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Q: Azure KeyVault Encrypt API Endpoint Not Returning Base64 Encoded String In PowerShell I'm calling the KeyVault Encrypt API endpoint to encrypt a connection string.
First I create a byte array of the connection string and then I create a base64 string. Then I call the /encrypt endpoint
[byte[]]$bytes = [System.Text.Encoding]::Unicode.GetBytes($targetConnectionString)

echo "Plain Bytes" $bytes

#$base64Array = [Convert]::ToBase64String($bytes)

#$EncryptedTargetConnectionString = Encrypt-ByteArray -accessToken $kvToken -vaultName $kvName -keyID $encryptionKeyId -plainArray $bytes

# converting the byte array to Base64 string
  # Note from [Chris Clayton (AzureCAT)].
  # "When performing the post methods there is a common theme of converting to Base64 strings which just makes sense so we do not introduce any invalid characters etc. and creating a JSON body as you can see here"
  $base64Array = [Convert]::ToBase64String($bytes)

  echo "Base 64 Array" $base64Array

  # constructing the url for the vault encrypt operation by specifying the KeyID and Key version. If key version is not specified, then the most recent key version will be used
  $queryUrl = "$encryptionKeyId" + '/encrypt?api-version=2016-10-01'

  #echo "Query URL: " $queryUrl

  # injecting the access token we obtained in the authorization header
  $headers = @{ 'Authorization' = "Bearer $kvToken"; "Content-Type" = "application/json" }

  echo "KV Token" $kvToken

  # construct the body with our data to be encrypted and specifying RSA as encryption alg.
  $bodyObject = @{ "alg" = "RSA-OAEP"; "value" = $base64Array }
  $bodyJson = ConvertTo-Json -InputObject $bodyObject

  # invoking a POST call to do the magic
  $encryptionResponse = Invoke-RestMethod -Method POST -Ur $queryUrl -Headers $headers -Body $bodyJson

That seems to be working, because when I use Postman to call the Decrypt endpoint I get the response I'd expect.
Then I call an Azure Function which will need to be able to decrypt this connection string.
$EncryptedTargetConnectionString = $encryptionResponse.value

$webRequestBody = @"
{
        "BlobUrl" : "$blobUrl",
        "Environment" : "$env",
        "State" : "$state",
        "EncryptedConnectionStrings" : [
            {
            "EncryptedConnectionString": "$EncryptedTargetConnectionString"
            }
        ]
    }
"@

$webRequest = Invoke-WebRequest http://localhost:7071/api/DisableReplication -Headers @{"Accept" = "application/json"} -Method 'POST' -ContentType 'application/json; charset=utf-8' -Body $webRequestBody -UseBasicParsing

What's crazy here, though, is that the RESPONSE from the /encrypt endpoint doesn't seem to be in base64 format.
When I try to run Convert.FromBase64String(encryptedString) in C# against the encrypted string, it doesn't work. It tells me about it not being it base64 format.
C# Code
byte[] inputAsByteArray = Convert.FromBase64String(encryptedString);

            // Encryption algorithm MUST match what was used in the PowerShell task in the Release Pipeline to encrypt the string
            // Ex: $bodyObject = @{ "alg" = "RSA-OAEP"; "value" = $base64Array }
            DecryptResult decryptResult = await cryptographyClient.DecryptAsync(JsonWebKeyEncryptionAlgorithm.RSAOAEP, inputAsByteArray);

            string decryptedString = Encoding.UTF8.GetString(decryptResult.Plaintext);

            return decryptedString;

None of the Key Vault SDK Clients that I've found for .NET Core support decrypting a string directly, and I can't seem to convert the string into a byte array the way I'd expect to be able to.
Any idea what kind of mind games Microsoft is playing here?

A: Key Vault uses Base64Url Encoding which is slightly different than Base64 encoding.
Base64URL encoding is nearly identical to Base64 encoding except for the fact that it substitutes characters that are not safe in a URL ('+','~', etc.) with other characters. In addition, it does not use padding ('=').

redpajama_set_name: RedPajamaStackExchange


text:
Rajasthan Chief Minister Ashok Gehlot. (File Photo)
Rajasthan CM meets Finance Minister, seeks timely release of funds
New Delhi [India], June 15 (ANI): Rajasthan Chief Minister Ashok Gehlot met Union Finance Minister Nirmala Sitharaman here on Saturday to discuss various financial issues related to the state.
He requested the centre to give approval to the pending proposals of drinking water projects and ensure timely release of funds of the centrally-sponsored schemes, an official statement said.
"As per the past practice the state's share in the central taxes should be released to the respective states on the first date of the month. As this practice had been changed since past few years, the states were facing financial issues. The states have to make payments towards salaries and pension on the first date, which becomes very difficult," the Rajasthan Chief Minister was quoted as saying.
The Chief Minister also requested the Finance Minister to approve the financial proposals of seven drinking water projects costing Rs 5473 crore at the earliest.
According to the official statement, Gehlot asked for approval for the proposal for external financial aid for the Phase 3 of Rajiv Gandhi Lift Canal Project costing Rs. 1,450 crore.
He a;sp sought swift approval of several drinking water projects pending with Union Ministries of Drinking Water & Sanitation and Urban Development.
"The Chief Minister asked the Union Minister to support the state's loan waiver scheme to help farmers of Rajasthan recover from the indebtedness. He informed that 24 lakh farmers affiliated to cooperative banks had been given the benefit of loan waiver in Rajasthan and the state had to bear a burden of Rs 15,679 crore," the official statement said.
Gehlot suggested that to ensure no dearth of funds for various development projects, the states should be allowed to take loans upto 4% instead of 3% of the Gross State Domestic Product (GSDP). He said that the Centre should make the procedure for the states to take loans from the open markets a stable one.
The delay on part of the Centre to give approval for taking loans was hampering the ongoing development projects, the Chief Minister said. (ANI)
Ashok Gehlot
Nirmala Sitharaman
Assam: Over 50 animals dead in Kaziranga National Park as floods...
Kaziranga (Assam) [India], July 18 (ANI): Over 50 animals, including five rhinos, have died in Kaziranga National Park as floods continue to wreak havoc in Assam.

redpajama_set_name: RedPajamaCommonCrawl


Blog | Lawson & Co.
LAWSON & CO shares what you need to know to be success as buyer or seller when it comes to farm auctions.
What is the Best Time for Selling Your Collectibles and Personal Property?
Selling your collectibles and antiques may be more or less successful depending on the time of year. Read the Lawson blog to find out when the best time of year is for selling your collections.
A specialty collection can be fun to build. Curate it in a specific manner to be able to trace provenance and establish value. When you want to sell your collection, we can help with a Specialty Auction aimed at specific buyers and group of collectors.
What Is the Best Option for Selling Your Home Sweet Home?
With so many options and complexities in the real estate market, it can be difficult to determine which method of selling your home is best for your situation. Selling on contract, FSBO, Realtor™ listing, or selling on auction are some of the options to consider.
There are many ways to sell your belongings, however, a consignment auction could be the best choice for many. How do you know? LAWSON & CO compares your options.


Rockwell Kent was both a perfect creature of his time and a complete original. His brawny, broad-shouldered, but sleekly elegant Art Deco illustrations are firmly planted in early 20th America, when books and print served as potent messengers of cultural meaning, capable of the sharp sting of reproach as well as the fine pleasure of a literate, well-composed page. He is one of a number of American artists, including some of those I wrote about two posts ago, who give us a clear picture of a very particular time.
Kent needed no paltry second-hand references for his churning waves, high seas, and angry nature - he'd been there. The list is long and impressive - Greenland (where he was shipwrecked and lived for some time) Tierra del Fuego, Alaska, Newfoundland, an island off the Maine coast - and the wilds of the Adirondack Mountains, his stateside home for many years.
Vanity Fair illustration by "Hogarth Jr."
Kent walked another tightrope of sorts in his career and life, moving from 'society illustrations' for Vanity Fair (he signed them Hogarth Jr., a nod to William Hogarth, the sly 18th c. English artist and satirist) and sleek ads for a luxury boat manufacturer, to high-toned book commissions, and finally to crisply critical social commentary that aligned with his deeply-held belief in the rights of the common man.
'Workers of the World, Unite' and 'Wake Up America,' both in a show at the Philadelphia Museum, are good examples of how he could temper his message from urgent to subtle. 'Workers of the World, Unite,' seen here on the cover of a 'red' journal, is a powerful call to action while 'Wake Up America' has more than a little Norman Rockwell folksiness - it takes a few moments to register the message of numbing destructive apathy - time is running out and the flag is being bent and broken overhead.
Kent was a gifted writer too, one who put his experiences and his philosophies into words as well as pictures. Like his art, his books are a slightly dated but earnest, authentic record of a fascinating period of time from a singular perspective. A passage from his book Voyaging spells out his identity: "it the reality of mountains and plains, the sea and the unfathomable heavens, unchangingly forever dominating man, cradling him in that remote hour of his awakening into consciousness, forever smiling, brooding, thundering upon him, that have imposed their nature upon man and made him what he is."
Along with many other creative intellectuals, Kent put time into the Communist Party in the hopes that it would provide a better answer; inevitably his involvement led to clashes with Joe McCarthy and his committee. It also, however, led to remarkable honors and actions. Kent, the first American artist to show his work in the U.S.S.R., was awarded a Lenin Peace Prize; he donated the money to help women and children in both Vietnams during the Vietnam War years.
The Philadelphia Museum is currently showing Rockwell Kent - Voyager: An Artist's Journey in Prints, Drawings, and Illustrated Books. Through July 29, 2012. The show is organized by Kent's longtime friend Carl Zigrosser, the founding curator of the museum's department of prints and drawings. The show includes a range of work, including woodcuts, pen and ink drawings, pencil sketches, lithography (a self-portrait with the stone that made the print is fascinating) and watercolors.
I recently heard Grayson Perry on a BBC podcast about traditions and was so struck by his interesting remarks that I had to find out more about him - and discovered a whole new world. I can't believe I've missed him until now - he's no shrinking violet. A new member of the Royal Academy, he currently has a big show at the British Museum called The Tomb of the Unknown Craftsman. Part of his contribution to the Traditions discussion was a comment about liking to work with mediums that require skill and relate to age-old processes, not only ceramics but lately also tapestry.
In his ceramics he sticks with traditional forms such as standing vases reminiscent of Greek amphora; he claims to like lulling people into a feeling of security with forms they think they recognize and with expectations of reassuring patterns - flowers or a simple landscape - and then hitting them over the head with his intricately drawn decorations full of wit, puns, historical references and social comment.
He says he has little use for the post-Duchamp conceptual idea of art - it's art if I say it's art - so goes out of his way to make his life as well as his art an exercise in complicated, elaborate craftsmanship. And he's having a great deal of fun doing it - he seems a bit like Cindy Sherman with a sense of humor. And there's his 50-year old teddy bear, Alan Measles, who is something of a god in his world.
I've always on the lookout for great use of drawing and here's a wonderful example of a good hand, plus mind and heart working together to create a sum much greater than the parts. An instant favorite for me is his Punters in the Snow - a witty homage to Bruegel's elegiac 16th century painting Hunters in the Snow. Grayson Perry, his work and life, brought to mind Shakespeare's line from The Tempest: O brave new world That has such people in't!


