text:
The Tolt Pipeline Trail is a  wide, unpaved equestrian, pedestrian and mountain bike trail in Seattle's Eastside suburbs.  It stretches   along the Tolt pipeline right-of-way from Bothell, Washington to the Snoqualmie Valley Regional Trail near Duvall.

References

See also

 List of rail trails

Parks in King County, Washington
Transportation in King County, Washington
Hiking trails in Washington (state)


text:
Meet Barmer's E-ASHA
While the typical ASHA moves around with pen and paper 'collecting information', Barmer's E-ASHA provides counselling with the help of fancy tablets. You heard it right!
In 2006, the Ministry of Panchayati Raj named Barmer among India's 250 most backward districts. In addition to economic challenges and the daily hardships imposed by a life in the desert, women and children in Barmer have limited access to quality health care and counseling services. The infant and maternal mortality rate in the district is among the highest in Rajasthan.
The people living in villages in this district, like any other district in India, rely heavily on ASHA workers to provide counselling on pregnancy and newborn care. In April this year, the community health mobilizers of Barmer witnessed a drastic transformation in their roles when they were equipped with tablets to support their work of counseling mothers and pregnant women on issues related to health, nutrition, sanitation and best hygiene practices. Besides significantly improving their counseling capacity, the tablets helped the health workers to keep a tab on the health of mothers, pregnant women and children.
A large credit for this innovation goes to the young students like Vivek Agarwal and Vaibhav Bahuguna from IIT Jodhpur who have worked hard for the idea of E-ASHA to be successful.
Join us in congratulating these students and also wishing Barmer's E-ASHAs like Pushpa Devi all the best for their work in Barmer.
Watch this video to see how ASHAs has incorporated the tablets as one key element in their day-to-day work.

redpajama_set_name: RedPajamaCommonCrawl


text:
Why We Like This 8% Yielding Preferreds CEF
Posted on July 22, 2022 by adsanalyticsllc
Within the broader fixed-income umbrella, PTA features a decent-quality profile, holding preferreds of primarily investment-grade household-name issuers.
The fund's preferred portfolio itself is about a third investment-grade with more than half allocated to securities rated BB+ (i.e., the highest sub-investment-grade rating) and above.
The fund trades at the widest discount in the sector – about 8% wider than the sector average, despite modest outperformance year-to-date.
Below we discuss in more detail why holding some higher-quality preferred exposure makes sense and why PTA looks attractive within its sector. Overall, our approach to the preferreds sector is to identify attractive fund families at any given time. Once identified, we tilt to the most attractive fund in that fund family.
The reason we view the Cohen and Steers preferreds funds positively has to do primarily with their leverage resilience as well as their duration and borrowing cost hedge.
The allocation profile of the three funds in the Cohen preferred CEF (PSF, LDP, PTA) suite is not identical, but it's not hugely different. All three funds have a credit rating profile of 66-72% in securities rated BB and above. All three are overweight Banks (54-60%). All three are overweight U.S. securities (52-64%). Perhaps the biggest difference is in the retail preferred allocation with PTA having the largest at 26% versus 14% and 9% for the other 2 funds.
Overall, this hasn't made a ton of difference to the performance of the three funds since 2021 (PTA launched in late 2020). LDP has outperformed by about 1% per annum since then, however that has only happened in the last few months – we suspect due to its slightly lower duration profile (which is literally in the name of the fund).
In short, the allocation differences do exist but they are not where we have particularly strong views. Where the real action is, in our view, is in the income profile.
The funds' income profile is a function of their valuation and features. The key features in our view is that PTA has a 1% management fee while the other two funds have a 0.7% fee. However, PTA has fixed 85% of its leverage cost at just 1.2% versus 1.7% for the other two funds. PTA also has a slightly higher leverage than the other two funds. Plus, PTA trades at a slightly wider discount than LDP and a significantly wider discount than PTA.
A lot of investors get hung up on discounts, but they are just part of the puzzle. What we need to do is line everything up on an apples-to-apples basis and see how the CEF income sausage factory looks.
LDP and Cohen & Steers Select Preferred and Income Fund, Inc. (PSF) are quite similar to each other – they have the same management fee, similar leverage level and the same leverage cost so we pick LDP to go up against PTA given LDP is trading at a significantly wider discount than PSF.
This is how the net yield picture stacks up between the two funds. We assume a portfolio yield of 6.5% – again, we don't need to get this exactly right since what we are after is a relative picture between the two funds.
PTA generates a bit more additional yield due to its slightly higher leverage level. LDP has a lower yield drag of 1.06% (versus 1.59% for PTA) due to its lower management fee. PTA, as expected, has a lower leverage cost and also generates more yield due to its slightly wider discount. Net net PTA comes out ahead by a higher net yield of 0.28% which is the reason why it's our pick in the suite at the moment. Outside of the Cohen suite, we estimate PTA to have the highest net yield in the sector due to its wide discount and low leverage cost.
We expect the macro and market environment to remain highly uncertain, at the very least until inflation convincingly peaks and begins to descend. This is why we view decent-quality fixed-income assets such as preferreds worth an allocation in diversified income portfolios. In addition, the margin-of-safety valuation of PTA as well as its mostly fixed and low leverage cost profile will allow it to maintain its income level at a stable and high level in the sector.


HOUSTON, Texas, Nov 30, 2018 (SEND2PRESS NEWSWIRE) -- EPIC Insurance Brokers and Consultants, a retail property, casualty insurance brokerage and employee benefits consultant, announced today that Senior Risk Control Consultant Chuck Simpson will present at OSHA Oil & Gas Safety and Health Conference on Tues., Dec. 4 at 3 p.m. at the Hilton Americas in Houston, Texas.
The OSHA Oil & Gas Safety and Health Conference gathers key decision-makers within management and executive positions to foster a better understanding of how industry and OSHA can work together to improve safety performance through cooperation, best practices, knowledge sharing and relationship building.
Simpson's presentation, "Beyond Hazard Recognition - Recognizing Hazard Precursors" will discuss how to effectively recognize clues in the work environment that have the potential to develop into a hazard, and how to mitigate the hazard before it is present.
Click here to see the full agenda: https://www.oshasafetyconference.org/Agenda.aspx.
Simpson has presented lectures at the ASSE Safety Professional Development Conference and Exposition, OSHA Oil and Gas Safety Conference, the International Conference on Offshore Safety, the Gulf Coast Safety and Training Group, and the Sulphur Institute's Sulphur World Symposium in the United Arab Emirates.
Simpson is an active participant in the ASSE, Gulf Coast Safety Training Group and serves as Chairman for the ANSI Z390 Hydrogen Sulfide Training Committee. He earned his bachelor's degree in international studies from the University of South Alabama.


text:
Мила Родино () є національним гімном Болгарії з 1964 року. Заснований на музиці і тексту пісні "Горда Стара планина", написана Цветаном Радославовим (1863–1931). Пісня була складена по дорозі до поля бою, під час сербсько-болгарської війни. Текст був змінений кілька разів, останній раз — в 1990 році, після прийняття першої демократичної Конституції країни. Пісня "Мила Родино" була знову затверджена як національний гімн Республіки Болгарії.

Гімн

Використовувані образи

Мультимедія

Див. також
 Герб Болгарії
 Прапор Болгарії

Посилання
 Історія болгарського гімну

Національні символи Болгарії
Болгарія
Болгарія
Музика Болгарії

redpajama_set_name: RedPajamaWikipedia


It's looking like Easter with Hallmark!
You are here: Home / Giveaways / It's looking like Easter with Hallmark!
Happy Spring! It's starting to look like Easter and what better way than to celebrate with Hallmark gifts and cards? As most of you already know…I am a big fan of Hallmark! These items are truly my favorite this Spring and Easter holiday. My absolute favorite is the Flappy Happy Chick! This one made me smile really big when it started dancing and singing Girls Just Wanna Have Fun. I remember this song fondly from my younger days and this is just adorable! At first, I thought this would be best for a really young child but when it stated singing..well, I realized that this would be great for all ages even Moms! I love how this chick bops to the beat…just too cool! You can find her online at Hallmark.com.
Features sound and movement when activated.
You really don't want to miss this deal!
Last but not least..there are Hallmark Cards for Easter! I received a collection and love them. What I like best about Hallmark cards is the variety and that you really can find the perfect card for anyone. They have some really cute ones this Easter!
We have teamed up with our friends at Hallmark for an Easter Giveaway! One winner will win the same set of itty bittys in a basket that I received and an Easter card pack (est. value $25). To enter, leave a message telling us your favorite itty bitty collection. (You can click on the links above to see the choices!) Deadline to enter is March 27th, 2015 at noon PST. One winner will randomly be selected via Random.org.
Self Disclosure: I received free samples of the products mentioned above to facilitate this holiday feature and giveaway.
I love all the Itty Bittys. They're so cute. But, looking through the Hallmark booklet, I was so excited to see the Nativity Itty Bittys. They're not available until Holiday 2015, but I'm sure I'll be buying them then. Until Christmas, though, I think our favorite is R2D2. My cat has claimed him!
I love all the itty bitty collections! The Disney Princess collection is probably the top of my list, but I really love those Disney Easter ones, too!
Awww…love the Itty Bittys…Snoopy is my favorite!
My daughter would love the Frozen characters.
I love the Star Wars itty bittys, especially Darth Vader!
I love the Wizard of Oz itty bittys, especially the Wicked Witch of the West!
I like the Star Wars collection!
I like the Star Wars Itty Bittys!
I love the Muppets! They're so adorable!


Konkona Sen Sharma's directorial debut, A Death In The Gunj, has been ready for a while now. The film, which has toured quite a few international film festivals, has received rave reviews.
A Death In The Gunj, which has an interesting ensemble cast, will be commercially released in India this Friday.
Meanwhile, Konkona organised a special screening of the film for her friends from the industry.
The film's star, Kalki Koechlin, was spotted at the screening.
Vikrant Massey, who has an important role in the film, was there with his director, Konkona Sen Sharma.
Gulshan Devaiah -- you might remember him from Hunterrr -- also stars in the film.
Tanuja, who will be seen in A Death In The Gunj after her last outing -- the 2013 Marathi film Pitruroon -- arrived with daughter Tanishaa Mukherji.
The gorgeous Aditi Rao Hydari will be seen playing Sanjay Dutt's daughter in Bhoomi.
Radhika Apte, who was there as well, will be seen in Vikramaditya Motwane's next, Bhavesh Joshi.
Bhumi Pednekar has three films in the pipeline -- Toilet -- Ek Prem Katha, Manmarziyan and Shubh Mangal Saavdhan. While she stars opposite Akshay Kumar in the first, the latter two will see her with Ayushmann Khurrana, opposite whom she made her debut in Dum Laga Ke Haisha.
Ira Dubey was last seen in Dear Zindagi.
Kunaal Roy Kapur dropped in with his wife, Shayonti.
Vikramaditya Motwane was accompanied by his mother and co-producer, Dipa De.
Tahir Bhasin, who will be seen in Manto next, was there as well.
Producer Siddharth Roy Kapur watched the film too.
Naseeruddin Shah arrived with wife Ratna Pathak Shah, whom you can now watch in the delightful Sarabhai Vs Sarabhai: Take 2.
The Ratna Pathak Shah interivew: It was deeply embarrassing to watch Sholay!
Their son, Imaad Shah, was there as well.
Sudhir Mishra -- who has some incredible films in his kitty -- watched the film too.
Rajkummar Rao has two releases next week, Raabta and Behen Hogi Teri.


Scotland will host Japan in a full cap International at Murrayfield in the Autumn, Best joins up with the Lions at Hartley's expense and Scotland's coaching talent will jet round the planet to pick up tips this summer.
The Scotland under-20s have already left for their World Cup in Italy. In their ranks they have Duncan Weir, Mark Bennett and Stuart Hogg. They also, it must be said, go with mixed expectations.


From time to time we would like to contact you with our monthly gardening tips newsletter, the details of other garden maintenance and landscaping services we provide and any offers we have available.
If you consent to us contacting you for this purpose please tick below to confirm and say how you would like us to contact you. You can change your mind at any time by emailing info@jhps-gardens.co.uk.
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Candye Kane has been playing with her son Evan as a drummer, for eight years now. But they decided it was time he persued his own musical career. Her new drummer is a good friend of guitarist Laura Chavez: Baldhead Fred Rautmann. Fred lives in Hollister, Ca and has played in a variety of bands.
YouTube for the modern day Pin Up?
Dee Ann writer and founder of Rockabilly-online.com from Rotterdam, The Netherlands.


text:
\section{Introduction}
Consider the following scenario.  A user Charlie has a secret $w$ that
he wants to use to encrypt and authenticate his hard drive.  However, $w$ is not a
uniformly random key; rather, it is a string with some amount of
entropy from the point of view of any adversary ${\A}$.  Naturally,
Charlie uses an extractor~\cite{NZ96}, which is a tool for converting
entropic strings into uniform ones.  An extractor $\mathsf{Ext}$ is an
algorithm that takes the entropic string $w$ and a uniformly random
seed $i$, and computes $R=\mathsf{Ext}(w;i)$ that is (almost) uniformly random
even given $i$.

