text:
Tunis to Tel Aviv to Tehran:
The US and Today's Middle East
L. Carl Brown
Garrett Professor in Foreign Affairs Emeritus, Princeton University
Minutes of the 15th Meeting of the 69th Year
The meeting was called to order at 10:15 AM by President Bob Varrin.Tom Fulmer led the invocation. Three guests were introduced: Charles Ascher by Luke Finlay; Michael Curtis, by Herb Abelson; and Micaella de Lingarolles, by Scott McVay. Claire Jacobus read the minutes of the previous meeting. Jack Reilly announced that Herb Abelson had attained Emeritus status. The speaker and program for next week's meeting were announced.
Jock McFarlane introduced the speaker, Emeritus Professor Carl Brown, an Old Guard member, who had graciously replaced the ailing scheduled speaker on very short notice. Prof. Brown is a graduate of Vanderbilt University, and he earned his doctorate at Harvard. He is an historian, a lifelong student of the countries and cultures of the Middle East. He is Director of the interactive Middle Eastern Studies Program at Princeton. He's also a member of and advisor to the Princeton Middle East Society. His talk was entitled "Tunisia to Tel Aviv to Tehran."
The first half of Dr. Brown's talk was about Tunisia, a favorite subject of his, dating back to the writing of his doctoral thesis. Tunisia, he said, is a fascinating country with a complicated history and a unique geo- graphical situation. It's the northernmost country of Africa, not far from Sicily and Italy. Originally populated by Berber Arabs, its history goes back to settlement by Phoenicians in the tenth century B.C., to Carthage which rivaled Rome as a Mediterranean power, and to successive occupation by Romans, Vandals, Byzantines, Arabs, and Ottomans. Under the Ottomans, Tunisia had a measure of independence. But after the French invaded from Algeria in 1881, Tunisia became a French protectorate.
In the Second World War, Tunisia was an important battleground. Most us remember Operation Torch, when Americans landed in North Africa in 1942, drove eastward and eventually met with British forces coming from Egypt and Libya; and after a series of hard-fought battles, the Allied Forces finally defeated Field Marshall Rommel's German-Italian forces who were forced to surrender in May 1943.
After the War, in 1956, the Tunisian Republic was established. Its first president was the charismatic nationalist Habib Bourguiba, for whom Prof. Brown had great admiration. Bourguiba led the modernization of his country. He gave increased rights and privileges to women, and resisted the Islamization of his country. He knew that Tunisia needed to stay away from religious fundamentalism. He developed good relationships with members of the European Union, especially France, as well as with other Arabic nations. He was a moderating voice in the continuing Arab-Israeli conflict. [Under Bourguiba,Tunisia was a member of the Arab League and the African Union.] The economy became more diversified, based on a meld of agriculture, mining, manufacturing (clothing, footwear, and machinery), and tourism. And although Islam was the official religion and 99 % of the people were Muslim, Tunisia has had a secular culture in which other religions are tolerated.
Bourguiba was regarded at one time as the most effective of third world leaders. He guided his country through the difficult but essential process of de-colonization. But, unfortunately, he didn't know when to stop; he hung on too long. But eventually, after 31 years in power, Bourguiba was declared incompetent and removed in a bloodless coup, in 1987. [Prof. Brown recalled what Marc Antony said after the death of Julius Caesar: "You all did love him once, not without cause."]
Bourguiba's successor as President was his Prime Minister Zenedine Ben Ali, who ruled for more than 23 years, from 1987 until January 2011. Ben Ali initially promised to maintain Bourguiba's reforms, but his regime gradually became corrupt, authoritarian, and repressive. With decline in the Tunisian economy, and the awakening of a more informed public, Ben Ali became progressively more unpopular, until he was forced out in a bloodless revolution two weeks ago, and Ben Ali had to seek asylum in Saudi Arabia.
Prof. Brown went on to consider what will be the impact of the current Tunisian revolution on other states in the region, what does it say about American actions in the past, and what guidance can the situation provide for our role in the future. As to the first question, it remains to be seen. But we can anticipate that the increasingly informed populations of the region will be progressively less tolerant of authoritarian regimes.
With regard to American actions in the Middle East in the past, Prof. Brown was unequivocal in condemning our invasion of Iraq; the results of our military intervention in Iraq were "abysmal," he said. He added, "An activist overthrow of a regime is not a good or useful way to proceed." Going on, although he had no doubt that Iran is seeking a nuclear capability, he said that a military action there would be counter-productive, and very bad. He emphasized the need for a prudent approach, and for recognizing the limits to what we can do.
Dr. Brown then discussed the effects of America's unequivocal support of Israel. He said that our commitment to the security of Israel is iron-clad and no longer subject to debate. But, he emphasized, the perception of our support of Israel is an important factor in the opposition to the United States. Our unyielding pro-Israel position undermines our efforts to be influential in the region, he said.
He closed by focusing on the continuing and unresolved Israeli-Palestinian conflict. He stated that the state of Israel needs to work for a separate Palestinian state, and to end their "occupation" of Palestinian territory. The need for this is "a matter of demographics," he said. Within Israel, there are 6.1 million Jews and 1.6 million Arabs, but in "Greater Israel," including Gaza and the West Bank, there are another 4 million Arabs. So within the next few years, Greater Israel will have a Jewish minority. Dr. Brown felt that the Israelis need to end their settlements policy, and move towards the recognition of a Palestinian state. But he noted the seeming inability of the Obama administration to move events in that direction. So no easy solutions seem to be in sight.
At the conclusion of Dr. Brown's talk, there were several provocative questions. In one of them, a member disputed the accuracy of some of Dr. Brown's statements, particularly with regard to Gaza and the West Bank settlements. Professor Brown conceded the accuracy of some of the members' statements, but he maintained that it is not in Israel's interest to continue its efforts to expand its occupation in the disputed territories.
The meeting was adjourned at 11:30 AM.
Harvey Rothberg, M D


If I'm not mistaken it could be saved/loaded without prompt via retainEntry/restoreEntry, etc.
Aside of browser local storage, Odoo doesn't have such kind of solution, Odoo Pos is very rudimentary, as pointed by Nhomar they used SAP for such 150 stores.
Does odoo have anything for pos local storage synchronization?
I think this struggle is a dead end. Losing any single record of an issued invoice cannot be tolerated for such an offline task, so you need a persistent storage. (For example pyhsical credit card POS devices can work offline and they are designed not to lose data when electrics are gone.) On the other hand, a browser's local storage is not persistent (http://stackoverflow.com/a/9729004). If something goes wrong with the browser or someone clears the cache intentionally, there might be no way of recovering this or recovering can be economically inefficient.
1) Use it online. Your server will be safe and get things done all synchronized.
2) Integrate it with a physical devices such as POS device, Cash register or probably Posbox. Keep invoice data on pyhsical device and get it synchronized with the server from there. You can use different series (prefix) on each device so that you don't need the second random number.
3) Use a cache server on POS site which will help POS locations with numbering and provide persistent storage which can be synchronized with server. You can again use different series (prefix) on each device or cache server.
Thank you, sir! And do we have any idea as to how many transactions local storage will hold? And do we have any idea as to what's typically behind users losing connectivity in the first place?


text:
El Puig Rodó és una muntanya de 2.007 metres que es troba al municipi d'Alins, a la comarca del Pallars Sobirà.

Referències

Muntanyes d'Alins

redpajama_set_name: RedPajamaWikipedia


text:
\section{\label{sec:level1}Introduction}

The field of game theory deals with situations involving strategic interdependence between a set of rational participants.  The study of classical game theory began around 1944 \citep{vonNeumanMorgenstern,Binmore,Rasmusen}, and was extended to the quantum regime, in 1999, by Meyer \citep{MeyerDavid} and Eisert et al \citep{Eisert1999} and has since been developed by many others \cite%
{Blaquiere,Wiesner,Mermin,Peres,Mermin1,Vaidman,BenjaminHayden,EnkPike2002,Johnson,MarinattoWeber,IqbalToor1,DuLi,Du,Piotrowski,IqbalToor3,FlitneyAbbott1,IqbalToor2,Piotrowski1,Shimamura1,FlitneyAbbott2,YongJianHan2002,IqbalWeigert,Mendes,CheonTsutsui,IqbalEpr:2005,Ozdemir2004,CheonAIP,Shimamura,IchikawaTsutsui,OzdemirShimamura,FlitneyGreentree,IqbalCheon,Ichikawa,Ramzan,FlitneyHollenberg2006,Aharon,Bleiler,ahmed2008three,Qiang,NFJP2010,IqbalAbbott,CIL,IqbalCheonAbbott,ChappellB,Chappell3Player,FlitneyAbbottRoyal,NawazToor,GuoZhang,iqbal2009quantum}.
Initially, studies in the arena of quantum games focused on two-player,
two-strategy non-cooperative games but has now been extended to multi-player games by various authors \citep{Popescu1995,BenjaminHayden,iqbalCheonConf,Broom_Cannings_Vickers_1997,QingChen2004,DuLi,du2002Multi,Mermin1,Flitney2009,Boyer2004quant}.  Quantum games have been reported in which players share Greenberger-Horne-Zeilinger
(GHZ) states and W states \citep{YongJianHan2002,Peres,Chappell3Player}, with analysis showing the benefits of players forming coalitions \citep{IqbalToor3,FlitneyGreentree} and also the effects of noise \citep{FlitneyAbbott2,Ramzan}. Such games can be used to describe multipartite situations, such as in the analysis of secure quantum communication \citep{NielsenChuang:2002}.

The usual approach to implementing quantum games involves players sharing a
multi-qubit quantum state with each player having access to an allocated qubit upon which they perform local unitary transformations; then a supervisor submits each qubit to measurement in order to determine the outcome of the game.
An alternative approach in constructing
quantum games uses an Einstein-Podolsky-Rosen (EPR) type setting \citep{IqbalWeigert,IqbalEpr:2005,IqbalCheon,EPR,Bohm,Bell,Bell1,Bell2,Aspect,ClauserShimony,Cereceda,IqbalCheonAbbott}, based on a framework developed by Mermin \cite{Mermin} in 1990.
In this approach, quantum games are described within an EPR apparatus, with the
players' strategies now being the classical choice between two possible measurement directions implemented when measuring their qubit. This thus becomes equivalent to the standard arrangement for playing a classical mixed-strategy game, in that in each run a
player has a choice between two pure strategies. Thus, as the players'
strategy sets remain classical, the EPR type setting avoids a well known
criticism \citep{EnkPike2002} of conventional quantum games, stemming from the fact that typically, in quantum game frameworks, players are given access to extended strategy sets consisting of local unitary transformations that can be interpreted as fundamentally changing the underlying classical game, and thus not being an authentic extension of it.

Recently \cite{CIL,ChappellB,Chappell3Player} the formalism of Clifford's geometric algebra
(GA) \citep{Hestenes111,GA,Doran2003,Venzo2007,Dorst:2002} has been
applied in the analysis of quantum games. These works demonstrate
that the formalism of GA facilitates analysis and improves
the geometric visualization of the game.
Multipartite quantum games are usually found significantly harder to
analyze, as we are required to define an $ N \times N $ payoff matrix and calculate measurement outcomes over $ N $-qubit states. In this regard, GA is identified as the most suitable formalism in order to allow
ease of analysis. This becomes particularly convincing in the case where $ N \rightarrow \infty $, where matrix methods become unworkable. As we will later show, an algebraic approach such as GA is both elegant and tractable as $ N \rightarrow \infty $.

Using an EPR type setting we firstly determine the probability distribution of measurement outcomes, giving the player payoffs, and then determine constraints that ensure a faithful embedding of the mixed-strategy version of the original classical game
within the corresponding quantum game.  We then apply our results to an $ N $ player prisoner dilemma (PD) game.