text:
Jibran Nasir Thankful to Fiance Mansha Pasha for Choosing Spouse
Social leader Jibran Nasir has praised his fiance actress Mansha Pasha and said that he is grateful to Mansha Pasha for choosing a person like him as his life partner. Mansha Pasha and Jibran Nasir got engaged in December 2019 and both are planning to get married soon. After the engagement, they have been seen together on social media and at public events, and the two continue to show love for each other. Recently, both the actresses participated in Meera Sethi's program, where they talked openly about politics and showbiz, including their private lives.
The two praised each other at the event and also thanked each other for their choice. In the program, Mansha Pasha revealed that Jibran Nasir gets very angry and the sooner he gets angry, the sooner he gets angry. In response to a question, Jibran Nasir praised his fiance and said that he was "thankful" to Mansha Pasha for choosing a person like him. Jibran Nasir admits that it is difficult to live with him as he is, but he is grateful to Mansha Pasha for choosing him as his life partner and he was not worthy of his love.
In the same program, Mansha Pasha, while answering a question, said that Jibran Nasir has given him complete freedom, sovereignty, and the power to make every decision regarding his caste and he understands everything about him. It becomes difficult for them to express. In response to another question, Jibran Nasir said that he does not care about how his future wife works. He said that their fields are different, so they will not be jealous of each other's development and admiration and he had thought before they got engaged that their field is different, so how did life go? Will you
A post shared by Mansha Pasha (@manshapasha)
Although the two expressed their intention to get married soon in the program, they did not specify how long they would get married. In addition to the aforementioned program, Mansha Pasha also talked about her marriage in a show called To Be Honest recently and said that she intends to get married soon but she could not announce the date and time. Mansha Pasha said that by announcing the date, they will be bound, so they do not want to be bound but intend to get married soon.
In the same program, Mansha Pasha revealed that he spent his childhood in Hyderabad, the second-largest city in Sindh province, and received his early education there. Mansha Pasha, while talking about Zamana Student, admitted that she was a sacrifice to a few boys in that period, but the boys to whom she was a sacrifice, did not know it. In response to another question, Mansha Pasha said that she was not obsessed with fair complexion since childhood, and boys used to die for her fair complexion as a child. However, when she came to showbiz, she was obsessed with a fair complexion for some time.
Mansha pasha Jibran nasir
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A locally produced Prince Pearl is being introduced this week in Pakistan
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text:
Former Iranian president defends Kaepernick
By Megan Keller - 09/04/18 03:33 PM EDT
Former Iranian President Mahmoud Ahmadinejad defended Colin Kaepernick Monday amid the former NFL quarterback's controversial role in a new Nike commercial.
"The #NFL season will start this week, unfortunately once again @Kaepernick7 is not on a NFL roster," Ahmadinejad tweeted.
The #NFL season will start this week, unfortunately once again @Kaepernick7 is not on a NFL roster. Even though he is one of the best Quarterbacks in the league.#ColinKaepernick #NFL
— Mahmoud Ahmadinejad (@Ahmadinejad1956) September 3, 2018
"The philosophy and spirit of sports is bringing people together, not causing more conflicts," Ahmadinejad wrote. "Politics must support the spirit of sports not destroy it."
Kaepernick, formerly with the San Francisco 49ers, gained national prominence with his kneeling protests during the national anthem, a form of activism that has been taken up by athletes across the league.
He has since filed a lawsuit against the NFL's league owners, alleging that they colluded to keep him from being signed to a new team amid the controversy. A arbitrator ruled that Kaepernick's case would go to trial last week.
Nike made Kaepernick the face of its new "Just Do It" ad that aired Monday, which stated, "Believe in something. Even if it means sacrificing everything."
ESPN reported that Nike signed an endorsement deal with Kaepernick in 2011 and has been paying him since.
Though the former Iranian president lauded Nike's move, it has not been well received by everyone.
Some people have destroyed their Nike products or remove Nike's logo in protest of the company's decision. And a NBC News/Wall Street Journal poll released last Friday showed that a majority of respondents disapprove of kneeling during the national anthem.
Cruz in a tweet slammed Ahmadinejad, Nike and Rep. Beto O'Rourke (D-Texas), his opponent in November's midterm elections, who says he supports the protest.
When a radical anti-Semite, anti-American Iranian dictator emphatically agrees with you, maybe that's a sign that Beto, the NFL, and Nike are all on the wrong side of the American people.... https://t.co/7MYcLfIsBz
— Ted Cruz Rafael (Ted) Edward CruzThe Hill's Morning Report - Presented by JUUL Labs - House to vote to condemn Trump tweet Cruz in 2016 said 'something fundamentally wrong' with Christians who back Trump: book Hillicon Valley: Twitter says Trump 'go back' tweet didn't violate rules | Unions back protests targeting Amazon 'Prime Day' | Mnuchin voices 'serious concerns' about Facebook crypto project | Congress mobilizes on cyber threats to electric grid MORE (@tedcruz) September 4, 2018
Tags Ted Cruz Colin Kaepernick Senator Ted Cruz Beto O'Rourke Nike Mahmoud Ahmadinejad


DRK and Company Realty is now proudly offering 71 Brice Road North for lease or sale to Greater Columbus-area businesses.
Situated on 2.95 acres, this industrial warehouse building includes 1,500 square feet of finished office space and 19,200 sf of warehouse space. The structure is on a secured, fenced lot and offers M-1 zoning.
Built in 2003, the property contains two docks and one drive-in door, and a dry pipe sprinkler system. Column spacing of 25' x 40' and a clear height of 20' make the property ideal for a variety of uses. The office area is fully conditioned, and the warehouse area has unit heaters and exhaust fans. In addition to the loading areas, the building includes 17 parking spaces.
The building is in a prime location on the east side of Columbus, located within one mile of the I-270 and East Broad Street interchange. This location is convenient in providing efficient access to the central Ohio region.
It is also just 11 minutes from the John Glenn International Airport (formerly Port Columbus International) and situated amongst a mix of industrial, retail and dining businesses.
For more information on 71 Brice Road North in Columbus, contact Jim Bain or Perry Smith today at 614.540.2404!


The Jackie WOD is one of the most popular, and challenging of the female named workouts of the day in the entire CrossFit stable. In this article, we'll dissect Jackie and then reconstruct her with winning strategies that will allow you to give your best performance next time you take this lady on.
Jackie is a timed workout. That means that you need to get through it as quickly as possible. To give you a gauge of what sort of time is possible, Jason Khalipa set a world record on this WOD in 2013 with a time of 5:04. You might not be able to get close to that mark, but if you get this thing done under ten and minutes, you'll be doing pretty darn good.
The rowing machine is a technical piece of equipment that requires proper form. Without it, you can and will waste a lot of your energy on bad form. That is why it is imperative that you get your form right from the start.
Start by strapping your feet into the foot stirrups securely.
There are three aspects to your rowing motions, and they involve three parts of your body - the legs, hips and arms. Start by keeping your back upright with your core tight. Lean slightly forward, making sure that your spine is in a neutral position so that you never hunch your back over.
A fierce fight erupted. The SEALS were outnumbered by an enemy force of more than 50 anti-coalition militia in rough terrain. One of the SEALS, Lt. Michael Murphy, risked his life to save his team-mates.
Your three step motion begins with the legs. Extend them out to open your hips, simultaneously leaning back and following through with the arms. This motion needs to happen fluidly as one smooth motion. You should be moving in a straight line. Each return should be an opportunity for a slight recovery.
On the way back, send your arms back in, then bend at the hips and then at the knees. So, it's legs, hips,and arms on the row, and arms, hips,and legs on the return.
You don't want to be rowing in short, fast bursts. Allow the turbine to run its complete course. This will actually allow you to cover more distance than going all out on short bursts. Think in terms of a long drive back and a recovery on the return.
Becca Borawski, editor of Breaking Muscle, offers the following advice on tackling the rowing portion of Jackie . .
"People are horrible rowers,and they ruin themselves first thing, going too fast, moving too inefficiently and not planning for the entire workout. And, before that, even, improve your rowing technique. So many CrossFitters spend so much time on other exercises and waste ridiculous amounts of energy on any workout containing rowing."
With the Thruster exercise, technique is, once more, imperative. The thruster, when done correctly, is a front squat which goes into a push press, culminating with a punch at the top. The key to success with this move is to join these three aspects in a fluid motion.
The placement of the bar is imperative. It needs to be resting across your shoulders as you perform the front squat and you need to keep your elbows high. This will keep your torso from tilting forward. It will also keep your elbows from getting in the way and hitting your knees when you descend into a deep squat. This can help to prevent some serious damage to the elbow joint.
Coming out of the squat, you should drive with your hips. This will cause a chain reaction of momentum that will transfer directly into the pressing movement. As you come up, squeeze your glutes and push up with your hips. Make sure that the bar is still resting on your shoulders, and not being held out in front of you.
Once you have done the push press, punch the bar up over your head. Do not grip the bar too tightly so that you can flip back your wrist as you punch up.
To get through your 30 pull ups in as fast a time as possible, you should opt for butterfly kipping pull ups. With this pull up you do not have to actually get your chin over the bar. You simply have to get it in line with the bar. The butterfly kip uses the energy that gravity gives you when you fall on the front portion and use that energy to create a circular motion.
The butterfly pull up needs to start at the shoulder. You want to roll your shoulders back and then forwards, which will start to elevate the elbows. You also want to create angle at the knee and the hips in order to create more power. To do this you should close up the hips and elevate the knee during the press down portion.
When you start the butterfly pull up, you need to understand the circular motion. Once you do, you will learn to kick harder with your legs, which will bring you higher into the butterfly. Think in turns of swooping down and following through as a fluid action. It is the leg swing that drives you up for the next rep.
You can practice the proper leg swoop motion by standing with your left hand cracked against an upright and doing leg circles with your right leg that mimic this popping motion. Get faster and faster as you swoop your leg down. Then repeat with the other leg.
Once you have this standing leg swoop down, jump up to a pull up bar and practice a hanging leg swoop. Once you are used to doing this, pull up a,little bit with your arms. Then keep pulling up as you drive with more momentum through the legs to bring yourself up to the level of the bar.
As with any WOD, pacing on the Jackie is critical. You don't won't to go all out in the first couple of minutes and then crash and burn before you're through. On the C2 Rowing Machine you should go at 90%. Try to hold a rating of 28, which should allow you to complete this part of the workout in around 3:20. Remember to focus on a long, smooth rowing motion. If you sprint, you will only end up building up lactic acid in the muscle, which is not a good thing. As you row, count off sets of a hundred rows and imagine yourself out on the water.
Your goal should be to get through all 50 thrusters in one set. It is usually here where people tend to slow the pace, so your goal should be to do the opposite. Build momentum and speed as you, without sacrificing your form. Do not allow yourself to stop short. Rather than counting all the way to 50, which is too mentally daunting, focus on knocking off 5 reps, ten times.
Here is Borawaski's thruster advice . . .
"Pick up your speed over the course of the thrusters. Try to get faster when others are getting slower. Don't let your brain stop your thrusters set short. This is even truer on the pull ups. Just get them done. Don't think, don't breathe, keep moving. This is where you make up your time."
If this is your first time at attempting the Jackie workout, you should consider doing a mini version first, to allow yourself to work up to the real thing. You might use half the weight in thrusters, use a resistance band and to help you with your pull ups or lower the resistance on the rowing machine. Once you have done this a few times, use the actual resistance required on the workout, take away the resistance and and put the rower resistance back up, but cut the number of reps in half.


