof, see~\cite{CnMr}.
 \begin{lemma}\label{simplicial}
   If $G$ is a 2-connected graph with a
   vertex $v$ of    degree $d$, then  $(2^d-2)\alpha^*(G-v) \leq
   \alpha^*(G)$.
 \end{lemma}

 We will prove that $\alpha^{*}(K_{n})\geq n^{n-2}$, for $n\geq
 4$. When $n=4$, we have $\alpha^{*}(K_{4})=24>16=\tau(K_{4})$. We
 proceed by induction on $n$.
\begin{equation*}
\begin{split}
 \tau(K_{n+1})= (n+1)^{n-1}&= \left( \frac{n+1}{n}\right)^{n}
                             \left( \frac{n}{n+1}\right)^{2}
                             (n+1)\tau(K_{n})\\
                           &\leq e (n+1) \tau(K_{n})
                             \leq (2^{n}-2) \tau(K_{n})\\
                           &\leq (2^{n}-2) \alpha^{*}(K_{n}).
\end{split}
\end{equation*}
 The last quantity is less than or equal $\alpha^{*}(K_{n+1})$ by the
 previous lemma.

\begin{thm}\label{thm:complete}
 For all $ n\geq 3$, $M(K_{n})$  satisfies  Conjecture~\ref{old}.
\end{thm}

 The technique used for complete graphs can be used to prove the
 Conjecture~\ref{old} in the case of threshold graphs, a type of
 chordal graphs, see~\cite{CnMr}.
 Also in~\cite{CnMr} complete bipartite graphs are considered and the
 authors prove the following

\begin{thm}\label{thm:bipartite}
 For all $m\geq n\geq 2$, $M(K_{n,m})$  satisfies
 Conjecture~\ref{old}.
\end{thm}

  \subsection{Catalan matroids}
   A \emph{Dyck path} of length $2n$ is a path in the plane from (0,0) to
   ($2n$,0), with steps (1,1), called \emph{up-steps}, and (1,-1),
   called \emph{down-steps}. It is well-known that the number of Dyck
   paths of length $2n$ is the Catalan number
   $C_n=\frac{1}{n+1}\binom{2n}{n}$. Each Dyck path $P$ defines an
   \emph{up-step set}, consisting of the integers $i$, $1\leq i\leq 2n$, for
   which the $i$th-step of $P$ is an up-step. The collection of up-step
   sets of all Dyck paths of length $2n$ forms the bases of a matroid $M_{n}$
   over $\{1,2,\ldots, 2n\}$. These matroids are called Catalan
   matroids and have recently been studied extensively,
   see~\cite{AnaBonin} or~\cite{Ardila}.

   We consider the matroids $N_{n}$, $n\geq 2$,  obtained form $M_{n}$ by
   deleting the elements 1 and $2n$. This corresponds to deleting the
   loop and isthmus of $M_{n}$. From the results in~\cite{AnaBonin} it
   follows that the matroid $N_{n}$ is
   self-dual, but not identically self-dual.  An expression for the Tutte
   polynomial of $N_{n}$ follows from Corollary~5.8
   of~\cite{AnaBonin}.
   \[
    T_{N_{n}}(x,y)=\sum_{i,j>0}\frac{i+j-2}{n-1}\binom{2n-i-j-1}{n-i-j+1}x^{i-1}y^{j-1}.
    \]
    After some algebraic manipulations we get a formula for the
    evaluation at (2,0) and (0,2).
    \[ T_{N_{n}}(2,0)=T_{N_{n}}(0,2)=\sum_{k=0}^{m} \frac{k}{m}
     \binom{2m-k-1}{m-k} 2^{k},
    \]
     where $m=n-1$. This quantity equals $\binom{2m}{m}$ by the
     following list of equalities
   \begin{equation*}
     \begin{split}
      \sum_{k=0}^m \frac km \binom {2m-k-1}{m-k} 2^k &=
      \sum_{k=0}^m \bigg( \binom{2m-k-1}{m-1} - \binom{2m-k-1}m \bigg) 2^k\\
      &= \sum_{k=0}^m \sum_{j=0}^k \bigg( \binom{2m-k-1}{m-1} -
      \binom{2m-k-1}m \bigg) \binom {k}{j} \\
      &= \sum_{j=0}^m \sum_{k=j}^m \bigg( \binom{2m-k-1}{m-1} -
      \binom{2m-k-1}m \bigg) \binom {k}{j} \\
      &= \sum_{j=0}^m \bigg( \binom{2m}{m+j} - \binom{2m}{m+j+1} \bigg)\\
      &=\binom{2m}{m}.
     \end{split}
   \end{equation*}
   The key step in the middle uses the convolution identity
   $\sum_{k=0}^{2m-1} \binom {2m-k-1}{q} \binom {k}{j}$   =
   $\binom{2m}{q+j+1}$ that is the basic identity (5.6)
   in~\cite{concrete}. The  value of $T_{N_{n}}(1,1)$ is clearly
   $C_n=\frac{1}{n+1}\binom{2n}{n}$.