It may be problematic for Charlie to memorize or store the uniformly
random $R$ (this is in contrast to $w$, which can be, for example, a
long passphrase already known to Charlie, his biometric, or a physical token,
such as a physical one-way function~\cite{PRTG02}).  Rather, in order
to decrypt the hard drive, Charlie can use $i$ again to recompute
$R=\mathsf{Ext}(w;i)$.  The advantage of storing $i$ rather than $R$ is that
$i$ need not be secret, and thus can be written, for example, on an
unencrypted portion of the hard drive.

Even though the storage of $i$ need not be secret, the authenticity of
$i$ is very important.  If ${\A}$ could modify $i$ to $i'$, then
Charlie would extract some related key $R'$, and any guarantee on the
integrity of the hard drive would vanish, because typical encryption
and authentication schemes do not provide any security guarantees
under related-key attacks.  To authenticate $i$, Charlie would need to
use some secret key, but the only secret he has is $w$.

This brings us to the problem of building \emph{robust} extractors:
ones in which the authenticity of the seed can be verified at
reconstruction time.  A robust extractor has two procedures: a
randomized $\mathsf{Gen}(w)$, which generates $(R, P)$ such that $R$ is
uniform even given $P$ (think of $P$ as containing the seed $i$ as
well as some authentication information), and $\mathsf{Rep}(w, P')$, which
reproduces $R$ if $P'=P$ and outputs $\bot$ with high probability for
an adversarially produced $P'\neq P$.

Note that in the above scenario, the adversary ${\A}$, before
attempting to produce $P'\neq P$, gets to see the value $P$ and how
the value $R$ is used for encryption and authentication.  Because we
want robust fuzzy extractors to be secure for a wide variety of
applications, we do not wish to restrict how $R$ is used and,
therefore, what information about $R$ is available to ${\A}$.  Rather,
we will require that ${\A}$ has low probability of getting $\mathsf{Rep}(w,
P')$ to not output $\bot$ \emph{even} if ${\A}$ is given both $P$ and
$R$.  This strong notion of security is known as
\emph{post-application} robustness.

An additional challenge may be that the value $w$ when $\mathsf{Gen}$ is run
is slightly different from the value $w'$ available when $\mathsf{Rep}$ is
run: for example, the user may make a typo in a long passphrase, or a
biometric reading may differ slightly.  Extractors that can tolerate
such differences and still reproduce $R$ exactly are called
\emph{fuzzy}~\cite{DORS08}.  Fuzzy extractors are obtained by adding
error-correcting information to $P$, to enable $\mathsf{Rep}$ to compensate
for errors in $w'$.  The specific constructions depend on the kinds of
errors that can occur (e.g., Hamming errors, edit distance errors,
etc.).

Robust (fuzzy) extractors are useful not only in the single-party
setting described above, but also in interactive settings, where two
parties are trying to derive a key from a shared (slightly different
in the fuzzy case) secret $w$ that either is nonuniform or about which
some limited information is known to the adversary ${\A}$. One party,
Alice, can run $\mathsf{Gen}$ to obtain $(R, P)$ and send $P$ to the other
party, Bob, who can run $\mathsf{Rep}$ to also obtain $R$.  However, if ${\A}$
is actively interfering with the channel between Alice and Bob and
modifying $P$, it is important to ensure that Bob detects the
modification rather than derives a different key $R'$.  Moreover,
unless Alice can be sure that Bob truly received $P$ before she starts
using $R$ in a communication, post-application robustness is
needed.

\mypar{Prior Work.}
Fuzzy extractors, defined in~\cite{DORS08}, are essentially the
noninteractive variant of privacy amplification
and information reconciliation protocols,
considered in multiple works, including~\cite{Wyner,BBR88,Mau93,BBCM95}.
Robust (fuzzy) extractors, defined in~\cite{BDKOS05,DKRS06},
are the noninteractive variant of privacy amplification
(and information reconciliation) secure against active
adversaries~\cite{Mau97,MauW97,Wolf98,MauW03c,RW03,RW04B}.

Let the length of $w$ be $n$ and the entropy of $w$ be $m$.
Post-application robust fuzzy extractors cannot extract anything out
of $w$ if $m<n/2$, because an extractor with post-application
robustness implies an information-theoretically secure message
authentication code (MAC) with $w$ as the key%
\footnote{The MAC is
  obtained by extracting $R$, using it as a key to any standard
  information-theoretic MAC (e.g.,~\cite{WC81}), and sending $P$ along with the tag
  to the verifier},
which is impossible if $m<n/2$ (see \cite{DS02} for impossibility of deterministic MACs
	if $m<n/2$
and its extension by \cite{Wic08} to randomized MACs). Without any set-up
assumptions, the only previously
known post-application robust extractor, due to \cite{DKRS06},
extracts $R$ of length $\frac{2}{3}(m-n/2-\log\tfrac{1}{\delta})$ (or even less
if $R$ is required to be very close to uniform), where $\delta$ is
the probability that the adversary violates robustness. Making it
fuzzy further reduces the length of $R$ by an amount related to the
error-tolerance. (With set-up assumptions, one can do much better:
the construction of~\cite{CDFPW08}
extracts almost the entire entropy $m$,
reduced by an amount related to security and, in the fuzzy case, to
error-tolerance.  However, this construction assumes that a nonsecret
uniformly random string is already known to both parties, and that the
distribution on $w$, including adversarial knowledge about $w$, is
independent of this string.)

\mypar{Our Results.}
The robust extractor construction of~\cite{DKRS06} is parameterized by
a value $v$ that can be decreased in order to obtain a longer $R$.  In
fact, as shown in~\cite{DKRS06}, a smaller $v$ can be used for \emph{pre-application} robustness
(a weaker security notion, in which ${\A}$ gets $P$ but not $R$).  We
show in Theorem~\ref{theorem-dkrs-lowerbound}
that the post-application-robustness analysis of~\cite{DKRS06} is
essentially tight, and if $v$ is decreased, the construction becomes insecure.

Instead, in Section~\ref{sec-rfe-errorless}, we propose a new construction of an extractor with
post-application robustness that extracts $R$ of length
$m-n/2-\log\tfrac{1}{\delta}$, improving the previous result by a factor of
$3/2$ (more if $R$ is required to be very close to uniform).
While this is only a constant-factor increase, in scenarios where
secret randomness is scarce it can make a crucial difference.
Like~\cite{DKRS06}, we make no additional set-up
assumptions. Computationally, our construction is slightly more
efficient than the construction of~\cite{DKRS06}.  Our improved robust
extractor translates into an improved robust fuzzy extractor using the
techniques of~\cite{DKRS06}, with the same factor of $3/2$
improvement.

In addition, we show (in Section~\ref{section-dkrs-uniformity-dominates-improvement})
a slight improvement for the pre-application robust version of the extractor
of~\cite{DKRS06}, applicable when the extracted string must be particularly close
to uniform.

\section{Preliminaries}
\label{section-prelims}
\mypar{Notation.}
For binary strings $a,b$, $a||b$ denotes their concatenation, $|a|$
denotes the length of $a$. For a binary string $a$, for
we denote by $[a]^j_i$, the substring $b = a_ia_{i+1}\ldots a_j$.
If $S$ is a set, $x\leftarrow S$ means that $x$ is chosen uniformly
from $S$. If $X$ is a probability distribution (or a random variable), then $x\leftarrow X$
means that $x$ is chosen according to distribution $X$.
If $X$ and $Y$ are two random variables, then $X\times Y$ denotes
the product distribution (obtained by sampling $X$ and $Y$
\emph{independently}). All logarithms are
base $2$.

\mypar{Random Variables, Entropy, Extractors.}
Let $U_l$ denote the uniform distribution on $\{0,1\}^l$.
Let $X_1, X_2$ be two probability distributions over some set $S$.
Their {\em statistical distance} is
\[\sd{X_1, X_2} \stackrel{\mbox{\tiny{def}}}{=} \max_{T\subseteq S}
\{ \Pr[X_1 \in T] - \Pr[X_2 \in T] \}=
\frac{1}{2}\sum_{s\in S} \left|\Pr_{X_1}[s]-\Pr_{X_2}[s]\right |\]
(they are said to be $\varepsilon$-close if $\sd{X_1, X_2}\le \varepsilon$).
We will use the following lemma on statistical distance
that was proven in~\cite{DKRS08}:
\begin{lemma}
\label{lemma-sdswitch}
For any joint distribution $(A, B)$ and distributions
$C$ and $D$ over the ranges of $A$ and $B$ respectively,
if $\sd{(A, B), C\times D}\le \alpha$, then
$\mathbf{SD}((A,B), \allowbreak C \times B)  \le 2\alpha$.
\end{lemma}

\mypar{Min-entropy.} The \emph{min-entropy} of a random variable $W$ is
\ifnum0=1 defined as
\fi
${\mathbf{H}_{\infty}}(W) = -\log(\max_{w}\Pr[W=w])$ (all logarithms are base 2,
unless
specified otherwise).
Following \cite{DORS08}, for a joint distribution $(W, E)$,
define the (average) conditional min-entropy of $W$
given $E$ as \[{\widetilde{\mathbf{H}}_{\infty}}(W\mid E) =\allowbreak -\log (\expe_{e\leftarrow E}(2^{-{\mathbf{H}_{\infty}}(W\mid E=e)}))\]
(here the expectation is taken over $e$ for which $Pr[E=e]$ is nonzero).
A computationally unbounded adversary who receives the value of $E$ cannot find the correct value of $W$
with probability greater than $2^{-{\widetilde{\mathbf{H}}_{\infty}}(W\mid E)}$.
We will use the following lemma from ~\cite{DORS08}:
\begin{lemma}
\label{lemma-avg-min-entropy}
Let $A,B,C$ be random variables. If $B$ has at most $2^\lambda$ possible values, then
\ifnum0=0
the following holds:
${\widetilde{\mathbf{H}}_{\infty}}(A|B,C) \allowbreak \geq {\widetilde{\mathbf{H}}_{\infty}}((A,B)|C) -\lambda \geq {\widetilde{\mathbf{H}}_{\infty}}(A|C) -\lambda$.
\else
${\widetilde{\mathbf{H}}_{\infty}}(A|B,C) \geq {\widetilde{\mathbf{H}}_{\infty}}((A,B)|C) -\lambda \geq {\widetilde{\mathbf{H}}_{\infty}}(A|C) -\lambda$.
\fi
In particular, ${\widetilde{\mathbf{H}}_{\infty}}(A|B) \geq {\mathbf{H}_{\infty}}((A,B)) -\lambda \geq {\mathbf{H}_{\infty}}(A) -\lambda$.
\end{lemma}

Because in this paper the adversary is sometimes assumed to have some external
information $E$ about Alice and Bob's secrets, we need the following variant,
defined in~\cite[Definition 2]{DORS08}, of the definition of strong extractors
of~\cite{NZ96}:

\begin{definition}
Let $\mathsf{Ext}:\{0,1\}^n \rightarrow \{0,1\}^l$ be a polynomial time
probabilistic function that uses $r$ bits of randomness.
We say that $\mathsf{Ext}$ is an
{\sf average-case $(n, m, l, \varepsilon)$-strong extractor}
if for all pairs of random variables
$(W,E)$ such that $w \in W$ is an $n$-bit string and
${\widetilde{\mathbf{H}}_{\infty}}(W\mid E) \geq m$, we have
\ifnum0=0
$\sd{(\mathsf{Ext}(W;X), X,E), (U_l,X,E)}\le \varepsilon$,
\else
$\mathbf{SD}((\mathsf{Ext}(W;X), X,E), \allowbreak (U_l,X,E))\le \varepsilon$,
\fi
where $X$ is the uniform distribution
over $\{0,1\}^r.$
\end{definition}

Any strong extractor can be made average-case with a slight increase in input
entropy~\cite[Section 2.5]{DORS08}. We should note that some strong extractors,
such as universal hash functions~\cite{CW79,HILL99} discussed next,
generalize without any loss to average-case.