RIO DE JANEIRO (AP) When Michael Phelps decided to come out of retirement, he circled one race in particular on the Olympic schedule.
At the London Games, in what was supposed to be his Olympic farewell, Phelps was leading as he came to the wall in the stroke he does better than anyone.
But he mistimed the finish, glided a little too long, and was edged by Chad le Clos.
That result has irked Phelps ever since.
On Tuesday night, he'll get a chance to make amends against the South African.
Call it the Rematch in Rio.
''Should be a fun race,'' Phelps said.
Le Clos knows he'll be facing the most decorated athlete in Olympic history, a highly motivated rival still at the top of his game as he showed in leading the United States to a stirring victory in the 4x100 freestyle relay.
But Le Clos isn't backing down.
A pair of Hungarians could also be part of the mix. Tamas Kenderesi posted the fastest time in the semifinals, edging out Phelps, and five-time Olympic medalist Laszlo Cseh put up the third-fastest time. Le Clos was fourth, not bad considering it was the back end of a grueling double, coming less than an hour after he claimed silver in the 200 freestyle.
Television cameras caught Phelps with a serious game face before the semifinal swim as he stared in the direction of Le Clos, a moment Twitter users immediately flipped into memes poking fun at Phelps' intensity.
Phelps will be going for the 20th gold medal of his career, but the South African has plenty of motivation, too. His parents, both battling cancer, are cheering him on in Rio.
Phelps will actually have two chances for gold on Tuesday. He will also be part of the 4x200 freestyle relay, a race the Americans have won at the last three Olympics.
The U.S. will have a good shot at another gold in the women's 200 free, with Katie Ledecky looking to make it 2-for-2 in the longer freestyle events. She already crushed the world record in the 400 free, in addition to anchoring the Americans to a silver medal in the 4x100 free relay.
Ledecky was second in the semifinals behind Sweden's Sarah Sjostrom, gold medalist in the 100 butterfly.
Ledecky and Sjostrom aren't the only women dominating at the pool.
Hungary's Katinka Hosszu became the first two-time swimming gold medalist at the Rio Games with a victory in the 100 backstroke.
Phelps also took aim at swimmers linked to doping, an issue that has been at the forefront on the heels Russian doping scandal. Harsh words have been directed as athletes such as Yulina Efimova and Sun Yang, who previously served doping suspensions.
Phelps joined the chorus of those demanding a clean competition.
''You're probably going to see a lot of people speaking up more. I think something needs to be done,'' he said. ''It's sad that today in sports in general, not just only swimming, there are people who are testing positive who are allowed back in the sport - and multiple times.


text:
Q: What is the most appropriate HTTP method to send info? I would like to know what is the most appropriate HTTP method to send info about an receive some logical result in a response.
I have to build an endpoint to receive some information about address and postal code, and this endpoint should process this data to respond information about the result of the process.
The info that I have to send is an address, then it could contain special characters, or whitespaces, and the postal code.
I was thinking how to do it, I thought in two different ways, but I would like to know from you what will be the best approach.
First:

*

*GET

*

*www.myapi.com/georefence?postal_code=33210&address=street 56 1254

*Send the information in the path as params

*POST

*

*www.myapi.com/georefence

*Send the information in the body as object

*body: { "postal_code": 33210, "address": "street 56 1254" }
Expected:
Information about the longitude, latitude, if the postal code was found successfully, the precision, and other things.
Thanks.


text:
Asia | Taiwan
Now for the rest of the banquet
Jul 10th 1997 | TAIPEI AND HONG KONG
Try boiling oil
AFTER Hong Kong, Taiwan? As China's leaders celebrate the return of Hong Kong to the "motherland", they have left little doubt that they see Hong Kong merely as the first course in the great reunification banquet. The main dish, Taiwan, remains to be consumed. In his speech celebrating Hong Kong's return early on July 1st, Jiang Zemin, China's president, made it clear that the 21m people in Taiwan—and especially their government, descended from the Nationalists who lost the civil war to China's Communists in 1949—are very much on China's menu. Indeed, for all the talk of ending the shame of colonialism, it seems to be the intra-Chinese quarrel with Taiwan that stirs the deepest emotions of the leadership in Beijing.
Taiwan's leaders are outwardly unconcerned. Lee Teng-hui, the island's sprightly 74-year-old president, dismisses China's hopes of extending the "one country, two systems" formula from Hong Kong to Taiwan as "wishful thinking". But, privately, Taiwan's leaders are worried that the return of Hong Kong will lead to ever-increasing pressure to succumb to China's unwelcome embrace.
If the "one country, two systems" idea is even half-successful in Hong Kong, Taiwan's leaders may find more and more voices in the West urging them to accept the "inevitable". Western leaders know that China is of greater economic and strategic significance than Taiwan. They have also been impressed, indeed intimidated, by China's stated readiness to go to war over Taiwan. One senior American visitor to Beijing was more or less told that China would be ready to nuke Los Angeles, if America fought to defend Taiwan. This may be bluster, but it still gives pause for thought.
Fortunately for Taiwan, it has four things going for it, in comparison with Hong Kong. The first and most important is nearly 170 kilometres (100 miles) of sea between Taiwan and the mainland. It was a cliché in Hong Kong that China could take back the British colony whenever it liked, "with a phone call". But invading Taiwan, as China often threatens to, would require a military operation that would make D-day look simple. Preparations for any invasion would take months and would be quickly spotted by spy satellites. There could be no element of surprise, and Taiwan is well armed.
The second feature to distinguish Taiwan's plight from Hong Kong's is the lack of a deadline. The expiry of the lease on Hong Kong's New Territories in 1997 obliged the British to negotiate with China over the future of the island itself. Taiwan's leaders, however, can express a general desire for reunification and then play for time. In the long term, their main hope must be the emergence of a regime in Beijing which would either be more attractive as a marriage partner or more willing to accept that Taiwan should decide its own fate.
Taiwan's third card is democracy. Hong Kong's colonial system was a hangover from the past, but Taiwan's emerging democracy looks like an attractive model for much of the rest of East Asia. As a result, it is more likely to win support in the West. Democracy also means that it will be much harder to negotiate a deal over the heads of ordinary Taiwanese. Hong Kong's people were never consulted on their fate by their colonial masters. (The British, quite deliberately, did not put their plan to a referendum, for fear that it would undermine their carefully constructed deal with China.) By contrast, the democratisation of Taiwan over the past decade has revealed very little local desire for reunification with China—certainly on the terms the mainland has in mind. Recent polls suggest only 5% of Taiwanese favour swift reunification.
Democratisation in turn has fortified Taiwan's fourth differentiating feature: the strength of its relations with the United States. America—an ally of Britain—was ready to take a back seat over Hong Kong. By contrast, the Taiwan Relations Act of 1979 commits the United States to "resist any resort to force" in settling the Taiwanese issue. Some Americans argue that the United States has a legal, as well as a moral, obligation to protect Taiwan.
Playing for time
Will the four things that set Taiwan apart from Hong Kong be enough to save it from Hong Kong's fate? Possibly, but not necessarily. China cannot plausibly mount an invasion of Taiwan, but it could probably mount an effective blockade, either with missiles or with submarines. For a trade-dependent island-nation like the Taiwanese, such a blockade would have devastating consequences. And any western attempt to break a blockade would be nerve-racking—a modern-day Cuban missile crisis.
It would also be a mistake to think that the lack of a deadline offers Taiwan adequate protection. China suspects, probably rightly, that President Lee's talk of eventual reunification is a way of stalling, while quietly advancing the cause of independence. As a result, China now says it reserves the right to use force not just if Taiwan declares independence, but also if it thinks Taiwan is preparing for independence. China's fears that Taiwan may be edging towards going it alone could lead the government in Beijing to set a deadline for reunification. This would be a high-risk strategy, perhaps leading to war. It would, however, greatly increase pressure on Taiwan.
The threat that China might resort to a blockade or other forms of force is likely only to increase, because of Taiwan's domestic politics. Constitutional reforms in Taiwan may soon lead to the abolition of the Taiwanese provincial government—a redundant layer of administration whose only purpose is to support the fiction that Taiwan is a province of mainland China. China is not amused.
An even likelier source of trouble is the increasing weakness of President Lee's ruling Kuomintang, which is still rhetorically committed to reunification. The Kuomintang is likely to lose its majority in parliament in the election due next year. That may force it into an alliance with the largest opposition group, the Democratic Progressive Party, which has a formal commitment to independence in its constitution. The DPP's appearance in government could send China literally ballistic.
If so, Taiwan might have to fall back on its other safeguards—membership of the club of democracies and support from America. When China staged threatening missile tests in the Taiwan Strait last year to coincide with Taiwan's presidential election, America's reaction was unexpectedly robust—the dispatch of two aircraft carrriers to hover meaningfully in the area.
But talk to policymakers in leading western democracies and you cannot be confident that Taiwan would be supported in the face of stepped up Chinese belligerence. One leading figure in Australia blusters that "the idea of two Chinas is absurd." When it is suggested that popular opinion in Taiwan looks increasingly pro-independence, he replies, "That's why the big powers have got to get together and say No." In Japan, a strategist who is regarded as something of a hawk says that his country could not support Taiwanese independence: "That would be like declaring suicide."
It is America, however, to which Taiwan would look above all. And America's position is ambiguous, even contradictory. In part this ambiguity is a deliberate effort to keep China guessing. But it also reflects real divisions among American policymakers. Newt Gingrich, the speaker of the House of Representatives, is one of the few American politicians to say openly that America would defend Taiwan from attack. Other officials and politicians are wary of making such commitments, and are resentful of what they see as Taiwanese adventurism. Charles Freeman, the American former official to whom the Chinese subtly hinted about nuking Los Angeles, has complained that some Taiwanese politicians are behaving as if they have "a blank cheque, to fill out in American blood."
One fellow-democracy that gives Taiwan a sympathetic hearing is the Czech Republic. But then President Vaclav Havel is an unusually principled fellow. And his country has twice this century had the bitter experience of being sacrificed by bigger states on the altar of realpolitik.
More from Asia
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It is about to discover that no two metals are alike
A repugnant mass crimping of human potential


