text:
Ignatius Piazza – In Las Vegas Mercury
Falling Send Video Game
The Tunnel Rush Video Game, or the Particle Video game as it's likewise been called, is a really simple idea. One component Microsoft Paint and one component sand art, the game begins with 3 or 4 different streams of sand fragments dropping from the top of the screen. It's the gamer's job to create shapes to capture the sand as well as to produce intriguing patches of colored blocks by filling up the vacant locations up with the falling granules.
The gameplay for the Run 3 Game is relatively basic as well as simple. As quickly as the game lots, the granulated sand starts to drop from the top of the screen. Utilizing the mouse gamers click and drag to produce lines. These lines could be turned into half circles, ramps, and even a flat line across the display to earn certain that the sand can take a trip no further once it drops. The result is that the sand accumulates, the colors mix with each other as well as gamers could create an interesting spectrum of forms loaded up with the granulated mixtures.
The most effective suggestion for gamers that determine to play the Falling Sand Video game is to find into it with the appropriate perspective and also state of mind. This is not a video game in the sense that there are levels and victory problems, as well as all Return Man 3 players should understand that. Instead, the Falling Sand Game is something that people could do for fun to produce unusual layouts with the falling sand and also the shapes they make use of to load the screen up. Understanding that you desperate may inhibit some players, yet it may liberate others to be much more creative with their sand styles.
Besides that, players need to understand that in some cases they're against the clock when making particular layouts. The sand drops as soon as the Impossible Quiz game lots, so players that have extremely certain layouts in mind for blending the shades have to understand where they're going to put them. There's additionally no saving and also returning, so if you make a blunder with a line, you'll need to rejuvenate the home window and also shed every little thing you made up to that factor. It might take some doing, yet obtaining it right the very first time is constantly the ideal effort.
Bet Free
For players that such as the idea of creating some unique looking photos with the Falling Sand Game, this website uses the possibility to bet totally free. Really cost-free too, no membership, no individual information offered, simply going to the site and also packing up the Cubefield Game. It's truly just that easy. As well as soon as the video game is loaded players could go as long as they intend to, or at the very least up until they've produced specifically the ideal picture they intended to with this granulated game.
For players that appreciate the Falling Sand Video game, along with other cost-free to play video Learn To Fly 3 offered below, please gotten the word out and allow other individuals know where they could most likely to bet totally free. The even more the merrier!
Author Sonia DixonPosted on January 17, 2018 August 26, 2019 Categories Play Flash Games
When you watch for an enjoyable as well as enjoyable method to invest your time, several alternatives will certainly jump out at you. You can go for a swim at the regional swimming pool, head out purchasing at the shopping center, go obtain a banana split at your favorite The Impossible Quiz shop, or go to the park for some Frisbee golf. These are all excellent choices when you wish to have some enjoyable, but just what is there to do when your fresh out of money and also stuck at home?
The solution is basic and also very easy, and probably something that you have actually believed regarding before. When you need to, or want to, remain at residence and also have a good time in a relaxing, smooth way, you could count on free online Run 3 Unblocked to obtain your repair of fun as well as enjoyment. There are thousands of totally free flash video games offered on the web, and also they are all very easy to obtain accessibility to.
Just go to the website where the game is hosted, wait for it to lots, and play away, late into the evening. There is no should download and install any kind of hazardous documents to your computer. This safeguards you and your computer system, and maximizes time given that downloads have the tendency to take a huge piece of time. These Return Man 3 games are available to anybody who has a good computer system and also broadband web. The video game is played right in your web browser window, it just couldn't be simpler.
If you are searching for a fantastic on-line flash Learn To Fly 3 video game that will assist you shave endless hours off your day, look no more than Balloon Tower Defense 5. This game further broadens on its remarkable predecessor, Balloon Tower Protection 4.
The object of the game is to stop the Cubefield from making it right across the map. This can be tricky, as these little devils work quick, and also it will certainly take you a while to get down the ropes. In order to stop them from getting to the end of the map you should use logic and reasoning abilities, placing different kinds of towers to aid quit them.
These towers each have an one-of-a-kind function that helps battle the bloons, as well as each growth of the Tetris game has actually included new towers. This newest development of the video game has actually added two new towers that truly amp up the power on your side.
The game begins with a specific quantity of loan that you utilize to build towers, and once the loan is gone, you will should generate much more in order to build extra towers, otherwise the bloons will make it throughout the map and win the Raze 3. Different trouble setups are offered for you to pick from, so as your ability level expands you could readjust the setups for even more of a difficulty.
If you are seeking a video Tunnel Rush that will actually check your abilities, compared to check out Balloon Tower Protection 5. This video game will certainly have you going incredibly for hrs. As always, if you enjoyed this write-up, share as well as like it so others could see it also!
Bera Has Currently Reached High Degree Of Success
Despite the fact that, Bera has currently reached high degree of success. He never unwinds. He is constantly ready for brand-new difficulties. His working day begins at 10 o'clock. The most valuing thing concerning Bera is that besides his work, he studies. It is very rare for people like him. Other on Bera's area would certainly consider that they are already grownups and as they have occupation, do not need to obtain any even more education and learning. Yet Bera likes the process of getting brand-new info. He suches as reading as well as recognizing points he questioned. As he says, he does not examine all the subjects, just the ones that will certainly serve in his life. Presently Bera is studying approach and also arts. He states that understanding of philosophy is crucial for every single person that has a passion of being a wise and also effective individual since; philosophy is a base of everything around us. One can never ever set his/her mind drivened only on material values, we should likewise establish our internal side. That is why we need the expertise of approach. The reason why Bera researches art is that every singer and musician typically, must be informed in this ball. As he obtains older, Bera aims to establish not just in sense of career and also vocal singing, yet additionally in a feeling of his way of thinking.
Bera says that life has lots of challenges. That's the factor why he regularly gets lessons from his life. Bera was only 13 years of ages when he was separated from his family as well as mosted likely to Paris. It was the very first time he began working about his task. While residing in Paris, far from his family members, Bera obtained made use of to self-reliance. He recognized that he needs to deal with whatever by himself as well as must never ever be relied on his family members as well as loved ones. After Paris, Bera went to the USA. In America Bera encountered completely different atmosphere. In the international country where he had no friends or family members no one supported him or assisted him out. Again, he was all alone as well as had to take care of every challenge by himself. At the starting it was extremely hard. He felt himself lonely and also separated. As his first tune was released, Bera thought that everyone around him would certainly become left and also delighted for his initial success. Yet unfortunately he encountered various fact. He realized that there was a huge competition around him and also nobody would be happy for his success. A lot of the people would come to be envious of such a young, full of life and point of view person.
When Bera discusses his individuality, he highlights that he is a little ambitious, however in a healthy means. Bera believes that every person ought to aspire. If he does not have a passion to overcome a person and also create something that no person else was able to, one will never get to success. We should frequently have a hard time between our obstacles and also desires. Bera discusses that he normally contends not with others but himself. He constantly attempts to create as well as develop something huger that previously. This is his passion, which does not leave him alone. When his one objective is achieved he asks himself, exactly what's next? Collections another objective as well as continues the process of battle which in some sense is pleasant. A young adult that contains life as well as has a daring spirit needs to never stand aside as well as see a person else doing something that he would do. This is the major objective of Bera. To do every little thing to satisfy himself and also to earn his family members as well as caring individuals around him honored.
Bera emphasizes that he is overly important towards himself. This critics consists of every little thing yet mostly his tracks. As he starts working with his brand-new song, Bera becomes totally committed to it. He thinks of this song 24/7 as well as tries to consider every possible method to earn it seem better. He says that he lives in addition to his tune every day as well as complying with weak after the song is released. He aims to be crucial In each and every single information in his track and also tries to make it as near to ideal as feasible. Bera likewise highlights that he does not only listen to individuals that constantly praise him, praises and also sing along. He believes that those people that consider his song in more vital way are better. Although Bera is amongst many popular stars worldwide, as well as is buddies with a lot of them, he states that he doesn't want to go to the celebrations. The reason he is a friend with such renowned people, is not that he attempts to become better to every celebrity, but since he needs to collaborate with exact same manufacturers as they do. Bera dislikes fakeness and also aims to remain himself. Also in such a fake career as singer, he still attempts to stand highly with ideology and precepts and also never violate them.
Author Sonia DixonPosted on January 17, 2018 January 17, 2018 Categories Best Pop Music And Songs
Bera Is Very Hard-working Person
There are two individuals who mainly work with Bera's cd "Georgian Dream", Rob Fusari as well as compositor of Michael Jackson, Rodney Jerkins. Even though Bera has a manager, still he places an optimal effort in his task. As he says, young singer always attempts to reach success individually due to the fact that the real champion is the one who does whatever without others help. Bera likewise highlights the fact that he has nobody around whom he could rely on all his work. Not that he is too addicted on his work or has some type of affection. He just feels secure and also tranquil when he has everything under his control.
This attribute is really useful but at the same time uncommon for such young vocalist as Bera. In one of the interviews Bera defines the duration when he released his very first album in United States and also returned back to his residence nation- Georgia. As Bera states it was extremely surprising for him when individuals stopped him on his means and asked to take images with him. Teenagers looked at Bera with the regard in their eyes and also proudly told him that they where listening his tunes 5 mins earlier. Bera might not hide his amazement he did not obtain utilized to this sort of popularity and also the reaction of these people was unanticipated for him.
Even though the listeners in the Unites States Bera quickly accepted Bera was not satisfied yet. He, as one of the most patriot people intended to be liked by Georgians. Inning accordance with vocalist, this was his primary strategy. The moment when he felt his goal reached was the time when he saw the grinning faces of his people. The moment when he really felt love and warm mindset from the strangers who where revealing just how much they loved young singer. Originally Bera began singing just for enjoyable. It was a leisure activity for a 15 years of ages young adult yet as time passed Bera understood that singing for him was something even more than hobby. It was a lot extra severe job. At the age of 16 Bera recognized that composing songs was something he intended to dedicate his life to; something that would certainly become his concern for the rest of his life. Now when this idea is not his dream anymore and it came to be fact Bera is glad to each and every single individual who played even a small function in his success. He's happy to his loving family, educators and also manufacturer. Is grateful to his loyal close friend that had actually never ever questioned his success.
A young singer began his job with the song "Georgian Dream". As Bera states, he never makes any forecasts regarding his tunes or messages. He states that it's impossible to forecast individuals's responses as well as emotions. You can never ever make people love you or make them become your fans. This comes naturally as well as you simply need to entrust your fait. When Bera was asked if he is a daydreamer or otherwise. His response was of course; he with confidence claimed that he has been a dreamer because childhood and the verses from his initial track came while he was resting on French language. Vocalist discusses that desire has 2 meanings, while resting as well as while thinking of something you desire or something you had and also shed. Both significances have a solid link as well as play substantial function in his life. Dream is the only location where he can be totally himself and get to a sensation of a complete perfection. Everyone has some type of dream; there are no unimportant or small dreams, each of the existed desires is something worth to be pleased with. Bera thinks that dream is one progression to the goal. When person does not have dreams, he has absolutely nothing to help. As soon as one begins fantasizing, he ends up being focused on this dream. The primary job is to transform the desire right into a goal and also begin helping it.
In among the interviews Bera discusses one of his tunes, "Gpirdebi". This track has Georgian verses. As young singer stresses he is always excited when he introduces tunes on Georgian to the public. The reason for his excitement is that individuals are extremely judgmental since they have bunch of skilled singers or author around them and also it's really hard for Bera to surprise the audience. Still, the a lot more challenging his work is, much more focused he is. Bera says that he suches as life packed with different obstacles and is 100% prepared to get rid of each and every single trouble on his way to success. As he recalls and also recognizes how a lot he went trough he feels satisfaction. Still he never ever boasts with all these success because he never ever likes to speak round his success, he takes into consideration that individual must not praise himself as well as jugging him in an excellent way simply like in a negative method is a work of other people.
Bera Is Georgian Songwriter
Bera is a Georgian singer, songwriter, hip-hop and R&B artist and worldwide pop celebrity. He's understood with his stage name as Bera. At the age of 5 he was currently able to play piano, drums and also violin. At the age of 11 he composed his first tune. One of the most appreciable feature of this young and also successful vocalist is that he is totally dedicated to his singing job. In among the interviews Bera stated that he has actually been brought in to the music from the day he was born. As he grew up, this tourist attraction came to be more powerful and also stronger. Because of this in his very early years he got to big achievements: dealt with some of the sector's biggest producers such as Rodney Jerkins as well as Rob Fusari, organized a 2014 VMA Pre-Party with Chris Brown as well as Pia Mia, launched his very first album "Gpirdebi". Bera is one of the few Georgian vocalists that had the ability to exit borders of Georgia and arrive globally; still, Bera always likes to remain in his nation near his household and close friends.
Besides Bera's successful career in his early ages, his generous personality is additionally worth of focus. He has very imaginative as well as enthusiastic personality, which plays a big duty in his success. In one of the meetings Bera defines just how he created among his most renowned tune "I look good on you". "I remained in one of the studios in the United States, functioning on my various other song, while, in order to have a rest I rested in front of the piano and also began playing. All of a sudden I understood that automatically I was developing a completely new track with harmonious sounds. As I listened to the melody, the idea promptly came and after half a hr verses were likewise ready". Bera has a very solid bond with his family. For him, being aside from his family members is associated with catastrophe and while investing time with his relative he does not assume about his career. "There is absolutely nothing and nobody I enjoy greater than my nation as well as my household." Mentions Bera, in his interview. While his good friends chat regarding Bera, they discuss that he is extremely easygoing loyal and also sensible person. Besides he is amusing and amusing. It is an excellent satisfaction to invest the moment with his kind of individual. You can constantly be honest to him since he is not one of that judgmental people. He will always pay attention meticulously to someone's problem as well as will do whatever to discover the method out.
Bera Ivanishvili is boy of the Georgian billionair Bidzina Ivanishvili. Inning accordance with Forbs, Bidzina is the country's wealthiest male. He's also an owner of the "Georgian Desire" event, which won the political elections versus the "United National Activity" led by then-ruling Head of state of Georgia. In his works, Bera has some political suggestions but none of them are straight for his daddy. He describes the truth of his nation from the eyes of his generation and talks what he assumes. Bera believes that there are some problems that bother him just like other people and aims to share these ideas in his songs. Therefore, he denies that his tracks are about national politics and also indicates that he always sings concerning the reality around him. No matter which government will remain in the head, he constantly talks what he takes into consideration is true. In his tracks Bera never slams the actions of government and never ever stresses that he's a household member of "Georgian Desire Event". Still he sustains his daddy and also his suggestions. In among his songs- "Qartuli ocneba" which suggests Georgian desire Bera sings about the unfairness in his nation. Underlines that Georgian people are tired of this unfairness and require the motivation and a push to stand as well as fight in order to safeguard their legal rights. With his songs Bera tries to awaken the sensation of nationalism as well as dedication inside residents of Georgia.
Bera doesn't call his audiences fans, he calls them his supporters who share his ideology. For him each audience is useful and appreciated. He claims that individuals like him since they feel his dedication towards the songs. Listeners understand that every word in Bera's tune comes from his heart and also every sentence in his rap is honest. This sincerity is the only trick of his success. When people comprehend that you actually respect their faith as well as you share same issues as they do, they begin fallowing you. The major issue is not to lead them in an incorrect direction as well as be extremely careful with your messages. That's the hardest job I'm facing. Most individuals use their power in an upside-down however all I attempt to do is making individuals understand that unity and stamina is the essential to freedom.
Bera Claims That Life Has Plenty Of Difficulties
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Technology that draws lessons from the vision of nocturnal animals could improve the luminance of electronic devices.
Researchers from the Daegu Gyeongbuk Institute of Science and Technology in Korea have developed electroluminescent film that is four times brighter than conventional film.
To create the technology, scientists studied how nocturnal animals are able to utilise light in an extremely efficient way.
The film that the researchers created was composed of a light transmitting luminescent film on a retroreflective electrode.
The authors contend that this technology holds potential for the creation of flexible and stretchable light-emitting devices with high brightness efficiency.
Research leader, Dr Byeongdae Choi, highlighted that the research has the potential to strengthen Korea's competitiveness in the global lighting market.
"This study is significant as it has applied the light reflection principles of nocturnal animal eyes, which have high light utilisation efficiency, to light emitting devices," he explained.
The research was published in Advanced Materials Technologies.
Are these spectacles the solution to travel sickness?