\section{EPR setting for playing multi-player quantum games}

The EPR setting \citep{IqbalWeigert,IqbalCheon,IqbalCheonAbbott} for a
multi-player quantum game assumes that players $P^i$ are
spatially-separated participants of a non-cooperative game, who are located
at the $ N $ arms of an EPR system \citep{Peres}, as shown in Fig.~\ref{eprFigure}.
In one run of the experiment, each player chooses one out of
two possible measurement directions. These two directions in space, along which spin or
polarization measurements can be made, are the players' strategies.
As shown in Fig.~\ref{eprFigure}, we represent the $ i^{\rm{th}} $ players' two measurement directions as $%
\kappa _{1}^{i},\kappa _{2}^{i}$, with a measurement returning $+1$ or $-1$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=3.5in]{eprN}
\end{center}
\caption{The EPR setup for an $N$-player quantum game. In this setup, each player $ i $ has a choice of two measurement directions  $ \kappa _{1}^{i} $ and $ \kappa _{2}^{i}$ for their qubit, allocated from a shared $N$-qubit quantum state.}
\label{eprFigure}
\end{figure}

Over a large number of runs consisting of a sequence of $N$-particle
quantum systems emitted from a source, upon which measurements are performed on each qubit, subject to the players choices of measurement direction, a record is maintained of the experimental outcomes from which players' payoffs can be determined. These payoffs depend on the $N$-tuples of the various
players' strategic choices made over a large number of runs and on the dichotomic outcomes (measuring spin-up or spin-down) from the measurements performed along those directions.

\subsection{\label{GAintro} Clifford's geometric algebra (GA) }

Typically in a quantum game analysis the tensor product formalism along with Pauli matrices are employed, however matrices become cumbersome for higher dimensional spaces, and so GA is seen as an essential substitute in this case, where the tensor product is replaced with the geometric product and the Pauli matrices are replaced with algebraic elements. The use of GA has also previously been developed in the context of quantum information processing \citep{GA4}.

To setup the required algebraic framework, we firstly denote $\{{e} _{i}\}$ as a basis for $\Re ^{3}$. Following \cite{ChappellB,Chappell3Player}, we can then form the bivectors $ e_i e_j $, which are non-commuting for $ i \ne j $, with $ e_i e_j = - e_j e_i $ but if $ i = j $ we have  $ e_i^2 = e_i e_i = 1 $.
We also have the trivector
\begin{equation}
\iota ={e} _{1}{e} _{2}{e} _{3},
\end{equation}%
finding $\iota ^{2}={e} _{1}{e} _{2}{e}
_{3}{e} _{1}{e} _{2}{e} _{3}=-1$ and furthermore, that $\iota $
commutes with each vector ${e} _{i}$, thus acting in a similar fashion to the unit imaginary $ \sqrt{-1} $.
We have ${e} _{1}{e} _{2}={e} _{1}{e} _{2}{e} _{3}{e}
_{3}=\iota {e} _{3}$ and so ${e} _{i}{e} _{j}=\iota {e} _{k}$
for cyclic $i,j,k$.  We can therefore summarize the algebra of the
basis elements $\{{e} _{i}\}$ by the relation
\begin{equation}
{e} _{i}{e} _{j}=\delta _{ij}+\iota \epsilon _{ijk}{e} _{k},
\label{eq:pauliGA}
\end{equation}%
which is isomorphic to  the algebra of the Pauli matrices \citep{Doran2003}, but now defined as part of $\Re ^{3}$.

In order to express quantum states in GA we use the one-to-one mapping \citep{Doran2003,Dorst:2002} defined as follows
\begin{equation}
|\psi \rangle =\alpha |0\rangle +\beta |1\rangle =%
\begin{bmatrix}
a_{0}+\mathrm{i}a_{3} \\
-a_{2}+\mathrm{i}a_{1}%
\end{bmatrix}%
\leftrightarrow \psi =a_{0}+a_{1}\iota {e} _{1}+a_{2}\iota {e}
_{2}+a_{3}\iota {e} _{3},  \label{eq:spinorMapping}
\end{equation}%
where $a_{i}$ are real scalars and $ \mathrm{i} = \sqrt{-1} $.

\subsection{\label{sharedStates} Symmetrical $ N $ qubit states}

For $N$-player quantum games an entangled state of $ N $ qubits is prepared, which for fair games should be symmetric with regard to the interchange of the $N$ players, and it is assumed that all information about the state once prepared is known by the players.
Two types of entangled starting states can readily be identified which are symmetrical with respect to the $ N $ players.
The GHZ-type state
\begin{equation}
| {\rm{GHZ}} \rangle_N = \cos \frac{\gamma}{2} |00 \dots 0 \rangle  + \sin \frac{\gamma}{2} |11 \dots 1 \rangle,
\end{equation}
where we include an entanglement angle $ \gamma \in [-\frac{\pi}{2},\frac{\pi}{2}] $ and the $ W $-type state
\begin{equation}
| {\rm{W}} \rangle_N = \frac{1}{\sqrt{N}}\left (|1000 \dots 00 \rangle + |0100 \dots 00 \rangle + |0010 \dots 00 \rangle + \dots + |0000 \dots 01 \rangle \right ).
\end{equation}
To represent these in geometric algebra, we start with the mapping for a single qubit from Eq.~(\ref{eq:spinorMapping}), finding
\begin{equation}
|0\rangle  \longleftrightarrow  1 , \,\, |1\rangle  \longleftrightarrow  -\iota {e}_2 ,
\end{equation}
so that for the GHZ-type state in GA we have
\begin{equation}
\psi_{{\rm{GHZ}}_N} = \cos \frac{\gamma}{2} + (-)^N \sin \frac{\gamma}{2} \iota {e}_2^1 \iota {e}_2^2 \dots \iota {e}_2^N,
\end{equation}
where the superscript on each bivector indicates which particle space it refers to. Also for the W-type state we have in GA
\begin{equation} \label{eq:psiWGA}
\psi_{{\rm{W}}_N} = -\frac{1}{\sqrt{N}} \left (\iota {e}_2^1 + \iota {e}_2^2 + \dots + \iota {e}_2^N \right ).
\end{equation}

\subsection{\label{generalUnitary} Unitary operations and observables in GA}

General unitary operations on a single qubit in GA can be represented as
\begin{equation} \label{rotorDefn}
R(\theta_1,\theta_2,\theta_3) = {\rm{e}}^{-\theta_3 \iota {e}_3/2} {\rm{e}}^{-\theta_1 \iota {e}_2/2} {\rm{e}}^{-\theta_2 \iota {e}_3/2},
\end{equation}
which is the Euler angle form of a rotation that can completely explore the space of a single qubit, and is equivalent to a general local unitary transformation.
We define $ U^i = R(\theta_1^i,\theta_2^i,\theta_3^i) $ for a general unitary transformation acting locally on each qubit $ i $, which the supervisor applies to the individual qubits that gives the starting state
\begin{equation} \label{eq:EisertFinal}
\left (U^1 \otimes U^2 \otimes \dots \otimes U^N \right ) |\psi \rangle,
\end{equation}
upon which the players now decide upon their measurement directions.

The overlap probability between two states $ \psi $ and $ \phi $, in the $N$-particle case \citep{Doran2003}, is
\begin{equation} \label{NParticleProbabilities}
P(\psi,\phi) = 2^{N-2} \langle \psi E \psi^{\dagger} \phi E \phi^{\dagger} \rangle_0 - 2^{N-2} \langle \psi J \psi^{\dagger} \phi J \phi^{\dagger} \rangle_0 ,
\end{equation}
where the angle bracket $ \langle \,\, \rangle_0 $ indicates that we retain only the scalar part of the product, and where
\begin{eqnarray} \label{EObservable}
E & = & \prod_{b=2}^N \frac{1}{2}(1 - \iota {e}_3^1 \iota {e}_3^b) =  \frac{1}{2^{N-1}}  \left (1 +  \sum_{r=1}^{\lfloor \frac{N}{2}\rfloor}  (-)^r C_{2r}^N (\iota {e}_3^i) \right ), \\ \nonumber
\end{eqnarray}
where $ \lfloor x \rfloor $ returns the nearest integer less than or equal to a given number $ x $, and where we define $ C_r^N(\iota {e}_3^i) $ to represent all possible combinations of $ N $ items taken $ r $ at a time, acting on the object inside the bracket. For example $ C_2^3(\iota {e}_3^i) = \iota {e}_3^1 \iota {e}_3^2 + \iota {e}_3^1 \iota {e}_3^3 + \iota {e}_3^2 \iota {e}_3^3 $. The number of terms produced being given by the standard combinatorial formula $ C_r^N = \frac{N!}{r! (N-r)!} $.

We also have
\begin{equation}
J = E \iota {e}_3^1 = \frac{1}{2^{N-1}}  \sum_{r=1}^{\lfloor  \frac{N+1}{2} \rfloor }  (-)^{r+1} C_{2r-1}^N (\iota {e}_3^i),
\end{equation}
where for simplicity, we initially assume that $ N $ is odd, which simplifies our derivation, and our results can easily be generalized later for all $ N $.

The supervisor now submits each qubit for measurement, through $ N $ Stern-Gerlach type detectors, with each detector being set at one of the two angles chosen by each player.  As mentioned, each player's choice, is a classical choice between two possible measurement directions, and hence each player's strategy set remains the same as in the classical game, with the quantum outcomes arising solely from the shared quantum state.

In order to calculate the measurement outcomes, we define a separable state $ \phi = A_1 A_2 \dots A_N $, to represent the players directions of measurement, where $ A_i $ is a rotor defined in Eq.~(\ref{rotorDefn}), with probabilistic outcomes calculated according to Eq.~(\ref{NParticleProbabilities}). The use of Eq.~(\ref{NParticleProbabilities}) gives the projection of the state $ \psi $ onto $ \phi $, and thus returns identical quantum mechanical probabilities conventionally calculated using the projection postulate of quantum mechanics. The set of $ | 0 \rangle $ and $ | 1 \rangle $ outcomes obtained from the measurement of each of the $ N $ qubits gives a reward to each player $ p $ according to a payoff matrix $ G^p $.  The expected payoff for each player then calculated from
\begin{equation} \label{eq:AlicePayoff}
\Pi_p = \sum_{i^1, \dots ,i^N=0}^1 G_{i^1 \dots i^N}^p P_{i^1 \dots i^N} = f(P_{i^1 \dots i^N}) ,
\end{equation}
where $ P_{i^1 \dots i^N} $ is the probability of recording the state $ |i^1 \rangle |i^2 \rangle \dots |i^N\rangle$ upon measurement, where $ i^1,\dots ,i^N \in \{0,1\} $, and $ G_{i^1 \dots i^N}^p $ is the payoff for this measured state. For large $ N $ it is preferable to calculate the payoff as some function $ f $ of the measured states, to avoid the need for large $ N \times N $ payoff matrices, as developed in Section E.2.