   \begin{thm}\label{thm:catalan}
     For all $n\geq 2$, $N_{n}$  satisfies equation~(\ref{eq:MW}).
   \end{thm}

Notice that in all of the classes that we have considered, either the
ground set contains two disjoint bases or is the union of two
bases. We therefore propose the following conjecture which is a weaker
form of Conjecture~\ref{old} and may turn out to be more tractable.

\begin{conjecture}\label{new}
 If  $M$ contains two disjoint bases or its ground set  is the union
 of two bases then $\max\{ T_{M}(2,0),\ T_{M}(0,2) \} \geq T_{M}(1,1)$.
\end{conjecture}

 \section{Conclusion and Discussion}

 We have proved that $T_M$ is convex along the portion of the line $x+y=p$
 lying in the positive quadrant, whenever $M$ is a coloopless paving
 matroid.  By Definition~\ref{def:activities},
 $T_M$  is convex along the semilines $y-mx+b$ for $m\geq 0$ and $b\in
 \mathbb{R}$ in the positive quadrant. It is natural to ask for which
 matroids is $T_M$ convex in the positive quadrant?

There is no clear link between convexity of the Tutte polynomial in
the positive quadrant and the classes of matroids that we have
considered. Coloopless paving matroids may or may not have Tutte
polynomials that are convex in the positive quadrant. For example, the
Tutte polynomials of uniform matroids and the graphic matroid $M(K_4)$
are convex in the positive quadrant; on the other hand the Tutte
polynomial $y^{l}(y^k+\ldots +y+x)$ of the
 paving matroid  $U_{1,k+1}\oplus U_{0,l}$, where $l\geq 1$  and
 $k\geq 1$ is not a convex or concave  function. There are also
 non-paving matroids whose
Tutte polynomial is convex, for example $U_{n,n}^2$, for $n\geq 3$,
the 2-thickening of $U_{n,n}$. The Tutte polynomial of this matroid is
$(x+y)^n$ which is clearly convex. Note however that this latter class
of matroids has two disjoint bases.

Establishing the convexity of the Tutte polynomials of matroids within
a given large class seems to be a difficult problem. The Tutte
polynomials of the graphs at the
top of Fig.~1 are convex functions while the Tutte
polynomial of the
graph at the bottom is neither convex nor concave. A similar situation holds
for the matroids in Fig.~2,  the Tutte polynomials
of the two matroids at
the top of the figure are convex functions while the polynomial for the
matroid at the bottom is neither convex nor concave.

   \begin{figure}
    \begin{center}
\psset{unit=0.6}
\pspicture(-5,-5)(15,3)

\qdisk(1,0){0.14}\qdisk(2,-0.1){0.14}\qdisk(3,-0.08){0.14}\qdisk(3,-0.32){0.14}
\qdisk(4,-0.3){0.14}\qdisk(4,-0.54){0.14}\qdisk(4,-0.06){0.14}\qline(1,0)(4,-0.3)

\qdisk(-1,0){0.14}\qdisk(-1,0.24){0.14}\qdisk(-1,-0.24){0.14}\qdisk(-2,0.1){0.14}
\qdisk(-3,0.2){0.14}\qdisk(-4,0.3){0.14}\qline(-1,0)(-4,0.3)