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UC Collaboration
Interactive Touch-Screens
Marketing Numbers
On-hold Marketing
Screening Cameras
Call Monitoring
Integrated Services for Digital Network, abbreviated to ISDN, is a set of worldwide communications standards for the digital transmission of voice, video and data services over the PSTN (public switched telephone network). ISDN is set to be switched off in 2025 by BT.
A PBX (private branch exchange) is a telephone system within an enterprise that switches calls between employees on local lines while allowing all users to share a certain number of external phone lines.
Public Switched Telephone Network abbreviated to PSTN, is the collection of worldwide circuit-switched telephone networks that carry analog voice and data over copper wire. This network will be shut down completely by BT by 2025.
Session Initiated Protocol (SIP) trunking is a replacement for traditional ISDN. It facilitates the connection of a PBX telephone system to the Internet. This means that the internet replaces the conventional telephone trunk/line, allowing businesses to communicate using the benefits of internet telephony.
Service Level Agreement, abbreviated to SLA, is a document in which the service provider promises to solve any problems with the service in a certain time-window.
VoIP (Voice over Internet Protocol) refers to the delivery of voice, video, and data communications over the internet, rather than over the legacy ISDN.
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About 2nd Primary School of Vrontados "Panagias Erithianis"
The 2nd Primary School of Vrontados "Panagias Erithianis" is located in the north-east part of Greece, next to the Greek –Turkish borders, in a small town (Vrontados), on an island called Chios.
The school was build in 1913 and is an old, high-ceiling, neoclassic building made of stone, which has some facilities problems: i.e there is no gymnastics facility and no recreation classroom.There are six classrooms, one teachers' office, a library, a classroom for children with learning difficulties, a computer room and one for the afternoon school.
The school was named after the near-by church called "Panagia Erythini"(which means Virgin Mary Erithiani) and it is one of the 3 primary schools of Vrontados.
Originally, there were only three teachers teaching in the school and only 3 classrooms, (2 class levels per classroom). Now there are 6 teachers, one for each classroom.
In our school attend 105 pupils, aged 6-12, among them children from other European countries (mainly Albania and Bulgaria), repatriates, as well as Roma children. There are approximately 20 students per classroom.
Moreover, the school staff in the morning school (08:15-14:00) consists of 13 teachers, among them the Head teacher, a teacher for children with learning difficulties, a Gymnastics teacher and three teachers of foreign languages (English, French and German).
Furthermore, at 12:30 starts the afternoon school, which finishes at 16:15. There are 23 pupils from all classes in the afternoon school. It is optional and its aim is to help children study their lessons. Moreover, children have the opportunity to attend English, Gym, Dance and Computer classes.
Additionally, our school is in the process of organising and presenting environmental and cultural projects every year.


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\section{Introduction}
Communication complexity studies the minimum amount of
communication needed to compute a function whose input variables
are distributed between two or more parties.  Since the
introduction by Yao \cite{Yao79} of an elegant mathematical model
to study this question, communication complexity has grown into a
rich field both because of its inherent mathematical interest and
also its application to many other models of computation.  See the
textbook of Kushilevitz and Nisan \cite{KN97} for a comprehensive
introduction to the field.

In analogy with traditional computational complexity classes, one
can consider different models of communication complexity based on
the resources available to the parties.  Besides the standard
deterministic model, of greatest interest to us will be a
randomized version of communication complexity, where the parties
have access to a source of randomness and are allowed to err with
some small constant probability, and a quantum model where the
parties share a quantum channel and the cost is measured in
qubits.

Several major open questions in communication complexity ask about
how different complexity measures relate to each other.  The log rank
conjecture, formulated by Lov\'{a}sz and Saks \cite{LS88}, asks if
the deterministic communication complexity of a Boolean function
$F : X \times Y \rightarrow \{0,1\}$ is upper bounded by a polynomial
in the logarithm of the rank of the matrix $[F(x,y)]_{x,y}$.
Another major open question is if randomized and quantum
communication complexity are polynomially related for all total
functions.  We should mention here that the assumption of the
function being total is crucial as an exponential separation is
known for a partial function \cite{Raz99}.

One approach to these questions has been to study them for
restricted classes of functions. Many functions of interest are
{\em block composed} functions.  For finite sets $X,Y$, and $E$, a
function $f: E^n \rightarrow \{-1,+1\}$, and a function $g : X \times Y \rightarrow
E$, the block composition of $f$ and $g$ is the function $f \circ
g^n: X^n \times Y^n \to \{-1,+1\}$ defined by $(f \circ
g^n)(x,y)=f(g(x^1, y^1), \ldots, g(x^n, y^n))$ where $(x^i,y^i)
\in X \times Y$ for all $i=1, \ldots, n$. For example, if $E=\{-1,+1\}$,
the inner product function results when $f$ is PARITY and $g$ is AND,
set-intersection when $f$ is OR and $g$ is AND, and the equality
function when $f$ is AND and $g$ is the function IS-EQUAL, which
is one if and only if $x=y$.

In a seminal paper, Razborov \cite{Raz03} gave tight bounds for
the bounded-error quantum communication complexity of block composed
functions where the outer function $f$ is symmetric and the inner
function $g$ is bitwise AND.  In particular, this result showed that
randomized and quantum communication complexity are polynomially
related for such functions.

More recently, very nice frameworks have been developed by
Sherstov \cite{She07,She09b} and independently by Shi and Zhu
\cite{SZ09b} to bound the quantum complexity of block composed
functions that goes beyond the case of symmetric $f$ to work for
any $f$ provided the inner function $g$ satisfies
certain technical conditions. When $g$ satisfies these conditions,
this framework allows one to lower bound the quantum communication
complexity of $f \circ g^n$ in terms of the approximate polynomial
degree of $f$, a classically well-studied measure.  Shi and Zhu are
able to get a bound on $f \circ g^n$ in terms of the approximate
degree of $f$ whenever $g$ is sufficiently ``hard''---unfortunately, the
hardness condition they need is in terms of ``spectral discrepancy,''
a quantity which is somewhat difficult to bound, and their bound
requires that $g$ is a function on at least $\Omega(\log(n/d))$ bits,  where $d$ is the
approximate polynomial degree of $f$. Because of this, Shi-Zhu are
only able to reproduce Razborov's results with a polynomially
weaker bound.

Sherstov developed so-called {\it pattern matrices} which are the
matrix representation of a block composed function when $g$ is a
fixed function of a particularly nice form.  Namely, in a pattern
matrix the inner function $g: \{-1,+1\}^k \times ([k] \times
\{-1,+1\}) \to \{-1,+1\}$ is parameterized by a positive integer
$k$ and defined by $g(x,(i,b))=x_i \cdot b$, where $x_i$ denotes
the $i^{th}$ bit of $x$. In other words, the first argument of $g$ is a $k$
bit string $x$, and the second argument selects a bit of $x$ or its negation.
So here $X=\{-1,+1\}^k$, $Y = [k] \times
\{-1,+1\}$ and the intermediate set $E$ is $\{-1,+1\}$.  With this $g$,
Sherstov shows that the approximate polynomial degree of $f$ is a lower
bound on the quantum communication complexity of $f \circ g^n$,
for any function $f$. Though seemingly quite special, pattern matrices
have proven to be an extremely useful concept.  First, they give a simple proof of
Razborov's tight lower bounds for $f(x\wedge y)$ for symmetric
$f$. Second, they have also found many other applications in
unbounded-error communication complexity \cite{She08b,RS08} and
have been successfully extended to multiparty communication
complexity \cite{Cha07,LS09,CA08,BHN09}.

A key step in both the works of Sherstov and Shi-Zhu is to bound
the spectral norm of a sum of matrices $\sn{\sum_i {B_i}}$.  This
is the major step where these works differ.  Shi-Zhu apply the
triangle inequality to bound this as $\sn{\sum_i {B_i}} \leq
\sum_i \sn{B_i}$. On the other hand, Sherstov observes that in the case of
pattern matrices the terms of this sum are mutually orthogonal,
\ie $B_i^\dag B_j = B_i B_j^\dag = 0$ for all $i\neq j$.  In this case,
one has a stronger bound on the spectral norm $\sn{\sum_i {B_i}} =
\max_i \sn{B_i}$.

In this paper, we extend both of the frameworks of Sherstov and Shi-Zhu.
In the case of Shi-Zhu, we are able to reprove their theorem with
the usual notion of discrepancy instead of the somewhat awkward spectral discrepancy they
use.  The main observation we make is that as all Shi-Zhu use in this step is the
triangle inequality, we can repeat the argument with any norm here, including discrepancy itself.