House Democrats this week are set to determine for the battle to lead Trump's potential impeachment, although with misleadingly huge outcomes: choosing who might lead potential indictment procedures against President Donald Trump should they reclaim the House.
Reps. Jerry Nadler (D-N.Y.) and Zoe Lofgren (D-Calif.), two long-term individuals who both hail from control focuses of the Democratic Party, are contending to be the best Democrat on the House Judiciary Committee. That is the board that verifiably holds the keys to the reprimand procedure for the battle to lead Trump's potential impeachment.
Both Nadler and Lofgren know they're joining to be Trump's central enemy in Congress. Be that as it may, in discrete meetings, they likewise accentuated the significance of Democrats not exaggerating their hand.
They're both intensely mindful that showing up excessively enthusiastic about arraigning the president could distance voters and powerless occupants in their gathering who recoil when the point is raised.
"I'm not running as an indictment competitor and I don't think Jerry Nadler is either," Lofgren said amid a meeting in her office.
She and Nadler have been moving for the best Judiciary spot for quite a long time. The inward decision was not expected until after the midterms next November, but rather was hurried by the current renunciation of Rep. John Conyers (D-Mich.), the long-lasting best Democrat on the council, over inappropriate behavior claims for the battle to lead Trump's potential impeachment. The full Democratic Caucus will hold a decision of the post Wednesday.
For Lofgren — a Democratic staff member on the board of trustees amid Richard Nixon's prosecution and an individual from the board amid the examination concerning Bill Clinton — the procedure would need to be bipartisan, with purchases in from the Republicans and the general population to be fruitful.
Lofgren indicated that sort of help would likely come just if Special Counsel Robert Mueller revealed, reported confirmation demonstrating Trump took activities that were a danger to the nation. Both Nadler and Lofgren as of late voted against a denunciation determination offered by Democratic Rep. Al Green (D-Texas), concurring with different Democrats that the move was dreadfully untimely.
Nadler, as well, said denunciation must have a generous purchase in from recent Trump supporters to be beneficial. Halfway, he stated, as a result of the math: Though the House can impugn a president, the Senate requires a 66% vote in favor of expulsion, which would be unimaginable without some Republican help.
"There's very little point in impugning a president and having him absolved in the Senate as occurred with Clinton," he said.
Besides, he stated, for Democrats seeking for the battle to lead Trump's potential impeachment after indictment, they ought to have a sensible conviction that some fragment of Trump voters will comprehend and bolster the push in light of proof.
Be that as it may, their more quick concern is an established emergency, should Trump endeavor to flame or ruin Mueller.


text:
{{Ficha de taxón
| name = Carabus caelatus
| image         = Carabus.caelatus.-.calwer.02.02.jpg
| image_width = 240px
| regnum        = Animalia
| phylum        = Arthropoda
| subphylum     = Hexapoda
| classis       = Insecta
| ordo          = Coleoptera
| subordo       = Adephaga
| familia       = Carabidae
| subfamilia = Carabinae| tribus = Carabini
| genus         = Carabus
| subgenus = Carabus (Megodontus)
| species       = C. caelatus| binomial = Carabus caelatus
| binomial_authority = Fabricius, 1801
| synonyms      =
Megadontus caelatus
|}}Carabus caelatus''''' es una especie de coleóptero adéfago de la familia Carabidae. Habita en Europe, donde han sido observados en Albania, Austria, Bosnia y Herzegovina, Croacia, Italia, Macedonia, Eslovenia y Yugoslavia.

Referencias

Enlaces externos

 Carabus caelatus on Fauna Europaea

caelatus

redpajama_set_name: RedPajamaWikipedia


Growing up in Zion, Mesa Verde, Petrified Forest and Grand Teton national parks, Fred J. Fagergren listened to his father, Fred C. Fagergren, his uncle, Russ Mahan, and many park rangers and naturalists as they shared stories of the early park rangers. They spoke of the importance of preserving park resources, as well as the need for rangers to stand fast in protecting those special places for future generations.
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In 1981, Fagergren became the first superintendent of Big Cypress National Preserve in South Florida. The enabling legislation for Big Cypress provided a strong legal basis for resource protection and directed that the pre-existing consumptive uses were to occur only to the extent proven to be compatible with protection of the preserve's resources. Using that authority, he made steady and significant changes. Through continuous attention to the issue of the 500 existing trespass camps and a court case which confirmed the camps were an adverse impact to preserve resources, Fagergren and his staff removed the remaining trespass camps. In addition, during his ten year tenure, he obtained the first major hunting regulation changes for the preserve, eliminated hunting dogs, and initiated major checks and balances such as quota permits. He and his staff also gave the ongoing oil and gas exploration and development increased attention. They gave the proposals they received complete and careful review, which resulted in more off-set drilling and completely self-contained operations to protect critical resources. When a major seismic survey was proposed over several hundred miles using off-road vehicles, his insistence on an environmental impact statement ultimately prevailed.
Fagergren recognized that the many challenges facing the preserve could not be resolved without an effective General Management Plan (GMP). Big Cypress would be managed with a priority on protection of the unique resources and consumptive uses would occur only as proven to be compatible with resource protection. The GMP was approved and although Fred moved on to his final superintendency at Bryce Canyon NP, his knowledge of the preserve, its legislation, and the GMP resulted in his continued regular involvement with Big Cypress issues. He repeatedly emphasized that Big Cypress was created to protect critical resources while allowing for certain uses to the extent they are compatible with resource protection.
Interestingly, Fagergren concluded his NPS career in 2002 at Bryce Canyon NP where his father began in 1934. He continued his emphasis on resource protection even in the face of ever increasing visitation and the demand for expanded development.


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

\begin{wrapfigure}{r}{40mm}
  \vspace{-2em}
  \begin{center}
    \includegraphics{BothConesVertical}
  \end{center}
  \vspace{-1em}
  \caption{(Top) The construction of the Yao graph. (Bottom) The construction of the $\theta$-graph.}
  \vspace{-1em}
  \label{fig:cones}
\end{wrapfigure}

A \emph{$t$-spanner} of a weighted graph $G$ is a spanning subgraph $H$ with the property that for all pairs of vertices, the weight of the shortest path between the vertices in $H$ is at most $t$ times the weight of the shortest path in $G$. The \emph{spanning ratio} of $H$ is the smallest $t$ for which it is a $t$-spanner. We say that a graph is a \emph{spanner} if it has a finite spanning ratio. The graph $G$ is referred to as the \emph{underlying graph}. In this paper, the underlying graph is the complete graph on a finite set of $n$ points in the plane and the weight of an edge is the Euclidean distance between its endpoints. A spanner of such a graph is called a \emph{geometric spanner}. For a comprehensive overview of geometric spanners, we refer the reader to the book by Narasimhan and Smid \cite{NS06}.

One simple way to build a geometric spanner is to first partition the plane around each vertex into a fixed number of cones and then add an edge between the vertex and the closest vertex in each cone (see Figure~\ref{fig:cones}, top). The resulting graph is called a Yao graph, and is typically denoted by $Y_k$, where $k$ is the number of cones around each vertex. If the number of cones is sufficiently large, we can find a path between any two vertices by starting at one and walking to the closest vertex in the cone that contains the other, then repeating this until we reach the destination. Intuitively, this results in a short path because we are always walking approximately in the right direction, and, since our neighbour is the closest vertex in that direction, never too far. However, when the number of cones is small, the path found in this way can be very long (see Section~\ref{sec:lbrouting}).

Yao graphs were introduced independently by Flinchbaugh and Jones~\cite{flinchbaugh1981strong} and Yao~\cite{yao1982constructing}, before the concept of spanners was even introduced by Chew in 1986~\cite{chew1986there}. To the best of our knowledge, the first proof that Yao graphs are geometric spanners was published in 1993, by Alth{\"o}fer~\emph{et al.}\xspace~\cite{althofer1993sparse}. In particular, they showed that for every $t > 1$, there exists a $k$ such that $Y_k$ is a $t$-spanner. It appears that some form of this result was known earlier, as Clarkson~\cite{clarkson1987approximation} already remarked in 1987 that $Y_{12}$ is a $1 + \sqrt{3}$-spanner, albeit without providing a proof or reference. In 2004, Bose~\emph{et al.}\xspace~\cite{bose2004approximating} provided a more specific bound on the spanning ratio, by showing that for $k > 8$, $Y_k$ is a geometric spanner with spanning ratio at most $1 / (\cos \theta - \sin \theta)$, where $\theta = 2\pi/k$. This bound was later improved to $1 / (1 - 2 \sin(\theta / 2))$, for $k > 6$~\cite{bose2012piArxiv}.

If we modify the definition of Yao graphs slightly, by connecting not to the closest point in each cone, but to the point whose projection on the bisector of that cone is closest (see Figure~\ref{fig:cones}, bottom), we obtain another type of geometric spanner, called a $\theta$-graph. These graphs were introduced independently by Clarkson~\cite{clarkson1987approximation} and Keil~\cite{keil1988approximating,keil1992classes}, who preferred them to Yao graphs because they are easier to compute. It turns out that $\theta$-graphs share most of the properties of Yao graphs: there is a constant $k$ for every $t > 1$ such that $\theta_k$ is a $t$-spanner, and the spanning ratio for $k > 6$ is $1 / (1 - 2 \sin(\theta / 2))$~\cite{ruppert1991approximating}. Very recently, the bounds on $\theta$-graphs were even pushed beyond those on Yao graphs~\cite{bose2013spanning}, including a matching upper- and lower bound of $1 + 2 \sin(\theta / 2)$ for all $\theta_k$-graphs with $k \geq 6$ and $k \equiv 2 \pmod 4$~\cite{bose2012optimal}.

Although most early research focused on Yao and $\theta$-graphs with a large number of cones, using the smallest possible number of cones is important for many practical applications, where the cost of a network is mostly determined by the number of edges. One such example is point-to-point wireless networks. These networks use narrow directional wireless transceivers that can transmit over long distances (up to 50km \cite{dist1,dist2}). The cost of an edge in such a network is therefore equal to the cost of the two transceivers that are used at each endpoint of that edge. In such networks, the cost of building a $\theta_6$-graph is approximately 29\% higher than the cost of building a \ensuremath{\theta_5}-graph\xspace if the transceivers are randomly distributed~\cite{morin2013average}. This leads to the natural question: for which values of $k$ are $Y_k$ and $\theta_k$ spanners? Kanj~\cite{kanj2013geometric} presented this question as one of the main open problems in the area of geometric spanners.

\begin{wrapfigure}{r}{40mm}
  \vspace{-2em}
  \begin{center}
    \includegraphics{ShortestPathCrosses}
  \end{center}
  \vspace{-1em}
  \caption{A $\theta_5$-graph where the shortest path between two vertices (in bold) crosses itself.}
  \vspace{-1em}
  \label{fig:cross}
\end{wrapfigure}

Surprisingly, this question was not studied until quite recently. In 2009, El~Molla~\cite{el2009yao} showed that, for $k < 4$, there is no constant $t$ such that $Y_k$ is a $t$-spanner. These proofs translate to $\theta$-graphs as well. Bonichon~\emph{et al.}\xspace~\cite{bonichon2010connections} showed that $\theta_6$ is a 2-spanner, and this is tight. This result was later used by Damian and Raudonis~\cite{damian2012yao} to show that $Y_6$ is a spanner as well. Bose~\emph{et al.}\xspace~\cite{bose2012pi} showed that $Y_4$ is a spanner, and a flurry of recent activity has led to the same result for both $\theta_4$~\cite{barba2013stretch} and $Y_5$~\cite{barba2014new}.

In this paper we present the final piece of this puzzle, by giving the first constant upper bound on the spanning ratio of the \ensuremath{\theta_5}-graph\xspace, thereby proving that it is a geometric spanner. Since the proof is constructive, it gives us a path between any two vertices, $u$ and $w$, of length at most $\ensuremath{\sqrt{50 + 22 \sqrt{5}} \approx 9.960}\xspace \cdot |uw|$. Surprisingly, this path can cross itself, a property we observed for the shortest path as well (see Figure~\ref{fig:cross}).

