text:
## Hunger

For intuitive and critical discussions, from spirituality to theological doctrines. Fair warning: because the subject matter is personal, moderation is strict.

Moderator: Dan~

### Re: Hunger

And to you!!
I will be bitchy, cranky, sweet, happy, kind, pain in the ass all at random times from now on. I am embracing my mentalpause until further notice. Viva lack of total control!!!!!!!!!!!!!!!!!!! This is not a test,,, this is my life right now. Have a good day and please buckle up for safety reasons,, All those in high chairs, go in the back of the room.

Kriswest
ILP Legend

Posts: 20554
Joined: Thu Dec 22, 2005 2:26 pm
Location: stuck in permanent maternal mode.

### Re: Hunger

As consumers, are we able to make the distinction between natural and unnatural hungers?
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Depends upon health and upbringing
I will be bitchy, cranky, sweet, happy, kind, pain in the ass all at random times from now on. I am embracing my mentalpause until further notice. Viva lack of total control!!!!!!!!!!!!!!!!!!! This is not a test,,, this is my life right now. Have a good day and please buckle up for safety reasons,, All those in high chairs, go in the back of the room.

Kriswest
ILP Legend

Posts: 20554
Joined: Thu Dec 22, 2005 2:26 pm
Location: stuck in permanent maternal mode.

### Re: Hunger

Even conservative "Christians" are opposed to socialized medicine or universal health care. Since their opinion is a majority one in the US, little will be done to insure proper health for those considered undeserving of help. The wealthy can afford the best health care money can buy. The poor cannot. It's economic eugenics. Let the weak die off so that only the strong survive.
The argument for feeding and taking care of the poor as a problem of overpopulation is lame. We have yet to farm the oceans. We can, with greenhouses. grow food in arid land. There are so many things we could do to make the life experience beneficial to all; but we choose not to do it because of the lie that helping others somehow is a deficit of our own sense of entitlement.
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Search your heart. What do you really think/feel about the poor in America? I am one. But what I have is more than a majority of humans have. What, if anything, do we owe to each other?
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Ierrellus wrote:Search your heart. What do you really think/feel about the poor in America? I am one. But what I have is more than a majority of humans have. What, if anything, do we owe to each other?

Define poor, by your standards, Ierrellus, without giving away your life.
You don't live out in the street, as far as I know. Do you go hungry everyday? Is your apartment (if you do not live out in the street) warm enough for you to be content?
Is your perspective one of having more than enough or of having enough or never enough?

You compare yourself to the poor in America -- relative to whom?
"Look closely. The beautiful may be small."

"Two things fill the mind with ever new and increasing admiration and awe, the oftener and more steadily we reflect on them: the starry heavens above me and the moral law within me."

“Whereas the beautiful is limited, the sublime is limitless, so that the mind in the presence of the sublime, attempting to imagine what it cannot, has pain in the failure but pleasure in contemplating the immensity of the attempt.”

Immanuel Kant

Arcturus Descending
Consciousness Seeker

Posts: 15689
Joined: Sat Sep 06, 2008 5:15 pm
Location: A state of unknowing

### Re: Hunger

Chakra Superstar
Philosopher

Posts: 1199
Joined: Sat Apr 07, 2012 10:42 am

### Re: Hunger

Kriswest wrote:I disagree about all being free and if you eat feces and drink urine that is all on you no one else.

I don't care that you disagree. You disagreeing means nothing to me. You're telling me that if everything was free, as in, you would still have personal belongings to make your own and give your unique touches to, but could get anything you wanted from a store for free, while being able to expect that what you made your own would not be stolen since everyone else would be able to do the same, that you would not choose to work for free if it were impressed upon you that certain jobs needed to be performed for you to keep having those free things? Keep in mind that such things do not last forever and eventually new ones would have to be made through production and manufacturing. Not to mention the jobs of cleaning up garbage, clearing septic systems, etc. Even if everyone went to college and received degrees of their education, those jobs would still need to be done.

At the point of you saying 'no, I would not work for free to have free things and I perceive myself to be too good for those jobs,' then you leave it open for others to say those things as well and create the same selfish system that we have now. At the point of saying that you disagree, all it really means is that you have not thought it out well at all and gave a customary and usual response; in bullshit fashion and thus causing me to eat more feces whether it is my choice to or not; and therefore anything you say after such a point of reasonable conjecture is lacking reason and sense, thus discrediting anything you might say of actual value.

Now, you want to say that if I eat feces and drink urine that it is all on me. Now, that was a metaphorical statement on my behalf and not a literal one, which you knew full-well before responding in the dismissive manner that you did. However, it is not all on me. I take my fair share of the responsibility and avoid it as much as possible, but it so widespread throughout society that even you, here and in this place, can not avoid serving me up shit on a platter for my mornings consumption as I digest it and seek to find a more suitable means of response other than of agitation, which I have done. So, my response is on me, since my most usual response to such bullshit is to give vehemence and such dismissive attitude in return; to act as a mirror of your own reaction. But, my response is still dictated somewhat by what I am given to digest and feed upon.

It is not my fault that I came here to discuss philosophy seriously and intelligently since that is what interests me in life. In fact, even while I take responsibility for that, it is not a bad thing and does not denote that I should be given such shit to sate my palate. It is similarly not my fault that you took my response as shallow and responded to it as you did, thus giving me the shit to sate my palate, which sate it did not. It is similarly not my priority to stop coming here nor to stop discussing philosophy or life itself, nor to stop putting on the table reasonable and sensible methods of solving societal concerns with actual things people could agree to if freed from the unreasonable and nonsensical viewpoints thus imported to them by society.

In fact, it is more your fault, in this instance, that I am given such shit to feast upon and I do hope you take your own medicine and advice and take your share of the responsibility, for I gave you something better to eat than you gave to me.

You once told me to work within this decadent system to fix it and I once told you that that's exactly what I was doing. You meant for me to get into politics, etc. and to actually make a difference along those terms and yet I am already part of the system as a conscientious and informed citizen; from birth I am enslaved to this system while told that I am free and the system has done little to actually take care of me. No, it has come down to the kindness of rare individuals that I am really coming to appreciate that gets me through most days. Since I am still a part of this system, regardless of choice, I have been working within it to make changes and fixes, yet what I meant at the point of that response was that I was working within the mental aspect of the system, which is also true.

I don't respect you as a philosopher nor do I respect the things that you have said or responded to me with, for they lack understanding and they lack so much more than mere understanding.

For me being homeless and without a job, I am richer than you; I have more things that can not be taken away and I will shred every single bit of every faulty argument that crosses my path in the pursuit of my goals to create a better world for future generations before I die, just to give my pain and suffering a reason for being, to have pushed me into a state of wisdom and sense to actually pay attention and listen to people instead of being just like everyone else.

I would still rather be a nothing than a piece-of-shit something.
(Reality isn't so kind. Everything doesn't work out the way you want it to. That's why...) As long as you don’t get your hopes up, you can take anything... You feel less pain.

(Right and wrong are not what separate us and our enemies. It's our different standpoints, our perspectives that separate us. Both sides blame one another. There's no good or bad side. Just two sides holding different views.)

What do you think? To tell you the truth... I worry too much about what others think of me. I hate that side of me... That's why I didn't want anyone to get to know me. I wanted to hide that side of myself. I hate it.

The Eternal Warrior
Philosopher

Posts: 2571
Joined: Fri Aug 01, 2014 1:26 am

### Re: Hunger

Phred the Phukhead wrote:I would still rather be a nothing than a piece-of-shit something.

Reminds me of the chorus of an old song ... something to the effect ..."Freedom's just another word for "nothing left to lose"
"Do not be influenced by the importance of the writer, and whether his learning be great or small; but let the love of pure truth draw you to read. Do not inquire, “Who said this?” but pay attention to what is said”

Thomas Kempis 1380-1471

pilgrim-seeker_tom
Philosopher

Posts: 1858
Joined: Sun Dec 17, 2006 11:16 am

### Re: Hunger

Arcturus Descending wrote:
Ierrellus wrote:Search your heart. What do you really think/feel about the poor in America? I am one. But what I have is more than a majority of humans have. What, if anything, do we owe to each other?