In the case of pattern matrices, special properties of the spectral norm are used,
namely the fact about the spectral norm of a sum of orthogonal matrices.
We step back to see what key features of a pattern matrix lead to this
orthogonality property.
We begin with the Boolean case, that is, where the intermediate set $E$ is taken
to be $\{-1,+1\}$.  In this case, a crucial concept is the notion of a {\em strongly
balanced} function.  We say that $g : X \times Y \rightarrow \{-1,+1\}$ is strongly
balanced if in the sign matrix $M_g[x,y]=g(x,y)$ all rows and all columns sum to zero.
We show that whenever the inner function $g$ is strongly balanced, the key orthogonality
condition holds; this implies that whenever $g$ is strongly balanced and the communication
matrix of $g$ has rank larger than one, the approximate degree of the outer function $f$ is a lower
bound on the quantum communication complexity of $f \circ g^n$.

The requirement that the communication matrix of $g$ has rank larger than one is necessary for
such a statement.  For example, when $g(x,y)=\oplus(x,y)$ is the
XOR function on one bit, then the communication complexity of $\mathrm{PARITY} \circ g^n$
is constant, while PARITY has linear approximate polynomial degree.  It turns out that when
$g$ is rank-one, the appropriate measure of the complexity of $f \circ g^n$ is no longer the
approximate degree of $f$, but the minimum $\ell_1$ norm of Fourier coefficients of a function
entrywise close to $f$; see the survey \cite{LS09d} for a description of this case.

We also consider the general case where the intermediate set is any group $G$. That is, we
consider functions $F(x,y) = f(g(x,y))$, where $g: X \times Y \rightarrow G$ for a group $G$ and
$f: G \rightarrow \{-1, +1\}$ is a class function on $G$. The case $E = \{-1,+1\}$ discussed above
corresponds to taking the group $G=\mathbb{Z}_2^n$.  When $G$ is a general Abelian group, the key
orthogonality condition requires more than that the matrix $M_g[x,y]=g(x,y)$ is strongly balanced;
still, it admits a nice characterization in terms of group invariance.  A multiset $T\in G\times G$ is
said to be $G$-invariant if $(s,s)T = T$ for all $s\in G$. The orthogonality condition will hold if and
only if all pairs of rows and all pairs of columns of $M_g$ (when viewed as multisets) are $G$-
invariant. One can generalize the results discussed above to this general setting with appropriate
modifications. In the case that $G = \mathbb{Z}_2^n$, the $G$-invariant condition degenerates to the
strongly balanced requirement of $M_g$.

\section{Preliminaries}
\label{sec:main_preliminaries}
All logarithms are base two.  For a complex number $z=a+ib$ we let
$\bar z=a- ib$ denote the complex conjugate of $z$ and $|z|=\sqrt{a^2+b^2}$
and $\mathrm{Re}(z)=a$.

\subsection{Complexity measures}
We will make use of several complexity measures of functions and
matrices. Let $f: \{-1,+1\}^n \rightarrow \{-1,+1\}$ be a
function. For $T \subseteq \{0,1\}^n$, the Fourier
coefficient of $f$ corresponding to the character $\chi_T$ is
$\hat f_T=\frac{1}{2^n} \sum_x f(x) \chi_T(x)=\frac{1}{2^n} \sum_x f(x) \prod_{i \in T} x_i.$
The {\em degree} of $f$ as a polynomial, denoted $\deg(f)$, is the
size of a largest set $T$ for which $\hat f_T \ne 0$.

We will need some notations for matrices. We reserve $J$ for the all ones matrix,
whose size will be determined by the context.  For a matrix $A$ let
$A^\dag$ denote the conjugate transpose of $A$.  We use $A \bullet B$ for the
entrywise product of $A,B$, and $A \otimes B$ for the tensor product.
If $A$ is an $m$-by-$n$ matrix then we say that $\mathrm{size}(A)=mn$.  We
use $\braket{A}{B}=\mathrm{Tr}(AB^\dag)$ for the inner product of $A$ and $B$.

Let $\|A\|_1$ be the $\ell_1$ norm of $A$, i.e.\ sum of the absolute values
of entries of $A$, and $\|A\|_\infty$ the $\ell_\infty$ norm.
For a positive semidefinite matrix $M$ let $\lambda_1(M) \ge \cdots \ge \lambda_n(M) \ge 0$ be
the eigenvalues of $M$.  We define the $i^{th}$ singular value of
$A$, denoted $\sigma_i(A)$, as
$\sigma_i(A)=\sqrt{\lambda_i(AA^\dag)}$.  The rank of $A$, denoted
$\mathrm{rk}(A)$ is the number of nonzero singular values of $A$.  We will
use several matrix norms. The spectral or operator norm is the largest singular value $\sn{A}=\sigma_1(A)$, the trace norm is the summation of all singular values $\trn{A}=\sum_i \sigma_i(A)$, and the Frobenius norm is the $\ell_2$ norm of the singular values $\fro{A}=\sqrt{\sum_i \sigma_i(A)^2}$.

When $AB^\dag =A^\dag B=0$ we will say that $A,B$ are {\em orthogonal}. Please note the difference with the common use of this term, which usually means $\braket{A}{B}=0$. The following facts are easily seen.
\begin{fact}\label{fact:ortho}
Let $A,B$ be two matrices of the same dimensions and suppose that
$AB^\dag=A^\dag B=0$.  Then
\begin{align*}
&\mathrm{rk}(A+B) =\mathrm{rk}(A)+\mathrm{rk}(B), \ \trn{A+B} =\trn{A}+\trn{B}, \ \sn{A+B} =\max \{\sn{A},\sn{B}\}.
\end{align*}
\end{fact}

Another norm we will use is the $\gamma_2$ norm, introduced to complexity theory
in \cite{LMSS07}, and familiar in matrix analysis as the Schur product operator norm.
The $\gamma_2$ norm can be viewed as a weighted version of the trace norm.
\begin{definition}
$$\gamma_2(A)=\max_{u,v: \|u\|=\|v\|=1} \trn{A \bullet uv^\dag}.$$
\end{definition}
Here $A \bullet B$ denotes the entrywise product of $A$ and $B$.
It is clear from this definition that $\gamma_2(A) \ge \trn{A}/\sqrt{mn}$ for a
$m$-by-$n$ matrix $A$.

For a norm $\Phi$, the dual norm $\Phi^*$ is defined as
$\Phi^*(v)=\max_{u: \Phi(u) \le 1} |\braket{u}{v}|$.  For example, the
$\ell_\infty$ norm is dual to the $\ell_1$ norm, and the
spectral norm is dual to the trace norm.

The norm $\gamma_2^*$, dual to the $\gamma_2$ norm, looks as follows.
\begin{definition}
$$
\gamma_2^*(A)=\max_{\substack{u_i,v_j \\ \|u_i\|=\|v_j\|=1}} \sum_{i,j} A[i,j] \braket{u_i}{v_j}.
$$
\end{definition}

Another complexity measure we will make use of is discrepancy
\begin{definition}
Let $A$ be an $m$-by-$n$ sign matrix and let $P$ be a probability distribution on the entries
of $A$.  The discrepancy of $A$ with respect to $P$, denoted $\mathrm{disc}_P(A)$, is defined as
$$
\mathrm{disc}_P(A)=\max_{\substack{x \in \{0,1\}^{m} \\ y \in \{0,1\}^{n}}} |x^\dag A\bullet P y|.
$$
\end{definition}

We will write $\mathrm{disc}_U(A)$ for the special case where $P$ is the uniform distribution.
It is easy to see from this definition that
$
\mathrm{disc}_U(A) \le \frac{\sn{A}}{\sqrt{\mathrm{size}(A)}}.
$
Shaltiel \cite{Sha03} has shown the deeper result that this bound is in fact polynomially tight:
\begin{theorem}[Shaltiel]\label{thm:shaltiel}
Let $A$ be a sign matrix. Then
$$\frac{1}{108} \left(\frac{\sn{A}}{\sqrt{\mathrm{size}(A)}}\right)^3 \le \mathrm{disc}_U(A).$$
\end{theorem}

Discrepancy and the $\gamma_2^*$ norm are very closely related.  Linial and Shraibman
\cite{LS09b} observed that Grothendieck's inequality gives the following.
\begin{theorem}[Linial-Shraibman]
For any sign matrix $A$ and probability distribution $P$
$$
\mathrm{disc}_P(A) \le \gamma_2^*(A \bullet P) \le K_G \; \mathrm{disc}_P(A)
$$
where $1.67\ldots \le K_G \le 1.78\ldots$ is Grothendieck's constant.
\end{theorem}

\subsubsection*{Approximate measures}
We will also use approximate versions of these complexity measures
which come in handy when working with bounded-error
models.  Say that a function $g$ gives an
$\epsilon$-approximation to $f$ if $|f(x) - g(x)| \le \epsilon$
for all $x \in \{-1,+1\}^n$.  The $\epsilon$-approximate
polynomial degree of $f$, denoted $\deg_\epsilon(f)$, is the
minimum degree of a function $g$ which gives an
$\epsilon$-approximation to $f$.

We will similarly look at the $\epsilon$-approximate version
of the trace and $\gamma_2$ norms.  We give the general definition with respect to
any norm.
\begin{definition}[approximation norm]
Let $\Phi: \mathbb{R}^n \rightarrow \mathbb{R}$ be an arbitrary norm.  Let
$v \in \mathbb{R}^n$ be a sign vector.  For $0 \le \epsilon < 1$ we define
the approximation norm $\Phi^\epsilon$ as
$$
\Phi^\epsilon(v)=\min_{\substack{u \\ \|v-u\|_\infty \le \epsilon}} \Phi(u).
$$
\end{definition}
Notice that an approximation norm $\Phi^\epsilon$ is not itself a norm---
we have only defined it for sign vectors, and it will in general not satisfy
the triangle inequality.

As a norm is a convex function, using the separating hyperplane theorem
one can quite generally give the following equivalent dual formulation
of an approximation norm.
\begin{proposition}
\label{prop:dual_approximate_norm} Let $v \in \mathbb{R}^n$ be a sign vector, and
$0 \le \epsilon < 1$
$$
\Phi^\epsilon(v) = \max_u
\frac{|\braket{v}{u}|-\epsilon\|u\|_1}{\Phi^*(u)}
$$
\end{proposition}
A proof of this can be found in the survey \cite{LS09d}.

\subsection{Communication complexity}
Let $X,Y,S$ be finite sets and $f: X \times Y \rightarrow S$ be a function.
We will let $D(f)$ be the deterministic communication complexity of $f$,
and $R_\epsilon(f)$ denote the randomized public coin complexity of
$f$ with error probability at most $\epsilon$.  We refer to the reader to
\cite{KN97} for a formal definition of these models.  We will also study
$Q_\epsilon(f)$ and $ Q_\epsilon^*(f)$, the $\epsilon$-error quantum communication
complexity of $f$ without and with shared entanglement, respectively.
We refer the reader to \cite{Raz03} for a nice description of these models.

For notational convenience, we will identify a function
$f: X \times Y \rightarrow \{-1,+1\}$ with its sign matrix $M_f = [f(x,y)]_{x,y}$.  Thus,
for example, $\sn{f}$ refers to the spectral norm of the sign matrix representation
of $f$.

