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## Charmed baryons at the physical point in 2+1 flavor lattice QCD    [PDF]

Y. Namekawa, S. Aoki, K. -I. Ishikawa, N. Ishizuka, K. Kanaya, Y. Kuramashi, M. Okawa, Y. Taniguchi, A. Ukawa, N. Ukita, T. Yoshié
We investigate the charmed baryon mass spectrum using the relativistic heavy quark action on 2+1 flavor PACS-CS configurations previously generated on $32^3 \times 64$ lattice. The dynamical up-down and strange quark masses are tuned to their physical values, reweighted from those employed in the configuration generation. At the physical point, the inverse lattice spacing determined from the $\Omega$ baryon mass gives $a^{-1}=2.194(10)$ GeV, and thus the spatial extent becomes $L = 32 a = 2.88(1)$ fm. Our results for the charmed baryon masses are consistent with experimental values, except for the mass of $\Xi_{cc}$, which has been measured by only one experimental group so far and has not been confirmed yet by others. In addition, we report values of other doubly and triply charmed baryon masses, which have never been measured experimentally.
View original: http://arxiv.org/abs/1301.4743


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I have issue in question named as C. Rumor -893C, link :- https://codeforces.com/problemset/problem/893/C. I think that i have understand the questions basic as we have to find number of component in graph with minimum price of component and simply we have to do addition i on them. I have come up with pretty good solution but test case 8 getting fail. can you help me where is the problem lies in my code.

include<bits/stdc++.h>

using namespace std;

typedef long long ll;

int solve(){ ll n,m; cin>>n>>m; ll a[n]; for(ll i=0;i<n;i++) cin>>a[i]; for(ll i=0;i<m;i++){ ll x,y; cin>>x>>y; x--;y--; ll t=min(a[x],a[y]); a[x]=t; a[y]=0; } ll res=0; for(ll i=0;i<n;i++) res+=a[i]; cout<<res<<endl; return 0; }

int main(){ ll t=1; //cin>>t; while(t--) solve(); return 0; }

• -2

 » 11 days ago, # |   0 Auto comment: topic has been updated by adityav664 (previous revision, new revision, compare).
 » 11 days ago, # |   +3 Format your code for christsake, no one will read that
 » 11 days ago, # | ← Rev. 2 →   0 What the heck!! Your code seems like a shit!!It is already unreadable if you don't know how to write a blog learn from someone or just stop shitting around you . The pattern is following: first you you write "  " without quotes in start and in end of code. example : note : quotes are for example." "print("Hello world!")" " it will look like this print("Hello world!")
 » 11 days ago, # | ← Rev. 4 →   0 The idea is correct but your implementation seems wrong to me.You have just considered every edge and made one of them equal to min(first, second) while making the other one $0.$ Do you think it will actually work? HintWhen you make vertex $x$ cost = $min($ $c_x$, $c_y$ $)$ and vertex $y$ cost = $0$, you are actually saying that the minimum cost vertex for this component must not be connected to $y$. (as you made $c_y$ $= 0$ which is less than any possible vertex cost. Test Case (Don't see before understanding the hint)3 2 5 6 1 1 2 2 3
•  » » 11 days ago, # ^ |   0 thanks i understand now
 » 11 days ago, # |   0 It's my first time for post so sorry for that
•  » » 11 days ago, # ^ |   0 I think you can just remove your code. The submission link is enough. Also if your doubt is resolved just write [SOLVED] or [RESOLVED]` at the start of topic so people know your doubt has been cleared.
•  » » » 11 days ago, # ^ |   0 Thanks
 » 11 days ago, # |   0 This is my submission: 167158337.
•  » » 11 days ago, # ^ |   0 Got It

format: plain


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# Deriving the $0$-component of 4-momentum using the relativistic Lagrangian

My question arises from Susskinds book on Special Relativity and Classical Field Theory. (page 102 equation 3.29 to 3.30 and page 105 equation 3.34 to 3.36.)

The relativistic Lagrangian for a free particle is given by the following equation. $$L = -mc^2\sqrt{1-\frac{v^2}{c^2}} = \frac{-mc^2}{\dot{X}^0},\tag{1}$$ where dot means differentiation with respect to the proper time. The $$i^{th}$$ component of momentum is given by ($$i=1, 2, 3$$), $$P_{i} = \frac{\partial L}{\partial \dot{X^{i}}}.\tag{2}$$

This definition works perfectly fine for the 3 spatial components of relativistic momentum and gives $$P_{i} = m\dot{X^{i}}.\tag{3}$$

However, for the time component of 4-momentum, Susskind uses the relativistic Hamiltonian to derive $$P_{0} = m\dot{X^{0}}.\tag{4}$$

I am aware that the time component of 4-momentum corresponds to the energy, but I would like to know why we can't use the Lagrangian definition: $$P_{0} = \frac{\partial L}{\partial \dot{X^{0}}}\tag{5}$$ here.

I am new to this subject and would be really grateful for any help or insights.

That's a good question.

