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# In the formula V = 1/(2r)^3, if r is halved, then V is multi

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In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]  15 Jul 2008, 11:39
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In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Feb 2014, 04:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: PS: formula V= 1/(2r)^3 [#permalink]  15 Jul 2008, 11:42

Make r = 1/2, V = 1

Make r = 1, v = 1/8
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Senior Manager Joined: 07 Jan 2008 Posts: 297 Followers: 1 Kudos [?]: 32 [0], given: 0 Re: PS: formula V= 1/(2r)^3 [#permalink] 15 Jul 2008, 12:00 I always mess up these kinds... Say r=2 => V1 = 1/64 when r=1; V2 = 1/8 V2 = 8*V1. B? SVP Joined: 30 Apr 2008 Posts: 1888 Location: Oklahoma City Schools: Hard Knocks Followers: 38 Kudos [?]: 505 [0], given: 32 Re: PS: formula V= 1/(2r)^3 [#permalink] 15 Jul 2008, 12:05 I totally agree with you. I mess them up to and about did this one. I saw that if I make 2r = 1, then r=1/2 so to the power of 3 doesn't matter, it's still 1/1. Then I did r =1 (double it) so then V = 1/8. So I was thinking oh, go from 1 to 1/8 multiply by 1/8...but I had worked the problem backwards. There are tricks the authors throw at us, and then there are tricks we bring to the test ourselves. We have to eliminate our own tricks. Easier said than done! mbawaters wrote: I always mess up these kinds... Say r=2 => V1 = 1/64 when r=1; V2 = 1/8 V2 = 8*V1. B? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: PS: formula V= 1/(2r)^3 [#permalink]  15 Jul 2008, 12:09
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???
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Re: PS: formula V= 1/(2r)^3 [#permalink]  15 Jul 2008, 14:30
vksunder wrote:
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???

I think its asking us to reflect the changes in the value of V if r is halved!
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Re: PS: formula V= 1/(2r)^3 [#permalink]  15 Jul 2008, 22:22
vksunder wrote:
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???

Not really, it's asking what happens to V when r is halved. In this type of problems, just pick an easy value for r and solve both equations.
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Re: In the formula V= 1/(2r)^3, if r is halved, then V is [#permalink]  21 Feb 2014, 02:14
Can someone solve this question with an explanation?

Thanks
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Re: In the formula V= 1/(2r)^3, if r is halved, then V is [#permalink]  21 Feb 2014, 04:28
Expert's post
benjamin09 wrote:
Can someone solve this question with an explanation?

Thanks

In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64

Given: $$V = \frac{1}{(2r)^3}=\frac{1}{8r^3}$$

Now, halved r is r/2, substitute r/2 instead of r: $$V' = \frac{1}{(2*\frac{r}{2})^3}=\frac{1}{r^3}$$.

$$V =\frac{1}{8r^3}$$ should be multiplied by 8 to get $$V'=\frac{1}{r^3}$$.

Or:
When r=2, then $$V = \frac{1}{(2r)^3}=\frac{1}{64}$$.
When r=1, then $$V' = \frac{1}{(2r)^3}=\frac{1}{8}$$.

1/64 should be multiplied by 8 to get 1/8.

Similar question to practice: if-the-function-q-is-defined-by-the-formula-q-5w-4x-z-131051.html

Hope it helps.
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]  21 Feb 2014, 06:34
Thanks a lot Bunuel!
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]  07 Apr 2015, 19:28
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi   [#permalink] 07 Apr 2015, 19:28
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C ++ set vector gives segmentation fault after doing push_back

I have created a vector of sets in my program and I need to go through each of the sets. If a specific element is found in the set, I need to add a new set to the vector. However, this gives me a segmentation fault as soon as my array counter reaches the elements that I inserted later (inside the loop). In the following code, including list.push_back (cS) gives me a segmentation fault.

int main(void)  {
set<int> cS;
vector<set<int> > list;

cS.insert(1);
list.push_back(cS);

cS.insert(2);
list.push_back(cS);

for (int ctr = 0; ctr < list.size(); ctr++)
{
for (set<int>::iterator itr = list[ctr].begin(); itr != list[ctr].end(); itr++)
{
if (*itr == 1 || *itr == 2)
{
cS.clear();
cS.insert(3);
//list.push_back(cS);
}
}
}

for (int ctr = 0; ctr < list.size(); ctr++)
{
for (set<int>::iterator itr = list[ctr].begin(); itr != list[ctr].end(); itr++)
{
cout << *itr << endl;
}
}

return 0;
}

I would be grateful if anyone can explain why this is giving an error (in gcc).

Thanks for visiting my post.

+3

source to share

When you are push_back

in your vector, you will invalidate all references to elements in it in the event that the vector needs to allocate more memory. In your case, the iterator itr

becomes invalid after push_back

. One solution would be to add the sets to a separate list (vector) and then add them right after the for loop:

vector<set<int> > add;
for (int ctr = 0; ctr < list.size(); ctr++)
{
for (set<int>::iterator itr = list[ctr].begin(); itr != list[ctr].end(); itr++)
{
if (*itr == 1 || *itr == 2)
{
cS.clear();
cS.insert(3);
}
}
}

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# $\aleph_1 = 2^{\aleph_0}$

The above definition was given to me by a friend of mine who introduced me to $\aleph$, and infinite sets. The argument he gave me was (paraphrased how I understand it).

While the set of natural numbers is infinite, each number has only a finite number of digits.
Each real number as an infinite number of digits. (The rest is filled up with 0s).
We can represent each real number using binary numbers.
Map each bit of the real numbers to an element of $\mathbb N$.
The cardinality of $\mathbb N$ is $\aleph_0$.
There are $2^{\aleph_0}$ possible combinations of the bits.
There are $2^{\aleph_0}$ real numbers.

He then sort of defined $2^{\aleph_0}$ as $\aleph_1$. I'm guessing the cardinality of the real numbers is taken to be $\aleph_1$?

Said friend also said he has a proof that $\aleph_0^k = \aleph_0$. I haven't yet read his proof as at the time of writing, and may update this question with it when I do. I've been told here that

you can't claim $2^{\aleph_0} = \aleph_1$

which surprised me, but I can't be so sure of the definition myself, so.
My question is:

Is $\aleph_1 = 2^{\aleph_0}$?

• en.wikipedia.org/wiki/Continuum_hypothesis – i707107 Jun 6 '17 at 17:39
• what do you ask? – Mirko Jun 6 '17 at 17:48
• Is k a cardinal or just a natural number? If it is a natural number then the result is obvious. – Jacob Wakem Jun 6 '17 at 17:50
• Coincidentally, the question proof of $2^{\aleph_{0}} = \mathfrak{c}$ was posted just yesterday (and spun off some discussion about the distinction between $\aleph_{1}$ and $\mathfrak{c}$). – Andrew D. Hwang Jun 6 '17 at 17:53

Your friend was incorrect when he defined $\aleph_1$ to be $2^{\aleph_0}$. It is possible that $\aleph_1=2^{\aleph_0}$ (this statement is called the "continuum hypothesis"), but it is also possible that $\aleph_1<2^{\aleph_0}$. Indeed, it is known that the usual axioms of set theory (ZFC) cannot decide this question: it is consistent with ZFC that CH holds, and consistent with ZFC that CH fails. In fact, essentially the only thing that ZFC can prove about $2^{\aleph_0}$ is that it has uncountable cofinality; for instance, it is consistent with ZFC that $2^{\aleph_0}=\aleph_{\omega^2\cdot 3+17}$, which is pretty wild!

So what are the right definitions?

• $2^{\aleph_0}$ is defined to be the cardinality of the set of functions from $\mathbb{N}$ (or, any set of size $\aleph_0$) to $\{0,1\}$ (or, any set of size $2$). It turns out that this is also the cardinality of $\mathbb{R}$; this isn't hard to show, but also isn't trivial (an important issue is the fact that a real number can have multiple decimal expansions). By Cantor's diagonal argument, $2^{\aleph_0}$ is uncountable.

• Meanwhile, $\aleph_1$ is defined to be the cardinality of the smallest uncountable ordinal; equivalently, it is the number of distinct ordertypes of countable ordinals. By an argument similar to that of the Burali-Forti paradox, $\aleph_1$ is uncountable, and (assuming a very mild version of the axiom of choice) $\aleph_1$ is the least uncountable cardinality.

Note that these two objects have very different definitions, so it shouldn't be surprising that they are not perfectly related. They are often conflated, possibly due to the fact that $2^{\aleph_0}$ is often assumed to be the next "nicely definable" cardinality after $\aleph_0$ (which is an interesting claim which is studied extensively in descriptive set theory); however, this conflation is indeed incorrect unless the author explicitly assumes CH (or some additional axiom implying CH).

