HUMIRA TAKING AGES TO KICK IN!!
I have been on Azathioprine and prednisilone for 2 months - STILL HAD upset stomach.
now i'm back on prednisilone ( which causes weight gain ) and also back on 2 azathioprine and 1 HUMIRA INJECTION 40mg every 2 wks... dr wants me to do this for 6 wks and then see what happens - if again stomach is bad he'll go to 3 azathioprines...!!
he didn't want me to increase my humira - why? well too expensive he said....and he thinkt its not working..
did tell him well i have tried azathioprine + prednisilone for 2 months mate and nothing...but he reckons try it again but this time with the injections at same time..
I know theres not much left after humira as theyve exhausted all options now except for surgery.
Has anyone else experienced anything like this, on humira or not.
My 9yr old neice was diagnosed with Crohn's Disease a few years ago.At times,she has had spells of passing blood.The doctors are now thinking of giving her Humira shots.My sister in law was told they can be painful.What are some of the side effects from the shot,and how does it help??
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i have been on the humira shots for almost a year. They have helped my Crohn's tremedously as well as my eczema on my hands is pretty much non existant since being on the shots. I have had no side effects until about 2 months ago I broke out in these bumps. They looked like mosquito bites and some were so large they looked like ringworm. They did a biopsie on a couple of spots and had to even send it off to another place for a second opinion. They said that all they could come up with was it was a side effect of Humira.. WIERD... Good luck!!!!
I just recently started on the Humira self injections and they aren't that bad. Its best to numb the sight where you inject with an ice pack for a couple minutes first. Good luck and keep healthy.
I just recently started taking the Humira shots. They are not any more painful then any other shots. I haven't had any side effects, in fact they seem to be doing me a lot of good. The only bad thing about Humira is that the first dose is 4 shots and you can only inject them on your stomach or on your thighs. The second dose is two weeks later and it is 2 shots. After that it is just one shot every two weeks.
I was diagnosed with moderate to severe Crohns about 3 years ago. spent the first year in out of hospital trying different combinations of meds. all would work for period of time then id have problems again. Finally went to dr who suggested Remicaid or Humira. Chose Humira for convenience and havent had any problems for last 2 years. Give myself a small injection every 2 weeks. recently had checkups and no sign of inflamation. in complete remmition. So suggest considering humira instead of remicade.
Anyone had this issue? Rash looks like a bunch of red dots.
Hi I've had Crohns Disease for 22 years but since giving birth 7 years ago has not been settled. I've been told humira is my only chance of not having surgery. I have been to the hospital 3 times for the injection now but have felt no difference, how long does it take til you see a difference?
Is anyone out there on Humira and if so how is it working for you?
Wow.. I will keep you in my thoughts and prayers. A lot has happened since I posted that. My body rejected Humira and I am now on my last option.. Cimzia... Hoping it works..
Hi Shawna -- I'm on Humira and have been for about two years now. It has provided relief of my symptoms and I can usually feel it "kick in" within a day or so of the shot! I was on Remicade before that, which was like a miracle drug for me, but I built up a resistance to it and eventually it stopped working for me. The Humira has been great! If your doctor is suggesting it, and willing to put you on it, and will write a letter to your insurance company (it's expensive!) I would say go for it. As a sidenote, I even continued to use it during pregnancy (which kept me from being able to breastfeed but did not effect anything otherwise so far).
I took Humira for Crohn's disease for 6 months. I didn't get better. I got worse and I gained weight. I am sure that it may be helpful for some people but not for me.
Hi i am 31 years old and have had chrons for the pas 12 years, i have already had 2 majour bowel operations and after spending 3 months in hospital last summer the 2nd operation hasnt worked, i have now been told i may have to have whats left of my colon totaly removed and a colostomy.
I have been put on the humira programme but dont know what to do due to already having a very low imune system due to steriods i am currently on 40mg daily along with a cocktail of medication, i have high bp and fibromalgia along with malabsorbtion of liver bile salts, I Have constant pain and pass motions aroun 10-12 times a day, as you can imagine i am very drained but i manage to work and have 2 lovely children that keep me going but i would love to do more with them and am at the point of not knowing what to do next for the best?
The humira is a option but the side affects are quite serious does anyone have any comments for me?
my body and joints ache, have had 3 surgeries for crohn's and wear and ostomy. Now I am giving myself Humira shots, and my vision and joints are bothersome. How do I get some relief without my doctors prescribing me the pain medicine?? I feel like I'm losing all capabilities at times., It's hard to deal with getting out of bed, but I must work to pay and afford the bills for doctors and ostomy supplies. Is there any known cases of people getting Disability for this condition??
I am losing my teeth, strength in my legs and arms as well as my hands.
i have had crohns for about 55 years and this stuff is not going to work. it looks like they say it will fix every thing. if you take it you could die. just live with crohns. the only thing that helps it dont get up set. nerves are what sets it off.
Im due to go on humria soon as nothing else has worked for me plus the refuse to-do anymore operations on me cause ive had so many already. So this is my last option really I don't know what to expect from having it I just hope it works for me.
I dont know if it is different in the USA but here in England some people have claimed Disibility Allowance Benefit, not all are successfull but I am one of the lucky ones, its not much but every little helps. I was turned down at first but I appealed the decision. I wish you well as I know how much this desease can affect your whole life.
Hi I am 22 years old and I've has Crohns for 5 years now. I've only ever web on Pentasa with some steroids when I am flaring. However they put me on Imuran two and a half weeks ago and I've been extremely Ill since. stomach pains, and nausea. I haven't vommitted but the pain and uncomfortableness is unbearable. I don't want to be sick on Christmas. I got married six months ago and I would very much like to spend my first Christmas with my husband healthy instead of in bed. My doctor told me to split up the dosage but it hasn't helped. I feel as though there is now sharp pains in my abdominal area. Almost like gas pain that I can't get rid of.
Is there a better alternative? Would humira be better and easier to take? I'm worried about cost but if I need it cost shouldn't be the reason not to get it. What are the side affects of humira? My doctor assured me that Imuran had no side affects other then the possibility of getting cancer. So obviously they lied and I would like to hear from everyone else what works from them.
That sounds good. With the exception of the holiday and weekends I usually give up sugar all together because of the side effects (weigh gain/stomach aches). And during the week I usually stick to a 1200 calorie diet. Ive been trying to work out 2-3 times a week for 45min at a time. I just hope continuing this or being more strict about my diet will help my weight stay as it is now. I am actually trying to lose about 5 pounds but it seems impossible as the holidays were filled with cakes and fatty foods. Thanks for all your advice this makes me feel better.
Pleased to hear that your doctor is doing something about it.
With respect to the weight gain. I would prepare for it, if you think that it is going to be a problem.
An idea would be not to buy anything sweet. So, no sweets, chocolates, colas and other fizzy drinks, no biscuits, no cakes hoarded at home. Ideally, you would cut out pure sugar from your diet completely. The likelihood of you gaining weight will be minimised if you also make sure that you include at least an hour's brisk walking/fitness into your day.
It is my experience that people who say that they don't eat much, are not counting all the calories they ingest and are kidding themselves. Don't let yourself be fooled that way. Keep a food and calorie diary and learn how many calories are in your food and adjust accordingly as you get better.
If you want cake then make carrot/banana/peach/apple/apricot cake at home. Halve the amount of sugar and use oil instead of butter or margarine - increase the bicarb.
I hope that the Humira works.
My doctor lowered my dosage but the side affects did not go away so I have stopped taking the medication and will be starting humira soon. However I am now scared that I will gain weight. My doctor says there's no weight gain side effects and it says online that it is not a side effect but on a few blogs I read that some people randomly gained 40 pounds while exercising and eating low calorie meals. If I gain 40 pounds I would look ridiculous!
You are having the same reaction to Imuran (aka Azathioprine, aka AZA) as I had.
Your doctor is irresponsible to tell you that it has not side-effects. It is a cytostatic, a toxic chemotherapy drug that works by killing off fast-growing cells. Granted the doses are lower than in cancer treatments, but how could it not have any other side-effects (except for potentially causing cancer)?
In addition, it is well known that there is a sub-set of patients that are not able to metabolise Imuran (prodrug) to 6-Mercaptopurine (actual drug/active agent). http://www.thedrugmonitor.com/azathioprine.html It is highly likely that you are one of these.
As I understand it a low level of TPMT will result in higher drug toxicity, i.e. low intolerance to the purine analogues AZA and 6-MP. Your doctor should be aware of this.
If you can't get your hands on some 6-MP prior to Xmas then I suggest that you ring or write to your doctor (e-mail, letter or Fax) and list all your side-effects with severity. He should suggest that you lower the dose to a level that reduces the side-effects, if at all possible.
Keep a patient food/mood/symptom/medication/supplement diary. This will help you better monitor your health and help you to communicate with your doctors.
I am having alot of the side effects and would like to know what others on Humira are dealing with??
I have had seizures and other neurological problems before I was on Humira but it is making these problems really flare up!!
quite common a little scary tho .
I have recently been diagnosed with Crohn's Disease and am having difficulty getting it into remission. It's been 12 weeks since the onset with little relief from Colazal (9 tablets-6750 mg per day) Have scheduled an appointment with my specialist to discuss the use of Prednisone. Not looking forward to the side effects of steroids and hoping there is another option. Has anyone had much luck with Humira? Please advise.
I totally agree with James. I have had Crohn's for 32 years and have been able to keep the flare-ups under control by taking a round of Predisone, which usually takes about 2-3 weeks to finish. I have not taken Humira so can't comment on it. However, at the onset of fever or diarrhea that cannot be controlled, I start the Prednisone. I won't take chances anymore by waiting to see if the symptoms worsen, because too many times I have been literally carried into the emergency room, so dehydrated and weak from diarrhea and fever that I couldn't walk on my own. I haven't been hospitalized in over 15 years now; and previously, I was in the hospital at least once a year with Crohn's. The complications from the Crohn's are far worse than the side effects of Prednisone - especially if you don't continually take the Prednisone but rather take a round and then stop. Which reminds me, when taking Prednisone or any steroid, do not suddenly stop taking it. You must decrease the dosage little by little, on a daily basis, until you're taking just one tablet per day, before totally discontinuing taking the tablets. Also, if you have been on Prednisone or other steriods recently and require surgery, you must tell the doctor that you are either on steroids or have recently taken them. The doctor will need to take that into consideration when determining what course of action to take.
When i was first diagnosed with CD(1989). I was put on a course of Prednisone. The effects of a flair up far outweighs the possible side-effects of taking this medicine. I was up and back to living normally within weeks of taking this. It is such a relief that life can resume again. I have had several courses over the years and have never had any side effects. There is a risk of side effects with any medication, regardless of how harmless it may be. Prednisone may not be the answer in every case but it worked for me and is a common anti-inflammatory drug used for a wide variety of illnesses i.e Asthma and skin conditions.
I am having a hard time getting information. I have Chron's disease and had sugery August 2nd 2011 for small intestine obstruction. 10 days post surgery I was back in surgery with a massive inflammation blockage. This is first time something like that has happened to me since I had 3 prior surgeries. The only difference is that I am taking Humira for past 5 months and was told not to take the injection before surgery. Now I cannot get straight answers from Dr's on how this inflammation formed so quickly after surgery. On doctor suspects that because I have not taken Humira injection for about a month that it might have played a role in this mystery, Has anyone had anything similar to this or have insight on success and failures from Humira and Remicade, I really appreciate any and all responses and insight.
My daughter started on Humira last week for Crohn's Disease ..She is also taking Imuran. She suffers from Seizures which was under control with Tegretol until she had a Grand Mal on Saturday June 5..2010. Can someone shed a light on this for me? Should she continue with Humira and have her Seizure Dosage increased? Or should she stop taking Humira completely? Any suggestions will be greatly appreciated.
Have had three sections of bowel removed no relief Doctor tried all drugs and now has me on humira still no relief I have gained weight all around my abdominal region still sit on toilet at least 10 times a day and am only eating twice a day pain is unbearable has any one else experienced this??
I started humira and didnt even get a chance to get my second dose when i was in the hospital with a really bad case of pneumonia which resulted in having to have a 2 inch tube surgicaly put into my ribs...the absolute most painful thing i have ever went through and i have had 3 kids! imhaving a hard time excepting the hu,mira and remicade type drugs because if the chrons doesnt make us sick then these immune suppresent drugs do!!
You symptoms sound just the same as mine, i have recently started humira but had a very bad reaction and it got stopped, i really dnt know what will be the next step lets just hope a cure found soon.
My name is Shelly and i am 25. I was diagnosed with crohns 5 yrs ago. 5 months ago i had my worst flare up. The previous 5 years i never really had a serious flare up and was on salofalk tablets.
My doctor put me on Entorcort (which didnt really help me at all). I kept trying to ignore the pain and tried to continue with life as best as i could. Until 2 weeks ago when i ended up in hospital.
I was put on hydro-cortisone (if thats how u spell it?) and then predisilone. Neither of which helped improve my blood results. My CPR (inflammation marker went up to 165 and my MRI scan showed significant inflammation.
I am still in hospital and my doctor has put me on Humira. As i have been in hospital for over a week, i have been reading all about it on the internet and am extremely scared about the side effects of this drug. I am stressing so much that i am finding it hard to sleep and am constantly thinking that it can cause cancer.
My doctor said that it is my only option at the moment. I would have liked to get a 2nd opinion, but when you are in so much pain you just go with whatever is recommended i guess.
Anyway i was hoping to hear from anyone else that is on this drug? How are you feeling? What reactions have you had?
I had the first dose of injections 2 days ago and my blood results are better but the thought of injecting myself again with something that may cause cancer scares the living daylights out of me.
Please any information of feedback on this drug would be appreciated.
My son was diagnosed with Crohn's disease last year. After a bowel resection in May, he has done quite well. He has been on Imuran (although a low dose of 50mg/day), it was stopped about 6 weeks ago to make sure it was not the cause of his anemia. His physician wants to try him on Humira. The only medication he is currently taking is an iron supplement.
Can anyone tell me about their experience with Humira?
My husband has severe Crohn's. He had his colon removed, went to the "Best" NY hospital, "Best" surgeon and had staph infection had to be hospitalized after the colon was removed, had to be on TPN,5 more surgeries later because when they removed colon, small intestine was not put back correctly. Had to have a graph on his abdomen.
He has been through hell and back. His joints are hurting him now. He has been off Humria for almost 1 year now because of all that has happened, and the gastro wants him back on. His Crohns is in remission, so why put him back on this heavy drug?
I am 27 years old and I have crohns disease andhave been living with it for years now... I had my surgery in 2008 and I recently had a colonoscopy showing that it is back... now I have to take the humaira shots as prescribed by my doctor... how does this drug work? Is it worth taking? Any input will be greatly appreciated and taken into consideration... just need answers!
My son will be starting Humira shortly. He failed on Immuran (caused anemia) and our insurance company would not approve Remicade without trying Humira first.
Would anyone care to comment on their experience with Humira such as effectiveness, side effects etc. What is the longest you can be on Humira without it losing it's effectiveness?
After going in and out of doctors surgeries with stomach problems at 3 months old, to having my first really serious bleeding attack at 12 years of age, followed by 28 years of medications, bowel removal operations, bowel re-sections and the constant threat of a colostomy bag or colon cancer, I can state today that I no longer have Crohn's and haven't since early June 2000.
I am now working with an IBD and mucosal specialist at the University of Tasmania, in order to provide an alternative way of looking at this and other gastrointestinal diseases.
For the past 13 years, I have been self-studying neuro biology, neuro plasticity and cell biology, in order to explain to others what the changes were that occurred with me.
During this time, I have written many emails to other sufferers, sharing my thoughts and understandings about this debilitating disease and have had some great results.
Last year, I decided that I wanted to share more to more people and wrote a book, which I finally finished two months ago and have only presently got it up on Amazon as an ebook. With this, I hope to enlighten more people who suffer, from anything, how much the mind has an effect on the body.
Before you dismiss me and what I want to share, check out my youtube channel, which is where I share most of what's in my book anyway.
I wish you all the best in your individual journey.