\subsection{GHZ-type state}

Firstly, we calculate the probability distribution of measurement outcomes from Eq.~(\ref{NParticleProbabilities}), from which we then calculate player payoffs from Eq.~(\ref{eq:AlicePayoff}).
For the GHZ-type state we have the first observable given by Eq.~(\ref{EObservable}) producing
\begin{eqnarray} \label{EObservableCalc}
\psi E \psi^{\dagger}
& = & \frac{1}{2^{N-1}} \left ( \prod_{i=1}^N U^i \right ) \left (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N (\iota {e}_3^i) \right ) \left ( \prod_{i=1}^N U^{i^{\dagger}} \right ) \\ \nonumber
& = & \frac{1}{2^{N-1}} \left (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N (V_3^i) \right ), \nonumber
\end{eqnarray}
where we define $ V_k^j = \iota U^j {e}_k U^{j^{\dagger}} $, and
\begin{equation} \label{JObservableCalc}
\psi J \psi^{\dagger} = \frac{1}{2^{N-1}} \cos \gamma \sum_{r=1}^{\lfloor \frac{N+1}{2} \rfloor }  (-)^{r+1} C_{2r-1}^N (V_3^i)  - \sin \gamma \Big (\sum_{r=0}^{\lfloor N/2 \rfloor} (-)^{r+\frac{N-1}{2} } C_{2r}^N ( V_2^i V_2^j) V_1^k \dots V_1^N \Big ).
\end{equation}
For the measurement settings with a separable wave function $ \phi = \prod_i A^i $, we deduce the observables by setting $ \gamma = 0 $ in Eq.~(\ref{EObservableCalc}) and Eq.~(\ref{JObservableCalc}) to be
\begin{eqnarray}
\phi J \phi^{\dagger} & = & \frac{1}{2^{N-1}}  \sum_{r=1}^{\lfloor \frac{N+1}{2} \rfloor }  (-)^{r+1} C_{2r-1}^N (M_3^i) \\ \nonumber
\phi E \phi^{\dagger} & = & \frac{1}{2^{N-1}} \left (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N (M_3^i) \right ), \nonumber
\end{eqnarray}
where $ M_k^j = \iota A^j {e}_k A^{j^{\dagger}} $.
For $ A^j = e^{-\iota \kappa {e}_2^j/2} $ that allows a rotation of the detectors by an angle $ \kappa $, we find
\begin{eqnarray} \label{MeasurementObs}
\phi J \phi^{\dagger} & = & \frac{1}{2^{N-1}}  \sum_{r=1}^{\lfloor \frac{N+1}{2} \rfloor }  (-)^{r+1} C_{2r-1}^N \left ( \iota {e}_3^i e^{\iota \kappa {e}_2^i} \right ) \\ \nonumber
\phi E \phi^{\dagger} & = & \frac{1}{2^{N-1}} \left (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N \left ( \iota {e}_3^i e^{\iota \kappa {e}_2^i} \right )  \right ).  \nonumber
\end{eqnarray}
It should be noted in Eq.~(\ref{MeasurementObs}) that we have defined the measurement angles with a simplified rotor, $ e^{-\iota \kappa {e}_2^i/2} $, and we assume no loss of generality, which is in accordance with the known result \citep{Peres} that Bell's inequalities
can still be maximally violated when the allowed directions of measurement are located in a single plane, as opposed to being
defined in three dimensions.

So, referring to Eq.~(\ref{NParticleProbabilities}), we find, through combining Eq.~(\ref{EObservableCalc}) and Eq.~(\ref{MeasurementObs})
\begin{eqnarray} \label{eq:Method1MeasureE}
2^{N-2} \langle \psi E \psi^{\dagger} \phi E \phi^{\dagger} \rangle_0 & = & \frac{1}{2^N} \Big \langle \Big (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N (V_3^i) \Big )  \Big (1 +  \sum_{r=1}^{\lfloor \frac{N}{2} \rfloor }  (-)^r C_{2r}^N (  \iota {e}_3^i e^{\iota \kappa {e}_2^i}) \Big ) \Big  \rangle_0 \\ \nonumber
& = & \frac{1}{2^N} \left (1+\sum_{r=1}^{\lfloor \frac{N}{2} \rfloor } C_{2r}^N (K^i ) \right ),  \nonumber
\end{eqnarray}
where $ K^i = V_3^i \iota e_3^i {\rm{e}}^{\iota \kappa e_2^i } = \cos \kappa^i \cos \alpha_1^i + \sin \kappa^i \sin \alpha_1^i \cos \alpha_3^i $, using the standard results listed in Appendix~A.
The cross terms in the expansion of the brackets in Eq.~(\ref{eq:Method1MeasureE}), do not contribute because we only  retain the scalar components in this expression.
We also have for the second part of Eq.~(\ref{NParticleProbabilities}), through combining Eq.~(\ref{JObservableCalc}) and Eq.~(\ref{MeasurementObs})
\begin{equation} \label{eq:Method1MeasureJ}
 - 2^{N-2} \langle \psi J \psi^{\dagger} \phi J \phi^{\dagger} \rangle_0 = \frac{1}{2^N} \Big (\cos \gamma \sum_{r=1}^{\lfloor \frac{N+1}{2} \rfloor }  C_{2r-1}^N (K^{i} )  +  \sin \gamma \Omega \Big ),
\end{equation}
where we define
\begin{eqnarray} \label{Xidefn}
\Omega & = &  \sum_{r=0}^{\lfloor N/2 \rfloor} (-)^{r} C_{2r}^N ( X_2^i X_2^j) X_1^k \dots X_1^N \\ \nonumber
X_1^i & = & V_1^i \iota e_3^i {\rm{e}}^{\iota \kappa e_2^i } = \left (-\sin \kappa (\cos \alpha
_{1}\cos \alpha _{2}\cos \alpha _{3}-\sin \alpha _{2}\sin \alpha _{3})+\sin
\alpha _{1}\cos \alpha _{2}\cos \kappa \right )^i  \\ \nonumber
X_2^i & = & V_2^i \iota e_3^i {\rm{e}}^{\iota \kappa e_2^i } =  \left ( \sin \kappa (\cos \alpha
_{2}\sin \alpha _{3}+\sin \alpha _{2}\cos \alpha _{3}\cos \alpha _{1})-\sin
\alpha _{1}\sin \alpha _{2}\cos \kappa \right )^i ,
\end{eqnarray}
also referring to Appendix A.

\subsubsection{Probability amplitudes for $N$ qubit state, general measurement directions}

So combining our last two results from Eq.~(\ref{eq:Method1MeasureE}) and Eq.~(\ref{eq:Method1MeasureJ}) using Eq.~(\ref{NParticleProbabilities}), we find the probability to find any outcome after measurement, which can be shown to be valid for all $ N $ not just $ N $ odd as initially assumed, is
\begin{equation} \label{eq:finalDensityGeneral}
P_{k^1 \dots k^N}  =  \frac{1}{2^N} \Big (1+\sum_{r=1}^{\lfloor \frac{N}{2} \rfloor } C_{2r}^N ( \epsilon^{i} K^{i} ) + \cos \gamma \sum_{r=1}^{ \lfloor \frac{N+1}{2} \rfloor } C_{2r-1}^N (\epsilon^{i} K^{i} )   + \epsilon^{1 \dots N}  \Omega \sin \gamma  \Big ) ,
\end{equation}
where we have included $ \epsilon^i = (-)^{k^i} \in \{+1,-1\} $, to select the probability to measure spin-up or spin-down on a given qubit.

If we take $ \gamma = 0 $, describing the classical limit, we have from Eq.~(\ref{eq:finalDensityGeneral})
\begin{eqnarray}
P_{k^1 \dots k^N} & = & \frac{1}{2^N} \left (1+\sum_{r=1}^{\left\lfloor N/2 \right \rfloor} C_{2r}^N (\epsilon^{i} K^{i} ) +  \sum_{r=1}^{\left\lfloor (N+1)/2 \right \rfloor} C_{2r-1}^N (\epsilon^{i} K^i ) \right ) \\ \nonumber
 & = & \frac{1}{2^N} \left (1+\sum_{r=1}^{ N} C_{r}^N ( \epsilon^{i} K^i) \right )  \\ \nonumber
 & = & \frac{1}{2^N} (1 + \epsilon^{1} K^1)(1 + \epsilon^{2} K^2) \dots (1 +\epsilon^{N} K^N), \nonumber
\end{eqnarray}
which shows that for zero entanglement we can form a product state as expected.
Alternatively  with general entanglement, but only for operations on the first two qubits, we have
\begin{equation} \label{eq:twoQubits}
P_{k^i k^j}  = \frac{1}{8} \left (1 + \epsilon^{k} \cos \gamma \right ) \left (1+ \sum_{r=2}^{ N} C_{r}^N ( \epsilon^{i}) \right ) \left (1 + \epsilon^{ik} K^i)(1 + \epsilon^{jk} K^j \right ),
\end{equation}
which shows that for the GHZ-type entanglement that each pair of qubits is mutually un-entangled, a well-known result for GHZ-type states.

\subsubsection{Player payoffs}

In general, to represent the permutation of signs introduced by the measurement operator we can define for the first player, say Alice,
\begin{equation} \label{aDefn}
a^{i^1 \dots i^N} = \frac{1}{2^N} \sum_{j^1 \dots j^N=0}^1 \epsilon^{i^1 \dots i^N} G_{j^1 \dots j^N}^1,
\end{equation}
so for example, $ a^{0 \dots 0}  =  \frac{1}{2^N} \sum_{j^1 \dots j^N=0}^1 G_{j^1 \dots j^N}^1 $, and we adopt the notation $ a^i a^j = a^{ij} $ etc., i.e.~we write $ a^{0 \dots 1 \dots 0} $ with a 1 in the $ i $th position as $ a^i $.

Using the payoff function we find for Alice
\begin{equation}
\Pi_A(\kappa_j^i) = a^{0 \dots 0}+ \sum_{r=1}^{\left\lfloor N/2 \right \rfloor} C_{2r}^N (a^{i} K^i)  + \cos \gamma \sum_{r=1}^{\left\lfloor (N+1)/2 \right \rfloor} C_{2r-1}^N( a^{i} K^i) + a^{k^1 \dots k^N} \Omega  \sin \gamma
\end{equation}
and similarly for the second player, say Bob, where we would use Bob's payoff matrix in place of Alice's.

\subsubsection{Mixed-strategy payoff relations}

For a mixed strategy game, players choose their first
measurement direction $\kappa _{1}^{i},$  with probabilities $x^i $, where $x^i \in
\lbrack 0,1]$ and hence choose the direction $\kappa _{2}^{i} $ with probabilities $(1-x^i)$,
respectively. Then Alice's payoff is now given as
\begin{eqnarray}
&&\Pi _{A}(x^1,x^2, \dots,x^N)  \notag \\
&=&x^1 \dots x^N \sum_{i,j,k=0}^{1}P_{i^1 \dots i^N}(\kappa _{1}^{1},\kappa _{1}^{2},\dots ,\kappa
_{1}^{3})G_{i^1 \dots i^N} \\ \nonumber
&+ & \dots + x^1 (1-x^2) \dots x^N \sum_{i,j,k=0}^{1}P_{i^1 \dots i^N}(\kappa _{1}^{1},\kappa
_{2}^{2}, \dots ,\kappa _{1}^{3})G_{i^1 \dots i^N}  \notag \\
&+& \dots + (1-x^1)(1-x^2) x^3 \dots x^N \sum_{i,j,k=0}^{1}P_{i^1 \dots i^N}(\kappa _{2}^{1},\kappa
_{2}^{2},\kappa
_{1}^{3}, \dots , \kappa _{1}^{N})G_{i^1 \dots i^N}  \notag \\
& + & \dots +  (1-x^1)(1-x^2)(1-x^3) \dots (1-x^N) \sum_{i,j,k=0}^{1}P_{i^1 \dots i^N}(\kappa
_{2}^{1},\kappa _{2}^{2},\kappa _{2}^{3}, \dots , \kappa_2^N)G_{i^1 \dots i^N}.  \notag \\
&&  \label{eq:AlicePayoffFourCoin}
\end{eqnarray}

\subsection{Embedding the classical game}

If we consider a strategy $N$-tuple $(x^1,x^2,x^3,\dots,x^N)=(0,1,0, \dots 0)$ for example, at zero
entanglement, then the payoff for Alice is obtained from Eq.~(\ref{eq:AlicePayoffFourCoin}) to be
\begin{eqnarray}
\Pi _{A}(x^1,\dots ,x^N) &=&\frac{1}{2^N}%
[G_{000 \dots 0}(1+K_{2}^1)(1+K_{1}^2)(1+K_{2}^3) \dots (1+K_{2}^N) \\ \nonumber
&  & + G_{100 \dots 0}(1-K_{2}^1)(1+K_{1}^2)(1+K_{2}^3) \dots (1+K_{2}^N)
\notag \\
&&+G_{010 \dots 0}(1+K_{2}^1)(1-K_{1}^2)(1+K_{2}^3) \dots (1+K_{2}^N) \\ \nonumber
& & +G_{110 \dots 0}(1-K_{2}^1)(1-K_{1}^2)(1+K_2^3) \dots (1+K_{2}^N)
\notag \\
& & + \dots +G_{111 \dots 1}(1-K_{2}^1)(1-K_{1}^2)(1-K_{2}^3) \dots (1-K_{2}^N)].
\end{eqnarray}%
Hence, in order to achieve the classical payoff of $G_{101 \dots 1}$, we can see that we require $K_{2}^1=-1$, $K_{1}^2=+1$ and $K_{2}^3 \dots K_{2}^N=-1$.