Incidentally, it is quite easy to show that $\aleph_0^k=\aleph_0$ for finite $k$, so your friend was correct there.

• Can you not show that $2^{\aleph_0}\geq\aleph_1$ using Cantor's theorem? I.e., the powers set of $\aleph_0$ should be at least as great as the next infinite cardinal. Thanks. With regards, – user12802 Oct 21 '17 at 12:40
• @Andrew Yes, of course. And one can do more: by Konig's theorem, $2^{\aleph_0}$ has uncountable cofinality, so we can't have e.g. $2^{\aleph_0}=\aleph_\omega$. And Solovay showed that that is all we can prove, in ZFC. – Noah Schweber Oct 21 '17 at 14:54
• Thanks for your insight. – user12802 Oct 21 '17 at 17:37


text:
# Mathematica rule to express exponentials as multiplication

I'm sorry if this has been asked before, but it is a very simple question and I would imagine it shouldn't be terribly hard for someone that knows to answer.

I'm essentially trying to tell mathematica to not simplify

Exp[x] Exp[y] -> Exp[x+y]

I am using an add on package which is having a hard time with some replacement rules. So, the question is, is it possible to force mathematica to express Exp[x] Exp[y] as is and not simplify it?

A secondary question, which should just be an extension of the first, I would like (Exp[x] Exp[y] )^2 to be replaced as Exp[x]^2 Exp[y]^2 not Exp[2x+2y]! Or even being left as is is ok.

• Would it be an acceptable solution to try to fix the package instead? Is it publicly and freely available? Feb 19, 2014 at 17:59
• Hold[Exp[x] Exp[y]] is supposed to solve this problem, but I've not had much success with Hold[] or HoldForm[] for some reason. Maybe others can address this. Feb 19, 2014 at 18:00
• I'm asking because I don't think it's possible to prevent this simplification without Holding the expression, as Steve suggests. But doing that might prevent the package from working with the expression. Feb 19, 2014 at 18:01
• Thanks for the suggestions. @Szabolcs you were right, the hold command restricts the package. The package is freely available, however, it interfaces in such a way to Singular which restricts this as a solution to the problem. Feb 19, 2014 at 18:16
• Is this this package? Feb 19, 2014 at 18:56

One possibility is to execute your entire code in some dynamic environment, where certain simplification rules are permanently or temporarily blocked. Here is the generic environment generator:

ClearAll[withBlockedSymbols];
withBlockedSymbols[syms : {__Symbol}] :=
Function[code, Block[syms, code], HoldAll];

We can now produce an environment generator specifically for Exp:

withBlockedExp = withBlockedSymbols[{Exp}]

Now, any code executed inside this wrapper, will not use the automatic simplification rules for Exp, which will be inert inside this wrapper. For example:

withBlockedExp[Hold[Evaluate[Exp[x] Exp[y]]]]

(* Hold[Exp[x] Exp[y]] *)

withBlockedExp[Hold[Evaluate[(Exp[x] Exp[y])^2]]]

(* Hold[Exp[x]^2 Exp[y]^2] *)

The only reason I needed to use Hold@Evaluate was to prevent the simplification happening after the execution exits withBlockedExp. Hold@Evaluate@expr allows us to preserve a final form of expr evaluated inside dynamic environment. If you have less trivial transformations, you will see the utility of withBlockedExp better, since you don't have to use Hold etc. for intermediate steps inside withBlockedExp.

• Thank you, this actually answered my original question, however, I was wrong in thinking it would solve the issue I'm having with the interface to my package! Feb 19, 2014 at 22:13
• @Richard Well, too bad then. But thanks for the accept. And you can always ask more questions (as separate questions), until your problem hopefully gets resolved. Feb 19, 2014 at 22:19


text:
# Eigenvectors and inner product

• I
Hi, what is the physical meaning, or also the geometrical meaning of the inner product of two eigenvectors of a matrix?

I learned from the previous topics that a vectors space is NOT Hilbert space, however an inner product forms a Hilbert space if it is complete.

Can two eigenvectors which are linearly independent form a Hilbert space in their inner product?

e_1 = [x_1, y_1]
e_2 = [x_2, y_2]

##\langle e_1, e_2 \rangle## ?

Thanks

Mentor
2022 Award
Hi, what is the physical meaning, or also the geometrical meaning of the inner product of two eigenvectors of a matrix?
An inner product of two vectors, let them be eigenvectors of some transformation or not, is an assignment which can be used to measure lengths and angles, physically and geometric. An explanation and a figure can be found here: https://en.wikipedia.org/wiki/Inner_product_space or more elaborated here: https://arxiv.org/pdf/1205.5935.pdf
I learned from the previous topics that a vectors space is NOT Hilbert space, ...
A Hilbert space is a vector space, but not every vector space is also a Hilbert space. That is: a Hilbert space has more properties.
... however an inner product forms a Hilbert space if it is complete.
Basically, yes, but the wording is a bit weird. Not the inner product forms anything, a Hilbert space is defined by being a real or complex vectorspace equipped with an inner product (additional property). The inner product defines measurements as angles and lengths, e.g. a distance: ##\operatorname{dist}(x,y)=||x-y||=\sqrt{\langle x-y,x-y \rangle}##, which is basically the theorem of Pythagoras here. This distance has to be complete, that is every Cauchy sequence (the elements of the sequence get closer and closer) converges, which means the infinite point towards the elements are narrowing down is actually part of the space. Completeness thus depends on how we measure distances. There is generally more than one way to define a distance. So Hilbert spaces are strictly speaking triplets: (vector space, field (real or complex), inner product).

E.g. with the rational numbers we'll get a problem. Let's define ##a_1=b_1=1\; , \;a_{n+1}=a_n+b_n\; , \;b_{n+1}=a_{n+1}+a_n\; , \;c_n = \dfrac{b_n}{a_n}##. Then ##(c_n)_{n\in \mathbb{N}}## defines a Cauchy sequence of rational numbers which converges to ##\sqrt{2}##. But this limit isn't part of the space ##\mathbb{Q}##, so ##\mathbb{Q}## is incomplete. If we take the exact same situation to the reals (or complex numbers), then this limit is part of the space - ##\mathbb{R}## and ##\mathbb{C}## are complete spaces, i.e. also Hilbert spaces.
Can two eigenvectors which are linearly independent form a Hilbert space in their inner product?

e_1 = [x_1, y_1]
e_2 = [x_2, y_2]

##\langle e_1, e_2 \rangle## ?
This has nothing to do with eigenvectors. Besides, any two vectors are always eigenvectors of some linear function. Two linear independent vectors over the real or complex numbers span a two-dimensional real or complex vector space. E.g. with the Euclidean metric we get a Euclidean space. And this is a two-dimensional Hilbert space, as the real or complex numbers guarantee that any sequence of vectors which is narrowing down by Euclidean distance (Pythagoras) also contains the vector towards the sequence is converging.

SeM
An inner product of two vectors, let them be eigenvectors of some transformation or not, is an assignment which can be used to measure lengths and angles, physically and geometric. An explanation and a figure can be found here: https://en.wikipedia.org/wiki/Inner_product_space or more elaborated here: https://arxiv.org/pdf/1205.5935.pdf
A Hilbert space is a vector space, but not every vector space is also a Hilbert space. That is: a Hilbert space has more properties.

Basically, yes, but the wording is a bit weird. Not the inner product forms anything, a Hilbert space is defined by being a real or complex vectorspace equipped with an inner product (additional property). The inner product defines measurements as angles and lengths, e.g. a distance: ##\operatorname{dist}(x,y)=||x-y||=\sqrt{\langle x-y,x-y \rangle}##, which is basically the theorem of Pythagoras here. This distance has to be complete, that is every Cauchy sequence (the elements of the sequence get closer and closer) converges, which means the infinite point towards the elements are narrowing down is actually part of the space. Completeness thus depends on how we measure distances. There is generally more than one way to define a distance. So Hilbert spaces are strictly speaking triplets: (vector space, field (real or complex), inner product).

E.g. with the rational numbers we'll get a problem. Let's define ##a_1=b_1=1\; , \;a_{n+1}=a_n+b_n\; , \;b_{n+1}=a_{n+1}+a_n\; , \;c_n = \dfrac{b_n}{a_n}##. Then ##(c_n)_{n\in \mathbb{N}}## defines a Cauchy sequence of rational numbers which converges to ##\sqrt{2}##. But this limit isn't part of the space ##\mathbb{Q}##, so ##\mathbb{Q}## is incomplete. If we take the exact same situation to the reals (or complex numbers), then this limit is part of the space - ##\mathbb{R}## and ##\mathbb{C}## are complete spaces, i.e. also Hilbert spaces.