\qdisk(-1.5,-1.5){0.14}\qdisk(-0.5,-1.6){0.14}\qdisk(0.5,-1.7){0.14}\qdisk(1.5,-2.16){0.14}
\qdisk(1.5,-1.92){0.14}\qdisk(1.5,-1.68){0.14}\qdisk(1.5,-1.44){0.14}\qline(1.5,-1.8)(-1.5,-1.5)
\rput(0,-4.5){Figure 1}

\qdisk(11,0){0.14}
\qdisk(14,0){0.14}
\qdisk(12.5,2.6){0.14}
\qline(11,0)(14,0)
\qline(14,0)(12.5,2.6)
\psbezier(12.5,2.6)(13.52,2.03)(14.02,1.17)(14,0)
\psbezier(12.5,2.6)(12.48,1.43)(12.98,0.57)(14,0)
\psbezier(12.5,2.6)(11.48,2.03)(10.98,1.17)(11,0)
\psbezier(12.5,2.6)(12.52,1.43)(12.02,0.57)(11,0)

\qdisk(9,0){0.14}
\qdisk(6,0){0.14}
\qdisk(7.5,2.6){0.14}
\qline(9,0)(6,0)
\qline(6,0)(7.5,2.6)
\psbezier(7.5,2.6)(8.52,2.03)(9.02,1.17)(9,0)
\psbezier(7.5,2.6)(7.48,1.43)(7.98,0.57)(9,0)

\qdisk(10,-1){0.14}
\qdisk(11.5,-3.6){0.14}
\qdisk(8.5,-3.6){0.14}
\qline(10,-1)(11.5,-3.6)
\qline(10,-1)(8.5,-3.6)
\qline(8.5,-3.6)(11.5,-3.6)
\psbezier(10,-1)(11.02,-1.57)(11.52,-2.43)(11.5,-3.6)
\psbezier(10,-1)(9.98,-2.17)(10.48,-3.03)(11.5,-3.6)

\rput(10,-4.5){Figure 2}
\endpspicture
\end{center}

    \end{figure}

  We proved  Conjecture~\ref{old} for some families of
  graphs and matroids.  There are some more families for which the
  conjecture holds: for example Marc Noy
  (private communication) proved that $\tau(G)\leq \alpha(G)$ when $G$
  is a maximal   outerplanar graph,
  using equation~(\ref{eq:deletion-contraction}).

\section{Acknowledgment}
  We thank Bill Jackson for helpful discussions.

redpajama_set_name: RedPajamaArXiv


Managing and dealing real estate worth billion dollars is not an easy issue. But one person has managed properties worth millions of dollars in the past with extreme ease and comfort and made everything as a profitable venture. Meet Mr. David, the Chief Executive Officer of Black Plum LLC a pioneer in real estate business. This well-educated and experienced guy has numerous commercial properties and residential properties in California. This Principal Officer of Black Plum LLC, New Port, California has more than two decades of experience in investment fund management, investment advisory, securities and real industries and real estate broker. Mr. Giunta is extremely versatile and flexible person who will adapt to any types of situations with extreme ease.
He will walk what he talks and show his customers world class residential and commercial properties. Investors those are planning to invest millions or billions of dollars in real estate business can approach this gentleman easily. This flamboyant gentleman and smart real estate baron will be at the desk before the scheduled time and invite the clients with open heart and mind. This guy is a genius right from his young age and toiled for several years to reach the top position in this celebrated Real-estate Company. This guy has extraordinary will power and guts to face any types of challenges and odds. Customers will get valuable and wonderful quote when they approach this soft spoken gentleman or his fast growing real estate company.
Mr. David Giunta has the ability to transform the ordinary stones into precious stones with his divine touch. Hundreds of investors those who believed the honest words of this guy are counting profits. Do not hesitate to meet this influential gentleman when there are plans to purchase million dollar properties. This guy is always found amidst a group of prominent builders and investors. Meet this guy in his sophisticated office room and discuss all the real estate requirements with an open mind. He will show right direction to all his clients which will make them extremely happy. Mr. Giunta has also worked in various prominent companies before taking charge of this company and also holds a Certified Public Account degree.
He has done lots of audits in some of the Fortune 500 companies in his young days. It will be a turning point in one's life when someone meets this smart individual who has years of experience in real estate business. Everyone can happily deal with this real estate kingpin since he holds valid certified license. Stop scouting for luxury houses and try to meet this person who sits in a posh office situated in California. He will show the best luxury apartment which will meet the customer's requirements and will come within the budget. This guy is also large hearted and helps the impoverished families by organizing events during free times. Customers will love to interact with this guy and learn many interesting things about real estate business. They will drive more confidence and energy when they meet him in his world class office.


