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More flash flooding problems in South Shields
Ocean Road suffers again.
The brief but intense thunderstorm which hit the north east of England this afternoon was not in any way an exceptional summer weather event, the storm lasted less than an hour and the volume of rain was far less than the last storm to hit us in South Tyneside. Yet once again a familiar story unfolded as parts of South Shields were dealt a miserable hand due mainly to inadequate drainage solutions. Ocean Road once more bore the brunt of the weather as shopkeepers and guest house owners reached for the sand bags and plastic sheeting in a hurried attempt to prevent damage to their flooring. A similar picture was emerging at Tyne Dock at its junction with Eldon Street.
After scores of promises of action, millions of pounds worth of investment from Northumbrian Water, a similar level of activity from South Tyneside Council, we still see scenes such as these in South Shields and other parts of the borough. People that I spoke to this afternoon were almost distraught and bitterly frustrated that their businesses could be put at risk by a simple summer rainstorm and the apparent inability of local officialdom and utility players to keep a few drains running clear. The residents know exactly which drains will block, how quickly, and can almost accurately count the minutes before the waters will subside, yet they still have to suffer.
We can count ourselves lucky that South Tyneside Council were only having three free concerts in the Bents Park as part of the Summer Festival this year, otherwise the usual final concert would have been an ordeal to say the least.
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Written by curly
Posted in Blogging, England, News, North-East, Rant, South Shields, South Tyneside, Weather
Tagged with stormy weather
How to encourage growth
Better cuts and more Keynesian spending programmes says Iain Malcolm
Following this post I posed the following questions to a number of our South Tyneside politicians of ALL parties, including the new "South Tyneside First" group of councillors, I also asked for opinions from some of our local businessmen and nationally recognised prominent bloggers and commentators:
If you were in a position to be advising the Chancellor and the Prime Minister, what would you be telling them about efforts to achieve deficit reduction and balance the budget?
How would you advise them to promote growth in the British economy to increase Treasury revenues?
What steps would you advise taking to help our local authority in South Tyneside promote the right conditions for job creation and improve consumer demand locally?
Sadly, some haven't even bothered to engage or contribute to the discussion, one decided he could only contribute if I gave him space to openly criticise officers of South Tyneside Council (which was not within the remit of the discussion because officers do not decide upon policies), another said he would contribute but failed to do so. The ONLY local politician who took the time to send some carefully considered comments was Labour's leader in South Tyneside Cllr. Iain Malcolm, who told me this:
The deficit, caused by the greed of the bankers not the last Labour Government's spending on infrastructure, needs to be reduced. However the Tory-led Government are using the deficit as an excuse for making deeply ideological cuts in public expenditure. The Government should be clawing back the unpaid tax by multi-national corporations and taking a leaf out of local government for ways of streamlining their civil service budgets. The Prime Minister has admitted that Councils are the most efficient part of the public sector, so send in the Local Government Association to review Whitehall expenditure with a aim to making drastic savings in their budgets.
But we also need growth in the economy and that means Government spending on economic infrastructure projects, I would rather pay people to work and receive their taxes than pay them for being unemployed. The Chancellor himself is starting to realise his austerity plans are not working and the deficit will need to be reduced over a longer period of time.
The greed of the banks nearly led to a total collapse of the economic system, if Gordon Brown had not stepped in (where other world leaders failed to lead) and reassured the public that their savings, pensions and homes were safe, we would have been in a far grave state than we are today. There would have been public panic (as witnessed over Northern Rock) on an unprecedented scale with possible social disorder. Whatever else one feels about our former Prime Minister remember it was his bold leadership that prevented a complete melt-down.
Finally, there needs to be greater regulation of the banking system – the light touch of the former Government (supported by the Conservatives at the time) clearly didn't work.
It's a fairly straight down the middle "follow the party line" sort of statement, but it does not show any real vision or initiatives to help the local economy in South Tyneside return to growth. there is no commitment. for example, to retain zero growth in Council Tax rises which would help maintain the little spending power that consumers have right now, and spending power is what lights the blue touch paper of economic growth.
However I'm very grateful and thank him for his participation.
In the table below are the combined thoughts and initiatives of the independents, Conservatives, UKIP, bloggers, and business people.
Not very encouraging is it?
Perhaps this is one part of the explanation as to why we have such a heavily Labour dominated council here.
This is from UKIP Cllr. David Potts
It never fails to amaze me that Labour politicians have the sheer nerve to blame the financial sector for the global economic situation. Coun Malcolm states that the crisis has been "caused by the greed of the bankers". This is a pathetic throw-away comment backed up with zero evidence, and Coun Malcolm is intelligent enough to know this. Gordon Brown sold UK Gold reserves in the $200-300/oz range. Today, it is trading at $1600/oz. The Labour Government spent money like a drunken sailor, and the mantra that they "didn't fix the roof while the sun was shining" could not be more accurate.
It really annoys me when governments blame the financial sector for operating within the regulatory boundaries that THEY create in the first place.
However, we must move away from national and global issues and focus upon what matters to us as councillors; South Tyneside. The highly respected think tank, IPPR North has warned that the South Tyneside jobless rate is set to skyrocket, particularly among our young people. It is deeply concerning that Coun Malcolm and his £170'000PA Chief Exec, Martin Swales publically comment that it is nothing to do with them and it is 'all the evil Tory Government's fault'. That is nonsense. II speak to local business people regularly. I know one energy broker who offered to save the council hundreds of thousands of pounds by switching their suppliers. His exact words were "I tried, but it was pointless, it's a closed shop". There is an active movement among the business community in South Tyneside to push ahead an agenda of progress and growth through directly influencing Council policy via peaceful protest and diplomatic pressure.
Now would councillor Malcolm like to tell us how many local firms are involved in the building of our new swimming pool complex? Would he also like to explain why suppliers of materials to South Tyneside Homes were based hundreds of miles away down South?
Look after the people of your Borough first.
Coun Malcolm and Mr Swales take people for fools and playing on this nonsense assertion that they are socialists. If they are socialists, I'm a Dutchman.
This from George Smith, President of South Shields Conservative Association.
I asked our Campaign Director (Dorothy) for a brief, but they "don't do Blogs".
Sorry about this, according to the Levenson inquiry they do Twitter, a bit too much for me.
Posted in Blogging, Conservative, David Cameron, Democracy, Economics, Employment, Gordon Brown, Independent, Labour, LibCon Coalition, North-East, politics, South Tyneside, Taxes, UKIP
Tagged with How to encourage ideas to grow in Opposition
Great North Dog Walk 2012
South Shields hosts another "record breaker!"
This is becoming a bad joke, little more than a publicity stunt, even CBBC cameras and crews were in attendance as the Great North Dog Walk organised by Tony Carlisle set about bursting the Guinness World Record once again. Am I the only one in South Tyneside who finds the figures rather less than credible? Am I the only one in South Shields asking the right questions at the wrong time?
I'd like to know just why these numbers are apparently going up and up year after year, yet the event looks only a little larger to me, and the crowds and tents have all dispersed by late afternoon. I was down on the Leas this morning to photograph the start of the Great North Dog Walk 2012 which now claims to have more dogs taking part than the first Great North Runs organised by Brendan Foster, yet I parked easily behind the Bamburgh, as far as I am aware there were no extra Metros or buses today, there were no road closures at all, and nobody to police such a huge gathering, no safety stewards or metal barriers. The start line did not stretch a mile and a half and the first participants passed by my camera in little more than twenty minutes. I'd like to know if anyone either in South Tyneside Council or close to Mr. Carlisle has any interest in trying to answer the questions that I have posed in previous years.
I guess someone thinks that the publicity factor is good to have at the start of summer, it puts South Shields on the map and makes a news story for a few outlets which lasts for a couple of days and if all of the money that is claimed to be raised is going to good local causes then hip hip hooray!
However I estimated that around 3000 dogs at most walked by in that first hour, and so to realise figures of over 24000 would need that rate of participation to be sustained until at least 6:00 pm and clearly it was not, do we really need to be accepting these claims without scrutiny?
The questions that I posed last year:
How do the organisers confirm how many dogs take part on the day?
Is there any involvement by any official from Guinness World Records Ltd on the day, and on site, to validate the "world record" claim?
How much does the event cost to organise and run?
How much sponsorship is raised and from whom?
How much money is raised for charities and which charities benefit?
Are any accounts published, and where can the public read them?
A comment left last year by Steve G:
Curly, originally, was querying whether the reportage of the Great North Dog Walk was accurate.
The claims made suggest there were over 22700 dogs on the Leas on Sunday over a three or four hour period.
This equates to 5600 dogs on the Leas at any given hour.
For the sake of argument there is one dog owner with each dog.
Curly is simply pointing out that given the amount traffic this mass attendance would create – are the claims made in the Gazette true?
Bearing in when I was jogging, a few years ago now, the figures for people running the Great North Run was around the 18000 figure – less than the dog walk. This involved road closures along the Leas, additional public transport, police attendance, stewarding etc.
The Great North Dog Walk should have been decidedly noticeable and quite frankly it was not. Not this year, not last year or previous years.
The video of last year on the Great North Dog Walk (on the website) does not appear to show the crowds that 16000 or so dogs would require last year.
If the number of dogs attending is exaggerated, then logically, so is the amount raised for charity. How can this sum be checked? It is only an estimate. Though £45 per dog per walk, since inception, seems achievable.
The Guinness Book of World Records website gives no indication of how the numbers involved are validated. In fact in the 2010 video neither the owners nor the dogs appeared to be allocated numbers as per the Great North Run. As the picture in tonight's (14th. June 2011) Gazette confirms – how are the numbers checked and entrance fees collected?
Does this event show the borough in a positive and record breaking way?
Well, it was not reported in the Sunderland Echo, nor the Evening Chronicle – I did not check the Journal. So only we know about it!!
I am not concerned over the relatively small amount this event received from the council, what I am concerned about is that there appears to be little or no accountability regarding this small and many other larger amounts distributed by our council.
Look after the pennies and the pounds will look after themselves.
Also, I am getting tired of the Gazettes inaccurate reporting or rather the reprinting of other peoples press releases. But that is another issue in itself.
I hope my pictures show that these numbers require answers to the questions, they were all shot between 10:00 and 11:00 am this morning.
A great day out for families and pooches, Deputy Mayor Ernie Gibson showed true bulldog spirit manfully and doggedly reading through his speech as the wind tried to whippet away, but you'd have to be pretty barking to believe that this is a Great North Run sized event which one day may end up spilling egg on our chins with some explaining to do!
This picture shows a small section of the crowd at the start line of the 1982 Great North Run when 20000 people competed, call me hard to please if you wish, but I'm not convinced that we saw anything like these numbers at the Great North Dog Walk.
Here is a section of the crowd at the Great North Dog walk
Double money canine cash give away
On dogs and water trends
Great North Dog Walk
Posted in Blogging, Bloopers, Charity, Events, News, North-East, South Shields, South Tyneside
Tagged with Great North Dog Walk
It's party time!
What is beneath the gloss?
So now that the Diamond Jubilee parties are over and we bask a little longer in the glow, the next party is upon us from tomorrow in the shape of a footballing feast known as the European Championships in Poland and Ukraine. Another few weeks of supermarket prodding will have us overfilled with lager, beer, and frozen pizzas, our homes and cars will be draped with the flag of St. George, the "red tops" will exhort the England football team to glories well beyond their reach (they have only ONCE reached the semi final of this tournament), the new flatscreen TVs will be overheating, and when it all ends……….well there are the Olympics and Paralympics to look forward to in London, along with commercially sponsored torch bearers. One only hopes that South Tyneside Council is NOT tempted to mount the giant screen in the park again, I fear the demand will not be there to sit and watch in a South Shields park!
Of course our politicians will probably be a little more sanguine than myself, especially having invested so heavily in the London Olympics, and I confess most of the Olympian sports hold little interest for me even though there will be some football in Newcastle, but for others this will indeed be a veritable feast of summer sports with Wimbledon and Test Match cricket thrown into the mix. David Cameron and News International will be hoping to share the glory of English and British sportsmen and women as the medal count increases the "feel good" factor and the national index of good cheer. Parliament will be enjoying its long summer break with only the Leveson inquiry interrupting the good news, yet quietly gurgling away beneath the veneer of good cheer and "gloss" will be ……………………the economy.
Let's not forget, that this is the main event!
After two years of a Conservative led coalition government the track record is not that impressive, the deficit continues to grow (albeit at a slower pace), the national debt is still woefully out of control, borrowing is still at record levels and getting higher, public expenditure is higher now than in the last years of Brown's disastrous administration, so much for the cuts eh? The slight saving grace is that unemployment has not yet broken past levels which might lead to the sort of civil unrest witnessed in Greece. The Eurozone crisis refuses to go away, banks still appear to need recapitalisng (again) and none of the major European players seem willing to take the bull by the horns and dismantle the single currency. All the while the pan European policy appears to be one of further austerity, less public spending, and increasing taxation, as near every nation in the EU other than Germany slumbers along on virtually zero growth.
Growth? Did I mention growth?
Ah yes, who has the secret elixir which will put more money into the hands of consumers or savers? It is they alone who can provide the demand and stimulus that economies need, and any person in South Tyneside who tells you that they couldn't use a few extra quid a month is madly deluded. Banks rely on savers to swell deposits, greater deposits create more capital, more capital creates greater liquidity which normally leads to easier lending and investment, yet the current economic cycle has brought us interest rates which creep closer to zero and discourage any saving at all. Similarly current personal and indirect taxation levels do nothing at all to inspire consumer confidence and help us to prioritise our spending on essentials such as mortgages and rents, fuel, and keeping our older cars on the road for even longer. It is almost a pan European deflationary cycle. Our own government is caught up in the very same frame of mind as Labour's outgoing Chancellor Alistair Darling who was equally determined to reduce the deficit whilst raising taxes and sending out hopeful signals on government expenditure. Neither Labour nor the current coalition appears to have had a "plan B" that might have included some major initiatives to promote economic growth and rekindle consumer demand.
So, sadly, whilst we are in the summer party of love mood we have to remind the politicians that "it's the economy stupid", and with that in mind I've asked some of our prominent politicians in South Tyneside, along with some business people and other bloggers for their ideas on what we need to do to get our local and national economy on the move. Of course your own ideas are very welcome too.
Posted in Blogging, Conservative, David Cameron, Economics, Employment, Energy, England, Europe, Events, Football, Gordon Brown, Labour, Labour's cuts, LibCon Coalition, Money, motoring, North-East, politics, Rant, Royalty, South Shields, South Tyneside, Sport, Taxes, Television
Former Mayor seeks end to "negative" politics.
John Anglin looking for positive solutions in South Tyneside.
South Tyneside's former Mayor Cllr. John Anglin won a handsome victory in South Shields last night when he ousted the Deputy Leader of the Independent Alliance Ahmed Khan in the Beacon and Bents ward of South Shields. Returning Labour Cllr. Anglin gave this reaction to the Corner Shop:
As far as Beacon and Bents is concerned I am delighted that the residents have allowed me to represent them once more and promise that I will concentrate on positives and positive actions that are so necessary in these gloomy times. Enough of the continually negative attitudes. Lets see what can be done to improve our Ward and our town for the wonderful people who live in it.
Posted in Blogging, Democracy, Events, Independent, Labour, News, North-East, politics, South Shields, South Tyneside
Tagged with Election reaction
Live election blog – South Tyneside
Results as they come in
Keep refreshing this page in your browser window (press f5)
Just keeping an eye on the local government elections here in South Tyneside as results are declared in South Shields and Jarrow.
First off a wrong prediction in the early declarations, Labour's Melanie Cartwright has taken Monkton from the Independent Alliance with a majority of 391 (didn't see that one coming).
Labour holds Whitburn and Marsden as predicted, Tracey Dixon romped home with more than a thousand votes to spare.
Labour holds Bede ward in Jarrow, as predicted, welcome to Margaret Peacock.
Ann Walsh holds Biddick and All Saints with an 877 majority.
Alan Smith (Labour) defeats the Independent Alliance councillor Geraldine White in Fellgate and Hedworth (another prediction comes right).
Jim Perry holds Primrose for Labour with a majority of over 1000.
John McCabe holds Hebburn South for Labour
Gladys Hobson makes a return to South Tyneside Council as Labour wins in West Park (not looking good for the Progressives).
Just after midnight and Labour has already picked up an extra three seats on South Tyneside Council.
At 45.5% Cleadon Village and East Boldon has the highest turnout in the borough again.
HUGE win for Labour in Beacon and Bents, Ahmed Khan the Independent Alliance defendant has been beaten by former Mayor John Anglin with a good workable majority of 203!
Joanne Bell holds Boldon Colliery for Labour
Pat Hay wins Harton for Labour, driving another nail into the Progressive's coffin.
Mark Walsh wins another seat for Labour, knocking out Gordin Finch (Independent Alliance) in Horsley Hill, if Westoe falls to Labour this could signal the end of the Alliance.
Westoe has gone Labour again! They must be jubilant as Sheila Stephenson records a 300 odd majority as the Independent Alliance is reduced to a solitary member. (is that right, just Jane Branley left?)
Former Conservatice Alex Donaldson (also a former Mayor) holds Cleadon Park for Labour.
Labour hold both seats in Simonside and Rekendyke
Just waiting for results in Whiteleas and Cleadon Village now, perhaps they are closer than expected?
Doreen Purvis eventually wins a seat on the Council as Labour takes Whiteleas from Independent Terry Haram
Jeff Milburn survives as the only Conservative councillor in South Tyneside holding his seat in Cleadon Village and East Boldon with a majority of only 249.
Mary Butler has taken Hebburn North for Labour to complete the evening's bruising for the opposition parties.
South Tyneside Council now comprises 48 Labour members (a huge increase of nine seats) and a paltry six seats for the opposition, the real opposition must now come from within the Labour Party. Looks like my predictions yesterday were not far off the mark.
Biggest sweat must have been on the brow of Alex Donaldson who saw his majority reduced to only 21 as June Elsom came very close to unseating him, in total the Labour Party has won 18 of the 19 seats contested, a pretty formidable victory by any standards!
Conservative Jeff Milburn was unable to attend the count after being badly bitten by a dog during election day, he spent the evening being treated at the hospital.
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Posted in Blogging, Democracy, Events, politics, Poll, South Shields, South Tyneside
Tagged with Local Election results
Back at last!
So, just what have I been up to?
Well, TWO new computers, both in the same case!
It has taken a while, but the hardware on the old Windows XP machine was deteriorating and could only have bits and pieces replaced so often, the operating system was outdated and not capable of keeping pace with the latest developments. So now I have a dual boot machine, i.e. when I switch it on it gives me a choice, do I wish to fire up Windows 7 Ultimate, or Linux Ubuntu 12? I've used the Windows machine to set up my home wireless networking solutions streaming content out and sharing files and folders between this desktop, a laptop, a Blackberry, a PS3, and a Nintedo, everyone is happy on that score! I also use it to edit my photographs, because Photoshop is simply the best. However, what has kept me away from here for so long is my determination to learn how to do things in Linux, how to code, work in the command line and "Terminal", it is challenging and fun, and seriously I cannot get enough of it.
I wanted to rise to the challenge of running my home computing on a machine using only software that is commonly shared and legitimately free, and as you can see from the picture of my two different desktops above, I have even managed to get the thing to look and behave just like a megabucks Mac!
I am determined to learn how to use the Gimp image editing software, it may never be as good as Photoshop, but I'm pretty sure I can make it produce good pictures, the example above shows that I have a long way to go.
So I'm ready to be back in the saddle to ride this vehicle wherever we want to go, political interests have had to take a back seat for a few weeks, the hunger for knowledge has been winning the battle.
I'll start tomorrow with a little piece on the local elections here in South Tyneside to get the ball rolling again – remember to go and use your vote, you really have no right at all to complain if you fail to mark a cross.
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text:
\section{Introduction}\label{intro}