For all of our lower bound results we will actually lower bound the approximate
trace norm or $\gamma_2$ norm of the function.  Razborov showed that the approximate
trace norm can be used to lower bound on quantum communication complexity, and
Linial and Shraibman generalized this to the $\gamma_2$ norm.
\begin{theorem}[Linial-Shraibman \cite{LS09c}] \label{thm: approx trace}
Let $A$ be a sign matrix and $0 \le \epsilon < 1/2$.  Then
$$
Q_\epsilon^*(A) \ge \log\left(\gamma_2^{2\epsilon}(A) \right) - 2.
$$
\end{theorem}

\subsubsection*{Composed functions} \vspace{-1em}
Before discussing lower bounds on a block composed function $f \circ g^n$,
let us see what we expect the complexity of such a function to be. A
fundamental idea going back to Nisan \cite{Nis94} and Buhrman, Cleve, and Wigderson
\cite{BCW98}, is that the complexity of $f \circ g^n$ can be related
to the {\em decision tree} complexity, also known as {\em query} complexity, of $f$ and the
communication
complexity of $g$. Let $\mathrm{DT}(f)$ be the query complexity of
$f$, that is the number of queries of the form $x_i=?$ needed to
evaluate $f(x)$ in the worst case.  Similarly, let $\mathrm{RT}_{\epsilon}(f),
\mathrm{QT}_{\epsilon}(f)$ denote the randomized and quantum query
complexity of $f$ respectively, with error probability at most
$\epsilon$.  For formal definitions of these measures and a survey of
query complexity we recommend Buhrman and de Wolf \cite{BW02}.
\begin{theorem}[Nisan \cite{Nis94}, Buhrman-Cleve-Wigderson \cite{BCW98}]\label{thm:composed_upper}
For any two Boolean functions $f: \{-1,+1\}^n \rightarrow \{-1,+1\}$ and
$g: X \times Y \rightarrow \{-1,+1\}$,
\begin{align*}
D(f \circ g^n)&=O(\mathrm{DT}(f) D(g)) \\
R_{1/4}(f \circ g^n)&=O(\mathrm{RT}_{1/4}(f) R_{1/4}(g) \log \mathrm{RT}_{1/4}(f)) \\
Q_{1/4}(f \circ g^n)&=O(\mathrm{QT}_{1/4}(f) Q_{1/4}(g) \log n).
\end{align*}
\end{theorem}

One advantage of working with block composed functions in light of this upper bound is that
query complexity is in general better understood than communication complexity.  In particular,
a polynomial relationship between deterministic query complexity and degree,
and randomized and quantum query complexities and approximate degree is known.
\begin{theorem}[\cite{NS94,BBCMW01}]
Let $f: \{0,1\}^n \rightarrow \{-1,+1\}$.  Then
\begin{align*}
\mathrm{DT}(f) =O(\deg(f)^4), \quad \mathrm{DT}(f) =O(\deg_{1/4}(f)^6)
\end{align*}
\end{theorem}

Using this result together with Theorem~\ref{thm:composed_upper} gives the following corollary:
\begin{corollary}\label{cor:deg}
\begin{align*}
D(f \circ g^n) &= O(\deg(f)^4 D(g)), \quad
R_{1/4}(f \circ g^n) = O(\deg_{1/4}(f)^6 R_{1/4}(g) \log
\deg_{1/4}(f))
\end{align*}
\end{corollary}

Our goal, then, in showing lower bounds on the complexity of a block composed function
$f \circ g^n$ is to get something at least in the ballpark of this upper bound.  Of course, this is not
always possible --- the protocol given by Theorem~\ref{thm:composed_upper} is not always
optimal.  For example, when $f$ is the PARITY function on $n$ bits, and $g(x,y)=\oplus(x,y)$ this
protocol just gives an upper bound of $n$ bits, when the true complexity is constant. See recent
results by Zhang \cite{Zha09} and Sherstov \cite{She10} for discussions on the tightness of the
bounds in Theorem \ref{thm:composed_upper}.

\section{Rank of block composed functions}\label{sec: strongly balanced, rank}
We begin by analyzing the rank of a block composed function $f \circ g^n$ when the inner function
$g$ is strongly balanced.  This case will illustrate the use of the strongly balanced assumption, and
is simpler to understand than the bounded-error situation treated in the next section.

Let us first formally state the definition of strongly balanced.
\begin{definition}[strongly balanced]
Let $A$ be a sign matrix, and $J$ be the all ones matrix of the
same dimensions as $A$.  We say that $A$ is {\em balanced} if
$\mathrm{Tr}(A J^\dag)=0$.  We further say that $A$ is {\em strongly
balanced} if $A J^\dag=A^\dag J=0$.  In words, a sign matrix is strongly
balanced if the sum over each row is zero, and similarly the sum over
each column is zero. We will say that a two-variable Boolean function is balanced or strongly
balanced if its sign matrix representation is.
\end{definition}

\begin{theorem}
\label{th:rank_compose} Let $f : \{-1,+1\}^n \rightarrow
\{-1,+1\}$ be an arbitrary function, and let $g$ be a strongly
balanced function. Then
$$
\mathrm{rk}(M_{f \circ g^n})=\sum_{\substack{T \subseteq [n],\ \hat f_T \ne 0}} \mathrm{rk}(M_g)^{|T|}.
$$
\end{theorem}

\begin{proof}
Let us write out the sign matrix for $\chi_T \circ g^n$ explicitly.  If we let $M_g^0=J$ be the all
ones matrix and $M_g^1=M_g$, then we can nicely write the sign matrix representing
$\chi_T(g(x^1, y^1), \ldots, g(x^n,y^n))$ as
$$
M_{\chi_T\circ g^n}=\bigotimes_i M_g^{T[i]}
$$
where $T[i]=1$ if $i \in T$ and $0$ otherwise.

We see that the condition on $g$ implies $M_{\chi_T \circ g^n}
M_{\chi_S \circ g^n}^\dag=0$ if $S \ne T$. Indeed,
\begin{align*}
M_{\chi_T \circ g^n} M_{\chi_S \circ g^n}^\dag & =
\left(\bigotimes_i M_g^{T[i]}\right) \left( \bigotimes_i M_g^{S[i]}\right)^\dag \\
 & =\bigotimes_i \left(M_g^{T[i]} (M_g^{S[i]})^\dag \right) =0.
\end{align*}
This follows since, by the assumption $S \ne T$, there is some $i$ for which $S[i] \ne T[i]$ which
means that this term is either $M_g J^\dag=0$ or $J M_g^\dag=0$ because $g$ is strongly balanced.
The other case follows similarly.

Now that we have established this property, we can use Fact~\ref{fact:ortho} to obtain
\begin{align*}
 \mathrm{rk}(M_{f \circ g^n}) & =  \mathrm{rk}\Big(\sum_{T \subseteq [n]} \hat f_T \chi_T(g(x^1, y^1), \ldots, g(x^n,y^n))\Big)  \\
 & = \sum_{\substack{T \subseteq [n] \\ \hat f_T \ne 0}} \mathrm{rk}(M_{\chi_T \circ g^n}) \\
 & =\sum_{\substack{T \subseteq [n] \\ \hat f_T \ne 0}} \mathrm{rk}(M_g)^{|T|}
\end{align*}
In the last step we used the fact that rank is multiplicative
under tensor product.
\end{proof}

Theorem~\ref{th:rank_compose} has the following implication for the log rank conjecture of the composed function with the assumption of the same conjecture for the inner function.
\begin{corollary}
Let $X,Y$ be finite sets, $g: X \times Y \rightarrow \{-1,+1\}$
be a strongly balanced function, and $M_g[x,y]=g(x,y)$ be the
corresponding sign matrix.  Let $f : \{-1,+1\}^n \rightarrow
\{-1,+1\}$ be an arbitrary function. Assume that $\mathrm{rk}(M_g) \ge 2$
and further suppose that there is a constant $c$ such that $D(g)
\le (\log \mathrm{rk}(M_g))^c$. Then
$$
D(f \circ g^n)=O( \log \mathrm{rk}(f \circ g)^{4+c}).
$$
\end{corollary}

\begin{proof}
By Corollary~\ref{cor:deg},
$D(f \circ g^n)=O(\deg(f)^4 D(g))=O(\deg(f)^4 (\log \mathrm{rk}(M_g))^c)$. Now, it follows from
Theorem~\ref{th:rank_compose} that $\log \mathrm{rk}(f \circ g) \ge
\deg(f)\log \mathrm{rk}(M_g)$ as by definition of degree there is some $T
\subseteq \{0,1\}^n$ with $|T|=\deg(f)$ and $\hat f_T \ne 0$.
\end{proof}

In particular, this Corollary means that whenever $g$ is a strongly balanced function on a constant
number of bits and $\mathrm{rk}(M_g) >1$, then the log rank conjecture holds for $f \circ g^n$.  If $g$ is strongly
balanced and $\mathrm{rk}(M_g)=1$ then, up to permutation of rows and columns, which does not change
the communication complexity, $M_g$ is a tensor product of the XOR function with an all ones matrix.
The log rank conjecture in the case
$(f \circ \oplus^n) (x,y)=f(x_1 \oplus y_1, \ldots, x_n \oplus y_n)$ remains an interesting open question.
Shi and Zhang \cite{SZ09} have recently resolved this question when $f$ is symmetric.

\section{A bound in terms of approximate degree}
\label{sec: strongly balanced, disc}

In this section, we will address the frameworks of Sherstov and Shi-Zhu.  We
extend both of these frameworks to give more general conditions on the inner function $g$
which still imply that the approximate degree of $f$ is a lower bound on the quantum query
complexity of the composed function $f \circ g^n$.  In outline, both of these frameworks
follow the same plan.  By Theorem~\ref{thm: approx trace} it suffices to lower bound the
approximate $\gamma_2$ norm (or even approximate trace norm) of $f \circ g^n$.  To do this,
they use the  dual formulation given by Proposition~\ref{prop:dual_approximate_norm}
and construct a witness matrix $B$ which has non-negligible correlation with the target function and small $\gamma_2^*$ (or spectral) norm.

A very nice way to construct this witness, used by both Sherstov and Shi-Zhu, is to use the
{\em dual polynomial} of $f$.  This is a polynomial $v$ which certifies that the approximate
polynomial degree of $f$ is at least a certain value.
More precisely, duality theory of linear programming
gives the following lemma.
\begin{lemma}[Sherstov
\cite{She09b}, Shi-Zhu
\cite{SZ09b}]
Let $f:\{-1,+1\}^n$ $ \rightarrow \{-1,+1\}$ and let $d=\deg_{\epsilon}(f)$.
Then there exists a function $v: \{-1,+1\}^n \rightarrow \mathbb{R}$ such that
\begin{enumerate}
  \item $\braket{v}{\chi_T}=0$ for every character $\chi_T$ with $|T|< d$.
  \item $\|v\|_1=1$.
  \item $\braket{v}{f} \ge \epsilon$.
\end{enumerate}
\label{dual_poly}
\end{lemma}
Items (2),(3) are used to lower bound the correlation of the witness matrix with the target
matrix and to upper bound the $\ell_1$ norm of the witness matrix.   In the most difficult step,
and where these works diverge, Item~(1) is used to upper bound the $\gamma_2^*$
(or spectral) norm of the witness matrix.

We treat each of these frameworks separately in the next two sections.

\subsection{Sherstov's framework}

The proof of the next theorem follows the same steps as Sherstov's proof for pattern matrices
(Theorem 5.1 \cite{She09b}). Our main contribution is to identify the strongly balanced condition
as the key property of pattern matrices which enables the proof to work.

\begin{theorem}
Let $X,Y$ be finite sets, $g: X \times Y \rightarrow \{-1,+1\}$
be a strongly balanced function, and $M_g[x,y]=g(x,y)$ be the
corresponding sign matrix.  Let $f : \{-1,+1\}^n \rightarrow
\{-1,+1\}$ be an arbitrary function.  Then
\begin{equation*}
    Q_{\epsilon}^*(f \circ g^n) \ge \deg_{\epsilon_0}(f) \log_2 \Big(\frac{\sqrt{|X| |Y|}}{\sn{M_g}}
    \Big) - O(1).
\end{equation*}
for any $\epsilon >0$ and $\epsilon_0 > 2\epsilon$.
\label{thm: block lb}
\end{theorem}
In particular, this result means that the quantum and randomized
complexities of $f \circ g^n$ are polynomially related whenever
$g$ is strongly balanced and $\log
\tfrac{\sqrt{\mathrm{size}(M_g)}}{\sn{M_g}}$ is polynomially related to
the randomized communication complexity of $g$.  While the complexity measure of $g$ used
here may look strange at first, Shaltiel \cite{Sha03} has shown that it is closely related
to the discrepancy of $g$ under the uniform distribution, as noted above in Theorem~\ref
{thm:shaltiel}.  This theorem strictly generalizes the case of pattern matrices,
but it could still be the case that the results of Shi-Zhu can
show bounds not possible with this theorem.

\begin{proof}[Proof of Theorem~\ref{thm: block lb}]
Let $d=\deg_{\epsilon_0}(f)$ and let $v$ be a dual polynomial for $f$ with properties as in
Lemma~\ref{dual_poly}.  We define a witness matrix as
$$
B[x,y]=\frac{2^n}{\mathrm{size}(M_g)^n} v(g(x^1, y^1), \ldots, g(x^n, y^n))
$$

Let us first lower bound the inner product $\braket{M_{f \circ g^n}}{B}$.  Notice that as $M_g$ is
strongly balanced, it is in particular balanced, and so the number of ones (or minus ones) in
$M_g$ is $\mathrm{size}(M_g)/2$.
\begin{align*}
 \braket{M_{f \circ g^n}}{B}
= \frac{2^n}{\mathrm{size}(M_g)^n}\sum_{z \in \{-1,+1\}^n} f(z) v(z)
\prod_{i=1}^n \Big( \sum_{\scriptsize x^i, y^i: \atop \scriptsize g(x^i,y^i)=z_i} 1\Big)
= \braket{f}{v} \ge \epsilon_0
\end{align*}
A similar argument shows that $\|B\|_1=1$ as $\|v\|_1=1$.

