'Srikalahasti' is one of the holy town situated on the banks of River Swarnamukhi which is about 28 kms from the city of Tirupathi , Andhra Pradesh . The word " Srikalahasti refers to Sri – Spider , Kala – Snake, Hasti – Elephant and it's believed since ages that spider , snake and elephant worshiped Lord Shiva in this divine place to attend penance (Moksha) . Many pilgrims visit this holy place when they complete Tirumala Pilgrimage to earn the blessings of SreeKalahasteeswara Swamy and Prasunambika Devi .
Srikalahasti temple is quite popular among devotees for it's Rahu – Kethu pooja . It's believed that this pooja is performed to remove obstacles in their lives . Huge number of devotees throng in to the temple premises during Rahu Kaalam and eclipse days either lunar or solar . There are special packages for performing pooja which varies from Rs. 500/- to Rs. 2500/- and it depends on the pilgrim's choice to select any package .
Rahu – Kethu pooja long last for 40 minutes and in one round let say 300 people would be performing this pooja under the supervision of temple priest . After this devotees would be asked to take the blessings of Lord Shiva & Mother Parvathi for successful pilgrimage. Under the temple premises a gigantic dining hall has been arranged for devotees to fulfill their hunger . Believe me guys this was one of the greatest place ever I visited in my lifetime and I really appreciate the whole staff , management for providing such an delicious and tasty food. Don't ever hesitate to visit such a beautiful and peaceful place . We applaud the ancient Chola King's & it's dynasty for giving us such an magnum opus architecture .


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ANZA-SIDM
An Affiliate of the Society to Improve Diagnosis in Medicine
Leading change in Australia and New Zealand to improve diagnosis and eliminate harm from diagnostic error.
The ANZA-SIDM affiliate is dedicated to leading change in Australia and New Zealand to improve diagnosis and eliminate harm from diagnostic error. Our mission is to advance the field of diagnostic safety and quality and to reduce the incidence of diagnostic error through improved collaboration and communication in the diagnostic process. Our members include patients, clinicians (doctors, nurses, paramedics, radiologists, pathologists, allied health), medical educators and researchers.
In 2017 and 2019 the Society to Improve Diagnosis in Medicine hosted the Australasian Diagnostic Error in Medicine conference. Both conferences were chaired by Dr. Carmel Crock and had more than 200 attendees. The outpouring of support in the region for improving diagnostic quality and safety led to the formation of ANZA-SIDM - the first affiliate of SIDM outside the United States.
If you are from Australia or New Zealand and would like to become part of this exciting new international effort, become a member of SIDM and you will automatically be joined to the ANZA-SIDM affiliate.
"Diagnostic errors are not just a problem in the United States. We are excited to bring focused attention to this problem throughout Australia and New Zealand. We have the active support of specialty colleges and indemnity providers and we look forward to driving research, teaching and practice improvement in diagnosis to Australasia."
Carmel Crock, MD
Chair of the Board of the new ANZA-SIDM Affiliate
ANZA-SIDM affiliate leadership were ratified by the SIDM Board of Directors in November of 2019 and will serve two-year terms. The leaders of the affiliate include:
Board Chair - Associate Professor Carmel Crock, MD, Director, Emergency Department, The Royal Victorian Eye and Ear Hospital
Vice-Chair - Associate Professor Amanda Walker, Senior Clinical Advisor, Australian Commission on Safety and Quality in Healthcare
Secretary - Professor Roger Wilson, MD, Chief Pathologist, NSW Health Pathology
Member - Associate Professor Julia Harrison, MD Emergency Physician, Senior Lecturer, Monash University
Member - Professor Jill Klein, Melbourne Business School, Professorial Fellow, Melbourne Medical School

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Q: Showing non-expansion for $x\rightarrow x+1, x\rightarrow 2x.$ Construct a graph having $V=\mathbb{Z}/p\mathbb{Z}$ as its set of vertices and $\{\{x,x+1\}: x \in V\}\cup \{\{x,2x\}: x \in V\}$ as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.
Question: what is the simplest way to show that this graph is not an expander?
An obvious strategy is to construct a set $A$ such that $|A \cup (A+1) \cup 2A| < (1+\epsilon) |A|$
(for $\epsilon$ arbitrary and $p$ large enough in terms of epsilon). How to construct a set A is less obvious.
Two possible constructions:
(a) If $p = 2^n+1$, or more generally $p = 2^n+O(1)$, then $A = $(reductions modulo $p$ of itnegers between $0$ and $p!$ with more $0$'s than $1$'s in their binary expansion) should work.
(b) For general p, J. Cilleruelo points out to me that the set $A$ constructed by Gonzalo Fiz in Proposition 3.2 of http://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s)
should give an answer, at least if 2 is replaced by 4 (or any other constant square).
Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.