This has nothing to do with eigenvectors. Besides, any two vectors are always eigenvectors of some linear function. Two linear independent vectors over the real or complex numbers span a two-dimensional real or complex vector space. E.g. with the Euclidean metric we get a Euclidean space. And this is a two-dimensional Hilbert space, as the real or complex numbers guarantee that any sequence of vectors which is narrowing down by Euclidean distance (Pythagoras) also contains the vector towards the sequence is converging.

Thanks for an excellent elucidation.

I basically want to state that two eigenvectors of a matrix form a Hilbert space, and use the following criteria to prove that:

The eigenvectors satisfy the condition

\langle x,y \rangle = \int_a^bx(t)\overline{y(t)}dt,

which is complete, and form the complex vector space ##L^2[a,b]## which satisfies the minimal condition:

x(t)\overline{x(t)}=|x(t)|^2.

However, I want to make sure I am not writing something odd here, or even connecting two things that are not connected. As from your reply, I understand that one can conclude that two vectors form a Hilbert space, but whether it is complete or not, requires the continuous nature of these, so to satisfy what you describe here as the Cauchy criteria. Is it correctly given here, with the "minimal condition"?

Thanks!


text:
# Finding the eigenvectors (& describing the eigenspace) of a Householder transformation matrix

If one is asked to find the eigenvector(s) for a Householder transformation matrix, but one is not given the values of or dimensions of the unit vector $u$.

So if $H = I_n - 2uu^T$ where $I_n$ is the n x n identity matrix and has length/norm $||u||^2 = 1$.

It can easily be shown that H is symmetric so that $H = H^T$ and that the eigenvalues are either (and only) of values $\lambda = 1$ and $\lambda = -1$.

Am I correct when assuming that the eigenvectors of $H$ would be the same as those for the identity matrix $I_n$. But is it possible to be more specific/detailed than that when describing the eigenvectors/eigenspaces of these eigenvalues?

• Every vector is an eigenvector of $I$, but not so for $H$. – kennytm May 19 '16 at 13:32

The geometric transformation encoded by $$R=I_n - 2 uu^T$$ is easy to understand: for any $v\perp u$ we have $Rv=v$, while $Ru=-u$, so $R$ encodes a reflection with respect to the orthogonal subspace of $u$. In your case $$H = I_n + 2uu^T$$ maps any $v\perp u$ into itself and $u$ into $3u$. In both cases, a base of eigenvectors is given by $u$ and a base of the orthogonal subspace of $u$.


text:
## Where communities thrive

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##### Activity
Dan Di Spaltro
@dispalt
Is there a way to read from the journal in reverse order, newest first?
Patrik Nordwall
@patriknw
@dispalt no, that is not supported
Released akka-persistence-cassandra 0.103 with a new Cleanup tool for deleting events and snapshots. https://doc.akka.io/api/akka-persistence-cassandra/0.103/akka/persistence/cassandra/cleanup/Cleanup.html
Dan Di Spaltro
@dispalt
@patriknw cool so I would have to basically copy the code (reuse as much as possible) to expose something like that?
Jacek Ewertowski
@jewertow
@patriknw @chbatey, when do you plan to release version 1.0?
Patrik Nordwall
@patriknw
we are about to release 1.0.0-RC1 later this week or early next week
anilkumble
@anilkumble

Hi @patriknw Patrik
I have used Cleanup tool to delete all the events related to persistence id : node1
It deleted all events successfully but it has one single entry with all null values. Why so ??
Below is that entry.

persistence_id | partition_nr | sequence_nr | timestamp | timebucket | used | event | event_manifest | message | meta | meta_ser_id | meta_ser_manifest | ser_id | ser_manifest | tags | writer_uuid
----------------+--------------+-------------+-----------+------------+------+-------+----------------+---------+------+-------------+-------------------+--------+--------------+------+-------------
node1 | 0 | null | null | null | True | null | null | null | null | null | null | null | null | null | null

Patrik Nordwall
@patriknw
@anilkumble ok, that is expected because used is a "static column". That column has been optionally removed/replaced in version 0.101+, but complete removal requires the steps described in the migration guide: https://doc.akka.io/docs/akka-persistence-cassandra/0.103/migrations.html#migrations-to-0-101
anilkumble
@anilkumble
Thank you @patriknw,
May I know the purpose of the below columns ?
1. event_manifest
2. message
3. ser_id
4. meta
5. meta_ser_id
6. meta_ser_manifest
7. ser_manifest
8. tags
9. writer_uuid
Patrik Nordwall
@patriknw
Latest documentation for this isn't published yet but you can see a preview at https://github.com/akka/akka-persistence-cassandra/blob/master/docs/src/main/paradox/journal.md#messages-table
anilkumble
@anilkumble
Thank you
anilkumble
@anilkumble
Anyone can help me to choose minimum hardware requirements for a typical production Cassandra server like RAM, cpu cores and etc ?
anilkumble
@anilkumble
Is there any limitation for persistence_id length ?
Can I have very lengthy persistence Id?
Also is there any size limitation for event(persistencable message) size ?
anilkumble
@anilkumble
@patriknw Can I use latest Akka Persistence Cassandra version 0.103(Latest) with Akka Persistence(All other Akka like cluster, ddata) version 2.5.23 ?
Patrik Nordwall
@patriknw
longer: 0.103 is built with Akka 2.5.29, which means that you can use it with 2.5.29 or later version of Akka (including 2.6.x) but not an earlier version. We are backwards binary compatible, but not forwards binary compatible.
persistence_id is a text column, so whatever limits Cassandra has to that
event size, it will be inefficient with too large events, and I think Cassandra has some limit, such as 16MB (but that is anyway way more than what I'd recommend)
anilkumble
@anilkumble
Okay, I can use scala 2.11 or 2.12. persistence Casandra latest version 0.103 is available for both scala version right ?
anilkumble
@anilkumble

@anilkumble ok, that is expected because used is a "static column". That column has been optionally removed/replaced in version 0.101+, but complete removal requires the steps described in the migration guide: https://doc.akka.io/docs/akka-persistence-cassandra/0.103/migrations.html#migrations-to-0-101

Here you mentioned the used column is optional in version 0.101+, So while creating message table i can remove used column right ?

p-andreas
@p-andreas

@patriknw regarding

longer: 0.103 is built with Akka 2.5.29, which means that you can use it with 2.5.29 or later version of Akka (including 2.6.x) but not an earlier version. We are backwards binary compatible, but not forwards binary compatible.

How can I use akka persistence 0.103 in combination with akka 2.6.x?

As far as I noticed, it is not possible to use akka peristence cassandra together with the current version of akka 2.6.x?
ManifestInfo.checkSameVerison does not allow such a scenario:

java.lang.IllegalStateException: Detected possible incompatible versions on the classpath. Please note that a given Akka version MUST be the same across all modules of Akka that you are using, e.g. if you use [2.6.4] all other modules that are released together MUST be of the same version. Make sure you're using a compatible set of libraries. Possibly conflicting versions [2.6.4, 2.5.29] in libraries [akka-persistence:2.6.4, akka-discovery:2.6.4, akka-coordination:2.6.4, akka-actor:2.6.4, akka-remote:2.6.4, akka-cluster:2.6.4, akka-protobuf-v3:2.6.4, akka-distributed-data:2.6.4, akka-cluster-typed:2.6.4, akka-cluster-tools:2.6.4, akka-actor-typed:2.6.4, akka-persistence-query:2.5.29, akka-slf4j:2.6.4, akka-persistence-typed:2.6.4, akka-stream:2.6.4]
[error]     at akka.util.ManifestInfo.checkSameVersion(ManifestInfo.scala:206)

Please let me know if I am missing something or if akka 2.6.x is not compatible with akka persistence cassandra

Patrik Nordwall
@patriknw
Scala version: use 2.12 then
used column: yes, you can skip creating it if you disable it with config as described in migration guide. Good to do that if you start now.
Akka 2.6: that’s fine, you just have to add all of the problematic dependencies explicitly in your build, with version 2.6.x
Patrik Nordwall
@patriknw
akka-persistence-query, from that error message
We are also releasing 1.0.0-RC1 on Monday, so you could start right there.
anilkumble
@anilkumble

used column: yes, you can skip creating it if you disable it with config as described in migration guide. Good to do that if you start now.

I am getting the below exception If I do not have used column in message column family

java.util.concurrent.ExecutionException: com.datastax.driver.core.exceptions.InvalidQueryException: Undefined column name used

I am using version 0.103
Why i am getting this ?