\mypar{The Leftover Hash Lemma}
We first recall the notion of universal hashing~\cite{CW79}:
\begin{definition}
A family of efficient functions ${\cal H}= \set{h_{i}:\{0,1\}^n \rightarrow
\{0,1\}^\ell}_{i \in I}$ is \emph{universal} if for all
distinct $x, x'$ we have $\Pr_{i \leftarrow I}[h_i(x) = h_i(x')] \leq
2^{-l}$.

${\cal H}$ is \emph {pairwise independent} if for all distinct
$x, x'$ and all $y, y'$ it holds that $\Pr_{i \in I}[h_i(x) = y \wedge
h_i(x') = y'] \leq 2^{-2\ell}$.
\hspace*{\fill}$\diamondsuit$
\end{definition}

\begin{lemma}[Leftover Hash Lemma, average-case version~\cite{DORS08}]
\label{lemma-leftover-hash}
For $\ell, m, \varepsilon>0$, ${\cal H}$ is a strong $(m,
\varepsilon)$ average-case extractor (where the index of the hash function is
the seed to the extractor)
if ${\cal H}$ is universal and $\ell \le m+2-2\log\tfrac{1}{\varepsilon}$.
\end{lemma}
\noindent This Lemma easily generalizes to the case
when ${\cal H}$ is allowed to depend on the extra
information $E$ about the input $X$. In other words,
every function in ${\cal H}$ takes an additional input $e$, and the family ${\cal H}$ is
universal for every fixed value of $e$.

\mypar{Secure Sketches and  Fuzzy Extractors.}\label{sec:fuzzy-defs}
We start by reviewing the
definitions of secure sketches and fuzzy extractors from
\cite{DORS08}. Let ${\cal M}$ be a metric space with distance function
$\mathsf{dis}$ (we will generally denote by $n$ the length of each element in
${\cal M}$).
Informally, a secure sketch enables recovery
of a string $w\in {\cal M}$ from any ``close'' string $w'\in {\cal M}$ without leaking too
much information about~$w$.

\begin{definition} \label{def:ss}
An {\sf $(m,\tilde{m},t)$-secure sketch} is a pair of efficient
randomized procedures ($\mathsf{SS},\mathsf{SRec}$) s.t.:
\begin{enumerate}
\item The sketching procedure $\mathsf{SS}$ on input $w\in {\cal M}$ returns a bit
string $s\in \bit{*}$.
The recovery procedure  $\mathsf{SRec}$ takes an element $w'\in {\cal M}$ and
$s\in \bit{*}$.

\item \emph{Correctness:} If $\dis{w}{w'}\le t$ then $\mathsf{SRec}(w',\mathsf{SS}(w))=w$.
\item \emph{Security:} For any
distribution $W$ over ${\cal M}$ with min-entropy $m$, the (average)
min-entropy of $W$ conditioned on $s$ does not decrease  very
much. Specifically, if ${\mathbf{H}_{\infty}}(W) \geq m$ then ${\widetilde{\mathbf{H}}_{\infty}}(W\mid
\mathsf{SS}(W))\geq \tilde{m}$.
\end{enumerate}
The quantity $m-\tilde{m}$ is called the {\em entropy
loss} of the secure sketch.
\hspace*{\fill}$\diamondsuit$
\end{definition}

In this paper, we will construct a robust fuzzy extractor
for the binary Hamming metric using secure sketches for the
same metric.
We will briefly review the syndrome construction from~\cite[Construction 3]{DORS08}
that we use
(see also references therein for its previous incarnations).
Consider an efficiently decodable $[n,n-k,2t+1]$
linear error-correcting code $C$.  The sketch $s=\mathsf{SS}(w)$ consists of
the $k$-bit syndrome $w$ with respect to $C$.
We will use the fact
that $s$ is a (deterministic) linear function of $w$ and that the
entropy loss is at most $|s|=k$ bits in the construction of our
robust fuzzy extractor for the Hamming metric.

We note that, as was shown in \cite{DKRS06}, the secure sketch
construction for the set difference metric of $\cite{DORS08}$ can
be used to extend the robust fuzzy extractor construction
in the Hamming metric to the set difference metric.

While a secure sketch enables recovery of a string $w$ from a close string $w'$,
a fuzzy extractor extracts a close-to-uniform string $\key$
and allows the precise reconstruction of $\key$
from any string $w'$ close to $w$.

\begin{definition} \label{def:fe}
An {\sf $(m,\ell,t,\varepsilon)$-fuzzy extractor} is a pair of efficient
randomized procedures ($\mathsf{Gen},\mathsf{Rep}$) with the following properties:
\begin{enumerate}
\item The generation procedure $\mathsf{Gen}$, on input $w\in {\cal M}$, outputs an
extracted string $\key\in \{0,1\}^\ell$ and a helper string $P\in
\bit{*}$. The reproduction procedure $\mathsf{Rep}$ takes an element
$w'\in {\cal M}$ and a string $P\in \bit{*}$ as inputs.

\item \emph{Correctness:}
If $\dis{w}{w'}\le t$ and
$(\key, P)
\leftarrow \mathsf{Gen}(w)$, then $\mathsf{Rep}(w',P)=\key$.

\item \emph{Security:}
For any distribution $W$ over ${\cal M}$ with min-entropy $m$, the
string $\key$ is close to uniform even conditioned on the value
of~$P$. Formally, if ${\mathbf{H}_{\infty}}(W) \geq m$ and $\tuple{\key,P} \leftarrow
\mathsf{Gen}(W)$, then we have
$\sd{(\key,P), U_\ell \times P }\le \varepsilon$.
 \hspace*{\fill}$\diamondsuit$
\end{enumerate}
\end{definition}

Note that fuzzy extractors allow the information $P$ to be revealed
to an adversary without compromising the security of the
extracted random string $\key$. However, they provide no guarantee when
the adversary is active.
Robust fuzzy extractors defined (and constructed) in ~\cite{DKRS06}
formalize the notion of security against active adversaries. We review the definition below.

If $W, W'$ are two (correlated) random variables over a metric space
${\cal M}$,
we say $\dis{W}{W'}\le t$ if
the distance between $W$ and $W'$ is at most $t$ with probability one.
We call $(W,
W')$ a {\em $(t, m)$-pair} if $\dis{W}{W'} \leq t$ and ${\mathbf{H}_{\infty}}(W)
\geq m$.

\begin{definition}\label{def:robust}
An $(m, \ell, t, \varepsilon)$-fuzzy extractor has {\sf post-application
\ifnum0=0(resp., pre-application)
\else (resp., pre-appli- \allowbreak cation)
\fi
    robustness $\delta$} if for all $(t,m)$-pairs $(W,
W')$ and all adversaries ${\A}$, the probability that the following
experiment outputs ``success'' is at most $\delta$: sample $(w, w')$ from $(W,
W')$; let $(\key, P) = \mathsf{Gen}(w)$; let $\tilde P = {\A}(\key, P)$ (resp.,
$\tilde P = A(P)$); output ``success'' if $\tilde P \neq P$ and $\mathsf{Rep}(w',
\tilde{P}) \neq \perp$.
\hspace*{\fill}$\diamondsuit$
\end{definition}

We note that the above definitions can be easily extended to give
\emph{average-case} fuzzy extractors (where the
adversary has some external information $E$ correlated with $W$), and that our
constructions satisfy those stronger definitions, as well.

\section{The New Robust Extractor}
\label{sec-rfe-errorless}
In this section we present our new extractor with post-application
robustness.  We extend it to a robust \emph{fuzzy} extractor in
Section~\ref{sec-rfe-errors}.  Our approach is similar to that
of~\cite{DKRS06}; a detailed comparison is given in
Section~\ref{sec-comparison-dkrs}.

\mypar{Starting point: key agreement secure against a passive adversary.}
Recall that a strong extractor allows extraction of
a string that appears uniform to an adversary even given
the presence of the seed used for extraction.
Therefore, a natural way of achieving key agreement in the
 errorless case is for Alice to pick a random seed $i$
  for a strong extractor and send it to Bob (in the clear).
They could then use $R=\mathsf{Ext}(w;i)$ as the shared key.
 As long as the adversary is passive, the shared key looks uniform to her.
 However, such a protocol can be rendered completely insecure
 when executed in the presence of an active adversary because
${\A}$ could adversarially modify $i$ to $i'$ such
that $R'$ extracted by Bob has no entropy.
To prevent such malicious modification of $i$ we will require
Alice to send an authentication of $i$ (along with $i$) to Bob.
 In our construction, we authenticate $i$ using $w$
 as the key and then extract from $w$ using $i$
as the seed. Details follow.

\mypar{Construction.}
For the rest of the paper we will let $w \in \{0,1\}^n$. We will
assume that $n$ is even (if not, drop one bit of $w$, reducing its entropy by at most 1).
To compute $\mathsf{Gen}(w)$, let
$a$ be the first half of $w$ and $b$ the second:
$a=[w]_1^{n/2}, b=[w]_{n/2+1}^n$.
 View $a$,$b$ as elements of ${\mathbb F}_{2^{n/2}}$.
Let $v=n-m+\log\tfrac{1}{\delta}$, where $\delta$ is the desired
robustness.
Choose a random $i \in {\mathbb F}_{2^{n/2}}$.
Compute $y = ia+b$.
Let $\sigma$ consist of the first
$v$ bits of $y$ and the extracted key $R$ consist of the rest of $y$:
$\sigma = [y]^v_1$, $R = [y]^{n/2}_{v+1}.$
Output $P = (i,\sigma).

\begin{tabbing}
\ifnum0=0
wwwwwwwwwwww\=www\=3. \=wwwwwwwww\=\kill
\else
wwwwww\=wwww\=3. \=wwwwwwwww\=\kill
\fi
\>\+
$\mathsf{Gen}(w)$:\\
1. Let $a=[w]_1^{n/2}, b=[w]_{n/2+1}^n$ \\
2. Select a random $i\leftarrow {\mathbb F}_{2^{n/2}}$\\
3. Set $\sigma = [ia+b]^{v}_{1}$, $R= [ia+b]^{n/2}_{v+1}$ and output $P=(i, \sigma)$\\
\ \\
$\mathsf{Rep}(w,P' = (i',\sigma'))$:\\
1. Let $a=[w]_1^{n/2}, b=[w]_{n/2+1}^n$ \\
2. If $\sigma' = [i'a+b]^{v}_{1}$ then compute $R' = [i'a+b]^{n/2}_{v+1}$ else output $\bot$
\end{tabbing}

\begin{theorem}
Let ${\cal M} = \{0,1\}^n.$ Setting $v = n/2 - \ell$,
the above construction is an $(m,\ell,0,\varepsilon)-\mathsf{fuzzy\ extractor}$
with robustness $\delta$, for any $m,\ell,\varepsilon,\delta$ satisfying
$\ell \leq m - n/2 -\log\frac{1}{\delta}$ as long as $m\geq n/2 + 2\log\frac{1}{\varepsilon}.$
\end{theorem}

If $\varepsilon$ is so low that the constraint $m\geq n/2+2\log\tfrac{1}{\varepsilon}$ is not satisfied,
then the construction can be modified as shown in Section~\ref{section-uniformity-dominates}.

\begin{proof}

\mypar{Extraction.}
Our goal is to show that $R$ is nearly uniform given $P$.
To do so, we first show that the function $h_i(a,b) = (\sigma, R)$
 is a universal hash family. Indeed, for $(a,b) \neq (a',b')$ consider

\begin{eqnarray*}
\Pr_i[h_i(a,b) = h_i(a',b')]
&=&\Pr_i[ia+b = ia'+b']\\
&=&\Pr_i[i(a-a') = (b-b')]\\
&\leq&2^{-n/2}\,.
\end{eqnarray*}
To see the last inequality recall that $(a,b) \neq (a',b').$
 Therefore, if $a = a'$, then $b\neq b'$ making
the $\Pr_i[i(a-a') = (b-b')]=0$.
If $a \neq a'$, then there is a unique $i=(b-b')/(a-a')$ that
satisfies the equality.
Since $i$ is chosen randomly from ${\mathbb F}_{2^{n/2}}$,
 the probability of the specific $i$ occurring
 is $2^{-n/2}$.