1. Note first of all that it is inconsistent to use proper time $$\tau$$ as the world-line (WL) parameter $$\lambda$$ for the principle of stationary action (PSA). The point is that the WL parameter $$\lambda$$ is never varied in the PSA, but the action $$S$$ happens to be proportional to $$\tau$$, which we are trying to maximize. In particular, the rightmost expression $$-m_0c^2\left(\frac{dx^{0}}{d\tau}\right)^{-1}$$ in OP's eq. (1) cannot be used as an off-shell formula for the Lagrangian $$L$$, although correct in value. The same issue is discussed in my Phys.SE answer here using slightly different words.

2. In Ref. 1 the WL parameter $$\lambda=t\equiv \frac{x^0}{c}$$ is instead the laboratory time, i.e. it uses the static gauge where $$\dot{x}^0=c$$. (In this answer dot means differentiation wrt. $$\lambda$$.) Conceptually this is the easiest route. However, this destroys manifest (but not actual) Lorentz covariance, so the derivative $$\frac{\partial L}{\partial \dot{x}^0}$$ does not make sense. Ref. 1 therefore obtains the 0-component $$p_0$$ in a roundabout manner, which is equivalent to my Phys.SE answer here.

3. Finally, let us return to OP's question: Yes, there exists a manifest Lorentz covariant formulation where $$p_0=\frac{\partial L}{\partial \dot{x}^0}$$, but it involves gauge symmetry and constraints, and is conceptually more challenging, cf. e.g. my Phys.SE answers here & here.

References:

1. L. Susskind & A. Friedman, Special Relativity and Classical Field Theory: The Theoretical Minimum, 2017; p. 102-106.
• I don't know enough to clearly understand point 1 and 2 (i hope to get there soon). However, point 3 satisfies my curiosity for now. That said, regarding the rightmost expression in equation 1, i don't see why its wrong because $\dot{X}^0$ is $\frac{1}{\sqrt{1-(v/c)^2}}$ – a_point_particle May 31 '20 at 5:08
• I updated the answer. – Qmechanic May 31 '20 at 6:31

$$X_i$$ is being differentiated with respect to the proper time $$X_0$$ only. So if you consider the derivative of $$X_0$$ with respect to $$X_0$$, that's one, and hence $$d(\dot{X_0})$$ is identically zero! However, if you wish to use the Lagrangian only to calculate energy, you can appeal to Noether's theorem and calculate the Noether's charge corresponding to time translations. I hope this helps.


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# Is there a version of Miller forcing "guided by" an ultrafilter?

It’s well known that Mathias forcing factors as a two-step iteration $P(\omega)/\mathrm{fin}\ast \mathbb{M}_{\dot U}$, where $\mathbb{M}_{\dot U}$ is Mathias forcing guided by the generic ultrafilter added by the first poset. That is, Mathias forcing is the two-step iteration of a countably closed poset and a $\sigma$-centered poset. In a similar way, Laver forcing completely embeds into a closed $\ast$ centered poset. It’s natural to wonder whether their cousin Miller forcing (aka rational perfect set forcing) has the same property. While Laver and Mathias forcings have natural ccc versions guided by an ultrafilter, it’s not clear (to me) how to guide Miller forcing by an ultrafilter. Here is my question:

Does Miller forcing completely embed into the two-step iteration of a $\sigma$-closed poset and a $\sigma$-centered poset?

This question allows for slightly more than the vague question of the title; I suppose the first closed poset might be something other than $P(\omega)/\mathrm{fin}$.

One reason to ask this is that a "Yes" would imply that finite products of Miller, Laver, and Mathias forcing are all proper. [Why? Every poset completely embeddable into a closed $\ast$ centered poset is proper, and the class of such posets is closed under finite products.] Note that Spinas has already established the properness of finite powers of Miller & Laver forcing.

Of course, I’d be grateful for whatever references—for negative or positive results—you’re willing to provide.

The answer is no. Like Sacks forcing, Miller forcing adds a minimal degree, meaning that if $G$ is the Miller generic and $x\in M[G]$, then either $x\in M$ or $G\in M[x]$. Hence we cannot have Miller forcing $\mathbb{Q}$ equivalent to a two-step iteration $\mathbb{P}_0*\mathbb{P}_1$ since the generic for $\mathbb{P}_0$ contradicts the minimality of the degree corresponding to the generic for $\mathbb{P}_0*\mathbb{P}_1$.

The minimality of the generic added by $\mathbb{Q}$ is a consequence of Theorem 2 in "Combinatorics on ideals and forcing with trees" by Marcia Groszek. I think this is also covered in Miller's original paper, Rational perfect set forcing, though I don't have access right now.

• But this doesn't preclude Miller forcing from embedding into a countably closed * ccc forcing, right? Mar 12, 2015 at 17:54
• @ToddEisworth No, this would be true if "minimal degree" meant "minimal real degree", as it often does, but Miller and Groszek consider minimal degrees in the general sense: if $X \in M\setminus M[G]$ then $G \in M[X]$. Mar 13, 2015 at 1:04

More of an extended comment than an answer:

Sabok and Zapletal have considered Miller forcing "guided" by $J^+$ for an ideal $J$ in their paper "Forcing properties of ideals of closed sets", J. Symbolic Logic 76 (2011) no 3, 1070-1095.

Also, you may want to take a look at Blass's paper "Applications of superperfect forcing and its relatives", Lecture notes in mathematics 1401, 1989, pp. 18-40, where the analogy "Miller forcing is to Matet forcing as Laver forcing is to Mathias forcing" is discussed.