RAHUL REDDY KATIKI REDDY
@katikireddy622
Hi all
osboxes@osboxes:~/jmx$nodetool status Exception in thread "main" java.lang.reflect.InvocationTargetException at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) at java.lang.reflect.Method.invoke(Method.java:498) at sun.instrument.InstrumentationImpl.loadClassAndStartAgent(InstrumentationImpl.java:386) at sun.instrument.InstrumentationImpl.loadClassAndCallPremain(InstrumentationImpl.java:401) Caused by: java.net.BindException: Address already in use at sun.nio.ch.Net.bind0(Native Method) at sun.nio.ch.Net.bind(Net.java:433) at sun.nio.ch.Net.bind(Net.java:425) at sun.nio.ch.ServerSocketChannelImpl.bind(ServerSocketChannelImpl.java:220) at sun.nio.ch.ServerSocketAdaptor.bind(ServerSocketAdaptor.java:85) at sun.net.httpserver.ServerImpl.bind(ServerImpl.java:133) at sun.net.httpserver.HttpServerImpl.bind(HttpServerImpl.java:54) at io.prometheus.jmx.shaded.io.prometheus.client.exporter.HTTPServer.<init>(HTTPServer.java:145) at io.prometheus.jmx.shaded.io.prometheus.jmx.JavaAgent.premain(JavaAgent.java:31) ... 6 more FATAL ERROR in native method: processing of -javaagent failed Aborted (core dumped) my Cassandra Status is ------------------------------------ osboxes@osboxes:~/jmx$ sudo service cassandra status
● cassandra.service - LSB: distributed storage system for structured data
Active: active (running) since Mon 2020-03-23 13:29:27 EDT; 24min ago
Docs: man:systemd-sysv-generator(8)
Process: 6803 ExecStop=/etc/init.d/cassandra stop (code=exited, status=0/SUCCESS)
Process: 6920 ExecStart=/etc/init.d/cassandra start (code=exited, status=0/SUCCESS)
Tasks: 44 (limit: 4915)
CGroup: /system.slice/cassandra.service
└─6998 java -javaagent:/home/osboxes/jmx/jmx_prometheus_javaagent-0.12.0.jar=8888:/home/osboxes/jmx/config.yaml -Xloggc:/var/log/cassandra/gc.log -ea -XX:+UseThreadPriorities -XX:ThreadPriority

Mar 23 13:29:27 osboxes systemd[1]: Starting LSB: distributed storage system for structured data...
Mar 23 13:29:27 osboxes systemd[1]: Started LSB: distributed storage system for structured data.
lines 1-12/12 (END)
osboxes@osboxes:~/jmx$Andrew @mcandre Who's running BOINC? Patrik Nordwall @patriknw @anilkumble Did you add akka.persistence.cassandra.journal-static-column-compat = off to your application.conf? anilkumble @anilkumble Yes I added this entry in application.conf Patrik Nordwall @patriknw Do you also have a stack trace in the logs that you can share? Patrik Nordwall @patriknw @/all akka-persistence-cassandra 1.0.0-RC1 has been released. See https://discuss.lightbend.com/t/akka-persistence-cassandra-1-0-0-rc1-released/ and make sure you follow the migration guide https://doc.akka.io/docs/akka-persistence-cassandra/current/migrations.html Let us know if you find any problems. anilkumble @anilkumble Do you also have a stack trace in the logs that you can share? I am using this version: akka-persistence-cassandra_2.11 » 0.103 Below find the stack trace [INFO] [03/24/2020 12:26:06.440] [ReactorCluster-akka.actor.default-dispatcher-31] [akka.cluster.Cluster(akka://ReactorCluster)] Cluster Node [akka.tcp://ReactorCluster@127.0.0.1:4001] - Leader is moving node [akka.tcp://ReactorCluster@127.0.0.1:4001] to [Up] [ERROR] [03/24/2020 12:26:06.476] [ReactorCluster-akka.actor.default-dispatcher-31] [akka.tcp://ReactorCluster@127.0.0.1:4001/user/supervisor] Failed to persist event type [java.util.HashMap] with sequence number [1] for persistenceId [node1]. java.util.concurrent.ExecutionException: com.datastax.driver.core.exceptions.InvalidQueryException: Undefined column name used at com.google.common.util.concurrent.AbstractFuture.getDoneValue(AbstractFuture.java:476) at com.google.common.util.concurrent.AbstractFuture.get(AbstractFuture.java:435) at com.google.common.util.concurrent.AbstractFuture$TrustedFuture.get(AbstractFuture.java:79)
at akka.persistence.cassandra.package$ListenableFutureConverter KaTeX parse error: Can't use function '$' in math mode at position 5: anon$̲2: anon$2
anonfun$run$2.apply(package.scala:50)
at scala.util.Try$.apply(Try.scala:192) at akka.persistence.cassandra.package$ListenableFutureConverter$$anon2.run(package.scala:50) at akka.dispatch.TaskInvocation.run(AbstractDispatcher.scala:41) at akka.dispatch.ForkJoinExecutorConfiguratorAkkaForkJoinTask.exec(ForkJoinExecutorConfigurator.scala:49) at akka.dispatch.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:260) at akka.dispatch.forkjoin.ForkJoinPoolWorkQueue.runTask(ForkJoinPool.java:1339) at akka.dispatch.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1979) at akka.dispatch.forkjoin.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:107) Caused by: com.datastax.driver.core.exceptions.InvalidQueryException: Undefined column name used at com.datastax.driver.core.ResponsesError.asException(Responses.java:181) at com.datastax.driver.core.SessionManager4.apply(SessionManager.java:250) at com.datastax.driver.core.SessionManager4.apply(SessionManager.java:219) at com.google.common.util.concurrent.FuturesAsyncChainingFuture.doTransform(Futures.java:1442) at com.google.common.util.concurrent.FuturesAsyncChainingFuture.doTransform(Futures.java:1433) at com.google.common.util.concurrent.FuturesAbstractChainingFuture.run(Futures.java:1408) at com.google.common.util.concurrent.Futures21.run(Futures.java:1177) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149) at java.util.concurrent.ThreadPoolExecutorWorker.run(ThreadPoolExecutor.java:624) at io.netty.util.concurrent.FastThreadLocalRunnable.run(FastThreadLocalRunnable.java:30) at java.lang.Thread.run(Thread.java:748) @/all akka-persistence-cassandra 1.0.0-RC1 has been released. See https://discuss.lightbend.com/t/akka-persistence-cassandra-1-0-0-rc1-released/ and make sure you follow the migration guide https://doc.akka.io/docs/akka-persistence-cassandra/current/migrations.html Let us know if you find any problems. what is difference between version 0.103 and 1.0.0-RC1 Johan Andrén @johanandren Patrik Nordwall @patriknw and the release announcement describes most of the changes at a high level: https://discuss.lightbend.com/t/akka-persistence-cassandra-1-0-0-rc1-released/ Patrik Nordwall @patriknw @anilkumble I have double checked and all statements where the used column is included are guarded by the cassandra-journal.write-static-column-compat. I think your config is not working as you intend. My previous comment was wrong for that version. The config property is cassandra-journal.write-static-column-compat=off, not akka.persistence.cassandra.journal-static-column-compat as I said previously. That must have been changed accidentally in the migration guide. I'll fix. anilkumble @anilkumble @anilkumble I have double checked and all statements where the used column is included are guarded by the cassandra-journal.write-static-column-compat. I think your config is not working as you intend. My previous comment was wrong for that version. The config property is cassandra-journal.write-static-column-compat=off, not akka.persistence.cassandra.journal-static-column-compat as I said previously. Thanks @patriknw It is working now anilkumble @anilkumble When working with persistence, Below is the recieve recover method @Override public Receive createReceiveRecover() { return receiveBuilder() .match(String.class, str -> { System.out.println(str); }) .match(RecoveryCompleted.class, recoveryCompleted -> { int i=10/0; }) .build(); } here it will throw an arithmetic exception while receive RecoverCompleted message In this situation it is automatically kills the actor and recreating it. Is this expected ? If yes, Why ? struijenID @struijenID To me this is expected because you are dividing 10 by 0 when a RecoveryCompleted event arrives 😁 -> int i=10/0; anilkumble @anilkumble i=10/0; Yes I am adding this statement purposefully, If there is an exception then it can kill the actor I am wondering why it is recreating ? @Override public Receive createReceiveRecover() { return receiveBuilder() .match(String.class, str -> { int i=10/0; }) .match(RecoveryCompleted.class, recoveryCompleted -> { }) .build(); } For the above case, It just kills the actor. But If there is any exception in RecoveryCompleted message then it is killing and recreating it. And it went to infinity loop (like kills and recreate infinitely) anilkumble @anilkumble @patriknw Kindly help for the above doubt Patrik Nordwall @patriknw The restart is by the default supervision for actors. You can add supervision strategy to handle aritmeticException differently. https://doc.akka.io/docs/akka/current/fault-tolerance.html Exceptions when replaying events (second example) are different. They are treated as fatal, stopping the actor. By the way, since it looks like you are starting a new project I would strongly recommend that you use the new Actor APIs. For persistence the EventSourcedBehavior https://doc.akka.io/docs/akka/current/typed/persistence.html anilkumble @anilkumble May I know reason for the below exception "" "127.0.0.1" "kumble-8174" "" "" "" "47" "com.adventnet.mfw.StarterSysLogStream" "log" "INFO" "31-03-2020 21:05:15:100" "78" "[ERROR] [03/31/2020 21:05:15.098] [CircuitExecutionCluster-akka.actor.default-dispatcher-2] [akka.tcp://CircuitExecutionCluster@127.0.0.1:2551/user/supervisorActor] Persistence failure when replaying events for persistenceId [supervisorActor]. Last known sequence number [0] java.lang.IllegalStateException: Sequence number [1] still missing after [10.00 s], saw unexpected seqNr [7] for persistenceId [supervisorActor]. at akka.persistence.cassandra.query.EventsByPersistenceIdStage KaTeX parse error: Can't use function '' in math mode at position 5: anon̲1.lookForMissin…: anon1.lookForMissingSeqNr(EventsByPersistenceIdStage.scala:377) at akka.persistence.cassandra.query.EventsByPersistenceIdStage anon1.onTimer(EventsByPersistenceIdStage.scala:359) at akka.stream.stage.TimerGraphStageLogic.akkastreamstageTimerGraphStageLogic$onInternalTimer(GraphStage.scala:1593) at akka.stream.stage.TimerGraphStageLogic$anonfunakkastreamstageTimerGraphStageLogic KaTeX parse error: Can't use function '' in math mode at position 22: …erAsyncCallback̲1.apply(GraphSt…: getTimerAsyncCallback1.apply(GraphStage.scala:1582) at akka.stream.stage.TimerGraphStageLogic anonfunakkastreamstageTimerGraphStageLogic KaTeX parse error: Can't use function '' in math mode at position 22: …erAsyncCallback̲1.apply(GraphSt…: getTimerAsyncCallback1.apply(GraphStage.scala:1582) at akka.stream.impl.fusing.GraphInterpreter.runAsyncInput(GraphInterpreter.scala:452) at akka.stream.impl.fusing.GraphInterpreterShellAsyncInput.execute(ActorGraphInterpreter.scala:481) at akka.stream.impl.fusing.GraphInterpreterShell.processEvent(ActorGraphInterpreter.scala:581) at akka.stream.impl.fusing.ActorGraphInterpreter.akkastreamimplfusingActorGraphInterpreter processEvent(ActorGraphInterpreter.scala:749) at akka.stream.impl.fusing.ActorGraphInterpreter$$anonfun$receive$1.applyOrElse(ActorGraphInterpreter.scala:764)
at akka.actor.Actor$class.aroundReceive(Actor.scala:539) at akka.stream.impl.fusing.ActorGraphInterpreter.aroundReceive(ActorGraphInterpreter.scala:671) at akka.actor.ActorCell.receiveMessage(ActorCell.scala:612) at akka.actor.ActorCell.invoke(ActorCell.scala:581) at akka.dispatch.Mailbox.processMailbox(Mailbox.scala:268) at akka.dispatch.Mailbox.run(Mailbox.scala:229) at akka.dispatch.Mailbox.exec(Mailbox.scala:241) at akka.dispatch.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:260) at akka.dispatch.forkjoin.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:1339)
at akka.dispatch.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1979)