Define poor, by your standards, Ierrellus, without giving away your life.
You don't live out in the street, as far as I know. Do you go hungry everyday? Is your apartment (if you do not live out in the street) warm enough for you to be content?
Is your perspective one of having more than enough or of having enough or never enough?

You compare yourself to the poor in America -- relative to whom?

I'm statistically poor. Otherwise, I have enough--decent, warm apartment, good food. Meanwhile there are people in our city who are homeless. I have much to be thankful for. So how does one decide what is more than enough?
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

"I'm just a soul whose intentions are good.
Oh Lord, don't let me be misunderstood."--Eric Burden
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

pilgrim_tom wrote:
Phred the Phukhead wrote:I would still rather be a nothing than a piece-of-shit something.

Reminds me of the chorus of an old song ... something to the effect ..."Freedom's just another word for "nothing left to lose"

I have plenty left to lose, but plenty more to gain through freedom. Freedom is more than just having nothing left to lose. It's looking at the world around you and how bullshit it is and deciding, 'you know what, I may not benefit from this, but if I can succeed, at least future generations can have better than I did and hopefully will be more appreciative for it if I truly do succeed.'

It's something to gain for more than just myself, which is the appeal to me. I didn't drag myself out of my darkened days for just myself, that would have been pointless, though I had to do it for myself to better be able to help others and the world around me. I did it for non-selfish means.
(Reality isn't so kind. Everything doesn't work out the way you want it to. That's why...) As long as you don’t get your hopes up, you can take anything... You feel less pain.

(Right and wrong are not what separate us and our enemies. It's our different standpoints, our perspectives that separate us. Both sides blame one another. There's no good or bad side. Just two sides holding different views.)

What do you think? To tell you the truth... I worry too much about what others think of me. I hate that side of me... That's why I didn't want anyone to get to know me. I wanted to hide that side of myself. I hate it.

The Eternal Warrior
Philosopher

Posts: 2571
Joined: Fri Aug 01, 2014 1:26 am

### Re: Hunger

Of course charity begins at home. You have to take care of your own needs before you can tend to those of anyone else.
Consumerism does not teach this; it depends on the bloating of your hungers in order to sell products you often do not need.
I'm reminded of the 1980s when kids demanded expensive designer jeans and tennis shoes in order to fit in with their peers. Our family couldn't afford that, so my wife ran up all credit cards out of peer pressure. The 80s reversed the 60s where individual expression determined what was bought. I don't know where minds are in century 21, except that consumerism still reigns supreme, that is taught hunger trumps actual hunger.
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Ierrellus,

So how does one decide what is more than enough?

Good question, Ierrellus.

The way I look at it, by taking an honest, in-depth look at the way other people live. We have to really "see" others, experience them. For instance, families living out in the street, and "feeling" and comparing their human experience to our own. Of course, I think it takes a sense of appreciation and gratitude for what we have. I don't really see a whole lot of that these days. The more we have, the more we want.
Their seems to be such a "hunger" out there for things because of the inner hungers which aren't satisfied or which were not often satisfied, because of the human fear which people experience and which they try to stifle and suppress with "things".

I think that it is human consciousness and fairness (compassion and empathy for those who do not have) which can also decide what is more than enough. Our human intuition can tell us this but we have to be willing to first "see" it then to affirm it. But gratitude I think is the "missing link" here. We're not often grateful because we are just not capable of seeing what holds "real" meaning in our lives.

Yes, it is a decision to "see" that we have more than enough, to voice that to ourselves, but the decision can only be made through conscious self-awareness and of those around us who have much less.

Maybe it's akin to pain. Some have a greater threshhold (sp) for pain than others - perhaps some have a greater threshhold for living with less - and it being MORE THAN ENOUGH. It's like the Charlie Brown christmas tree for me - simplicity in something is a beautiful thing - it's in the eye of the beholder - sometimes less is more in a way I can't quite figure out.

I also have a sense that when we can live with less we appreciate it more. Can we stuff and stuff and stuff our faces with good food and appreciate that more or less than when we eat more of what is good but eat it slowly and be more discriminate about what we're eating? okay, I'm getting carried away here.

We need to learn to be more epicurean when it comes to everyting and I sense that having less allows us a better appreciate for what we do have but that's just me.
"Look closely. The beautiful may be small."

"Two things fill the mind with ever new and increasing admiration and awe, the oftener and more steadily we reflect on them: the starry heavens above me and the moral law within me."

“Whereas the beautiful is limited, the sublime is limitless, so that the mind in the presence of the sublime, attempting to imagine what it cannot, has pain in the failure but pleasure in contemplating the immensity of the attempt.”

Immanuel Kant

Arcturus Descending
Consciousness Seeker

Posts: 15689
Joined: Sat Sep 06, 2008 5:15 pm
Location: A state of unknowing

### Re: Hunger

I actually had and still have quite a bit to lose. When I first started down this path, I knew the Devil's game already; I already knew the family member against family member, the friend against friend. Even closer to our perceived reality, I already knew that what I was getting myself into on the internet was attracting attention from various agencies around the world, both government and secret and I knew that they wouldn't hesitate to make me 'vanish', if they thought they could, that they would seek to rip me away from those who loved me, ruin my life, drive me into homelessness and drive me into reacting in ways that would make people stop caring about me or to drop communication with me.

I knew that by making the choices I was making that I was putting my family and friends at risk just the same as myself and made my decision anyway. I live everyday of my life with the knowledge that the woman that I love; that I am en route to go sweep off her feet if I can, is in more danger than any other person I could ever care about or love because of my infatuation and my enemies.

And, I kept myself. I kept contact with people, I still have people that care about me and miss me and it becomes too widespread for these agencies to actually do what they want to do since enough people care about those people and would miss them, too, that it becomes too much clean-up to do and they're not ready to play openly like that, at least not yet.

And then there's you all, here, that I keep in contact with while homeless and recount my experiences with and who I continue conversations with as if my life hasn't changed all that much and just taken on a fresh path; I made it align; and don't think that wasn't without intent on my part.

You all sit here, watching me go through life; through hardship; and not raising a finger to help. Not to judge, I didn't ask any of you for help and I don't intend to start. You also haven't hindered my progress and I look at that; it's not your responsibility to help me, either. But the fact remains that you sit here and watch and pay attention to more than most of you will admit to; not just to this message board, but to life in general, though certainly some members here do pay more attention to certain members and threads than they care to admit. Regardless, it keeps a line of communication open and steady enough to where if anything happened to me, you'd know about it. Whether friend or enemy, it would alert you that something happened. Either I died on the road in the cold or in nature due to some wild animal or some random thug, which is entirely possible but I also seem capable of handling myself through such situations already, or what you fear more actually happened, and I got wiped out for running my mouth against a system that hates to have loose lips around or loose ends flapping in the breeze. And, it chafes them.

I have more to lose than any person here, because I understand the gravity of my plight and appreciate all of the good given to me in my life to the fullest, more than any person here can truly understand. I've already lost everything, even the things I don't have, yet; at least in vision and emotion and feeling. Something tried to kill me all away and failed substantially in the past year and a half; or multiple somethings, which is far more likely.

Also, in response to the question even though it wasn't asked of me; when you react naturally to something in a violent seeming manner, then you know when it is more than enough. If you had ever been in such a situation where something has been more than you can handle without resorting to anger, impatience and violence, you wouldn't even have to ask the question. If you've ever even pushed someone else to a breaking point or been there your self, you wouldn't have to ask the question.

And, we only truly appreciate things more when we live with less because we have to. People today live in such excess that they lost sight of what it was like just 130 years ago. I just read a book about Butch Cassidy last night that highlighted some of the differences in what people go through then and now. They didn't even truly know that what they had was less and lived off the land to an extent; weren't afraid to go hungry or possibly freeze to death or get a bullet; at least not the people who had yet to be conformed by society into false living.

Realistically, when you have to live with less while so many others have more, or when you know you could have more, it does make you appreciate the more more, but then you realize that you're not even living bare minimum like people did 130 years ago and nobody truly is living with bare minimum and yet everyone is 'making only enough to make ends meet' because of what they choose to spend their money on, bills, etc. That makes me appreciate it more coupled with the fact that I have to do without the luxuries that I should have based on the resources available to the world at large. I should have the same things as everyone else, regardless of my appreciation for them.