Suppose that $G=\{g_1,\ldots ,g_n\}$ is a finite group.  Assigning variables $x_g$, $g\in G$, one defines the
{\em group determinant,} $\mathscr{D}_G(x_{g_1},\ldots ,x_{g_{n}})$, to be  the determinant of the $n\times n$ group matrix whose $ij$th entry is  $x_{g_i g_j^{-1}}$. Plainly $\mathscr{D}_G(x_{g_1},\ldots ,x_{g_{n}})$ will be a homogeneous polynomial of degree $n$ in the $x_g$.

 For $G=\mathbb Z_n$, the cyclic group of order $n$,  this is a circulant determinant, where the next row in the group matrix is obtained from the previous row by a cyclic shift of one to the right.
An old problem of Olga Taussky-Todd is to determine which  integers can be achieved as an $n\times n$ circulant determinant when the entries in the matrix are integers. We can of course ask the same question for an arbitrary finite  group $G$ and
define $\mathcal S(G)$ to be this set of integers:
$$ \mathcal S (G) :=\{ \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})  \, : \, x_{g_1},\ldots ,x_{g_n}\in \mathbb Z\}. $$
Notice that $\mathcal S(G)$ is closed under multiplication
$$ \mathscr{D}_G(a_{g_1},\ldots ,a_{g_n})\mathscr{D}_G(b_{g_1},\ldots ,b_{g_n})=\mathscr{D}_G(c_{g_1},\ldots ,c_{g_n}), \;\;  \; c_{g}:=\sum_{\substack{u,v\in G\\ uv=g}} a_ub_v,$$
corresponding to multiplication $\left(\sum_{g\in G} a_g g\right)\left(\sum_{g\in G} b_g g\right) =\sum_{g\in G} c_g g$ in the group ring.
While it has been proved \cite{Formanek} that the group determinant polynomial determines the group it is not known whether the set of integer values $\mathcal S (G)$ determines the group.

For $G=\mathbb Z_n$  Newman \cite{Newman1} and Laquer\cite{Laquer} gave divisibility restrictions on the elements in $\mathcal S (G)$ and some  attainable values. For example, $n^2$ and any $m$ with $\gcd(m,n)=1$
will be in $\mathcal S(\mathbb Z_n)$, and if $m$ is in $\mathcal S(\mathbb Z_n)$
then so is $-m$ and if $p\mid m$ and $p^{\alpha}\parallel n$ then $p^{\alpha+1}\mid m$. Here and throughout $p$ denotes a prime. Newman and Laquer both  obtained a complete description for  $G=\mathbb Z_p$  when $p\geq 2$, and Laquer for $G=\mathbb Z_{2p}$ when $p\geq 3$:
\begin{align*}
\mathcal S (\mathbb Z_p) & = \{ p^a m\, : \, \gcd(m,p)=1,\;\; a=0 \hbox{ or } a\geq 2\},\\
\mathcal S (\mathbb Z_{2p}) & = \{ 2^a p^b m\, : \, \gcd(m,2p)=1,\;\; a =0 \hbox{ or } a\geq 2,\;\; b =0 \hbox{ or } b\geq 2\},
\end{align*}
cases where the divisibility conditions are necessary and sufficient.
Newman \cite{Newman2} similarly showed that
$$\mathcal S(\mathbb Z_{9})=\{3^am\, : \, \gcd(m,3)=1,\; a=0 \hbox{ or } a\geq 3\},$$
but that $p^3$ is not in $\mathcal S(\mathbb Z_{p^2})$ for any  $p\geq 5$ (in particular the basic divisibility
conditions are not sufficient). Currently no $\mathcal S(\mathbb Z_{p^2})$ with $p\geq 5$  has  been fully determined, though there are upper and lower set inclusions.