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 0 0 0 0.000%

## 문제

Dr. Extreme experimentally made an extremely precise telescope to investigate extremely curious phenomena at an extremely distant place. In order to make the telescope so precise as to investigate phenomena at such an extremely distant place, even quite a small distortion is not allowed. However, he forgot the influence of the internal gas affected by low-frequency vibration of magnetic flux passing through the telescope. The cylinder of the telescope is not affected by the low-frequency vibration, but the internal gas is.

The cross section of the telescope forms a perfect circle. If he forms a coil by putting extremely thin wire along the (inner) circumference, he can measure (the average vertical component of) the temporal variation of magnetic flux:such measurement would be useful to estimate the influence. But points on the circumference at which the wire can be fixed are limited; furthermore, the number of special clips to fix the wire is also limited. To obtain the highest sensitivity, he wishes to form a coil of a polygon shape with the largest area by stringing the wire among carefully selected points on the circumference.

Your job is to write a program which reports the maximum area of all possible m-polygons (polygons with exactly m vertices) each of whose vertices is one of the n points given on a circumference with a radius of 1. An example of the case n = 4 and m = 3 is illustrated below.

In the figure above, the equations such as “p1 = 0.0” indicate the locations of the n given points, and the decimals such as “1.000000” on m-polygons indicate the areas of m-polygons.

Parameter pi denotes the location of the i-th given point on the circumference (1 ≤ i ≤ n). The location p of a point on the circumference is in the range 0 ≤ p < 1, corresponding to the range of rotation angles from 0 to 2π radians. That is, the rotation angle of a point at p to the point at 0 equals 2πp radians. (π is the circular constant 3.14159265358979323846....)

You may rely on the fact that the area of an isosceles triangle ABC (AB = AC = 1) with an interior angle BAC of α radians (0 < α < π) is (1/2)·sin α, and the area of a polygon inside a circle with a radius of 1 is less than π.

## 입력

The input consists of multiple subproblems followed by a line containing two zeros that indicates the end of the input. Each subproblem is given in the following format.

n m
p1 p2 ··· pn

n is the number of points on the circumference (3 ≤ n ≤ 40). m is the number of vertices to form m-polygons (3 ≤ m ≤ n). The locations of n points, p1, p2, ..., pn, are given as decimals and they are separated by either a space character or a newline. In addition, you may assume that 0 ≤ p1 < p2 < ··· < pn < 1.

## 출력

For each subproblem, the maximum area should be output, each in a separate line. Each value in the output may not have an error greater than 0.000001 and its fractional part should be represented by 6 decimal digits after the decimal point.

## 예제 입력 1

4 3
0.0 0.25 0.5 0.666666666666666666667
4 4
0.0 0.25 0.5 0.75
30 15
0.00 0.03 0.06 0.09 0.12 0.15 0.18 0.21 0.24 0.27
0.30 0.33 0.36 0.39 0.42 0.45 0.48 0.51 0.54 0.57
0.61 0.64 0.66 0.69 0.72 0.75 0.78 0.81 0.84 0.87
40 20
0.351 0.353 0.355 0.357 0.359 0.361 0.363 0.365 0.367 0.369
0.371 0.373 0.375 0.377 0.379 0.381 0.383 0.385 0.387 0.389
0.611 0.613 0.615 0.617 0.619 0.621 0.623 0.625 0.627 0.629
0.631 0.633 0.635 0.637 0.639 0.641 0.643 0.645 0.647 0.649
0 0

## 예제 출력 1

1.183013
2.000000
3.026998
0.253581


text:
# How do you factor p^2+4p+4-q^2?

$\left(p + 2 + q\right) \left(p + 2 - q\right)$
${p}^{2} + 4 p + 4 - {q}^{2}$
$= {\left(p + 2\right)}^{2} - {q}^{2}$
Since ${a}^{2} - {b}^{2} - \left(a + b\right) \left(a - b\right)$
Similarly the above equation $= {\left(p + 2\right)}^{2} - {q}^{2}$ can be written as
$\left(p + 2 + q\right) \left(p + 2 - q\right)$


text:
# Can't see the preview at the bottom of a long post

Dynamically-updated previews are wonderful. But because the preview appears below the post, I can only see the top of it. If I'm adding to the bottom of a long post, the browser is working and working to update the preview, but I can't actually see the relevant part of the preview, because it is off the bottom of the screen!

Possible solutions would include scroll bars on the preview area, preview side-by-side with editing area instead of underneath it, or some sort of windowing effect where the preview area would show only the part of the preview around the part of the post that was most recently edited.

Here is a screenshot showing the typical situation on my (1680×1050 pixel) laptop display when editing this answer:

As you can see, only the top of the preview window is visible.