And when you realize that appreciating things better does give better life satisfaction, you also realize that others don't truly appreciate what they have and the only way to share your brand of wealth is to take it all away.

At least, that's what I've, personally, come to realize. And, when you already appreciate life better than other people, even in luxury, you begin to get a little bit too angry enough to burst out of being angry to become calculating, instead, and thoughtful. And, if you have half a brain, you put off a cheery air, seem simple and stupid enough; maybe crazy; until the actual action begins to hit and then you become as dangerous as you actually are, and surprise the people who sorely underestimated you and that's what makes you more than dangerous; it makes you perilous.

Also, that deep appreciation of life and what exactly it is that you stand to lose, makes you lot more dangerous than your average idiot when you use your intellect correctly; and infinitely more dangerous than any evil thing could be. It's what makes legends like The Lord of the Rings and Aragorn seem perfectly plausible; that one man can be worth the weight of thousands of warriors in battle; that just by name alone, can be feared more than the might of an army. One man.
(Reality isn't so kind. Everything doesn't work out the way you want it to. That's why...) As long as you don’t get your hopes up, you can take anything... You feel less pain.

(Right and wrong are not what separate us and our enemies. It's our different standpoints, our perspectives that separate us. Both sides blame one another. There's no good or bad side. Just two sides holding different views.)

What do you think? To tell you the truth... I worry too much about what others think of me. I hate that side of me... That's why I didn't want anyone to get to know me. I wanted to hide that side of myself. I hate it.

The Eternal Warrior
Philosopher

Posts: 2571
Joined: Fri Aug 01, 2014 1:26 am

### Re: Hunger

Phred is such a Jesus, and Kris is such a Jew.
trogdor

Ultimate Philosophy 1001
BANNED

Posts: 8311
Joined: Tue Oct 20, 2015 10:57 pm

### Re: Hunger

From "Bear Creek 1998"

The best way to stop a revolution
Is to put it up for sale.
The market place solution
Is that fads will never fail
To bring in the customers daily
To buy what they cannot do
While the video, voyeur security
Sucks the life right out of you.

This is what happened to the youth movements of the 1960s. It's also what century 21 is all about in the USA.

"Buying and spending, we lay waste our powers;
Little we see in Nature that is ours."--Wordsworth

Without a sense of being part of Nature, we are imprisoned in our heads. Commerce gives us that sense of belonging by having, not by being. It is no wonder that lust for having creates have nots.
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

I admit I'm an old hippie. But I haven't seen here in the USA a better, or even an equal, sense of brotherhood than that expressed by the young people of the 60s. Consumerism dulls all ideals eventually. It is taught hunger that fits the ego better than natural hunger.
The wealth disparity now in the USA is 98 % poor, 2% wealthy. Doesn't that prove there is something wrong with a system that allows this to happen?
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Ierrellus wrote:I admit I'm an old hippie. But I haven't seen here in the USA a better, or even an equal, sense of brotherhood than that expressed by the young people of the 60s. Consumerism dulls all ideals eventually. It is taught hunger that fits the ego better than natural hunger.
The wealth disparity now in the USA is 98 % poor, 2% wealthy. Doesn't that prove there is something wrong with a system that allows this to happen?

"98% poor, 2% wealthy" - really?

Arminius
ILP Legend

Posts: 5732
Joined: Sat Mar 08, 2014 10:51 pm
Location: Saltus Teutoburgiensis

### Re: Hunger

Arminius wrote:
Ierrellus wrote:I admit I'm an old hippie. But I haven't seen here in the USA a better, or even an equal, sense of brotherhood than that expressed by the young people of the 60s. Consumerism dulls all ideals eventually. It is taught hunger that fits the ego better than natural hunger.
The wealth disparity now in the USA is 98 % poor, 2% wealthy. Doesn't that prove there is something wrong with a system that allows this to happen?

"98% poor, 2% wealthy" - really?

Yes. This amount of income inequality has not been experienced for 100 years.
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Ierrellus wrote:
Arminius wrote:
Ierrellus wrote:I admit I'm an old hippie. But I haven't seen here in the USA a better, or even an equal, sense of brotherhood than that expressed by the young people of the 60s. Consumerism dulls all ideals eventually. It is taught hunger that fits the ego better than natural hunger.
The wealth disparity now in the USA is 98 % poor, 2% wealthy. Doesn't that prove there is something wrong with a system that allows this to happen?

"98% poor, 2% wealthy" - really?

Yes. This amount of income inequality has not been experienced for 100 years.
Well, you can draw the line wherever you want so you can get practically any number ... 98%, 50%, 15%.
But by saying that 98% of the US is poor, you lose all credibility because even a cursory investigation shows that much more than 2% of the population is well-off and comfortable.

Also : Poor is the US is much different than poor in Bangladesh.
phyllo
ILP Legend

Posts: 12386
Joined: Thu Dec 16, 2010 1:41 am

### Re: Hunger

I'm not discussing Bangladesh.
About the wealth disparity in the USA, just google "wealth inequality in the US"; you'll find my claim not exorbitant.
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Ierrellus wrote:I'm not discussing Bangladesh.
About the wealth disparity in the USA, just google "wealth inequality in the US"; you'll find my claim not exorbitant.
There is a difference between wealth inequality and wealth.

For example, let's say there are 100 people ... one person has a yearly income of $1 billion and the others each have have an income of$100,000. All of them can live comfortably on their incomes. There is income inequality but the 99 are not poor.
phyllo
ILP Legend

Posts: 12386
Joined: Thu Dec 16, 2010 1:41 am

### Re: Hunger

About 20% of Am. families make a six figure income. One hundred thousand dollars per year would make a rich man of a single individual, but it reduces a family of four to ability to pay for the bare essentials. In the USA this family of four or more with a six figure income (100,000 per year) would be considered poor. The % of persons, families who make less than this is more than 20. So, what happened to the American dream--that anyone can pull themselves up by their own bootstraps and, with true grit and practical husbandry, get the rewards of prosperity?
"We must love one another or die." W.H.Auden
I admit I'm an asshole. Now, can we get back to the conversation?
From the mad poet of McKinley Ave.
Ierrellus
ILP Legend

Posts: 12994
Joined: Sat Jun 10, 2006 12:52 pm
Location: state of evolving

### Re: Hunger

Americans should have 2 kids, no more, anything else is habitat destruction.
trogdor

Ultimate Philosophy 1001
BANNED

Posts: 8311
Joined: Tue Oct 20, 2015 10:57 pm

PreviousNext

Return to Religion and Spirituality

### Who is online

Users browsing this forum: Majestic-12 [Bot]


text:
American Institute of Mathematical Sciences

June  2017, 14(3): 607-624. doi: 10.3934/mbe.2017035

Flow optimization in vascular networks

 1 Department of Mathematics, University of Colorado Colorado Springs, Colorado Springs, CO 80919, USA 2 Dipartimento di Ingegneria dell'Informazione ed Elettrica e Matematica Applicata, Universita degli Studi di Salerno, Fisciano (SA), 84084, Italy

The first author would like to thank University of Salerno for its hospitality

Received  June 2015 Accepted  November 06, 2016 Published  December 2016

The development of mathematical models for studying phenomena observed in vascular networks is very useful for its potential applications in medicine and physiology. Detailed $3$D studies of flow in the arterial system based on the Navier-Stokes equations require high computational power, hence reduced models are often used, both for the constitutive laws and the spatial domain. In order to capture the major features of the phenomena under study, such as variations in arterial pressure and flow velocity, the resulting PDE models on networks require appropriate junction and boundary conditions. Instead of considering an entire network, we simulate portions of the latter and use inflow and outflow conditions which realistically mimic the behavior of the network that has not been included in the spatial domain. The resulting PDEs are solved numerically using a discontinuous Galerkin scheme for the spatial and Adam-Bashforth method for the temporal discretization. The aim is to study the effect of truncation to the flow in the root edge of a fractal network, the effect of adding or subtracting an edge to a given network, and optimal control strategies on a network in the event of a blockage or unblockage of an edge or of an entire subtree.