As shown by Vipismakul \cite{Cid1} there is a very close  relationship between the group determinant for a finite abelian group and Lind's generalization  \cite{Lind} of the Mahler measure  for that group; in particular the corrresponding  Lind-Lehmer problem of  finding the minimal positive logarithmic
Lind-Mahler  measure for the group,   corresponds to finding the smallest non-trivial group determinant
$$ \lambda(G) :=\min \{ |m| \, : \, m\in \mathcal S (G), \; |m|\geq 2\}. $$
Kaiblinger \cite{Norbert2} used the Lind measure approach to find
$$\mathcal S(\mathbb Z_4)=\{ 2m+1,\;\; 2^4m\, : \, m\in \mathbb Z\}, $$
and $\mathcal S(\mathbb Z_8),$ and obtain
upper and lower inclusions for other $\mathcal S(\mathbb Z_{2^k})$. The Lind-Lehmer constant  $\lambda (G)$ is known for a number of groups, see for example   \cite{Norbert,Pigno1,pgroups,Cid2}, including the cyclic groups $G=\mathbb Z_n$ with $892371480\nmid n$.
Writing $D_{2n}$ for the dihedral group of order $2n$,  the sets
$\mathcal S(D_{2p})$, $\mathcal S(D_{4p}),$
$$\mathcal S(D_{8})=\{4m+1,\;\; 2^8m \, : \, m\in \mathbb Z\}, $$
and $\mathcal S(D_{16})$ were determined in \cite{dihedral}, and $\lambda(D_{2n})$ found for $n< 1.89\times 10^{47}$.
In particular the cases $p=2$ and $3$ give us  the Klein $4$-group
$$ \mathcal S(\mathbb Z_2\times \mathbb Z_2)=\{4m+1,\; \; 2^4(2m+1),\; \; 2^6m\, : \, m\in \mathbb Z\}, $$
and the symmetric group of degree $3$
$$   \mathcal S(S_3)= \{  2^a3^bm \, : \, \gcd(m,6)=1,\; a=0 \hbox{ or } a\geq 2,\; b=0 \hbox{ or } b\geq 3    \}. $$

In \cite{smallgps} a complete description of the integer group determinants was obtained for the remaining groups of order at most $14$, including   the alternating group $A_4$:
$$\mathcal S(A_4)=\mathcal S(A_4)_{odd} \cup \mathcal S(A_4)_{even},$$ with
\begin{align*} \mathcal S(A_4)_{odd} & =\{m\equiv 1 \text{ mod } 4 \, : \, 3\nmid m \hbox{ or } 3^2\mid m\},\\
\mathcal S(A_4)_{even} & =  \{ 2^a3^bm\, : \, \gcd(m,6)=1,\; a=4 \hbox{ or } a\geq 8, \; b=0 \hbox{ or } b\geq 2\}. \end{align*}
Although $\mathcal S(S_3)$ and $\mathcal S(A_4)$ were obtained without too much difficulty, other small groups  in \cite{smallgps}, for example $G= \mathbb Z_2\times \mathbb Z_6$, were considerably more complicated; making it clear  that obtaining a description for  general  $\mathcal S (G)$ is probably not feasible. Indeed, even in the case of circulant determinants,  we are yet to obtain a complete description   of $\mathcal S (\mathbb Z_{15})$ and $\mathcal S(\mathbb Z_{16})$. Our goal here is to show that $G=S_4$ is one of those rare cases where we can completely  determine $\mathcal S(G)$:

\begin{theorem}\label{main}
For $G=S_4$ the odd group  determinants in $\mathcal S(G)$  are the integers $m=1$ mod $4$ with the property that $3\nmid m$ or $3^3\mid m$.

The even determinants are the  integers of the form
$$ 2^8m, \;\; m\equiv 1 \text{ mod }4,\;\;\;  \hbox{ or }\; \;\;2^{10}m, \;\; m\equiv -1 \text{ mod }4,\;\;\; \hbox{ or }\;\;\;  2^{12}m ,\;\;  $$
where $m$ is an integer with $3\nmid m$ or $3^3\mid m$.

\end{theorem}
Comparing  $\mathcal S(S_4)$ to $\mathcal S(G)$ for $G=\mathbb Z_2,\mathbb Z_3, \mathbb Z_4,\mathbb Z_2\times \mathbb Z_2,D_8,S_3,A_4$
it is  tempting to ask whether in general $\mathcal S(G)\subseteq \mathcal S(H)$ whenever $H\leq G$.

\section{Factoring the group determinant for  $S_4$}

Dedekind \cite{Dedekind} observed that for a finite abelian group the group characters $\hat{G}$ couuld be used to factor
the group determinant into linear factors
\begin{equation} \label{abelian}  \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n}\}=\prod_{\chi \in \hat{G}} \left(\chi(g_1) x_{g_1}+\cdots +\chi(g_n)x_{g_n}\right). \end{equation}
For non-abelian groups the group determinant will contain non-linear factors and, as  shown by Frobenius \cite{Frobenius}, the
counterpart to \eqref{abelian} is to  use  the set $\hat{G}$ of irreducible  group representations for $G$,
$$  \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n}\}=\prod_{\rho \in \hat{G}}\det  \left(x_{g_1}\rho(g_1)+\cdots +x_{g_n}\rho(g_n)\right)^{\deg(\rho)},    $$
see Conrad \cite{Conrad} for a historical survey.

The irreducible representations for $S_4$ are discussed in Serre \cite[$\S$5.8]{Serre}. One can also use GAP or a similar computer algebra system to generate them  (although we have reduced the four generators used  there to two and also reindexed).

We take as our two generators for $S_4$
$$ \alpha=(1234),\;\;\;\beta=(12), $$
and order the even permutations, with the coefficients $a_1,\ldots a_{12}$, by
\begin{align*} & \{ 1,(13)(24),(14)(23),(12)(34),(134),(243),(142),(123),(143),(132),(124),(234)\}\\
 & =\{1,\alpha^2,\beta\alpha^2\beta,\alpha^2\beta\alpha^2\beta,\alpha\beta,\alpha^3\beta,\alpha\beta\alpha^2,\alpha^3\beta\alpha^2,\beta\alpha^3,\alpha^2\beta\alpha,\alpha^2\beta\alpha^3,\beta\alpha\}\end{align*}
and the odd permutations, with coefficients $b_1,\ldots ,b_{12},$ by
\begin{align*} & \{ (1234),(1432),(24),(13),(14),(23),(1243),(1342),(12),(34),(1324),(1423)\}\\
 & = \{ \alpha,\alpha^3,\alpha\beta\alpha^2\beta,\beta\alpha^2\beta\alpha,\alpha^3\beta\alpha,\alpha\beta\alpha^3,\alpha\beta\alpha,\alpha^3\beta\alpha^3,\beta,\alpha^2\beta\alpha^2,\beta\alpha^2,\alpha^2\beta\}.
\end{align*}
For $S_4$ we have two linear representations, $\chi_0(x)=1$ and $\chi_1(x)=\hbox{sgn}(x)$, giving two linear factors
\begin{align*}
\ell_1:=(a_1+\cdots + a_{12})+(b_1+\cdots +b_{12}),\\
\ell_2:=(a_1+\cdots + a_{12})-(b_1+\cdots +b_{12}).
\end{align*}
We have one degree two representation with
$$ \rho_1(\beta)=\left(\begin{matrix} 0 & \omega^2 \\ \omega & 0\end{matrix}\right),\;\;\;\rho_1(\alpha)=\left(\begin{matrix} 0 & 1 \\ 1 & 0\end{matrix}\right),\;\;\;    \omega := e^{2\pi i/3}, $$
giving
\begin{align*}  \rho_1(1)  =\rho_1(\alpha^2)=\rho_1(\beta\alpha^2\beta)=\rho_1(\alpha^2\beta\alpha^2\beta) &=\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right), \\
\rho_1(\alpha\beta) = \rho_1(\alpha^3\beta)=\rho_1(\alpha\beta\alpha^2)=\rho_1(\alpha^3\beta\alpha^2)&=\left(\begin{matrix} \omega & 0 \\ 0 & \omega^2 \end{matrix}\right), \\
\rho_1(\beta\alpha^3) =\rho_1(\alpha^2\beta\alpha)=\rho_1(\alpha^2\beta\alpha^3)=\rho_1(\beta\alpha)&=\left(\begin{matrix} \omega^2 & 0 \\ 0 & \omega \end{matrix}\right), \\
 \rho_1(\alpha)=\rho_1(\alpha^3)=\rho_1(\alpha\beta\alpha^2\beta)=\rho_1(\beta\alpha^2\beta\alpha)&=\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right), \\
\rho_1(\alpha^3\beta\alpha)=\rho_1(\alpha\beta\alpha^3)=\rho_1(\alpha\beta\alpha)=\rho_1(\alpha^3\beta\alpha^3)&=\left(\begin{matrix} 0 & \omega \\ \omega^2 & 0 \end{matrix}\right), \\
\rho_1(\beta)=\rho_1(\alpha^2\beta\alpha^2)=\rho_1(\beta\alpha^2)=\rho_1(\alpha^2\beta)&=\left(\begin{matrix} 0 & \omega^2 \\ \omega & 0 \end{matrix}\right),
\end{align*}
and, writing
\begin{align*}
u_1 & =a_1+a_2+a_3+a_4,\;\;\; u_2=a_5+a_6+a_7+a_8,\;\;\; u_3=a_9+a_{10}+a_{11}+a_{12},\\
v_1 & =b_1+b_2+b_3+b_4,\;\;\; v_2=b_5+b_6+b_7+b_8,\;\;\; v_3=b_9+b_{10}+b_{11}+b_{12},
\end{align*}
the quadratic factor
\begin{align*}  q_1:& = \det \left(\begin{matrix} u_1+u_2\;\omega+u_3\; \omega^2 & v_1+v_2\;\omega +v_3\;\omega^2 \\ v_1+v_2\;\omega^2+v_3\;\omega & u_1+u_2\;\omega^2+u_3\;\omega \end{matrix} \right) \\
 & =N(u_1+u_2\; \omega +u_3\; \omega^2) - N(v_1+v_2\; \omega +v_3\; \omega^2),
\end{align*}
 where $N(x)=|x|^2$.