• Some context is missing. Are you using a screen so small (smartphone?) that it cannot vertically fit the height of the full preview and a few lines of the editor? Or, perhaps, are you wishing that the previewer forces scrolling? May 13 '12 at 21:01
• I am not using an unusually small screen, but I think the context is irrelevant. A sufficiently long post will exceed the size of any screen. I will post a typical screenshot.
– MJD
May 13 '12 at 23:10
• The screenshot says it all. I too have encountered that in long answers, but only rarely since my primary display is large (2560x1600). I agree it is a nontrivial problem. May 13 '12 at 23:20
• Here's the feature request on meta.SE to enable side-by-side Mardkwown preview. Apr 9 '15 at 22:09
• Your browser may allow two separate windows showing the same web page. Maybe this is what you want. Apr 10 '15 at 13:56

There is an old feature request Preview on the right please, optionally? which was never implemented by SE. The top answer there offered jQuery code which hides the right sidebar and moves the preview in its place. Because of some tweaks in SE page markup it no longer works, but here is an updated version as a bookmarklet. I tested it in Chrome and Firefox under Windows 7. Requirements:

• horizontal display resolution of at least 1400
• user must be logged in to SE (the code seems to be incompatible with SE interface for anonymous users)

Let me know if it fails to work for you. For completeness, the code is below.

javascript:(function(){
$("#sidebar").hide();$("#content").width(1360);
$("#post-editor").removeClass("post-editor");$("#post-editor").width(1360);
$(".community-option").css("float","left");$(".wmd-container").css("float","left");
\$("#wmd-preview").css({"clear":"none","margin-left":"20px","float":"left"});
})()

• Thanks. The JS function looks plausible, but isn't working for me; I tried tinkering with it with no success. Does it work for you?
– MJD
Jun 5 '13 at 15:01
• @MJD At the time it didn't, but after my edits, it does. Jun 6 '13 at 7:15
• @MJD: Are there any other recourses available now? Mar 6 '14 at 8:01
• Doesn't work. Just tried it on gis.stackexchange.com/posts/141770/edit. The width does increase, but the preview is still under the edit area. Apr 9 '15 at 21:51
• @DanDascalescu Indeed, seems broken. Something may have changed in the markup of SE pages since 2013.
– user147263
Apr 9 '15 at 22:03
• @pizza: I've posted a feature request on meta.SE to enable side-by-side Mardkwown preview. Apr 9 '15 at 22:05
• A few tweaks of this gets it working again on answer boxes: gist.github.com/shu8/3f854c2e0bcb5468b5ac Jul 12 '15 at 20:00
• @ᔕᖺᘎᕊ Thanks for fixing it (the user formerly known as ˈjuː.zɚ79365 :)
– user147263
Jul 12 '15 at 20:36
• @1999 wait what?? The owner of this answer is you!? Jul 12 '15 at 20:38
• For more info, ᔕᖺᘎᕊ's script has a post about it on StackApps. Oct 20 '15 at 16:23

I've run into the same problem while posting answers with tall screenshots.

I think there is a relatively easy solution available: adopt the same side-by-side fluid preview layout that Discourse uses. Discourse is forum software created by Jeff Atwood and others after he left StackOverflow.

Here's the feature request posted on meta.SE to enable side-by-side Mardkwown preview.

• This version of ˈjuː.zɚ79365's script I made seems to work on answer boxes... Jul 12 '15 at 20:04

Are there any other recourses available to this question? I tried user79365's bookmarklet using Windows 7 and Firefox 27.0. The result is the following. There is a preview on the RHS, but the input box ceases to function. To wit, after I click on the bookmarklet for that same window, I can't type anything in the input box.

• The bookmarklet works for me in Chrome (Windows7). Here is a screenshot; I even posted this answer, which was written for testing purpose. :) What browser are you using? Mar 6 '14 at 18:46
• Oh, I guess that's Firefox. I tested in Firefox 27, also Win7. Works. Mar 6 '14 at 18:52
• @ 127.0.9.6: Thanks for your response. I've retested it but the input box still registers nothing? Mar 10 '14 at 7:01
• Looks like we need official support for side-by-side previews. I'm posted the feature request on meta.SE to enable side-by-side Mardkwown preview. Apr 9 '15 at 22:08
• @DanDascalescu Thanks for the info. I upvoted your feature request there! Apr 10 '15 at 4:07

Addendum: This doesn't work; I am leaving it here for reference.

Why not: Tile Tabs can't tile a tab with itself. What it does it open a new tab to the same URL, and tile the old and new tabs together. But then editing in one tab doesn't affect the preview in the other tab.

I have finally found a good solution to this problem.

Firefox has an add-on called Tile Tabs. It lets you split the browser's viewport in half, horizontally or vertically, and place one of its tab views in each half. You can tile the contents of two different tabs into one window, which is useful if you are copying information from one tab to the other. And you can tile the same tab into the two halves of a window, with different views in each:

Here the edit area is in the left viewport, and the display area is in the right viewport. There is plenty of room in both. The screenshot is of a browser window that does not even take up the full width of my laptop screen.


text:
# Stone-Čech compactification of a discrete space

I would like to know:

Why is the Stone-Čech compactification of a discrete space the set of ultrafilters on that space?

We can show this by checking the following universal property for the Stone-Čech compactification:

If $X$ is a topological space, then the Stone-Čech compactification of $X$ consists of a compact Hausdorff space $\beta X$, together with a continuous map $\iota: X \to \beta X$, such that given any other compact Hausdorff space $Y$ and a continuous map $f : X \to Y$, there exists a unique continuous map $\tilde{f} : \beta X \to Y$ such that $\tilde{f} \circ \iota = f$.

So let $X$ be a discrete space. Note that an arbitrary function from $X$ to any other space is continuous, so we don't have to worry about checking continuity for maps coming out of $X$.

Let $F(X)$ be the set of ultrafilters of $X$. We claim that $F(X)$ is the Stone-Čech compactification of $X$. The topology on $F(X)$ is defined as follows. For any $S \subseteq X$, let ${F_{S}}(X)$ be the set of ultrafilters containing $S$. Then the sets ${F_{S}}(X)$ are closed under finite union: ${F_{S}}(X) \cup {F_{T}}(X) = {F_{S \cup T}}(X)$. (This is equivalent to the following property of ultrafilters: An ultrafilter contains $S \cup T$ iff it contains either $S$ or $T$.) Thus the sets ${F_{S}}(X)$ form a closed basis of a topology (closed sets are defined to be arbitrary intersections of sets of the form ${F_{S}}(X)$).

The map $\iota: X \to F(X)$ is defined by sending $x$ to the principal ultrafilter associated to $x$.

Now let $Y$ be any compact Hausdorff space, and consider a map $f: X \to Y$. How will we extend $f$ to $F(X)$? If $U$ is an ultrafilter, then for any $S \in U$, consider the subset of $Y$ defined by $f(S) \stackrel{\text{df}}{=} \{ f(s) \mid s \in S \}$. Now the properties of ultrafilters imply that for any finite number of elements $S_{1},\ldots,S_{n} \in U$, the intersection $\bigcap_{i} S_{i}$ is nonempty. Hence $\bigcap_{i} f[S_{i}]$ is nonempty as well. So any finite number of $f[S]$ have nonempty intersection. By compactness of $Y$, the intersection of all the closures, $\bigcap_{S \in U} \overline{f[S]}$, is nonempty.

We claim this intersection cannot have more than one element. If $y_{1},y_{2} \in Y$ and $y_{1} \neq y_{2}$, then the preimages (under $f$) of $y_{1}$ and $y_{2}$ are disjoint subsets of $X$, and so every ultrafilter of $X$ contains an element meeting one of the preimages but not the other. Hence the intersection $\bigcap_{S \in U} f[S]$ (before taking closures) does not contain more than one element. The intersection of closures $\bigcap_{S \in U} \overline{f[S]}$ also cannot contain more than one element, since if $y_{1}$ and $y_{2}$ were distinct elements in $\bigcap_{S \in U} \overline{f[S]}$, we use the Hausdorff property to find disjoint open subsets $Y_{1},Y_{2}$ of $Y$ separating $y_{1}$ and $y_{2}$, and reach a similar contradiction by considering the disjoint preimages of $Y_{1}$ and $Y_{2}$ under $f$.

Hence, for every ultrafilter $U$, the intersection $\bigcap_{S \in U} \overline{f[S]}$ contains a single element $y \in Y$. We define $\tilde{f}(U) \stackrel{\text{df}}{=} y$.

Now I will leave it to you to check all of the following:

1. $F(X)$ is a compact Hausdorff space;
2. $\tilde{f}$ extends $f$ (where $X$ is identified with its image under $\iota$);
3. $\tilde{f}$ is continuous;
4. $\tilde{f}$ the unique extension satisfying (2) and (3).

That is just one of many ways to define the Čech-Stone compactification $\beta D$ of a discrete space $D$. The Čech-Stone compactification of a Tikhonov space is characterized by the fact that it is the maximal compactification of that space, or by the fact that it has a certain universal property. Any construction that yields the maximal compactification of a space $X$, or that yields a compactification with that universal property, is a construction of $\beta X$, the Čech-Stone compactification of $X$. However, in the case of a discrete space $D$, the ultrafilter construction is in many ways the simplest.