Citation: Radu C. Cascaval, Ciro D'Apice, Maria Pia D'Arienzo, Rosanna Manzo. Flow optimization in vascular networks. Mathematical Biosciences & Engineering, 2017, 14 (3) : 607-624. doi: 10.3934/mbe.2017035
References:

show all references

References:
The Riemann Problem. $A_L$, $U_L$ ($A_R$, $U_R$) represent the cross section and flow velocity on the left (right) side of the interface, while $W_f$ ($W_b$) are the forward (backward) characteristic information
Types of junctions used in the simulations
Temporal oscillations of pressure and flow velocity for moderately high resistance ($R_t=0.8$) during 40 second simulation of the 15 edge fractal tree, as recorded in the middle an edge. After reaching steady state, slow oscillations ($\sim$ 0.4 Hz) are generated
Temporal oscillations of pressure and flow velocity for maximum resistance ($R_t = 1$) during 28 second simulation of the 15 edge fractal tree, as recorded in the middle an edge. Slow oscillations ($\sim$ 0.1 Hz) are generated during the pressure build-up
Temporal recordings for pressure (top) and flow velocity (bottom) in the zero generations (blue) and two generations (red) fractal trees
Pressure (left) and flow velocity (right) distributions in the network at a fixed time. The color scales correspond to the units used for pressure (kPa) and for flow velocity (m/s)
Pressure and flow velocity at the inflow (top) and outflow (bottom) in the two networks
Pressure and flow before and after blockage removal in edges 1, 3 and 4
Physical lengths and radii used in the junctions generating the fractal tree
 Edge Length (m) Radius (mm) 1 1 10 2 0.9 9 3 0.8 8
 Edge Length (m) Radius (mm) 1 1 10 2 0.9 9 3 0.8 8
Physical lengths and radii used in the truncated tree
 Edge Length (m) Radius (mm) 1 1 10 2 2.439 9 3 1.952 8
 Edge Length (m) Radius (mm) 1 1 10 2 2.439 9 3 1.952 8
 [1] Zexuan Liu, Zhiyuan Sun, Jerry Zhijian Yang. A numerical study of superconvergence of the discontinuous Galerkin method by patch reconstruction. Electronic Research Archive, 2020, 28 (4) : 1487-1501. doi: 10.3934/era.2020078 [2] Leilei Wei, Yinnian He. A fully discrete local discontinuous Galerkin method with the generalized numerical flux to solve the tempered fractional reaction-diffusion equation. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020319 [3] Predrag S. Stanimirović, Branislav Ivanov, Haifeng Ma, Dijana Mosić. A survey of gradient methods for solving nonlinear optimization. Electronic Research Archive, 2020, 28 (4) : 1573-1624. doi: 10.3934/era.2020115 [4] Alberto Bressan, Sondre Tesdal Galtung. A 2-dimensional shape optimization problem for tree branches. Networks & Heterogeneous Media, 2020  doi: 10.3934/nhm.2020031 [5] Shenglan Xie, Maoan Han, Peng Zhu. A posteriori error estimate of weak Galerkin fem for second order elliptic problem with mixed boundary condition. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020340

2018 Impact Factor: 1.313

Metrics

• HTML views (59)
• Cited by (3)

• on AIMS

format: plain


text:
# Chapter 5 Polynomials and Polynomial Functions - 5.3 Add, Subtract, and Multiply Polynomials - 5.3 Exercises - Skill Practice - Page 349: 10

$$4t^3-4t^2+9t-8$$

#### Work Step by Step

In order to solve this problem, it is necessary to combine like terms. Recall, like terms must (1) be raised to the same power and (2) consist of the same variable (i.e. x and y are not like terms). It is also important to note that numbers are all like terms with each other when not multiplied by anything. Also, note that with subtraction problems, it is necessary to distribute the negative to all terms within the parenthesis. Thus, combining like terms, we obtain: $$4t^3+(-11+7)t^2 +(4+5)t -8 \\ 4t^3-4t^2+9t-8$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Courses

# Integral Calculus MCQ Level - 1

## 10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Integral Calculus MCQ Level - 1

Description
This mock test of Integral Calculus MCQ Level - 1 for Physics helps you for every Physics entrance exam. This contains 10 Multiple Choice Questions for Physics Integral Calculus MCQ Level - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Integral Calculus MCQ Level - 1 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this Integral Calculus MCQ Level - 1 exercise for a better result in the exam. You can find other Integral Calculus MCQ Level - 1 extra questions, long questions & short questions for Physics on EduRev as well by searching above.
QUESTION: 1

### The area of the region bounded by the curves y = |x – 1| and y = 3 – |x| is :

Solution:

The required area can be divided into, three regions

= 2 – 1 + 2 + 8 – 4 – 4 +1
= 4 sq. units
The correct answer is: 4 sq. units

QUESTION: 2

### The area enclosed between the curve y = log(x + e) and the coordinate axes is :

Solution:

The area will be given by

e – e + 1
= 1 sq. unit
The correct answer is: 1 sq. unit

QUESTION: 3

### The area of the plane, region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to :

Solution:

The required area is

The correct answer is 4/3 sq.units

QUESTION: 4

The value of

Solution:

After changing the order, we get

QUESTION: 5

The area bounded by the curve y = 2x – x2 and the straight line y = –x is given by :

Solution:

The required area is bounded by the curves

∴  The area will be given by

The correct answer is:  9/2 sq.units

QUESTION: 6

The change of order of integration of the integral

Solution:

The region of integration is

Hence, changing the order of integration, we get,

QUESTION: 7

The area bounded by the curves |x| + |y| ≥ 1 and x2 + y2 ≤ 1 is :

Solution:

The required area can be evaluated by

Aliter: Required area = Area of circle – Area of square.

= (π - 2)

sq. units

The correct answer is: (π - 2)  sq. units

QUESTION: 8

The area of the region bounded by the curves y = |x – 2|, x = 1, x = 3 and the x–axis is :

Solution:

The required area is

The correct answer is: 1 sq. unit

QUESTION: 9

The integral   becomes    after reversing the order of integration.

Then the incorrect option would be :

Solution:

The given region of integration is bounded by
plotting all these curves, we have

Reversing the order of integration, we have the domain of integration divided into 2 regions – I and II.

∴ The integral becomes:

The correct answer is:  Q = e

QUESTION: 10

is equivalent to :

Solution:

The region of integration is bounded by  and

Changing the order of integration,


text:
## Vectors

Vectors, concretely, are arrows, with a head and a tail. If two arrows share a tail, then you can measure the angle between them. The length of the arrow represents the magnitude of the vector.

The modern abstract view is much more interesting but let’s start at the beginning.

#### Force vectors

Originally vectors were conceived as a force applied at a point.

As in, “That lawn ain’t mowing itself, boy. Now you git over there and apply a continuous stream of vectors to that lawnmower, before I apply a high-magnitude vector to your bee-hind!”

##### Thanks Galileo, totally gonna get you back, man

The Galilean idea of splitting a point into its x-coordinate, y-coordinate and z-coordinate works with vectors as well. “Apply a force that totals 5 foot-pounds / second² in the x direction and 2 foot-pounds / second² in the y direction”, for instance.

Therefore, both points and vectors benefit from adding more dimensions to Galileo’s “coordinate system”. Add a w dimension, a q dimension, a ξ dimension — and it’s up to you to determine what those things can mean.

If a vector can be described as (5, 2, 0), then why not make a vector that’s (5, 2, 0, 1.1, 2.2, 19, 0, 0, 0, 3)? And so on.

###### 4th Dimension Plus

So that’s how you get to 4-D vectors, 13-D vectors, and 11,929-D vectors. But the really interesting stuff comes from considering -dimensional vectors. That opens up functional space, and sooooo many things in life are functions.

(Interesting stuff also happens when you make vectors out of things that are not traditionally conceived to be “numbers”. Another post.)

### Abstractions

In the most general sense, vectors are things that can be added together. The modern, abstract view includes as vectors:

### Things you can do with vectors

Given two vectors, you should be able to take their outer product or their inner product.

The inner product allows you to measure the angle between two vectors. If the inner product makes sense, then the space you are playing in has geometry. (Not all spaces have geometry — some just have topology.)

And — this is weird — if the concept of angle applies, then the concept of length applies as well. Don’t ask me why; the symbols just work that way.

##### Magnitude

But the “length” of a song (one of my for-instances above) would not be something like 2:43. The magnitude of a song vector would be the total amount of energy in the sound wave | compression wave.