text:
eLaws | eCases | Florida State | Florida Courts | Counties & Cities of Florida | Code of Federal Regulations |
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Home » 2011 Issues » Vol. 37, No. 37 (09/16/2011) » a permanent waiver of subsection 11B-27.002(4), F.A.C., from Diego Abel Colon-Batiz. Subsection 11B-27.002(4), F.A.C., requires recruits to complete basic recruit training, pass the SOCE, and gain employment within four years of beginning basic ...
a permanent waiver of subsection 11B-27.002(4), F.A.C., from Diego Abel Colon-Batiz. Subsection 11B-27.002(4), F.A.C., requires recruits to complete basic recruit training, pass the SOCE, and gain employment within four years of beginning basic ...
DEPARTMENT OF LAW ENFORCEMENT
Criminal Justice Standards and Training Commission
RULE NO.: RULE TITLE:
11B-27.002: Certification, Employment or Appointment, Reactivation, and Terminating Employment or Appointment of Officers
NOTICE IS HEREBY GIVEN that on August 30, 2011, the Criminal Justice Standards and Training Commission, received a petition for a permanent waiver of subsection 11B-27.002(4), F.A.C., from Diego Abel Colon-Batiz. Subsection 11B-27.002(4), F.A.C., requires recruits to complete basic recruit training, pass the SOCE, and gain employment within four years of beginning basic recruit training. The petition states that the Petitioner began basic on April 2, 2007, and paid his own way through basic by working full time. Petitioner further states that he was injured while completing basic and that prevented him from graduating until August 13, 2008. Petitioner states that he will suffer a substantial hardship if he is not permitted an extra six months to find employment because he will have to retake basic recruit training and retake the SOCE. The Petitioner requests an extra six months to find employment.
A copy of the Petition for Variance or Waiver may be obtained by contacting: Grace A. Jaye, Assistant General Counsel, Florida Department of Law Enforcement, P. O. Box 1489, Tallahassee, FL 32302 or by telephoning (850)410-7676.
Grace A. Jaye, Assistant General Counsel, Florida Department of Law Enforcement, P. O. Box 1489, Tallahassee, FL 32302 or by telephoning (850)410-7676.
Related Rules: (1)
11B-27.002. Certification, Employment or Appointment, Reactivation, and Terminating Employment or Appointment of Officers
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redpajama_set_name: RedPajamaCommonCrawl


text:
LASIKA Digital sport watch related knowledge
Categories:Industry News
(Summary description)Generally, the outer ring of a Digital sport watch will be engraved with 500/400/350 and other successively decreasing numbers. The physical quantity represented by these numbers is the speed, and the unit is kilometers/hour; and this bezel is also called a speedometer
LASIKA Digital sport watch
Generally, the outer ring of a Digital sport watch will be engraved with 500/400/350 and other successively decreasing numbers. The physical quantity represented by these numbers is the speed, and the unit is kilometers/hour; and this bezel is also called a speedometer, which generally has this Most of the watches with a similar outer ring are sports watches, which are also the most common calculation scale system in chronographs. The speedometer application is used together with the timing function of the Digital sport watch itself.
How does the LASIKA Digital sport watch measure the speed?
Take a car driving on the highway as an example. Generally, there is a distance indicator on the highway. When the starting point is passed, the stopwatch is started, and the second hand starts timing at 12 o'clock. When it moves 1 km, it stops timing. The number referred to is the average speed of the previous kilometer. For example, if there is 60 at 12 o'clock, then one kilometer is walked in one minute, so the speed is 60 kilometers per hour.
How to calculate the numbers?
So how is the position corresponding to the LASIKA Digital sport watch bezel number calculated? Actually, it is not difficult, as long as you know where the number 500 is engraved, and then do a simple division: Calculate 1 km at a speed of 500 km/h How much time is needed, in seconds.
1 hour = 3600 seconds, which means that 500 kilometers need 3600 seconds to complete, and 1 kilometer requires 3600 divided by 500 equals 7.2 seconds, so 500 must be engraved at 7.2 seconds on the LASIKA Digital sport watch disk. Other digital scales can be deduced by analogy.
Digital sport watch
Cheap Digital sport watch
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AVIS Macau welcomes inquiries on long term rental and/or leasing programs. We tackle each individual project, being for companies or individual users, to provide tailor made, competitive long term leasing programs.
• Long-term leasing of brand new vehicles for company use could be the best choice of car allowance. Many companies today would own a few company vehicles for the purpose of providing convenient transportation services to its clients and/or for its staff's usage. Having a company car could also be an additional benefit for employees. However, in order to obtain the balance between gaining staff's satisfaction and retention, and the costs involved in maintaining a transportation department, is not an easy task.


text:
package de.fu_berlin.agdb.crepe.json.algebra;

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonSubTypes;
import com.fasterxml.jackson.annotation.JsonTypeInfo;
import de.fu_berlin.agdb.crepe.algebra.Operator;
import de.fu_berlin.agdb.crepe.json.algebra.operators.JSONMatchToStation;
import de.fu_berlin.agdb.crepe.json.algebra.operators.logic.JSONBinaryOp;
import de.fu_berlin.agdb.crepe.json.algebra.operators.logic.JSONNot;
import de.fu_berlin.agdb.crepe.json.algebra.operators.numeric.JSONComparisonOperation;
import de.fu_berlin.agdb.crepe.json.algebra.operators.numeric.JSONNumericOperation;

import static com.fasterxml.jackson.annotation.JsonSubTypes.Type;

@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, property = "@class")
@JsonSubTypes({
        @Type(JSONMatch.class),
        @Type(JSONMatchToStation.class),
        @Type(JSONBinaryOp.class),
        @Type(JSONNot.class),
        @Type(JSONNumericOperation.class),
        @Type(JSONComparisonOperation.class)
})
public abstract class JSONOperator<T extends Operator> extends JSONAlgebraElement<T> {
}

redpajama_set_name: RedPajamaGithub


I bet you don't tuck your shirt into your underpants either.
It's no good, Ronin83, people whose legs conform to the average trouser don't understand the problem.
I regularly grit my teeth in outdoor shops when contemplating the inches of trouser leg sweeping the floor. And it's no good saying cut the bottoms off the legs, as walking trousers usually have some kind of shaping at the knees, and so even if you altered the legs to the right length the knees are in the wrong place.
There do seem to be trends in this, a few years ago, most walking trousers could be cinched in at the ankle, and I remember being told that it was precisely to accommodate different leg lengths, but it seems less common now.
I tend to wear Paramo trousers, they don't need overtrousers. I have both cascada and velez, and both have ankle adjustment. They aren't cheap though. I justify them to myself on the grounds that they are walking trousers and overtrousers in one, and you get 50 quid back on the empties.
Have you considered an alternative? A Rain skirt for example, with gaiters on, the skirt works well, when needed, on and off in 5-10 seconds. You don't even have to stop walking.
I have a pair of Marmot PreCip full length zip "pants", which have adjustable ankles.
..but you end up with wet feet!