And the answer to your question is that $\beta D$ is the set of ultrafilter on $D$ (equipped with the appropriate topology) precisely because the space that you get from that construction can be shown to have the universal property that characterizes $\beta D$.


text:
Can a metric conformal to a Kahler metric be Kahler? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:57:00Z http://mathoverflow.net/feeds/question/69792 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler Can a metric conformal to a Kahler metric be Kahler? Gunnar Magnusson 2011-07-08T13:01:27Z 2013-01-05T05:53:29Z
Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?

If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.

If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?
 http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69796#69796 Answer by Francesco Polizzi for Can a metric conformal to a Kahler metric be Kahler? Francesco Polizzi 2011-07-08T13:27:08Z 2011-07-08T13:34:17Z
There are examples in real dimension $4$ of manifolds having two conformally equivalent Kahler metrics, inducing the same conformal structure but with opposite orientation.

See the paper Ambikahler geometry, ambitoric surfaces and Einstein 4-orbifolds by Apostolov, Calderbank and Gauduchon.
 http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69798#69798 Answer by BS for Can a metric conformal to a Kahler metric be Kahler? BS 2011-07-08T13:49:59Z 2011-07-08T13:57:30Z
You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$.

This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$.

The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).

This said, as in Francesco's answer, you can have non proportional conformal riemannian metrics that are Kähler with respect to different complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.
 http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69799#69799 Answer by Spiro Karigiannis for Can a metric conformal to a Kahler metric be Kahler? Spiro Karigiannis 2011-07-08T14:08:44Z 2011-07-08T14:08:44Z
The paper by Apostolov, Calderbank, and Gauduchon that Francesco mentions find different Kaehler structures whose associated Riemannian metrics are conformal to each other. But they correspond to different complex structures $J_+$ and $J_-$.

I believe what Gunnar is asking is whether or not one can have $f \omega$ be closed and thus Kaehler with respect to the same complex structure $J$ associated to $\omega$. The answer is no, and this has nothing at all to do with compactness or the hard Lefschetz theorem. On any almost Hermitian manifold $(M, J, \omega, g)$, it is a fact that the wedge product with the Kaehler form $\omega$ on the space of $1$-forms is injective, regardless of the compactness of $M$, the integrability of $J$, or the closedness of $\omega$. This follows, for example, from the identity

$$\ast( \omega \wedge (\ast ( \omega \wedge \alpha) ) = - (m-1) \alpha$$

where $\alpha$ is any $1$-form on $M$, where $\ast$ is the Hodge star operator, and the real dimension of $M$ is $2m$. (One sees that the only requirement is that $m>1$.)


text:
Settings help

## 1Purpose

e04ab is the AD Library version of the primal routine e04abf. Based (in the C++ interface) on overload resolution, e04ab can be used for primal, tangent and adjoint evaluation. It supports tangents and adjoints of first order.

## 2Specification

Fortran Interface
 Subroutine e04ab_AD_f ( ad_handle, funct, e1, e2, a, b, maxcal, x, f, iuser, ruser, ifail)
 Integer, Intent (Inout) :: maxcal, iuser(*), ifail ADTYPE, Intent (Inout) :: e1, e2, a, b, ruser(*) ADTYPE, Intent (Out) :: x, f Type (c_ptr), Intent (Inout) :: ad_handle External :: funct
Corresponding to the overloaded C++ function, the Fortran interface provides five routines with names reflecting the type used for active real arguments. The actual subroutine and type names are formed by replacing AD and ADTYPE in the above as follows:
#include <dco.hpp>
namespace nag {
}
}
The function is overloaded on ADTYPE which represents the type of active arguments. ADTYPE may be any of the following types:
double,
dco::ga1s<double>::type,
dco::gt1s<double>::type

## 3Description

e04ab is the AD Library version of the primal routine e04abf.
e04abf searches for a minimum, in a given finite interval, of a continuous function of a single variable, using function values only. The method (based on quadratic interpolation) is intended for functions which have a continuous first derivative (although it will usually work if the derivative has occasional discontinuities). For further information see Section 3 in the documentation for e04abf.

## 4References

Gill P E and Murray W (1973) Safeguarded steplength algorithms for optimization using descent methods NPL Report NAC 37 National Physical Laboratory

## 5Arguments

In addition to the arguments present in the interface of the primal routine, e04ab includes some arguments specific to AD.
A brief summary of the AD specific arguments is given below. For the remainder, links are provided to the corresponding argument from the primal routine. A tooltip popup for all arguments can be found by hovering over the argument name in Section 2 and in this section.
On entry: a handle to the AD configuration data object, as created by x10aa.
2: funct – Subroutine External Procedure
The specification of funct is:
Fortran Interface
 Subroutine funct ( ad_handle, xc, fc, iuser, ruser)
 Integer, Intent (Inout) :: iuser(*) ADTYPE, Intent (In) :: xc ADTYPE, Intent (Inout) :: ruser(*) ADTYPE, Intent (Out) :: fc Type (c_ptr), Intent (Inout) :: ad_handle
On entry: a handle to the AD configuration data object.
2: Input
3: Output
4: iuser($*$) – Integer array User Workspace
5: ruser($*$)ADTYPE array User Workspace
3: Input/Output
4: Input/Output
5: Input/Output
6: Input/Output
7: maxcal – Integer Input/Output
8: Output
9: Output
10: liuser Input
User workspace dimension (C++ only), see x10af to specify the dimension from Fortran.
11: iuser($*$) – Integer array User Workspace
User workspace.
12: lruser Input
User workspace dimension (C++ only), see x10ae to specify the dimension from Fortran.
13: ruser($*$) – ADTYPE array User Workspace
User workspace.
14: ifail – Integer Input/Output

## 6Error Indicators and Warnings

e04ab preserves all error codes from e04abf and in addition can return:
${\mathbf{ifail}}=-89$
See Section 4.8.2 in the NAG AD Library Introduction for further information.
${\mathbf{ifail}}=-199$
The routine was called using a mode that has not yet been implemented.
${\mathbf{ifail}}=-443$
This check is only made if the overloaded C++ interface is used with arguments not of type double.
${\mathbf{ifail}}=-444$
A C++ exception was thrown.
The error message will show the details of the C++ exception text.
${\mathbf{ifail}}=-899$
Dynamic memory allocation failed for AD.
See Section 4.8.1 in the NAG AD Library Introduction for further information.

Not applicable.

## 8Parallelism and Performance

e04ab is not threaded in any implementation.

None.

## 10Example

The following examples are variants of the example for e04abf, modified to demonstrate calling the NAG AD Library.
Description of the primal example.
A sketch of the function
 $F(x)=sinxx$
shows that it has a minimum somewhere in the range $\left[3.5,5.0\right]$. The following program shows how e04ab can be used to obtain a good approximation to the position of a minimum.

Language Source File Data Results
Fortran e04ab_a1w_fe.f90 None e04ab_a1w_fe.r
C++ e04ab_a1w_hcppe.cpp None e04ab_a1w_hcppe.r

### 10.2Tangent modes

Language Source File Data Results
Fortran e04ab_t1w_fe.f90 None e04ab_t1w_fe.r
C++ e04ab_t1w_hcppe.cpp None e04ab_t1w_hcppe.r

### 10.3Passive mode

Language Source File Data Results
Fortran e04ab_p0w_fe.f90 None e04ab_p0w_fe.r
C++ e04ab_p0w_hcppe.cpp None e04ab_p0w_hcppe.r


text:
# MATLAB: Grouping values of 3 columns

grouping values of 3 columns

hello all, i have one question which is related to grouping columns. for instance:
data=[2 0 2; 0 0 0; 0 2 0; 0 0 12];
what i want are
202 as one row and one col000 as one row and one col020 as one row and one col0012 as one row and one col

#### Best Answer

• Not clear what you mean: Do you want a numeric array or a string? If it's numeric, there's no 020 but only 20. If you want string, you'll have to use cells. One possibility is:
cellfun(@(c)sprintf('%d',c),mat2cell(data,ones(size(data,1),1),size(data,2)),'uniform',false)ans =       '202'      '000'      '020'      '0012'
If it's still numeric, you could do this:
sum(bsxfun(@times,data,10.^(size(data,2)-1:-1:0)),2)ans =     202       0      20      12


text:
# Security Implications of chroot

DONG Yuxuan @ Jan 07, 2020 Asia/Shanghai

The famous FTP server program vsftpd has an option chroot_local_user. If it’s set to YES local users will be (by default) placed in a chroot() jail in their home directory after login. Intuitively this would make the server more secure but the document of vsftpd notes that there could be a security implication. This article explains the reason.

When a user connects to the server, the FTP daemon will, of course, call some system-calls to function. Thus it may load things from a dynamic library, for example, libc. The user can put a faked library in $HOME/lib/x86_64-linux-gnu/libc.so.6. There is no damage if chroot is not called. However, if it does, the FTP daemon will think the faked libc is the real one because $HOME becomes the root path, and of course, you’re hacked.