This shows that we can select any required classical payoff by the
appropriate selection of $K_{j}^i =\pm 1$. We therefore have the conditions for obtaining
the classical mixed-strategy payoff relations as
\begin{equation}
K_{j}^i = \cos \alpha _{1}^i \cos \kappa _{j}^{i}+\sin \alpha_{1}^i \cos \alpha
_{3}^i \sin \kappa_{j}^{i}=\pm 1.
\end{equation}

We find two classes of solution: If $\alpha
_{3}^i \neq 0$, then for the equations satisfying $K_{2}^i=-1$ we
have for Alice in the first equation $\alpha_{1}^i =0$, $\kappa _{2}^{i}=\pi $
or $\alpha _{1}^i=\pi $, $\kappa _{2}^{i}=0$ and for the equations satisfying $%
K_{1}^i=+1$ we have $\alpha_{1}^i =\kappa_{1}^{i}=0$ or $\alpha
_{1}^i=\kappa_{1}^{i}=\pi $, which can be combined to give either $\alpha
_{1}^i=0$, $\kappa _{1}^{i}=0$ and $\kappa _{2}^{i}=\pi $ or $\alpha _{1}^i=\pi $%
, $\kappa _{1}^{i}=\pi $ and $\kappa _{2}^{i}=0$. For the second class with $%
\alpha _{3}=0$ we have the solution $\alpha _{1}^i-\kappa _{2}^{i}=\pi $ and
for $K_{1}^i=+1$ we have $\alpha _{1}^i-\kappa _{2}^{i}=0$.

So in summary, for both cases we can deduce that the two measurement directions are
$\pi $ out of phase with each other, and for the first case ($\alpha
_{3}^i \neq 0$) we can freely vary $\alpha _{2}^i $ and $\alpha _{3}^i $, and for the
second case ($\alpha _{3}^i=0$), we can freely vary $\alpha_{1}^i $ and $\alpha
_{2}^i $ to change the initial quantum quantum state without affecting the game
Nash equilibrium~(NE) or payoffs \citep{Rasmusen,Binmore}. These results can be shown to imply in both cases that  $ \Omega = 0 $.

The associated payoff for Alice therefore becomes
\begin{eqnarray}
\Pi _{A}(x^1,x^2,\dots x^N) & = & a_{00 \dots 0}  -\cos \gamma  \sum_{r=1}^{\left\lfloor (N+1)/2 \right \rfloor} C_{2r-1}^N[ a^{i0} (1-2 x^i ) +  a^{0i} ( 1 -2 x^i) ]
\\ \nonumber
&&+  \sum_{r=1}^{\left\lfloor N/2 \right \rfloor}  C_{2r}^N [a^{1i} (1-2x^1)(1- 2 x^i)+ a^{0ij}(1-2 x^i)(1-2x^j)] .
\label{eq:AlicePayoffQuantumGameClassical}
\end{eqnarray}%

For example, for three players this will reduce to
\begin{eqnarray} \label{eq:AlicePayoffQuantumGameClassicalThreePlayer}
& & \Pi _{A}(x^1,x^2,x^3) \\ \nonumber
&=& a_{000} + a_{011} (1-2 x^2)(1-2 x^3)+ a_{110}(1-2x^1)((1- 2 x^2)+ (1-2x^3)) \notag \\
&-&\cos \gamma \left ( a_{111}(1-2x^1)(1-2x^2)(1-2x^3) + a_{100} (1 - 2 x^1 ) + a_{001}(2 - 2 x^2 -2 x^3)  \right ) , \nonumber
\end{eqnarray}%
in agreement with previous results for three-player games \citep{Chappell3Player}.
Now, we can write the equations governing the NE for the first player as
\begin{eqnarray} \label{eq:NEEPREmbeddedStagHuntThree}
&&\Pi _{A}(x^{i \ast },x^{2 \ast }, \dots x^{N \ast })-\Pi _{A}(x^i,x^{2 \ast },\dots ,x^{N \ast })
\notag \\
&=&  ( x^{1*} - x^1 ) \left ( - \sum_{r=1}^{\left\lfloor N/2 \right \rfloor}  C_{2r}^N ( a^{1i} I^1 (1- 2 x^{i*})) +\cos \gamma  \sum_{r=1}^{\left\lfloor (N+1)/2 \right \rfloor} C_{2r-1}^N ( a^{i0} I^1 (1- 2 x^{i*}) ) \right ) \geq 0 . \notag
\end{eqnarray}%
We are using $ I^1 $ as a placeholder, which has a value one, but ensures that the correct number of terms are formed from $ C_r^N () $.
For example, for three players we find the NE governed by
\begin{eqnarray} \label{eq:NEEPREmbeddedStagHuntThreePlayer}
&&\Pi _{A}(x^{1\ast },x^{2\ast },x^{3\ast })-\Pi _{A}(x^1,x^{2\ast },x^{3\ast })
 \\
&=&(x^{1\ast }-x^1)[a_{110}(2x^{2 \ast }-1)+a_{101}(2x^{3 \ast }-1)+\cos \gamma
\{a_{100}+a_{111}(2x^{2\ast }-1)(2x^{3 \ast }-1)\}]\geq 0 , \notag
\end{eqnarray}%
in agreement with previous results \citep{Chappell3Player}.

\subsubsection{Symmetric game}

For a symmetric game we have $ a^{1 \dots 1} = b^{1 \dots 1} = \rm{etc} $, $ a^{0 \dots 0} =b^{0 \dots 0} = \rm{etc} $ and $ a_{11000 \dots 0} = a_{10100 \dots 0} = a_{10010 \dots 0} = \dots $, and similarly for other symmetries, and using these conditions for a symmetric game, we can find the NE for other players, such as Bob, from the constraint
\begin{eqnarray} \label{eq:NEEPREmbeddedThreeReduced}
&&\Pi _{A}(x^{i \ast },x^{2 \ast }, \dots x^{N \ast })-\Pi _{A}(x^{i*},x^{2 },\dots ,x^{N \ast })
 \\
&=&  ( x^{2*} - x^2 ) \left (-\sum_{r=1}^{\left\lfloor N/2 \right \rfloor}  C_{2r}^N ( a^{1i} I^2 (1- 2 x^{i*})) +\cos \gamma  \sum_{r=1}^{\left\lfloor (N+1)/2 \right \rfloor} C_{2r-1}^N ( a^{i0} I^2 (1- 2 x^{i*}) ) \right ) \geq 0 .  \notag
\end{eqnarray}
We can see that the new quantum behavior is governed solely by the payoff
matrix and by the entanglement
angle $\gamma $, and not by other properties of the quantum state.

\subsubsection{Linear payoff relations}

We can see that as $ N \rightarrow \infty $, that we need to define an infinite number of components of the payoff matrix as shown by Eq.~(\ref{aDefn}).  Hence in order to proceed to solve specific games for large $ N $, we need to write the payoff matrix as some functional form of the measurement outcomes, as shown in Eq.~(\ref{eq:AlicePayoff}).
The simplest approach is to define linear functions over the set of player choices, as developed in \cite{FlitneyHollenberg2006}, defining the following general payoff function
\begin{eqnarray} \label{eq:nonFlippingPlayerEqn}
\$_0 & = &  a n + b , \,\,\,\, \$_1  =  c n + d ,
\end{eqnarray}
where $ \$_0 $ is the payoff for players which choose their first measurement direction and $ \$_1 $ is the payoff for the players which choose their second measurement direction, and  where $ n $ is the number of players choosing their first direction and $ a,b,c,d \in \Re $.

This approach enables us to simply define various common games.  For example the prisoner dilemma (PD), which has the essential feature that a defecting player achieves a higher payoff, is represented if we have $ c \ge a $, $ d > a + b $ and $ a>0 $.  These conditions ensure that if a cooperating player decides to defect, then his payoff rises as determined by Eq.~(\ref{eq:nonFlippingPlayerEqn}).
For example for $ a =3, b=-3,c=4,d=1 $ we have defined an $ N $ player PD, and for $ N =2 $ we find
\begin{equation} \label{StdPDPayoffMatrix}
G_{ij}^A =\begin{bmatrix} 3 & 0 \\
5 & 1 \end{bmatrix} ,
\end{equation}
which gives us the typical payoff matrix for two-player PD game.  In the EPR setting for the quantum game, a cooperating player is defined as the player who chooses their first measurement direction and a defecting player as one who chooses their second measurement direction.

For the Chicken game (also called the hawk-dove game) \cite{Rasmusen}, which involves the situation where the player that does not yield to the other is rewarded, but if neither player yields then they are both severely penalized, in this case we require $ c \ge a $ , $ d < a + b $ and $ a > 0 $ and for the minority game, an implementation would be $ c = -a $, $ a < 0 $ and $ d = b + a N $ which rewards a minority choice and punishes a majority one.
Hence we are led to define
\begin{equation} \label{defnp1p2}
 p_1 = d-(a+b)  , \,\,\,\ p_2 = c - a ,
\end{equation}
as two key determinants of quantum games, and we will find that the NE is indeed a function of $p_1 $ and $ p_2 $ alone, see Eq.~(\ref{eq:NEConditionsGHZ}).  With this definition the PD game is selected if $p_1>0$ and $ p_2 \ge 0$ and the minority game with $ p_1<0 $ and $ p_2 > 0$ for example.

It should be noted that while the definition in Eq.~(\ref{eq:nonFlippingPlayerEqn}) can generally define an infinite set of PD games through simply putting conditions on $ p_1 $ and $ p_2 $, it is still only a subset of the space of all possible PD games defined over $ N \times N $ payoff matrices.

Using the linear functions defined in Eq.~(\ref{eq:nonFlippingPlayerEqn}) we find
\begin{eqnarray} \label{aseriesCalc}
a^{0 \dots 0} & = & \frac{1}{4} ( N(c+a)-p_2 + 2(b+d)) \\ \nonumber
a^{10 \dots 0} & = & -\frac{1}{4} ( (N-1)(c-a) + 2(d-(a+b))) = -\frac{1}{4} \left ( (N-1)p_2 + 2 p_1 \right )\\ \nonumber
a^{110 \dots 0} & = & -\frac{c-a}{4} = -\frac{p_2}{4} \\ \nonumber
a^{1110 \dots 0} , a^{11110 \dots 0} \dots & = & 0  \nonumber
\end{eqnarray}
and
\begin{eqnarray} \label{aseriesCalc0Type}
a^{010 \dots 0} & = & \frac{c+a}{4} \\ \nonumber
a^{011 \dots 0} , a^{0111 \dots 0} , \dots & = & 0.  \nonumber
\end{eqnarray}
If required, Eq.~(\ref{eq:nonFlippingPlayerEqn}) can be extended with quadratic terms in $ n $ to allow a greater variety of PD games to be defined, and we find that if this is done that one extra term is added to the series in Eq.~(\ref{aseriesCalc}) and Eq.~(\ref{aseriesCalc0Type}).

\subsubsection{NE and payoff for linear payoff relations}

We can see that the series in Eq.~(\ref{aseriesCalc}) and Eq.~(\ref{aseriesCalc0Type}) terminates, which thus allows us to simplify the NE conditions, for the first player to
\begin{equation} \label{eq:NEConditionsGHZ}
  ( x^{1*} - x^1 ) \left (p_2 \sum_{i=2}^N (1- 2 x^{i*}) -\cos \gamma  \left ( (N-1) p_2+ 2 p_1 \right )  \right ) \geq 0
\end{equation}
and similarly for the other $ N-1$ players, which thus determines the available NE for all games, defined as linear functions, in terms of the two parameters $ p_1 $ and $ p_2 $.