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The ASEAN Committee on Disaster Management (ACDM), and the International Federation of Red Cross and Red Crescent Societies (IFRC) officially launched a report entitled "ASEAN Disaster Law Mapping – Implementing AADMER: A Regional Stocktake" on 2 May 2018 in Jakarta, Indonesia.
The report contains the first ever mapping of legal provisions supporting national and regional preparedness and response mechanisms in the domestic laws of 10 ASEAN member states. The legal mapping was guided by the minimum standards set in the ASEAN Agreement on Disaster Management and Emergency Response (AADMER). It is accompanied by an online platform hosted on the Red Cross and Red Crescent Resilience Library.
Mr. Vilayphong Sisomvang, Deputy Director General of the Lao PDR Ministry of Labor, speaking on behalf of the ACDM Prevention and Mitigation Working Group which his Ministry co-chairs, was pleased to see the strong collaboration between ASEAN member states, the ASEAN Secretariat and the IFRC in this initiative and encouraged participating agencies to use the mapping as a basis for cross-regional learning to strengthen preparedness and response for future disasters in the region. He also expressed the importance of this legal mapping to the ongoing legal review of Lao PDR's disaster management law.
He together with Giorgio Ferrario, IFRC Head to Indonesia and Timor Leste, thanked the Government of Canada and Canadian Red Cross, who supported this initiative under it Regional Resilience Initiative. The Government of Canada representative, Ms. Sharon Armstrong, Director for Development of the Embassy of Canada to Indonesia and ASEAN, emphasized the importance of strong disaster laws in resilience building, as well as the importance of gender equality in development.
At the close of the launch, Mr. Ferrario reiterated IFRC's commitment to building a more resilient ASEAN, and underscored the important foundation that inclusive laws and policies play in regard to this community resilience building. This institutional resilience component will form as an important part of the ASEAN Coalition for Resilience which aims at making 100 million people more resilient in ASEAN countries by 2025, as a direct contribution to the IFRC global initiative of the One Billion Coalition for Resilience.


text:
New Disney Cruise Line Pin – October 2014 Halloween on the High Seas
Disney Theme Park Merchandise provided a preview of the 2014 Halloween on the High Seas trading pin. The new Disney trading pin is scheduled to launch in October across the Disney Cruise Line fleet. This release date falls late into the Halloween on the High Seas cruises which begin in early September.
Disney Cruise Line 2014 – Halloween on the High Seas
This all new trading pin will be released to commemorate the 2014 Halloween on the High Seas cruises. The limited edition pin will only be sold onboard while supplies last. The pin is inspired by the Pumpkin Trees that will be featured in atrium of each cruise ship.
Release Date: October 2, 2014 October 16, 2014 • SKU: 400008894647 • Retail Price: $14.95 • Limited Edition: 1000
UPDATE: The release date has changed from 10/16 to 10/2.
This entry was posted in Merchandise and tagged Disney Pin Trading, Halloween on the High Seas by Scott Sanders. Bookmark the permalink.
6 Replies to "New Disney Cruise Line Pin – October 2014 Halloween on the High Seas"
Tom August 23, 2014 at 12:09 pm
We will be on the Disney Magic October 4 Southern Caribbean cruise. Once again DCL under estimates demand for these pins on special cruises! 500 per ship and the Halloween cruises start in September!
Statler August 24, 2014 at 5:08 pm
That does seem like a small number. It may be even smaller per ship if all four ships participate. I imagine all ships will have a party.
Scott Sanders Post author August 24, 2014 at 5:21 pm
The 2013 Very MerryTime pin was release around Thanksgiving on all four ships with a LE of 2000. The Halloween Pin is half the edition, and is listed to be released mid-October rather than in early September when the Halloween on the High Seas cruises begin.
Larrel August 31, 2014 at 8:45 am
🙁 we will be the Dream until 10/2 🙁
Stephanie Everman October 24, 2014 at 10:21 pm
We just sailed and they had no pins. Is there anyway to purchase them?
Scott Sanders Post author October 24, 2014 at 11:52 pm
There were only 1000 pins made and divided amongst the fleet. If you are set on getting the pin you can try eBay, but the last one I saw was around $50.


text:
What Languages Do Syrian Refugees Speak?
Since there are more and more refugees that find themselves seeking asylum on foreign soil, there are increased chances that you will meet one of them while in the grocery store or other public areas. Even if you have never experienced it yourself, you may be able to remotely imagine how stressful it can be to uproot yourself and run away from everything you know and be whisked away to another country.
If you are one who wants to make the transition easier for refugees, you may be wondering how to interact with them. What languages do Syrian refugees speak? This is one of the first questions you need to ask to be able to communicate with them and make them feel more at ease in their overall situation. This blog post will shed some light on this topic and help you better understand and talk to the refugees you meet.
There are many people who came to live in Syria when it was a young civilization. This is why a number of different languages are spoken in the country today. Refugees may speak any one of the languages mentioned below.
The official language of the Syrian Arab Republic is Arabic. The language that is used in school textbooks, printed materials, official documents, and books is Traditional Modern Arabic. However, the language that is spoken between people is slightly different. Arabic dialects that are spoken include Mesopotamian Arabic, Levantine Arabic, Najdi Arabic, Bedawi Arabic.Mesopotamian Arabic, Levantine Arabic, Najdi Arabic and Bedawi Arabic.
This is the second-most widely spoken language in Syria. People who live in the Eastern portion of Syria speak Kurdish but also pride themselves in being fluent in Arabic as well.
In the olden days, Syriac was one of the prominent Christian languages that were widely spoken. This is the language that was the one that gave Syria its name. It is still widely spoken today.
In the 19th and 20th centuries, many Armenians flocked to Syria to permanently migrate and seek a new home. The language is taught in schools in Aleppo, where most of the Armenian Syrians live.
English and French
Foreign languages like English and French are also widely used and spoken. Syrian schools start teaching English beginning the first grade of elementary school and French is taught starting the seventh grade. This is why most Syrians are able to carry a conversation in English quite easily. Another foreign language that is offered in schools is Russian. It is also now being offered in some schools as an alternative to French.
Now that you have some basic knowledge of the languages refugees speak, you may be able to try and talk to them. If you know about any Syrian refugees who have joined your religious organization, workplace, or even parents organization in your children's school, it may be a good idea to try and make their new home feel more welcoming by trying to talk to them.
Should you want to do what you can to help refugees, come to UOSSM! The Union of Medical Care and Relief Organizations-USA (UOSSM USA) is a nonprofit, 501(c)(3) charitable, independent, non-government, medical humanitarian organization, incorporated in the state of Texas. Contact us today to donate to the Syrian relief fund and contribute to other initiatives to help!


text:
\section{Introduction}\label{xs:1}

\mypar{Background.} The study of theoretical and practical aspects of highly dynamic distributed systems has received a great deal of attention~\cite{CFQS12,KO,MS18}. These models involve a constantly changing network of computational devices called \emph{processes} (sometimes referred to as ``processors'' or ``agents''). This dynamism is typical of modern real-world systems and is the result of technological innovations, such as the spread of mobile devices, software-defined networks, and wirelesses sensor networks, wearable devices or smartphones, etc.

There are several models of dynamism~\cite{CFQS12}; a popular choice is the \emph{1-interval-connected} network model~\cite{KLO10, DW05}. Here, a fixed set of $n$ processes communicate through links forming a time-varying graph, i.e., a graph whose edge set changes at discrete time units called \emph{rounds} (thus, the system is synchronous); such a graph changes unpredictably, but is assumed to be connected at all times.

A large number of research papers have considered dynamic systems where each process has a distinct identity (\emph{unique IDs})~\cite{KLO10}. In this setting, there are efficient algorithms for consensus~\cite{KOM11}, broadcast~\cite{CFMS15}, counting~\cite{KLO10, DW05}, and many other problems~\cite{KLO11,MS18}.

The study of dynamic 1-interval-connected networks with unique IDs began with a seminal paper by Kuhn et al.~\cite{KLO10}, who showed that knowing the size of the system (i.e., the number $n$ of processes) is useful for non-trivial computations. For example, to find out if there is at least one process with input $x$ in the system, every process can start broadcasting its input and wait $n-1$ rounds to see if it receives $x$. Note that this technique is ineffective if nothing is known about $n$.

\smallskip
\mypar{Counting problem.}  There are real-world scenarios in which the individual processes may be unaware of the size of the system; an example are large-scale ad-hoc sensor networks~\cite{DW05}. In such scenarios, the problem of determining $n$ is called \emph{Counting problem}. In~\cite{KLO10} it is shown how to solve the Counting problem in at most $n+1$ rounds in a 1-interval-connected network with unique IDs.

\smallskip
\mypar{Anonymous systems.} The known landscape changes if we consider networks without IDs, which are called \emph{anonymous systems}. In this model, all processes have identical initial states, and may only differ by their inputs. In the last thirty years, a large body of works~\cite{BV01,CDS06,CGM08,JMM12,FPP00,SUW15,YK88} have investigated the computational power of anonymous static networks, giving a characterizations of what can be computed in various settings~\cite{BV01,BV02}. Studying anonymous systems is not only important from a theoretical perspective, but also for their practical relevance. In a highly dynamic system, IDs may not be guaranteed to be unique due to operational limitations~\cite{DW05}, or may compromise user privacy. Indeed, users may not be willing to be tracked or to disclose information about their behavior; examples are COVID-19 tracking apps~\cite{SM20}, where a threat to privacy was felt by a large share of the public even if these apps where assigning a rotating random ID to each user. As a matter of fact, an adversary can easily track the continuous broadcast of a fixed random ID tracing the movements of a person~\cite{LAOOO20}.