$\dpi{300} \bg_white \| \text{song} \| = \int \text{compression wave}$

What is the angle between two songs, two spike-trains, two security prices? What is the angle between two heartbeats? It’s the correlation between them.

#### Linear Algebra

Also, you can do linear algebra on vectors — provided they’re coming out of the same point. Some might say that the ability to do linear algebra on something is what makes a vector.

That can mean different things in different spaces — like maybe you’re superposing wave-forms, or maybe you’re converting bitmap images to JPEG. Or maybe you’re Photoshopping an existing JPEG. Oh, man, Photoshop is so math-y.

Shearing the mona lisa (linear algebra on an image — from the Wikipedia page on eigenvectors, one of which is the red arrow)

23 notes

1. clazzjassicalrockhop reblogged this from isomorphismes
2. anticancerdiet reblogged this from isomorphismes
3. trotuarnajaplitka reblogged this from isomorphismes
4. cellular-automaton reblogged this from isomorphismes
5. evanmarble reblogged this from isomorphismes
6. isomorphismes posted this


text:
One of the following principles of management is adhered in standard costing:

One of the following principles of management is adhered in standard costing:

a. management by objective

b. management by exception

c. delegation of authority

d. maximizing profits


text:
# Faster Algorithms for some Optimization Problems on Collinear Points

We propose faster algorithms for the following three optimization problems on n collinear points, i.e., points in dimension one. The first two problems are known to be NP-hard in higher dimensions. 1- Maximizing total area of disjoint disks: In this problem the goal is to maximize the total area of nonoverlapping disks centered at the points. Acharyya, De, and Nandy (2017) presented an O(n^2)-time algorithm for this problem. We present an optimal Θ(n)-time algorithm. 2- Minimizing sum of the radii of client-server coverage: The n points are partitioned into two sets, namely clients and servers. The goal is to minimize the sum of the radii of disks centered at servers such that every client is in some disk, i.e., in the coverage range of some server. Lev-Tov and Peleg (2005) presented an O(n^3)-time algorithm for this problem. We present an O(n^2)-time algorithm, thereby improving the running time by a factor of Θ(n). 3- Minimizing total area of point-interval coverage: The n input points belong to an interval I. The goal is to find a set of disks of minimum total area, covering I, such that every disk contains at least one input point. We present an algorithm that solves this problem in O(n^2) time.

## Authors

• 20 publications
• 29 publications
• 15 publications
• 24 publications
• 2 publications
• 17 publications
• ### Range Assignment of Base-Stations Maximizing Coverage Area without Interference

We study the problem of assigning non-overlapping geometric objects cent...
05/25/2017 ∙ by Ankush Acharyya, et al. ∙ 0

• ### Submodular Clustering in Low Dimensions

We study a clustering problem where the goal is to maximize the coverage...
04/11/2020 ∙ by Arturs Backurs, et al. ∙ 0

• ### Algorithms for the Line-Constrained Disk Coverage and Related Problems

Given a set P of n points and a set S of m weighted disks in the plane, ...
04/29/2021 ∙ by Logan Pedersen, et al. ∙ 0

• ### On the Minimum Consistent Subset Problem

Let P be a set of n colored points in the plane. Introduced by Hart (196...
10/22/2018 ∙ by Ahmad Biniaz, et al. ∙ 0

• ### Smallest k-Enclosing Rectangle Revisited

Given a set of n points in the plane, and a parameter k, we consider the...
03/15/2019 ∙ by Timothy M. Chan, et al. ∙ 0

• ### Faster Minimization of Tardy Processing Time on a Single Machine

This paper is concerned with the 1||∑ p_jU_j problem, the problem of min...
03/16/2020 ∙ by Karl Bringmann, et al. ∙ 0

• ### To Close Is Easier Than To Open: Dual Parameterization To k-Median

The k-Median problem is one of the well-known optimization problems that...
11/16/2020 ∙ by Jaroslaw Byrka, et al. ∙ 0

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

Range assignment is a well-studied class of geometric optimization problems that arises in wireless network design, and has a rich literature. The task is to assign transmission ranges to a set of given base station antennas such that the resulting network satisfies a given property. The antennas are usually represented by points in the plane. The coverage region of an antenna is usually represented by a disk whose center is the antenna and whose radius is the transmission range assigned to that antenna. In this model, a range assignment problem can be interpreted as the following problem. Given a set of points in the plane, we must choose a radius for each point, so that the disks with these radii satisfy a given property.

Let be a set of points in the -dimensional Euclidean space. A range assignment for is an assignment of a transmission range (radius) to each point . The cost of a range assignment, representing the power consumption of the network, is defined as for some constant . We study the following three range assignment problems on a set of points on a straight-line (-dimensional Euclidean space).

Problem 1

Given a set of collinear points, maximize the total area of nonoverlapping disks centered at these points. The nonoverlapping constraint requires to be no larger than the Euclidean distance between and , for every .

Problem 2

Given a set of collinear points that is partitioned into two sets, namely clients and servers, the goal is to minimize the sum of the radii of disks centered at the servers such that every client is in some disk, i.e., every client is covered by at least one server.

Problem 3

Given a set of points on an interval, find a set of disks covering the entire interval such that the total area of disks is minimized and for every the disk contains the point .

In Problem 1 we want to maximize , in Problem 2 we want to minimize , and in Problem 3 we want to minimize . These three problems are solvable in polynomial time in 1-dimension. Both Problem 1 and Problem 2 are NP-hard in dimension , for every , and both have a PTAS [2, 3, 4].

Acharyya et al. [2] showed that Problem 1 can be solved in time. Eppstein [7] proved that an alternate version of this problem, where the goal is to maximize the sum of the radii, can be solved in time for any constant dimension . Bilò et al. [4]

showed that Problem 2 is solvable in polynomial time by reducing it to an integer linear program with a totally unimodular constraint matrix. Lev-Tov and Peleg

[9] presented an -time algorithm for this problem. They also presented a linear-time 4-approximation algorithm. Alt et al. [3] improved the ratio of this linear-time algorithm to 3. They also presented an -time 2-approximation algorithm for Problem 2. Chambers et al. [6] studied a variant of Problem 3—on collinear points—where the disks centered at input points; they showed that the best solution with two disks gives a -approximation. Carmi et al. [5] studied a similar version of the problem for points in the plane.

### 1.1 Our Contributions

In this paper we study Problems 1-3. In Section 2, we present an algorithm that solves Problem 1 in linear time, provided that the points are given in sorted order along the line. This improves the previous best running time by a factor of . In Section 3, we present an algorithm that solves Problem 2 in time; this also improves the previous best running time by a factor of . In Section 4, first we present a simple algorithm for Problem 3. Then with a more involved proof, we show how to improve the running time to .

## 2 Problem 1: Disjoint Disks with Maximum Area

In this section we study Problem 1: Let be a set of points on a straight-line that are given in sorted order. We want to assign to every a radius such that the disks with the given radii do not overlap and their total area, or equivalently , is as large as possible. Acharyya et al. [1] showed how to obtain such an assignment in time. We show how to obtain such an assignment in linear time.

###### Theorem 1.

Given collinear points in sorted order in the plane, in time, we can find a set of nonoverlapping disks centered at these points that maximizes the total area of the disks.

With a suitable rotation we assume that is horizontal. Moreover, we assume that is the sequence of points of in increasing order of their -coordinates. We refer to a set of nonoverlapping disks centered at points of as a feasible solution. We refer to the disks in a feasible solution that are centered at as , respectively. Also, we denote the radius of by ; it might be that . For a feasible solution we define . Since the total area of the disks in is , hereafter, we refer to as the total area of disks in . We call a full disk if it has or on its boundary, a zero disk if its radius is zero, and a partial disk otherwise. For two points and , we denote the Euclidean distance between and by .

We briefly review the -time algorithm of Acharyya et al. [1]. First, compute a set of disks centered at points of , which is the superset of every optimal solution. For every disk , that is centered at a point , define a weighted interval whose length is , where is the radius of , and whose center is . Set the weight of to be . Let be the set of these intervals. The disks corresponding to the intervals in a maximum weight independent set of the intervals in forms an optimal solution to Problem 1. By construction, these disks are nonoverlapping, centered at , and maximize the total area. Since the maximum weight independent set of intervals that are given in sorted order of their left endpoints can be computed in time [8], the time complexity of the above algorithm is essentially dominated by the size of . Acharyya et al. [1] showed how to compute such a set of size and order the corresponding intervals in time. Therefore, the total running time of their algorithm is .