text:
superrocketcar
About superrocketcar
Super Rocket Car is an indie pop rock band from Mountainside, New Jersey. Despite growing up on the same block, JR Pacifico, Bob Adashev, Dave Boyce, and Sagar Desai waited until their senior year at Governor Livingston High School to establish themselves as a band. Beginning with a few local shows in friends' backyards, the band created enough buzz to rise swiftly through the ranks of rock n' roll stardom, going from completely unknown to relatively unknown. Previously known as Camping Equipment, the band decided to change their name to Super Rocket Car following the departure of two former members. Their brand of sunny rock n' roll pop and their energetic live performances will keep you dancing/partying/love-making into the wee hours of the morning. The band has managed to attract themselves a modest following of devoted fans, and are extremely excited to expose their music to some fresh ears with the release of their eponymous debut EP, recorded with Tom Suhey at Portrait Studios.


Extensive kitchen remodel with a dining room addition in Hartland. We expanded the kitchen & reconfigured the design to open up into the dining room addition. What a great space to entertain friends & family.


text:
Marines experimenting with defensive cyber teams for reconnaissance
II Marine Expeditionary Force Information Group is experimenting with how to use its defensive cyber teams for reconnaissance.
Marines with Marine Corps Forces Cyberspace Command conduct offensive and defensive cyber operations in support of United States Cyber Command and operate, secure and defend the Marine Corps Enterprise Network. )Marine Corps photo)
One Marine Corps information warfare unit is experimenting with how to use its defensive cyber teams for reconnaissance, according to the organization's commander.
The Marine Expeditionary Force Information Groups (MIGs), which were created in 2017 and support each MEF within the Corps, integrate electronic warfare with intelligence, communications, military information support operations, space, cyber and communication strategy to provide MEF commanders with an information advantage.
Each of these units incorporates what is known as a defensive cyber operations-internal defense measures (DCO-IDM) company, which protect networks and hunt adversaries on friendly systems at the tactical edge.
II MIG is the lead within the MIGs for experimenting with reconnaissance and counter reconnaissance, its commander Col. Brian Russell said in a podcast hosted by the Brute Krulak Center for Innovation & Future Warfare at Marine Corps University, Wednesday.
Through that experimentation, Russell said they are looking to use their DCO-IDM companies as a reconnaissance force.
"As we lay out our network terrain and determine what's critical from a reverse targeting methodology, certainly, malicious cyber actors are coming after our kill chains," he said. "We draw essentially named areas of interest around those critical nodes and that's where we apply our resource with a sensor [that] I call the electronic version of binoculars to look and confirm that adversary presence so we can do something about it. That to me is a form of cyber reconnaissance that we're experimenting here at II MEF."
Russell has previously discussed the need to reimagine how these defensive cyber teams can be used in the gray zone against adversaries, or the competition space that exists below the threshold of armed conflict.
"We can employ this capability to influence adversary decision-making by combining DCO-IDM operations with any other element of the Fleet Marine Force," he said in 2020. "These operations, below the level of armed conflict (gray zone), enable us to understand the adversary, condition their behavior in advance of conflict, and even impose costs on their operations and strategic intent."
These DCO-IDM teams are trained to the same standards as U.S. Cyber Command's high end defensive cyber protection teams that respond to and defend against malicious activity on enterprise networks, Russell said, adding they are essentially interoperable.