\smallskip
\mypar{Unique leader.} In order to deterministically solve non-trivial problems in anonymous systems, it is necessary to have some form of initial ``asymmetry''~\cite{A80,BV02,MCS13,YK88}. The most common assumption is the existence of a \emph{leader}, i.e., a single special process that starts in a different and unique initial state. The presence of a leader is a realistic assumption: examples include a base station in a mobile network, a gateway in a sensors network, etc. For these reasons, the computational power of anonymous systems enriched with a leader has been extensively studied in the classic model of static networks~\cite{FPP00,Sa99,YK96}, as well as in population protocols~\cite{AAE08,ABBS16,BBCD15,BBK11,DFIISV19}.

\smallskip
\mypar{State of the art.} A long series of papers have studied the Counting problem in anonymous 1-interval-connected networks with a leader~\cite{CMM16,DB15,DB16,DBBC14a,DBBC14b,DBCB13,KM18,KM19,KM22,MCS13}. These works have shown better and better upper bounds, leading to the recent paper~\cite{KM22}, which solves the Counting problem in $O({n^{4+ \epsilon}} \log^{3} (n))$ rounds (for $\epsilon>0$). Almost all of these works share the same basic approach of implementing a mass-distribution mechanism similar to the local averaging used to solve the average consensus problem~\cite{C11,OT09,T84}. (In \cref{as:4} we give a comprehensive survey.) We point out that the mass-distribution approach requires processes to exchange numbers whose representation size grows at least linearly with the number of rounds. Thus, all previous works on counting in anonymous 1-interval-connected networks need messages of size at least $\Omega(n^4)$.

In spite of the technical sophistication of this line of research, there is still a striking gap in terms of running time---a multiplicative factor of $O(n^3 \log ^{3}(n))$---between the best algorithm for anonymous networks and the best algorithm for networks with unique IDs. The same gap exists with respect to static networks, where the Counting problem is known to be solvable in $2n$ rounds~\cite{MCS13}. Given the current state of the art, solving non-trivial problems in large-scale dynamic networks is still impractical.

\subsection{Our Contributions}
\mypar{Main results.} In this paper, we close the aforementioned gap by showing that counting in 1-interval-connected anonymous dynamic networks with a leader is linear: we give a deterministic algorithm for the Counting problem that terminates in $3n-2$ rounds (\cref{xth:term}), as well as a non-terminating algorithm that stabilizes on the correct count in $2n-2$ rounds (\cref{xth:stab}).

We also prove a lower bound of roughly $2n$ rounds, both for terminating and stabilizing counting algorithms (\cref{xth:lower}); this is the first non-trivial lower bound for anonymous networks (better than $n-1$), and it shows that our stabilizing algorithm is optimal up to a small additive constant.

In addition, our algorithms support network topologies with multiple parallel links and self-loops (i.e., \emph{multigraphs}, as opposed to the simple graphs used in traditional models).

\medskip
\mypar{Significance.} Our algorithms actually solve a generalized version of the Counting problem: when processes are assigned inputs, they are able to count how many processes have each input. Solving such a \emph{Generalized Counting problem} allows us to solve a much larger class of problems, called \emph{multi-aggregation problems}, in the same number of rounds (\cref{xth:compl}). On the other hand, these are the only problems that can be solved deterministically in anonymous networks (\cref{xth:imp}).

Thus, we come to an interesting conclusion: In 1-interval-connected anonymous dynamic networks with a leader, \ul{any problem that can be solved deterministically has a solution in at most $3n-2$ rounds}. Our lower bounds show that there is an overhead to be paid for counting in anonymous dynamic networks compared to networks with IDs, but the overhead is only linear. This is much less than what was previously conjectured~\cite{KM20,KM22}; in fact, our results make computations in anonymous and dynamic large-scale networks possible and efficient in practice.

We remark that the local computation time and the amount of processes' internal memory required by our algorithms is only polynomial in the size of the network. Also, like in previous works, processes need to send messages of polynomial size.

\medskip
\mypar{Technique.} Our algorithms and lower bounds are based on a novel combinatorial structure called \emph{history tree}, which completely represents an anonymous dynamic network and naturally models the idea that processes can be distinguished if and only if they have different ``histories'' (\cref{xs:3}). Thanks to the simplicity of our technique, this paper is entirely self-contained, our proofs are transparent and easy to understand, and our algorithms are elegant and straightforward to implement.\footnote{A C\# implementation can be found here: \url{https://github.com/anonymityDistComp/anonymity}. The repository includes a dynamic network simulator that can be used to run tests and visualize the history trees of custom networks.}

We argue that history trees are of independent interest, as they could help in the study of different network models, as well as networks with specific restricted dynamics.

\section{Definitions and Preliminaries}\label{xs:2}
Here we give some informal definitions; the interested reader may find formal ones in \cref{as:1}.

\medskip
\mypar{Dynamic network.} Our model of computation involves a system of $n$ \emph{anonymous} processes in a \emph{1-interval-connected} dynamic network whose topology changes unpredictably at discrete time units called \emph{rounds}. That is, at round~$t\geq 1$, the network's topology is modeled by a connected undirected multigraph $G_t$ whose edges are called \emph{links}. Processes can send \emph{messages} through links and can update their \emph{internal states} based on the messages they receive.

\medskip
\mypar{Round structure.} Each round is subdivided into two phases: in the \emph{message-passing phase}, each process broadcasts a message containing (an encoding of) its internal state through all the links incident to it. After all messages have been exchanged, the \emph{local-computation phase} occurs, where each process updates its own internal state based on a function $\mathcal A$ of its current state and the multiset of messages it has just received. Processes are \emph{anonymous}, implying that the function $\mathcal A$ is the same for all of them. We stress that all local computations are deterministic.

\medskip
\mypar{Input and output.} At round~$0$, each process is assigned an \emph{input}, which is deterministically converted into the process' initial internal state. Furthermore, at every round, each process produces an \emph{output}, which is a function of its internal state. Some states are \emph{terminal}; once a process enters a terminal state, it can no longer change it.

\medskip
\mypar{Stabilization and termination.} A system is said to \emph{stabilize} if the outputs of all its processes remain constant from a certain round onward; note that a process' internal state may still change even when its output is constant. If, in addition, all processes reach a terminal state, the system is said to \emph{terminate}.

\medskip
\mypar{Problems.} A \emph{problem} defines a relationship between processes' inputs and outputs: an algorithm $\mathcal A$ \emph{solves} a given problem if, whenever the processes are assigned a certain multiset of $n$ inputs, and all processes execute $\mathcal A$ in their local computations, the system eventually stabilizes on the correct multiset of $n$ outputs. Note that the same algorithm $\mathcal A$ must work for all $n\geq 1$ (i.e., the system is unaware of its own size) and regardless of the network's topology (as long as it is 1-interval-connected). A stronger notion of solvability requires that the system not only stabilizes but actually terminates on the correct multiset of outputs.

\medskip
\mypar{Unique leader.} A commonly made assumption is the presence of a \emph{unique leader} in the system. This can be modeled by assuming that each process' input includes a \emph{leader flag}, and an input assignment is valid if and only if there is exactly one process whose input has the leader flag set.

\medskip
\mypar{Multi-aggregation problems.} A \emph{multi-aggregation problem} is a problem where the output to be computed by each process only depends on the process' own input and the multiset of all processes' inputs. In the special case where all processes have to compute the same output, we have an \emph{aggregation problem}. Notable examples of aggregation problems include computing statistical functions on input numbers, such as sum, average, maximum, median, mode, variance, etc.

As we will see, the multi-aggregation problems are precisely the problems that can be solved in 1-interval-connected anonymous dynamic networks with a unique leader.
Specifically, in \cref{xs:4} we will show that all multi-aggregation problems can be solved in a linear number of rounds. On the other hand, in \cref{xs:5} we will prove that no other problem can be solved at all.

\medskip
\mypar{Counting problem.} An important aggregation problem is the \emph{Generalized Counting problem}, where each process must output the multiset of all processes' inputs. That is, the system has to count how many processes have each input. In the special case where all (non-leader) processes have the same input, this reduces to the \emph{Counting problem}: determining the size of the system, $n$.

\medskip
\mypar{Completeness.} The Generalized Counting problem is \emph{complete} for the class of multi-aggregate problems (no matter if the network is connected or has a unique leader), in the following sense:

\begin{theorem}\label{xth:compl}
If the Generalized Counting problem can be solved (with termination) in $f(n)$ rounds, then every multi-aggregate problem can be solved (with termination) in $f(n)$ rounds, too.
\end{theorem}
\begin{proof}
Once a process with input $x$ has determined the multiset $\mu$ of all processes' inputs, it can immediately compute any function of $(x, \mu)$ within the same local-computation phase.
\end{proof}

\section{History Trees}\label{xs:3}
We introduce \emph{history trees} as a natural tool of investigation for anonymous dynamic networks. An example of a history tree is illustrated in \cref{xfig:hg}, and a formal definition is found in \cref{as:2}.

\medskip
\mypar{Indistinguishable processes.} Since processes are anonymous, they can only be distinguished by their inputs or by the multisets of messages they have received. This leads to an inductive definition of \emph{indistinguishability}: Two processes are indistinguishable at the end of round~$0$ if and only if they have the same input. At the end of round~$t\geq 1$, two processes $p$ and $q$ are indistinguishable if and only if they were indistinguishable at the end of round~$t-1$ and, for every equivalence class $A$ of processes that were indistinguishable at the end of round~$t-1$, both $p$ and $q$ receive an equal number of (identical) messages from processes in $A$ at round~$t$.

\medskip
\mypar{Levels of a history tree.} A \emph{history tree} is a structure associated with a dynamic network. It is an infinite graph whose nodes are subdivided into \emph{levels} $L_{-1}$, $L_0$, $L_1$, $L_2$, \dots, where each node in level $L_t$, with $t\geq 0$, represents an equivalence class of processes that are indistinguishable at the end of round~$t$. The level $L_{-1}$ contains a unique node $r$, representing all processes in the system. Each node in level $L_0$ has a label indicating the (unique) input of the processes it represents.

\medskip
\mypar{Black and red edges.} A history tree has two types of undirected edges; each edge connects nodes in consecutive levels. The \emph{black edges} induce an infinite tree rooted at $r$ and spanning all nodes. A black edge $\{v, v'\}$, with $v\in L_{t}$ and $v'\in L_{t+1}$, indicates that the \emph{child node} $v'$ represents a subset of the processes represented by the \emph{parent node} $v$.

The \emph{red multiedges} represent messages. A red edge $\{v, v'\}$ with multiplicity $m$, with $v\in L_{t}$ and $v'\in L_{t+1}$, indicates that, at round~$t+1$, each process represented by $v'$ receives a total of $m$ (identical) messages from processes represented by $v$.

\medskip
\mypar{Anonymity of a node.} The \emph{anonymity} $a(v)$ of a node $v$ of a history tree is defined as the number of processes represented by $v$. Since the nodes in a same level represent a partition of all the processes, the sum of their anonymities must be $n$. Moreover, by the definition of black edges, the anonymity of a node is equal to the sum of the anonymities of its children.