Matet forcing factors as $\sigma$-closed * $\sigma$-centered, so maybe you can prove something analogous to what we have for Laver.

• Thanks for the references. It didn't occur to me to look at Matet forcing. I'll think about it. Mar 10, 2015 at 2:05


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# Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both numbers are odd.

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Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both numbers are odd.

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by (17.1k points)
selected by

The integers from 1 through 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Out of these, there are 5 even and 6 odd integers.

Let A: Both numbers chosen are odd

B: Sum of numbers is even at random

S: Choosing 2 numbers from 11 numbers.

Then, n(S) = 11C2

n(A) = 6C2 (∴  There are 6 odd integers)

As the sum of both chosen integers can be even if both are even or both are odd, so

n(B) = 6C2 + 5C2

and n(A ∩ B) = 6C2

∴                               P(A) = $\frac{n(A)}{n(S)} = \frac {^6C_2}{^{11}C_2}= \frac{6\times 5}{11 \times 10} =\frac{3}{11}$

P(B) = $\frac{n(B)}{n(S)} = \frac {^6C_2+^5C_2}{^{11}C_2}= \frac{6\times 5+ 5\times 4}{11 \times 10} =\frac{5}{11}$

$P(A\,\cap\,B) = \frac{n(A\,\cap\,B)}{n(S)} =\frac{^6C_2}{^{11}C_2} =\frac{5}{11}$


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# Dividend Payout Ratio

Dividend payout ratio is the percentage of a company’s earnings that it pays out to investors in the form of dividends. It is calculated by dividing dividends paid during a period by net earnings for that period.

Dividend payout ratio is reciprocal of retention ratio (or plowback period) which measures the percentage of earnings a company reinvests in projects to generate future growth.

Dividend payout ratio is an important indicator of a company’s performance from an investor’s point of view.

## Formula

 Dividend Payout Ratio = Dividends Paid = Dividend per Share Net Income Earnings per Share

## Analysis

People invest in a company expecting a return on their investment which comes from two sources: capital gains and dividends. The return from these two sources is inter-related. A high dividend payout ratio means that the company is reinvesting less earnings in future projects, which in turn means less capital gains in future periods. Similarly, low payout ratio today may result in higher capital gains in future.

Some investors prefer companies that provide high potential for capital gains while others prefer companies that pay high dividends. Dividend payout ratio helps each class of investors identify which companies to invest in.

Dividend payout ratio also provides an indication of a company’s future growth potential. For investors that are interested in high growth companies, a high dividend payout ratio may not be a good sign.

Dividend payout ratio should be analyzed in the context of industry and other ratios such as dividend yield ratio, P/E ratio, etc.

A more meaningful dividend analysis would also include analysis of any share buyback by a company. A share buyback is a way of returning cash to investors which involves purchase of own shares by companies. Such share buybacks increase share price of the company, which in turn results in capital gains for the shareholders.

## Example

Based on the information given below, calculate and analyze dividend payout ratio for Apple, Inc. and ExxonMobil Corp.

All amounts are USD in million.

 2011 2012 2013 2014 Apple (NYSE: AAPL) Dividends - 2,488 10,564 11,126 Net income 25,922 41,733 37,037 39,510 Cash and investments 25,952 29,129 40,590 25,158 ExxonMobil (NYSE:XOM) Dividends 9,020 10,092 10,875 11,568 Net income 41,060 44,880 32,580 32,520 Cash and investments 12,664 9,582 4,644 4,616

Solution

 Dividend Payout Ratio for AAPL for 2012 = $2,488 = 5.96%$41,733

The table below shows dividend payout ratios for AAPL and XOM from 2011 till 2014.

 Industry 2011 2012 2013 2014 AAPL Technology 0 5.96% 28.52% 28.16% XOM Oil and Gas 21.97% 22.49% 33.38% 35.57%

Apple, Inc. did not pay any dividends in 2011 because the management believed that higher return for investors can be achieved if the earnings generated are reinvested in projects that will generate future growth. This is supported by the company’s exceptional revenue growth in 2012. However, during the period from 2012 till 2014 the company’s cash pile was way above the level needed to avail the feasible new projects, so the management paid generous dividends in these years. However, in case of a technology company, high dividend payouts are an exception and not a rule.

ExxonMobil Corp. on the other hand is in a mature industry which it is expected to maintain a steady dividend payout and even increase it over periods. XOM payout ratio has hovered in the range of 23-36% in the four years which is pretty stable.

Written by Obaidullah Jan, ACA, CFA <--- Hire me on Upwork


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# How would a neutron star affect the habitability of planets orbiting a companion star?

A star system consisting of two stars forms, with one being more massive. As such, the larger star soon exhausts its fuel and ends up as a neutron star. The secondary star has a semi-major axis of 700 au, with the closest approach being 438 au.

Based on the information, would the secondary star be able to harbor life-supporting planets? The two issues are whether or not it would have been distant enough to survive the death of the primary star, and whether or not the resulting neutron star would produce too much radiation.

• There are different kinds of lifeforms. You can imagine all sorts of things, some people talk about silicon based life, there is the possibility of life in stars. There are lifeforms deep in earth's crust. I think you should specify what you mean by life. – Raditz_35 Jun 20 '17 at 9:22

Not habitable at first, but possibly later.