This interoperability "opens some doors that lets you work on other people's networks with allies and partners who trust you because you're trained to a certain standard," he said, noting this is the same approach they'll take with offensive teams abiding by Cyber Command standards.
II MIG primarily supports European Command but also services as the Marines' global response MEF. They have previously conducted exercises and experiments in the theater, learning that building and winning narratives before a conflict is important and everything takes place in the information environment.
Officials have long maintained that the MIGs would not be built overnight. Exercises and experimentation help the Marine Corps shape the direction of these forces and better understand what needs to be tweaked.
In fact, based upon lessons learned from exercises and experimentation, the Marine Corps made alterations to the MIGs three years after they were formed.
Additionally, officials noted that despite the tactical nature of these teams and a lot of what the Marine Corps does, units must also be tied into the operational and strategic trends to be successful.
"In order for you to be tactically successful, you have to be aware of strategic and operational level effects, usually non lethal, that shape the environment to allow your tactical action to occur," Col. Ray Gerber, commander of III MIG, said on the same podcast. "I would say that the Marine Corps is struggling with this because we have grown up in a world where the tactical action is the thing that everything centers around."
This force, however, can project power globally from one location, a departure from the traditional domains that is unique to the information environment.
"I've got Marines in this building right now supporting operations in the USEUCOM AOR, supporting the conflict in USEUCOM AOR from either analysis or capability provision perspective," Russell said. "I think that is a growth industry. [Continental U.S.] base operational support … I don't need to deploy forward to provide the operational value. I can do that from home station or other locations that aren't necessarily quote unquote, in the conflict zone. That's what this modern information environment enables us to do."
defensive cyber
MEF Information Group
Senate wants tighter cyber-electronic warfare integration, clarity on organizations for cyber ops
Capt. Seth Hayden of the Army Cyber Protection Brigade monitors exercise progress in the white team operations area for exercise Operation Tiger Stance, at the Indiana National Guard's Muscatatuck Urban Training Center in Butlerville, Ind., Aug. 23, 2018. (Photo by Bill Roche)
New Army intel unit having big impact on recently established 'triad'
The Army's Cyber Military Intelligence Group is providing important insights for a new triad between Army Cyber Command, Army Space and Missile Defense Command and Army Special…
A U.S. Army Multi Domain Task Force operates from the Tactical Command Post as a part of their premier appearance at Valiant Shield 2018 Sept. 20, 2018. (U.S. Navy photo by Mass Communication Specialist 1st Class Danica M. Sirmans)
Army set to field new cyber tool to improve situational awareness
Then-Maj. Gen. Kevin B. Kennedy, Assistant Deputy Chief Information Officer, Digital Transformation, and Assistant Deputy Chief of Staff for Cyber Effects Operations, Headquarters US Air Force, participates in the Enterprise IT-as-a-Service (EITAAS) Vendors Executive panel discussion during the 2019 Air Force Information Technology & Cyberpower Conference, Aug. 27, 2019. (U.S. Air Force photo by William Birchfield/Released)
New Air Force information warfare commander wants to increase partnerships
Top Marine expects unmanned technology to spur change in designs of amphibs
Marine Corps releases new doctrine on information
House Armed Services Committee concerned with state of Navy cyber readiness
SASC bill authorizes budget boost for several Cyber Command initiatives