Now we turn to evaluate $\sn{B}$.  As shown above, the strongly balanced property of $g$
implies that the matrices  $\chi_T \circ g^n$ and $\chi_S \circ g^n$ are
orthogonal for distinct sets
$S,T \subseteq \{0,1\}^n$.  We can thus use Fact~\ref{fact:ortho} to compute as follows.
\begin{align*}
\sn{B}&=\frac{2^n}{\mathrm{size}(M_g)^n}\sn{\sum_{T \subseteq [n]} \hat v_T M_{\chi_T \circ g^n}} \\ & =\frac{2^n}{\mathrm{size}(M_g)^n} \max_{T} |\hat v_T| \sn{M_{\chi_T \circ g^n}} \\
&=\max_{T} \ 2^n |\hat v_T| \prod_{i} \frac{\sn{M_g^{T[i]}}}{\mathrm{size}(M_g)} \\
&\le \max_{T: \hat v_T \ne 0} \prod_{i} \frac{\sn{M_g^{T[i]}}}{\mathrm{size}(M_g)} \\
& = \left(\frac{\sn{M_g}}{\sqrt{\mathrm{size}(M_g)}}\right)^d \left(\frac{1}{\mathrm{size}(M_g)} \right)^{n/2}
\end{align*}
In the second to last step we have used that $|\hat v_T| \le 1/2^n$ as $\|v\|_1=1$, and in the
last step we have used the fact that $\sn{J}=\sqrt{\mathrm{size}(M_g)}$.

Now putting everything together we have
\begin{align*}
\frac{\trn{M_{f \circ g^n}}^{\epsilon_0}}{\sqrt{\mathrm{size}(M_{f \circ g^n})}} \ge
\frac{1}{12} \left(\frac{\sqrt{\mathrm{size}(M_g)}}{\sn{M_g}}\right)^d
\end{align*}
The lower bound on quantum communication complexity now follows from
Theorem~\ref{thm: approx trace}.
\end{proof}

Using the theorem of Shaltiel relating discrepancy to the spectral norm
Theorem~\ref{thm:shaltiel}, we get the following corollary:
\begin{corollary}\label{cor:disc}
Let the quantities be defined as in Theorem~\ref{thm: block lb}.
$$
Q_{1/8}^*(f \circ g^n) \geq
\frac{1}{3}\deg_{1/3}(f) \big(\log \big(\frac{1}{\mathrm{disc}_U(M_g)}\big)-7 \big) -O(1).
$$
\end{corollary}

\noindent {\bf Comparison to Sherstov's pattern matrix}: As mentioned in \cite{She09b}, Sherstov's pattern matrix method can prove quantum lower bound of $\Omega(\deg_{\epsilon}(f))$ for block composed functions $f\circ g^n$ if the matrix $M_g$ contains the following $4\times 4$ one as a submatrix:
\begin{equation*}
	S_4 =
	\begin{bmatrix}
	 1 & -1 & 1 & -1 \\
   1 & -1 & -1 & 1 \\
   -1 & 1 & 1 & -1 \\
   -1 & 1 & -1 & 1
	\end{bmatrix}
\end{equation*}
In this paper we show that the same lower bound holds as long as $M_g$ contains a
strongly balanced submatrix or rank greater than one. Are there strongly balanced matrices not
containing $S_4$ as a submatrix? It turns out that the answer is yes: we give the following
$6\times 6$ matrix as one example.
\begin{equation*}
	S_6 =
	\begin{bmatrix}
		1 & 1 & 1 & -1 & -1 & -1  \\
		1 & 1 & -1 & 1 & -1 & -1  \\
		1 & -1 & -1 & -1 & 1 & 1  \\
		-1 & -1 & 1 & 1 & 1 & -1  \\
		-1 & 1 & -1 & -1 & 1 & 1  \\
		-1 & -1 & 1 & 1 & -1 & 1
	\end{bmatrix}
\end{equation*}

\subsection{Shi-Zhu framework}
The method of Shi-Zhu does not restrict the form of the inner function $g$, but
rather works for any $g$ which is sufficiently ``hard.''  The hardness condition they
require is phrased in terms of a somewhat awkward measure they term spectral discrepancy.

\begin{definition}[spectral discrepancy]
Let $A$ be a $m$-by-$n$ sign matrix.
The spectral discrepancy of $A$, denoted $\rho(A)$, is the smallest $r$ such that there is
a submatrix $A'$ of $A$ and a probability distribution $\mu$ on the entries of $A'$ satisfying:
\begin{enumerate}
  \item $A'$ is balanced with respect to $\mu$, i.e.\ the distribution which gives equal weight to
  $-1$ entries and $+1$ entries of $A'$.
   \item The spectral norm of $A' \bullet \mu$ is small:
   $$\sn{A' \bullet \mu} \le \frac{r}{\sqrt{\mathrm{size}(A')}}$$
  \item The entrywise absolute value of the matrix $A' \bullet \mu$ should also have a bound on its
  spectral norm in terms of $r$:
  $$\sn{|A' \bullet \mu|} \le \frac{1+r}{\sqrt{\mathrm{size}(A')}}$$
\end{enumerate}
\end{definition}
While conditions (1),(2) in the definition of spectral discrepancy are quite natural,
condition~(3) can be complicated to verify.  Note that condition~(3) will always
be satisfied when $\mu$ is taken to be the uniform distribution.
Using this notion of spectral discrepancy, Shi-Zhu show the following theorem.
\begin{theorem}[Shi-Zhu \cite{SZ09b}]
Let $f : \{-1,+1\}^n \rightarrow \{-1,+1\}$, and $g: X \times Y
\rightarrow \{-1,+1\}$.  For any $\epsilon$ and $\epsilon_0 > 2\epsilon$,
$$
Q_{\epsilon}(f \circ g^n) \ge \Omega(\deg_{\epsilon_0}(f)).
$$
provided $\rho(M_g) \le \tfrac{\deg_{\epsilon_0}(f)}{2en}$.
Here $e=2.718\ldots$ is Euler's number.
\end{theorem}

Chattopadhyay \cite{Cha08} extended the technique of Shi-Zhu to the case of multiparty
communication complexity, answering an open question of Sherstov \cite{She08c}.  In doing
so, he gave a more natural condition on the hardness of $g$ in terms of an upper bound on
discrepancy frequently used in the multiparty setting and originally
due to Babai, Nisan, and Szegedy \cite{BNS92}.
As all that is crucially needed is subadditivity, we do the argument here with $\gamma_2^*$,
which is essentially equal to the discrepancy.

\begin{theorem}
Let $f : \{-1,+1\}^n \rightarrow \{-1,+1\}$, and $g: X \times Y
\rightarrow \{-1,+1\}$.
Fix $0 < \epsilon < 1/2$, and let $\epsilon_0 > 2\epsilon$.
Then
$$
Q_\epsilon^*(f \circ g^n) \ge \deg_{\epsilon_0}(f) -O(1).
$$
provided there is a distribution $\mu$ which is balanced with respect to $g$ and
for which $\gamma_2^*(M_g \bullet \mu) \le \tfrac{\deg_{\epsilon_0}(f)}{2en}$.
\label{thm:shi-zhu}
\end{theorem}

\begin{proof}
We again use Proposition~\ref{prop:dual_approximate_norm}, this time with the
$\gamma_2$ norm instead of the trace norm.
$$
\gamma_2^{\epsilon_0}(M_{f \circ g^n}) = \max_B
\frac{\braket{M_{f \circ g^n}}{B} -\epsilon_0 \|B\|_1}{\gamma_2^*(B)}.
$$
To prove a lower bound we choose a witness matrix $B$ as follows
$$
B[x,y]=2^n \cdot v(g(x^1,y^1), \ldots, g(x^n,y^n)) \cdot \prod_{i=1}^n \mu(x^i,y^i).
$$
where $v$ witnesses that $f$ has approximate degree at least $d=\deg_{\epsilon_0}(f)$.
This definition is the same as in the previous section where $\mu$ was
simply the uniform distribution.  As argued before, we have
$\braket{M_{f \circ g^n}}{B} \ge \epsilon_0$ and $\|B\|_1=1$ because
$M_g \bullet \mu$ is balanced.

We again expand $B$ as
$$
B=2^n \sum_{T: |T| \ge d} \hat v_T \bigotimes_{i=1}^n (M_g \bullet \mu)^{T(i)},
$$
where $(M_g \bullet \mu)^1=M_g \bullet \mu$ and $(M_g \bullet \mu)^0=\mu$.

Now comes the difference with the previous proof.  As we do not have special knowledge
of the function $g$, we simply bound $\gamma_2^*(B)$ using the triangle inequality.
\begin{align*}
\gamma_2^*(B) &\le 2^n
\sum_{T: |T| \ge d} |\hat v_T| \ \gamma_2^*\left(\bigotimes_{i=1}^n (M_g \bullet \mu)^{T(i)} \right)\\
&= 2^n \sum_{T: |T| \ge d} |\hat v_T| \ \gamma_2^*(M_g \bullet \mu)^{|T|} \gamma_2^*(\mu)^{n-|T|} \\
&\le \sum_{T: |T| \ge d} \gamma_2^*(M_g \bullet \mu)^{|T|},
\end{align*}
where in the last step we have used that $\gamma_2^*(\mu) \le 1$ as $\mu$ is a probability
distribution and that $|\hat v_T| \le 2^{-n}$. In the second step (equality) we used the fact
that $\gamma_2^*$ is multiplicative with respect to tensor product, a property proved
in \cite{LSS08}.
We continue with simple arithmetic:
\begin{align*}
\gamma_2^*(B) &\le \sum_{i=d}^n {n \choose i} \gamma_2^*(M_g \bullet \mu)^i \\
&\le \sum_{i=d}^n \left( \frac{en \gamma_2^*(M_g \bullet \mu)}{d} \right)^i \\
&\le 2^{-d}
\end{align*}
provided that $\gamma_2^*(M_g \bullet \mu) \le \frac{d}{2en}$.
\end{proof}

\section{A general framework for functions composed through a group} \label{sec:group}
In this section we begin the study of more general function composition
through a group $G$.  In this case the outer function $f: G \rightarrow \{-1,+1\}$
is a class function, i.e.\ invariant on conjugacy classes, and the inner function
$g: X \times Y \to G$ has range $G$.  We define the composed function as
$F(x,y)=f(g(x,y))$.  In previous sections of the paper we have just dealt with the case
$G=\mathbb{Z}_2^n$.

Let us recall the basic idea of the proof of Theorem~\ref{thm: block lb}.  To prove a lower bound on
the quantum communication complexity for a composed function $f \circ g$, we constructed a
witness
matrix $B$ which had non-negligible correlation with $f \circ g$ and small spectral norm.
To do this, following the work of Sherstov and
Shi-Zhu \cite{She09b,SZ09b}, we considered the dual polynomial $p$ of $f$ using LP duality.  The
dual polynomial has two important properties, first that $p$ has non-negligible correlation with $f$
and
second that $p$ has no support on low degree polynomials.  We can then use the first property to
show that the composed function $p \circ g$ will give non-negligible inner
product with $f \circ g$ and the second to upper bound the spectral norm of $p \circ g$.
The second of these tasks is the more difficult. In the case of $G = \{-1,+1\}^n$, the degree of a
character $\chi_T$ is a natural
measure of how ``hard'' the character is --- the larger $T$ is, the smaller the spectral norm of
$\chi_T \circ g$ will be.  In the general group case, however, it is less clear what the corresponding ``hard'' and ``easy'' characters should be. In Section \ref{sec: general functions}, we will show that this framework actually works for an arbitrary partition of the basis functions into Easy and Hard.  That is for any arbitrary partition of the basis functions into Easy and Hard sets we can follow the plan outlined above and look for a function with support on the Hard set which has non-negligible correlation with $f$.