Finally we have two degree three representations
$$ \rho_2(\alpha)=  \left( \begin{matrix}   0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right),\;\;\;    \rho_2(\beta)= \left( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0    \end{matrix} \right), $$
and $\rho_3(x)=\hbox{sgn}(x)\rho_2(x)$. Here $\rho_2$ comes from the natural representation of $S_4$ in $\mathbb C^3$. That is, we take the $3$-dimensional subspace
$$V=\{x_1e_1+x_2e_2+x_3e_3+x_4e_4\, : \, x_1+x_2+x_3+x_4=0\} $$
of a  $4$-dimensional vector space and let $S_4$ permute its  basis vectors $e_1,e_2,e_3, e_4$, although to produce  our  $\rho_2$  we take a  less obvious  basis for $V$:
$$ e_1+e_2-e_3-e_4,\;\; -e_1+e_2+e_3-e_4,\;\; e_1-e_2+e_3-e_4. $$

 For the even permutations we have
\begin{align*}
  \rho_2(1) &=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right), &
\rho_2((13)(24))&=\left(\begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix}\right), &
\rho_2((14)(23))&=\left(\begin{matrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}\right),\\
 \rho_2((12)(34))& =\left(\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{matrix}\right), &
\rho_2((134)) &=\left(\begin{matrix} 0 & 0 & -1 \\ 1 & 0  & 0 \\ 0 & -1 & 0 \end{matrix}\right), &
\rho_2((243)) &=\left(\begin{matrix} 0 & 0 & 1 \\ -1 & 0  & 0 \\ 0 & -1 & 0 \end{matrix}\right),\\
\rho_2((142))& =\left(\begin{matrix} 0 & 0 & -1 \\ -1 & 0  & 0 \\ 0 & 1 & 0 \end{matrix}\right),&
\rho_2((123)) &=\left(\begin{matrix} 0 & 0 & 1 \\ 1 & 0  & 0 \\ 0 & 1 & 0 \end{matrix}\right), &
\rho_2((143)) &=\left(\begin{matrix} 0  & 1 & 0 \\ 0  & 0 & -1 \\ -1 & 0 & 0 \end{matrix}\right),\\
 \rho_2((132)) &=\left(\begin{matrix} 0  & 1 & 0 \\ 0  & 0 & 1 \\ 1 & 0 & 0 \end{matrix}\right),&
\rho_2((124)) &=\left(\begin{matrix} 0  & -1 & 0 \\ 0  & 0 & 1 \\ -1 & 0 & 0 \end{matrix}\right),&
\rho_2((234)) &=\left(\begin{matrix} 0  & -1 & 0 \\ 0  & 0 & -1 \\ 1 & 0 & 0 \end{matrix}\right),
\end{align*}
and for the odd permutations
\begin{align*}
\rho_2((1234))&=\left( \begin{matrix}   0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right), &
\rho_2((1432))&=\left( \begin{matrix}   0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right), &
\rho_2((24))&=\left( \begin{matrix}   0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right), \\
\rho_2((13))&=\left( \begin{matrix}   0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right), &
\rho_2((14))&=\left( \begin{matrix}   0 & 0 & -1 \\  0 & 1 & 0  \\ -1 & 0 & 0  \end{matrix} \right), &
\rho_2((23))&=\left( \begin{matrix}   0 & 0 & 1 \\  0 & 1 & 0  \\ 1 & 0 & 0  \end{matrix} \right), \\
\rho_2((1243)) &=\left( \begin{matrix}   0 & 0 & 1 \\  0 & -1 & 0  \\ -1 & 0 & 0  \end{matrix} \right),&
\rho_2((1342)) &=\left( \begin{matrix}   0 & 0 & -1 \\  0 & -1 & 0  \\ 1 & 0 & 0  \end{matrix} \right),&
\rho_2((12)) &=\left( \begin{matrix}  1 &  0 & 0  \\  0 & 0 & 1  \\  0 & 1 & 0   \end{matrix} \right),\\
\rho_2((34)) &=\left( \begin{matrix}  1 &  0 & 0  \\  0 & 0 & -1  \\  0 & -1 & 0   \end{matrix} \right), &
\rho_2((1324)) &=\left( \begin{matrix}  -1 &  0 & 0  \\  0 & 0 & 1  \\  0 & -1 & 0   \end{matrix} \right), &
\rho_2((1423)) &=\left( \begin{matrix}  -1 &  0 & 0  \\  0 & 0 & -1  \\  0 & 1 & 0   \end{matrix} \right).
\end{align*}

Thus we have two cubic factors
$$ d_1:=\det \left( A+B\right),\;\;\; d_2:=\det\left(A-B\right), $$
where
$$ A=\left(\begin{matrix}    a_1-a_2-a_3 +a_4 & a_9+a_{10}-a_{11}-a_{12}  & -a_5   +a_6-a_7+a_8    \\  a_5-a_6-a_7+a_8     &  a_1-a_2+a_3-a_4 & -a_9+a_{10}+a_{11}-a_{12}  \\  -a_9+a_{10}-a_{11}+a_{12}   & -a_5-a_6+a_7+a_8  &  a_1+a_2-a_3-a_4        \end{matrix}\right) $$
and
$$ B= \left( \begin{matrix}   b_9+b_{10}-b_{11}-b_{12}  & -b_1+b_2-b_3+b_4 &  -b_5+b_6+b_7-b_8\\ b_1-b_2-b_3+b_4   & b_5+b_6-b_7-b_8 & b_9-b_{10}+b_{11}-b_{12} \\ -b_5+b_6-b_7+b_8 & b_{9}-b_{10}-b_{11}+b_{12} & -b_1-b_2+b_3+b_4  \end{matrix} \right)    .$$
So for $G=S_4$  the group determinant takes the form
$$ \mathscr{D}_{G}\left(a_1,\ldots ,a_{12},b_{1},\ldots ,b_{12}\right) =\ell_1 \ell_2 \;q_1^2 \;d_1^3d_2^3. $$

\begin{comment}
\begin{theorem}
For $G=S_4$ the odd group determinants are the integers $m=1$ mod $4$ with the property that $3\nmid m$ or $3^3\mid m$.

The even determinants are the  integers of the form
$$ 2^8m, \;\; m\equiv 1 \text{ mod }4,\;\;\;  \hbox{ or }\; \;\;2^{10}m, \;\; m\equiv -1 \text{ mod }4,\;\;\; \hbox{ or }\;\;\;  2^{12}m ,\;\;  $$
where $m$ is an integer with $3\nmid m$ or $3^3\mid m$.

\end{theorem}

\end{comment}

We can think of the determinant as the unormalized Lind  measure of the `polynomial' (really an element in the group ring $\mathbb Z [S_4]$)
\begin{align*} a_1 & +a_2x^2+a_3yx^2y+a_4x^2yx^2y+a_5xy+a_6x^3y +a_7xyx^2+a_8x^3yx^2 \\ & +a_9yx^3+a_{10}x^2yx+a_{11}x^2yx^3+a_{12}yx + b_1x+b_2x^3+b_3 xyx^2y + b_4yx^2yx \\ & + b_5x^3yx + b_6xyx^3+b_7xyx + b_8 x^3yx^3+ b_9y+b_{10}x^2yx^2+b_{11}yx^2+b_{12}x^2y,
\end{align*}
where  monomials do not commute but we can reduce a polynomial in $\mathbb Z[x,y]$ to this form using the group relations $y^2=1$, $x^4=1$, $yxy=x^3yx^3$, $yx^3y=xyx$, $yx^2yx^2=x^2yx^2y$, $x^3yx^2y=yx^2yx,$ $x^2yx^2yx=xyx^2y$ etc.

\section{Proof of Theorem \ref{main}}
\begin{proof}
We first show that we can achieve the stated values.

We begin with the values coprime to $3$ that are $1$ mod $4$.

Taking $a_1=1+k$ and the remaining values equal to $k$ gives
$$\mathscr{D}_G=(1+24k)\cdot 1 \cdot (1-0)^2\cdot |I_3|^3\cdot |I_3|^3=1+24k. $$

Taking $a_2=a_5=a_9=1+k$, $b_3=b_5=1+k$ and the remaining values $k$ we get
$$ \mathscr{D}_G=(5+24k)\cdot 1 \cdot (0-1)^2\cdot \left| \begin{matrix} -1 & 0 & -2 \\ 0 & 0 & -1 \\ -2 & -1 & 2 \end{matrix}\right|^3  \cdot   \left|\begin{matrix} -1 & 2 & 0 \\ 2 & -2 & -1 \\ 0 & -1 & 0 \end{matrix}\right|^3=5+24k. $$

With $a_1=a_3=1+k$, $a_5=a_6=a_7=1+k$, $a_9=a_{10}=1+k$, and $b_1=b_3=1+k$, $b_5=b_6=1+k$, $b_{11}=b_{12}=1+k$ and the other values $k$ we obtain
$$\mathscr{D}_G=(13+24k)\cdot 1 \cdot (1-0)^2 \cdot \left| \begin{matrix}   -2 & 0 & -1 \\ -1 & 4 & 0 \\ 0 & -1 & 0 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   2 & 4 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{matrix} \right|^3 =13+24k. $$

From $a_1=a_2=a_3=1+k$, $a_6=a_7=a_8=1+k$, $a_9=a_{10}=a_{11}=1+k$ and $b_1=b_3=b_4=1+k$, $b_5=b_7=b_8=1+k$, $b_9=b_{10}=1+k$, the others $k,$ we get
$$\mathscr{D}_G=(17+24k)\cdot 1 \cdot (0-1)^2 \cdot \left| \begin{matrix}   1 & 0 & 0 \\ 0 & 0 & 1 \\ -2 & 1 & 2 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   -3 & 2 & 2 \\ -2 & 2 & 1 \\ 0 & 1 & 0 \end{matrix} \right|^3 =17+24k. $$

Next we  get  the powers of three, $\pm 3^j$ with $j\geq 3$, that are $1$ mod $4$.

With $a_1=a_3=1+k$, $b_3=1+k$ and the others $k$ we have
$$\mathscr{D}_G=(3+24k)\cdot 1 \cdot (4-1)^2 \cdot \left| \begin{matrix}   0 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   0 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & -1 \end{matrix} \right|^3 =-3^3(1+8k), $$
and with $a_1=a_2=1+k$ and $a_5=1+k$ and the other values $k$
$$\mathscr{D}_G=(3+24k)\cdot 3 \cdot (3-0)^2 \cdot \left| \begin{matrix}   0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & -1 & 2 \end{matrix} \right|^6 =3^4(1+8k). $$

Finally we deal with the powers of two.

Taking $a_1$, $a_5=1$ and the others zero  gives
$$\mathscr{D}_G=2\cdot 2 \cdot (1-0)^2 \cdot \left| \begin{matrix}   1 & 0 & -1 \\ 1 & 1 & 0 \\ 0& -1 & 1 \end{matrix} \right|^6 =2^8, $$
while $a_1=-1$, $a_5=a_6=1$ and $b_5=1$, $b_{10}=-1$ with the others zero  has
$$\mathscr{D}_G=1\cdot 1 \cdot (7-3)^2 \cdot \left| \begin{matrix}   -2 & 0 & -1 \\ 0 & 0 & 1 \\ -1 & -1 & -1 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   0 & 0 & 1 \\ 0 & -2 & -1 \\ 1 & -3 & -1 \end{matrix} \right|^3 =-2^{10}, $$
and $a_2=1$,$a_5=1$,$a_9=1$ and $b_{11}=1$ the rest zero
$$\mathscr{D}_G=4\cdot 2 \cdot (0-1)^2 \cdot \left| \begin{matrix}   -2 & 1 & -1 \\ 1 & -1 & 0 \\ -1 & -2 & 1 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   0 & 1 & -1 \\ 1 & -1 & -2 \\ -1 & 0 & 1 \end{matrix} \right|^3 =2^{12}, $$
and $a_1=a_2=1$, $a_5=-1$, $a_9=-1$, $b_1=1$ and the rest zero
$$\mathscr{D}_G=1\cdot -1  \cdot (9-1)^2 \cdot \left| \begin{matrix}  0 & -2 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   0 & 0 & 1 \\ -2 & 0 & 1 \\ 1 & 1 & 3 \end{matrix} \right|^3 =-2^{12}.$$
Taking $a_1=a_2=1+k$, $a_6=1+k$, $a_{10}=a_{11}=1+k$, $b_4=1+k$,  $b_6=1+k$, $b_{10}=1+k$
and the rest $k$
$$\mathscr{D}_G=(8+24k)\cdot 2  \cdot (1-0)^2 \cdot \left| \begin{matrix}  1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   -1 & -1 & 0 \\ -2 & -1 & 3 \\ -1 & 0 & 1 \end{matrix} \right|^3 =2^{13}(1+3k),$$
and $a_2=a_3=a_4=1+k$, $a_5=1+k$,  $a_9=1+k$, $b_4=1+k$, $b_5=b_6=1+k$,  and the rest $k$
$$\mathscr{D}_G=(8+24k)\cdot 2  \cdot (4-3)^2 \cdot \left| \begin{matrix}  -1 & 2 & -1 \\ 2 & 1 & -1 \\ -1 & -1 & 0 \end{matrix} \right|^3 \cdot    \left| \begin{matrix}   -1 & 0 & -1 \\ 0 & -3 & -1 \\ -1 & -1 & -2 \end{matrix} \right|^3 =-2^{13}(1+3k),$$
giving  the remaining powers of two. Products of these achieve the stated values.