text:
Top 100 Events
Plastic & Packaging
plastprintpack Kenya 2023
fairtrade launches agrofood & plastprintpack Kenya Annual business platform for Kenyan and Swahili speaking East & Central African professionals Based on their vast experience in organizing professional trade shows in North and sub-Saharan Africa, the German trade show specialists fairtrade...
IPF Bangladesh 2023
The Largest Industrial Trade Exhibition of Plastic, Rubber, Printing and Packaging in Bangladesh Last year IPF gathered over 460 exhibitors from 19 countries and areas, namely Australia, Austria, Bangladesh, China, Egypt, Hong Kong, India, Indonesia, Italy, Japan, Pakistan, Singapore, South...
Media Expo Mumbai 2023
The international trade exhibition on indoor and outdoor advertising and signage solutions. Backed by a successful legacy of over two decades, Media Expo is undoubtedly the most trusted business platform that brings together leading global brands and creative professionals to...
Plastic, Packaging & Print Asia 2023
"The Packaging Material, Supplies & Accessories Industry Expo" The Plastic, Packaging & Print Asia is the most promising packaging exhibition that showcases products and services like Packaging & Printing Machinery & Equipment, Plastic, Packaging & Printing Plant Systems, Paper &...
3P – Plas Print Pack Pakistan 2023
"International Plastic, Printing & Packaging Industry Exhibition." 3P Plas Print Pack Pakistan will bring the world's best Plastic, Printing, and Packaging suppliers to Pakistan. It will also have companies involved in Rubber and Chemical industries. 3P Pakistan will be an...
Nepal 5P International Expo
Nepal's only International Expo for Plastic, Printing, Packaging, Paper & Petrochemicals Industries Nepal 5P International Expo is distinguished exclusively as the most-ranging industrial trade event of its kind in Nepal; Nepal 5P is without question, the most crucial trading &...
Dhaka International Plastic, Packaging & Printing Expo 2023
"International Exhibition on Plastic, Packaging & Printing" Dhaka International Plastic, Packaging & Printing Expo will be an excellent sourcing platform for international manufacturers, professional buyers, and decision-makers in the plastics, printing, and packaging industries, but also brings the opportunity for...
OMAN PLAST 2023
"International Plastics, Rubber, Petrochemicals, Chemicals, Fertilizers, Printing & Packaging Industry Exhibition & Conference" OMAN PLAST is an exclusive International Plastics, Rubber, Petrochemicals, Chemicals, Fertilizers, Plastics Recycling, Printing and Packaging Industry Exhibition & Conference. The objective of this prestigious exhibition is...
PPPEXPO KENYA 2023
ABOUT THE EVENT Exhibitors from over 30 countries participating at Kenya's Biggest Int'l. Trade Exhibition PPPEXPO Africa is the only dedicated industry specific international trade show for Plastics, Printing & Packaging Industry in the entire East African region that promises...
BANGLADESH 3P INTERNATIONAL EXPO
INTERNATIONAL PAPER, PLASTIC, & PACKAGING TRADESHOW OF BANGLADESH Bangladesh 3P International Expo 2023 is an Interconnected Sectors trade show focused on the Materials, Machinery and Applied Products. Bangladesh, growing rapidly over a decade, is on its way to becoming the...


text:
package com.google.cloud.solutions.mobilepushnotification;

import com.google.appengine.api.datastore.Cursor;
import com.google.appengine.datanucleus.query.JPACursorHelper;

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

import javax.persistence.EntityManager;
import javax.persistence.Query;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * HttpServlet for preprocessing push notification requests. In particular it takes a request for
 * sending push notification to all registered devices, and uses a cursor to process it in batches
 * with as much transactional consistency as possible.
 *
 * It is intended to be called by Push Task Queue, so the request is retried if it fails.
 *
 */
public class PushPreProcessingServlet extends HttpServlet {
  private static final long serialVersionUID = 1L;

  /*
   * BATCH_SIZE indicates how many device tokens will be included in a payload of a single task.
   * Typically the larger the batch, the higher the throughput as fewer RPC calls need to be made.
   * However, the payload of a single task has a size limit. Also, it may be impractical to track
   * the notifications from a single task on per device token basis. When an instance
   * is restarted, it is given a finite amount of time to complete processing (e.g., 30 seconds) and
   * the batch size should be small enough to guarantee that the whole batch will be processed
   * within the allotted time to avoid duplicated notifications while still guaranteeing that the
   * notification is sent to each device in a batch.
   */
  private static final int BATCH_SIZE = 250;

  @Override
  public void doPost(HttpServletRequest req, HttpServletResponse resp) throws IOException {
    if (!Configuration.isRequestFromTaskQueue(req, resp)) {
      return;
    }
    String alertMessage = req.getParameter("alert");
    String queryType = req.getParameter("queryType");
    if (queryType.equalsIgnoreCase("allDevices")) {
      String cursorString = req.getParameter("cursorString");
      preprocessBatchOfDevices(alertMessage, cursorString);
    }
  }

  private void preprocessBatchOfDevices(String alertMessage, String cursorString) {
    EntityManager mgr = null;
    Cursor cursor = null;

    try {
      mgr = getEntityManager();