Life would definitely not survive if one star in a binary system would go supernova. Even in a loose binary with suggested (438+ AU) distance, radiation would be just too strong.

However, after the neutron star is born, things are looking better for the distant companion star and its planetary system. Because the neutron companion is fairly remote, the remaining star would not be suffering from an ablation process. Neutron stars are known for being X-ray and gamma-ray sources, but this kind of activity is caused by accretion process. If there is not much accretion material, neutron star should become very quiet after a while.

So, in a few hundred million years, "normal" star system may produce life and even become suitable for human habitation.

A neutron star always results from the collapse of massive stars (from eight solar masses upwards). Massive stars have very short lifespans in the main sequence. An estimation I've found is that the time in the main sequence for a star is about $10^{10} {M^{-2.5}}$ years (where $M$ = star mass in solar masses) which for an 8-solar-mass star means about 55 million years. At this point an Earth-like planet orbiting the other star would have probably not even cooled enough to have a solid surface or liquid water on it, let alone life. So if the stars were born together, the more massive one will be already a neutron star long before the least massive one has a mature planetary system.

• Would the force of the supernova disrupt planetary formation, or at least delay or alter it? – Dark Martin Jun 20 '17 at 2:45
• I was thinking about that. Wikipedia says "After between three and ten million years, the young Sun's solar wind would have cleared away all the gas and dust in the protoplanetary disc, blowing it into interstellar space, thus ending the growth of the planets." So there should be time enough for planets to form before the supernova goes off. But I don't know, really. – pablodf76 Jun 20 '17 at 14:28

This is surprisingly feasible. In a recent paper about alien civilizations building Dyson-rings around pulsars Zaza Osmanov found their habitable range "should be of the order of (10^−4 to 10^−1) AU with temperature interval (300–600) K for relatively slowly rotating pulsars and (10–350) AU with temperature interval (300–700) K for rapidly spinning neutron stars, respectively." Obviously, for the temperature range specified. So, thermal habitability is fine for the parameters of the secondary star and its planets even for rapidly rotating pulsars.

Provided the secondary star wasn't in the plane of the emitted radiation and, therefore, subject to the pulsar's pulsing it should be safe in terms of radiation.

Since a habitable planet needs a strong geomagnetic field to protect its biosphere from radiation, if the pulsar beam glanced off the planet's magnetic field this produce the most magnificent auroras in the galaxy.

Any life evolving on the planets of the secondary star will be adapted for higher levels of radiation than those of most other planets. Somewhat more extremophile, but definitely more radiation adapted.

Essentially its not implausible that a habitable planet could in the secondary star's planetary system. The odds against it being a safe place to live long term are high. However, if the conditions are right, then it is possible. This means the distance is great enough and its position relative to the pulsar is out of the path of its radiation beam.

Unfortunately, the birth of a neutron star would cause a terrifying radiation pulse.

From one AU, the supernova explosion would be as bright as a hydrogen bomb exploding in front of you... by nine orders of magnitude. From a distance of 700 AU, the explosion would be about six orders of magnitude less powerful than that, which still equals one thousand hydrogen bombs going off in front of you.

While it is possible (if not likely) that the planet may survive, I fear that habitability is out of the question: the planet's atmosphere is going to be shorn away from either the flash or the ensuing super-stellar winds and become part of the scattering planetary nebula.

It is more likely that the planet might form afterwards (the supernova would have happened very early in its companion's life). I seem to remember reading something of the kind in a commentary to Larry Niven's Smoke Ring, where you have a vaguely similar situation - a supernova remnant, LeVoy's Star, with at least one planet (Gold), orbiting around a yellow star.

It might be possible for a part of the supernova's ejecta to condense in an accretion disk around its companion, and finally give birth to a planet. After a couple of billion years you might then get a habitable planet around a star with a neutron companion. The neutron star's jets, if still present, would likely be directed axially (the two stars, being formed from the same original nebula, would have parallel axes of rotation) and not trouble the planet at all.

• Are you sure it's enough to vaporise or otherwise destroy the planet? That kind of assertion could use some math to back it up. (no I didn't do the math to check) – M i ech Jun 20 '17 at 0:38
• The planet can regenerate its atmosphere over millions of years from volcanic outgassing. If the Earth lost its atmosphere tomorrow, it would quickly (by geological standards) rebuild a CO2 atmosphere. – Mike Scott Jun 20 '17 at 6:58


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The Arrow C++ library includes a generic filesystem interface and specific implementations for some cloud storage systems. This setup allows various parts of the project to be able to read and write data with different storage backends. In the arrow R package, support has been enabled for AWS S3 and Google Cloud Storage (GCS). This vignette provides an overview of working with S3 and GCS data using Arrow.

In Windows and macOS binary packages, S3 and GCS support are included. On Linux when installing from source, S3 and GCS support is not always enabled by default, and it has additional system requirements. See vignette("install", package = "arrow") for details.

## Creating a FileSystem object

One way of working with filesystems is to create ?FileSystem objects. ?S3FileSystem objects can be created with the s3_bucket() function, which automatically detects the bucket’s AWS region. Similarly, ?GcsFileSystem objects can be created with the gs_bucket() function. The resulting FileSystem will consider paths relative to the bucket’s path (so for example you don’t need to prefix the bucket path when listing a directory).