It remains to show that the determinants can only take the values claimed.  Plainly we have the congruences
\begin{equation} \label{basic}   \ell_1\equiv \ell_2 \text{ mod } 2,\;\;\;  d_1\equiv d_2 \text{ mod } 2,\;\;\;   q_1\equiv \ell_1\ell_2 \text{ mod } 3, \end{equation}
and, as can be checked on Maple,
\begin{equation} \label{productmod4} d_1d_2 \equiv \ell_1\ell_2 q_1^2 \text{ mod } 4 \end{equation}
and
\begin{equation} \label{summod4}   d_1+d_2 \equiv (\ell_1+\ell_2 )q_1 \text{ mod } 4. \end{equation}

Suppose that $3\mid \mathscr{D}_G=\ell_1\ell_2\; q_1^2\; (d_1d_2)^3.$ If $3\mid d_1d_2$ then  $3^3\mid (d_1d_2)^3$. If  $3\mid \ell_1\ell_2$  or $3\mid q_1$
then  $3$ divides both by \eqref{basic} and $3^3\mid \ell_1\ell_2 q_1^2$. Hence $3\nmid \mathscr{D}_G$
or $3^3\mid \mathscr{D}_G$.

If $\mathscr{D}_G$ is odd then by \eqref{productmod4} we have $\mathscr{D}_G\equiv (\ell_1\ell_2 q_1^2)^4\equiv 1$ mod $4$.

Suppose that $\mathscr{D}_G$ is even. From \eqref{productmod4} we have $2\mid d_1d_2$ and $2\mid \ell_1\ell_2q_1$.
Hence by \eqref{basic} we have $2\mid d_1$ and $2\mid d_2,$ and either $2\mid q_1$ or $2\mid \ell_1$ and $2\mid \ell_2$.
Hence $2^2\mid (\ell_1\ell_2)q_1^2$, $2^6\mid  (d_1d_2)^3$ and $2^8\mid \mathscr{D}_G$. It remains to rule out $2^9\parallel \mathscr{D}_G$ and $2^{11}\parallel \mathscr{D}_G$, and show that the $2^8\parallel \mathscr{D}_G$ or $2^{10}\parallel \mathscr{D}_G$
are of the stated forms.

As can be checked on Maple, expanding the determinant $d_1$ gives
\begin{equation} \label{d1mod4}  d_1=\ell_1 \left( q_1 +2uv+ 2w\right)+4C(\vec{a},\vec{b}) \end{equation}
where $C(\vec{a},\vec{b})$ is a homogeneous cubic integer  polynomial  in the $a_1,\ldots ,a_{12}$,
$b_{1},\ldots ,b_{12}$,
$$ u:=u_1+u_2+u_3, \;\;\; v:=v_1+v_2+v_3, $$
and
$$ w:=u_1B_1+u_2B_2+u_3B_3+v_1A_1+v_2A_2+v_3A_3, $$
where
\begin{align*}
A_1:&= a_1+a_2+a_5+a_8+a_{9}+a_{10}, & B_1 & = b_1+b_2+b_7+b_8+b_{11}+b_{12}, \\
A_2 :&=a_1+a_3+a_6+a_8+a_{10}+a_{12}, &  B_2&=b_2+b_3+b_5+b_8+b_{10}+b_{11}, \\
A_3:&= a_1+a_4+a_7+a_8+a_{10}+a_{11}, & B_3& =b_1+b_3+b_5+b_7+b_{10}+b_{12}.\end{align*}
Similarly, replacing the the $b_i$ by $-b_i$
$$ d_2=\ell_2 \left( q_1 -2uv-2w\right)+4C(\vec{a},-\vec{b}) $$
and we get
\begin{align}
d_1+d_2&  \equiv 2uq_1   +4uv^2 +4vw \text{ mod } 8,\label{mod8} \\
d_1-d_2 & \equiv 2v q_1 +4u^2 v  +4uw \text{ mod } 8. \label{mod8minus}
\end{align}

Suppose that $2^2\mid \ell_1$  or $2\mid \ell_1$ and $2\mid q_1$, then from \eqref{d1mod4} we have $d_1\equiv 0$ mod $4$ and
and, since  $2\mid \ell_2,d_2$, we get $2^{12}\mid \mathscr{D}_G$.
Similarly if $2^2\mid \ell_2$ using $d_2$.
Hence if $2^{12}\nmid \mathscr{D}_G$ we can assume that $2\nmid \ell_1\ell_2$ and $2\mid q_1$ or $2\parallel \ell_1,\ell_2$ and $2\nmid q_1$.  If $2\parallel \ell_1,\ell_2$ then $4\mid (\ell_1+\ell_2)$
and if $2\nmid \ell_1\ell_2$ then $2\mid q_1$  and $2\mid (\ell_1+\ell_2)$. In either case \eqref{summod4} gives
$d_1\equiv d_2$ mod $4$, and  if $2^2\mid d_1$ or $d_2$ then $2^2$ divides both and $2^{14}\mid \mathscr{D}_G$. Hence we have $2\parallel d_1,d_2$ and so an even number of 2's divide $(d_1d_2)^2$, $q_1^2$ and $\ell_1\ell_2$, ruling out $2^9$ or $2^{11}\parallel \mathscr{D}_G$.

Suppose that $2^8$ or $2^{10}\parallel \mathscr{D}_G$. So $d_1=2\delta_1$, $d_2=2\delta_2$ with $\delta_1,\delta_2$ odd, and for $2^{10}\mid \mathscr{D}_G$ we must have $2\nmid \ell_1\ell_2$ and $2^2\parallel q_1$, and for  $2^8\parallel \mathscr{D}_G$ we have either $2\nmid \ell_1,\ell_2$ and $2\parallel q_1$, or $2\parallel \ell_1,\ell_2$ and $2\nmid q_1$. Suppose that $2\nmid \ell_1,\ell_2$. If $\ell_1\ell_2=u^2-v^2\equiv 1$  mod $4$  then $2\nmid u$, $2\mid v$ and from \eqref{mod8}
we have $\delta_1+\delta_2 \equiv uq_1$ mod $4$. For $\mathscr{D}_G=2^{10}m$ we have $q_1=2^2\rho$ with $\rho$ odd, and $\delta_1\equiv -\delta_2$ mod $4$ and $m=\ell_1\ell_2 \rho^2 (\delta_1\delta_2)^3 \equiv 1\cdot 1\cdot (-1)^3 \equiv -1$ mod $4$.
For $\mathscr{D}_G=2^8m$ we have $q_1=2\rho$ with $\rho$ odd and $\delta_1+\delta_2 \equiv 2$ mod $4$ and
$\delta_1\equiv \delta_2$ mod $4$ and $m=\ell_1\ell_2 \rho^2 (\delta_1\delta_2)^3 \equiv 1\cdot 1\cdot 1^3 \equiv 1$ mod $4$.
Similarly when $\ell_1\ell_2\equiv -1$ mod $4$ we have $2\mid u$, $2\nmid v$, and  use \eqref{mod8minus} to get $\delta_1\equiv \delta_2$ mod $4$ for $2^{10}$
and $\delta_1\equiv  -\delta_2$ mod $4$ for $2^8$, and $m\equiv -1$ and $1$ mod $4$ respectively as before.

This leaves the case $\mathscr{D}_G=2^8m$ where $\ell_1=2\xi_1$, $\ell_2=2\xi_2$ with $\xi_1$,$\xi_2$ odd. If $\xi_1\xi_2\equiv 1$ mod $4$ then $4\mid v=(\xi_1-\xi_2)$ and $2\parallel u=(\xi_1+\xi_2)$ and from \eqref{mod8minus} we get $\delta_1\equiv \delta_2$ mod $4$ and $m=\xi_1\xi_2 q_1^2 (\delta_1\delta_2)^3 \equiv 1$ mod $4$. Likewise if $\xi_1\xi_2\equiv -1$ mod $4$
then $4\mid u$ and $2\parallel v$ and from \eqref{mod8} we have $\delta_1\equiv -\delta_2$ and again $m\equiv 1$ mod $4$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


I'm glad I bought this refrigerator it is very spacious and works great!
Bought this refrigerator in November 2017 because the one I had went bad.
Had it 2 months. Great features. Excellent cooling. Ice maker is small but work for us. Second unit can be added. Water/ice in door does not drip! Holds a LOT both in trig and freezer. Like that temps are controlled by degree not 1-6.
-Pro: Big space, DOOR-IN-DOOR feature-Con: a little noisy when it starts the cooler. Does Not have a controllable food drawer.
We are enjoying all of the conveniences that our new GE refrigerator offers. Plenty of room in the refrigerator AND the bottom freezer. ZERO regrets on this purchase!!!


text:
Client Login Contact Sitemap
WEALTH INSIGHTS
Trustplutus Team
Relationship Team
TrustPlutus Story
People - Trustplutus Team
Chetan Pandya, FRM
Head - Investment Advisory, TrustPlutus Family Office and Investment Advisers (India) Pvt. Ltd.
Chetan has over 20 years of experience in Banking and Finance Industry. He has successfully handled assignment in Wealth Management, Transaction Banking, Retail Distribution & Institutional Sales, having a versatile functional experience of Investment Advisory, Institutional Sales, Product & Strategy role.
Prior to joining Trustplutus, Chetan was with RBS N.V., where he was working in the capacity of Investment Head (South). Prior to RBS N.V., he was associated with organizations like ICICI Bank, Kotak Mahindra Bank, Jayant Sanghvi Enterprises & Motilal Oswal Securities Ltd.
Chetan has done his MBA in Finance from Bhavnagar University. He has done Financial Risk Manager Certified Course by Global Association of Risk Professional, New York.
Vishal Chandiramani, CFA CFP
Head – Products
Vishal has over 10 years of experience in Financial Services and Consulting. Vishal has handled assignments in sales, investment advisory and portfolio management and has also worked as a research analyst.
Prior to joining TrustPlutus, Vishal has worked with Religare Macquarie Wealth Management Ltd., ASK Wealth Advisors Pvt. Ltd. and McKinsey & Company.
Vishal is an MBA from SP Jain Center of Management, Dubai/Singapore. Vishal is also a Certified Financial Planner from FPSB, India and has successfully completed all three levels of CFA from the CFA Institute, USA.
Suresh Babu Gunakala
Head – Operations, Information Technology (IT) & Client Servicing
Suresh has over 18 years of experience in the financial services Operations domain comprising Registrars and Transfer Agents (RTAs), Distributors & Wealth Management Houses. He is instrumental in IT Implementations for Wealth Houses.
Prior to joining TrustPlutus, Suresh has worked with Tata Capital Financial Services Ltd., Karvy Computershare Pvt. Ltd., Deutsche Investor Services Private Limited, CAMS, etc.,
Suresh is a Commerce graduate, an MFM (Financial Management) from Pondicherry University.
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Reshma has over 10 years of experience in Financial and Accounting functions across leading organizations.
Prior to joining TrustPlutus, Reshma was associated with Axis Private Equity Limited, a fully-owned subsidiary of Axis Bank
Reshma is a Commerce Graduate, an MBA - Finance from Welingkar Institute of Management Development and Research
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Deepak has 28 years of work experience in the field of Banking Operations, Audit, and Credit Management.
Prior to joining TrustPlutus, Deepak was associated with Punjab National Bank.
Deepak is a Commerce graduate from Mumbai University and holds Post Graduate as Certified Associate of Indian Institute of Bankers.
Disclaimer | Disclosures | Trust Group
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Created by id labs