Because $|(R, \sigma)|=n/2$,
Lemma~\ref{lemma-leftover-hash} gives us $\sd{(R,P),U_{|R|}\times U_{|P|}}\le \varepsilon/2$
as long as $n/2\le m+2-2\log\frac{2}{\varepsilon}$, or, equivalently,
$(R,P)$ is $2^{(n/2-m)/2-1}$-close to $U_{|R|}\times U_{|P|}$.
Applying Lemma~\ref{lemma-sdswitch} to $A=R$, $B=P$, $C=U_{\frac{n}{2}-v}$,
$D=U_{\frac{n}{2}}\times U_{v}$, we get that $(R, P)$ is $\varepsilon$-close
to $U_{(\frac{n}{2})-v}\times P$, for $\varepsilon = 2^{(n/2-m)/2}$.
From here it follows that for extraction to be possible, $m\geq n/2+ 2\log\tfrac{1}{\varepsilon}.$

\mypar{Post-Application Robustness.}
In the post-application robustness security game,
the adversary ${\A}$ on receiving $(P=(i, \sigma), R)$
(generated
according to procedure $\mathsf{Gen}$) outputs $P' = (i',\sigma')$,
and is considered successful if $(P' \neq P) \wedge [i'a+ b]^v_1 = \sigma'$.
 In our analysis, we will assume that $i' \neq i$. We claim that this
 does not reduce ${\A}$'s success probability.
Indeed, if $i' = i$ then, for $P'\neq P$ to hold,
${\A}$ would have to output $\sigma' \neq \sigma$.
 However, when $i'=i$, $\mathsf{Rep}$ would output $\bot$ unless $\sigma' = \sigma$.

In our analysis, we allow ${\A}$ to be deterministic.
This is without loss of generality since we allow an unbounded adversary.
We also allow ${\A}$ to arbitrarily fix $i$. This
makes the result only stronger since we demonstrate robustness for
a worst-case choice of $i$.

Since $i$ is fixed and ${\A}$ is deterministic,
$(\sigma, R)$ determines
the transcript
$\mathsf{tr}=(i,\sigma,R,i',\sigma')$.
For any particular $\mathsf{tr}$, let $\mathsf{Succ}_{\mathsf{tr}}$ be the event
that the transcript is $\mathsf{tr}$ and ${\A}$ wins, i.e., that
$ia+b = \sigma||R \wedge [i'a+b]^v_1 = \sigma'$.
We denote by ${\mathsf{Bad}}_{\mathsf{tr}}$
the set of $w=a||b$ that make $\mathsf{Succ}_{\mathsf{tr}}$
true.
For any $\mathsf{tr}$, $\Pr_w[\mathsf{Succ}_{\mathsf{tr}}] \leq |{\mathsf{Bad}}_{\mathsf{tr}}|2^{-m}$, because each $w$
in ${\mathsf{Bad}}_{\mathsf{tr}}$ occurs with probability at most $2^{-m}$.
We now partition the set ${\mathsf{Bad}}_{\mathsf{tr}}$ into $2^{\ell}$ disjoint
sets, indexed by $R' \in \{0,1\}^{\ell}$:
\begin{eqnarray*}
{\mathsf{Bad}}^{R'}_{\mathsf{tr}} &\stackrel{\mbox{\tiny{def}}}{=}& \{w\,|\, w\in {\mathsf{Bad}}_{\mathsf{tr}} \wedge [i'a+b]^\ell_{v+1} = R'\}\\
&=& \{w\,|\,  (ia+b = \sigma||R) \wedge (i'a+b = \sigma'||R')\}
\end{eqnarray*}
\noindent
For a particular value of $(\mathsf{tr},R')$, $w = a||b$ is uniquely
determined by the constraints that define the above set
\ifnum0=0 Therefore,
\else i.e;
\fi
$|{\mathsf{Bad}}^{R'}_{\mathsf{tr}}| = 1$. Since
${\mathsf{Bad}}_{\mathsf{tr}} = \bigcup_{R' \in \{0,1\}^\ell}{\mathsf{Bad}}^{R'}_{\mathsf{tr}}$, we get
$|{\mathsf{Bad}}_{\mathsf{tr}}| \le 2^{\ell} = 2^{{n/2}-v}$.
 From here it follows that
\[\Pr[\mathsf{Succ}_{\mathsf{tr}}] \le |{\mathsf{Bad}}_{\mathsf{tr}}|2^{-m} \le 2^{{n/2}-v-m}\,.\]

$\Pr[\mathsf{Succ}_{\mathsf{tr}}]$ measures the probability that
the transcript is $\mathsf{tr}$ and ${\A}$ succeeds.
To find out the probability that ${\A}$ succeeds,
we need to simply add $\Pr[\mathsf{Succ}_\mathsf{tr}]$ over all possible $\mathsf{tr}$.
Since a transcript is completely determined by $\sigma, R$,
the total number of possible transcripts is $2^{|\sigma|+|R|} = 2^{n/2}$
and, therefore,
${\A}$'s probability of success is at most $2^{n-v-m}$.

To achieve $\delta$-robustness, we need to set $v$ to at least $n-m+\log\tfrac{1}{\delta}$.
From here it follows that $\ell = \frac{n}{2} - v \leq \frac{1}{2}(2m - n  - 2\log\tfrac{1}{\delta})$.
\ifnum0=1\qed\fi
\end{proof}

\subsection{Getting Closer to Uniform}
\label{section-uniformity-dominates}
If $\varepsilon$ is so low that the constraint $m\geq n/2+2\log\tfrac{1}{\varepsilon}$ is not satisfied,
then in our construction we can simply
shorten $R$ by $\beta = n/2+2\log\tfrac{1}{\varepsilon}-m$ bits,
as follows:
keep  $v=n-m+\log\tfrac{1}{\delta}$ (regardless of $\ell$), and let
$R=[ia+b]_{v+1}^{\ell+v}$, for any $\ell \le 2m-n-\log\tfrac{1}{\delta}-2\log\tfrac{1}{\varepsilon}$.
This keeps $\sigma$ the same, but shortens $R$ enough for the leftover hash lemma to work.
The proof remains essentially the same,
except that to prove robustness, we will give the remaining bits $[ia+b]_{\ell+v+1}^{n/2}$
for free to ${\A}$.

\subsection{Improving the construction of~\cite{DKRS06} When the Uniformity Constraint Dominates}
\label{section-dkrs-uniformity-dominates-improvement}
The construction of Dodis et al.~\cite{DKRS06} parses $w$ as two strings
$a$ and $b$ of lengths $n-v$ and $v$, respectively.
The values $\sigma,R$ are computed as $\sigma = [ia]^v_1+b$ and $R = [ia]^n_{v+1}$;
$P = (i,\sigma)$. In order to get $R$ to be uniform given
$P$, the value $v$ is increased until the leftover hash lemma can be applied to $(R, \sigma)$.  However, we observe that this unnecessarily increases the length of $\sigma$ (i.e., for every bit added to $v$, two bits are subtracted from $R$).    Instead, we propose to improve this construction with essentially
the same technique as we use for our construction in Section~\ref{section-uniformity-dominates}.  The idea
is to simply shorten $R$  without
increasing the length of $\sigma$.
This improvement applies to both pre- and post-application robustness.

For post-application robustness, suppose the uniformity constraint dominates, i.e.,
$2\log\tfrac{1}{\varepsilon}>(2m-n+\log\tfrac{1}{\delta})/3$.
Modify the construction of~\cite{DKRS06} by setting
$v=(2n-m+\log\tfrac{1}{\delta})/3$ and $R=[ia]_{v+1}^{n-v-\beta}$, where $\beta=2\log\tfrac{1}{\varepsilon}-(2m-n-\log\tfrac{1}{\delta})/3$.
This will result in an extracted key of length $\ell=(4m-2n-\log\tfrac{1}{\delta})/3-2\log\tfrac{1}{\varepsilon}$.  However,
even with the improvement, the extracted key will be always shorter than the key extracted by our scheme, as explained
in Section~\ref{section-dkrs-uniformity-dominates-comparison}

In contrast, this improvement seems useful in the case of pre-application robustness.
Again, suppose the uniformity constraint dominates, i.e.,
$2\log\tfrac{1}{\varepsilon}>\log\tfrac{1}{\delta}$.
Modify the construction of~\cite{DKRS06} by setting $v = n - m +\log\tfrac{1}{\delta}$ and
$R = [ia]_{v+1}^{n-v-\beta}$, where $\beta=2\log\tfrac{1}{\varepsilon}-\log\tfrac{1}{\delta}$. This will
result in an extracted key of length $\ell=2m - n -2\log\tfrac{1}{\varepsilon} -\log\tfrac{1}{\delta}$, which is
$2\log\tfrac{1}{\varepsilon}-\log\tfrac{1}{\delta}$ longer than the key extracted without this modification.

\section{Comparison with the construction of \cite{DKRS06}}
\label{sec-comparison-dkrs}
\subsection{When the Robustness Constraint Dominates}
Recall that the construction of Dodis et al.~\cite{DKRS06} parses $w$ as two strings
$a$ and $b$ of lengths $n-v$ and $v$, respectively.
The values $\sigma,R$ are computed as $\sigma = [ia]^v_1+b$ and $R = [ia]^n_{v+1}$;
$P = (i,\sigma)$.
Notice that, like in our construction, increasing $v$ improves robustness and
decreases the number of extracted bits.  For pre-application robustness,
setting $v = n-m+\log\tfrac{1}{\delta}$ suffices, and thus
the construction extracts nearly $(2m-n)$ bits.
However, for post-application robustness, a much higher $v$ is needed, giving
only around $\frac{1}{3}(2m-n)$ extracted bits.

The post-application robustness game reveals more information to ${\A}$ about
$w$ than the pre-application robustness game. This additional
information---namely, $R$---may make it easier for ${\A}$ to guess
$\sigma'$ for a well-chosen $i'$.
The key to our improvement is in the pairwise independence of the function
$ia+b$ that computes both $\sigma$ and $R$: because of pairwise independence,
the value $(\sigma, R)$ of the function
on input $i$ tells ${\A}$ nothing about the value $(\sigma',R')$ on another
input $i'$.  (This holds, of course, for uniformly chosen key $(a, b)$;
when $(a, b)$ has entropy $m$, then ${\A}$ can find out $n-m$ bits of information
about $\sigma'$.)

In contrast, in the construction of~\cite{DKRS06}, only $\sigma$ is
computed using a pairwise independent hash function.  This works well
(in fact, better than our construction, because $b$ can be shorter)
for pre-application robustness, where ${\A}$ does not find out $R$.
But it makes it possible for $R$ to decrease ${\A}$'s uncertainty
about $\sigma'$ by as much as $\ell=|R|$, thus necessitating the
length $v$ of $\sigma'$ (and hence $\sigma$) to be $v>\ell+(n-m)$ (the
$(n-m)$ term is the amount of entropy already potentially ``missing''
from $\sigma'$ because of the nonuniformity of $w$).  See
Section~\ref{sec-dkrs-lowerbound} for a detailed description of an
adversarial strategy that utilizes $R$ to obtain $\sigma'$ in
the~\cite{DKRS06} construction.

Another way to see the differences between the two constructions is
through the proof.  In the proof of post-application robustness, the
transcript $\mathsf{tr}$ includes $R$, which makes for $2^\ell$ times more
transcripts than in the proof of pre-application robustness.  However,
the fact that this $R$ imposes an additional constraint of $w$, thus
reducing the size of the set ${\mathsf{Bad}}_\mathsf{tr}$, can compensate for this
increase.  It turns out that for the construction of~\cite{DKRS06},
this additional constraint can be redundant if the adversary is clever
about choosing $i'$ and $\sigma'$, and the size of ${\mathsf{Bad}}_\mathsf{tr}$ doesn't decrease.
Using a pairwise-independent function for computing $R$ in our
construction ensures that this additional constraint decreases the
size of ${\mathsf{Bad}}_\mathsf{tr}$ by $2^\ell$.  Thus, our construction achieves the
same results for pre- and post-application robustness.

\subsection{When the Uniformity Constraint Dominates}
\label{section-dkrs-uniformity-dominates-comparison}
It should be noted that there may be reasonable cases when the
uniformity constraint $\varepsilon$ on $R$ is strong enough that the
construction of \cite{DKRS06} extracts even fewer bits, because it
needs to take $v\ge n-m+2\log\tfrac{1}{\varepsilon}$ to ensure near-uniformity of $R$
given $P$. In that
case, as long as $m\ge n/2+2\log\tfrac{1}{\varepsilon}$,
our construction will extract the same amount of bits as before,
thus giving it an even bigger advantage.
And when $m<n/2+2\log\tfrac{1}{\varepsilon}$,
our construction still extracts at least $3/2$ times more bits than the
construction of~\cite{DKRS06}, even with the improvement of
Section~\ref{section-dkrs-uniformity-dominates-improvement} applied
(this can be seen by algebraic manipulation of the relevant parameters
for the post-application robustness case).

\subsection{Why the construction of \cite{DKRS06} cannot extract more bits}
\label{sec-dkrs-lowerbound}
Recall that the robust fuzzy extractor of~\cite{DKRS06} operates as
follows: parse $w$ as two strings $a, b$ of lengths $n-v,v$
respectively and compute $\sigma =
[ia]^v_1+b$ and $R = [ia]^n_{v+1}$; $P = (i,\sigma)$.