We show how to improve the running time to . In fact we show how to find a set of size and order the corresponding intervals in time, provided that the points of are given in sorted order.

### 2.1 Computation of D

In this section we show how to compute a set with a linear number of disks such that every disk in an optimal solution for Problem 1 belongs to .

Our set is the union of three sets , , and of disks that are computed as follows. The set contains disks representing the full disks and zero disks that are centered at points of . We compute by traversing the points of from left to right as follows; the computation of is symmetric. For each point with we define its signature as

 s(pi)={+if~{}~{}|pi−1pi|⩽|pipi+1|−if~{}~{}|pi−1pi|>|pipi+1|.

Set and . We refer to the sequence as the signature sequence of . Let be the multiset that contains all contiguous subsequences of , with , such that , and for all ; if , then there is no . For example, if , then . Observe that for every sequence in we have that

 |pipi+1|⩽|pi+1pi+2|⩽|pi+2pi+3|⩽⋯⩽|pj−1pj|,and|pj−1pj|>|pjpj+1|.

Every plus sign in belongs to at most one sequence in , and every minus sign in belongs to at most two sequences in . Therefore, the size of (the total length of its sequences) is at most . For each sequence in we add some disks to as follows. Consider the full disk at . Iterate on . In each iteration, consider the disk that is centered at and touches . If does not contain and its area is not larger than the area of , then add to and proceed to the next iteration, otherwise, terminate the iteration. See Figure 1. This finishes the computation of . Notice that contains at most disks. The computation of is symmetric; it is done in a similar way by traversing the points from right to left (all the signatures become and vice versa).

The number of disks in is at most . The signature sequence can be computed in linear time. Having , we can compute the multiset , the disks in as well as the corresponding intervals, as in [1] and described before, in sorted order of their left endpoints in total time. Then the sorted intervals corresponding to circles in can be computed in linear-time by merging the sorted intervals that correspond to sets , , and . It remains to show that contains an optimal solution for Problem 1. To that end, we first prove two lemmas about the structural properties of an optimal solution.

###### Lemma 1.

Every feasible solution for Problem 1 can be converted to a feasible solution where and are full disks and .

###### Proof.

Recall that . We prove this lemma for ; the proof for is similar. Since is a feasible solution, we have that . Let be a solution that is obtained from by making a full disk and a zero disk. Since we do not increase the radius of , it does not overlap , and thus, is a feasible solution. In , the radius of is , and we have that . This implies that . ∎

###### Lemma 2.

If , with , is a partial disk in an optimal solution, then .

###### Proof.

The proof is by contradiction; let be such an optimal solution for which . First assume that touches at most one of and . By slightly enlarging and shrinking its touching neighbor we can increase the total area of . Without loss of generality suppose that touches . Since ,

 (ri+ϵ)2+(ri−1−ϵ)2=r2i+r2i−1+2(riϵ−ri−1ϵ+ϵ2)>r2i+r2i−1>0,

for any . This contradicts optimality of . Now, assume that touches both and , and that . See Figure 2. We obtain a solution from by enlarging as much as possible, and simultaneously shrinking both and . This makes a zero disk, a full disk, a zero or a partial disk, and does not change the other disks. The difference between the total areas of and is

 ((ri+ri−1)2+(ri+1−ri−1)2)−(r2i−1+r2i+r2i+1)=r2i−1+2ri−1(ri−ri+1)>0;

this inequality is valid since . This contradicts the optimality of . ∎

###### Lemma 3.

The set contains an optimal solution for Problem 1.

###### Proof.

It suffices to show that every disk , which is centered at , in an optimal solution belongs to . By Lemma 1, we may assume that both and are full disks. If is a full disk or a zero disk, then it belongs to . Assume that is a partial disk. Since is optimal, touches at least one of and , because otherwise we could enlarge .

First assume that touches exactly one disk, say . We are going to show that belongs to (If touches only , by a similar reasoning we can show that belongs to ). Notice that , because otherwise we could enlarge and shrink simultaneously to increase , which contradicts the optimality of . Since is partial and touches , we have that is either full or partial. If is full, then it has on its boundary, and thus . By our definition of , for some , the sequence belongs to . Then by our construction of both and belong to , where plays the role of . Assume that is partial. Then touches , because otherwise we could enlarge and shrink simultaneously to increase . Recall that . Lemma 2 implies that . This implies that , and thus . Since is partial and touches , we have that is either full or partial. If is full, then it has on its boundary, and thus . By a similar reasoning as for based on the definition of and , we get that , , and are in . If is partial, then it touches and again by Lemma 2 we have and consequently . By repeating this process, we stop at some point , with , for which is a full disk, , and ; notice that such a exists because is a full disk and consequently is a zero disk. To this end we have that , , and is a plus sequence. Thus, is a subsequence of some sequence in . Our construction of implies that all disks belong to .

Now assume that touches both and . By Lemma 2 we have that is strictly smaller than the largest of these disks, say . By a similar reasoning as in the previous case we get that . ∎

## 3 Problem 2: Client-Server Coverage with Minimum Radii

In this section we study Problem 2: Let be a set of points on a straight-line that is partitioned into two sets, namely clients and servers. We want to assign to every server in a radius such that the disks with these radii cover all clients and the sum of their radii is as small as possible. Bilò et al. [4] showed that this problem can be solved in polynomial time. Lev-Tov and Peleg [9] showed how to obtain such an assignment in time. Alt et al. [3] presented an O(n log n)-time 2-approximation algorithm for this problem. We show how to solve this problem optimally in time.

###### Theorem 2.

Given a total of collinear clients and servers, in time, we can find a set of disks centered at servers that cover all clients and where the sum of the radii of the disks is minimum.

Without loss of generality assume that is horizontal, and that is the sequence of points of in increasing order of their -coordinates. We refer to a disk with radius zero as a zero disk, to a set of disks centered at servers and covering all clients as a feasible solution, and to the sum of the radii of the disks in a feasible solution as its cost. We denote the radius of a disk by , and denote by a disk that is centered at the point with the point on its boundary.

We describe a top-down dynamic programming algorithm that maintains a table with entries . Each table entry represents the cost of an optimal solution for the subproblem that consists of points . The optimal cost of the original problem will be stored in ; the optimal solution itself can be recovered from . In the rest of this section we show how to solve a subproblem . In fact, we show how to compute recursively by a top-down dynamic programming algorithm. To that end, we first describe our three base cases:

• There is no client. In this case .

• There are some clients but no server. In this case .

• There are some clients and exactly one server, say . In this case is the radius of the smallest disk that is centered at and covers all the clients.

Assume that the subproblem has at least one client and at least two servers. We are going to derive a recursion for .

###### Observation 1.

Every disk in any optimal solution has a client on its boundary.

###### Lemma 4.

No disk contains the center of some other non-zero disk in an optimal solution.

###### Proof.

Our proof is by contradiction. Let and be two disks in an optimal solution such that contains the center of . Let and be the centers of and , respectively, and and be the radii of and , respectively. See Figure 3(a). Since contains , we have . Let be the disk of radius that is centered at . Notice that covers all the clients that are covered by . By replacing and with we obtain a feasible solution whose cost is smaller than the optimal cost, because . This contradicts the optimality of the initial solution. ∎

Let be the rightmost client in . For a disk that covers , let be the smallest index for which the point is in the interior or on the boundary of , i.e., is the index of the leftmost point of that is in . See Figure 3(b).

We claim that only one disk in an optimal solution can cover , because, if two disks cover then if their centers lie on the same side of , we get a contradiction to Lemma 4, and if their centers lie on different sides of , then by removing the disk whose center is to the right of we obtain a feasible solution with smaller cost. Let be an optimal solution (with minimum sum of the radii) that has a maximum number of non-zero disks. Let be the disk in that covers . All other clients in are also covered by , and thus, they do not need to be covered by any other disk. As a consequence of Lemma 4, the servers that are in and the servers that lie to the right of cannot be used to cover any clients in . Therefore, if we have , then the problem reduces to a smaller instance that consists of the points to the left of , i.e., . See Figure 3(b). Thus, the cost of the optimal solution for the subproblem can be computed as .