After completion of this research, we discovered that some form of this result appears to have been known already in 1991, as Ruppert and Seidel~\cite{ruppert1991approximating} mention that they could prove a bound near 10 on the spanning ratio of $\theta_5$. However, their paper does not include a proof and, to the best of our knowledge, they have not published one since.

In addition to the upper bound on the spanning ratio, we prove two lower bounds. We give an example of a point set where the \ensuremath{\theta_5}-graph\xspace has a spanning ratio of \ensuremath{\frac{1}{2}(11\sqrt{5} -17) \approx 3.798}\xspace, and we show that the traditional $\theta$-routing algorithm (follow the edge to the closest vertex in the cone that contains the destination) can result in very long paths, even though a short path exists.

\section{Connectivity}
\label{sec:Connectivity}

In this section we first give a more precise definition of the \ensuremath{\theta_5}-graph\xspace, before proving that it is connected.

Given a set $P$ of points in the plane, we consider each point $u \in P$ and partition the plane into 5 cones (regions in the plane between two rays originating from the same point) with apex $u$, each defined by two rays at consecutive multiples of $\theta=2\pi/5$ radians from the negative $y$-axis. We label the cones $C_0$ through $C_4$, in clockwise order around $u$, starting from the top (see~Figure~\ref{fig:theta5cones}a). If the apex is not clear from the context, we use $C_i^u$ to denote cone $C_i$ with apex $u$. For the sake of brevity, we typically write ``a cone \emph{of} $u$'' instead of ``a cone with apex $u$''.

\begin{figure}[ht]
  \centering
  \includegraphics{Cones}
  \caption{(a) The cones of a vertex $u$. (b) The vertex $v$ is closest to $u$. The shaded region is the canonical triangle $\T{u}{v}$.}
  \label{fig:theta5cones}
\end{figure}

The \ensuremath{\theta_5}-graph\xspace is then built by considering each vertex $u$ and connecting it with an edge to the `closest' vertex in each of its cones, where distance is measured by projecting each vertex onto the bisector of that cone (see Figure~\ref{fig:theta5cones}b). We use this definition of \emph{closest} in the remainder of the paper.

For simplicity, we assume that no two points lie on a line parallel or perpendicular to a cone boundary. This guarantees that each vertex connects to at most one vertex in each cone, and thus that the graph has at most $5n$ edges. For any set of points that does not satisfy this assumption, there exists a tiny angle such that the assumption holds if we rotate all cones by this angle. In terms of the graph, this rotation is equivalent to a tie-breaking rule that always selects the candidate that comes last in clockwise order. Thus, our conclusions about the spanning ratio hold in either case, even though our proofs rely on the general position assumption.

Given two vertices $u$ and $v$, we define their \emph{canonical triangle} $\T{u}{v}$ to be the triangle bounded by the cone of $u$ that contains $v$ and the line through $v$ perpendicular to the bisector of that cone. For example, the shaded region in Figure~\ref{fig:theta5cones}b is the canonical triangle $\T{u}{v}$. Note that for any pair of vertices $u$ and $v$, there are two canonical triangles: $\T{u}{v}$ and $\T{v}{u}$. We equate the size $|\T{u}{v}|$ of a canonical triangle to the length of one of the sides incident to the apex $u$. This gives us the useful property that any line segment between $u$ and a point inside the triangle has length at most $|\T{u}{v}|$.

To introduce the structure of the main proof, we first show that the \ensuremath{\theta_5}-graph\xspace is connected.

\begin{theorem}
 \label{thm:connected}
 The \ensuremath{\theta_5}-graph\xspace is connected.
\end{theorem}
\begin{myproof}
 We prove that there is a path between any (ordered) pair of vertices in the \ensuremath{\theta_5}-graph\xspace, using induction on the size of their canonical triangle. Formally, given two vertices $u$ and $w$, we perform induction on the rank (relative position) of $\T{u}{w}$ among the canonical triangles of all pairs of vertices, when ordered by size. For ease of description, we assume that $w$ lies in the right half of $C_0^u$. The other cases are analogous.

 If $\T{u}{w}$ has rank 1, it is the smallest canonical triangle. Therefore there can be no point closer to $u$ in $C_0^u$, so the \ensuremath{\theta_5}-graph\xspace must contain the edge $(u, w)$. This proves the base case.

 If $\T{u}{w}$ has a larger rank, our inductive hypothesis is that there exists a path between any pair of vertices with a smaller canonical triangle. Let $a$ and $b$ be the left and right corners of $\T{u}{w}$. Let $m$ be the midpoint of $ab$ and let $x$ be the intersection of $ab$ and the bisector of $\angle mub$ (see~Figure~\ref{fig:canon}a).

 \begin{figure}[ht]
  \centering
  \includegraphics{Canonical.pdf}
  \caption{(a) The canonical triangle $\T{u}{w}$. (b) If $w$ lies between $m$ and $x$, $\T{w}{u}$ is smaller than $\T{u}{w}$.}
  \label{fig:canon}
 \end{figure}

 If $w$ lies to the left of $x$, consider the canonical triangle $\T{w}{u}$. Let $m'$ be the midpoint of the side of $\T{w}{u}$ opposite $w$ and let $\alpha = \angle muw$ (see~Figure~\ref{fig:canon}b). Note that $\angle uwm' = \frac{\pi}{5} - \alpha$, since $um$ and the vertical border of $\T{w}{u}$ are parallel and both are intersected by $uw$. Using basic trigonometry, we can express the size of $\T{w}{u}$ as follows.
 \[ |\T{w}{u}|
~~=~~\frac{|wm'|}{\cos \frac{\pi}{5}}
~~=~~\frac{\cos \angle uwm' \cdot |uw|}{\cos \frac{\pi}{5}}
~~=~~\frac{\cos \left( \frac{\pi}{5} - \alpha \right) \cdot \frac{|um|}{\cos \alpha}}{\cos \frac{\pi}{5}}
~~=~~\frac{\cos \left( \frac{\pi}{5} - \alpha \right)}{\cos \alpha} \cdot |\T{u}{w}|
 \]
 Since $w$ lies to the left of $x$, the angle $\alpha$ is less than $\pi/10$, which means that $\cos ( \frac{\pi}{5} - \alpha ) / \cos \alpha$ is less than 1. Hence $\T{w}{u}$ is smaller than $\T{u}{w}$ and by induction, there is a path between $w$ and $u$. Since the \ensuremath{\theta_5}-graph\xspace is undirected, we are done in this case. The rest of the proof deals with the case where $w$ lies on or to the right of $x$.

 If $\T{w}{u}$ is empty, there is an edge between $u$ and $w$ and we are done, so assume that this is not the case. Then there is a vertex $v_w$ that is closest to $w$ in $C_3^w$ (the cone of $w$ that contains $u$). This gives rise to four cases, depending on the location of $v_w$ (see Figure~\ref{fig:cases}a). In each case, we will show that $\T{u}{v_w}$ is smaller than $\T{u}{w}$ and hence we can apply induction to obtain a path between $u$ and $v_w$. Since $v_w$ is the closest vertex to $w$ in $C_3$, there is an edge between $v_w$ and $w$, completing the path between $u$ and $w$.

 \begin{figure}[ht]
  \centering
  \includegraphics{Cases.pdf}
  \caption{(a) The four cases for $v_w$. (b) Case 1: The situation that maximizes $|\T{u}{v_w}|$ when $v_w$ lies in $C_2^u$. (c) Case 4: The situation that maximizes $|\T{u}{v_w}|$ when $v_w$ lies in $C_4^u$.}
  \label{fig:cases}
 \end{figure}

 \paragraph{Case 1.} $v_w$ lies in $C_2^u$. In this case, the size of $\T{u}{v_w}$ is maximized when $v_w$ lies in the bottom right corner of $\T{w}{u}$ and $w$ lies on $b$. Let $y$ be the rightmost corner of $\T{u}{v_w}$ (see~Figure~\ref{fig:cases}b). Using the law of sines, we can express the size of $\T{u}{v_w}$ as follows.
 \[
  |\T{u}{v_w}|
~~=~~|uy|
~~=~~\frac{\sin \angle uv_wy}{\sin \angle uyv_w} \cdot |uv_w|
~~=~~\frac{\sin \frac{3\pi}{5}}{\sin \frac{3\pi}{10}} \cdot \tan \frac{\pi}{5} \cdot |\T{u}{w}|
~~<~~|\T{u}{w}|
 \]
 \paragraph{Case 2.} $v_w$ lies in $C_1^u$. In this case, the size of $\T{u}{v_w}$ is maximized when $w$ lies on $b$ and $v_w$ lies almost on $w$. By symmetry, this gives $|\T{u}{v_w}| = |\T{u}{w}|$. However, $v_w$ cannot lie precisely on $w$ and must therefore lie a little closer to $u$, giving us that $|\T{u}{v_w}| < |\T{u}{w}|$.
 \paragraph{Case 3.} $v_w$ lies in $C_0^u$. As in the previous case, the size of $\T{u}{v_w}$ is maximized when $v_w$ lies almost on $w$, but since $v_w$ must lie closer to $u$, we have that $|\T{u}{v_w}| < |\T{u}{w}|$.
 \paragraph{Case 4.} $v_w$ lies in $C_4^u$. In this case, the size of $\T{u}{v_w}$ is maximized when $v_w$ lies in the left corner of $\T{w}{u}$ and $w$ lies on $x$. Let $y$ be the bottom corner of $\T{u}{v_w}$ (see~Figure~\ref{fig:cases}c). Since $x$ is the point where $|\T{u}{w}| = |\T{w}{u}|$, and $v_wyuw$ forms a parallelogram, $|\T{u}{v_w}| = |\T{u}{w}|$. However, by general position, $v_w$ cannot lie on the boundary of $\T{w}{u}$, so it must lie a little closer to $u$, giving us that $|\T{u}{v_w}| < |\T{u}{w}|$.

 \bigskip
 Since any vertex in $C_3^u$ would be further from $w$ than $u$ itself, these four cases are exhaustive.
\end{myproof}

\section{Spanning ratio}
\label{sec:SpanningRatio}

In this section, we prove an upper bound on the spanning ratio of the \ensuremath{\theta_5}-graph\xspace.

\begin{lemma}
 \label{lem:spanningPath}
 Between any pair of vertices $u$ and $w$ of a \ensuremath{\theta_5}-graph\xspace, there is a path of length at most $c \cdot |\T{u}{w}|$, where $c = \ensuremath{2 \left( 2 + \sqrt{5} \right) \approx 8.472}\xspace$.
\end{lemma}
\begin{myproof}
 We begin in a way similar to the proof of Theorem~\ref{thm:connected}. Given an ordered pair of vertices $u$ and $w$, we perform induction on the size of their canonical triangle. If $|\T{u}{w}|$ is minimal, there must be a direct edge between them. Since $c > 1$ and any edge inside $\T{u}{w}$ with endpoint $u$ has length at most $|\T{u}{w}|$, this proves the base case. The rest of the proof deals with the inductive step, where we assume that there exists a path of length at most $c \cdot |T|$ between every pair of vertices whose canonical triangle $T$ is smaller than $\T{u}{w}$. As in the proof of Theorem~\ref{thm:connected}, we assume that $w$ lies in the right half of $C_0^u$. If $w$ lies to the left of $x$, we have seen that $\T{w}{u}$ is smaller than $\T{u}{w}$. Therefore we can apply induction to obtain a path of length at most $c \cdot |\T{w}{u}| < c \cdot |\T{u}{w}|$ between $u$ and $w$. Hence we need to concern ourselves only with the case where $w$ lies on or to the right of $x$.