With a FileSystem object, you can point to specific files in it with the $path() method and pass the result to file readers and writers (read_parquet(), write_feather(), et al.). For example, to read a parquet file from the example NYC taxi data (used in vignette("dataset", package = "arrow")): bucket <- s3_bucket("voltrondata-labs-datasets") # Or in GCS (anonymous = TRUE is required if credentials are not configured): bucket <- gs_bucket("voltrondata-labs-datasets", anonymous = TRUE) df <- read_parquet(bucket$path("nyc-taxi/year=2019/month=6/data.parquet"))

Note that this will be slower to read than if the file were local, though if you’re running on a machine in the same AWS region as the file in S3, the cost of reading the data over the network should be much lower.

You can list the files and/or directories in a bucket or subdirectory using the $ls() method: bucket$ls("nyc-taxi")
# Or recursive:
bucket$ls("nyc-taxi", recursive = TRUE) NOTE: in GCS, you should always use recursive = TRUE as directories often don’t appear in $ls() results.

See help(FileSystem) for a list of options that s3_bucket()/S3FileSystem$create() and gs_bucket()/GcsFileSystem$create() can take.

The object that s3_bucket() and gs_bucket() return is technically a SubTreeFileSystem, which holds a path and a file system to which it corresponds. SubTreeFileSystems can be useful for holding a reference to a subdirectory somewhere (on S3, GCS, or elsewhere).

One way to get a subtree is to call the $cd() method on a FileSystem june2019 <- bucket$cd("2019/06")
df <- read_parquet(june2019$path("data.parquet")) SubTreeFileSystem can also be made from a URI: june2019 <- SubTreeFileSystem$create("s3://voltrondata-labs-datasets/nyc-taxi/2019/06")

## URIs

File readers and writers (read_parquet(), write_feather(), et al.) also accept a URI as the source or destination file, as do open_dataset() and write_dataset(). An S3 URI looks like:

s3://[access_key:secret_key@]bucket/path[?region=]

A GCS URI looks like:

gs://[access_key:secret_key@]bucket/path
gs://anonymous@bucket/path

For example, one of the NYC taxi data files used in vignette("dataset", package = "arrow") is found at

s3://voltrondata-labs-datasets/nyc-taxi/year=2019/month=6/data.parquet
# Or in GCS (anonymous required on public buckets):
gs://anonymous@voltrondata-labs-datasets/nyc-taxi/year=2019/month=6/data.parquet

Given this URI, you can pass it to read_parquet() just as if it were a local file path:

df <- read_parquet("s3://voltrondata-labs-datasets/nyc-taxi/year=2019/month=6/data.parquet")
# Or in GCS:
df <- read_parquet("gs://anonymous@voltrondata-labs-datasets/nyc-taxi/year=2019/month=6/data.parquet")

### URI options

URIs accept additional options in the query parameters (the part after the ?) that are passed down to configure the underlying file system. They are separated by &. For example,

s3://voltrondata-labs-datasets/?endpoint_override=https%3A%2F%2Fstorage.googleapis.com&allow_bucket_creation=true

is equivlant to:

fs <- S3FileSystem$create( endpoint_override="https://storage.googleapis.com", allow_bucket_creation=TRUE ) fs$path("voltrondata-labs-datasets/")

Both tell the S3FileSystem that it should allow the creation of new buckets and to talk to Google Storage instead of S3. The latter works because GCS implements an S3-compatible API–see File systems that emulate S3 below–but for better support for GCS use the GCSFileSystem with gs://. Also note that parameters in the URI need to be percent encoded, which is why :// is written as %3A%2F%2F.

For S3, only the following options can be included in the URI as query parameters are region, scheme, endpoint_override, access_key, secret_key, allow_bucket_creation, and allow_bucket_deletion. For GCS, the supported parameters are scheme, endpoint_override, and retry_limit_seconds.

In GCS, a useful option is retry_limit_seconds, which sets the number of seconds a request may spend retrying before returning an error. The current default is 15 minutes, so in many interactive contexts it’s nice to set a lower value:

gs://anonymous@voltrondata-labs-datasets/nyc-taxi/?retry_limit_seconds=10

## Authentication

### S3 Authentication

To access private S3 buckets, you need typically need two secret parameters: a access_key, which is like a user id, and secret_key, which is like a token or password. There are a few options for passing these credentials:

• Include them in the URI, like s3://access_key:secret_key@bucket-name/path/to/file. Be sure to URL-encode your secrets if they contain special characters like “/” (e.g., URLencode("123/456", reserved = TRUE)).

• Pass them as access_key and secret_key to S3FileSystem$create() or s3_bucket() • Set them as environment variables named AWS_ACCESS_KEY_ID and AWS_SECRET_ACCESS_KEY, respectively. • Define them in a ~/.aws/credentials file, according to the AWS documentation. • Use an AccessRole for temporary access by passing the role_arn identifier to S3FileSystem$create() or s3_bucket().

### GCS Authentication

The simplest way to authenticate with GCS is to run the gcloud command to setup application default credentials:

gcloud auth application-default login

To manually configure credentials, you can pass either access_token and expiration, for using temporary tokens generated elsewhere, or json_credentials, to reference a downloaded credentials file.