In the rest of this section we compute a set of disks each of them covering . Then we claim that belongs to . Having , we can compute by the following recursion:

 T(k)=min{T(ψ(D)−1)+r(D):D∈Dk}.

Now we show how to compute . Recall from Observation 1 that every disk in the optimal solution (including ) contains a client on its boundary. Using this observation, we compute in two phases. In the first phase, for every server we add the disk to . In the second phase, for every client , with , we add a disk to , where is the first server to the right side of the midpoint of segment . Since for every server and for every client (except for ) we add one disk to , this set has at most disks. The disks that we add in phase one can be computed in time by sweeping the servers from right to left. The disks that we add in phase two can also be computed in time by sweeping the clients from right to left, using this property that the server associated with the next client is on or to the left side of the server associated with the current client. Hence, the set , and consequently the entry , can be computed in time. Therefore, our dynamic programming algorithm computes all entries of in time.

One final issue we need to address is the correctness of our algorithm, which is to show that belongs to . Let be the server that is the center of and let be the client on the boundary of (such a client exists by Observation 1). Recall that covers the rightmost client . If , then has been added to in the phase one. Assume that . In this case is the left intersection point of the boundary of with because is to the left side of . Let be the mid point of the line segment , and let be the first server to the right of . The server cannot be to the left side of because otherwise could not cover . Also, cannot be to the right side of because otherwise the disk , which is smaller than , covers the same set of clients as does, in particular it covers and . Therefore, we have , and thus , which has been added to in phase two. This finishes the proof of correctness of our algorithm.

## 4 Problem 3: Point-Interval Coverage with Minimum Area

Let be an interval on the -axis in the plane. We say that a set of disks covers if is a subset of the union of the disks in this set. Let be a set of points on , that are ordered from left to right, and such that and . A point-interval coverage for the pair is a set of disks that cover such that for every the disk contains the point , i.e., is in the interior or on the boundary of . See Figure 4. The point-interval coverage problem is to find such a set of disks with minimum total area. In this section we show how to solve this problem in time.

###### Theorem 3.

Given points on an interval, in time, we can find a set of disks covering the entire interval such that every disk contains at least one point and where the total area of the disks is minimum.

If we drop the condition that should contain , then the problem can be solved in linear time by using Observation 2 (which is stated below). First we prove some lemmas about the structural properties of an optimal point-interval coverage. We say that a disk is anchored at a point if it has on its boundary. We say that two intersecting disks touch each other if their intersection is exactly one point, and we say that they overlap otherwise.

###### Lemma 5.

There is no pair of overlapping disks in any optimal solution for the point-interval coverage problem.

###### Proof.

Our proof is by contradiction. Consider two overlapping disks and , with , in an optimal solution. Since contains and contains , there exists a point on the line segment that is in . Let and be the leftmost and the rightmost points of the interval that is covered by ; see Figure 5(a). Let and be the disks with diameters and , respectively. The areas of and are smaller than the areas of and , respectively. Moreover, contains , contains , and covers the same interval as does. Therefore, by replacing and with and we obtain a solution whose total area is smaller than the optimal area, which is a contradiction.

###### Lemma 6.

In any optimal solution, if the intersection point of and does not belong to , then and have equal radius.

###### Proof.

Let be the intersection point of and . Let be the left intersection point of the boundary of with the -axis, and let be the right intersection point of the boundary of with the -axis; see Figure 5(b). We proceed by contradiction, and assume, without loss of generality, that is smaller than . We shrink (while anchored at ) and enlarge (while anchored at ) simultaneously by a small value. This gives a valid solution whose total area is smaller than the optimal area, because our gain in the area of is smaller than our loss from the area of . This contradicts the optimality of our initial solution. ∎

The following lemma and observation play important roles in our algorithm for the point-interval coverage problem, which we describe later.

###### Lemma 7.

Let be a real number, and be a sequence of positive real numbers such that . Then

 k∑i=1r2i≥k∑i=1(R/k)2=R2/k, (1)

i.e., the sum on the left-hand side of (1) is minimum if all are equal to .

###### Proof.

If is a convex function, then—by Jensen’s inequality—we have

 f(k∑i=1rik)⩽k∑i=1f(ri)k.

Since the function is convex, it follows that

 (Rk)2=f(Rk)=f(k∑i=1rik)⩽k∑i=1r2ik,

which, in turn, implies Inequality (1). ∎

The minimum sum of the radii of a set of disks that cover is . The following observation is implied by Lemma 7, by setting and .

###### Observation 2.

The minimum total area of disks covering is obtained by a sequence of disks of equal radius such that every two consecutive disks touch each other; see Figure 6.

We refer to the covering of that is introduced in Observation 2 as the unit-disk covering of with disks. Such a covering is called valid if it is a point-interval coverage for .

### 4.1 A Dynamic-Programming Algorithm

In this subsection we present an -time dynamic-programming algorithm for the point-interval coverage problem. In Subsection 4.2 we show how to improve the running time to .

First, we review some properties of an optimal solution for the point-interval coverage problem that enable us to present a top-down dynamic programming algorithm. Let be the sequence of disks in an optimal solution for this problem. Recall that as a consequence of Lemma 5, the intersection of every two consecutive disks in is a point. If there is no for which the intersection point of and belongs to , then Lemma 6 implies that all disks in have equal radius, and thus, is a valid unit-disk covering. Assume that for some the intersection point of and is a point . Notice that either or . In either case, is the union of the optimal solutions for two smaller problem-instances and where , , and .

We define a subproblem and represent it by four indices where and . The indices and indicate that . The set contains the points of that are on provided that belongs to if and only if and belongs to if and only if . For example, if and , then . We define to be the cost (total area) of an optimal solution for subproblem . The optimal cost of the original problem will be stored in . We compute as follows. If the unit-disk covering is a valid solution for , then by Observation 2 it is optimal, and thus we assign its total area to . Otherwise, as we discussed earlier, there is a point of with that is the intersection point of two consecutive disks in the optimal solution. This splits the problem into two smaller subproblems, one to the left of and one to the right of . The point is assigned either to the left subproblem or to the right subproblem. See Figure 7 for an instance in which the unit-disk covering is not valid, and is assigned to the right subproblem. In the optimal solution, is assigned to the one that minimizes the total area, which is

 T(i,j,i′,j′)=min{T(i,k,i′,1)+T(k,j,0,j′),T(i,k,i′,0)+T(k,j,1,j′)}.

Since we do not know the value of , we try all possible values and pick the one that minimizes .

There are three base cases for the above recursion. (1) No point of is assigned to the current subproblem: we assign to , which implies this solution is not valid. (2) Exactly one point of is assigned to the current subproblem: we cover with one disk of diameter and assign its area to . (3) More than one point of is assigned to the current subproblem and the unit-disk covering is valid: we assign the total area of this unit-disk covering to .

The total number of subproblems is at most , because and take different values, and each of and takes two different values. The time to solve each subproblem is proportional to the time for checking the validity of the unit-disk covering for this subproblem plus the iteration of from to ; these can be done in total time . Thus, the running time of our dynamic programming algorithm is .

In the next section we present a more involved dynamic-programming algorithm that improves the running time to . Essentially, our algorithm verifies the validity of the unit-disk coverings for all subproblems in time.

### 4.2 Improving the running time

We describe a top-down dynamic programming algorithm that maintains a table with entries where and . Each entry represents the cost of an optimal solution for the subproblem that consists of interval and a point set . If , then , whereas, if , then . The optimal cost of the original problem will be stored in ; the optimal solution itself can be recovered from . In the rest of this section we show how to solve subproblem . If the unit-disk covering is a valid solution for , then we assign its total area to . Otherwise, there must be some point with that is the intersection point of two consecutive disks in the optimal solution. Let be the largest such . This choice of implies that the interval is covered by unit disks, and thus, we only need to solve the subproblem to the left of optimally for two cases where and . Let denote the cost of a unit-disk covering for the problem instance (that is defined in the previous section). Then

 T(j,j′)=min{T(i,1)+U(i,j,0,j′),T(i,0)+U(i,j,1,j′)}.

Since we do not know the value of , we try all possible values and pick the one that minimizes .