The payoff can then also be simplified for the first player to
\begin{eqnarray} \label{payoffFirstPlayer}
\Pi_A  & = & \frac{1}{4} \Big ( 2(b+d) -p_2 + (c+a) \big (N - \cos \gamma \sum_{i=2}^N (1- 2 x^{i}) \big )  \\ \nonumber
&+ &(1-2 x^1) \big (\cos \gamma  ((N-1)p_2 +2 p_1) - p_2 \sum_{i=2}^N (1- 2 x^{i}) \big ) \Big ) . \nonumber
\end{eqnarray}
For the minority game defined previously, we find $ (N-1) p_2+ 2 p_1 = 0 $, which gives an interesting result for this game that both the NE and the payoff are unaffected by the entanglement of the state.

\subsubsection{Prisoner dilemma (PD)}

For the PD, having $p_2 \ge 0$ and $p_1>0$, and we find from the equation for Nash equilibrium in Eq.~(\ref{eq:NEConditionsGHZ}) that in order to produce the classical outcome we require $ \sum_{i=2}^N (1- 2 x^{i*}) < \cos \gamma ( N - 1 + 2 p_1/p_2 ) $ which thus requires $  \cos \gamma > \frac{N-1}{N-1+2 p_1/p_2} $  and hence the phase transitions, in terms of $ \cos \gamma $, are given by
\begin{equation}  \label{eq:PDTransitions}
 \frac{N-1-2n}{N-1+\delta} < \cos \gamma < \frac{N+1-2n}{N-1+\delta} = \lambda_n ,
\end{equation}
where $ \delta  = \frac{2 p_1 }{p_2}  $, and with the PD $ \delta \in (0,\infty) $, and hence the above inequality will hold for $ N \ge 2 $.
So in summary, at the classical limit we have all players defecting, and then we have the transition to the non-classical region at $ \lambda_1 $ and we then have equally spaced transitions as entanglement increases down to maximum entanglement where we have the number of players cooperating $ n = \lfloor N/2 \rfloor $.  That is, we always have the same number of transitions for a given number of players, but they concertina closer together as the first transition $ \lambda_1 $, moves towards zero, through changing the game parameters, $ p_1 $ and $ p_2 $.

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\begin{figure}[htb]
\begin{picture}(75,80)
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\put(11,50){\line(1,-1){6}}
\put(29,35){\line(3,-1){8}}
\put(29,37){\line(1,-1){8}}
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\thicklines
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\put(2,56){Payoff} \put(64,2){$ \cos \gamma $}
\put(9,2){$ 0 $} \put(58,2){$ 1 $}
\put(3,11.5){$ 1 $} \put(44,2){$ \lambda_1 $} \put(36,2){$ \lambda_2 $}  \put(28,2){$ \lambda_3 $}
\put(13,2){$ \lambda_{\lfloor N/2 \rfloor} $} \put(24,2){$ ... $}
\put(10,51){ $ \lfloor N/2 \rfloor$} \put(30,36){n=2}\put(38,30){n=1}\put(48,17){n=0, all players defect}
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\end{picture}
\caption{Phase structure for $N $-player Prisoner dilemma.  For $ \cos \gamma > \lambda_1 $ we identify the classical regime, where all players defect, and as entanglement increases we find an increasing number of players cooperating, up to $ \left \lfloor N/2 \right \rfloor $ near maximum entanglement.  The left and right hand edges of the boundaries each form an inverted parabola in $ \cos \gamma $ given by Eq.~(\ref{PDExampleParabolic}). \label{GA3Space}}
\end{figure}

The maximum payoff, close to maximum entanglement, can be found from Eq.~(\ref{payoffFirstPlayer}) as
\begin{eqnarray}
\Pi_A^c & = & \frac{1}{4}(2(b+d)+(c+a)N+(c-a)_{N \in {\rm{Odd}}}) \\ \nonumber
\Pi_A^d & = & \frac{1}{4}(2(b+d)+(c+a)N-(c-a)_{N \in {\rm{Odd}}}) , \nonumber
\end{eqnarray}
where the final $ (c-a) $ term only occurs for odd $ N $.  So for $ N $ even the payoffs are equal, but for odd $ N $, the cooperating player receives a higher or equal payoff to the defecting player.
The payoff rises linearly with $ N $, whereas without entanglement, we have the payoff fixed at $d$ units from Eq.~(\ref{eq:nonFlippingPlayerEqn}).

\subsubsection{The conventional prisoner dilemma (PD) game for all $ N $}

For the special case with the PD settings shown in Eq.~(\ref{StdPDPayoffMatrix}), which gives the conventional PD game for two players, we find from Eq.~(\ref{defnp1p2}), $ p_1 = 1 $ and $ p_2 = 1 $, and so we can then simplify the general NE conditions in Eq.~(\ref{eq:NEConditionsGHZ}), for the first player to
\begin{equation} \label{eq:NEConditionsGHZPDExample}
  ( x^{1*} - x^1 ) \left (\sum_{i=2}^N (1- 2 x^{i*}) -(N+1) \cos \gamma \right ) \geq 0
\end{equation}
and similarly for the other $ N-1$ players.
The left and right edges of each NE zone, shown in Fig.~\ref{GA3Space}, can now be written from Eq.~(\ref{eq:PDTransitions}) as
\begin{equation}  \label{eq:PDTransitionsExample}
 \frac{N-1-2n}{N+1} < \cos \gamma < \frac{N+1-2n}{N+1} .
\end{equation}
In each zone we find the payoff for cooperation and defection, from Eq.~(\ref{payoffFirstPlayer}), now given by
\begin{eqnarray} \label{payoffPlayerExample}
\Pi^c & = & \frac{1}{2} \left ( 4 N -2 - n - ( 4 + 4 N -7 n) \cos \gamma  \right ) \\  \nonumber
\Pi^d & = & \frac{1}{2} \left ( 3 N -2 + n + ( 4 - 3 N+ 7 n) \cos \gamma  \right ) , \nonumber
\end{eqnarray}
which defines the payoff diagram for an $ N $ player PD, and which produces the classical PD at $ N =2 $ at zero entanglement.

At each left hand boundary, for the defecting player, we have from Eq.~(\ref{eq:PDTransitionsExample}), $ \frac{N-1-2 n}{N+1} = \cos \gamma $ or $ n = \frac{1}{2} \left ( N -1 - (N+1) \cos \gamma \right ) $.  Substituting this into the defecting player payoff in Eq.~(\ref{payoffPlayerExample}), we find
\begin{equation} \label{PDExampleParabolic}
\Pi^d = -3 + \frac{7}{4} (N+1) (1 - \cos^2 \gamma) = -3 + \frac{7}{4} (N+1) \sin^2 \gamma ,
\end{equation}
for the defecting players' payoff.
We thus see that the payoff at each boundary follows a downwards parabolic curve in $ \cos \gamma $, if drawn on Fig.~\ref{GA3Space}.  If we allow $ N $ to increase without limit, then the boundaries would concertina infinitesimally close together, and in the limit as $ N \rightarrow \infty $, the payoff's would form a continuous downward parabolic curve in $ \cos \gamma $ given by Eq.~(\ref{PDExampleParabolic}).  The special case of the PD selected here with $ p_1 = 1 $ and $ p_2 = 1 $ forms a parabola, whereas for the general case of a PD game with $p_2 \ge 0$ and $p_1>0$ from Eq.~(\ref{defnp1p2}), we will produce a quadratic curve in $ \cos \gamma $ for the payoff.  We can also see that this will be a general feature for all games defined using linear functions as both the NE in Eq.~(\ref{eq:NEConditionsGHZ}) and the payoffs in Eq.~(\ref{payoffFirstPlayer}) are linear in $ \cos \gamma $, therefore typically producing a payoff diagram quadratic in $ \cos \gamma $.

We can also note that Eq.~(\ref{payoffPlayerExample}) indicates a different payoff for the defecting and cooperating player at the NE.  If a player decides to try to change their choice in order to improve their payoff, often a lower payoff will be the outcome, because overall the player's choices have now moved away from the NE.  This then illustrates the value of coalitions and in aligning one's choices with the coalition with the higher payoff \citep{IqbalToor3,FlitneyGreentree}.

\subsection{W entangled state}

Following the same procedure as used for the GHZ-type state, we find the probability distribution for the W-type state
\begin{equation} \label{eq:finalDensityW}
P_{k^1 \dots k^N} = \frac{1}{N 2^{N}}  ( N  + \sum_{r=1}^{N} (N-2r) C_{r}^N (\epsilon^{i} K^i) +2 \sum_{r=2}^{N} C_{r}^N (\epsilon^{i} \epsilon^{j} \epsilon^{k} ( X_2^i X_2^j + X_1^i X_1^j) K^k ) ).
\end{equation}
We can then find the payoff function for the first player, Alice
\begin{equation} \label{WtypePrisonerDilemma}
\Pi_A (\kappa^1,\dots ,\kappa^N) =  N a^{0 \dots 0} + \sum_{r=1}^{N} (N-2r) C_r^N (a^{i} K^i) +2 \sum_{r=2}^{N}  C_{r}^N  (a^{i j k} ( X_2^i X_2^j + X_1^i X_1^j) K^k )
\end{equation}
and similarly for other players.
However with the W-type state it is impossible to turn off the entanglement, and so it will not be possible to embed the classical game, as we have done with the GHZ-type state.  Hence we will not proceed any further except to show the result of maximizing the payoff function in Eq.~(\ref{WtypePrisonerDilemma}) for the PD.

\subsubsection{Prisoner Dilemma (PD)}

For the PD we can maximize the payoff function, and we find that we require all players to defect, for all $N$ and the resultant payoff for the first player Alice and hence all players is
\begin{equation}
\Pi_A = c+d - \frac{c+d-(a+b)}{N}.
\end{equation}
So as $ N \rightarrow \infty $, then the payoff approaches $ c + d $ from below.

\section{Conclusion}

Using Clifford's geometric algebra, the probability distribution is found for general measurement directions on a general $ N $ qubit entangled state, for the GHZ-type state shown in Eq.~(\ref{eq:finalDensityGeneral}) and for the W-type state shown in Eq.~(\ref{eq:finalDensityW}).

Linear functions parameterized by the number of players selecting their first measurement direction for an $ N $ player game are then defined as shown in Eq.~(\ref{eq:nonFlippingPlayerEqn}), from which games can then be easily defined for general $ N $. Using these linear functions, the Nash equilibrium and payoff relations are then determined for general $ N $ as shown in Eq.~(\ref{eq:NEConditionsGHZ}) and Eq.~(\ref{payoffFirstPlayer}) respectively.  We also find a general feature for these games of producing a payoff diagram with phase transition boundaries quadratic in $ \cos \gamma $, as shown in Fig.~\ref{GA3Space}.  If the linear functions are increased in order, then we would expect the payoff diagram to become a higher order polynomial in $ \cos \gamma $.

As a specific example the PD is solved for a general $ N $ and we find an interesting feature, that the payoffs at the Nash equilibrium are equal for the defecting and cooperating player only for even $N$ and also in the limit of large $ N $ the payoff rises linearly with $ N $ given by $ (c+a) N/4 $ for the GHZ-type state.

At maximum entanglement the payoff for the GHZ-type and W-type states for the PD become equal at $ N = 2 $, producing the formula from the parameters of the linear functions as
\begin{equation}
\Pi_{\rm{GHZ}} =  \Pi_{\rm{W}} = \frac{a+b+c+d}{2} .
\end{equation}
This equality is to be expected at $ N = 2 $, because these two states are equivalent under local operations.

In summary, we have produced a general quantum game environment, with the number of players $ N \ge 2 $, which will embed the classical game at zero entanglement, and using linear functions we determine the NE and player payoffs for general $ N $. These general results thus subsume previous analyses for two-player and three-player games in an EPR setting \citep{ChappellB,Chappell3Player}.