 If $u$ is the vertex closest to $w$ in $C_3^w$ or $w$ is the closest vertex to $u$ in $C_0^u$, there is a direct edge between them and we are done by the same reasoning as in the base case. Therefore assume that this is not the case and let $v_w$ be the vertex closest to $w$ in $C_3^w$. We distinguish the same four cases for the location of $v_w$ (see~Figure~\ref{fig:cases}a). We already showed that we can apply induction on $\T{u}{v_w}$ in each case. This is a crucial part of the proof for the first three cases.

 Most of the cases come down to finding a path between $u$ and $w$ of length at most $(g + h \cdot c) \cdot |\T{u}{w}|$, for constants $g$ and $h$ with $h < 1$. The smallest value of $c$ for which this is bounded by $c \cdot |\T{u}{w}|$ is $g/(1-h)$. If this is at most \ensuremath{2 \left( 2 + \sqrt{5} \right) \approx 8.472}\xspace, we are done.

 \paragraph{Case 1.} $v_w$ lies in $C_2^u$. By induction, there exists a path between $u$ and $v_w$ of length at most $c \cdot |\T{u}{v_w}|$. Since $v_w$ is the closest vertex to $w$ in $C_3^w$, there is a direct edge between them, giving a path between $u$ and $w$ of length at most $|wv_w| + c \cdot |\T{u}{v_w}|$.

 Given any initial position of $v_w$ in $C_2^u$, we can increase $|wv_w|$ by moving $w$ to the right. Since this does not change $|\T{u}{v_w}|$, the worst case occurs when $w$ lies on $b$. Then we can increase both $|wv_w|$ and $|\T{u}{v_w}|$ by moving $v_w$ into the bottom corner of $\T{w}{u}$. This gives rise to the same worst-case configuration as in the proof of Theorem~\ref{thm:connected}, depicted in Figure~\ref{fig:cases}b. Building on the analysis there, we can bound the worst-case length of the path as follows.
 \[
     |wv_w| + c \cdot |\T{u}{v_w}|
~~=~~\frac{|\T{u}{w}|}{\cos \frac{\pi}{5}} + c \cdot \frac{\sin \frac{3\pi}{5}}{\sin \frac{3\pi}{10}} \cdot \tan \frac{\pi}{5} \cdot |\T{u}{w}|
 \]
 This is at most $c \cdot |\T{u}{w}|$ for $c \geq 2 \left( 2 + \sqrt{5} \right)$. Since we picked $c = 2 \left( 2 + \sqrt{5} \right)$,  the theorem holds in this case. Note that this is one of the cases that determines the value of $c$.

 \begin{figure}[ht]
  \centering
  \includegraphics{WorstCases.pdf}
  \caption{(a) Case 2: Vertex $v_w$ lies on the boundary of $C_3^w$ after moving it down along the side of $\T{u}{v_w}$. (b) Case 3: Vertex $v_w$ lies on the boundary of $C_0^u$ after moving it left along the side of $\T{u}{v_w}$. (c) Case 4: Vertex $v_w$ lies in $C_4^u \cap C_3^b$.}
  \label{fig:worst}
 \end{figure}

 \paragraph{Case 2.} $v_w$ lies in $C_1^u$. By the same reasoning as in the previous case, we have a path of length at most $|wv_w| + c \cdot |\T{u}{v_w}|$ between $u$ and $w$ and we need to bound this length by $c \cdot |\T{u}{w}|$.

 Given any initial position of $v_w$ in $C_1^u$, we can increase $|wv_w|$ by moving $w$ to the right. Since this does not change $|\T{u}{v_w}|$, the worst case occurs when $w$ lies on $b$. We can further increase $|wv_w|$ by moving $v_w$ down along the side of $\T{u}{v_w}$ opposite $u$ until it hits the boundary of $C_1^u$ or $C_3^w$, whichever comes first (see~Figure~\ref{fig:worst}a).

 Now consider what happens when we move $v_w$ along these boundaries. If $v_w$ lies on the boundary of $C_1^u$ and we move it away from $u$ by $\Delta$, $|\T{u}{v_w}|$ increases by $\Delta$. At the same time, $|wv_w|$ might decrease, but not by more than $\Delta$. Since $c > 1$, the total path length is maximized by moving $v_w$ as far from $u$ as possible, until it hits the boundary of $C_3^w$. Once $v_w$ lies on the boundary of $C_3^w$, we can express the size of $\T{u}{v_w}$ as follows, where $a'$ is the top corner of $\T{u}{v_w}$.
 \[
  |\T{u}{v_w}|
  ~~=~~ |\T{u}{w}| - |wa'|
  ~~=~~ |\T{u}{w}| - |wv_w| \cdot \frac{\sin \angle wv_wa'}{\sin \angle wa'v_w}
  ~~=~~ |\T{u}{w}| - |wv_w| \cdot \frac{\sin \frac{\pi}{10}}{\sin \frac{7\pi}{10}}
 \]
Now we can express the length of the complete path as follows.
 \[
  |wv_w| + c \cdot |\T{u}{v_w}|
  ~~=~~ |wv_w| + c \cdot \left( |\T{u}{w}| - |wv_w| \cdot \frac{\sin \frac{\pi}{10}}{\sin \frac{7\pi}{10}} \right)
  ~~=~~ c \cdot |\T{u}{w}| - \left( c \cdot \frac{\sin \frac{\pi}{10}}{\sin \frac{7\pi}{10}} - 1 \right) \cdot |wv_w|
 \]
Since $c > \sin \frac{7\pi}{10} / \sin \frac{\pi}{10} \approx 2.618$, we have that $c \cdot (\sin \frac{\pi}{10} / \sin \frac{7\pi}{10}) - 1 > 0$. Therefore $|wv_w| + c \cdot |\T{u}{v_w}| < c \cdot |\T{u}{w}|$.

 \paragraph{Case 3.} $v_w$ lies in $C_0^u$. Again, we have a path of length at most $|wv_w| + c \cdot |\T{u}{v_w}|$ between $u$ and $w$ and we need to bound this length by $c \cdot |\T{u}{w}|$.

 Given any initial position of $v_w$ in $C_0^u$, moving $v_w$ to the left increases $|wv_w|$ while leaving $|\T{u}{v_w}|$ unchanged. Therefore the path length is maximized when $v_w$ lies on the boundary of either $C_0^u$ or $C_3^w$, whichever it hits first (see~Figure~\ref{fig:worst}b).

 Again, consider what happens when we move $v_w$ along these boundaries. Similar to the previous case, if $v_w$ lies on the boundary of $C_0^u$ and we move it away from $u$ by $\Delta$, $|\T{u}{v_w}|$ increases by $\Delta$, while $|wv_w|$ might decrease by at most $\Delta$. Since $c > 1$, the total path length is maximized by moving $v_w$ as far from $u$ as possible, until it hits the boundary of $C_3^w$. Once there, the situation is symmetric to the previous case, with $|\T{u}{v_w}| = |\T{u}{w}| - |wv_w| \cdot (\sin \frac{\pi}{10} / \sin \frac{7\pi}{10})$. Therefore the theorem holds in this case as well.

 \paragraph{Case 4.} $v_w$ lies in $C_4^u$. This is the hardest case. Similar to the previous two cases, the size of $\T{u}{v_w}$ can be arbitrarily close to that of $\T{u}{w}$, but in this case $|wv_w|$ does not approach $0$. This means that simply invoking the inductive hypothesis on $\T{u}{v_w}$ does not work, so another strategy is required. We first look at a subcase where we \emph{can} apply induction directly, before considering the position of $v_u$, the closest vertex to $u$ in $C_0$.

 \paragraph{Case 4a.} $v_w$ lies in $C_4^u \cap C_3^b$. This situation is illustrated in Figure~\ref{fig:worst}c. Given any initial position of $v_w$, moving $w$ to the right onto $b$ increases the total path length by increasing $|wv_w|$ while not affecting $|\T{u}{v_w}|$. Here we use the fact that $v_w$ already lies in $C_3^b$, otherwise we would not be able to move $w$ onto $b$ while keeping $v_w$ in $C_3^w$. Now the total path length is maximized by placing $v_w$ on the left corner of $\T{w}{u}$. Since this situation is symmetrical to the worst-case situation in Case 1, the theorem holds by the same analysis.

 \bigskip
 Next, we distinguish four cases for the position of $v_u$ (the closest vertex to $u$ in $C_0$), illustrated in Figure~\ref{fig:closestcases}a. The cases are: (4b) $w$ lies in $C_4^{v_u}$, (4c) $w$ lies in $C_0^{v_u}$, (4d) $w$ lies in $C_1^{v_u}$ and $v_u$ lies in $C_3^w$, and (4e) $w$ lies in $C_1^{v_u}$ and $v_u$ lies in $C_4^w$. These are exhaustive, since the cones $C_4$, $C_0$ and $C_1$ are the only ones that can contain a vertex above the current vertex, and $w$ must lie above $v_u$, as $v_u$ is closer to $u$. Further, if $w$ lies in $C_1^{v_u}$, $v_u$ must lie in one of the two opposite cones of $w$.

 We can solve the first two cases by applying our inductive hypothesis to $\T{v_u}{w}$.

 \begin{figure}[ht]
  \centering
  \includegraphics{ClosestCases.pdf}
  \caption{(a) The four different cases for the position of $v_u$. (b) The worst-case configuration with $w$ in $C_4^{v_u}$. (c) A configuration with $w$ in $C_0^{v_u}$, after moving $v_u$ onto the right side of $C_0^u$.}
  \label{fig:closestcases}
 \end{figure}

 \paragraph{Case 4b.} $w$ lies in $C_4^{v_u}$. To apply our inductive hypothesis, we need to show that $|\T{v_u}{w}| < |\T{u}{w}|$. If that is the case, we obtain a path between $v_u$ and $w$ of length at most $c \cdot |\T{v_u}{w}|$. Since $v_u$ is the closest vertex to $u$, there is a direct edge from $u$ to $v_u$, resulting in a path between $u$ and $w$ of length at most $|uv_u| + c \cdot |\T{v_u}{w}|$.