In carrying out this plan, one is still left with upper bounding $\|M_{p\circ g}\|$.  Here, as in the Boolean case, it as again very convenient to have an orthogonality condition which can greatly
simplify the computation of $\sn{M_{p \circ g}}$ and give good bounds.
In the Boolean case we have shown that $M_g$ being strongly balanced implies this key orthogonality
condition. In Section \ref{sec: group} and \ref{sec: abelian group}, we will show that for the general group, the condition is not only about each row and column of matrix $M_g$, but all pairs of rows and pairs of columns.  In the Abelian group case, this reduces to a nice group invariance condition.

Even after applying the orthogonality condition to use the maximum bound instead of the triangle inequality for $\|M_{p\circ g}\|$, the remaining term $\|M_{\chi_i \circ g}\|$ (where $\chi_i$ is a ``hard" character) is still not easy to upper bound. For block composed functions, fortunately, the tensor structure makes it feasible to compute. Section \ref{sec: block} gives a generalized version of Theorem \ref{thm: block lb}.

\subsection{General framework}\label{sec: general functions}
For a multiset $T$, $x\in T$ means $x$ running over $T$. Thus $T = \{a(s): s\in S\}$ means the multiset formed by collecting $a(s)$ with $s$ running over $S$.

For a set $S$, denote by $L_{\mathbb C}(S)$ the $|S|$-dimensional vector space over the field $\mbC$ (of complex numbers) consisting of all linear functions from $S$ to $\mbC$, endowed with inner product $\langle \psi, \phi \rangle = \frac{1}{|S|}\sum_{s\in S} \psi(s)\overline{\phi(s)}$. The distance of a function $f\in L_{\mathbb C}(S)$ to a subspace $\Phi$ of $L_{\mathbb C}(S)$, denoted by $d(f,\Phi)$, is defined as $\min\{\delta: \|f'-f\|_\infty \leq \delta, f'\in \Phi\}$, \ie the magnitude of the least entrywise perturbation to turn $f$ into $\Phi$.

In the above setting, Theorem~\ref{thm: block lb} generalizes to the following.

\begin{theorem}\label{thm: lb, general function}
    Consider a sign matrix $A = [f(g(x,y))]_{x,y}$ where $g: X\times Y \rightarrow S$ for a set $S$, and $f: S\rightarrow \{-1, +1\}$. Suppose that we can find an orthogonal basis functions $\Psi = \{\psi_i: i\in [|S|]\}$ for $L_{\mathbb C}(S)$. For any hardness partition $\Psi = \Psi_{Hard} \uplus \Psi_{Easy}$, let $\delta = d(f, span(\Psi_{Easy}))$. If
    \begin{enumerate}
        \item {\text{\bf(regularity)}} The multiset $\{g(x,y): x\in X, y\in Y\}$ is a multiple of $S$, \ie $S$ repeated for some number of times.
        \item {\text{\bf(orthogonality)}} for all $x,x',y,y'$ and all distinct $\psi_i,\psi_j \in \Psi_{Hard}$,
        \begin{align*}
            \sum_y \psi_i(g(x,y))\overline{\psi_j(g(x',y))} =  \sum_x \psi_i(g(x,y))\overline{\psi_j(g(x,y'))} = 0,
        \end{align*}
    \end{enumerate}
    then
    \begin{align*}
        Q_\epsilon(A) \geq \log_2 \frac{\sqrt{MN}\cdot (\delta -2\epsilon)}{\max_{\psi_i\in \Psi_{Hard}} (\max_g |\psi_i(g)| \cdot \|[\psi_{i}(g(x,y))]_{x,y}\|) } - O(1).
    \end{align*}
\end{theorem}
Using the idea of finding a certificate of the high approximate degree by duality \cite{She09b,SZ09b}, we have the following fact analogous to Lemma \ref{dual_poly}.
\begin{lemma}\label{lem: duality}
For a function $f: S\rightarrow \mbC$ and a subspace $\Phi$ of $L_{\mathbb C}(S)$, if $d(f,span(\Phi)) = \delta$, then there exists a function $h$ \st
\begin{align}
    &\label{eq: dual fn, hard} \hat h_i = 0, \ \forall \psi_i\in \Phi
    \\
    &\label{eq: dual fn, 1-norm} \sum_{g\in G} |h(g)| \leq 2,
    \\
    &\label{eq: dual fn, correlation} |\sum_{g\in G} f(g) \overline{h(g)}| > \delta
\end{align}
\end{lemma}
Using the lemma, we can prove the Theorem \ref{thm: lb, general function}.
\begin{proof} (of Theorem \ref{thm: lb, general function})
    By the regularity property, we know that when $(x,y)$ runs over $X\times Y$, $g(x,y)$ runs over $S$ exactly $K$ times where $K = MN/|G|$. Consider $B = \frac{1}{K}[h(g(x,y))]_{x,y}$, $h$ obtained by Lemma \ref{lem: duality}; we want to apply Proposition \ref{prop:dual_approximate_norm} and Theorem \ref{thm: approx trace} by using this $B$. First,
\begin{equation}\label{eq: 1-norm of B}
    \|B\|_1 = \frac{1}{K}\sum_{x,y}|h(g(x,y))| = \sum_{g\in G} |h(g)| \leq 1.
\end{equation}
Also,
\begin{align*}
    |\langle A, B \rangle| & = \frac{1}{K} |\sum_{x,y} f(g(x,y))\overline{h(g(x,y))}| \\
    & = |\sum_{g\in G} f(g)\overline{h(g)}| > \delta
\end{align*}
Now we need to compute \[\|B\| = \frac{1}{K}\big\|[\sum_{\chi_i\in Hard} \hat h_i \chi_i(g(x,y))]_{x,y}\big\|.\] Note that
\begin{align*}
    & [\psi_i(g(x,y))]_{x,y}^\dag [\psi_j(g(x,y))]_{x,y}
    =  [\sum_x \overline{\psi_i(g(x,y))}\psi_j(g(x,y'))]_{y, y'}
\end{align*}
and
\begin{align*}
    & [\psi_i(g(x,y))]_{x,y} [\psi_j(g(x,y))]_{x,y}^\dag
    =  [\sum_y \psi_i(g(x,y))\overline{\psi_j(g(x',y))}]_{x, x'}.
\end{align*}
Thus the orthogonality condition implies that
\begin{align*}
    & [\psi_i(g(x,y))]_{x,y}^\dag [\psi_j(g(x,y))]_{x,y}
    =  [\psi_i(g(x,y))]_{x,y} [\psi_j(g(x,y))]_{x,y} ^\dag = 0
\end{align*}
for all $i\neq j$. Now as in \cite{She09b}, we can use the max bound
\begin{align*}
    \|B\| & = \frac{1}{K}\max_{i: \psi_i\in \Psi_{Hard}} \big\|\hat h_{i} [\psi_{i}(x,y)]_{x,y}\big\| \\
    & \leq \frac{1}{K}\max_{i: \psi_i\in Hard} |\hat h_{i}| \max_{\psi_i\in Hard} \big\| [\psi_{i}(x,y)]_{x,y}\big\| \\
    & \leq \frac{1}{K|G|} \max_{\psi_i\in Hard} \big( \max_{g} |\psi_i(g)| \cdot \big\|[\psi_{i}(x,y)]_{x,y} \big\|\big).
\end{align*}
where the last inequality is due to Eq. \eqref{eq:l-infty bound of F(f)} and Eq. \eqref{eq: 1-norm of B}. Finally note $K|G| = MN$ to complete the proof.
\end{proof}

In the Boolean block composed function case, the regularity condition reduces to the matrix $[g(x,y)]$ being balanced, and later we will prove that the orthogonality condition reduces to the strongly balanced property. From this theorem we can see that the way to partition $\Psi$ into $\Psi_{Easy}$ and $\Psi_{Hard}$ does not really matter for the lower bound proof passing through. However, the partition does play a role when we later bound the spectral norm in the denominator.

\subsection{Functions with group symmetry}\label{sec: group}
For a general finite group $G$, two elements $s$ and $t$ are conjugate, denoted by $s\sim t$, if there exists an element $r\in G$ \st $rsr^{-1} = t$. Define $H$ as the set of all class functions, \ie functions $f$ \st $f(s) = f(t)$ if $s\sim t$. Then $H$ is an $h$-dimensional subspace of $L_{\mathbb C} (G)$, where $h$ is the number of conjugacy classes. The irreducible characters $\{\chi_i: i\in [h]\}$ form an orthogonal basis of $H$. For a class function $f$ and irreducible characters $\chi_i$, denote by $\hat f_i$ the coefficient of $\chi_i$ in expansion of $f$ according to $\chi_i$'s, \ie $\hat f_i = \langle \chi_i, f \rangle =  \frac{1}{|G|}\sum_{g\in G} \chi_i(g)\overline{f(g)}$. An easy fact is that for any $i$, we have
\begin{align}\label{eq:l-infty bound of F(f)}
	 |\hat f_i| & = \frac{1}{|G|}\left|\sum_{g\in G} \chi_i(g) \overline{f(g)}\right| \leq \frac{1}{|G|}\sum_{g\in G} |f(g)| |\chi_i(g)| \leq \Big(\frac{1}{|G|} \sum_{g\in G}|f(g)|\Big) \cdot \max_{g} |\chi_i(g)|.
\end{align}
If $G$ is Abelian, then it always has $|\chi_i(g)| = 1$, thus $\max_i |\hat f_i| \leq \frac{1}{|G|} \sum_{g\in G}|f(g)|$. For general groups, we have $|\chi_i(g)| \leq \deg(\chi_i)$, where $\deg(\chi_i)$ is the degree of $\chi_i$, namely the dimension of the associated vector space.

In this section we consider the setting that $S$ is a finite group $G$. The goal is to exploit properties of group characters to give better form of the lower bound. In particular, we hope to see when the second condition holds and what the matrix operator norm $\|[\psi_(g(x,y))]_{x,y}\|$ is  in this setting.

The standard orthogonality of irreducible characters says that $\sum_{s\in G} \chi_i(s) \overline{\chi_j(s)} = 0$. The second condition in Theorem \ref{thm: lb, general function} is concerned with a more general case: For a multiset $T$ with elements in $G\times G$, we need
\begin{equation}\label{eq: orthogonal graph requirement}
    \sum_{(s, t)\in T} \chi_i(s) \overline{\chi_j(t)} = 0, \qquad \forall i\neq j.
\end{equation}
The standard orthogonality relation corresponds to the special that $T = \{(s,s): s\in G\}$. We hope to have a characterization of a multiset $T$ to make Eq. \eqref{eq: orthogonal graph requirement} hold.

We may think of the a multiset $T$ with elements in set $S$ as a function on $S$, with the value on $s\in S$ being the multiplicity of $s$ in $T$. Since characters are class functions, for each pair $(C_k, C_l)$ of conjugacy classes, only the value $\sum_{g_1\in C_k, t\in C_l} T(g_1, t)$ matters for the sake of Eq. \eqref{eq: orthogonal graph requirement}. We thus make $T$ a class function by taking average within each class pair $(C_k, C_l)$. That is, define a new function $T'$ as
\begin{equation*}
    T'(s, t) = \sum_{s\in C_k, t\in C_l} T(s, t) / (|C_k||C_l|), \ \forall s\in C_k,\ \forall t\in C_l.
\end{equation*}

\begin{proposition}\label{prop: general group condition}
    For a finite group $G$ and a multiset $T$ with elements in $G\times G$, the following three statements are equivalent:
    \begin{enumerate}
        \item $\sum_{(s, t)\in T} \chi_i(s)\overline{\chi_j(t)} = 0, \ \forall i\neq j$
        \item $T'$, as a function, is in $span\{\chi_i \otimes \overline{\chi_i}: i\in[h]\}$
        \item $[T'(s,t)]_{s,t} = C^\dagger D C$ where $D$ is a diagonal matrix and $C = [\chi_i(s)]_{i,s}$. That is, $T'$, as a matrix, is normal and diagonalized exactly by the irreducible characters.
    \end{enumerate}
\end{proposition}
\begin{proof}
    Let $H_2$ be the subspace consisting functions $f: G\times G \rightarrow \mbC$ \st $f(s, t) = f(s', t')$ if $s \sim s'$, $t \sim t'$. Note that for direct product group $G\times G$, $\{\chi_i\otimes \overline{\chi_j}: i,j\}$ form an orthogonal basis of $H_2$:
    \begin{align*}
        & \sum_{s, t\in G} \chi_{i}(s)\overline{\chi_{j}(t)}\overline{\chi_{i'}(s)\overline{\chi_{j'}(t)}} =  \big(\sum_{s\in G}\chi_{i}(s)\overline{\chi_{i'}(s)}\big) \big(\sum_{t\in G}\overline{\chi_{j}(t)}\chi_{j'}(t)\big) = 0
    \end{align*}
    unless $i = i'$ and $j = j'$. Note that by viewing $T$ as a function from $G\times G$ to $\mbC$, the Eq. \eqref{eq: orthogonal graph requirement} and the definition of $T'$ imply that
    \begin{equation*}
        \langle \chi_i\otimes \overline{\chi_j},\ T \rangle = 0, \forall i\neq j
    \end{equation*}
    Thus the first two statements are equivalent.