However, after a discussion with Gogdizzy, I have realized this case is very hard to trigger. For the above example about libc, the FTP daemon usually has loaded libc before chroot is called. The more important point is that if an FTP daemon loads any dynamic libraries after calling chroot, it will not work because it can’t find the library if the user doesn’t fake it.

Besides faking dynamic libraries, faking config files like /etc/hosts is also very hard. The reason is the same. If an FTP daemon reads a config file after calling chroot, it won’t work if the user doesn’t fake the file.

So, to trigger the case the FTP daemon must have a “load if it exists” or “read if it exists” strategy. It’s hard to find a reason why a secure-conscious FTP daemon will perform the strategy. However, the whole system is very complicated. It will be too harsh to ask the programmer developing the FTP daemon to figure out every strategy of every function he called. God knows if there’s a library that takes the strategy.


text: 2. $\displaystyle (x-a)^2+(y-b)^2+(z-c)^2=r^2$ is the general form of a sphere with center $\displaystyle (a,b,c)$ .


text:
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# A certain right triangle has sides of length x, y, and z, wh

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

Problem Solving
Question: 157
Category: Geometry; Algebra Triangles; Area; Inequalities
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Re: A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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SOLUTION

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$
(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$
(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$
(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$
(E) $$y<\frac{\sqrt{3}}{4}$$

The area of the triangle is $$\frac{xy}{2}=1$$ ($$x<y<z$$ means that hypotenuse is $$z$$) --> $$x=\frac{2}{y}$$. As $$x<y$$, then $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

Also note that max value of $$y$$ is not limited at all. For example $$y$$ can be $$1,000,000$$ and in this case $$\frac{xy}{2}=\frac{x*1,000,000}{2}=1$$ --> $$x=\frac{2}{1,000,000}$$.
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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20 Nov 2015, 14:11
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

This was a tough problem for me, to be honest I couldn't find a solution like Bunuel did, but I have tested aswer choices to derive at the correct answer.

We know that $$x*y=2$$ actually let's square this expression $$x^2*y^2=4$$ so let's test the values given in the answer choices with Min/Max approach (we ca also square values in the answer choices, as sides of a triangle can not be -ve)

(B) so $$y^2$$ can be max 2 and $$x^2$$= max $$\frac{3}{4}$$--> multiply $$x^2*y^2$$=1,5. So B is out because we need a 4 when $$x^2 and y^2$$ are multiplied
(C) $$y^2$$ max = $$\frac{3}{4}$$, $$x^2$$ max=2/9 --> $$\frac{2}{9}*\frac{3}{4} < 4$$ , C is out
(D) $$\frac{2}{9}*\frac{3}{16} < 4$$
(E) $$y^2$$ < $$\frac{3}{16}$$, so we know that y > x, thus when y < $$\frac{3}{16}$$ $$x^2$$ is also < $$\frac{3}{16}$$ and their product is < 4, E is also out and we are left with Answer Choice A
(A)
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Re: A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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27 Apr 2016, 17:59
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

Thank you![/textarea]

Hi,
from the area of triangle we know x*y = 2.
So there is one possibility that x could be 1 and y could be 2, but none of the answer choices ,except A confirms to y =2 possibility.

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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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27 Apr 2016, 20:00
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

You can think about it in terms of transition points too.

x < y< z so this means that x and y are the legs of the right triangle and z is the hypotenuse.
The area of the triangle will be (1/2)*xy = 1
xy = 2

Now, if x = y, then both x and y would be equal to $$\sqrt{2}$$.
But y is greater than x, so y would be at least a slight bit greater than $$\sqrt{2}$$ and x would be a slight bit less than $$\sqrt{2}$$. In all options other than (A), y takes values less than 1.414.
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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27 Apr 2016, 20:12
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

First step will be to breakdown the options into recognizable decimal representations (assuming $$\sqrt{2} \approx 1.4$$, $$\sqrt{3} \approx 1.7$$)

A) y>1.4
B) 0.8<y<1.4
C) 0.5<y<0.8
D) 0.4<y<0.5
E) y<0.4

We are given that x<y<z and that 0.5*x*y=1 --> x*y=2

Now from the relation xy=2 --> go back to the options and test for y=1. You get x=2 but we are given that x<y ---> y MUST be > $$\approx$$1.4 such that x < y

For any value of y < 1.4 , you will end up getting x>y (try with y=0.5 or 0.75 etc).

Only A satisfies this condition and is hence the correct answer.

Hope this helps.
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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29 Apr 2018, 01:25
VeritasPrepKarishma wrote:
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

(A) $$y >\sqrt{2}$$

(B) $$\frac{\sqrt{3}}{2}<y<\sqrt{2}$$

(C) $$\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}$$

(D) $$\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}$$

(E) $$y<\frac{\sqrt{3}}{4}$$

You can think about it in terms of transition points too.

x < y< z so this means that x and y are the legs of the right triangle and z is the hypotenuse.
The area of the triangle will be (1/2)*xy = 1
xy = 2

Now, if x = y, then both x and y would be equal to $$\sqrt{2}$$.
But y is greater than x, so y would be at least a slight bit greater than $$\sqrt{2}$$ and x would be a slight bit less than $$\sqrt{2}$$. In all options other than (A), y takes values less than 1.414.

Hi can you Please help me understand why have you assumed if x= y then the value for both is equal to root 2 ?
I understood that x < y< z and if z is the Hypotenuese and therefore in case its a 30 60 90 triangle then hypotenuse coud hold the value of root 2
But how Y could hold the value higher than root 2 and x an even smaller value than root 2 ? also in the equation above we dervided xy=2 don't we use this finding ?
A certain right triangle has sides of length x, y, and z, wh &nbs [#permalink] 29 Apr 2018, 01:25
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text:
# How to draw resource allocation graph in the picture?

I have found some similar solutions by tikz. Here is the link.

https://github.com/MartinThoma/LaTeX-examples/blob/master/tikz/resource-allocation-graph/resource-allocation-graph.tex

https://github.com/MartinThoma/LaTeX-examples/tree/master/tikz/resource-allocation-graph

But the graph in the book which has some variations.

It have some dots in the rectangles and some lines are from the dots not from rectangles.

So if there is some solution based on that one to meet the requirements, it is nice. Thanks.

Not sure if I'm really answering your question, but here you go. The shading on the circles isn't quite right, but I think the rest is pretty close.

Code:

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows}
\usepackage{xcolor}

\definecolor{sqBlue}{RGB}{218,234,233}
\definecolor{ballBlue}{RGB}{40,156,156}

\begin{document}
\begin{tikzpicture}[x=0.6in,y=0.6in]

%% Draw the rectangles
\draw[fill = sqBlue] (-0.5,0) rectangle (0.5,1);
\node at (0,0.75) {Ra};
\draw[fill = sqBlue] (-0.5,-3.5) rectangle (0.5,-2.5);
\node at (0,-3.25) {Rb};

%% Draw the big circles
\fill[ball color = ballBlue] (1.6875,-1.25) circle[radius=0.5] node {P2};
\fill[ball color = ballBlue] (-1.6875,-1.25) circle[radius=0.5] node {P1};

%% Draw the small dots
\coordinate (a) at (-0.344,0.19);
\coordinate (b) at (-0.0625,0.19);
\coordinate (c) at (0.25,0.19);
\coordinate (d) at (-0.0625,-2.69);
\coordinate (e) at (0.22,-2.69);

%% Draw the arrows
\draw[-latex', very thick] (b) to node[sloped, above, inner sep = 2pt] {\footnotesize\hspace{1em}Held by} (1.37,-0.88);
\draw[latex'-, very thick] (0.375,-2.5) to node[sloped, above, inner sep = 2pt] {\footnotesize Requests} (1.37,-1.63);
\draw[-latex', very thick] (d) to node[sloped, above, inner sep = 2pt] {\footnotesize Held by} (-1.37,-1.63);
\draw[-latex', very thick] (-1.37,-0.88) to node[sloped, above, inner sep = 2pt] {\footnotesize\hspace{-2em}Requests} (-0.19,0);

\end{tikzpicture}
\end{document}

• It couldn't be compiled successfully. Here is the error message: "l.37 \draw[latex'-, very thick] (0.375,-2.5) to node[sloped, above, inner se..." – Christophe Mar 9 '16 at 2:27
• @Christophe Can you share the entire log file? The snippet you gave is not enough to diagnose the issue. – Tony Mottaz Mar 9 '16 at 2:29
• paste.ubuntu.com/15332111 – Christophe Mar 9 '16 at 2:33
• Ah! I somehow lost a square bracket when copying my code. My answer has been updated, and it should work now. – Tony Mottaz Mar 9 '16 at 2:37
• Thanks a lot! I use your example to make my own graph successfully. – Christophe Mar 9 '16 at 3:49