      // Retrieve entities (and not just deviceToken property) in order to paginate using cursor
      Query query = mgr.createQuery("select from DeviceRegistration as DeviceRegistration");
      if (cursorString != null && cursorString != "") {
        cursor = Cursor.fromWebSafeString(cursorString);
        query.setHint(JPACursorHelper.CURSOR_HINT, cursor);
      }

      query.setMaxResults(BATCH_SIZE);

      @SuppressWarnings("unchecked")
      List<DeviceRegistration> deviceRegistrations = query.getResultList();
      cursor = JPACursorHelper.getCursor(deviceRegistrations);
      if (cursor != null) {
        cursorString = cursor.toWebSafeString();
      }

      List<String> deviceTokens = new ArrayList<String>();
      for (DeviceRegistration deviceRegistartion : deviceRegistrations) {
        deviceTokens.add(deviceRegistartion.getDeviceToken());
      }

      if (deviceTokens.isEmpty()) {
        // no more matching device tokens matching this query.
        return;
      }

      mgr.getTransaction().begin();

      try {
        PushNotificationUtility.enqueuePushAlert(alertMessage, deviceTokens);

        if (deviceRegistrations.size() == BATCH_SIZE) {
          PushNotificationUtility.continueEnqueueingPushAlertToAllDevices(
              alertMessage, cursorString);
        }

        mgr.getTransaction().commit();
      } catch (RuntimeException e) {
        if (mgr.getTransaction().isActive()) {
          mgr.getTransaction().rollback();
        }
        throw e;
      }

    } finally {
      mgr.close();
    }
  }

  private static EntityManager getEntityManager() {
    return EMF.get().createEntityManager();
  }
}


text:
Helga Esteb / Shutterstock.com
Faith: Atheist
Career: Actor
Accomplishments: Nickelodeon Kids' Choice Awards
Rider strong was born in San Francisco, and began his acting career at 9 when he starred as Gavroche in the San Francisco production of Les Misérables. His career progressed to smaller television roles until he received his first regular role as Julie Andrew's son on Julie, followed by his debut as Amy Irving's son in Benefit of the Doubt. In 2004, Strong graduated magna cum laude from Columbia University, and completed his Master of Fine Arts degree in June 2009 at Bennington College.


What if you could easily build the equivalent of a Turtlebot for under $300? I'm going to show you how easy it is.
Here's a quick tutorial on running ROS with a BeagleBone or Arduino.
What is a TurtleBot? According to [url=http://www.turtlebot.com]http://www.turtlebot.com[/url]: "TurtleBot is a low cost personal robot kit with open source software.
The rocker bogey suspension system is an amazingly simple design with proven effectiveness.
Harness the rays from the sun to power your outdoor bots.
The idea behind this homebrew build is to develop an open source hardware platform capable of performing useful household tasks, plus be expandable.


text:
Q: Closed Formulas in a First Order Logic Theory According to wiki a theory (im instered in FOL) is

is a set of sentences in a formal language. Usually a deductive system
  is understood from context

It's clearly specificied that we are considering only sentences ( = closed formulas) but we can't build a theory using open formulas. But why ???

A: We can, but open axioms must be read as universally quantified.
See the well-known example of Tarski's axioms for (elementary) geometry, formulathed in first-order logic with identity, and requiring no set theory (1959):

the axioms should be read as universal closures; hence any free variables should be taken as tacitly universally quantified.

See examples:

Congruence axioms: Reflexivity of Congruence
$xy\equiv yx$,

and so on.

redpajama_set_name: RedPajamaStackExchange


This module aims to introduce students to the main components and concepts of marketing a business. It will serve as a conceptual core to which more specialist modules, such as entrepreneurship and accounting, can be related.
The content of this module includes the fundamental concepts and process of marketing; how environmental factors (situation analysis) impact the marketing of a business; buying behaviour; market segmentation, targeting, positioning and differentiation. The elements of marketing (7Ps, including online and social elements); services marketing and customer relationship management will also be covered.