    Note that
    \begin{align*}
        T'\in span\{\chi_i \otimes \overline{\chi_i}: i\in[h]\}
        \quad \Leftrightarrow  \quad T'(s, t) = \sum_i \alpha_i \chi_i(s) \overline{\chi_i(t)} \quad \text{ for some  $\alpha_i$'s}
    \end{align*}
    Denote by $C_{h\times |G|} = [\chi_i(g)]_{i,g}$ the matrix of the character table. Then observe that the summation in the last equality is nothing but the $(s, t)$ entry of the matrix $C^\dag diag(\alpha_1, \cdots, \alpha_h) C$. Therefore the equivalence of the second and third statements follows.
\end{proof}

\subsection{Abelian group}\label{sec: abelian group}
When $G$ is Abelian, we have further properties to use. The first one is that $|\chi_i(g)| = 1$ for all $i$. The second one is that the irreducible characters are homomorphisms of $G$; that is, $\chi_i(st) = \chi_i(s)\chi_i(t)$. This gives a clean characterization of the orthogonality condition by group invariance. For a multiset $T$, denote by $sT$ another multiset obtained by collecting all $st$ where $t$ runs over $T$. A multiset $T$ with elements in $G\times G$ is \emph{$G$ invariant} if it  satisfies $(g,g)T = T$ for all $g\in G$. We can also call a function $T: G\times G\rightarrow \mbC$ $G$ invariant if $T(s,t) = T(rs,rt)$ for all $r,s,t\in G$. The overloading of the name is consistent when we view a multiset $T$ as a function (counting the multiplicity of elements).

\begin{proposition}\label{prop: Abelian group invariant condition}
    For a finite Abelian group $G$ and a multiset $T$ with elements in $G\times G$,
    \begin{equation}\label{eq: orthogonal graph, Abelian}
        \text{$T$ is $G$ invariant} \Leftrightarrow \sum_{(s, t)\in T} \chi_i(s)\overline{\chi_j(t)} = 0, \quad \forall i\neq j.
    \end{equation}
\end{proposition}
\begin{proof}
    $\Rightarrow$: Since $T$ is $G$ invariant, $T = (r,r)T$ and thus,
    \begin{align}
        \sum_{(s, t)\in T} \chi_i(s)\overline{\chi_j(t)} & = \sum_{(s, t)\in (r,r)T} \chi_i(s)\overline{\chi_j(t)} \\
        & = \sum_{(s', t')\in T} \chi_i(rs')\overline{\chi_j(rt')}
    \end{align}
    Now using the fact that irreducible characters of Abelian groups are homomorphisms, we have
    \begin{align*}
        \sum_{(s', t')\in T} \chi_i(rs')\overline{\chi_j(rt')}
        = & \sum_{(s', t')\in T} \chi_i(r) \chi_i(s')\overline{\chi_j(r)}\overline{\chi_j(t')} \\
        = & \chi_i(r)\overline{\chi_j(r)} \Big(\sum_{(s', t')\in T} \chi_i(s')\overline{\chi_j(t')}\Big)
    \end{align*}
    But note that this holds for any $r\in G$, thus also for the average of them. That is,
    \begin{align*}
        \sum_{(s, t)\in T} \chi_i(s)\overline{\chi_j(t)} = & \frac{1}{|G|} \Big(\sum_{r\in G} \chi_i(r)\overline{\chi_j(r)}\Big) \Big(\sum_{(s', t')\in T} \chi_i(s')\overline{\chi_j(t')}\Big) = 0,
    \end{align*}
    by the standard orthogonality property of different irreducible characters.

    $\Leftarrow$: Since $\sum_{(s, t)\in T} \chi_i(s)\overline{\chi_j(t)} = 0$, $\forall i\neq j$, we know that $T$ as a function is in $span\{\chi_i \otimes \overline{\chi_i}: i\}$. Note that any linear combination of $G$ invariant functions is also $G$ invariant. Thus it remains to check that each basis $\chi_i \otimes \overline{\chi_i}$ is $G$ invariant, which is easy to see:
    \begin{equation*}
        \chi_i(rs) \overline{\chi_i(rt)} = \chi_i(r)\chi_i(s)  \overline{\chi_i(r)}\overline{\chi_i(t)} =\chi_i(s) \overline{\chi_i(t)}.
    \end{equation*}
    This finishes the proof.
\end{proof}

Another nice property of Abelian groups is that the orthogonality condition condition implies the regularity one.
\begin{proposition}\label{prop: Abelian ortho reg}
    For an Abelian group $G$, if either $T^{y,y}$ is $G$ invariant for all $y$ or $S^{x,x}$ is $G$ invariant for all $x$, then $G|\{g(x,y): x\in X, y\in Y\}$.
\end{proposition}
\begin{proof}
    Note that $T^{y,y}(s,s) = |\{x:g(x,y) = s\}|$, thus $T^{y,y}$ being $G$ invariant implies that $|\{x: g(x,y) = s\}| = |\{x: g(x,y) = t\}|$ for all $s,t\in G$. Thus the column $y$ in matrix $[g(x,y)]_{x,y}$, when viewed as a multiset, is equal to $G$ repeated $|Y|/|G|$ times. Therefore the whole multiset $\{g(x,y): x\in X, y\in Y\}$ is a multiple of $G$ as well.
\end{proof}

What we finally get for Abelian groups is the following.
\begin{corollary}
    For a sign matrix $A = [f(g(x,y))]_{x,y}$ and an Abelian group $G$, if $d(f,span(Ch_{Easy})) = \Omega(1)$, and the multisets $S^{x,x'} = \{(g(x,y), g(x',y)): y\in Y\}$ and $T^{y,y'} = \{(g(x,y), g(x,y')): x\in X\}$ are $G$ invariant for any $(x,x')$ and any $(y,y')$, then
    \begin{equation*}
        Q(A) \geq \log_2 \frac{\sqrt{MN}}{\max_{i\in Hard} \|[\chi_i(g(x,y))]_{x,y}\|} - O(1).
    \end{equation*}
\end{corollary}

\subsection{Block composed functions}\label{sec: block}
We now consider a special class of functions $g$: block composed functions. Suppose the group $G$ is a product group $G = G_1 \times \cdots \times G_t$, and $g(x,y) = (g_1(x^1, y^1), \cdots, g_t(x^t, y^t))$ where $x = (x^1, \cdots, x^t)$ and $y = (y^1, \cdots, y^t)$. That is, both $x$ and $y$ are decomposed into $t$ components and the $i$-th coordinate of $g(x,y)$ only depends on the $i$-th components of $x$ and $y$. The tensor structure makes all the computation easy. Theorem \ref{thm: block lb} can be generalized to the general product group case for arbitrary groups $G_i$.
\begin{definition}
	The $\epsilon$-approximate degree of a class function $f$ on product group $G_1\times \cdots \times G_t$, denoted by $d_\epsilon(f)$, is the minimum $d$ \st $\|f-f'\|_\infty \leq \epsilon$, where $f'$ can be represented as a linear combination of irreducible characters with at most $d$ non-identity component characters.
\end{definition}

\begin{theorem}
    For sign matrix \[A = [f(g_1(x^1,y^1), \cdots, g_t(x^t,y^t)]_{x,y}\] where all $g_i$ satisfy their orthogonality conditions, we have
    \begin{equation*}
        Q(A) \ge \min_{\{\chi_i\}, S} \sum_{i\in S} \log_2 \frac{\sqrt{\mathrm{size}(M_{g_i})}}{\deg(\chi_i)\|M_{\chi_i \circ g_i}\|} - O(1)
    \end{equation*}
    where the minimum is over all $S\subseteq [n]$ with $|S| > \deg_{1/3}(f)$, and all non-identity irreducible characters $\chi_i$ of $G_i$.
\end{theorem}

\begin{proof}
Recall that an irreducible character $\chi$ of $G$ is the tensor product of irreducible characters $\chi_i$ of each component group $G_i$. Let Hard be the set of irreducible characters $\chi$ with more than $d$ non-identity component characters. Fix a hard character $\chi$, and denote by $S$ the set of coordinates of its non-identity characters.
\begin{align*}
	\|[\chi (g(x,y))]\| & = \|\bigotimes_{i\in [t]} [\chi_i (g_i(x^i,y^i))]\| \\
	& = \prod_{i\in [t]} \| [\chi_i (g_i(x^i,y^i))]\| \\
	& = \prod_{i\in S} \| [\chi_i (g_i(x^i,y^i))]\| \times \prod_{i\notin S} \|J_{|X_i|\times |Y_i|}\| \\
	& = \prod_{i\in S} \| M_{\chi_i \circ g_i}\| \times \prod_{i\notin S} \sqrt{\mathrm{size}(M_{g_i})}
\end{align*}
Thus by Theorem \ref{thm: lb, general function} and Eq. \eqref{eq:l-infty bound of F(f)}, we have
\begin{align}
	Q(A) \geq \log_2 \prod_{i\in S} \frac{\sqrt{\mathrm{size}(M_{g_i})}}{\deg(\chi_i)\cdot \|M_{\chi_i\circ g_i}\|} - O(1)
\end{align}
proving the theorem.
\end{proof}

Previous sections as well as \cite{SZ09b} consider the case where all $g_i$'s are the same and all $G_i$'s are $\mbZ_2$. In this case, the above bound is equal to the one in Theorem \ref{thm: block lb}, and the following proposition says that the group invariance condition degenerates to the strongly balanced property.
\begin{proposition}\label{prop: degeneration}
    For $G = \mbZ_2^{\times t}$, the following two conditions for $g = (g_1, \cdots, g_t)$ are equivalent:
    \begin{enumerate}
        \item The multisets $S^{x,x'} = \{(g(x,y), g(x',y)): y\in Y\}$ and $T^{y,y'} = \{(g(x,y), g(x,y')): x\in X\}$ are $G$ invariant for any $(x,x')$ and any $(y,y')$,
        \item Each matrix $[g_i(x^i,y^i)]_{x^i, y^i}$ is strongly balanced.
    \end{enumerate}
\end{proposition}
\begin{proof}
    1 $\Rightarrow$ 2: $S^{x,x'}$ being $G$ invariant implies that for all $\{z_i\}, \{u_i\}, \{v_i\}$,
    \begin{align*}
        & |\{y: z_i g_i(x^i, y^i) = u_i,\ z_i g_i(x'^i, y^i) = v_i, \forall i\}| \\
        = & |\{y: g_i(x^i, y^i) = u_i,\ g_i(x'^i, y^i) = v_i, \forall i\}|.
    \end{align*}
    Take $x' = x$ and $u = v$. Now for each $i$ and each row $x^i$, take $z_i = -1$ (where the group $\mbZ_2$ is represented by $\{\pm 1\}$). For all other $i'\neq i$, take $z_{i'} = 1$. This assignment will show that
    \begin{equation*}
        |\{y: g_i(x^i, y^i) = - u_i\}| = |\{y: g_i(x^i, y^i) = u_i\}|.
    \end{equation*}
    That is, the row $x^i$ in matrix $[g_i(x^i,y^i)]$ is balanced. Similarly we can show the balance for each column.