For post-application robustness, the concern is that $R$ can reveal
information to the adversary about $\sigma'$ for a cleverly chosen
$i'$.  Because the length of $\sigma'$ is $v$ and $\ell+(n-m)$ bits of
information about $\sigma'$ may be available (the $\ell$ term comes
from $|R|$, and $(n-m)$ term comes from the part of $w$ which has no
entropy), this leads to the requirement that $v\ge \ell+n-m+\log\tfrac{1}{\delta}$
to make sure the adversary has to guess at least $\log\tfrac{1}{\delta}$ bits about
$\sigma'$.  Plugging in $\ell=n-2v$, we obtain $\ell\le
\frac{2}{3}(m-n/2-\log\tfrac{1}{\delta})$, which is the amount extracted by the
construction.

Here we show an adversarial strategy that indeed utilizes $R$ to
obtain information about $\sigma'$ to succeed with probability
$\delta/2$.  This demonstrates that the analysis in~\cite{DKRS06} is
tight up to one bit.  To do so we have to fix a particular (and
somewhat unusual) representation of field elements.  (Recall that any
representation of field elements works for constructions here and
in~\cite{DKRS06}, as long as addition of field elements corresponds to
the exclusive-or of bit strings.)  Typically, one views ${\mathbb F}_{2^{n-v}}$
as ${\mathbb F}_2[x]/(p(x))$ for some irreducible polynomial $p$ of degree
$n-v$, and represents elements as ${\mathbb F}_2$-valued vectors in the basis
$(x^{n-v-1}, x^{n-v-2}, ..., x^2, x, 1)$.  We will do the same, but
will reorder the basis elements so as to separate the even and the odd
powers of $x$:
\ifnum0 =0 $(x^{n-v-1}, x^{n-v-3}, \dots, x, x^{n-v-2}, x^{n-v-4},
\dots, 1)$
\else $(x^{n-v-1}, x^{n-v-3}, \dots, x, \allowbreak x^{n-v-2}, x^{n-v-4},
\dots, 1)$
\fi
(assuming, for concreteness, that $n-v$ is even).  The
advantage of this representation for us is that the top half of
bits of some value $z\in F_{2^{n-v}}$ is equal to the bottom half of the
bits of $z/x$, as long as the last bit of $z$ is 0.

Now suppose the distribution on $w$ is such that the top $n-m$ bits of
$b$ are 0 (the rest of the bits of $w$ are uniform).  Then by
receiving $\sigma$ and $R$, the adversary gets to see the top
$\ell+(n-m)$ bits of $ia$.  Therefore, the adversary knows
$\ell+(n-m)$ bits from the bottom half of $ia/x$ as long as the last
bit of $ia$ is 0, which happens with probability $1/2$.  To use this
knowledge, the adversary will simply ensure that the difference
between $\sigma'$ and $\sigma$ is $[ia/x]_1^v$, by letting $i'=i+i/x$.

Thus, the adversarial strategy is as follows: let $i'=i+i/x$; let
$\tau$ consist of the $\ell$ bits of $R$, the top $n-m$ bits of
$\sigma$, and $\log\tfrac{1}{\delta}=v-\ell-(n-m)$ randomly guessed bits, and let
$\sigma'=\sigma+\tau$.  The adversary wins whenever $\tau=[ia/x]_1^v$,
which happens with probability $2^{v-\ell-(n-m)}/2=\delta/2$, because all but
$\log\tfrac{1}{\delta}$ bits of $\tau$ are definitely correct as long as the last
bit of $ia$ is 0.

The above discussion  gives us the following result.
\begin{theorem}
\label{theorem-dkrs-lowerbound}
There exists a basis for $GF(2^{n-v})$ such that
for any integer $m$ there exists a distribution $W$ of min-entropy $m$ for which
the post-application robustness of the construction from \cite[Theorem 3]{DKRS06}
can be violated with probability at least $\delta/2$,
where $v$ is set as required for robustness $\delta$
by the construction (i.e., $v=(n-\ell)/2$ for $\ell=(2m-n-2\log\tfrac{1}{\delta})/3$).
\end{theorem}

Note that our lower bound uses a specific representation of field
elements, and hence does not rule out that for some particular
representation of field elements, a lower value of $v$ and, therefore,
a higher value of $\ell$ is possible.  However, a security proof for a
lower value of $v$ would have to then depend on the properties of that
particular representation and would not cover the
construction of~\cite{DKRS06} in general.

\section{Tolerating Binary Hamming Errors}
\label{sec-rfe-errors}

We now consider the scenario where Bob has
a string $w'$ that is close to Alice's input $w$ (in the Hamming metric).
In order for them to agree on a random string, Bob
would first have recover $w$ from $w'$. To
this end, Alice could send the secure sketch $s=\mathsf{SS}(w)$
to Bob along with $(i,\sigma)$.
To prevent an undetected modification of $s$ to $s'$, she could
send an authentication of $s$ (using $w$ as the key) as well.
The nontriviality of making such an extension work
arises from the fact that modifying $s$ to $s'$ also
gives the adversary the power to influence
Bob's verification key $w^* = \mathsf{SRec}(w',s')$. The adversary
could perhaps exploit this circularity
to succeed in an active attack (the definition of
standard authentication schemes only guarantee security
when the keys used for authentication and verification are the same).

We break this circularity by exploiting
the algebraic properties of the Hamming metric space,
and using authentication secure against algebraic
manipulation~\cite{DKRS06,CDFPW08}.
The techniques that we use are essentially the same as
used in ~\cite{DKRS06}, but adapted to our construction.
We present the construction here and then
discuss the exact properties that we use in the proof of security.

\mypar{Construction.}
Let ${\cal M}$ be the Hamming metric space on $\{0,1\}^n$.
Let $W$ be a distribution of min-entropy $m$ over ${\cal M}$.
Let $s = \mathsf{SS}(w)$ be a deterministic,
linear secure sketch; let $|s| = k$, $n' = n-k$. Assume that $\mathsf{SS}$ is a surjective linear
function (which is the case for the syndrome construction for the Hamming metric mentioned in Section~\ref{section-prelims}). Therefore,
there exists a $k \times n$ matrix ${\mathsf{S}}$ of rank $k$ such that $\mathsf{SS}(w) = {\mathsf{S}} w$.
Let ${\mathsf{S}}^\bot$ be an $n' \times n$ matrix such that
 $n \times n$ matrix $\left(\frac{{\mathsf{S}}}{{\mathsf{S}}^\bot}\right)$
has full rank. We let $\mathsf{SS}^\bot(w) = {\mathsf{S}}^\bot(w)$.

To compute $\mathsf{Gen}(w)$, let $s = \mathsf{SS}(w)$, $c = \mathsf{SS}^\bot(w)$; $|c| = n'$.
We assume that $n'$ is even (if not, drop one bit of $c$,  reducing its entropy
by at most 1). Let $a$ be the first half of $c$ and $b$ the second. View $a,b$
as elements of ${\mathbb F}_{2^{n'/2}}$. Let $L = 2\lceil \frac{k}{n'} \rceil$ (it will
important for security that $L$ is even).
Pad $s$ with 0s to length $Ln'/2$, and then
split it into $L$ bit strings $s_{L-1}, \dots, s_0$
of length $n'/2$ bits each, viewing
each bit string as an element of ${\mathbb F}_{2^{n'/2}}$.
 Select $i\leftarrow {\mathbb F}_{2^{n'/2}}.$
Define $f_{s,i}(x) = x^{L+3}+x^2(s_{L-1}x^{L-1} + s_{L-2}x^{L-2}+\dots+s_0) + ix.$
Set $\sigma = [f_{s,i}(a)+b]^{v}_{1}$, and output
$P=(s,i,\sigma)$ and $R= [f_{s,i}(a)+b]^{n'/2}_{v+1}$.

\begin{tabbing}
\ifnum0=0
\\
wwwwwwwwwwww\=www\=3. \=wwwwwwwww\=\kill
\else
ww\=wwww\=3. \=wwwwwwwww\=\kill
\fi
\>\+
$\mathsf{Gen}(w)$:\\
1. Set $s = \mathsf{SS}(w)$, $c = \mathsf{SS}^\bot(w)$, $k = |s|$, $n' =|c|$. \\
\> - Let $a = [c]^{n'/2}_1$, $b=[c]^{n'}_{n'/2+1}$\\
\> - Let $L = 2\lceil \frac{k}{n'} \rceil.$ Pad $s$ with 0s to length $Ln'/2$.\\
\> - Parse the padded $s$ as $s_{L-1}||s_{L-2}|| \ldots || s_0$ for $s_i\in {\mathbb F}_{2^{n'/2}}$.\\
2. Select $i\leftarrow {\mathbb F}_{2^{n'/2}}.$\\
3. Set $\sigma = [f_{s,i}(a)+b]^{v}_{1}$, and output $R= [f_{s,i}(a)+b]^{n'/2}_{v+1}$ and $P=(s,i, \sigma).$\\
\ \\
$\mathsf{Rep}(w',P' = (s',i',\sigma'))$:\\
1. Compute $w^* = \mathsf{SRec}(w',s')$\\
\> - Verify that $\mathsf{dis}(w^*,w') \leq t$ and $\mathsf{SS}(w^*) = s'$. If not, output $\bot$. \\
2. Let $c'=\mathsf{SS}^{\perp}(w^*)$. Parse $c'$ as $a'||b'$. \\
3. Compute $\sigma^* = [f_{s',i'}(a')+b']^{v}_{1}$. \\
\> - Verify that $\sigma^* = \sigma'$. If so, output $R= [f_{s',i'}(a')+b']^{n'/2}_{v+1}$, else output $\bot$.
\end{tabbing}

In the theorem statement below, let $B$ denote the volume of a Hamming ball or radius $t$ in $\{0,1\}^n$ ($\log B\le nH_2(t/n)$~\cite[Chapter 10, \S 11, Lemma 8]{MW77} and $\log B \le t \log (n+1)$~\cite{DKRS06}).

\begin{theorem}
Assume $\mathsf{SS}$ is a deterministic linear $(m,m-k,t)-$secure sketch of output length $k$ for the Hamming metric
on $\{0,1\}^n$. Setting $v =(n-k)/2-l$, the above construction
is an $(m,l,t,\varepsilon)$ fuzzy extractor with robustness $\delta$ for any $m,l,t,\varepsilon$ satisfying
$l\leq m-n/2-k-\log B - \log\left(2\left\lceil\frac{k}{n-k}\right\rceil +2\right)-\log\tfrac{1}{\delta}$
as long as $m\geq \frac{1}{2}(n+k)+2\log\tfrac{1}{\varepsilon}.$
\end{theorem}

Again, if $m<\frac{1}{2}(n+k)+2\log\tfrac{1}{\varepsilon}$, the construction can be modified,
as shown in Section~\ref{section-closer-to-uniform-fuzzy}.

\begin{proof}
\mypar{Extraction.}
Our goal is to show that $R$ is nearly uniform given $P = (i,s,\sigma)$.
To do so, we first note that for every $s$,
the function $h_i(c) = (\sigma, R)$
is a universal hash family. Indeed for $c \neq c'$ there is a unique $i$ such that $h_i(c) = h_i(c')$
(since $i(a-a')$ is fixed, like in the errorless case).
We also note that ${\widetilde{\mathbf{H}}_{\infty}}(c\mid \mathsf{SS}(W)) \ge {\widetilde{\mathbf{H}}_{\infty}}(c, \mathsf{SS}(W))-k={\mathbf{H}_{\infty}}(W)-k=m-k$
by Lemma~\ref{lemma-avg-min-entropy}. Because $|(R,\sigma)| = n'/2$,
Lemma~\ref{lemma-leftover-hash} (or, more precisely,
its generalization mentioned in the paragraph following the lemma,
needed here because $h_i$ depends on $s$)
gives us
\[\sd{(R,P),U_{|R|} \times \mathsf{SS}(W) \times U_{n'/2}\times U_{v}}\le \varepsilon/2\] for
$n'/2 \leq m -k +2 -2\log(2/\varepsilon)$.  This is equivalent to saying that
$(R,P)$ is $2^{(n'/2-m+k)\frac{1}{2}-1}$-close to $U_{|R|} \times \mathsf{SS}(W) \times U_{n'/2}\times U_{v}$.