\newpage

redpajama_set_name: RedPajamaArXiv


This year marks one of the most influential elections in our nation's history. On the national stage it looks as if Hillary Clinton and Donald Trump will be fighting it out for the nation's top office. While this is a very important election, it's also important that we are voting in the local and state elections. While many think it may be hard – but not impossible – for Donald Trump to win the White House, Republicans can still come out on top if they win local and state elections. No matter what side of the aisle you are on, it's important for your voice to be heard. Here's what you need to know to ensure that your voice is heard in the 2016 presidential, state and local races.
Voter Registration Deadline :: The last day to register to vote in Virginia is Monday, October 17.
Still need to register? Download the proper registration forms and get details on how to return your voter registration document.
Unsure of you if you are registered? Check your voter registration status.
Virginia law requires all voters to provide an acceptable form of photo identification (photo ID) when voting in person at their polling place. There are many different types of photo IDs that voters can use when they vote. All of the acceptable forms of photo ID can be used up to a year after ID has expired. Find out which IDs are accepted for voting.
Learn more about absentee voting.
Find your Election Day polling place. Polls will be open 6:30 a.m. – 7:30 p.m. on Election Day.


text:
Will We Stop Trump Before It's Too Late
Fascism poses a more serious threat now than at any time since the end of World War II.
By MADELEINE ALBRIGHT
APRIL 6, 2018
On April 28, 1945 — 73 years ago — Italians hung the corpse of their former dictator Benito Mussolini upside down next to a gas station in Milan. Two days later, Adolf Hitler committed suicide in his bunker beneath the streets of war-ravaged Berlin. Fascism, it appeared, was dead.
To guard against a recurrence, the survivors of war and the Holocaust joined forces to create the United Nations, forge global financial institutions and — through the Universal Declaration of Human Rights — strengthen the rule of law. In 1989, the Berlin Wall came down and the honor roll of elected governments swelled not only in Central Europe, but also Latin America, Africa and Asia. Almost everywhere, it seemed, dictators were out and democrats were in. Freedom was ascendant.
Today, we are in a new era, testing whether the democratic banner can remain aloft amid terrorism, sectarian conflicts, vulnerable borders, rogue social media and the cynical schemes of ambitious men. The answer is not self-evident. We may be encouraged that most people in most countries still want to live freely and in peace, but there is no ignoring the storm clouds that have gathered. In fact, fascism — and the tendencies that lead toward fascism — pose a more serious threat now than at any time since the end of World War II.
Warning signs include the relentless grab for more authority by governing parties in Hungary, the Philippines, Poland and Turkey — all United States allies. The raw anger that feeds fascism is evident across the Atlantic in the growth of nativist movements opposed to the idea of a united Europe, including in Germany, where the right-wing Alternative für Deutschland has emerged as the principal opposition party. The danger of despotism is on display in the Russia of Vladimir Putin — invader of Ukraine, meddler in foreign democracies, accused political assassin, brazen liar and proud son of the K.G.B. Putin has just been re-elected to a new six-year term, while in Venezuela, Nicolás Maduro, a ruthless ideologue, is poised to triumph in sham balloting next month. In China, Xi Jinping has persuaded a docile National People's Congress to lift the constitutional limit on his tenure in power.Around the Mediterranean, the once bright promise of the Arab Spring has been betrayed by autocratic leaders, such as Abdel Fattah el-Sisi of Egypt (also just re-elected), who use security to justify the jailing of reporters and political opponents. Thanks to allies in Moscow and Tehran, the tyrant Bashar al-Assad retains his stranglehold over much of Syria. In Africa, the presidents who serve longest are often the most corrupt, multiplying the harm they inflict with each passing year. Meanwhile, the possibility that fascism will be accorded a fresh chance to strut around the world stage is enhanced by the volatile presidency of Donald Trump.
If freedom is to prevail over the many challenges to it, American leadership is urgently required. This was among the indelible lessons of the 20th century. But by what he has said, done and failed to do, Mr. Trump has steadily diminished America's positive clout in global council

redpajama_set_name: RedPajamaCommonCrawl


The large-scale structures observed at or above the logarithmic-layer of wall-turbulence were investigated using Large Eddy Simulation. We have mainly focused on the Reynolds-number scaling of the large-scales and the influence of thermal stratification on these motions. The grid resolution required for the reproduction of the large-scale structures in plane channels was intensively studied and it was found that rather higher grid resolution that was sufficient to reproduce the near-wall structures was required to properly reproduce the large-scale motions. This fact indicates that there would be a strong relation between the small-scales near the wall and the large-scales in the outer-layer. The large-scales in the neutral state basically obey the outer-scaling and their spanwise size was about twice larger than the boundary-layer thickness independent of the Reynolds number tested here. The large-scales were also found to be strongly affected by the thermal stratification and were suppressed at a certain weakly stable stratification tested here, contrarily to the fine-scale structures being rather insensitive to the thermal stratification.


text:
package org.apache.subversion.javahl.types;

import org.apache.subversion.javahl.ClientException;
import org.apache.subversion.javahl.NativeResources;

import java.util.List;

/**
 * Object that describes a revision range list, including operations on it.
 * Returned from new accessors in {@link Mergeinfo}.
 * @since 1.9
 */
public class RevisionRangeList implements java.io.Serializable
{
    // Update the serialVersionUID when there is a incompatible change
    // made to this class.  See any of the following, depending upon
    // the Java release.
    private static final long serialVersionUID = 1L;

    /**
     * Load the required native library.
     */
    static
    {
        NativeResources.loadNativeLibrary();
    }

    private List<RevisionRange> ranges;

    /**
     * Wrap a list of revision ranges.
     */
    public RevisionRangeList(List<RevisionRange> ranges)
    {
        this.ranges = ranges;
    }

    /**
     * @return The wrapped list of revision ranges.
     */
    public List<RevisionRange> getRanges()
    {
        return ranges;
    }

    /**
     * Remove revisions in eraser from the current object
     * and return the resulting difference.
     * @param eraser The list of revisions to remoove.
     * @param considerInheritance Determines how to account for the
     *   {@link RevisionRange#isInherited} property when comparing
     *   revision ranges for equality.
     */
    public native List<RevisionRange> remove(List<RevisionRange> eraser,
                                             boolean considerInheritance)
        throws ClientException;

    /**
     * Remove revisions in eraser from the current object
     * and return the resulting difference.
     * @param eraser The list of revisions to remoove.
     * @param considerInheritance Determines how to account for the
     *   {@link RevisionRange#isInherited} property when comparing
     *   revision ranges for equality.
     */
    public RevisionRangeList remove(RevisionRangeList eraser,
                                    boolean considerInheritance)
        throws ClientException
    {
        return new RevisionRangeList
            (remove(eraser.ranges, considerInheritance));
    }

    // TODO: More svn_rangelist_t operations
}

redpajama_set_name: RedPajamaGithub


The folks at Dengarden have a lot of conduit bending resources. If you're familiar with the way that conduit bending is usually taught, it's via shortcuts and rote memorization of how to do things like offsets, kicks, bends, etc.
Suppose you studied trigonometry in high school, as most of us did. Maybe you want to apply math that you already know to conduit bending.
Dengarden's got you covered for that too. And I've got you covered with a PDF local copy in case the original goes away.
The only thing I have to add to their documentation is "don't overthink this". Don't start thinking about lines through the centerline of the conduit and try to hang your calculations in the middle of the air - the only line that really matters is one that is tangent to the tube where the EMT touches the wall longitudinally. Run all your calculations based on that and everything else falls into place.