A: Harald, I certainly will not claim that this is simpler, but there is a spectral approach to your question, which has been considered. Consider the subgroup $BS(1,2)$ of the affine group of the real line, generated by $a:x\mapsto 2x$ (dilation by 2) and $b:x\mapsto x+1$ (translation by 1). If $p$ is an odd prime, reducing modulo $p$ we have a homomorphism $BS(1,2)\rightarrow Aff_1(p)$ (the affine group of $Z/pZ$) and your graphs can be viewed as Schreier graphs of $BS(1,2)$. In that paper
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.em/1045952352
F. Martin and I prove that the spectrum of the adjacency matrix of the Cayley graph of $BS(1,2)$ w.r.t. $\{a^{\pm 1},b^{\pm 1}\}$is the interval $[-3,4]$, and that it is the closure of the union of the spectra of your graphs. This implies that your graphs cannot have a spectral gap.

A: Wait, this isn't that hard. Let a positive integer $\lambda\ll 1$ be given. Let
$A=[0,1/2n]\subset \mathbb{R}/\mathbb{Z}$. Let $\phi$ be the multiplication-by-$\lambda $ map; then $\phi^{-k}(A)$ is a union of $\lambda^k$ intervals with total measure $1/2n$.
We let $B$ be the union of the sets $\phi^{-k}(A)$ for $k$ going from $0$ to $n-1$; we show that there isn't too much overlap, so that $1\ll |B|\leq 1/2$. Then the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)$.
Now let $f$ be the natural homomorphism of abelian groups $f:\mathbb{Z}/p\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$. Of course, multiplication by $\lambda$ (i.e., addition $\lambda$ times) gets taken to multiplication by $\lambda$. Because $B$ is the union of $O_{\lambda,n}(1)$ intervals,

*

*we have
$p\ll_\lambda |f^{-1}(B)|\leq p/2 + O_{\lambda,n}(1)$,

*the boundary of $B$ under $x\mapsto \lambda x$ is of size $O(1/2n)+O_\lambda(1)$,

*the boundary of $B$ under $x\mapsto x+1$ is of size $O_\lambda(1)$,

and so the problem is solved.
(Note: this is very close to what Fiz does.)

A: Just noticed this.  The problem has been already solved in two ways,
but not the generalization suggested by H A Helfgott where the multiplier $2$
is replaced by an arbitrary constant (though the solution may be
implicit in Alain Valette's answer).  In fact, for any fixed $r$, $s$
and any $a_j$ ($1 \leq j \leq r$) and $m_k$ ($1 \leq k \leq s$)
the graph of degree $2(r+s)$ on ${\bf Z} / p {\bf Z}$ obtained by joining
every $x \bmod p$ to the $2(r+s)$ residues $x \pm a_j$ and $m_k^{\pm 1} x$
is not an expander as $p \rightarrow \infty$, no matter how the
addends $a_j$ and multiplicands $m_k$ are chosen.  This is basically
what I gave as the "Exercise" at the end of my accepted answer to
question
MO.125251 on the analogous graphs on finite fields of $2^n$ elements.
(The exercise concerned only $r=s=1$, but a comment noted that the
same argument applies in for any fixed $r,s$).
To spell it out: Let $A = \sum_j A_j$ and $M = \sum_k M_k$ be the
corresponding operators on ${\bf C}^p$, so that $A_j$ (respectively $M_k$)
sends the unit vector $e_x$ to $e_{x+a_j} + e_{x-a_j}$
(resp. $e_{m_k x} + e_{m_k^{-1} x}$
The all-$1$ vector ${\bf 1}$ has eigenvalue $2(r+s)$ for $A+M$.
We show that there is no spectral gap by finding a vector $v$
orthogonal to ${\bf 1}$ such that the Rayleigh quotient
$\langle v, (A+M) v \rangle / \langle v, v \rangle$ is $2(r+s) - o(1)$
as $p \rightarrow \infty$, uniformly over all choices of $a_j$ and $m_k$.
(The inner product is
$\langle v, w \rangle = \frac1p \sum_{x \in {\bf Z}/p{\bf Z}} v_x \overline w_x$.)
Fix $N$.  We'll construct $v$ such that
$\langle v, (A+M) v \rangle / \langle v, v \rangle > 2(r+s) - 2s/N - o(1)$.
For $u \in {\bf Z} / p{\bf Z}$ let $\chi_u$ be the character
$x \mapsto e^{2\pi i u x/p}$ on ${\bf Z}/p{\bf Z}$, so that the
$\chi_u$ form an orthonormal basis for ${\bf C}^p$.  Set
$$
v = \sum_c \chi_{m^c u_0}
$$
for some nonzero $u_0 \in {\bf Z} / p{\bf Z}$ to be chosen later;
here the sum extends over all integer vectors $c = (c_1,c_2,\ldots,c_s)$
with each $c_k \in [1,N]$, and $m^c := m_1^{c_1} m_2^{c_2} \cdots m_s^{c_s}$.
Then
$$
\langle v, M v \rangle \geq \frac{N-1}{N} 2s \langle v, v \rangle
$$
(the inequality may be strict if there are small multiplicative
relations among the $m_k$).  Now each $\chi_u$ is an eigenvector for $A$
with eigenvalue $2\sum_{j=1}^r \cos (2 a_j u/p)$.
We want to choose $u_0$ so that each of the $N^r s$ multiples
$a_j m^c u_0$ is $o(p)$, so that $A \chi_{m^c u_0}$ has eigenvalue
$2r-o(1)$ and $\langle v, A v \rangle = (2r-o(1)) \langle v, v \rangle$.
By a standard pigeonhole argument such $u_0$ exists that makes
each $a_j m^c u_0 / p$ within $O(p^{-1/N^r s})$ of an integer,
uniformly over choices of $a_j$ and $m_k$.
This gives the desired estimate as $p \rightarrow \infty$.
Since $N$ can be taken arbitrarily large, we are done.
The special case $r=s=1$, $m_1=2$ is a bit easier:
we can always take $u_0=1$ and obtain a bound $O(2^N/p)$
that decreases faster than $O(p^{-1/N})$.