If you haven’t configured credentials, then to access public buckets, you must pass anonymous = TRUE or anonymous as the user in a URI:

bucket <- gs_bucket("voltrondata-labs-datasets", anonymous = TRUE)
fs <- GcsFileSystem$create(anonymous = TRUE) df <- read_parquet("gs://anonymous@voltrondata-labs-datasets/nyc-taxi/year=2019/month=6/data.parquet") ## Using a proxy server If you need to use a proxy server to connect to an S3 bucket, you can provide a URI in the form http://user:password@host:port to proxy_options. For example, a local proxy server running on port 1316 can be used like this: bucket <- s3_bucket("voltrondata-labs-datasets", proxy_options = "http://localhost:1316") ## File systems that emulate S3 The S3FileSystem machinery enables you to work with any file system that provides an S3-compatible interface. For example, MinIO is and object-storage server that emulates the S3 API. If you were to run minio server locally with its default settings, you could connect to it with arrow using S3FileSystem like this: minio <- S3FileSystem$create(
scheme = "http",
endpoint_override = "localhost:9000"
)

or, as a URI, it would be

s3://minioadmin:minioadmin@?scheme=http&endpoint_override=localhost%3A9000

(note the URL escaping of the : in endpoint_override).

Among other applications, this can be useful for testing out code locally before running on a remote S3 bucket.

## Disabling the use of environment variables

As mentioned above, it is possible to make use of environment variables to configure access. However, if you wish to pass in connection details via a URI or alternative methods but also have existing AWS environment variables defined, these may interfere with your session. For example, you may see an error message like:

Error: IOError: When resolving region for bucket 'analysis': AWS Error [code 99]: curlCode: 6, Couldn't resolve host name

You can unset these environment variables using Sys.unsetenv(), for example:

Sys.unsetenv("AWS_DEFAULT_REGION")
Sys.unsetenv("AWS_S3_ENDPOINT")

By default, the AWS SDK tries to retrieve metadata about user configuration, which can cause conficts when passing in connection details via URI (for example when accessing a MINIO bucket). To disable the use of AWS environment variables, you can set environment variable AWS_EC2_METADATA_DISABLED to TRUE.

Sys.setenv(AWS_EC2_METADATA_DISABLED = TRUE)


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# CMI2012-B-05b

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Given an undirected weighted graph $G = (V, E)$ with non-negative edge weights, we can compute a minimum cost spanning tree $T = (V, E')$. We can also compute, for a given source vertex $s \epsilon V$ , the shortest paths from s to every other vertex in $V$. We now increase the weight of every edge in the graph by $1$. Are the following true or false, regardless of the structure of $G$? Give a mathematically sound argument if you claim the statement is true or a counterexample if the statement is false.

All the shortest paths from $s$ to the other vertices are unchanged.

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The given statement  "All the shortest paths from s to the other vertices are unchanged."  is false . From the above graph it is clear that the shortest path from $S$ to $D$ is $S\implies A\implies B\implies C\implies D$  and the cost is $6$.

Now, we increment the edge cost of  all the edges by $1$.

After the increment, the shortest  path from S to D gets changed. Now the shortest path becomes $S \implies E \implies D$  and shortest path cost is $9$. The above graph is the proof.

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i think statement should be true because in this question it says weight of every edge in graph is increased by 1 . so spanning tree and shortestpath will remain same.

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if spanning tree is unique then it will not change ,,,,but , shortest path will change as mentioned solution is right
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Is this a valid min spanning tree? with 6 vertices it should not any cycle and therefore not more than 5 edges.

## Related questions

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Given an undirected weighted graph $G = (V, E)$ with non-negative edge weights, we can compute a minimum cost spanning tree $T = (V, E')$. We can also compute, for a given source vertex $s \epsilon V$ , the shortest paths from s to every other vertex in ... argument if you claim the statement is true or a counterexample if the statement is false. $T$ is still a minimum cost spanning tree of $G$.
Let $A$ be an array of $n$ integers, sorted so that $A[1] \leq A[2] \leq \dots A[n]$. Suppose you are given a number $x$ and you wish to find out if there exist indices $k$ and $l$ such that $A[k]+A[l] = x$. Design an $O(n)$ algorithm for this problem.
A certain string-processing language offers a primitive operation which splits a string into two pieces. Since this operation involves copying the original string, it takes $n$ units of time for a string of length $n$, regardless of the location of the cut. Suppose, now, that you ... $m + 1$ pieces. You may assume that all m locations are in the interior of the string so each split is non-trivial.
Let $A$ be an array of $n$ integers, sorted, so that $A[1] \leq A[2] \leq \dots A[n]$. Suppose you are given a number $x$ and you wish to find out if there are indices $k$ and $l$ such that $A[k]+A[l] = x$. Design an $O(n \log n)$ time algorithm for this problem.