text:
Related Expertise: Technology Industry, Media Industry, Marketing and Sales
Augmented Reality: Is the Camera the Next Big Thing in Advertising?
03 April 2018 By Carmen Bona, Martin Kon, Lara Koslow, David Ratajczak, and Michael Robinson
The ubiquity of the smartphone and its integrated camera has brought about a new era in consumer behavior. Users are increasingly engaging with the physical world through a digital lens, spending a large portion of their time with their smartphone and camera and sharing pictures and videos with their friends. An estimated 1.3 trillion photos were taken worldwide in 2017, nearly four times the number taken in 2010, most of them on smartphones. As Snap's CEO, Evan Spiegel, puts it, "The camera is evolving from being a barrier, separating users from the world, to becoming an aperture, for users to explore the world." (Snap commissioned this report; see "About This Report.")
AR technology provides the ability to overlay or superimpose digital images and representations (such as furniture or apparel) onto real-world environments. AR is a fundamentally different technology from virtual reality (VR), which immerses users in a completely artificial and digital environment, typically through a head-mounted display. This report focuses exclusively on AR.
Given the fast-rising adoption of AR by users and the increasing interest by advertisers, Snap commissioned BCG to undertake a study of AR's impact on the consumer-advertising market. As part of this BCG Marketing Executive Benchmarking Study, we surveyed more than 50 CMOs and senior marketing executives from the top 200 advertisers in the US, spanning multiple industries, on their current usage of AR and their outlook for future usage. We interviewed a sample of these executives directly to provide further details and insights.
All market projections and usage figures are specific to the US. We have discussed our work with Snap executives, but the analysis and conclusions are entirely the perspective of BCG.
This massive shift of eyeballs provides marketers with a powerful new canvas on which to create advertising experiences. "Camera marketing" could be the next wave of digital marketing, similar to the advent of web advertising in the mid-1990s or the development of mobile-first ads at the beginning of this decade. One specific form of camera marketing, the use of augmented reality (AR), is quickly gaining traction, as demonstrated by Pokémon Go's meteoric rise to 50 million users in the US in a matter of weeks.
By blending the physical and digital worlds and enabling users to actively shape their experiences, AR offers unique ways for marketers to interact with consumers. These can range from brand overlays, to navigation and wayfinding, to virtual showrooming. (See Exhibit 1.) Our survey of over 50 of the top 200 advertisers in the US shows that marketers recognize the opportunities in AR. More than four out of five respondents said that AR marketing is a differentiated way to engage with customers. "There is high potential in AR to help transform the way we market," a senior retail marketer told us. "The current traditional forms of digital marketing someday will be looked at as old school."
Although they recognize the opportunity, few companies have yet moved beyond experimentation with AR, and only a small number of them consider the technology an integral part of their marketing strategy. Our research and client experience indicate that this is set to change. This report looks at what companies need to know—and do—to catch the fast-rising AR wave.
The AR Opportunity
The AR opportunity is already bigger than many marketers realize. On the basis of 2017 data from Snapchat, Pokémon Go, and other AR applications, we estimate that more than 80 million people in the US, or roughly one-third of all smartphone users, engage with AR at least monthly. (See Exhibit 2.) The consumers using AR applications are a particularly attractive segment for advertisers: most are millennials (aged 19 to 34) or Generation Z consumers (aged 18 or younger). They use their smartphones and cameras continually throughout the day, communicating via images and video as much as they do through text and voice. This behavior opens up an entirely new method for engaging with consumers—an interactive approach where the advertiser is a participant in the "conversation" or activity rather than a company delivering a message.
As a result of rapid consumer adoption, a dynamic ecosystem is developing around AR and—in particular—around smartphone-based AR. Advertisers have two distinct pathways to reach consumers. First-party AR marketing uses applications or platforms owned and managed by brands, such as proprietary mobile apps or brickand- mortar store experiences. Third-party AR marketing involves advertising units bought on AR platforms owned or managed by social media, messaging, gaming, or digital-publishing companies, among others. First-party advertising so far is fragmented, with a growing number of bespoke mobile applications from major advertisers touting AR features but very few reaching significant scale. In comparison, third-party AR is concentrated across a few well-developed platforms, enabling advertisers to reach audiences at scale. Examples include Snapchat, which reports 70 million monthly active AR users in the US, and Pokémon Go.
Moving forward, we expect the AR ecosystem will continue to develop quickly as tech giants such as Facebook, Google, Apple, and Amazon make their own big bets on AR. In parallel, other players such as ad agencies, app and software developers, and ad networks are staking out their own roles in the value chain. Marketers can expect to have access to a wide array of AR-marketing options in the future as well as support and expertise from a host of external partners for the development and execution of campaigns.
Given the traction AR is gaining with consumers, and its fast-developing ecosystem, we expect it will reach more than 120 million regular users in the US by 2020. This growth will both drive and be fueled by an explosion of offerings across a broadening array of use cases, as well as by further penetration across all demographics. As Apple's CEO Tim Cook told the BBC, "I'm incredibly excited by AR because I can see uses for it everywhere."
How Advertisers See and Use AR Today
Marketers are moving quickly to capitalize on the opportunities in AR. Nine out of ten large consumer advertisers in our survey are already using, or are planning to use, AR in their marketing campaigns. Most of this usage, however, is still at the experimental stage; only one out of ten companies indicates that AR is well integrated into its marketing strategy. (See Exhibit 3.)
Companies in industries such as retail, fashion and beauty, and gaming and entertainment are leading adoption—and for good reason. In these verticals, AR offers a unique advantage over other channels in that it enables consumers to visualize products in use in a specific context, such as makeup on a potential purchaser's face, furniture in an actual room, or video game or movie characters in a real-world setting. While these are the most intuitive use cases for AR, many other creative uses already exist and many more will emerge in a variety of sectors.
Advertisers regard AR as a marketing tool that can help engage consumers along all stages of the marketing funnel, from awareness to purchase. Currently, most are focused on the top and middle stages (awareness and consideration), but they expect more bottom-of-funnel activity to develop. In our research, fewer than one in five advertisers considers "incremental sales" a primary objective for AR marketing today, but more than a third expect it to be a primary objective 24 months from now. This prioritization of bottom-of-funnel objectives is also reflected in the metrics that advertisers use in AR campaigns. Today, advertisers look to AR to boost reach, engagement, and awareness, but the metrics they most want to use AR for two years from now are engagement, purchase intent, and sales. (See Exhibit 4.)
Against these objectives, AR advertising has already started to demonstrate measurable success, and it has done so across a wide variety of industries, including automotive, consumer goods, retail, food and beverage services, entertainment and gaming, and fashion, luxury, and beauty.
At the top of the funnel, marketers have seen double-digit increases in brand awareness and favorability. For example, Nissan drove a 15-percentage-point improvement in brand favorability with its See the Unseen campaign, which showcased advanced vehicle safety technologies through interaction with Star Wars movie characters. (See "Nissan Uses New Tech to Show Off New Tech.") According to Jeremy Tucker, vice president, marketing at Nissan North America, "We got results that you don't see in traditional media." A number of third-party AR campaigns have shown an average 19-percentage-point increase in ad awareness, according to measurement company Nielsen.
Nissan Uses New Tech to Show Off New Tech
Nissan was looking for a new way to showcase its Intelligent Mobility technology to consumers. The automaker initiated a first-of-its-kind AR-marketing campaign, which it called See the Unseen. (See the exhibit.) Using mobile devices that Nissan set up in dealerships around the US, the company let consumers interact with characters from the movie Star Wars: The Last Jedi, which was being released at the same time. The movie characters helped educate viewers about automobile technologies that were otherwise invisible to the naked eye.
Nissan came up with the See the Unseen campaign by first articulating the clear business objective of developing an interactive, educational means of communicating the benefits of the Intelligent Mobility technology. To execute the concept—given the complexity inherent in developing a first-of-its-kind campaign—Nissan mobilized significant internal and external capabilities. Internally, the automaker used learning from a previous third-party AR campaign and an experienced team. Externally, the automaker tapped into a network of creative and development agencies well versed in AR marketing applications. Underlying the entire effort was Nissan's culture of innovation, which is fostered by its marketing leadership. The company empowered the team to develop and iterate the campaign in the face of some significant obstacles and bring the concept to life.
See the Unseen was embraced by consumers and outperformed similar campaigns using both traditional and digital media. In a survey of those who had been exposed to the campaign, Nissan found a 67-percentage-point rise in familiarity with the Intelligent Mobility technology and a 15-percentage- point lift in brand favorability— both sizable increases in important metrics for an automaker.
Owing in part to the success of See the Unseen, the automaker plans to step up its use of AR and further integrate the medium into its marketing strategy. Nissan believes that the technology will, in addition to making it possible to tell richer product stories, allow the company to improve its sales process. The marketing team is currently engaged in rolling out new AR applications, and it is focused on scaling these experiences to a wider audience.
At the bottom of the funnel, marketers have seen equally powerful results. Retailer Williams-Sonoma is using AR in many ways, including a mobile app that enables customers of Pottery Barn (a subsidiary) to visualize how furniture will look in their homes. Laura Alber, CEO of Williams-Sonoma, told Forbes, "Augmented reality and 3D rendering are fundamental for driving and changing the shopping experience." Numerous third-party AR campaigns for consumer goods companies have delivered an average 10% increase in sales with a 3.1-fold return on ad spending, as measured by Nielsen Catalina Solutions. In the food and beverage services space, McDonald's ran a particularly successful AR campaign on Snapchat that drove a significant lift in foot traffic and sales with a 10.5-fold return on ad spending, as measured by Placed. (See "McDonald's Deploys a New Way to Feel Good.")
McDonald's Deploys a New Way to Feel Good
McDonald's is driven by an aggressive test-and-learn culture that led the company to experiment with and innovate in AR marketing. Among its various AR executions to date, McDonald's has developed a first-party application to support its World Cup partnership, teamed up with a third-party platform for its 2016 Christmas campaign, and executed a Big Mac promotion with Snapchat.
In 2017, McDonald's wanted to support a cold-beverages promotional campaign and create feel-good moments for customers (see the exhibit). It decided AR was a differentiated way to do so. The campaign had a clear business objective: to promote an offer of $1 cold beverages in order to drive foot traffic to McDonald's restaurants, with the expectation that consumers would purchase a larger basket once in the restaurant.
McDonald's used Snapchat because the app provided engaging AR features as well as the reach necessary for a campaign whose success depended on quickly driving mass scale. To ensure successful execution, McDonald's tapped into its prior learning and worked alongside its media and creative partners as well as a Snapchat team to execute the cold-beverages AR campaign.
The campaign achieved exceptional results. A poll by Nielsen showed an 11-percentage-point increase in message awareness against the control group, twice McDonald's normal campaign experience. In an attribution study that Placed conducted of the ad's impact on in-store visits, McDonald's observed a 6% behavioral lift—that is, an increase in the likelihood of visiting a McDonald's among consumers exposed to the campaign. The campaign generated a 10.5-fold return on advertising spending, three to four times better than McDonald's norms.
The Barriers to More AR Advertising
Despite the size of the opportunity and the encouraging results from early movers, big advertisers report several challenges that are preventing them from scaling up their AR efforts. Specifically, companies are wrestling with obstacles related to market maturity, impact and measurement, and internal capabilities and capacity. (See Exhibit 5.) In some cases, these barriers appear to be rooted in a lack of knowledge or understanding, so we expect they will recede over time. Advertisers may need to make a concerted effort to overcome others, however.
Market maturity is the biggest challenge reported by marketers today. Advertisers in our survey pointed to such factors as "unable to reach my audience at scale," "lacks a track record of results, is too new and experimental," and "requires too large an investment" as barriers. In our judgment, some of these are more perceived than actual challenges; for example, scale is not an issue because there is already a substantial audience of AR users. And as the market matures and standard tools and platforms—such as Snap's Lens Studio, Facebook's AR Studio, and Amazon's Sumerian—become more widespread, deployment costs will drop and a track record of results will build.
Impact and measurement is the second-most-cited challenge, particularly the uncertainty around return on investment (ROI). While this is understandable given how new AR is, a body of evidence already exists to demonstrate that AR marketing can be effective at driving business outcomes, which will only continue to improve with greater adoption. In this context, it will be critical for advertisers to determine how to clearly measure the efficacy of their AR marketing activities—and encourage their partners to do so—in order to demonstrate the impact of AR compared with that of other marketing channels.
Internal capabilities are another barrier to increased AR adoption, which is hardly surprising given the newness of the technology. Advertisers need to gain experience, and as is the case with most advanced technologies, the supply of expert talent needs to catch up with fast-rising demand. In the meantime, advertisers can accelerate their capability development by deploying limited-cost test-and-learn approaches and by tapping the experience of external partners, such as agencies and advertising platforms.
Stepping Up Your AR Game
We expect the AR-advertising ecosystem to continue to mature quickly, overcoming barriers and pushing AR into the mainstream. For marketers, the user numbers are too large, the engagement and results too promising, and the potential brand halo too compelling to ignore. As one retail CMO stated, "AR is a powerful new technology that can elevate customer experiences, and it's incumbent upon marketing leadership to push involvement."
To lay the foundation for their AR strategies and to prepare their organizations to execute successfully, marketers need to take action in five areas. (See Exhibit 6.)
Elevate AR in your marketing strategy. Given AR's current usage and expected growth, advertisers should start considering this new channel a core component of their marketing portfolio. When setting up campaigns, advertisers should clearly articulate their business objectives and determine which marketing tool will serve those objectives most effectively and efficiently. In this decision, companies should consider AR advertising alongside other channels such as TV, social media, search, and email.
The actual campaigns in which AR will be the best tool in the toolkit will vary by advertiser and industry because of AR's many uses and its efficacy at different stages of the marketing funnel. In some cases, AR will be a clear fit (such as a cosmetic brand showcasing new makeup and hair-coloring products), while in others the execution of AR campaigns will require a more thorough assessment.
As companies ramp up their AR efforts, the guiding principle remains axiomatic: allocate spending to AR to the extent that it benefits your business and marketing objectives. AR will follow a journey similar to that of digital, mobile, and social advertising to become a core element of the marketing portfolio.
Pick the right platform for your AR strategy. Advertisers need to decide whether to develop their own first-party AR apps (as Ikea, Nike, and others have done) or to place AR advertising on third-party platforms such as Snapchat or Facebook—or use a combination of both approaches. This choice has substantial implications for the required resourcing and outcomes, and, as with AR strategy, it should be made in the context of overall marketing objectives.
Advertisers seeking a more bespoke experience for a highly targeted audience will be best served by a first-party execution. This high-touch approach can be very effective, especially when users are given a compelling reason to download and engage with an application. But this tactic is also more costly and resource intensive; it requires marketers to collaborate closely with the product team and manage customer acquisition as well as retention.
Conversely, advertisers aiming for lighter-touch executions that can achieve reach and scale might opt for standardized AR ads on third-party platforms. These campaigns generally require more limited internal capabilities because they can lean more heavily on the expertise of digital-media agencies, creative studios, and the third-party platforms themselves.
Of course, advertisers might also adopt hybrid strategies that involve both their own first-party apps and campaigns on third-party platforms. This approach will likely become more common over time as the leading third-party platforms facilitate the sharing of creative assets between first- and third-party participants.
Augment your AR capabilities. Advertisers recognize that their lack of capabilities and expertise prevents them from scaling up their AR advertising. At the same time, our survey revealed a strong link between the strength of internal capabilities and the effectiveness of AR campaigns (as seen in Exhibit 6). As has been the case with other innovations, advertisers may feel trapped in a difficult conundrum: they execute unsuccessful AR campaigns because they don't have the requisite capabilities, and at the same time, they don't invest in building or acquiring capabilities because they don't see positive returns from their AR campaigns.
Developing AR capabilities requires a conscious and active effort on the part of the CMO, and it is better to start now while the game is still in its early innings. Advertisers will be well served by small targeted investments such as hiring or training one or two people to become internal AR specialists (not necessarily dedicated exclusively to AR). These employees can champion when and how to use AR, centralize learning, interact with industry leaders and external partners, and train others as AR becomes more mainstream.
Internal resources are only half the equation. Developing relationships with partners across the fast-growing ecosystem also will help advertisers increase their capabilities, expertise, and bandwidth, without necessarily making large investments. Specialized AR media agencies, or AR teams within larger agencies, can help educate internal resources; AR development studios can do the heavy lifting to create AR campaigns, and third-party platforms are keen to sponsor test campaigns and impact studies.
Be disciplined about measurement. As with any digital-marketing channel, advertisers' success in AR will ultimately hinge on their ability to measure results and adjust campaigns for optimal impact against set objectives. Our research shows that the measurement-related processes we included in our survey (such as "capture performance data," "incorporate AR data into existing marketing reporting," and "conduct rigorous analytics on AR to compare against objectives") are key differentiators for advertisers that consider their AR campaigns effective.
Advertisers must start by setting metrics for AR that are aligned with the business objectives of a specific marketing campaign, such as building brand awareness, improving brand equity, or increasing sales. Although AR is still a new marketing tool for most, the approach to measuring its efficacy need not be. Advertisers can use existing measurement tools and metrics for AR, ensuring that AR is assessed on its own merits and compared with other channels transparently and on a level playing field. CMOs should demand a high level of clarity and hold their teams and themselves accountable for results and ROI. They should also establish recurring reporting (including both internal and partner studies) to optimize AR campaigns and make informed budget allocation decisions between AR and other marketing channels.
Embrace a test-and-learn culture. Marketers have been hearing for years that digital advertising allows for experimentation like no medium before. Although many companies have made progress toward developing a test-and-learn culture, BCG research has shown that measurement, and in particular testing, remains a capability that plenty of marketers (and their agencies) struggle to build. (See A Disconnect and a Divide in Digital-Marketing Talent, BCG Focus, March 2017.)
CMOs should instill a mindset of test and learn throughout their organizations by setting up the right processes, rewarding new ideas (and not punishing failed ones), and incorporating test and learn into the operating model with partners (and even into contracts when possible). Only by taking risks and continually improving will advertisers get the most out of a new marketing tool like AR.
No one can predict how new technologies such as camera marketing and AR will evolve or how they will shape consumer behavior and the marketing landscape. But the tailwinds for AR advertising are too strong for CMOs to ignore. Now that the pioneers have paved the way, other marketers should move quickly if they don't want to miss the boat.
Carmen Bona
Lara Koslow
David Ratajczak
Michael Robinson