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Django retrieving users from tuples  AUTH_USER_MODEL = getattr(settings, 'AUTH_USER_MODEL', 'auth.User')

 class Friend(models.Model):
     """ Model to represent Friendships """
     to_user = models.ForeignKey(AUTH_USER_MODEL, related_name='friends')
     from_user = models.ForeignKey(AUTH_USER_MODEL, related_name='_unused_friend_relation')
     created = models.DateTimeField(default=timezone.now)

     objects = FriendshipManager()

     class Meta:
         verbose_name = _('Friend')
         verbose_name_plural = _('Friends')
         unique_together = ('from_user', 'to_user')

     def __unicode__(self):
         return "User #%d is friends with #%d" % (self.to_user_id, self.from_user_id)

 class FriendshipManager(models.Manager):

     def unread_requests(self, user):
         """ Return a list of unread friendship requests """
         key = cache_key('unread_requests', user.pk)
         unread_requests = cache.get(key)

         if unread_requests is None:
             qs = FriendshipRequest.objects.select_related(depth=1).filter(
                 to_user=user,
            viewed__isnull=True).all()
             unread_requests = list(qs)
             cache.set(key, unread_requests)

         return unread_requests

 class Person(models.Model):

     user = models.ForeignKey(User))
     name = models.CharField(max_length=30, blank=True)
     image = models.ImageField(upload_to='photos', blank=True)

     def __unicode__(self):
    return self.user.username

I am using this friendship model  . This model allow you to create friends between users.
I also have a UserProfile model called Profile which is above .
The below command shows how to view all the friendship request for a particular user
 >>> user = User.objects.get(pk=1)
 >>> user1 = User.objects.get(pk=1)
 >>> user2 = User.objects.get(pk=2)
 >>> user3 = User.objects.get(pk=3)
 >>> Friend.objects.add_friend(user1,user3)
 <FriendshipRequest: User #1 friendship requested #3>
 >>> Friend.objects.add_friend(user2,user3)
 <FriendshipRequest: User #2 friendship requested #3>
 >>> requests = Friend.objects.unread_requests(user=user3)
 >>> requests
 [<FriendshipRequest: User #1 friendship requested #3>, <FriendshipRequest: User
 #2 friendship requested #3>]
 >>> requests[0].to_user
 <User: hello>

I'm trying to show all the friendship request for a particular user with the images But
I can't do at the moment that because the requests are in tuples ( below is an example) .
I want to able to retrieve all the users who requested a friendship for a particular user
and then use those users to retrieve their users UserProfile model referring to Person so I can show their images in a template
in it's simplest form
How can I retrieve all the users who request a friendship ( below example is I can view the user name of one
     >>> requests = Friend.objects.unread_requests(user=user3)
     >>> requests
     [<FriendshipRequest: User #1 friendship requested #3>, <FriendshipRequest: User
     #2 friendship requested #3>]
      >>> requests[0].to_user

and then collect their UserProfile ( Person )
   Person.objects.filter ( ? )

A: Create a list of Friends
list_of_friend_requests = [r.to_user for r in requests]

Filter using django's filtering
persons = Person.objects.filter(user__in=list_of_friend_requests)

redpajama_set_name: RedPajamaStackExchange


Government's answer to all fiscal issues is to get more money. It really does target the small business owners. Pathetic.
Stephanie, I agree with you for two reasons, (1) actions have consequences and (2) the City government's greed and inefficiency.
Actions have consequences, and the consequence of the three tax increases you mention will be decreased income for the restaurants and bars, and for their patrons. The patrons only have so much money to spend and, the more they spend on taxes, the less they will spend on beer, wine and spirits.
The City government is monumentally inefficient and unable to operate as it would like on its present income. Rather than correct its inefficiencies, the greedy City wants the taxpayers to pay more – gotta pay for those two new Denalis.
I'd like to point out that Bennett doesn't mention any of the previous fees or costs the City has slathered on restaurants, bars, delis, cafes, brewpubs, etc. He writes about the proposed tax increase as though it's just one hike out of the blue. Taken in the context of a myriad of other financial pressures the city has placed on these businesses during Reed's tenure, the increase is far more significant. I realize PARKAtlanta was approved prior to his taking office, and the water rate increase is part of an ongoing deluge of costs for rate-payers partially because of the federal consent order to fix Atlanta's water system, but what this says, on top of the other fees etc., is that the only way Atlanta knows how to fix its budget is to squeeze those least able to absorb the hit.
didn't they also increase the cost of the liquor licenses?
Its really kind of tough to fire folks and get rid of positions, when all of them are filled with friends and family. So, like the two previous posts, lets not fix what's wrong and go from there, let's just tax the hell out of everyone and run them all out of business. Way to work. Awesome.
One thing that really hit me personally about the article is the fact that I had always perceived Atlanta to be a "restaurant town". Being somewhat of a "foodie" who really enjoys dining out, it was one of the key reasons I chose Atlanta as a place to retire. I also remember my parents taking weekend trips to Atlanta from Pensacola in the 60's and 70's for no other reason that to dine at great restaurants. My point is, that Atlanta's thriving restaurant culture had always been a big tourist (and retirement) draw in the past, and now the city is going to destroy that "restaurant town" reputation by taxing these establishments right out of business.
Just think, in Midtown alone, how many restaurants have folded in the last two years. Some, as usual, deserved to fold due to poor quality food or service, but many decent places called it quits too. If Reed and his rubber stamp council have their way, we won't have much left intown except grocery stores, drugstores, hardware stores, and gas stations!
There are wonks. Then there's Bennett. Taking *wonk* to a whole new level.
Here's another option. Instead of increasing taxes, let's decreases services in the areas that vote most heavily against said tax increases.
The level of city services, such as police presence, street repair, etc is already so abysmally poor that you would think the city has NO tax income.
This conveniently ignores the doubling of the liquor license fees Sir Kasim slapped on bars and restaurants earlier this year. That put a serious hurt on small outfits struggling with a still-down economy. We were able to get a property tax roll back for a bar building we own (but they pay the property taxes on) – which only off-set the doubling of the license fees.
This also doesn't address that soon after Sir Kasim took office he ordered a sweep of all small businesses to shake down folks into getting business licenses. Granted many of them should have them, but it provides a new source of revenue and future revenue to try to shake businesses down in the future. And shortly after that, a fire station crew, for a station we had no idea covered our buildings, came thru doing a code inspection. We fully expected them to return the next week to try to issue fines for minor infractions, but alas maybe the fire department indicated it had better things to do with its staff and equipment.
Coupled with parking enforcement, as now APD is doing enforcement trollings thru hot spots (if you hadn't witnessed that yet), it's getting to be a very anti-small-business climate within COA.
Oh, the Council used to be called "the nod squad" under Shirley; with dear CT at the whip, need to come up with a new moniker.
Actually, my column (the con- side) was largely about that license fee increase. Bennett did not mention it, though, as you point out. That's the thing, this increase can't be viewed as an isolated hike. It's only the latest. I also agree with you that the business climate here is getting increasingly tougher for small businesses. Bennett's reasoning seems to be "we haven't raised drink taxes since the 1970s, so we were due." That's kind of like the reasoning behind the license fee increase: We're increasing them because we haven't in a while and no, you won't be getting any additional services for the additional money you pay.
A name? Hmmm. The "Stamp Camp"?


Discover Gengis Khan in its original 1991 bottle, hand lacquered and available in limited batch.
Conqueror of the 13th Century. Son of the Blue Wolf.
Top : Bergamot, Juniper berries, Thyme, Lavender, Spearmint.
Heart : Rose, Lily of the valley.
Base : Vetyver, Pine needles, Benjoin, Incense, Patchouli.


text:
Setting the record straight on Mount Everest mountain biking
Sports | October 13, 2015
Phil Lindeman
plindeman@summitdaily.com
Patrick Sweeney shows Tashi Lama his mountain bike during a recent mountain bike trip to the south Mount Everest base camp. Most people Sweeney encountered on the trip had never seen a bicycle before. Sweeney maintains that he is the first modern rider to bike the Nepali route to base camp, although several others (including Summit locals) have made the ride to the Tibetan base camp in the past few decades.
Special to the Daily | Provided
Patrick Sweeney celebrates at Mount Everest base camp on Feb. 24, 2015, nearly 100 miles further and 9,000 feet higher than when he started. Sweeney maintains that he is the first modern rider to bike the Nepali route to base camp, although several others (including Summit locals) have made the ride to the Tibetan base camp over the past few decades.
Summit resident Brian Brill with monks at a Tibetan monastary near the north Mount Everest base camp in 1998.
Special to the Daily |
Brian Brill pauses with his bike during a 1998 mountain bike trek to north Mount Everest base camp, accessed through military checkpoints in Tibet.
Brian Brill with a Chinese soldier at a military checkpoint outside of north Mount Everest base camp. The checkpoint was the highest Brill was allowed to travel on his 1998 mountain bike ride.
A Tibetan monk rides on Brian Brill's mountain bike for the first time during the Summit local's 1998 ride to north Mount Everest base camp. Like the Nepali monks encountered by Patrick Sweeney, another local adventurer who rode to the south base camp, Brill says the monks had never seen a bike until he arrived.
Tibetan children play on Brian Brill's bike during the Summit local's 1998 ride to the north Mount Everest base camp.
Summit local Brian Brill at north Mount Everest base camp in 1998. He believes he was one of the first people to make the trip on the Tibetan side.
Brian Brill poses on his bike at north Mount Everest base camp in 1998, with the peak looming in the background.
Brian Brill at the monastary just outside of the north Mount Everest base camp in Tibet. Brill believes he became one of the first mountin bikers to make the trip when he completed it in 1998.
On Sept. 12, the Summit Daily printed an article about Patrick Sweeney, a local adventurer who rode his mountain bike to the south Mount Everest base camp shortly before Nepal was ravaged by a magnitude 7.8 earthquake in April.
His experience — covering 100 miles and 9,000 vertical feet in less than a week, meeting monks who had never before seen a bicycle — spurred him to launch a support campaign for the Nepali people, known as 1,000 Tents in 100 Days. He's sending all tents and proceeds directly to Tashi Lama, one of the monks he met on the ride.
Lama also runs a campaign for local sherpa families, the Little Sherpa Foundation, and he's personally distributing Sweeney's tents to nearby Himalayan villages that were devastated by the quake. Nearly nine months later, Lama says the situation on the ground is still dire. The country is plagued by corrupt officials and too few resources, and what little global support the Nepali received in the weeks after the quake has essentially vanished.
Sweeney's story is incredible and his nonprofit work is commendable, readers said in comments on the SDN website and Facebook page. But, the article by writer Leo Wolfson claimed Sweeney was the first person to reach Mount Everest base camp on a bicycle. Shortly after the article went to print, we received several messages from readers who claimed they have also biked to a Mount Everest base camp. In one message, longtime Summit resident Brian Brill said he was impressed by Sweeney's journey and respected his nonprofit work, but that his claim of "first ever" just wasn't true. That's because Brill had biked to the Tibetan base camp in 1998.
I called Brill to get his side of the story. He set his sights on Mount Everest as part of a three-month, 1,1000-kilometer bike tour through central Asia, which came with rides through dust storms and packs of wild dogs. The trip eventually took him to Kathmandu, but he opted to reach base camp from the Tibetan side and paid $2,500 for a visa.
Like Sweeney, Brill says mountain bikers in the Himalayas are few and far between. Tibetan monks he encountered had also never seen a bicycle, and many had been at the same monastery on the Mount Everest route for decades.
"At the time, I'd been researching ways to get to the Everest base camp, and I was always told you couldn't go biking — it's impossible," Brill told me. "I just didn't come across any stories of anyone else."
Brill reached base camp on a K2 hardtail in a Keystone Resort jacket, then decided to see just how far he could go. He was stopped within a half-mile at the next military checkpoint, where the soldiers "weren't as much fun as everyone else."
'The goal was to see how high on Mount Everest I could go," Brill said. "There is a Chinese guard station up there and they make it a point that you shouldn't pass it."
What happens if he had? He would have been immediately sent to a Chinese-Tibetan prison. Instead, he stuck around the base camp area for five days, riding several nearby peaks and topping out at 19,500 vertical feet before he left.
"You have an incredible view of Everest, and then I went up a few subsidiary peaks on the fringes," Brill said. "As a Summit County person, I just wanted to see how high I could get."
A new MTB Mecca?
In Wolfson's article, Sweeney also talks about just how difficult it was to secure permits as a mountain biker. The Nepali government relies on climbing and trekking traffic, he said, and mountain bikes have been illegal for decades in Sagarmatha National Park, the southeastern gateway to Mount Everest.
"That (southern) route is so narrow that when you have yaks or horses or whatever, it's hard to get by on a bike," Sweeney told me later. "They don't want cyclists to get in the way of their bread and butter, the climbing community."
It took Sweeney a total of 14 months and roughly $1,200 to finally receive the go-ahead from Nepali officials. When he did, they also believed he was the first adventurer to officially make the journey, permits and all. A few small groups had tried it in the past, but they were quickly stopped — and, in at least one case, shot at — by the Nepali military at several checkpoints along the trail.
I sent an email to the Nepali consulate for further information on past bike trips and haven't received a reply. Sweeney maintains that he is the first modern mountain biker to reach base camp on the Nepali side, although he admits that plenty of individuals and tour groups have successfully reached the Tibetan base camp.
Sweeney claims that the northern route is considerably easier than the southern route, which was yet another reason he decided to deal with the headache of permits and corrupt officials. (It's also less popular as a climbing destination. The Tibetan side sees less than 10,000 trekkers per year, thanks in large part to extremely expensive visas, while the Nepali side sees upwards of 40,000 per year, according to a Fox News report after the April quake.) In terms of terrain and elevation, he wanted the ride to be as difficult as possible. It was part of the challenge.
"This was something I wanted to do no matter what," Sweeney said. "I would have done it as a personal goal, even if someone else had done it."
Still, it's odd to think that Mount Everest isn't more popular with singletrack junkies, particularly as mountain biking gets bigger and wilder and crazier every year. Very few people make the ride on either side of the peak, and there's hardly any concrete data on the number of adventurers who have successfully made the trip.
That could all change in the next few years. Along with the tent campaign in Nepal, Sweeney wants to launch a mountain bike festival in Kathmandu. He sees it as a potential tourism draw in spring, when climbing is usually too dangerous. Kathmandu is already the starting point for several base camp bike tours, including a combination bike and drive program through Bike Adventure Nepal, which claims to be "the first mountain bike company run by top professional Nepali mountain bikers."
Sweeney will start building interest in Himalayan biking this March, when he and Red Bull mountain biker Rebecca Rush attempt to set three MTB world records: Most nonstop ascent, most nonstop descent and fastest human time for the "Mail Run," the 200-mile trek from south Mount Everest base camp to the Kathmandu airport. They're all personal goal, he admits, but he also hopes it will show the Nepali government that mountain biking can be a tourism draw.
"In the off-season, the non-climbing season, biking can be a source of tourist revenue," he told me. "They're really hurting for that now, because people really are afraid to head out there."
Meet Your Mountains: Hiking Peak 1 in Frisco
Colorado Mountain Club to host meet and greet for current and prospective members
Breckenridge local Teague Holmes describes skiing exposed East Thorn line in Gore Range
Copper Mountain announces extended summer hours on Fridays
Sports ―
Liberty lines: Droves of freedom riders descend on Arapahoe Basin Ski Area for sunny July 4
Arapahoe Basin Ski Area slopes maintenance manager describes what it takes to open for the Fourth of July
Cinco de Mayo Celebration raises nearly $30,000 for Breckenridge Outdoor Education Center