 Given any intial positions for $v_u$ and $w$, moving $w$ to the left increases $|\T{v_u}{w}|$ while leaving $|uv_u|$ unchanged. Moving $v_u$ closer to $b$ increases both. Therefore the path length is maximal when $w$ lies on $x$ and $v_u$ lies on $b$ (see~Figure~\ref{fig:closestcases}b). Using the law of sines, we can express $|\T{v_u}{w}|$ as follows.
 \[
  |\T{v_u}{w}|
  ~~=~~ \frac{\sin \frac{3\pi}{5}}{\sin \frac{3\pi}{10}} \cdot |wv_u|
  ~~=~~ \frac{\sin \frac{3\pi}{5}}{\sin \frac{3\pi}{10}} \cdot \frac{\sin \frac{\pi}{10}}{\sin \frac{3\pi}{5}} \cdot |\T{u}{w}|
  ~~=~~ \frac{\sin \frac{\pi}{10}}{\sin \frac{3\pi}{10}} \cdot |\T{u}{w}|
  ~~=~~ \frac{1}{2} \left( 3 - \sqrt{5} \right) \cdot |\T{u}{w}|
 \]
 Since $\frac{1}{2} \left( 3 - \sqrt{5} \right) < 1$, we have that $|\T{v_u}{w}| < |\T{u}{w}|$ and we can apply our inductive hypothesis to $\T{v_u}{w}$. Since $|uv_u| = |\T{u}{w}|$, the complete path has length at most $c \cdot |\T{u}{w}|$ for
 \[
  c
  ~~\geq~~ \frac{1}{1 - \frac{1}{2} \left( 3 - \sqrt{5} \right)}
  ~~=~~ \frac{1}{2} \left( 1 + \sqrt{5} \right)
  ~~\approx~~ 1.618.
 \]

 \paragraph{Case 4c.} $w$ lies in $C_0^{v_u}$. Since $v_u$ lies in $C_0^u$, it is clear that $|\T{v_u}{w}| < |\T{u}{w}|$, which allows us to apply our inductive hypothesis. This gives us a path between $u$ and $w$ of length at most $|uv_u| + c \cdot |\T{v_u}{w}|$. For any initial location of $v_u$, we can increase the total path length by moving $v_u$ to the right until it hits the side of $C_0^u$ (see~Figure~\ref{fig:closestcases}c), since $|\T{v_u}{w}|$ stays the same and $|uv_u|$ increases. Once there, we have that $|uv_u| + |\T{v_u}{w}| = |\T{u}{w}|$. Since $c > 1$, this immediately implies that $|uv_u| + c \cdot |\T{v_u}{w}| \leq c \cdot |\T{u}{w}|$, proving the theorem for this case.

 \bigskip
 To solve the last two cases, we need to consider the positions of both $v_u$ and $v_w$. Recall that for $v_w$, there is only a small region left where we have not yet proved the existance of a short path between $u$ and $w$. In particular, this is the case when $v_w$ lies in cone $C_4^u$, but not in $C_3^b$.

 \begin{figure}[ht]
  \centering
  \includegraphics{ComplexCases.pdf}
  \caption{(a) The regions where $v_u$ (light) and $v_w$ (dark) can lie. (b) The worst case when $v_u$ lies on a given line $\ell$. (c) The worst case for a fixed position of $w$.}
  \label{fig:complex}
 \end{figure}

 \paragraph{Case 4d.} $w$ lies in $C_1^{v_u}$ and $v_u$ lies in $C_3^w$. We would like to apply our inductive hypothesis to $\T{v_u}{v_w}$, resulting in a path between $v_u$ and $v_w$ of length at most $c \cdot |\T{v_u}{v_w}|$. The edges $(w, v_w)$ and $(u, v_u)$ complete this to a path between $u$ and $w$, giving a total length of at most $|uv_u| + c \cdot |\T{v_u}{v_w}| + |v_ww|$.

 First, note that $v_u$ cannot lie in $\T{w}{v_w}$, as this region is empty by definition. Since $v_w$ lies in $C_4^u$, this means that $v_w$ must lie in $C_4^{v_u}$. We first show that $\T{v_u}{v_w}$ is always smaller than $\T{u}{w}$, which means that we are allowed to use induction. Given any initial position for $v_u$, consider the line $\ell$ through $v_u$, perpendicular to the bisector of $C_3$ (see~Figure~\ref{fig:complex}a). Since $v_w$ cannot be further from $w$ than $v_u$, the size of $\T{v_u}{v_w}$ is maximized when $v_w$ lies on the intersection of $\ell$ and the top boundary of $\T{w}{u}$. We can increase $|\T{v_u}{v_w}|$ further by moving $v_u$ along $\ell$ until it reaches the bisector of $C_3^w$ (see~Figure~\ref{fig:complex}b). Since the top boundary of $\T{w}{u}$ and the bisector of $C_3^w$ approach each other as they get closer to $w$, the size of $\T{v_u}{v_w}$ is maximized when $v_u$ lies on the bottom boundary of $\T{w}{u}$ (ignoring for now that this would move $v_u$ out of $\T{u}{w}$). Now it is clear that $|\T{v_u}{v_w}| < |\T{u}{v_w}|$. Since we already established that $\T{u}{v_w}$ is smaller than $\T{u}{w}$ in the proof of Theorem~\ref{thm:connected}, this holds for $\T{v_u}{v_w}$ as well and we can use induction.

 All that is left is to bound the total length of the path. Given any initial position of $v_u$, the path length is maximized when we place $v_w$ at the intersection of $\ell$ and the top boundary of $\T{w}{u}$, as this maximizes both $|\T{v_u}{v_w}|$ and $|wv_w|$. When we move $v_u$ away from $v_w$ along $\ell$ by $\Delta$, $|uv_u|$ decreases by at most $\Delta$, while $|\T{v_u}{v_w}|$ increases by $\sin \frac{3\pi}{5} / \sin \frac{3\pi}{10} \cdot \Delta > \Delta$. Since $c > 1$, this increases the total path length. Therefore the worst case again occurs when $v_u$ lies on the bisector of $C_3^w$, as depicted in Figure~\ref{fig:complex}b. Moving $v_u$ down along the bisector of $\T{w}{u}$ by $\Delta$ decreases $|uv_u|$ by at most $\Delta$, while increasing $|wv_w|$ by $1 / \sin \frac{3\pi}{10} \cdot \Delta > \Delta$ and increasing $|\T{v_u}{v_w}|$. Therefore this increases the total path length and the worst case occurs when $v_u$ lies on the left boundary of $\T{u}{w}$ (see~Figure~\ref{fig:complex}c).

 Finally, consider what happens when we move $v_u$ $\Delta$ towards $u$, while moving $w$ and $v_w$ such that the construction stays intact. This causes $w$ to move to the right. Since $v_u$, $w$ and the left corner of $\T{u}{w}$ form an isosceles triangle with apex $v_u$, this also moves $v_u$ $\Delta$ further from $w$. We saw before that moving $v_u$ away from $w$ increases the size of $\T{v_u}{v_w}$. Finally, it also increases $|wv_w|$ by $1 / \sin \frac{3\pi}{10} \cdot \Delta > \Delta$. Thus, the increase in $|wv_w|$ cancels the decrease in $|uv_u|$ and the total path length increases. Therefore the worst case occurs when $v_u$ lies almost on $u$ and $v_w$ lies in the corner of $\T{w}{u}$, which is symmetric to the worst case of Case~1. Thus the theorem holds by the same analysis.

 \paragraph{Case 4e.} $w$ lies in $C_1^{v_u}$ and $v_u$ lies in $C_4^w$. We split this case into three final subcases, based on the position of $v_u$. These cases are illustrated in Figure~\ref{fig:complex2}a. Note that $v_u$ cannot lie in $C_2$ or $C_3$ of $v_w$, as it lies above $v_w$. It also cannot lie in $C_4^{v_w}$, as $C_4^{v_w}$ is completely contained in $C_4^u$, whereas $v_u$ lies in $C_0^u$. Thus the cases presented below are exhaustive.

 \paragraph{Case 4e-1.} $|\T{w}{v_u}| \leq \frac{c - 1}{c} \cdot |\T{u}{w}|$. If $\T{w}{v_u}$ is small enough, we can apply our inductive hypothesis to obtain a path between $v_u$ and $w$ of length at most $c \cdot |\T{w}{v_u}|$. Since there is a direct edge between $u$ and $v_u$, we obtain a path between $u$ and $w$ of length at most $|uv_u| + c \cdot |\T{w}{v_u}|$. Any edge from $u$ to a point inside $\T{u}{w}$ has length at most $|\T{u}{w}|$, so we can bound the length of the path as follows.
 \[
  |uv_u| + c \cdot |\T{w}{v_u}|
  ~~\leq~~ |\T{u}{w}| + c \cdot \frac{c - 1}{c} \cdot |\T{u}{w}|
  ~~=~~ |\T{u}{w}| + (c - 1) \cdot |\T{u}{w}|
  ~~=~~ c \cdot |\T{u}{w}|
 \]

 \bigskip
 In the other two cases, we use induction on $\T{v_w}{v_u}$ to obtain a path between $v_w$ and $v_u$ of length at most $c \cdot |\T{v_w}{v_u}|$. The edges $(u, v_u)$ and $(w, v_w)$ complete this to a (self-intersecting) path between $u$ and $w$. We can bound the length of these edges by the size of the canonical triangle that contains them, as follows.
 \[
  |uv_u| + |wv_w|
  ~~\leq~~ |\T{u}{w}| + |\T{w}{u}|
  ~~\leq~~ |\T{u}{w}| + \frac{1}{\cos \frac{\pi}{5}} \cdot |\T{u}{w}|
  ~~=~~ \sqrt{5} \cdot |\T{u}{w}|
 \]
 All that is left now is to bound the size of $\T{v_w}{v_u}$ and express it in terms of $\T{u}{w}$.

 \begin{figure}[ht]
  \centering
  \includegraphics{ComplexCases2.pdf}
  \caption{(a) The three subcases for the position of $v_u$. (b) The situation that maximizes $\T{v_w}{v_u}$ when $v_u$ lies in $C_0^{v_w}$. (c) The worst case when $v_u$ lies in $C_1^{v_w}$.}
  \label{fig:complex2}
 \end{figure}

 \paragraph{Case 4e-2.} $v_u$ lies in $C_0^{v_w}$. In this case, the size of $\T{v_w}{v_u}$ is maximal when $v_u$ lies on the top boundary of $\T{u}{w}$ and $v_w$ lies at the lowest point in its possible region: the left corner of $\T{b}{u}$ (see~Figure~\ref{fig:complex2}b). Now we can express $|\T{v_w}{v_u}|$ as follows.
 \[
  |\T{v_w}{v_u}|
  ~~=~~ \frac{\sin \frac{\pi}{10}}{\sin \frac{7\pi}{10}} \cdot |bv_w|
  ~~=~~ \frac{\sin \frac{\pi}{10}}{\sin \frac{7\pi}{10}} \cdot \frac{1}{\cos \frac{\pi}{5}} \cdot |\T{u}{w}|
  ~~=~~ 2 \left( \sqrt{5} - 2 \right) \cdot |\T{u}{w}|
 \]
 Since $2 \left( \sqrt{5} - 2 \right) < 1$, we can use induction. The total path length is bounded by $c \cdot |\T{u}{w}|$ for
 \[
  c
  ~~\geq~~ \frac{\sqrt{5}}{1 - 2 \left( \sqrt{5} - 2 \right)}
  ~~=~~ 2 + \sqrt{5}
  ~~\approx~~ 4.236.
 \]