Observe that the Generalized Counting problem can be rephrased as the problem of determining the anonymities of all the nodes in $L_0$.

\begin{figure}
\centering
\includegraphics[width=\linewidth]{pics/hg.pdf}
\caption{The first rounds of a dynamic network with $n=12$ processes and the corresponding levels of the history tree. The letters L, A, B indicate processes' inputs (the process with input L is the leader). At each round, the network's links are represented by solid red edges. Sets of indistinguishable processes are indicated by dashed blue lines and unique labels. The same labels are also reported in the history tree: each node corresponds to a set of indistinguishable processes. It should be noted that labels other than L, A, B are not part of the history tree itself, and have been added for the reader's convenience. The dashed red edges in the history tree represent messages received by processes through the links; the numbers indicate their multiplicities (when greater than $1$). For example, the edge $\{c_2, b_4\}$ has multiplicity~$2$ because, at round~3, each of the two processes in the class~$c_2$ receives two identical messages from the (indistinguishable) processes that were in the class~$b_4$ in the previous round. On the other hand, the node labeled $c_5$ represents four processes (i.e., its \emph{anonymity} is $4$) at round~3; among them, only one receives a message from $c_6$ in the next round. Thus, this process is disambiguated, which causes the node $c_5$ to branch into two nodes: $d_7$, with anonymity~$3$, and $d_8$, with anonymity~$1$. The \emph{view} of the node labeled $b_3$ is the subgraph of the history tree induced by the nodes with labels in the set $\{b_3, a_3, a_5, \mbox{L}, \mbox{A}, \mbox{B}, r\}$.}
\label{xfig:hg}
\end{figure}

\mypar{History of a process.} A \emph{monotonic path} in the history tree is a sequence of nodes in distinct levels, such that any two consecutive nodes are connected by a black or a red edge. We define the \emph{view} of a node $v$ in the history tree as the finite subgraph induced by all the nodes spanned by monotonic paths with endpoints $v$ and $r$. The \emph{history} of a process $p$ at round~$t$ is the view of the (unique) node in $L_t$ that represents a set of processes containing $p$.

\medskip
\mypar{Fundamental theorem.} Intuitively, the history of a process at round~$t$ contains all the information that the process can use at that round for its local computations. This intuition is made precise by the following fundamental theorem (a formal proof is found in \cref{as:2.3}):
\begin{theorem}\label{xth:view}
The state of a process $p$ at the end of round~$t$ is determined by a function $\mathcal F_\mathcal A$ of the history of $p$ at round~$t$. The mapping $\mathcal F_\mathcal A$ depends entirely on the algorithm $\mathcal A$; it does not depend on $p$, on $t$, on the network's topology, nor on the inputs assigned to the processes at round~$0$.\qed
\end{theorem}

\mypar{Locally constructing the history.} The significance of \cref{xth:view} is that it allows us to shift our focus from dynamic networks to history trees. As shown in \cref{as:2.4}, there is a local algorithm $\mathcal A^\ast$ that allows processes to construct and update their history at every round. Thanks to \cref{xth:view}, processes are guaranteed not to lose any information if they simply execute $\mathcal A^\ast$, regardless of their goal, and then compute their outputs as a function of their history. Thus, in the following, we will assume without loss of generality that a process' internal state at every round, as well as all the messages it broadcasts, always coincide with its history at that round.

\section{Linear-Time Computation}\label{xs:4}
As discussed at the end of \cref{xs:3}, the following algorithms assume that all processes have their current history as their internal state and broadcast their history through all available links at every round. We will show how to solve the Generalized Counting problem in such a setting.

\subsection{Stabilizing Algorithm}\label{xs:4.1}
In a history tree (or in a view), a node $v$ is said to be \emph{non-branching} if it has exactly one child, which we denote as $c(v)$. Recall that the parent-child relation is determined by black edges only.

A pair of non-branching nodes $(v_1, v_2)$ in a history tree is said to be \emph{exposed} with multiplicity $(m_1, m_2)$ if the red edge $\{c(v_1), v_2\}$ is present with multiplicity $m_1\geq 1$, while the red edge $\{c(v_2), v_1\}$ has multiplicity $m_2\geq 1$ (see~\cref{xfig:mix}, left). Note that $v_1$ and $v_2$ must be on the same level.

Thanks to the following lemma, if we know the anonymity of a node in an exposed pair, we can determine the anonymity of the other node. (Recall that we denote the anonymity of $v$ by $a(v)$.)

\begin{lemma}\label{xl:guess1}
If $(v_1, v_2)$ is an exposed pair with multiplicity $(m_1, m_2)$, then $a(v_1)\cdot m_1=a(v_2)\cdot m_2$.
\end{lemma}
\begin{proof}
Let $v_1, v_2\in L_t$, and let $P_1$ and $P_2$ be the sets of processes represented by $v_1$ and $v_2$, respectively. Since $v_1$ is non-branching, we have $a(c(v_1))=a(v_1)$, and therefore $c(v_1)$ represents $P_1$, as well. Hence, the number of links between $P_1$ and $P_2$ in $G_{t+1}$ (counted with their multiplicities) is $a(c(v_1))\cdot m_1 = a(v_1)\cdot m_1$. By a symmetric argument, this number is equal to $a(v_2)\cdot m_2$.
\end{proof}

\begin{lemma}\label{xl:propamain} Let $P$ be a set of processes in a 1-interval-connected dynamic network of size $n$, such that $1\leq |P|\leq n-1$, and let $t\geq 0$. Then, at every round~$t'\geq t+|P|$, in the history of every process there is a node at level $L_t$ representing at least one process not in $P$.
\end{lemma}
\begin{proof}
Let $Q$ be the complement of $P$ (note that $Q$ is not empty), and let $Q_{t+i}$ be the set of processes represented by the nodes in $L_{t+i}$ whose view contains a node in $L_t$ representing at least one process in $Q$. We will prove by induction that $|Q_{t+i}|\geq |Q|+i$  for all $0\leq i\leq |P|$. The base case holds because $Q_t=Q$. The induction step is implied by $Q_{t+i}\subsetneq Q_{t+i+1}$, which holds for all $0\leq i < |P|$ as long as $|Q_{t+i}|<n$. Indeed, because $G_{t+i+1}$ is connected, it must contain a link between a process $p\in Q_{t+i}$ and a process $q\notin Q_{t+i}$. Thus, the history of $q$ at round~$t+i+1$ contains the history of $p$ at round~$t+i$, and so $Q_{t+i}\subsetneq Q_{t+i}\cup\{q\}\subseteq Q_{t+i+1}$.

Now, plugging $i:=|P|$, we get $|Q_{t+|P|}|=n$. In other words, the history of each node in $L_{t+|P|}$ (and hence in subsequent levels) contains a node at level $L_t$ representing a process in $Q$.
\end{proof}

It is well known that, in a 1-interval-connected network, any piece of information can reach all processes in $n-1$ rounds~\cite{KLO10}. We can translate this observation into the language of history trees:

\begin{corollary}\label{xl:propa}
In the history tree of a 1-interval-connected dynamic network, every node at level $L_t$ is in the view of every node at level $L_{t'}$, for all $t'\geq t+n-1$.
\end{corollary}
\begin{proof}
Let $v\in L_t$, and let $P$ be the set of processes not represented by $v$. If $P$ is empty, then all nodes in $L_{t'}$ are descendants of $v$, and have $v$ in their view. Otherwise, $1\leq|P|\leq n-1$, and \cref{xl:propamain} implies that $v$ is in the view of all nodes in $L_{t'}$.
\end{proof}

\begin{theorem}\label{xth:stab}
There is an algorithm that solves the Generalized Counting problem in 1-interval-connected anonymous dynamic networks with a leader and stabilizes in at most $2n-2$ rounds.
\end{theorem}
\begin{proof}
Let $n>1$, and let $L_T$ be the first level of the history tree whose nodes are all non-branching. Since $|L_{0}|\geq 2$ and $|L_{t-1}|\leq |L_t|\leq n$ for all $t\geq 0$, we have $T\leq n-2$ by the pigeonhole principle.

Let $\ell\in L_T$ be the node corresponding to the leader; we know that $a(\ell)=1$. Since $G_{T+1}$ is connected, the exposed pairs of nodes in $L_T$ must form a connected graph on $L_T$, as well. Hence, thanks to \cref{xl:guess1}, any process that has complete knowledge of $L_T$ and $L_{T+1}$ can use the information that $a(\ell)=1$ to compute the anonymities of all nodes in $L_T$. In fact, by \cref{xl:propa}, every process is able to do so by round~$(T+1)+n-1\leq 2n-2$.

Summarizing, the local algorithm for each process is as follows: Find the first level in your history whose nodes are all non-branching. If such a level exists and contains a node corresponding to the leader, assume this is $L_T$ and compute the anonymities of all its nodes. Then add together the anonymities of nodes having equal inputs, and give the result as output. Even if some outputs may be incorrect at first, the whole system stabilizes on the correct output by round~$2n-2$.
\end{proof}
If we were to use the same strategy of \cref{xth:stab} to devise a terminating algorithm, we would be bound to fail, as shown by the counterexample in \cref{as:3.1}. The problem is that a process has no easy way of knowing whether the first level in its history whose nodes are all non-branching is really $L_T$, and so it may end up terminating too soon with the wrong output. To find a correct terminating condition, we will have to considerably develop the theory of history trees.

\begin{figure}
\centering
\includegraphics[scale=0.75]{pics/ff1.pdf}\qquad\quad
\includegraphics[scale=0.75]{pics/ff2.pdf}\qquad\quad
\includegraphics[scale=0.75]{pics/ff3.pdf}
\caption{Left: an \emph{exposed} pair of nodes. Center: if the anonymities of $u$, $u_1$, $u_2$, \dots, $u_k$ are known, then $v$ is \emph{guessable} by $u$. Right: if the colored nodes are counted, the blue ones form a \emph{counting cut}, and the orange ones define a non-trivial \emph{isle} with root $s$, where the nodes with a dot are internal.}
\label{xfig:mix}
\end{figure}

\subsection{Terminating Algorithm}\label{xs:4.2}

\mypar{Guessing anonymities.} Inspired by \cref{xl:guess1}, we now describe a more sophisticated way of estimating the anonymity of a node based on known anonymities (see~\cref{xfig:mix}, center). Let $u$ be a node of a history tree, and assume that the anonymities of all its children $u_1$, $u_2$, \dots, $u_k$ are known: such a node $u$ is called a \emph{guesser}. If $v$ is not a child of $u$ and the red edge $\{v, u\}$ is present with multiplicity $m\geq 1$, we say that $v$ is \emph{guessable} by $u$. In this case, we can make a \emph{guess} $g(v)$ on the anonymity of $v$:
\begin{equation}\label{xe:guess}
g(v)=\left\lceil \frac{a(u_1)\cdot m_1+a(u_2)\cdot m_2+\dots+a(u_k)\cdot m_k}m\right\rceil,
\end{equation}
where $m_i$ is the multiplicity of the red edge $\{u_i, v'\}$ for all $1\leq i\leq k$, and $v'$ is the parent of $v$ (possibly, $m_i=0$). Although a guess may be inaccurate, it never underestimates the anonymity:

\begin{lemma}\label{xl:guess2}
If $v$ is guessable, then $g(v)\geq a(v)$. Moreover, if $v$ has no siblings, $g(v)=a(v)$.
\end{lemma}
\begin{proof}
Let $u,v'\in L_t$, and let $P_1$ and $P_2$ be the sets of processes represented by $u$ and $v'$, respectively. By counting the links between $P_1$ and $P_2$ in $G_{t+1}$ in two ways, we have $\sum_i a(u_i)\,m_i = \sum_i a(v_i)\,m'_i$, where the two sums range over all children of $u$ and $v'$, respectively (note that $v=v_j$ for some $j$), and $m'_i$ is the multiplicity of the red edge $\{v_i,u\}$ (so, $m=m'_j$). Our lemma easily follows.
\end{proof}

\mypar{Heavy nodes.} Even if a node is guessable, it is not always a good idea to actually assign it a guess. For reasons that will become apparent in \cref{xl:limit}, our algorithm will only assign guesses in a \emph{well-spread} fashion, i.e., in such a way that at most one node per level is assigned a guess.