Dataplas Ltd is able to manufacture star wheels, guides and associated componentry in all engineering plastics, providing you with a one-stop shop for all your processing and packaging line parts.
Because we manufacture everything on-site we can provide you with the convenience of working with one supplier who understands your needs and can ensure all parts work together as required.
Using our CNC router we can manufacture star wheels and complex guide shapes efficiently and accurately, with complete repeatability.
Star wheels and guides can be manufactured from sample, drawing or DXF file. They can be made in a range of engineering plastics. Polyethylene and nylon are often used for starwheels and guides because of their high wear resistance and excellent sliding properties.


It will be listed here in Discreet Magazine for FREE!
i was in iloilo last month and i stayed at Grand Tower along Isnart St going to YMCA. I had a great massage service at Royal Thai located near my hotel ky dula gd tana ko nga sakit sa lawas. Tanda gd cla sa accupoints and acrobatic stretching. They have variety of massage styles, f indi mo gusto nga permi ka gina butong, swedish ang mas relaxing,, dasig ka matologan, pagbugntaw mo, tawhay na. You may call them at 509-7203, they do home and hotel service as well. Kag in fairness, ka damo gwapo nga massuers. best service gd c sputnik kag si berto, i doubt if their telling their real names, cause it sounds funny. Massage was perfect and satisfaction guaranteed.
I will be in Iloilo city next wednesday and thursday. Can anyone recommend me a nice spa with hot male massuer, esp with big tool? Thanks!


text: Q: Is there version control software with commit messages that preserves macOS tags? As I mentioned in my answer on tags best practices, the distributed version control software (VCS) Git does not preserve macOS tags (or other extended file attributes). Is there version control software (distributed or centralized) that features commit messages (like Git) and that preserves macOS tags? (The built-in versioning system in macOS does not include commit messages.) I am especially interested in software that can be easily installed through one of the package managers for macOS, but any software that meets the criteria is welcome.

redpajama_set_name: RedPajamaStackExchange


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Employee who survived shooting sues Walmart
Norfolk, Va. ( Associated Press) – A Walmart employee who survived last week's massacre at a Virginia store has filed a $50 million lawsuit against the company for hiring the shooter – a store supervisor – "who used violence, threats, and bizarre behavior."
The lawsuit, apparently the first stemming from the shooting, was filed Tuesday in Chesapeake Circuit Court by Donya Priolue.
Walmart, which is based in Bentonville, Arkansas, said in a statement it was reviewing the lawsuit and would respond "as appropriate to the court."
The company said, 'The entire Walmart family is broken by the departure of important members of our team. "We express our deepest condolences to all those affected, including their loved ones and those injured. We are focused on supporting all of our partners with critical resources, including medical attention.
Prioleau's lawsuit claims that he experienced post-traumatic stress disorder, including physical and emotional distress, from witnessing the November 22 incident in the store's break room. The complaint provides new details about the horrific episode and a long list of troubling signs displayed by the shooter that it says managers failed to address.
The lawsuit states, "The bullet passed through the face and left side of plaintiff Donia Priolue, who narrowly escaped injury." "He saw how many of his co-workers were brutally murdered on either side of him."
The suit states that "right after being shot in the throat, Prioleo looked squarely into the eyes of one of her co-workers. Mrs. Prioleo saw the bullet wound on her co-worker's neck, bleeding, and her co-worker's helpless face Look at the expressions of wonder."
continue reading story
According to police, the store's supervisor, Andre Bing, 31, fatally shot six employees and wounded several others before dying of an apparent self-inflicted gunshot wound.
The lawsuit alleges that Bing "had personal feuds with several Walmart employees and maintained a 'list of potential targets' prior to the shooting."
The list refers to a "death note" found on Bing's phone that authorities released Friday. The note contained specific references to people he had worked with, but officials blacked out their names.


The convenience-store Icee machine is now behind dozens of America's best bars, helping mixologists create a new breed of cocktail—the sophisticated frozen drink.
To wrap your head around how the low-brow slushie machine is fast becoming a fixture in America's high-brow drinking dens you need to understand one of the key bars fueling this frozen-treat trend—the ever-paradoxical Mother's Ruin in New York City.
It's a dive bar—raucous, open until 4 am, and thick with the musty smell of stale beer. Yet it sits in the heart of Nolita, one of Manhattan's quietest and most stylish neighborhoods. Its main beer is canned, el cheapo Tecate, but everything else it serves up, from the cocktails and bar bites to the music it blasts, is often several steps ahead of the trends.
So it's fitting, perfect perhaps, that at the end of this dichotomous bar sits a 7-Eleven-style Slurpee machine swirling icy, top-shelf concoctions of mixology wonderment.
Necessity was the, ahem, mother of Mother's Ruin's invention. In 2011, T.J. Lynch, co-owner, wanted to put a frozen Paloma—with five-chili molasses—on the brunch menu.
Since then, Mother's, as regulars call it, has batched more than 230 different slushies, offering a new high-end brain freeze every week. Two past hits include the Paloma and a Banana's Foster slushie, made with two different rums, lime juice, bananas, white Creme de Cacao, dry Curaçao and burnt-sugar syrup.
Now, bars across the country are following suit, buying into the very simple idea of using quality ingredients to create complex, boozed-up Slurpees.
The Gibson | Washington, D.C.


The ancient Ionia (Ionia) consisted of 12 cities. These are Miletos, Priene, Melia, Myus, Ephesos, Colophon, Lebedos, Teos Klazomenai, Phokaia, Khios and Erythraidir. One of the two most important cities of Ionian, the Miletans (the other Ephesus) is from the word 'Didyma', which means the Greek 'twin sister' for the name of the god Apollonian island Didim.
Temple of Ephesus Artemis and Didymada are the temple of Artemisin's twin brother Apollo. Apollo paganist is a god of faith in the sun, prophecy, music and art. According to the legend, God Apollon, Didymade shepherd Brankhosa came to him, and he liked him very much and gave him the secrets of his miracle (prophecy). Brankhos built the Temple of Apollo in the place where the bay forest and the water source that he encountered were located.
Later, people from the line of Brankhos are referred to as 'Brankitler' and the temple is ruled. It was even called Didme 'Brankhidai'. Later, the Miletos made a giant temple in the name of the god Apollo in Didymada. They also built a Sacred Way to reach this temple. The gigantic Temple of Apollo has made Didymay the center of ancient antiquity.
In the middle of April-May, the convoy formed to join the Miletos of Didymadaki, reached religious festivals in four days giving religious ceremonies and breaks in the so-called stations. Before entering Didym, they were waiting in the sanctuary of Artemis, next to the temple. The Sacred Way was 24 km long and 6 m wide. Both sides were surrounded by Brankhos and lion sculptors. In 1858 a British named Newton took most of these sculptures to the British Museum.
Today, a great part of the Sacred Way is underground. The Miletos architecture was of great importance and the scale of the Temple of Apollo was so great in response to the Temple of Artemis, one of the Seven Wonders of the World, that it was not enough to finish the temple completely, neither the Miletos, nor the Great, nor the Romans. It is one of the three largest temples of the Hellenistic period, along with the Apollo Temple Artemision and the Hera Temple in Sisam. The temple measures 109x51 meters.
The inner part of the cult statue, the naos or cella area, is 53x21 meters in size. The temple was surrounded by a column of 120 ions of 19.5 meters in length. Today only three of them are standing.


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Microsoft/
Microsoft trolls Internet Explorer haters with latest browser ad
Source Internet Explorer (YouTube)
internet explorer 9 advert
Microsoft is tackling anonymous anti-Internet Explorer commenters head-on this week. The software maker launched the latest browser you loved to hate video today, showcasing the latest part of its attempts to keep Internet Explorer relevant. The video sets the stage for a comment war with an anonymous IE hater who leaves comments on articles, replies to Facebook posts, and sends tweets to the IE Twitter account.
It's a comical account of real life comments that can be found across various sites, including "IE sucks" and "IE is only good for downloading other browsers." Described as a comeback, the campaign shows that Microsoft "gets it" and is clearly aware of the feedback around Internet Explorer. The company has won praise for its recent attempts to adhere to web standards and push out solid releases for Windows 7 and Windows 8, but can't shake off the perception and memories of Internet Explorer 6 — a product it's trying hard to kill off.
The approach is fresh, but Microsoft faces some significant challenges against an increasingly popular Chrome browser and offerings from Mozilla and Apple, all of which offer varied and additional features. Internet Explorer is hovering at around 50 percent market share worldwide, so it's certainly time that Microsoft attempted to reverse the bleed it has experienced over the past several years. But, will it work?
More from Microsoft
Computing 101: how to zip and unzip a file
No, Microsoft DirectStorage isn't slashing RTX 4090 performance by 10 percent
Microsoft strikes 'strategic alliance' with solar manufacturer


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2012 bowl game projections: Virginia Tech picked for Music City Bowl
Examining the most recent bowl game projections for Virginia Tech and the rest of the nation's top programs.
By Scott Coleman@scottcoleman55 Oct 22, 2012, 7:15pm EDT
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Following another busy week of college football action, Jerry Palm of CBS Sports has updated his bowl game predictions for the 2012 season.
The Virginia Tech Hokies were picked by Palm to reach the Music City Bowl on Dec. 31 in Nashville. The Hokies would play the Vanderbilt Commodores in a battle of ACC and SEC foes.
As for the national championship game, he has the Alabama Crimson Tide facing the Kansas St. Wildcats. In the other BCS games, the Oklahoma Sooners and Boise St. Broncos would meet in Glendale for the Fiesta Bowl. The Florida Gators would take on the Notre Dame Fighting Irish in New Orleans for the Sugar Bowl.
The Nebraska Cornhuskers and USC Trojans would play for the Rose Bowl in the annual Big 10 vs. Pac-12 showdown. The Orange Bowl would pit ACC champion Florida State against Big East champion Louisville.
You can find all of Palm's predictions here.


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Individuals with surname Alves
Alves, Adair (I1949) -- Birth 28 August 1960
Alves, João Batista (I71) -- Birth 12 April 1960 -- Aparecida do Taboado - MS
Alves, Moiseniel (I1588) -- Birth 14 April 1974 -- Inocência - MS
Alves, Rafaela (I2680)
Alves, Rosa (I2679)
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