Applying Lemma~\ref{lemma-sdswitch} to $A=R$, $B=P$, $C=U_{n'/2-v}$,
$D=\mathsf{SS}(w)\times U_{n'/2}\times U_v$, we get that $(R, P)$ is
$\varepsilon$-close
to $U_{\frac{n'}{2}-v}\times P$, for $\varepsilon =
2^{(\frac{n'}{2}-m+k)/2}.$

From here it follows that for extraction to be possible,  $m\geq \frac{1}{2}(n+k)+2\log\tfrac{1}{\varepsilon}\,.$

\mypar{Post-Application Robustness.}
In the post-application robustness security game,
the adversary ${\A}$ on receiving $(P=(s,i, \sigma), R)$
(generated according to procedure $\mathsf{Gen}$) outputs
$P' = (s',i',\sigma')$,
and is considered successful if
$(P' \neq P) \wedge \mathsf{Rep}(w',s') \neq \bot$.
In our analysis, we will assume that $(i',s') \neq (i,s)$.
We claim that this does not reduce ${\A}$'s success probability.
Indeed, if $(i',s') = (i,s)$ then,
$c'$ computed within $\mathsf{Rep}$ will equal $c$.
So, for $P'\neq P$ to hold,
${\A}$ would have to output $\sigma' \neq \sigma$.
However, when $(i',c' ,s')=(i,c,s)$, $\mathsf{Rep}$ would compute
$\sigma^*=\sigma$, and therefore
would output $\bot$ unless $\sigma' = \sigma$.

In our analysis, we allow ${\A}$ to be deterministic.
This is without loss of generality since we allow an unbounded adversary.
We also allow ${\A}$ to arbitrarily fix $i$. This
makes the result only stronger since we demonstrate robustness for
a worst-case choice of $i$.

Since $i$ is fixed and ${\A}$ is deterministic,
the $\mathsf{tr} = (i,s,\sigma,R,i',s',\sigma')$
is determined completely by $(s,\sigma, R)$.
Recall that the prime challenge in constructing
a robust fuzzy extractor was that ${\A}$
could somehow relate the key used by $\mathsf{Rep}$ to verify $\sigma'$ to the
authentication key that was used by $\mathsf{Gen}$ to come up with $\sigma$.
As was done in ~\cite{DKRS06},
we will argue security of our construction
by showing that the
MAC
scheme implicitly used in our construction remains unforgeable
even when ${\A}$ could force the verification
key to be at an offset (of her choice) from the authentication key.
We will formalize such an argument by assuming that ${\A}$
learns $\Delta = w' - w$. Recall that $w^* = \mathsf{SRec}(w',s')$ and $c'=a'||b'=\mathsf{SS}^{\perp}(w^*) $.
The following claim that was proven in~\cite{DKRS06}
states that given $(\Delta, s)$, ${\A}$ can compute the
offsets $\Delta_a =a'-a, \Delta_b =b'-b$ induced
by her choice of $s'$.

\begin{claim}
Given $\Delta = w'-w,$  and the sketches $s,s',{\A}$ can compute $\Delta_a = a' - a$ and
$\Delta_b = b' - b$, or determine that $\mathsf{Rep}$ will reject before computing $a',b'$.
\end{claim}

In other words, she can compute the
offset between the authentication key that
$\mathsf{Gen}$ used to come up with $\sigma$
and the verification key that $\mathsf{Rep}$ will use
to verify $\sigma'$. We will now argue that as long as $W$ has sufficient
min-entropy, even knowing the offset does not help ${\A}$ succeed
in an active attack. Recall that since $i$ is arbitrarily fixed by ${\A}$,
${\A}$'s success depends on $w,w'$, or, alternatively, on $w,\Delta$.
Fix some $\Delta$.  For any particular $\mathsf{tr}$,
let $\mathsf{Succ}_{\mathsf{tr},\Delta}$ be the event that
the transcript is $\mathsf{tr}$ and ${\A}$ wins,
i.e., that
$f_{s,i}(a)+b = \sigma||R \wedge [f_{s',i'}(a')+b']^v_1 = \sigma' \wedge \mathsf{SS}(w)=s$,
conditioned on the fact that $w'-w$ is $\Delta$.
We denote by ${\mathsf{Bad}}_{\mathsf{tr},\Delta}$
the set of $w$ that make $\mathsf{Succ}_{\mathsf{tr},\Delta}$
true. We now partition the set ${\mathsf{Bad}}_{\mathsf{tr},\Delta}$ into $2^{\ell}$ disjoint
sets, indexed by $R' \in \{0,1\}^{\ell}$:
\begin{eqnarray*}
{\mathsf{Bad}}^{R'}_{\mathsf{tr},\Delta} &\stackrel{\mbox{\tiny{def}}}{=}& \{w\,|\, w\in {\mathsf{Bad}}_{\mathsf{tr},\Delta} \wedge [f_{s',i'}(a')+b']^\ell_{v+1} = R'\}\\
&=& \{w\,|\,  (f_{s,i}(a)+b = \sigma||R) \wedge (f_{s',i'}(a')+b' = \sigma'||R')\wedge \mathsf{SS}(w)=s\}.
\end{eqnarray*}
By Claim 1, fixing $(\mathsf{tr},\Delta)$, also fixes $\Delta_a,\Delta_b$.
It follows that every $w\in {\mathsf{Bad}}^{R'}_{\mathsf{tr},\Delta}$ needs to satisfy
\[f_{s,i}(a) - f_{s',i'}(a+\Delta_a) = (\Delta_b+\sigma-\sigma')||(R-R') \wedge \mathsf{SS}(w)=s.\]
For a given $\mathsf{tr},\Delta,R'$, the right hand side of the first equation takes a fixed value.
Let us now focus on the polynomial $f_{s,i}(a) - f_{s',i'}(a+\Delta_a)$.
We will consider two cases:
\begin{itemize}
\item $\Delta_a = 0$: In this case,
$f_{s,i}(x) - f_{s',i'}(x)$ is a polynomial in which
a coefficient of degree 2 or higher is nonzero if $s \neq s'$ and
a coefficient of degree 1 or higher is nonzero if $i \neq i'$.
\item $\Delta_a \neq 0$:
Observe that the leading term of the polynomial is $((L+3)\bmod 2)\Delta_ax^{L+2}$.
Since we forced $L$ to be even, the coefficient of the leading term is
nonzero, making $f_{s,i}(x) - f_{s',i'}(x+\Delta_a)$
a polynomial of degree $L + 2$.
\end{itemize}
Therefore, in either case, the $f_{s,i}(x) - f_{s',i'}(x+\Delta_a)$
is a nonconstant polynomial of degree at most $L+2$.
A nonconstant polynomial of degree $d$ can take on a fixed value
at most $d$ times. It, therefore, follows that there are at most
$L+2$ values of $a$ such that
$f_{s,i}(a) - f_{s',i'}(a+\Delta_a) = (\Delta_b+\sigma-\sigma')||(R-R')$.
Each such $a$ uniquely determines $b=(\sigma||R)-f_{s,i}(a)$. And $w$ is uniquely determined by $c=a||b=\mathsf{SS}^{\bot}(w)$ and $s=\mathsf{SS}(w)$.
Therefore, there are at most $L+2$
values of $w$ in the set ${\mathsf{Bad}}^{R'}_{\mathsf{tr},\Delta}$ i.e, $|{\mathsf{Bad}}^{R'}_{\mathsf{tr},\Delta}| \leq L+2$.
Since ${\mathsf{Bad}}_{\mathsf{tr},\Delta} = \bigcup_{R' \in \{0,1\}^\ell}{\mathsf{Bad}}^{R'}_{\mathsf{tr},\Delta}$, we get
$|{\mathsf{Bad}}_{\mathsf{tr},\Delta}| \le (L+2)2^{\ell} = (L+2)2^{{n'/2}-v}$.
Thus, $\Pr_w[\mathsf{Succ}_{\mathsf{tr},\Delta}] \leq |{\mathsf{Bad}}_{\mathsf{tr}}|2^{-{\mathbf{H}_{\infty}}(w|\Delta)} \leq (L+2)2^{n'/2-v-{\mathbf{H}_{\infty}}(w|\Delta)}$.

To find out the probability $\Pr_w[\mathsf{Succ}_\Delta]$
that ${\A}$ succeeds conditioned on a
particular $\Delta$, we need to add up $\Pr_w[\mathsf{Succ}_{\mathsf{tr},\Delta}]$ over
all possible transcripts.  Recalling that each transcript is determined
by $\sigma, R$ and $s$ and hence there are $2^{n'/2+k}$ of them, and that
$n'+k=n$,
we get $\Pr_w[\mathsf{Succ}_\Delta]\leq (L+2)2^{n-v-{\mathbf{H}_{\infty}}(w|\Delta)}$.

Finally, the probability of adversarial success it at most
\[
\expe_{\Delta}\Pr_w[\mathsf{Succ}_\Delta] \le  (L+2)2^{n-v-{\widetilde{\mathbf{H}}_{\infty}}(w|\Delta)}\,.
\]
In particular, if the errors $\Delta$ are independent of $w$,
then ${\widetilde{\mathbf{H}}_{\infty}}(w|\Delta)={\mathbf{H}_{\infty}}(w)=m$, and the probability of adversarial success is at most $(L+2)2^{n-v-m}$.
In the worst case, however, the entropy of $w$ may decrease at most by the number of bits needed to represent $\Delta.$
Let $B$ be the volume of the hamming ball of radius $t$ in $\{0,1\}^n$. Then, $\Delta$ can be represented in
$\log B$ bits and ${\widetilde{\mathbf{H}}_{\infty}}(w|\Delta) \geq m - \log B$, by Lemma~\ref{lemma-avg-min-entropy}. From here it follows that
\[
\Pr[{\cal A}'s\ success] \leq B(L+2)2^{n-v-m}
\]
To achieve $\delta-$robustness, we want  $B(L+2)2^{n-v-m} \leq \delta$
i.e.,  $v \geq n-m +\log B +\log(L+2) + \log\tfrac{1}{\delta}$. Setting $v = n-m+\log B+\log(L+2) +\log\tfrac{1}{\delta}$,
and using $L = 2\lceil\frac{k}{n-k}\rceil$
it follows that \[\ell \leq m - n/2 - k -\log B - \log\left(2\left\lceil\frac{k}{n-k}\right\rceil+2\right)  - \log\tfrac{1}{\delta}\,. \]
\ifnum0=1\qed\fi
\end{proof}

\subsection{Getting Closer to Uniform}
\label{section-closer-to-uniform-fuzzy}

If $\varepsilon$ is so low that $m \ge \frac{1}{2}(n+k) + 2\log\tfrac{1}{\varepsilon}$ does not hold,
we can modify our construction
just as we did in section~\ref{section-uniformity-dominates}, by
shortening $R$ by $\beta=\frac{1}{2}(n+k)+2\log\tfrac{1}{\varepsilon}-m$.
That is,  keep  $v = n-m +\log B+\log(L+2)+\log\tfrac{1}{\delta}$ fixed and
let $R = [f_{s,i}(a)+b]_{v+1}^{\ell+v}$, where
$\ell \le
n/2 -v- \beta$.