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# Séminaires : Séminaire Groupes Réductifs et Formes Automorphes

 Equipe(s) : fa, tn, Responsables : Alexis Bouthier, Benoît Stroh Email des responsables : alexis.bouthier@imj-prg.fr, benoit.stroh@imj-prg.fr Salle : Adresse : Description

 Orateur(s) Dmitry Gourevitch - Weizmann Institute of Science, Titre Generalized and degenerate Whittaker models Date 12/10/2015 Horaire 10:30 à 12:00 Diffusion Résume The Whittaker model is a very useful tool in the representation theory of reductive groups and in automorphic forms. However, it is known that only the largest representations have Whittaker models. In order to overcome this problem, for other representations various kinds of degenerate or generalized Whittaker models are considered since the 80s. Over non-archimedean fields, Moeglen and Waldspurger characterized the existence and multiplicities of these models in terms of the wave-front set. For $\mathrmGL(n, F)$ they can be also described in terms of the Bernstein-Zelevinsky derivatives. Over the Archimedean fields, only partial results are known in this direction. I want to talk about these partial results, including my recent works with Siddhartha Sahi (and sometimes others) on degenerate Whittaker models, Archimedean Bernstein-Zelevinsky derivatives and the connection between them. Some of these results are new also over $p$-adic fields. Some of the results also have adelic analogues. Salle Adresse


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# Optimizers¶

Curve fitting is an iterative process of finding parameters of mathematical function that has the best fit to a series of experimental data points. In ESCAPE optimizer objects of type optimizer_obj do curve fitting in terms of several algorithms:

• Levenberg-Marquardt 1, 2 which belongs to non-linear least-square algorithms,

• Differential evolution 3 - stochastic evultionary based algorithm which optiizes a model by maintaining a population of candidate solutions, iteratively combining them with a simple formulae and keeping a candidate with the best fitness.

1

Levenberg, Kenneth (1944). A Method for the Solution of Certain Non-Linear Problems in Least Squares, Quarterly of Applied Mathematics, 2 (2): 164-168.

2

Marquardt, Donald (1963). An Algorithm for Least-Squares Estimation of Nonlinear Parameters, SIAM Journal on Applied Mathematics, 11 (2): 431-441.

3

Storn, R and Price, K, Differential Evolution - a Simple and Efficient Heuristic for Global Optimization over Continuous Spaces, Journal of Global Optimization, 1997, 11, 341 - 359.

escape.core.optimizer.levmar(name, stack, ftol=1e-10, xtol=1e-10, gtol=1e-10, maxiter=300, maxfev=5000, nupdate=1)

Creates Levenberg-Marquardt optimizer.

Parameters
stack: modelstack_obj, list of models or model_obj

Models to be optimized.

ftol: non-negative double value

Measures the relative error desired in the sum of squares, i.e. cost function.

xtol: non-negative double value

Measures the relative error desired in the approximate solution.

gtol: non-negative double value

Measures the orthogonality desired between the cost function vector and the columns of the jacobian.

maxiter: non-negative integer

Maximum number of iterations.

maxfev: non-negative integer

Maximum number of cost function evaluations.

nupdate: non-negative integer

Controls calling of ‘on_iterate’ slot.

Returns

instance of optimizer_obj

escape.core.optimizer.diffevol(name, stack, popsize=5, strategy='best1bin', maxiter=300, mincost=1e-05, minconv=0.001, mutation=0.5, crossover=0.5, polish_candidate_maxiter=0, polish_final_maxiter=300, nupdate=1)

Differential Evolution is stochastic algorithm to find the minimum of cost function. It can search large areas of candidate space, but often requires larger numbers of function evaluations than conventional gradient-based techniques.

Parameters
stack: modelstack_obj, list of models or model_obj

Models to be optimized.

popsizeint, optional

A multiplier for setting the total population size. The population has popsize * num_params individuals.

strategystr, optional
The differential evolution strategy to use. Should be one of:
• ‘best1bin’

• ‘best1exp’

• ‘rand1exp’

• ‘randtobest1exp’

• ‘best2exp’

• ‘rand2exp’

• ‘randtobest1bin’

• ‘best2bin’

• ‘rand2bin’

• ‘rand1bin’

The default is ‘best1bin’.

maxiterint, optional

The maximum number of generations over which the entire population is evolved.

mincost: double, optional

When cost function reaches this value, optimizer will be terminated.

minconvfloat, optional

Relative tolerance for convergence.

mutationfloat or tuple(float, float), optional

The mutation constant. In the literature this is also known as differential weight, being denoted by F. If specified as a float it should be in the range [0, 2]. Larger values increase searching radius, but slow down convergence.

crossoverfloat, optional

The recombination constant, should be in the range [0, 1]. In the literature this is also known as the crossover probability, being denoted by CR. Increasing this value allows a larger number of mutants to progress into the next generation, but at the risk of population stability.

polish_candidate_maxiterint, optional

Maximum number of iterations for LM algorithm after every iteration. LM algorithm improves quality of candidates in the population matrix bringing them closer to the minimum. If zero, no polishing is applied for each candidate.

polish_final_maxiterint, optional

Maximum number of iterations for final polishing using LM optimizer.

nupdate: non-negative integer

Controls calling of ‘on_iterate’ slot.