Are you thinking about moving and you aren't sure where you want to head out? Are you being relocated to a lively tourist destination? There are lots of places throughout the U . s . that lure the crowds from all through the world. Just we travel the world and find new exciting places, people come to America and travel the actual country seeking to see just as many wonders as they possibly can. Moving into a tourist town can along with an advantage of knowing once the best time for go towards the attractions is and could be an exciting time for you and all your family. This article uses a take a some belonging to the top tourist towns individuals move each and every year.
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Give the little one's wardrobe a playful twist with this super cool Star Wars themed Toyshirt. Boasting short sleeves and a cartoon image of Darth Vader and Obi-Wan Kenobi caught in a battle and when the tab is pulled it reveals the clash of their lightsabers, they will be entertained for hours.
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text:
Q: TikZ - Placing several numbers in one cell of a Sudoku grid I've put at the end of this post one code coming from this page which is a pretty easy way to display Sudoku grid.
I would like to have help so as to add the possibility to indicate several entries in one cell like in the following screenshot ? This is to indicate how to solve the sudoku without giving the answer.

My question is about one TiKz way to produce at least the first preceding case with nine possibilities.
Code, from TeXample.net, that displays a Sudoku grid
\documentclass{article}
    \usepackage{tikz}

    \newcounter{row}
    \newcounter{col}

    \newcommand\setrow[9]{
        \setcounter{col}{1}
        \foreach \n in {#1, #2, #3, #4, #5, #6, #7, #8, #9} {
            \edef\x{\value{col} - 0.5}
            \edef\y{9.5 - \value{row}}
            \node[anchor=center] at (\x, \y) {\n};
            \stepcounter{col}
        }
        \stepcounter{row}
    }

\begin{document}

\begin{tikzpicture}[scale=.5]
  \begin{scope}
    \draw (0, 0) grid (9, 9);
    \draw[very thick, scale=3] (0, 0) grid (3, 3);

    \setcounter{row}{1}
    \setrow { }{2}{ }  {5}{ }{1}  { }{9}{ }
    \setrow {8}{ }{ }  {2}{ }{3}  { }{ }{6}
    \setrow { }{3}{ }  { }{6}{ }  { }{7}{ }

    \setrow { }{ }{1}  { }{ }{ }  {6}{ }{ }
    \setrow {5}{4}{ }  { }{ }{ }  { }{1}{9}
    \setrow { }{ }{2}  { }{ }{ }  {7}{ }{ }

    \setrow { }{9}{ }  { }{3}{ }  { }{8}{ }
    \setrow {2}{ }{ }  {8}{ }{4}  { }{ }{7}
    \setrow { }{1}{ }  {9}{ }{7}  { }{6}{ }

    \node[anchor=center] at (4.5, -0.5) {Unsolved Sudoku};
  \end{scope}

  \begin{scope}[xshift=12cm]
    \draw (0, 0) grid (9, 9);
    \draw[very thick, scale=3] (0, 0) grid (3, 3);

    \setcounter{row}{1}
    \setrow { }{2}{ }  {5}{ }{1}  { }{9}{ }
    \setrow {8}{ }{ }  {2}{ }{3}  { }{ }{6}
    \setrow { }{3}{ }  { }{6}{ }  { }{7}{ }

    \setrow { }{ }{1}  { }{ }{ }  {6}{ }{ }
    \setrow {5}{4}{ }  { }{ }{ }  { }{1}{9}
    \setrow { }{ }{2}  { }{ }{ }  {7}{ }{ }

    \setrow { }{9}{ }  { }{3}{ }  { }{8}{ }
    \setrow {2}{ }{ }  {8}{ }{4}  { }{ }{7}
    \setrow { }{1}{ }  {9}{ }{7}  { }{6}{ }

    \node[anchor=center] at (4.5, -0.5) {Solved Sudoku};

    \begin{scope}[blue, font=\sffamily]
      \setcounter{row}{1}
      \setrow {4}{ }{6}  { }{7}{ }  {3}{ }{8}
      \setrow { }{5}{7}  { }{9}{ }  {1}{4}{ }
      \setrow {1}{ }{9}  {4}{ }{8}  {2}{ }{5}

      \setrow {9}{7}{ }  {3}{8}{5}  { }{2}{4}
      \setrow { }{ }{3}  {7}{2}{6}  {8}{ }{ }
      \setrow {6}{8}{ }  {1}{4}{9}  { }{5}{3}

      \setrow {7}{ }{4}  {6}{ }{2}  {5}{ }{1}
      \setrow { }{6}{5}  { }{1}{ }  {9}{3}{ }
      \setrow {3}{ }{8}  { }{5}{ }  {4}{ }{2}
    \end{scope}
  \end{scope}
\end{tikzpicture}

\end{document}

A: I introduced some new commands to help explaining sudokus:
UPDATE: Highlights are put in background as suggested by Claudio Fiandrino. Also a new \highligthrectangle was added to highlight arbitrary rectangular sections, such as quadrants. Note how several highligths can overlap.
    % Highlight some cells
    \highlightcell{2}{8}
    \highlightrow{8}
    \highlighcolumn{5}
    \highlightrectangle{4}{1}{6}{3}

    % Other marks
    \circlecell{5}{2}
    \crosscell{8}{1}

    % Hints
    \hintcell{5}{5}{2,7,8}
    \hintcell{1}{1}{3,4,6,7}

Resulting in:

The code is as follows.
\usetikzlibrary{backgrounds}
% Some customizable styles
\tikzset {
  highlight/.style = {yellow, opacity=0.3},
  digit/.style = { minimum height = 5mm, minimum width=5mm, anchor=center },
  circle/.style = {draw=green!80!black, dotted, very thick},
  cross/.style = {red, opacity=.5, shorten >=1mm, shorten <=1mm, very thick, line cap=round},
  hint/.style={blue, font=\sf, minimum width=3mm, minimum height=3mm}
}

% Original code-----------------------------------------------------------
% Modified the \node to give a unique name to each one, which is the
% row number, a dash and the column number. E.g: 1-1, 4-5, etc.
\newcounter{row}
\newcounter{col}

\newcommand\setrow[9]{
    \setcounter{col}{1}
    \foreach \n in {#1, #2, #3, #4, #5, #6, #7, #8, #9} {
        \edef\x{\value{col} - 0.5}
        \edef\y{9.5 - \value{row}}
        \node[digit,name={\arabic{row}-\arabic{col}}] at (\x, \y) {\n};
        \stepcounter{col}
    }
    \stepcounter{row}
}

% New code -------------------------------------------------------------
\def\highlightcell#1#2{
\fill[highlight] (#1-#2.north west) rectangle (#1-#2.south east);
}

\def\circlecell#1#2{
\draw[circle] (#1-#2) circle(4mm);
}

\def\crosscell#1#2{
\draw[cross] (#1-#2.north west) -- (#1-#2.south east);
\draw[cross] (#1-#2.north east) -- (#1-#2.south west);
}

\def\highlightrow#1{
\fill[highlight] (#1-1.north west) rectangle (#1-9.south east);
}

\def\highlighcolumn#1{
\fill[highlight] (1-#1.north west) rectangle (9-#1.south east);
}

\def\hintcell#1#2#3{
\node at (#1-#2) {\hintbox{#3}};
}

\def\highlightrectangle#1#2#3#4{
\begin{pgfonlayer}{background}
\fill[highlight] (#1-#2.north west) rectangle (#3-#4.south east);
\end{pgfonlayer}
}