\ifnum0=0
\section*{Acknowledgements}
This work was supported in part by the U.S. National Science Foundation grants CCF-0515100 and CNS-0546614.
\fi

\newcommand{\etalchar}[1]{$^{#1}$}

redpajama_set_name: RedPajamaArXiv


text:
Navy And NASA Complete Recovery Test
Written by Patrick Boniface (Naval Correspondent)
Category: Naval Focus
In what might be seen as a return to the heady days of NASA's Apollo moon program, the US Navy is relearning the skills necessary to recover spacecraft from the ocean.
In the 1960s and 1970s, NASA's Mercury, Gemini, Apollo and Skylab space launches all used reentry capsules that landed in the ocean and were recovered by warships. By the end of this decade, the Orion launch system will, once again, return to this tried and tested technique.
The San Antonio-class amphibious transport dock USS John P Murtha was, in late-2018, deployed off southern California for what has been dubbed Underway Recovery Test, or URT. The nature of the tests involve the recovery from the ocean of what is known as a boiler Plate Test Article, BTA, essentially a replica Orion spacecraft.
The first launch into space of a complete, but unmanned, Orion spacecraft will take place when Exploration Mission 1 (EM-1) blasts off from Cape Canaveral in December 2019.
Norwegian News
The Norwegian frigate KNM Helge Instad was in danger of sinking after the Maltese-flagged oil tanker Sola TS ripped a huge hole in her side.
The frigate was returning home following its participation in the NATO Trident Juncture exercise when the incident occurred while docked in Sture, north of Bergen. Eight crew members were injured and two taken to hospital.
The frigate took on a large list as water rushed in, before it was intentionally grounded to prevent it sinking completely. Sadly, despite having heavy hawsers and anchors planted ashore, the frigate subsequently slipped back into deeper water and sank until only the mast and part of her flight deck was above water.
Indian News
_The Indian Government signed a US$950 million contract with Russia's United Shipbuilding Corporation for the delivery of a pair of upgraded Krivak III class stealth frigates to the Indian Navy.
The deal comes in the face of strong opposition from the United States who has imposed strong sanctions against Russian military sales. The Indians and Russians have undertaken a rupee-rouble transaction to circumvent the sanctions imposed under the Countering America's Adversaries Through Sanctions Act (CATSA).
The two frigates will be built at Yantar Shipyard and delivered towards the end of 2022.
The French Flagship and only aircraft carrier FS Charles de Gaulle completed her latest 18 month refit in Toulon on 9 November.
The midlife overhaul was undertaken by the Naval Group and is aimed at allowing the carrier to continue in service until at least 2038.
During the refit work was undertaken on the combat system, aircraft maintenance facilities, a new longerrange 3D radar and her deck landing systems were modifi ed to an 'all Rafale' configuration following the withdrawal from service of the Super Etendard aircraft.
More on this and other news in Sea Breezes Magazine - January 2019 Issue


I decided to start this blog as an outlet for my bad days to deal with my copious maladies, lol. So that my peeps didn't have to hear me whine is really what I trying to say, ha! My family and I were going through so much for years, I didn't know how to handle it on my own, how to process my feelings, how to be/try to be "normal" again. That one's tough. I struggle most days still, even after all of the medication changes, the brain surgery, weight loss, nutrition and healthy lifestyle changes, to be my new "normal" self and accepting that this is it after all the dust settled.
Albeit, better than dead, of course, I am living with hydrocephalus, narcolepsy, trigeminal neuralgia , 3 types of migraines: hemiplegic, ice pick, and classic migraine. I had an ETV surgery for my hydrocephalus, where an opening is created in the floor of the third ventricle using an endoscope placed within the ventricular system through a burr hole. Anyhoo, now I can officially say I have a hole in my head, lol! I This blog I hope will help someone else, will make you laugh, I hope it will make you think wtf, I tend to have that effect on people, lmao, that's cool too! Please feel free to add me, share my blog with a friend:) Thanx!
Poet/Writer/Food Lover/Music Lover/Animal Lover/"EMOTION HOARDER"


Jacobs Ladder is the most effective commercial cardio device on the market today.
It is designed specifically for high intensity, low impact exercise but, it goes much deeper than that. Because it is a functional exercise, it is used in rehabilitation, circuit training, personal training, high intensity intervals and endurance training. In addition, it burns more calories with a lower perceived exertion rate.
Also known as the ladder mill, this machine features ladder-like "rungs" on a 40-degree angle, which allow you to "climb" in place. The resistance is controlled by the user, who wears a belt that's tethered to the machine, rather than some preprogrammed speed setting. Unlike with the treadmill, you'll stimulate your entire body as you climb on all fours on a steep incline—no muscle group will be unscathed.


text:
Q: Creating .pptx file with OpenXML I want to create a PowerPoint presentation based on a template in C#.
I am using the OpenXML 2.0 SDK.
My first step.
I took my .pptx template, changed it in C# code and copied it in my project.

*

*I was able to create slides based on my templates.

*Now i have some placeholders in my template i want to replace.
I found the following method, on the msdn side
void SwapPlaceholderText(SlidePart slidePart, string placeholder, string value)
{
   //Find and get all the placeholder text locations.
   List<Drawing.Text> textList = slidePart.Slide.Descendants<Drawing.Text>().Where(t => t.Text.Equals(placeholder)).ToList();
   //Swap the placeholder text with the text from DB
   foreach (Drawing.Text text in textList) text.Text = value;
}

But the size of the list is 0.
Does this method call even searching in the right method of my template?
The information of the placeholder i am searching for is in the following method of my template.
private void GenerateUserDefinedTagsPart5Content(UserDefinedTagsPart userDefinedTagsPart5)
{
 ...
 Tag tag71 = new Tag(){ Name = "FIELD.CHAPTER.CONTENT", Val = "#header#" };
 Tag tag72 = new Tag(){ Name = "FIELD.CHAPTER.VALUE", Val = "#header#" };
 ... }

What i am doing wrong? and how do i need to modify the placeholder-method?
Best regards!
EDIT : I found out some strange thing. I declared in my PowerPoint template a placeholder called "#header#" and in my program i was searching for "#header#" but after i ran the program once, the "#header#" in the PowerPoint file changed to "header" i don't know why but when i search now for "header" my list size is 1 and the header will be replaced.
But after running my program and want to start the PowerPoint presentation, it says the data is corrupted. Maybe you can see some mistakes:
This is my main method.
    Template template = new Template();
    template.CreatePackage("templatesource");

    PresentationDocument doc = PresentationDocument.Open("templatesource", true);
    PowerPoint powerPoint = new PowerPoint(doc, proj);
    powerPoint.AddNewSlides();

PowerPoint-class:
public void addSlides()
{
    for (int i = 0; i < proj.projects.Capacity; i++  )
    {
        Console.WriteLine("Slide number: " + i);
        AddNewSlide(_document.PresentationPart);
    }
    _document.PresentationPart.Presentation.Save();
}

private void AddNewSlide(PresentationPart parent)
{
    if (_slideTemplate == null) return;

    var newSlidePart = parent.AddNewPart<SlidePart>("newSlide" + _slideId);

    newSlidePart.FeedData(_slideTemplate.GetStream(FileMode.Open));
    newSlidePart.AddPart(_slideTemplate.SlideLayoutPart, _slideTemplate.GetIdOfPart(_slideTemplate.SlideLayoutPart));

    SetPlaceholder(newSlidePart, "#header#", "My new header");

    newSlidePart.Slide.Save();

    SlideIdList listOfSlidesIDs = parent.Presentation.SlideIdList;
    uint maxSlideId = 1;

    foreach (SlideId slideId in listOfSlidesIDs.ChildElements)
    {
        if (slideId.Id > maxSlideId) maxSlideId = slideId.Id;
    }

    SlideId newSlideId = new SlideId { Id = ++maxSlideId, RelationshipId = parent.GetIdOfPart(newSlidePart) };
    listOfSlidesIDs.Append(newSlideId);

    _slideId++;
}

Thanks for your help :)

redpajama_set_name: RedPajamaStackExchange


Bell Group Carlisle branch opened in 1999 and is ideally positioned to operate on both sides of the border, covering the whole of Cumbria and the Lake District, Dumfries & Galloway in South West Scotland and Roxburghshire in the Scottish Borders.
We are proud to employ over sixty staff based throughout the area, giving a true local service backed with the strength of a national company. Based on our core values, we have built long standing relationships with our loyal customer base, to offer all aspects of joinery, painting and maintenance works.


text:
(function (app) {
    'use strict';

    app.factory('membershipService', membershipService);

    membershipService.$inject = ['apiService', 'notificationService', '$http', '$base64', '$cookieStore', '$rootScope'];

    function membershipService(apiService, notificationService, $http, $base64, $cookieStore, $rootScope) {

        var service = {
            login: login,
            register: register,
            saveCredentials: saveCredentials,
            getLoggedInUser:getLoggedInUser,
            removeCredentials: removeCredentials,
            isUserLoggedIn: isUserLoggedIn,
            hasRolePermission: hasRolePermission
        }

        function login(user, completed) {
            apiService.post('/api/account/authenticate', user,
            completed,
            loginFailed);
        }

        function register(user, completed) {
            apiService.post('/api/account/register', user,
            completed,
            registrationFailed);
        }

        function saveCredentials(user, roles) {
            var membershipData = $base64.encode(user.username + ':' + user.password);

            $rootScope.repository = {
                loggedUser: {
                    username: user.username,
                    authdata: membershipData,
                    roles: roles
                }
            };

            $http.defaults.headers.common['Authorization'] = 'Basic ' + membershipData;
            $cookieStore.put('repository', $rootScope.repository);
        }

        function removeCredentials() {
            $rootScope.repository = {};
            $cookieStore.remove('repository');
            $http.defaults.headers.common.Authorization = '';
        };

        function loginFailed(response) {
            notificationService.displayError(response.data);
        }

        function registrationFailed(response) {

            notificationService.displayError('Registration failed. Try again.');
        }

        function isUserLoggedIn() {
            return $rootScope.repository.loggedUser != null;
        }
        function hasRolePermission(role) {
            return (
                $rootScope.repository !== undefined &&
                $rootScope.repository.loggedUser !== undefined &&
                $rootScope.repository.loggedUser.roles !== undefined &&
                $rootScope.repository.loggedUser.roles.indexOf(',' + role + ',') > -1
        )
        }
        function getLoggedInUser() {
            return $rootScope.repository.loggedUser;
        }
        return service;
    }

})(angular.module('common.core'));

redpajama_set_name: RedPajamaGithub


There's a Japanese cafe exclusively for stuffed toys!
Now there's a bed and breakfast style cafe where you can send your stuffed toys for food, a game of cards and ghost stories!
It works like this: you mail your stuffed toys to Yawarakan's, where your stuffed toy will eat, play card games and have a story read to it before bedtime. The stuffed toy is then posted back to you the next day.
It's reported that Yawarakan's will serve your stuffed toy a real berry smoothie, followed by a rice omelette and a pancake stack for dessert – which are cooked by chefs Hebi-chan the snake and Karei the flounder. The food is served by the monkey waiter Saru Hasegawa.
The chefs and waiter are also stuffed toys.
The 4,968 Yen ($40) basic package includes food, lodging, souvenirs and a photo album to mark the occasion.


text:
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J. F. Splettstoesser
https://doi.org/10.1130/MEM170
© 1992 Geological Society of America
ISBN print:
Chapter 21: Structure of the Sentinel Range, Ellsworth Mountains, West Antarctica
Campbell Craddock
K. B. Spörli
John J. Anderson
https://doi.org/10.1130/MEM170-p393
Campbell Craddock, K. B. Spörli, John J. Anderson, 1992. "Chapter 21: Structure of the Sentinel Range, Ellsworth Mountains, West Antarctica", Geology and Paleontology of the Ellsworth Mountains, West Antarctica, G. F. Webers, C. Craddock, J. F. Splettstoesser
Refmanager
The 5,500+-m Paleozoic succession of the Sentinel Range includes the Heritage Group (Cambrian), the Crashsite Group (Cambrian and younger, including Devonian), the Whiteout Conglomerate (Carboniferous?), and the Polarstar Formation (Permian). All these predominantly clastic rocks were strongly folded during the early Mesozoic (?) Ellsworth (Gondwanide) Orogeny, and the folds formed mainly by a flexural-slip mechanism. Axial planar cleavage is abundant throughout the range; it may be a solution cleavage in the massive diamictite beds of the Whiteout Conglomerate. One major, eastward-dipping reverse (?) fault cuts Crashsite quartzite beds along the west side of range. The Sentinel Range succession was subjected...
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text:
package sun.nio.cs.ext;

import java.nio.charset.Charset;
import java.nio.charset.CharsetDecoder;
import java.nio.charset.CharsetEncoder;

public class MSISO2022JP extends ISO2022_JP
{
    public MSISO2022JP() {
        super("x-windows-iso2022jp",
              ExtendedCharsets.aliasesFor("x-windows-iso2022jp"));
    }

    public String historicalName() {
        return "windows-iso2022jp";
    }

    public boolean contains(Charset cs) {
      return super.contains(cs) ||
             (cs instanceof MSISO2022JP);
    }

    public CharsetDecoder newDecoder() {
        return new Decoder(this, CoderHolder.DEC0208, null);
    }

    public CharsetEncoder newEncoder() {
        return new Encoder(this, CoderHolder.ENC0208, null, true);
    }

    private static class CoderHolder {
        final static DoubleByte.Decoder DEC0208 =
            (DoubleByte.Decoder)new JIS_X_0208_MS932().newDecoder();
        final static DoubleByte.Encoder ENC0208 =
            (DoubleByte.Encoder)new JIS_X_0208_MS932().newEncoder();
    }
}


text:
Q: Add bare git repositories to gitlab Today I learnt that the rake import task for bare git repositories is soon to be deprecated (and will be removed later). I have not yet understood what I should do instead:

*

*The web project import is for a single project, but I need something I can do programmatically (via REST, ssh or alike).

*The automated import only mentions other vendors (like GitHub, BitBucket etc.), but I have bare git repositories, which are not on any git server.
So how do import a large number of bare git repositories into GitLab (self-managed)?