Suppose now that a node $v$ has been assigned a guess. We define its \emph{weight} $w(v)$ as the number of nodes in the subtree hanging from $v$ that have been assigned a guess (this includes $v$ itself). Recall that subtrees are determined by black edges only. We say that $v$ is \emph{heavy} if $w(v)\geq g(v)$.

\begin{lemma}\label{xl:limit}
In a well-spread assignment of guesses, if $w(v)>a(v)$, then some descendants of $v$ are heavy (the \emph{descendants} of $v$ are the nodes in the subtree hanging from $v$ other than $v$ itself).
\end{lemma}
\begin{proof}
Our proof is by well-founded induction on $w(v)$. Let $v_1$, $v_2$, \dots, $v_k$ be the ``immediate'' descendants of $v$ that have been assigned guesses. That is, for all $1\leq i\leq k$, no internal nodes of the black path with endpoints $v$ and $v_i$ have been assigned guesses.

By the basic properties of history trees, $a(v)\geq \sum_i a(v_i)$. Also, the induction hypothesis implies that $w(v_i) \leq a(v_i)$ for all $1\leq i\leq k$. Therefore, $w(v)-1 = \sum_i w(v_i) \leq \sum_i a(v_i)\leq a(v) \leq w(v)-1$. It follows that $w(v_i)=a(v_i)$ and $a(v)=\sum_i a(v_i)$.

Let $v_d$ be the deepest of the $v_i$'s, which is unique, since the assignment of guesses is well spread. Note that $v_d$ has no siblings at all, otherwise we would have $a(v)>\sum_i a(v_i)$. Due to \cref{xl:guess2}, we conclude that $g(v_d)=a(v_d)=w(v_d)$, and so $v_d$ is heavy.
\end{proof}

\mypar{Correct guesses.} We say that a node $v$ has a \emph{correct} guess if $v$ has been assigned a guess and $g(v)=a(v)$. The next lemma gives a criterion to determine if a guess is correct.

\begin{lemma}\label{xl:crit}
In a well-spread assignment of guesses, if a node $v$ is heavy and no descendant of $v$ is heavy, then $v$ has a correct guess.
\end{lemma}
\begin{proof}
Because $v$ is heavy, $g(v)\leq w(v)$. Since $v$ has no heavy descendants, \cref{xl:limit} implies $w(v)\leq a(v)$. Also, by \cref{xl:guess2}, $a(v)\leq g(v)$. We conclude that $g(v)\leq w(v)\leq a(v)\leq g(v)$, and therefore $g(v)=a(v)$.
\end{proof}
When the criterion in \cref{xl:crit} applies to a node $v$, we say that $v$ has been \emph{counted}. So, counted nodes are nodes that have been assigned a guess, which was then confirmed to be correct.

\medskip
\mypar{Cuts and isles.}  Fix a view $\mathcal{V}$ of a history tree $\mathcal H$. A set of nodes $C$ in $\mathcal{V}$ is said to be a \emph{cut} for a node $v\notin C$ of $\mathcal{V}$ if two conditions hold: (i)~for every leaf $v'$ of $\mathcal{V}$ that lies in the subtree hanging from $v$, the black path from $v$ to $v'$ contains a node of $C$, and (ii)~no proper subset of $C$ satisfies condition~(i). A cut for the root $r$ whose nodes are all counted is said to be a \emph{counting cut} (see~\cref{xfig:mix}, right).

Let $s$ be a counted node in $\mathcal{V}$, and let $F$ be a cut for $v$ whose nodes are all counted. Then, the set of nodes spanned by the black paths from $s$ to the nodes of $F$ is called \emph{isle}; $s$ is the \emph{root} of the isle, while each node in $F$ is a \emph{leaf} of the isle (see~\cref{xfig:mix}, right). The nodes in an isle other than the root and the leaves are called \emph{internal}. An isle is said to be \emph{trivial} if it has no internal nodes.

If $s$ is an isle's root and $F$ is its set of leaves, we have $a(s)\geq \sum_{v\in F} a(v)$, because $s$ may have some descendants in the history tree $\mathcal H$ that do not appear in the view $\mathcal{V}$. If equality holds, then the isle is said to be \emph{complete}; in this case, we can easily compute the anonymities of all the internal nodes by adding up anonymities starting from the nodes in $F$ and working our way up to $s$.

\medskip
\mypar{Algorithm overview.} Our counting algorithm repeatedly assigns guesses to nodes based on known anonymities (starting from the nodes corresponding to the leader). Eventually some nodes become heavy, and the criterion in \cref{xl:crit} causes the deepest of them to become counted. In turn, counted nodes eventually form isles; the complete isles are marked as counted, which gives rise to more guessers, and so on. In the end, if a counting cut has been created, a simple condition determines whether the anonymities of its nodes add up to the correct count, $n$.

\medskip
\mypar{Algorithm details.} The algorithm takes as input a view $\mathcal V$ (which, we recall, is the history of a process), and uses flags to mark nodes as ``guessed'' or ``counted''; initially, no node is marked. Thanks to these flags, we can check if a node $u\in \mathcal V$ is a guesser: let $u_1$, $u_2$, \dots, $u_k$ be the children of $u$ that are also in $\mathcal V$ (recall that a view does not contain all nodes of a history tree); $u$ is a \emph{guesser} if and only if it is marked as counted, all the $u_i$'s are marked as counted, and $a(u)=\sum_i a(u_i)$.

The algorithm will ensure that nodes marked as guessed are well-spread at all times; if a level of $\mathcal V$ contains a guessed node, it is said to be \emph{locked}. A level $L_t$ is \emph{guessable} if it is not locked and has a guessable non-counted node $v$, i.e., there is a guesser $u$ in $L_{t-1}$ and the red edge $\{v,u\}$ is present in $\mathcal V$ with positive multiplicity. The counting algorithm is as follows:

\medskip
\mypar{Step~1.} Assign anonymity~$1$ to all nodes corresponding to the leader, and mark them as counted.

\medskip
\mypar{Step~2.} If there are no guessable levels, go to Step~6. Otherwise, let $v$ be a guessable non-counted node of smallest depth. Assign $v$ a guess $g(v)$ as in \cref{xe:guess}, and mark $v$ as guessed.

\medskip
\mypar{Step~3.} Scan the black path from $v$ to $r$, incrementing the weight of the guessed nodes encountered. If along the way a heavy node $v'$ is created, immediately go to Step~4; otherwise, go to Step~2.

\medskip
\mypar{Step~4.} Assign the anonymity $a(v')=g(v')$ to $v'$, and mark it as counted (and no longer guessed). If $v'$ has now become the root or a leaf of a complete isle $I$, go to Step~5; otherwise, go to Step~2.

\medskip
\mypar{Step~5.} Compute the anonymities of all the internal nodes of the isle $I$, and mark them as counted. If some of them are guessed, unmark them and update all weights accordingly. Go to Step~2.

\medskip
\mypar{Step~6.} If there are no counting cuts, output ``N/A'' without entering a terminal state. Otherwise, let $C$ be a counting cut such that the deepest node $v^\ast\in C$ has minimum depth among all counting cuts. Let $L_t\ni v^\ast$, let $n'=\sum_{v\in C}a(v)$, and let $t'$ be the current round (note that $t'$ can be inferred from the height of $\mathcal V$). If $t'\geq t+n'$, output the anonymities of the nodes in $L_0$ and terminate; otherwise, output ``N/A'' without entering a terminal state.

\medskip
\mypar{Invariants.} We argue that the above algorithm maintains some \emph{invariants}, i.e., conditions that are satisfied whenever Step~2 is reached. Namely, (i)~the nodes marked as guessed are well spread, (ii)~there are no heavy nodes, and (iii)~all complete isles are trivial.

These can be verified by induction: (i)~the algorithm never makes a guess in a locked level; (ii)~as soon as a new guess in Step~2 creates some heavy nodes, the deepest one becomes counted, making all other nodes non-heavy (in Step~5, weights may only decrease, and no heavy nodes are created); (iii)~as soon as a complete isle $I$ is created due to a node $v'$ being marked as counted in Step~4, it is immediately reduced to a set of trivial isles.

The invariants also imply that Steps~3 and~4 are correct: if no nodes are heavy to begin with, the new guessed node $v$ may create heavy nodes only on the black path from $v$ to $r$. Thus, the first heavy node along this path has a correct guess due to \cref{xl:crit}. Furthermore, since the anonymities assigned in Step~4 are correct, then so are those assigned in Step~5.

\medskip
\mypar{Terminating condition.} We will now prove that Step~6 is correct: that is, the algorithm indeed gives the correct output if the terminating condition is met. We already know that the anonymities computed for the nodes of the counting cut $C$ are correct, and hence we only have to prove that the set $P$ of processes represented by the nodes of $C$ includes all processes. Assume the contrary; \cref{xl:propamain} implies that, if $t'\geq t+|P|=t+n'$, there is a node $z\in L_t$ representing some process in $P$. Thus, the black path from $z$ to the root $r$ does not contain any node of $C$, contradicting the fact that $C$ is a counting cut whose deepest node is in $L_t$. So, the terminating condition is correct.

\medskip
\mypar{Running time.} We have proved that the algorithm is correct; we will now study its running time.

\begin{lemma}\label{xl:bound1}
Whenever Step~2 is reached, at most $n-1$ levels are locked.
\end{lemma}
\begin{proof}
We will prove that, if the subtree hanging from a vertex $v$ of $\mathcal V$ contains more than $a(v)$ guessed nodes, then it contains a guessed node $v'$ such that $w(v')>a(v')$. The proof is by well-founded induction based on the subtree relation in $\mathcal V$. If $v$ is guessed, then we can take $v'=v$. Otherwise, by the pigeonhole principle, $v$ has at least one child $u$ whose hanging subtree contains at least $a(u)$ guessed nodes. Thus, $v'$ is found in this subtree by the induction hypothesis.