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It's A Match: 4,000 Miles Failed To Stop Scotsman's Tinder Date
Paddy made the 4,000 mile trip to meet Bridget in Chicago on May 17.
By Anamarija Brnjarchevska May 24, 2022
A romantic Scot flew 4,000 miles to meet an American girl he matched with on Tinder after only two months of chatting - after mistakenly believing she lived in Edinburgh. (Simon Galloway/Zenger)
A romantic Scot flew 4,000 miles to meet an American girl he matched with on Tinder after only two months of chatting – after mistakenly believing she lived in Edinburgh.
Paddy Campbell, 32, only used the app for the first time when a friend downloaded it for him when he became single in January.
On March 21 he matched with Bridget Kelly, 28, who was using Tinder Passport to search for a Scottish date after becoming bored of going out with Americans.
When Bridget asked 'What's the weather like over there?' Paddy realized she wasn't in Edinburgh, which he had assumed from the tartan dress she was wearing.
The couple began chatting nonstop despite a six-hour time difference and Bridget jokingly invited him to be her date at a friend's wedding on May 21.
Paddy made the 4,000 mile trip to meet Bridget in Chicago on May 17 – and they spent a magic few days going out for dinner and breakfast, and walking around the city.
Bridget accidentally told him she loved him a few days before he got on the plane but Paddy chose not to say anything until he had met her in person, and they are now telling each other all the time.
But they have both kept the app so they can nostalgically look through their messages, which they plan to screengrab and print out.
Bridget is planning to come to Scotland for two weeks in July, while Paddy, from Alloa, Clackmannshire, will go over to America again in September.
The couple said that when it works, it works, despite the massive constraints.
A romantic Scot flew 4,000 miles to meet an American girl he matched with on Tinder after only two months of chatting – after mistakenly believing she lived in Edinburgh. (Simon Galloway/Zenger)
Paddy said: "A friend got if for me back in January, I'd no idea about Tinder but I wanted to move on.
"I got a message from Bridget and thought she was in Edinburgh, she had a tartan dress on.
"She said 'How's the weather over there?' and I looked properly and realized she was 4,000 miles away.
"We had a lot in common and carried on chatting on Snapchat.
"Bridget said she had a wedding to go to and I offered my services."
Wedding dress salesman Bridget was specifically hoping to meet a hot Scottish guy of Irish heritage – and matched with a Celtic fan.
The couple snapped selfies together in Chicago with Paddy wearing his Hoops top.
Paddy believes Scottish men are best for dating because they are direct and funny – although he said Bridget needed to be a "translator" when they went out in Wisconsin, where she lives.
He said: "We've got the best patter in the world, we say it how it is, no beating about the bush. We always speak our minds.
"Bridget has been my translator."
He often gets up at 3 a.m. to chat with Bridget before starting work at 6 a.m., and Bridget sometimes starts work at 2 p.m. which makes the time difference easier to manage.
Paddy will fly home on May 31 and they have a packed itinerary including watching the play of To Kill a Mockingbird in Chicago.
On Friday evening, May 20, Bridget introduced her parents to her new beau.
Bridget said: "They make them different in Scotland, I was just intrigued about the idea.
"I'm of Irish heritage and I was curious about what it would be like to live in Ireland or Scotland.
"I've dated a couple of Irish guys in America, I thought 'let's try Scotland'."
She added: "We were on the phone and I was saying something I really like about him and I accidentally said 'I love you'.
"We have definitely said we loved each other."
Despite their connection, both of them realized it may not click when they met in person and were open to being "best friends" if that was the case.
Paddy said: "We were respectful of that but we got lucky."
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