A: Interesting that it doesn't mention it as part of the deprecation notice, but you could write a simple script that loops through all your repositories and creates the project through git push.
If you take a look at the Create a new project with Git push documentation, that should cover the use case you've described.
If you need to create the projects in different namespaces and those don't exist, then you'll need to create a group via API as part of your script.


https://www.ecpms.net/g8fzwsh9t?key=09ead7d42d83f9cfabf60cee8dd5512a Wonderfully Models Of Best Small Bathroom Designs 2018 - Through the thousand Figure on the web regarding best small bathroom designs 2018, we all filter the very best series along with excellent image resolution truly for our visitor, and of course,this photographs is usually among stocks selections within our greatest images gallery regarding Wonderfully Models Of Best Small Bathroom Designs 2018. I feel you may enjoy it.
posted by means of admin at 2019-02-10 13:49:50. To come up with many pics with Wonderfully Models Of Best Small Bathroom Designs 2018 pics gallery please comply with this specific website hyperlink. If you love this post, then please share with your friends.


American Political Science Association (Chair, Federalism and Intergovernmental Relations Section); Midwest Political Science Association; Canadian Political Science Association; Association for Canadian Studies in the United States. Editorial Boards: Political Research Quarterly; Publius: The Journal of Federalism, Governance: An International Journal of Policy, Administration and Institutions; Review of Policy Studies, American Review of Canadian Studies.
Intergovernmental climate policy development and implementation, with particular focus on policy diffusion and regionalization; comparative federalism and climate policy in North America and Europe; renewable energy politics and policy in North America; federalism and environmental policy development in the United States and Canada; comparative governance of nanotechnology; public opinion on climate, energy and environmental protection.


What can Chris Nolte Concreting do for you?
Whether you're building or renovating, a concrete patio, path or driveway can add value and appeal to your property.
Chris Nolte Concreting can help you go about these projects in the most affordable, professional way possible.
We work with many different types of concrete and can use a range of techniques to suit your project.
We do our own quotes for each of our services. We pay careful attention to detail to ensure our jobs are finished to the highest standards every time and we bring experience and technical knowledge to each – whether it's a big new build or a relatively minor renovation.
Get in touch with Chris Nolte Concreting today.


Nov. 6, 2018 - Professor Zhenan Bao, the K.K. Lee Professor in Chemical Engineering, with courtesy appointments in Chemistry and Materials Science and Engineering, at Stanford University was the featured guest speaker for the 17th Annual Fredrickson Lecture on Tuesday, November 6. Bao's seminar, "Skin-Inspired Organic Electronic Materials and Devices" described how skin, as the body's largest organ and a conformable, stretchable and biodegradable material simultaneously collects signals from external stimuli that translate into information such as pressure, pain, and temperature. The development of electronic materials, inspired by the complexity of this organ is a tremendous, unrealized materials challenge. However, the advent of organic-based electronic materials may offer a potential solution to this longstanding problem. Bao described the design of organic electronic materials to mimic skin functions. These new materials enabled unprecedented performance or functions in medical devices, energy storage and environmental applications.
Bao (center) received a commemorative Fredrickson Lecture plaque and is shown with CEMS bio-engineering faculty members after her lecture including (from left) Samira Azarin, Robert Tranquillo, Ben Hackel, Kevin Dorfman, and Friedrich Srienc.
The Fredrickson Lecture was established in 2001 to honor Professor Emeritus Arnold (Arnie) Fredrickson upon his retirement and foster broader intellectual interaction between undergraduate and graduate students.


text:
#include <stdlib.h>
#include <unistd.h>
#include <limits.h>
#include "sysrepo.h"
#include "sysrepo/trees.h"

int
main(int argc, char **argv)
{
    sr_conn_ctx_t *conn = NULL;
    sr_session_ctx_t *sess = NULL;
    const char *xpath = NULL;
    sr_node_t *subtree = NULL;
    int rc = SR_ERR_OK;

    /* turn on debug logging to stderr - to see what's happening behind the scenes */
    sr_log_stderr(SR_LL_DBG);

    /* connect to sysrepo */
    rc = sr_connect("app1", SR_CONN_DEFAULT, &conn);
    if (SR_ERR_OK != rc) {
        goto cleanup;
    }

    /* start session */
    rc = sr_session_start(conn, SR_DS_STARTUP, SR_SESS_DEFAULT, &sess);
    if (SR_ERR_OK != rc) {
        goto cleanup;
    }

    /* get one subtree */
    xpath = "/ietf-interfaces:interfaces/interface[name='eth0']";
    rc = sr_get_subtree(sess, xpath, SR_GET_SUBTREE_DEFAULT, &subtree);
    if (SR_ERR_OK != rc) {
        goto cleanup;
    }

    /* print the subtree content */
    printf("\n\nSubtree on xpath: %s =\n", xpath);
    sr_print_tree(subtree, INT_MAX);
    printf("\n\n");

cleanup:
    if (NULL != subtree) {
        sr_free_tree(subtree);
    }
    if (NULL != sess) {
        sr_session_stop(sess);
    }
    if (NULL != conn) {
        sr_disconnect(conn);
    }
    return rc;
}


> FAQs > Composite Fillings > Why does my child have so many cavities?
Provide a healthy diet. Having a healthy, well-balanced diet that limits sugary and sticky foods will help protect the enamel covering your child's teeth. Bacteria cause decay and they use sugary and starchy foods as fuel to reproduce and cause more damage.
Avoid bottle use at bedtime. After you brush your child's teeth before bed, they should only have water in their bottle or sippy cup. Not allowing milk or sugary drinks after bedtime (and during naps) will help protect your child's teeth at any age. Some of the most damaging drinks are fruit juices and soda. For older kids, energy drinks may be the culprit.
Be sure to use fluoride toothpaste. Fluoride strengthens protective enamel and protects teeth from the softening effects of bacteria. While many toothpastes on the market claim to fight cavities, check out the label and make sure it contains fluoride.


In this free online course Diploma in Python Programming you will learn about the concepts of programming using the Python language. You will learn about creating variables and storing different types of data in them. You will learn about working with dates and time, making decisions with your code and using loops. You will learn about saving and reading files with your program. You will learn about creating functions and handling errors with your code.
The course begins by introducing you to the concepts of programming. You will learn about creating your first Python program with Visual Studio. You will learn about commenting in your code and displaying text on screen. You will learn about how Python accepts multiple ways of containing a text string. You will learn about variables and the rules and guidelines for creating variables. You will learn about manipulating variables using built in functions, and briefly what functions are. You will learn about Visual Studio's IntelliSense feature and its helpfulness when programming.
Next, you will learn about storing numeric variables and their difference to string variables. You will learn about performing mathematical operations in your code. You will learn about formatting numbers and converting string variables to numeric variables. You will learn about working with date and time, displacing dates and getting today's date using functions. You will learn about formatting dates and importing libraries to help with the formatting. You will learn about working with time and how to reuse the same functions to format time as you did dates.
Next, you will learn about how you code check conditions and make decisions with the if statement. You will learn about the constraints for checking an if statement and how there are generally two different ways to write every if statement. You will learn about using the elif statement for checking multiple conditions. You will learn about combining conditions using logic operations and nesting if statements.
Next, you will learn about the concepts of loops, and using the turtle library to draw on screen. You will learn about creating and using for loops. You will also learn about creating and using while loops, and the differences between for and while loops. You will learn about common mistakes made when using loops.
Next, you will learn about using and creating lists to store multiple values. You will learn about how to modify and delete from your list. You will learn about searching a list and sorting it. You will learn about creating and writing to files. You will learn about closing the file after your finished using it. You will learn about opening files and reading data from them. You will learn about csv files and what they are.
Finally, you will learn about what functions are and how to create functions. You will learn about passing parameters in to your functions and why you should use functions. You will learn what types of errors can occur and how to handle errors. You will learn how to use the try/except statement to handle possible errors and get information on errors that occurred. You will learn how to save functions in a separate file and import them into your main program file as you need them.


The Fortress Hohensalzburg ist the town´s landmark; No Salzburg visitor can elude this mighty castle. The fortress Hohensalzburg is the largest preserved building of this kind all over Europe. The history of the fortress goes back until the year 1077 when archbishop Gebhard von Salzburg gave the instructions to build this strong fortress as a military stronghold.
However, the fortress was not as huge and its interior not as well-equipped in these days; the fortress was gradually enlarged within the following 400 years; Even in the 19th century, there were still amendments made at the fortress.
The fortress is so mighty that no conqueror could ever defeat it. Only in the 19th century this fortress was given over to Napoleon´s troups without conflict.
The fortress was built by an archbishop and from then on served as a residence to the archbishops and a military stronghold. For some time, it also accommodated barracks and a prison.
Being the city´s landmark, the fortress Hohensalzburg can be seen from almost every corner of Salzburg. Already upon your arrival, you see the mighty fortress at the Southern side of the town. The promenade at the river Salzach offers an especially beautiful view of the fortress. The view of the fortress from the Mirabellgarten is as magnificent.
You also have a wonderful view of fortress Hohensalzburg and the town from Salzburg´s second mountain, the Kapuzinerberg; When you walk up a few metres, you are rewarded with an unforgettable view of numerous church towers and of course the fortress.
When you wish to visit the fortress Hohensalzburg, you have two possibilities: you can either take cablecar or walk up by foot. Taking the cablecar one-way will cost you 6 €, a return ticket is 12 € - quite a price for a 15-minute walk. But the footpath can also be quite unpleasant when it rains or when it had rained . The paths leading to the fortress are quite steep and they are neither stepped nor tarred. And in bad weather, the paths can get as slippery as a mud slope.
The admission fee is 4 € for the exterior are and an additional 4 for the interior rooms, therefore a total of 8 €. When you already visited some inner areas of other castles before, you will not find anything new. Pompous rooms as the Goldene Stube or the Goldene Saal are always impressive, but in the end they are always the same; Gold, marble and precious materials as far as the eye can see.
You can obtain information about the history of the fortress in the Burgmuseum which also displays medieval weapons and equipments.
The inside of the fortress might be very fascinating for an American or Japanese tourist, but for somebody like me who has already visitied a large number of castles, the interior area of the fortress Hohensalzburg is rather standard and can be compared to the interior areas in Neuschwanstein or Schloss Herrenchiemsee.
But the fortress offers a fascinating and unbeatable view of the town; As the fortress is located on a hilltop, you can enjoy a panoramic view from nearly all spots. But the view of the Old Town is most impressive. You might get tempted to count all church towers of the town, but you will surely lose track very soon.
Looking westwards and southwards is also gigantic. While you see the medium-sized, wooded hilltops in the West, the South offers a magnificent view of alpine mountain peaks, many measuring a few thousand metres. They seem even more magnificent on a clear day with the sun behind your back. Surely then, you will understand why Salzburg and its vicinity attract so many tourist all year long.
Visiting the interior and exterior area of the fortress including ascent and descend, takes approximately 3 hours. The fortress offers numerous places for a snack and good meals in the panoramic restaurant.


I read the other day about the beauty of your holiness.
And I think I heard my spirit cry.
I heard the other day about Ezekiel's holy vision.
All that he saw in the raging cloud of fire.
Set my spirit spinning, wheels within wheels.
What is it that keeps the elders falling on their faces?
What is it that keeps the creatures crying, holy, holy holy?
What is it that keeps the cherubim,seraphim, and the angels in order?
Oh what someone turn on the light. Reveal your glory.
Out of the North, Ezekiel saw it coming.
Four living creatures in a whirlwind of a cloud.
Wheels filled with eyes. The Spirit within them.
Engulfed in a raging fire.
Coals burn a blaze. The appearance of torches.
Oh lightning flashing fire flying all around.
Tumultuous soundings like wings of an army, raging worship I gotta know.
You know we were born to bask in your beauty.
To be immersed in the feirceness of the worship.
The craving of our souls. The craving of our souls.


(KFVS) - The following closings and events cancellations for Monday, January 21 are due to weather.
The Ste. Genevieve Co. Nutrition Center is closed.
The membership meeting and exercise class at the AARP chapter 1270 in Piedmont has been canceled.
The Good Samaritan Food Pantry and the University Baptist Church Clothes closet in Carbondale is closed.
Clark and Son Salvabe in Cobden, Ill. will be closed on Monday, Jan. 21 due to weather conditions.


On Friday, July 26th, GRI will be doing an overnight investigation of the Gold Hill Hotel. We will be conducting on-scene investigations of the hotel both inside and externally near some of the old mining digs and have room for 4 or 5 investigators to join in.
This is an opportunity to join in at a classic and famous haunt site. GRI will treat the Spirits there on-site with respect and bring our traditional Whiskey offering in a shot-glass to honor those who have passed in the many mining accidents in the area and most specificially, in the 1850s accident at the Yellow Jacket mine which is mere feet away from where we will be staying.