Returns

instance of optimizer_obj

class escape.core.optimizer.optimizer_obj

Wrapper class for ESCAPE optimizers.

best_cost
Returns

Best cost for stochastic methods or final cost for levnar

cost_history

Returns cost history as a python list.

finalization_method

Used to set custom finalization method.

initial_parameters

Returns initial parameters values.

initialization_method

Used to set custom initialization method.

iteration_method

Used to set custom iteration method.

load_parameters(self, fn)

Load parameters from ASCII file with name ‘fn’.

modelstack
Returns

Modelstack instance.

name
Returns

Name of optimizer object.

num_of_evaluations
Returns

Number of function evaluations used for curve fitting.

num_of_iterations
Returns

Number of iterations used for curve fitting.

num_of_params
Returns

Number of parameters.

on_finalized(self)

Slot which is activated after optimization.

on_initialized(self)

Slot which is activated during optimizer’s initialization.

on_iteration(self)

Slot which is activated every n-th iteration. Iteration number ‘n’ is set with ‘nupdates’ setting.

parameter(self, size_t i)
reset_to_initial(self)

Reset parameters values to initial values.

save_parameters(self, fn)

Saves parameters to ASCII file with name ‘fn’.

shake(self)

Sets random values to all non-fixed independent parameters.

status_code
Returns

Status code.

status_msg
Returns

Status message.

stop(self)

Interrupts optimization process.

wait(self)

Waits with blocking for the optimizer to finish optimization process. Works if optimizer has been started asynchronously.


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# Download all files from SharePoint Online

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Update (2020-01-28):  I have updated this script to include the -Force parameter for Get-PnpFile, allowing you to willy-nilly overwrite destination files.

Yesterday, I came across a request for a script or method to download all files from SharePoint Online.

At first, I balked because CSOM isn’t my favorite and kind of finicky, but then remembered that we’d released the SharePoint PnP PowerShell cmdlets a while back. So, I decided to give those a peek and see what was in there, and was delighted to discover a few cmdlets–namely, Find-PnpFile and Get-PnpFile.

And thus my mini-project began.

After testing them for a while, I did find that they don’t have quite the flexibility for locating files.  In CSOM, we can use a CAML query (although I’ve had mixed results with it and generally perform some regular expression filtering/matching afterwards).  For example, perhaps I want to do something with all of the .MP3 files in a given site:

[regex]$Pattern = ‘\.mp3$’
$Library =$ClientContextSource.Web.Lists.GetByTitle(“Documents”)
$ClientContextSource.Load($Library)
$ClientContextSource.ExecuteQuery()$CamlQuery = New-Object Microsoft.SharePoint.Client.CamlQuery
$CamlQuery.ViewXml = “<View Scope=’RecursiveAll’ />”$AllItems = $Library.GetItems($camlQuery)
$ClientContextSource.Load($AllItems)
$ClientContextSource.ExecuteQuery() foreach ($obj in $allItems) { If ($obj[“FileRef”] -match $Pattern) { #write-host “File$($obj[“FileRef”]) matches.$FilesToUpdate += $obj } }$FilesToUpdate.<PickYourMethod>()
$ClientContextSource.ExecuteQuery() It’s like a ton of code to do something. Find-PnpFile seems like it has much easier syntax: Find-PnpFile -Match <string> That’s it. Unfortunately, string is literally just that–a single string (not a match in the regular expression kind of way). So, we can’t do multiple file filters or anything with it, which makes it harder to include or exclude things. That being said, I plowed on gave it the old college try. I don’t know how it will scale for sites or document libraries with hundreds of thousands of files, but I guess that’s what the user community is for. 🙂 I pass the values submitted for ListFolder, and Match to Find-PnpFile. The only one that’s required is Match, and if you don’t put anything in, the script assumes you want everything. Everything will include all of the HTML, CSS, and everything else in the site. Find-PnpFile also throws errors if you don’t have a matching folder or list (if you specify a name), and then seems to not honor the Match parameter values, so YMMV. At any rate, go ahead and give it a whirl–let me know what you think! https://gallery.technet.microsoft.com/All-SharePoint-Online-Files-adae7db1 ### Published by Aaron Guilmette Helping companies conquer inferior technology since 1997. I spend my time developing and implementing technology solutions so people can spend less time with technology. Specialties: Active Directory and Exchange consulting and deployment, Virtualization, Disaster Recovery, Office 365, datacenter migration/consolidation, cheese. ## Reader Comments 1. DarylGraves says: Hey Aaron, thank you very much for this! I got this working but I did have to make some changes first: – On Line 25 it states “$Directory = ($filename.DirectoryName)[0]”. I had to remove the “[0]” as this stopped it from passing the full path to the$Directory variable, causing errors.

– On Line 137 “Filename” is spelt wrong (you’ve got ‘Filname’) which caused another error.

After I made these changes it worked perfectly though thank you very much!

2. JaketheSnake9 says:

Hi Aaron, thank you very much for the script! It is the only one I could find that downloads all files from a given Site Collection for SPO. I am trying to use this script daily, and I want any updated files to replace the previous version, but it doesn’t seem to be doing that. Is there anywhere in the script you can point me to try and change that behavior?

1. JaketheSnake9 says:

Nevermind, I figured out I was able to use the -Force switch on Get-PNPFile. Thanks again for the great script.

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## Idea Factory: Some Of My Latest Musings

jeffro2pt0 has a little contest going on his blog, giving out coupons to wpdesigners.com which lets you get atleast 12 WordPress themes for just 5$. Only thing you have to do is give a unique idea that could help improve wordpress. Now, I don’t want anymore themes (actually I’m a cheap-skate who can’t spend even 5$ for a theme, unless one of you decides to get me one ), but I got a few ideas instantly. [Read More]