The total number of subproblems is , and the time to solve each subproblem is proportional to the total time for the iterations of from to plus the time for computing the unit-disk covering for and checking its validity. Let denote the time for computing and checking validity of unit-disk coverings for all . Then the time to compute is . Therefore, the running time of our algorithm, i.e. the time to compute , is

 n∑j=1O(j)+u(j)=O(n2)+n∑j=1u(j)=O(n2)+n∑j=1j−1∑i=2u(i,j,i′,j′),

where denotes the time of computing the unit-disk covering for and checking its validity. In the rest of this section we will show how to do this, for all , in time. This implies that the total running time of our algorithm is .

Take any . We show how to check the validity of the unit-disk covering for , where iterates from to . We are going to show how to do this in total time, for all values of . This will imply that we can check the validity of the unit-disk coverings for all and in time. We describe the procedure for the case when and ; the other three cases can be handled similarly. Recall that is the interval and is the point set that are associated with . Notice that the length of is , and the number of points in is . Then the diameter of each disk in the unit-disk covering of is . In order to have a valid unit-disk covering for , the following conditions are necessary and sufficient (see Figure 8)

 |pipi+1|  ⩽dij dij⩽ |pipi+2|  ⩽2⋅dij 2⋅dij⩽ |pipi+3|  ⩽3⋅dij 3⋅dij⩽ |pipi+4|  ⩽4⋅dij ⋮ (nij−1)⋅dij⩽ |pipj|   ⩽nij⋅dij.

The above inequalities are equivalent to the following two inequalities

 dij⩾ max{|pipi+1|,|pipi+2|2,|pipi+3|3,…,|pipj|nij}, (2) dij⩽ min{|pipi+2|,|pipi+3|2,|pipi+4|3,…,|pipj|nij−1}. (3)

Let denote the maximum value in Inequality (2), and let denote the minimum value in Inequality (3). To check the validity of the unit-disk covering for , it suffices to compare with these two values. Recall that is fixed and iterates from to . Now we show how to check the validity of the unit-disk covering for subproblem , in time. In this subproblem, the diameter of the unit disks is and the number of points is ; these values can be computed in time. In order to have a valid unit-disk covering for , the following two inequalities are necessary and sufficient

 di(j+1)⩾ max{|pipi+1|,|pipi+2|2,|pipi+3|3,…,|pipj|nij,|pipj+1|nij+1}, di(j+1)⩽ min{|pipi+2|,|pipi+3|2,|pipi+4|3,…,|pipj|nij−1,|pipj+1|nij}.

Thus, we can compute

 M(i,j+1)=max{M(i,j),


text:
# Fitting a long equation in tcolorbox

I am trying to fit a long equation in a tcbtheorem environment. The problem is, the equation sticks out the box on the right side. How do I fix this? I'd prefer to not use multline to break the equation into multiple lines (it just isn't aesthetically pleasing to me). I also tried using \scriptstyle, but then the math becomes too small compared to the text in the theorem box. Ideally, either I'd like to expand the width of the tcbtheorem box a bit on either side, and slightly shrink the font size of the equation so that it fits. However, I am open to other suggestions that look nice.

\documentclass{beamer}
\mode<presentation>
{
%  \usecolortheme{default} % or try albatross, beaver, crane, ...
%  \usefonttheme{default}  % or try serif, structurebold, ...
\setbeamertemplate{caption}[numbered]
}

\usepackage{tcolorbox}
\usepackage{tikz}
\usetikzlibrary{matrix,overlay-beamer-styles}

%%%%%%%%%
\usetikzlibrary{calc}
%%%%%%%%
\tcbuselibrary{theorems}

\newtcbtheorem[]{mylemmma}{Lemma}{colframe=green,colback=white, width=\textwidth}{lem}
\tcbset{colframe=green, colback=white}
\resetcounteronoverlays{tcb@cnt@mylemmma}

\begin{document}
\section{Proofs}
\begin{frame}

\begin{mylemmma}{Name of the Lemma}{lem}
The function changes as follows.
$-f(\sum_i g(-\psi - b)) \geq -f (\sum_i g(-\psi)) + \sum_i (b_i - h_i)^\top \left( \frac{g(-b)}{\sum_i g(-b)} \right) - error$
%       ${\scriptstyle -f(\sum_i g(-\psi - b)) \geq -f (\sum_i g(-\psi)) + \sum_i (b_i - h_i)^\top \left( \frac{g(-b)}{\sum_i g(-b)} \right) - error}$
%\begin{multline*}
%   -f(\sum_i g(-\psi - b)) \\
% \geq -f (\sum_i g(-\psi)) + \sum_i (b_i - h_i)^\top \left( \frac{g(-b)}{\sum_i g(-b)} \right) - error
%   \end{multline*}
\end{mylemmma}

\end{frame}

\end{document}

• I had already upvoted the question but I have no good proposal. Personally I would go with multline and company. – user194703 May 24 at 21:43
• Thank you! What is "company"? – lazulikid May 24 at 21:54
• For instance, align* with or without \MoveEqLeft from mathtools. – user194703 May 24 at 21:55

Reducing the left margin in the \tcolorbox, some spacing adjustments, and a medium sized formula; is it better?

\documentclass{beamer}
\mode<presentation>
{
% \usecolortheme{default} % or try albatross, beaver, crane, ...
% \usefonttheme{default} % or try serif, structurebold, ...
\setbeamertemplate{caption}[numbered]
}

\usepackage{tcolorbox}
\usepackage{tikz}
\usetikzlibrary{matrix,overlay-beamer-styles}
\usepackage{nccmath}

%%%%%%%%%
\usetikzlibrary{calc}
%%%%%%%%
\tcbuselibrary{theorems}

\newtcbtheorem[]{mylemmma}{Lemma}{colframe=green,colback=white, width=\textwidth, left=0pt}{lem}
\tcbset{colframe=green, colback=white}
\resetcounteronoverlays{tcb@cnt@mylemmma}

\begin{document}
\section{Proofs}
\begin{frame}

\begin{mylemmma}{Name of the Lemma}{lem}
The function changes as follows.
$\medmath{-f(\sum_i g(-\psi - b)) \geq -f \Bigl(\sum_i g(-\psi)\!\Bigr) + \sum_i (b_i - h_i)^{\!\top}\! \biggl( \frac{g(-b)}{\sum_i g(-b)} \biggr) - error}$
\end{mylemmma}

\end{frame}

\end{document}

• Thank you, @Bernard! I'll accept your answer. Unfortunately, the actual f and g I'm using in my slide are functions like exp and log, due to which the above expression is in fact even longer than now. This leads the equation to still spill out of the box. – lazulikid May 24 at 22:11
• One followup question: is it possible to use medmath with multline? I am trying to do so, but it's throwing compilation errors. I think that might be the nicest option for me. – lazulikid May 24 at 22:15
• I found it! So, thanks to Schrodinger's cat's comment above and your answer here, I was able to find the perfect way for me: I used \equation*, then \medmath, then \multlined (in mathtools package). It looks much better! – lazulikid May 24 at 22:19
• I guess \medmath{...} cannnot have line breaks. But there's also a medsize environment. If you're interested, you should have a look at the package documentation. I often use the \mfrac command, when I have a numerical fraction as coefficient and I feel it shouldn't have the same size as fraction expressions. – Bernard May 24 at 22:19


text:
# Two Connected Masses

Juggernaut06

## Homework Statement

Two masses are tied together by a string over a massless pulley so that they can both move vertically. Their masses are 1 kg and 3 kg. What is the acceleration of the 1 kg mass?

F=ma
W=mg

## The Attempt at a Solution

I have the 3 kg mass's summation of forces as ma-T. (In my free body diagram, the 3kg mass is the one closest to the pulley.)

The 1 kg mass's summation of forces are mg-T.

I'm unsure if my diagram is correct and I don't know where to go from here.

Homework Helper
Welcome to PF!

Hi Juggernaut06 ! Welcome to PF!
Two masses are tied together by a string over a massless pulley so that they can both move vertically. Their masses are 1 kg and 3 kg. What is the acceleration of the 1 kg mass?

I have the 3 kg mass's summation of forces as ma-T. (In my free body diagram, the 3kg mass is the one closest to the pulley.)

The 1 kg mass's summation of forces are mg-T.

No, both wrong

both masses have both a and mg.

And of course a is equal (and opposite) for both.