 \paragraph{Case 4e-3.} $v_u$ lies in $C_1^{v_w}$. Since $|\T{w}{v_u}| > \frac{c - 1}{c} \cdot |\T{u}{w}|$, $\T{v_w}{v_u}$ is maximal when $v_w$ lies on the left corner of $\T{w}{u}$ and $v_u$ lies on the top boundary of $\T{u}{w}$, such that $|\T{w}{v_u}| = \frac{c - 1}{c} \cdot |\T{u}{w}|$ (see~Figure~\ref{fig:complex2}c). Let $y$ be the intersection of $\T{v_w}{v_u}$ and $\T{w}{u}$. Note that since $v_w$ lies on the corner of $\T{w}{u}$, $y$ is also the midpoint of the side of $\T{v_w}{v_u}$ opposite $v_w$. We can express the size of $\T{v_w}{v_u}$ as follows.
 \[
  |\T{v_w}{v_u}|
  ~~=~~\frac{|v_wy|}{\cos \frac{\pi}{5}}
  ~~=~~\frac{|wv_w| - |wy|}{\cos \frac{\pi}{5}}
  ~~=~~\frac{\displaystyle\frac{|\T{u}{w}|}{\cos \frac{\pi}{5}} - \cos {\textstyle \frac{\pi}{10}} \cdot |wv_u|}{\cos \frac{\pi}{5}}
 \]
 \[
  =~~\frac{\displaystyle\frac{|\T{u}{w}|}{\cos \frac{\pi}{5}} - \cos {\textstyle \frac{\pi}{10}} \cdot \frac{\sin \frac{3\pi}{10}}{\sin \frac{3\pi}{5}} \cdot |\T{w}{v_u}|}{\cos \frac{\pi}{5}}
  ~~=~~\frac{\displaystyle\frac{|\T{u}{w}|}{\cos \frac{\pi}{5}} - \cos {\textstyle \frac{\pi}{10}} \cdot \frac{\sin \frac{3\pi}{10}}{\sin \frac{3\pi}{5}} \cdot \frac{c - 1}{c} \cdot |\T{u}{w}|}{\cos \frac{\pi}{5}}
 \]
 \[
  =~~\left( \frac{1}{c} + 5 - 2 \sqrt{5} \right) \cdot |\T{u}{w}|
 \]
 Thus we can use induction for $c > 1 / \left(2 \sqrt{5} - 4\right) \approx 2.118$ and the total path length can be bounded by $c \cdot |\T{u}{w}|$ for
 \[
  c
  ~~\geq~~\frac{\sqrt{5} + 1}{2 \sqrt{5} - 4}
  ~~=~~ \frac{1}{2}\left( 7 + 3 \sqrt{5} \right)
  ~~\approx~~ 6.854.
 \]
\end{myproof}

\noindent Using this result, we can compute the exact spanning ratio.

\begin{theorem}
 \label{thm:spanner}
 The \ensuremath{\theta_5}-graph\xspace is a spanner with spanning ratio at most \ensuremath{\sqrt{50 + 22 \sqrt{5}} \approx 9.960}\xspace.
\end{theorem}
\begin{myproof}
 Given two vertices $u$ and $w$, we know from Lemma~\ref{lem:spanningPath} that there is a path between them of length at most $c \cdot \min\left(|\T{u}{w}|, |\T{w}{u}|\right)$, where $c = \ensuremath{2 \left( 2 + \sqrt{5} \right) \approx 8.472}\xspace$. This gives an upper bound on the spanning ratio of $c \cdot \min\left(|\T{u}{w}|, |\T{w}{u}|\right) / |uw|$. We assume without loss of generality that $w$ lies in the right half of $C_0^u$. Let $\alpha$ be the angle between the bisector of $C_0^u$ and the line $uw$ (see~Figure~\ref{fig:canon}b). In the proof of Theorem~\ref{thm:connected}, we saw that we can express $|\T{w}{u}|$ and $|uw|$ in terms of $\alpha$ and $|\T{u}{w}|$, as $|\T{w}{u}| = (\cos (\frac{\pi}{5} - \alpha) / \cos \alpha) \cdot |\T{u}{w}|$ and $|uw| = (\cos \frac{\pi}{5} / \cos \alpha) \cdot |\T{u}{w}|$, respectively. Using these expressions, we can write the spanning ratio in terms of $\alpha$.
 \[
  \frac{c \cdot \min\left(|\T{u}{w}|, |\T{w}{u}|\right)}{|uw|}
  ~~=~~ \frac{c \cdot \min\left(|\T{u}{w}|, \frac{\cos \left( \frac{\pi}{5} - \alpha \right)}{\cos \alpha} \cdot |\T{u}{w}|\right)}{\frac{\cos \frac{\pi}{5}}{\cos \alpha} \cdot |\T{u}{w}|}
  ~~=~~ \frac{c}{\cos \frac{\pi}{5}} \cdot \min\left(\cos \alpha, \cos \left( {\textstyle \frac{\pi}{5}} - \alpha \right)\right)
 \]
 To get an upper bound on the spanning ratio, we need to maximize the minimum of $\cos \alpha$ and $\cos \left(\frac{\pi}{5} - \alpha\right)$. Since for $\alpha \in [0, \pi / 5]$, one is increasing and the other is decreasing, this maximum occurs at $\alpha = \pi/10$, where they are equal. Thus, our upper bound becomes
 \[
  \frac{c}{\cos \frac{\pi}{5}} \cdot \cos {\textstyle \frac{\pi}{10}}
  ~~=~~\sqrt{50 + 22 \sqrt{5}}
  ~~\approx~~9.960.
 \]
 \vspace{-2em}
\end{myproof}

\section{Lower bound}

\begin{wrapfigure}{r}{0.4\textwidth}
  \vspace{-2em}
  \begin{center}
    \includegraphics{LowerBoundPath}
  \end{center}
  \vspace{-1em}
  \caption{A path with a large spanning ratio.}
  \label{fig:lbpath}
\end{wrapfigure}

In this section, we derive a lower bound on the spanning ratio of the \ensuremath{\theta_5}-graph\xspace.

\begin{theorem}
 \label{thm:lowerbound}
 The \ensuremath{\theta_5}-graph\xspace has spanning ratio at least $\ensuremath{\frac{1}{2}(11\sqrt{5} -17) \approx 3.798}\xspace$.
\end{theorem}
\begin{myproof}
 For the lower bound, we present and analyze a path between two vertices that has a large spanning ratio. The path has the following structure (illustrated in Figure~\ref{fig:lbpath}).

 The path can be thought of as being directed from $w$ to $u$. First, we place $w$ in the right corner of $\T{u}{w}$. Then we add a vertex $v_1$ in the bottom corner of $\T{w}{u}$. We repeat this two more times, each time adding a new vertex in the corner of $\T{v_i}{u}$ furthest away from $u$. The final vertex $v_4$ is placed on the top boundary of $C_1^{v_3}$, such that $u$ lies in $C_1^{v_4}$. Since we know all the angles involved, we can compute the length of each edge, taking $|uw| = 1$ as baseline.
 \[
  |wv_1| = \frac{1}{\cos \frac{\pi}{5}}~~~~~|v_1v_2| = |v_2v_3| = 2 \sin {\textstyle \frac{\pi}{5}} \tan {\textstyle \frac{\pi}{5}}~~~~~|v_3v_4| = \frac{\sin \frac{\pi}{10}}{\sin \frac{3\pi}{5}} \tan {\textstyle \frac{\pi}{5}}~~~~~|v_4u| = \frac{\sin \frac{3\pi}{10}}{\sin \frac{3\pi}{5}} \tan {\textstyle \frac{\pi}{5}}
 \]
 Since we set $|uw| = 1$, the spanning ratio is simply $|wv_1| + |v_1v_2| + |v_2v_3| + |v_3v_4| + |v_4u| = \textstyle\frac{1}{2}(11\sqrt{5} -17) \approx 3.798$. Note that the \ensuremath{\theta_5}-graph\xspace with just these 5 vertices would have a far smaller spanning ratio, as there would be a lot of shortcut edges. However, a graph where this path is the shortest path between two vertices can be found in Appendix~\ref{app:lbgraph}.
\end{myproof}

\section{Lower bound on $\theta$-routing}
\label{sec:lbrouting}

\input{thetarouting}

\section{Conclusions}
\label{sec:Conclusions}

We showed that there is a path between every pair of vertices in the \ensuremath{\theta_5}-graph\xspace, of length at most \ensuremath{\sqrt{50 + 22 \sqrt{5}} \approx 9.960}\xspace times the straight-line distance between them. This is the first constant upper bound on the spanning ratio of the \ensuremath{\theta_5}-graph\xspace, proving that it is a geometric spanner. We also presented a \ensuremath{\theta_5}-graph\xspace with spanning ratio arbitrarily close to \ensuremath{\frac{1}{2}(11\sqrt{5} -17) \approx 3.798}\xspace, thereby giving a lower bound on the spanning ratio. There is still a significant gap between these bounds, which is caused by the upper bound proof mostly ignoring the main obstacle to improving the lower bound: that every edge requires at least one of its canonical triangles to be empty. Hence we believe that the true spanning ratio is closer to the lower bound.

While our proof for the upper bound on the spanning ratio returns a spanning path between the two vertices, it requires knowledge of the neighbours of both the current vertex and the destination vertex. This means that the proof does not lead to a local routing strategy that can be applied in, say, a wireless setting. This raises the question whether it is possible to route \emph{competitively} on this graph, i.e. to discover a spanning path from one vertex to another by using only information local to the current vertices visited so far.

\bibliographystyle{unsrt}

redpajama_set_name: RedPajamaArXiv


2007 Children's Moonbeam Book Award - Multi-Cultural Nonfiction - Ray Charles: Find Another Way!
SUSAN SLOATE is the author of more than 20 published books, including the 2013 bestseller STEALING FIRE, FORWARD TO CAMELOT: 50th ANNIVERSARY EDITION (with KEVIN FINN) and REALIZING YOU (with RON DOADES). All 3 novels--different genres, different subjects, different characters--were published within a single 90-day period.
STEALING FIRE was published by Drake Valley Press in August 2013 after being named a Quarter-Finalist in the 2012 Amazon Breakthrough Novel Award Contest (top 5%). It went to #2 in its category on Amazon in its first 48 hours and was a Hot New Release for its first 90 days. In 2014, it was honored in the Readers' Favorite Literary Competition (Women's Fiction). The newest edition of STEALING FIRE, which introduced Susan's own publishing imprint, Covfefe Press, was published in spring 2018.
FORWARD TO CAMELOT: 50th ANNIVERSARY EDITION is a re-edited version of the original 2003 time-travel thriller that became a #6 Amazon bestseller, took honors in 3 literary competitions and was optioned for film production by a Hollywood company. The 50th Anniversary Edition contains a new Afterword by the authors.
REALIZING YOU, published in September 2013, launched a new genre: the self-help novel, and has received laudatory reviews from both critics and readers.
Susan has also written 17 published young-adult fiction and nonfiction books, two produced stage plays and two optioned film works. In 2009, she appeared on The History Channel in a Mystery Quest special on Alcatraz, in support of her book, MYSTERIES UNWRAPPED: THE SECRETS OF ALCATRAZ. Her children's biography of Ray Charles won the silver medal in the 2007 Children's Moonbeam Book Awards competition.
Susan has managed two recent political campaigns and founded a local authors festival to promote literacy among students.
Susan lives in Mount Pleasant, SC.