Assume for a contradiction that at least $n$ levels of $\mathcal V$ are locked; hence, $\mathcal V$ contains at least $n$ guessed nodes. None of the nodes representing the leader is ever guessed, because all of them are marked as counted in Step~1. Hence, all of the guessed nodes must be in the subtrees hanging from the nodes in $L_0$ representing non-leader nodes, whose total anonymity is $n-1$. Thus, by the pigeonhole principle, the subtree of one such node $v\in L_0$ contains more than $a(v)$ guessed nodes. The node $v$ must be in $\mathcal V$, and therefore the subtree hanging from $v$ contains a guessed node $v'$ such that $w(v')>a(v')$.

Since the algorithm's invariant~(i) holds, we can apply \cref{xl:limit} to $v'$, which implies that there exist heavy nodes. In turn, this contradicts invariant~(ii). We conclude that at most $n-1$ levels are locked.
\end{proof}

\begin{lemma}\label{xl:bound2}
Assume that all the levels of the history tree up to $L_t$ are entirely contained in the view $\mathcal V$. Then, whenever Step~2 is reached and a counting cut has not been created yet, there are at most $n-2$ levels in the range from $L_1$ to $L_{t}$ that lack a guessable non-counted node.
\end{lemma}
\begin{proof}
Step~1 creates guessers in every level from $L_0$ to $L_{t-1}$ (these are the nodes representing the leader); hence, these levels must have a non-empty set of guessers at all times. Consider a level $L_i$ with $1\leq i\leq t$, and assume that all the guessable nodes in $L_i$ are already counted. Let $S$ be the set of guessers in $L_{i-1}$; note that not all nodes in $L_{i-1}$ are guessers, or else they would form a counting cut. The network is 1-interval-connected, so there is a red edge $\{u,v\}$ (with positive multiplicity) such that $u\in S$ and the parent of $v$ is not in $S$. By definition, the node $v$ is guessable; therefore, it is counted. Also, since the parent of $v$ is not a guesser, $v$ must have a non-counted parent or a non-counted sibling; note that such a non-counted node is in $\mathcal V$.

We have proved that every level (up to $L_{t}$) lacking a guessable non-counted node contains a counted node $v$ having a parent or a sibling that is not counted: we call such a node $v$ a \emph{bad} node. To conclude the proof, it suffices to show that there are at most $n-2$ bad nodes up to $L_{t}$.

We will prove by induction that, if a subtree $\mathcal W$ of $\mathcal V$ contains the root $r$, no counting cuts, and no non-trivial isles, then $\mathcal W$ contains at most $f-1$ bad nodes, where $f$ is the number of leaves of $\mathcal W$ not representing the leader. The base case is $f=1$, which holds because any counted non-leader node in $\mathcal W$ gives rise to a counting cut. For the induction step, let $v$ be a bad node of maximum depth in $\mathcal W$. Let $(v_1, v_2,\dots, v_k)$ be the black path from $v_1=v$ to the root $v_k=r$, and let $1<i\leq k$ be the smallest index such that $v_i$ is a branching node ($i$ must exist, because $v_k$ branches into leader and non-leader nodes). Let $\mathcal W'$ be the tree obtained by deleting the black edge $\{v_{i-1},v_i\}$ from $\mathcal W$, as well as the subtree hanging from it. Notice that the induction hypothesis applies to $\mathcal W'$: since $v_1$ is counted, and none of the nodes $v_2$, \dots, $v_{i-1}$ are branching, the removal of $\{v_{i-1},v_i\}$ does not create counting cuts or non-trivial isles. Also, $v_2$ is not counted (unless perhaps $v_2=v_i$), because $v_1$ is bad. Furthermore, none of the nodes $v_3$, \dots, $v_{i-1}$ is counted, or else $v_2$ would be an internal node of a (non-trivial) isle. Therefore, $\mathcal W'$ has exactly one less bad node than $\mathcal W$ and at least one less leaf; the induction hypothesis now implies that $\mathcal W$ contains at most $f-1$ bad nodes.

Observe that the subtree $\mathcal V'$ of $\mathcal V$ formed by all the levels up to $L_{t}$ satisfies all of the above conditions, as it contains the root $r$ and has no counting cuts, because a counting cut for $\mathcal V'$ would be counting cut for $\mathcal V$, as well. Also, invariant~(iii) ensures that $\mathcal V'$ contains no non-trivial complete isles. However, since the levels up to $L_{t}$ are contained in $\mathcal V'$, all isles in $\mathcal V'$ are complete, and thus must be trivial. We conclude that, if $\mathcal V'$ has $f$ non-leader leaves, it contains at most $f-1$ bad nodes. Since the non-leader leaves of $\mathcal V$ induce a partition of the $n-1$ non-leader processes, we have $f\leq n-1$, implying that the number of bad nodes up to $L_{t}$ is at most $n-2$.
\end{proof}

\begin{theorem}\label{xth:term}
There is an algorithm that solves the Generalized Counting problem in 1-interval-connected anonymous dynamic networks with a leader and terminates in at most $3n-2$ rounds.
\end{theorem}
\begin{proof}
It suffices to prove that, if a process $p$ executes the above algorithm at round~$t'=3n-2$, it terminates. By \cref{xl:propa}, the levels of the history tree up to $L_{2n-1}$ completely appear in $p$'s history. Due to \cref{xl:bound1,xl:bound2}, as long as a counting cut has not been created, at least one level between $L_1$ and $L_{2n-2}$ is guessable (because at most $n-1$ levels are locked, and at most $n-2$ levels lack a guessable non-counted node). Hence, when Step~6 is reached, a counting cut $C$ has been found whose deepest node is in $L_t$, with $t\leq 2n-2$. The level $L_t$ is completely contained in $p$'s history, and therefore $n'=\sum_{v\in C}a(v)=n$. Thus, $t'=3n-2\geq t+n'$, and $p$ terminates.
\end{proof}

Our analysis of the algorithm is tight: In \cref{as:3.2}, we will show that there are 1-interval-connected networks where the algorithm terminates in exactly $3n-3$ rounds.

\section{Negative Results}\label{xs:5}
\mypar{Unsolvable Problems.}
We will now prove that the multi-aggregation problems introduced in \cref{xs:2} are the only problems that can be solved deterministically in anonymous networks.

\begin{theorem}\label{xth:imp}
No problem other than the multi-aggregation problems can be solved deterministically, even when restricted to simple connected static networks with a unique leader.
\end{theorem}
\begin{proof}
Let us consider the (static) network whose topology at round~$t$ is the complete graph $G_t=K_n$, i.e., each process receives messages from all other processes at every round. We can prove by induction that all nodes of the history tree other than the root have exactly one child. This is because any two processes with the same input always receive equal multisets of messages, and are therefore always indistinguishable. Thus, the history tree is completely determined by the multiset $\mu$ of all processes' inputs; moreover, a process' history at any given round only depends on the process' own input and on $\mu$. By \cref{xth:view}, this is enough to conclude that if a process' output stabilizes, that output must be a function of the process' own input and of $\mu$, which is the defining condition of a multi-aggregation problem.
\end{proof}

\mypar{Lower Bound.} We now prove a lower bound of roughly $2n$ rounds on the Counting problem. In \cref{as:3.3}, we will provide additional (albeit weaker) lower bounds for several other problems.

We first introduce a family of 1-interval-connected dynamic networks. For any $n\geq 1$, we consider the dynamic network $\mathcal G_n$ whose topology at round~$t$ is the graph $G^{(n)}_t$ defined on the system $\{p_1,p_2,\dots,p_n\}$ as follows. If $t\geq n-2$, then $G^{(n)}_t$ is the path graph $P_n$ spanning all processes $p_1$, $p_2$, \dots, $p_n$ in order. If $1\leq t\leq n-3$, then $G^{(n)}_t$ is $P_n$ with the addition of the single edge $\{p_{t+1}, p_n\}$. We assume $p_1$ to be the leader and all other processes to be anonymous.

\begin{theorem}\label{xth:lower}
No deterministic algorithm can solve the Counting problem in less than $2n-6$ rounds, even  in a simple 1-interval-connected dynamic network, and even if there is a unique leader in the system. If termination is required, the bound improves to $2n-4$.\footnote{A slightly better bound can be obtained if we allow networks with double links. If we double the edge $\{p_{n-1}, p_n\}$ in $G^{(n)}_{n-2}$ and we add a double self-loop on $p_n$ in $G^{(n)}_t$ for all $t\geq n-1$, we obtain a lower bound of $2n-3$ rounds for stabilization and $2n-1$ rounds for termination.}
\end{theorem}
\begin{proof}
Let us consider the network $\mathcal G_n$ as defined above. It is straightforward to prove by induction that, at round~$t\leq n-3$, the process $p_{t+1}$ gets disambiguated, while all processes $p_{t+2}$, $p_{t+3}$, \dots, $p_n$ are still indistinguishable. So, the history tree of $\mathcal G_n$ has a very regular structure, which is illustrated in \cref{xfig:lower1,xfig:lower2}. By comparing the history trees of $\mathcal G_n$ and $\mathcal G_{n+1}$, we see that the leaders of the two systems have isomorphic histories from round~$1$ up to round~$2n-5$. Thus, by \cref{xth:view}, both leaders must have equal internal states and give equal outputs up to round~$2n-5$.

It follows that, if the leader of $\mathcal G_n$ were to output the number $n$ and terminate in less than $2n-4$ rounds, then the leader of $\mathcal G_{n+1}$ would do the same, producing the incorrect output $n$ instead of $n+1$. Assume now that the leaders of $\mathcal G_n$ and $\mathcal G_{n+1}$ could stabilize on the correct output in less than $f(n)=2n-6$ and $f(n+1)=2(n+1)-6=2n-4$ rounds, respectively. Then, at round~$2n-5$, the two leaders would output $n$ and $n+1$ respectively, contradicting the fact that they should give the same output.
\end{proof}

\mypar{Acknowledgments.} The authors would like to thank Gregory Schwartzman for useful comments.

\section{Conclusion}
We have presented an algorithm for the Generalized Counting problem that terminates in $3n-3$ rounds, which allows the linear-time computation of all deterministically computable functions in 1-interval-connected anonymous dynamic networks with a leader.

We also proved a lower bound of roughly $2n$ rounds for the Counting problem, both for terminating and for stabilizing algorithms, and we gave a stabilizing algorithm for Generalized Counting that matches the lower bound (up to a small additive constant). Determining whether the $2n$ lower bound is tight for terminating algorithms is left as an open problem.

We introduced the novel concept of \emph{history tree} as our main investigation technique. History trees are a powerful tool that completely and naturally captures the concept of symmetry in anonymous dynamic networks. We have demonstrated the effectiveness of our methods by optimally solving a wide class of fundamental problems, and we argue that our technique can be used in similar settings, such as the case with multiple leaders, as well as to obtain better bounds for networks with special topologies or randomly generated networks.