% UGLY code. Do not read :-)
\def\hintbox#1{
\resizebox{4.5mm}{4.5mm}{%
\tikz[scale=0.3]{%
  \def\auxc{0}
  \foreach \m in {1,...,9} {
    \pgfmathparse{mod(\auxc,3)}
    \xdef\x{\pgfmathresult}
    \pgfmathparse{-floor(\auxc/3)}
    \xdef\y{\pgfmathresult}
    \xdef\hintprinted{0}
    \foreach \n in {#1} {
      \ifnum\n=\m
       \node[hint] at (\x,\y) {\n};
       \xdef\hintprinted{1}
      \fi
     }
   \ifnum\hintprinted=0
      \node[hint, opacity=0.1] at (\x,\y) {\m};
   \fi
   \pgfmathparse{\auxc+1}
   \xdef\auxc{\pgfmathresult}   }
  }%
 }
}

\begin{tikzpicture}[scale=.5]
  \begin{scope}
    \draw (0, 0) grid (9, 9);
    \draw[very thick, scale=3] (0, 0) grid (3, 3);

    \setcounter{row}{1}
    \setrow { }{2}{ }  {5}{ }{1}  { }{9}{ }
    \setrow {8}{ }{ }  {2}{ }{3}  { }{ }{6}
    \setrow { }{3}{ }  { }{6}{ }  { }{7}{ }

    \setrow { }{ }{1}  { }{ }{ }  {6}{ }{ }
    \setrow {5}{4}{ }  { }{ }{ }  { }{1}{9}
    \setrow { }{ }{2}  { }{ }{ }  {7}{ }{ }

    \setrow { }{9}{ }  { }{3}{ }  { }{8}{ }
    \setrow {2}{ }{ }  {8}{ }{4}  { }{ }{7}
    \setrow { }{1}{ }  {9}{ }{7}  { }{6}{ }

    \node[anchor=center] at (4.5, -0.5) {Unsolved Sudoku};

    % Highlight some cells
    \highlightcell{2}{8}
    \highlightrow{8}
    \highlighcolumn{5}
    \highlightrectangle{4}{1}{6}{3}

    % Other marks
    \circlecell{5}{2}
    \crosscell{8}{1}

    % Hints
    \hintcell{5}{5}{2,7,8}
    \hintcell{1}{1}{3,4,6,7}
  \end{scope}
\end{tikzpicture}


text:
Московский областной государственный камерный театр — московский областной театр.

История
Московский областной Камерный театр был создан согласно Постановлению № 219 от 15 марта 1934 года Народного Комиссариата просвещения РСФСР (НКП РСФСР) и получил первоначальное название: Колхозный художественный театр Московского областного Управления театрально-зрелищными предприятиями.
 1942 год — театр переименовали в Первый Московский областной драматический театр.
 1992 — театр получил название, под которым проработал более двух десятков лет: Московский областной государственный камерный театр.
 1993 — театр стал работать на территории Юго-Восточного округа (Кузьминки), на Малой сцене Московского областного Дома Искусств (МОДИ).
В разное время коллектив возглавляли: С. А. Бенкендорф, М. Е. Веснин; В 1987 году  Главным Режиссёром театра стал 32-х летний Пётр Олев, предложивший слияние областных театров в один, настаивавший на строительстве помещения для бездомных тогда московских областных театров. Пётр Олев активно выступал  за улучшение условий работы артистов областных театров, за поднятие художественного  уровня областных театров. Он  выступал против контроля театров со стороны партийных структур и областного комитета КПСС. В 1988 году он был уволен, якобы подав заявление "по собственному желанию", которого никогда не подавал. На смену ему пришел В.И. Якунин, приглашенный  им на постановку пьесы Э. Олби "Всё в саду". С 1989 года — Валерий Иванович Якунин — заслуженный артист РФ, заслуженный деятель искусств РФ.

Московский областной государственный камерный театр (1934 — 2013) — ныне не существующий московский театр.

В 1968—1973 гг. главным художником театра был А. М. Авербах.

При Камерном театре также действовал экспериментальный актёрский курс.

Конец истории
В 2013 году произошло слияние двух областных театров: Московского областного драматического театра им. А. Н. Островского и Московского областного государственного Камерного театра. Новосозданный театр получил название Московский губернский театр; его художественный руководитель — народный артист РФ Сергей Витальевич Безруков.

Труппа
После 2013 года труппы Московского областного государственного камерного театра не существует.
До этого:

 Амелин, Александр Александрович
 Багров, Григорий Анатольевич
 Басканчин, Николай Владимирович
 Блинова, Валентина Александровна
 Бояринова, Марина Владимировна
 Веретин, Алексей Юрьевич
 Вершинин, Сергей Евгеньевич
 Гомоной, Евгений Юрьевич
 Губанов, Сергей Владимирович
 Гусаров, Алексей Алексеевич
 Доронина, Елена Евгеньевна
 Захарова, Алла Александровна
 Значков, Роман Николаевич
 Исаенков, Андрей Александрович
 Исаенкова, Елена Александровна
 Качалкина, Наталья Владимировна
 Кисса, Сергей Юрьевич
 Коноваленко, Валентина Николаевна
 Кудрина, Людмила Михайловна
 Кукушкина, Наталья Александровна
 Курлов, Олег Константинович
 Михеева, Елена Александровна
 Пестова, Юлия Геннадьевна
 Полозова, Анна Сергеевна
 Попов, Александр Григорьевич
 Ремезов, Павел Николаевич
 Роганова, Анна Сергеевна
 Рогов, Михаил Николаевич
 Рудова, Татьяна Кузьминична
 Серебренникова, Евгения Евгеньевна
 Фирсов, Григорий Николаевич
 Хаустова, Оксана Валерьевна
 Цагина, Елена Георгиевна
 Шутов, Виктор Дмитриевич
 Шелягин, Константин Андреевич

Примечания

Ссылки
 Московский областной государственный камерный театр, официальный сайт.
 На Яндекс-картах
 Московский областной государственный камерный театр на WikiData

Областные театры
Исчезнувшие театры Москвы
Камерные театры России

redpajama_set_name: RedPajamaWikipedia


Lab:One Recordings brings to you a revered classic effect as pert of the new PATRIOT range. This unit can help give that extra weight and depth to your current sounds. Immerse yourself with deeper harmonics, all generated from your original sound signal.
As with all Patriot range units, Stimulant comes with a dual function display, allowing you to monitor your signals pre and post processing, plus the extra function to select the source to generate your harmonics.
Stimulant comes with not one but two dedicated sub harmonics generators. With mono and dual operation (source selection dependent), Stimulant can give your existing sounds that extra low end that most sound designers aspire to find and create. All within a neat simple 1U rack unit.


Suits:The Greatest Episode Of All Time!
This week's episode of Suits was amazing! I was literally holding my breath for the entire hour! Travis Tanner's case against Harvey continues as Travis decides to take Harvey to court. They can't agree on a settlement, because the only offer Travis makes is to disbar Harvey.
Harvey brings in Louis to act as Travis during their mock trial. Rachel is acting as Donna, and Daniel Hardman plays the judge. Harvey brings in new girl Zoe, who apparently has shares an interesting past with him. She joins the firms team to assist them in the building of their court case. Anyways, Mike finds a way to convince Donna to help out Harvey even after she said she would plead the 5th. Harvey confronts Donna on the streets of Toronto, I mean New York City, she is literally fuming at him, she hates him but not really, he feels really bad but at the same time all he asks of her is to admit she made a mistake. She's mad and hurt that Harvey would let her go so easy.
So when Mike convinces Donna to help Harvey out, Donna shows up at Pearson and Hardman for the mock trial. Instead of Rachel playing Donna, it's Donna playing Donna! The mock trial gets intense when Louis refuses to follow the script- since Donna refuses to answer any questions. Louis starts asking personal questions, insinuating Donna has feelings for Harvey and that she's in love with him! Donna's panicking, Harvey feels bad for her, Louis keeps pushing and demanding the answer from Donna whether she loves Harvey or not. She doesn't answer, but it's obvious there are feelings between the two, Harvey stands up and demands for the mock trial to be over with. Donna gets up and leaves, Harvey follows her but he can't do anything to save her. I love the story between these 2, I cannot wait for the day until they're together!! Even if the writers give us a tiny inkling of hope for their romance, it's enough to keep us watching.
In the end, Mike and Hardman find some dirt on Travis. They decide the best option is to blackmail Travis and agree on a settlement. When Hardman brings the news to Jessica and Harvey they are weary of his intentions, since they don't fully trust Hardman. When Harvey confronts Mike, and lets him know they won't be settling, Mike has a hissy fit and reminds Harvey how hard they worked on this case, and how hard things will be if they go to court. Jessica and Hardman bring the decision to the partners, settle/trial, the vote is at a tie so the final decision is left for Harvey, he decides to settle and Jessica is completely shocked.
Just then, Hardman approaches the partners and brings up the issue of Jessica having trouble managing her firm. Hardman says they should have a vote about who has power him or her. Basically, he screwed them over.
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text:
CELEBRATE OSCAR NOMINATED CARTOON SALOON WITH NEW #IMAKEMOVIES CAMPAIGN FROM THE INDUSTRY TRUST
DUBLIN. 14th FEBRUARY 2019. The Industry Trust for IP Awareness Ireland and the Motion Picture Association (MPA) release a brand new #IMAKEMOVIES campaign, focused on the talent it takes to bring great stories to global audiences.
This #IMAKEMOVIES ANMITATION campaign represents the growth and success of Kilkenny based Cartoon Saloon, Academy Award, Golden Globe, BAFTA and Emmy nominated animation studio. From shorts to feature films and TV series they have carved a special place in the international Animation industry. This campaign highlights how much passion and hard work they have for their craft and piracy has been a constant threat to their business.
Due to the success of the first three campaigns online, #IMAKEMOVIES ANIMATION will receive a cinema release from Friday February 15th.
Alison Crinion, Communications Director Irish Industry Trust says 'It has been a pleasure to work with Cartoon Saloon on this #IMAKEMOVIES campaign. Piracy is a real threat to the growth of the Irish Film Industry. Paul and his team have illustrated with great honesty their passion for their craft and what the ongoing piracy challenge means to them in very real terms. The Irish Industry Trust feels that this campaign will resonate with Irish audiences the impact in making legitimate viewing choices'
Stan McCoy, President and Managing Director of the MPA EMEA says "We are proud to work with the Irish Industry Trust on this powerful campaign to encourage respect for the hard work of the Irish creative sector. This campaign reminds us that piracy impacts real people and local businesses. Through partnerships we can build a future where creativity thrives, where storytellers prosper, and audiences around the world celebrate their creations."
The Industry Trust's consumer education campaign continues to deliver this core message; inspiring audiences to choose the big screen experience and choose legal content when watching online. The #IMAKEMOVIES campaign directs audiences to an industry-supported film search engine, LOVEMOVIES.IE, which offers exclusive updates, news and real-life experiences from those working in film. There are also legal content sources allowing the user to book cinema tickets and buy or rent content at their convenience.
Online from 14/02/10
In Irish Cinemas 15/02/19
https://www.mpa-emea.org/wp-content/uploads/2019/02/imakemovies-ANIMATION.mp4
About The Industry Trust for IP Awareness
The Industry Trust is the film, TV and video industry's consumer education body, promoting the value of copyright and creativity. Its consumer awareness programmes address the on-going challenge of film and TV copyright infringement by inspiring audiences to value great movie moments and choose to watch film, TV and video via legitimate sources. Industry funded, their work aims to engage three distinct audience groups: young people, 16-34-year-old men; and parents, providing extensive insight around audience attitudes and understanding of intellectual property. For more information on the Industry Trust's work, www.lovemovies.ie
About The Motion Picture Association
The Motion Picture Association (MPA) is the international trade association that serves as the voice and advocate of the major international producers and distributors of films, home entertainment and television programs across Europe, the United States and around the world. The MPA is a subsidiary of the Motion Picture Association of America (MPAA), whose members are: Walt Disney Studios Motion Pictures, Netflix Inc., Paramount Pictures Corporation, Sony Pictures Entertainment Inc., Twentieth Century Fox Film Corporation, Universal City Studios LLC, and Warner Bros. Entertainment Inc. For more information about the MPA, www.mpa-emea.org