\clearpage
\begin{figure}
\centering
\includegraphics[scale=0.575]{pics/c2.pdf}
\caption{The first rounds of the dynamic network $\mathcal G_n$ used in \cref{xth:lower} (left) and the corresponding levels of its history tree (right), where $n=6$; the process in blue is the leader. The white nodes and the dashed edges in the history tree are not in the history of the leader at round~$7$. The labels $p_1$, \dots, $p_6$ have been added for the reader's convenience, and mark the processes that get disambiguated, as well as their corresponding nodes of the history tree, which have anonymity~$1$.}
\label{xfig:lower1}
\end{figure}

\begin{figure}
\centering
\includegraphics[scale=0.575]{pics/c3.pdf}
\caption{The first rounds of the dynamic network $\mathcal G_{n+1}$ with $n=6$. Observe that the history of the leader at round~$7$ is isomorphic to the history highlighted in \cref{xfig:lower1}. The intuitive reason is that, from round~$1$ to round~$n-3$, both networks have a cycle whose processes are all indistinguishable (and are therefore represented by a single node in the history tree), except for the one process with degree~$3$. Thus, the history trees of $\mathcal G_n$ and $\mathcal G_{n+1}$ are identical up to level~$n-3$. After that, the two networks get disambiguated, but this information takes another $n-3$ rounds to reach the leader. Therefore, if the leader of $\mathcal G_n$ and the leader of $\mathcal G_{n+1}$ execute the same algorithm, they must have the same internal state up to round~$2n-5$, due to \cref{xth:view}. In particular, they cannot give different outputs up to that round, which leads to our lower bounds on stabilization and termination for the Counting problem.}
\label{xfig:lower2}
\end{figure}

\clearpage

\begin{center}
\begin{LARGE}
\textbf{APPENDIX}
\end{LARGE}
\end{center}

redpajama_set_name: RedPajamaArXiv


text:
Daily readings about people, places, and events in American history.
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May 3rd
How the D.C. Got in Washington, D.C.
In 1790, a year after George Washington took office as president, Congress authorized him to find a site along the Potomac River for the new nation's capital. It was the first time a country had ever established its permanent capital by legislative action. The president ended up choosing a spot just a few miles upstream from his home at Mount Vernon, Virginia.
Surveyors staked out an area of one hundred square miles straddling the river. The idea was to create a special territory, not part of any state, to contain the capital city. The land came from Maryland and Virginia, and the territory was named the "District of Columbia" ("D.C." for short) in honor of Christopher Columbus.
George Washington hired French engineer Pierre L'Enfant to plan the city that would lie within the new District. In 1791, the District's commissioners decided to name that city "Washington" in honor of the first president. The federal government moved there in 1800.
On May 3, 1802, Washington was incorporated as a city, with a city council elected by local residents, and a mayor appointed by the president. People began to refer to the capital city inside the District of Columbia as "Washington, D.C." – just as they might write "Albany, N.Y." or "Charleston, S.C."
For a long time Washington remained a relatively small town, and much of the land inside the District of Columbia lay undeveloped. In 1846 Congress decided it would never need the District's land on the south side of the Potomac River, so it returned that portion to the state of Virginia. But of course the city did eventually grow, especially after World War II. Today it fills virtually the entire District of Columbia.
American History Parade
The first U.S. medical school is established at the College of Philadelphia.
Washington, D.C., is incorporated as a city.
West Virginia becomes the first state to impose a sales tax.
Air Service Lts. Oakley Kelly and John Macready land at Coronado Beach, California, completing the first transcontinental nonstop flight in 26 hours, 50 minutes.
Margaret Mitchell wins the Pulitzer Prize for her novel Gone with the Wind.
Lt. Cols. William Pershing Benedict and Joseph Otis Fletcher pilot the first plane to land at the North Pole, an Air Force ski-wheeled C-47.
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text:
Christians of the Holy Land – Exodus, Disintegration, and Ideological Necrophilia
Franck Salameh
Zurich, May 22, 2017
Franck Salameh: Christians in Holy Land Face "Existential Dangers"
Boston College Professor Sees Oldest Christian Community "Caught in Crossfire"
In a recent lecture here, Professor Franck Salameh of Boston College described the history and current plight of Christians in the "Holy Land" (Israel and the Palestinian Territories.) Along with other Christians in the Middle East, Holy Land Christians, the oldest Christian communities in the world, today face "existential dangers," Salameh said. "Their problems preceded the Arab–Israeli conflict," he argued. "Their plight is the sequel and the natural continuation of a long process of erasure and annihilation."
In his lecture, provocatively entitled, "Christians of the Holy Land – Exodus, Disintegration, and Ideological Necrophilia," Salameh drew attention to the tendency in the media and academia to impose new identities on Holy Land dwellers, using exclusively the language of modern nationalism, and thereby negating historic identities that are still alive, well and politically potent. In the case of Christians, the now-obsolete terms of Arab nationalism – "Arabs" or "Palestinians" – are still used for them collectively, whereas they historically identified themselves in terms of religious community.
Today, there is a tendency among Holy Land Christians to cast off these names and revive the ethnoreligious ones that predate the failed project of Arab nationalism. Salameh suggested that those writing and speaking about the Christians of the region should use the national appellations that these religious communities used for themselves – Greek Orthodox, Copts, Chaldeans, Maronites, Syriacs, etc. – long before Arab nationalist ideology was imposed on them. Doing so, he argued, will help make sense of their current plight.
Salameh went on to describe the history of the Holy Land's Christians from their origins in the time of the New Testament, through the Islamic conquests, the various Islamic empires, the Crusades, British rule, and finally, the Arab-Israeli conflict. "Under Muslim rule," Salameh said, "Christians of the Holy Land gradually became a minority in their ancestral homeland, suffering the exactions of dhimmitude –the legal restrictions that Islam placed on non-Muslims – causing their status and numbers to deteriorate over the course of 14 centuries."
As a result, "Christians of the Holy Land are today a mere shadow of their pre-Muslim Conquest existence," Salameh said. "They are barely two percent of the total population – roughly 160,000 strong in Israel proper and 40,000 in the West Bank and Gaza."
Within the borders of Israel, Salameh said, Christians have a "tenuous," "vulnerable" status, their relations with the Jewish majority being "crippled by intransigence and resentments of the Arab–Israeli conflict." Nevertheless, Israel is a "safe haven" for Christians in a dangerous region. "Israel has since its founding recorded an increase in its Christian population," Salameh noted, "even as the numbers of
Christians continue to dwindle elsewhere in the Middle East." Today, they are "the most enterprising, open-minded, and prosperous Arabophone element in Israeli society."
In contrast, in territory controlled by the Palestinian Authority and Hamas, dhimmitude conditions persist, and Christians "may indeed be on the verge of perdition," Salameh said. "Faced with daily offenses ranging from the wanton defacement of their institutions, to public displays of contempt, to intimidation, extortion, and pressure to abide by Islamic law," thousands of Christians have chosen to emigrate from the land altogether. The town of Bethlehem, Salameh said, was 90% Christian through the 1960s; today, under Palestinian Authority rule, Christians are less than 30%.
It is common today to attribute the crisis for Holy Land Christians entirely to the troubles caused by the Arab-Israeli conflict. Salameh resoundingly rejected this approach, pointing out that Middle Eastern Christians were dispossessed long before the creation of Israel, and still face extermination in countries where the conflict is not a factor, like Iraq, Syria and Egypt. Instead, he said, the crisis for Christians in the Holy Land and the rest of the Middle East calls for "straightforward, honest inquiries into their history of dispossession."
"Beyond the platitudes, and beyond the perfunctory expressions of concern and sympathy," Salameh concludes, Holy Land Christians "need answers as to whether or not their cultures, their civilizations, their histories, and their ancestral languages are worth saving."


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News briefs:May 17, 2010
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Wikinews interviews a Restore the Fourth organizer
A grassroots movement known as Restore the Fourth, dedicated to the protection of the 4th Amendment to the US Constitution, are to hold protests countrywide on July 4. The planned protests come in the wake of information about NSA surveillance leaked last month, notably the PRISM surveillance program and the collection of Verizon phone records. Wikinews interviewed Jett, a national organizer from this recently created movement.
((Wikinews)) First of all, could you explain what Restore the Fourth is all about?
Jett: At its core, RestoreTheFourth is about protecting citizens' constitutional rights. Specifically, we're dedicated to bringing awareness and action to the expanding overreach and elimination of the 4th Amendment to the US Constitution.
((WN)) What is your role at Restore the Fourth?
Jett: My job at RestoreTheFourth could be summarized as 'project coordinator'. Every person who wants to help can help in a tremendous way. I simply make sure that their skills get used in a way that would be most beneficial to the movement: web development, public relations, etc. I also field questions from the press and promote knowledge of our cause.
((WN)) What are your plans for direct action, outreach, etc.?
Jett: Our press release includes a list of 'demands' for what we want to see in order to restore our privacy rights, including reform of the PATRIOT Act and increased accountability for public officials. In the very short term, these protests and demonstrations bring awareness to the issue, something that's really important in enacting reform. In the long term, however, we expect to create a legal organization dedicated to restoring these rights inherent to every American. By partnering with various other organizations that share our moral values, we can further these goals.
On July 4, we will have over 100 protests in all 50 states, showing that the citizens of America are truly serious about protecting their rights.
((WN)) By what means do you hope to achieve such change?
Jett: This movement started only a few weeks ago, and since then we've experienced exponential growth and progress. Since the movement is still very young, plans diverge in the long term on what we hope to achieve. Personally, I'd like to see a combination of legislative and litigative action (something like what the ACLU does), and others want to see further plans of action. With organizations such as the BORDC, stopwatching.us and the EFF behind us, I feel that we can achieve all of this and much more.
What do you think is the right balance between surveillance and privacy?
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((WN)) Is the movement US-only or will it extend to other jurisdictions as well? Do you think it would be fair for the US to spy on non-citizens?
Jett: I believe that rights are inherent to all humans, not only United States citizens. In the long term I'd certainly like to see people of all nations protected from the slow elimination of privacy that we're all experiencing.
He's [Edward Snowden] being treated as a 'martyr' of sorts. It seems to distract from what he truly believed in.
((WN)) What do you think about Edward Snowden's whistleblowing?
Jett: I think that too much attention is being given to his personality instead of what he fought for. He's being treated as a 'martyr' of sorts. It seems to distract from what he truly believed in — transparency for the government and inherent privacy for all Americans.
((WN)) What do you think about his future, given the legal grey zone in which he currently is?
Jett: Hard to say. He may be captured by any number of agencies, or he may live a free man. Whatever happens, he has the eyes of millions of people on him, all of whom will yell very loudly if anything occurs.
((WN)) Thank you very much for your time.
Jett: Thanks for the opportunity.
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