    2 $\Rightarrow$ 1: It is enough to show that for each $i$ and each $\{z_i\}, \{u_i\}, \{v_i\}$,
    \begin{align*}
        & |\{y^i: z_i g_i(x^i, y^i) = u_i,\ z_i g_i(x'^i, y^i) = v_i\}| \\
        = & |\{y^i: g_i(x^i, y^i) = u_i,\ g_i(x'^i, y^i) = v_i\}|.
    \end{align*}
    First consider the case $x'^i = x^i$. If $u_i \neq v_i$ then both numbers are 0; if $u_i = v_i$ then both numbers are $|\{y^i\}|/2$ by the balance of row $x^i$. Now assume $x'^i \neq x^i$. Denote $a_{bb'} = |\{y^i: g_i(x^i, y^i) = b,\ g_i(x^i, y^i) = b'\}|$, then the above requirement amounts to $a_{00} = a_{11}$ and $a_{01} = a_{10}$.

    Note that we have
    \begin{align*}
        a_{00} + a_{01} & = |\{y^i: g_i(x^i, y^i) = 0\}| = |\{y^i: g_i(x^i, y^i) = 1\}| = a_{10} + a_{11}
    \end{align*}
    where the second equality is due to the balance of row $x^i$. And similarly we have $a_{00} + a_{10} = a_{01} + a_{11}$ by balance of row $x'^i$. Combining the two, we get $a_{00} = a_{11}$ and $a_{01} = a_{10}$ as desired.
\end{proof}

It is worth noting that the conclusion does not hold if any group $G_i$ with size larger than two. We omit the counterexamples here.

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


It's still hot during the day, but the flavor of autumn is stealing into the days. Personally, since I'm not a summer's child, I'm ready and looking forward to cooler weather.
To me, autumn means not just cooler weather, but a slower pace with more room for contemplation and more time to read or indulge another hobby. Here's hoping that you, too, enjoy the turning of the Wheel!
The September 4 Open Full Moon will be on Friday, September 4, 1998, at 7:30 PM, at the First Unitarian Church, 1400 Lafayette, Denver, CO.
"I intend to share the basic format of Dragonmoon worship with those attending. Worship is the focus of Dragonmoon, worship as mystical union with the divine. The format tends toward "high mass" since I feel the "Bells & Whistles" help focus the deep mind (which thinks in images and pictures) while lulling the thinking mind to sleep with repetition.
"In clan Dragcuin tradition, the season of Mabon is one of balance and decision making. At the Equinoxes of Mabon and Lady Day, the flow of power is in balance, thus it only takes a little push in any direction to start the ball rolling and once started, the momentum takes over. This means that one must choose carefully the direction in which to push. The ritual will include a meditation that will lead each participant on a journey through the elements to the realm of spirit where s/he will meet with (yep, you guessed it) the Mother of Dragons. The purpose is to gain wisdom in making decisions for the approaching season of Mabon."
Catherine Mock (you may know her as Fionnula, or as Catrion) has lived in Denver most of her life and is a graduate of D.U. She's a musician, dancer, legal secretary, historian, priestess (a solitaire, interested in a variety of Celtic and other practices), martial artist, voracious reader, and has been spotted nearly everywhere carrying her harp.
"I'm honored to be asked to do this for Hearthstone- but I am not going to learn that Churchlady dance..."
Welcome to the board, Catherine!
Send questions or suggestions for tips to: kwikbeam@earthlink.net or give them to me (Blaze) at OFM. You're also welcome to write an entire "guest" column.
Well, this month I got Dem Ol' Post-Dragonfest Blues and am feelin' lazy, so this is a lazy column. (From my point of view only. For you, it may turn out to be work.) [That's one of the best ways to lose your reader I've ever thought of.] Since one of my R's is Remember, I've decided to remind you, and myself as well, of some of the things I've covered in earlier columns. Then you get a quiz--you knew there'd be one sooner or later. So here goes.
**Remember the importance of Reducing: it's the number one priority.
**"Turn off your lights. In the silence of your darkened room, you can hear a thousand trees whispering 'Thanks'." The Whole Earth Catalog.
**Remember the ants. No one of them does very much, but there are so many of them that if most of them do a little, they accomplish a lot. So, if you are new to this, do what you can. One thing worked into your routine so it gets done, rather than forgotten, will eventually lead to something else.
**Remember one minute at one o'clock. Give a minute every day to meditation or prayer for the Goddess. We can build a pretty big anthill, here.
**Target stores will properly dispose of lithium batteries. It's one of their PR services.
**Resources: Choose to Reuse by Nikki and David Goldbeck. The Green Consumer, by Elkington, et al. Coop America's website: www.greenpages.org/ .
Now for the quiz. This quiz was put together with the idea that we need to know the areas we live in with a much deeper knowledge than simply being able to get from here to there. This holds true for everyone. Doubly so for Pagans, I should think. (Yes, I would flunk this quiz today. I will get an A on it in six months or so.) Check it out, and see how much (or little) you know about the world you live in.
What sort of soil is predominant in this place? What "soil series" is it?
What were the primary subsistence techniques of the culture that lived in your area before you?
Name five native edible plants in your region and the season(s)that they are available.
From what direction do the winter storms generally come in your region?
Where does your garbage ultimately go?
What is the land-use history of where you live?
This quiz was originally published in CoEvolution Quarterly 32 (Winter 1981) and was compiled by Leonard Charles, Jim Dodge, Lynn Milliman, and Victoria Stockley. It is also in Stephanie Fox's Whatever Happened to Ecology? (San Francisco: Sierra Club Books, 1989).
I didn't make it to Dragonfest and, since I wasn't frantically packing food and gear, I wasn't thinking of a good camping recipe. But just because I missed Dragonfest for one year doesn't mean I've forgotten how much fun it is to create a family feast on a two-burner propane stove.
Actually, I do sometimes really enjoy that kind of stressful cooking. There's something magickal about it. But what I like even more is cooking that involves a campfire. It's not that it's any neater or easier than cooking on the camp stove; in all honesty, cooking on or in the fire is one of the least convenient ways of preparing food. But there is something dramatically primal in this style of cooking. It's almost decadent.
) and reading a new cookbook.
Spirit of the Harvest: North American Indian Cooking is a lovely volume filled with recipes that are alleged by the authors to be authentic Native American dishes. So far, I'm only up to the segment on East Coast tribes, but it seems to be a reasonable representation of native cooking. The following is a Cherokee baked apple recipe to be made in a campfire. Just because Dragonfest is over doesn't mean you can't use this recipe. If you aren't planning on any more camping this summer, there's always next summer. Not to mention your barbecue grill. I'm thinking of putting this in the fire for my Mabon celebration.
Combine sugar, pecans, currants, and spice. Core apples from the top; the hole in center is for filling, not for cutting the skin at the bottom.
Place 1 tsp butter in each apple, followed by a tablespoon of the filling. Wrap tightly in aluminum foil and place, top down, directly in hot coals. After 5 minutes, using long tongs, turn apples right side up and continue to bake 3 to 5 minutes longer.
A central component of historical Paganism was sacrifice. Indeed, the giving of things considered valuable seems to have been a universal expression of commitment and gratitude to those powers regarded as divine. Today, the ritualized slaughtering of animals would probably hold little meaning to many of us, but there is still something very important which many of us can offer; a gift of life! The thought of blood donation may seem unimportant or even unpleasant to some reading this article, but this modern sacrifice of ourselves means hope and health to those in need.
To give some basic facts; nearly 97 percent of us living to the age of 72 will require blood or blood products at some point in our lives, but only 4 percent donate blood on a regular basis. It's up to us to support 170,000 transfusions each year at more than 80 hospitals and health facilities in Colorado. One unit of your blood can help as many as three patients, because whole blood is usually separated into its three main components of red cells, plasma, and platelets. If your blood is by some chance used for pooled serum medication or for research, the effects of your sacrifice can touch countless lives!
The act of giving blood takes less than an hour and with minimal discomfort. Due to the use of sterile equipment, it is impossible to contract communicable diseases from donating blood. However, due to concerns regarding the spread of HIV, those who have been tattooed or pierced within a year's time (something rather common in our community) and gay males are asked not to donate. Because of this, I believe that those of us who can donate have an increased responsibility to give of ourselves for the benefit of others.
What could be more appropriate than to perform this act of self-sacrifice in gratitude for the life our deities have given us, at the most holy times of the year. This is why I began the Pagan Blood Drives, which take place near the cusps of each season. As I sit in the phlebotomist's chair, I silent ask my patron god and godde4ss (Freyr and Freya) to accept my personal gift for the good of my fellow human beings. After having made five donations in this past year, I have become aware that now any stranger on the street may have benefited from these sacrifices, and this awareness in itself is truly a form of magick.
The next Pagan Blood Drive will be held at noon on September 26 at the Belle Bonfils Donation Center on 717 Yosemite. We will also be collecting canned food for the Food Bank of the Rockies. This fall, let's all think of how we can make sacrifices and save the lives of others!


Power demands for USB are ever-increasing as we get more and more battery life in our smart devices. This cigarette lighter adaptor provides 4 USB charging ports, for up to 7.2A combined charging power - that's a lot! Tablets, smartphones, and other high demand devices can then charge at their desired rate, up to 2.4A per port. Works on 12 or 24V systems, so suitable for trucks and boats too.


text:
Tag: kay walsh
London-born English actor and dancer (1911-2005), born Kathleen Walsh (Wikipedia entry)
Criterion Sunday 154: The Horse's Mouth (1958)
Published on 22 October 2017 16 August 2019 by Ewan MLeave a comment
Having never heard of it before it popped up on our Criterion watching project, this is a perfectly likeable colour film about a colourful character who paints colourful works of art and injects a bit of épater into those bourgeois lives he drifts through (well, more upper-class really), but I'm not sure what deeper meaning it really captures. The one the filmmakers presumably intend — that art is valuable, damn everything — comes through clearly though, and Alec Guinness in the lead as dishevelled painter Gulley Jimson is as ever reliable, not unlike the Meryl Streep of his day, all accents and imposture in the service of wit and well-crafted journeyman material. It has its diversions, and is pleasing on the eye.
Criterion Extras: There's a short interview with Ronald Neame from before he died (around when the DVD was released, presumably), who is a genial host and tells of the film's production. There's also a trailer. However, the standout extra is a short film which was shown with the feature at its original New York run in the late-1950s, a short film by D.A. Pennebaker called Daybreak Express. For all its five minutes running time, it is far the superior work. It's a jaunting work of jazzy cinematic propulsion, like a city symphony made my Soviet constructivists with a penchant for Duke Ellington. Rich and resonant colours, bold modern architecture, a train ride from the city to suburbs both exceeding that experience but also encapsulating it.
Director Ronald Neame; Writer Alec Guinness (based on the novel by Joyce Cary); Cinematographer Arthur Ibbetson; Starring Alec Guinness, Kay Walsh; Length 97 minutes.
Seen at a friend's home (DVD), London, Sunday 23 April 2017.
Categories Classification: PG, Country: UK, Criterion Sunday, Rating: GOOD•Tags 1958, alec guinness, arthur ibbetson, character study, comedy, criterion collection, film reviews, films seen at home, kay walsh, london, painting, ronald neame, uk
Criterion Sunday 32: Oliver Twist (1948)
After Great Expectations of a few years earlier (reviewed last week), comes another David Lean adaptation from the works of Charles Dickens and it is far superior. Guy Green outdoes his earlier cinematographic efforts with vast inky pools of blackness from which characters emerge into a shadowy, grey lost world, immured deep in England's memory, as orphan Oliver is born at a workhouse to a mysterious woman who dies in childbirth, and inducted into a life of poverty and hard grind. It all resolves itself neatly by the end, but it's given vivacity by the acting, particularly Robert Newton as a frantic Bill Sikes, Kay Walsh as his moll girlfriend, and of course the young John Howard Davies as Oliver. Nowadays the film is most known for Alec Guinness's creepy comedic turn as Fagin, though I feel Criterion's liner notes suggesting it's not anti-semitic because he's never called a Jew within the film is somewhat disingenuous: it's clearly a caricature and a fairly unflattering one at that. Still, Fagin is a fairly small element within the whole film, which remains impressive most of all for its beautifully filmed vision of a world that must have felt within reach in 1948 given the ravages of the era in which it was made.
Criterion Extras: Aside from those liner notes, there's just a trailer and English subtitles, so this is a bare bones package.
Director David Lean; Writers Lean and Stanley Haynes (based on the novel by Charles Dickens); Cinematographer Guy Green; Starring John Howard Davies, Kay Walsh, Alec Guinness, Robert Newton; Length 116 minutes.
Seen at home (DVD), London, Sunday 5 July 2015.
Categories Classification: U, Country: UK, Criterion Sunday, Rating: GOOD•Tags 1948, alec guinness, black-and-white, charles dickens, criterion collection, david lean, drama, england, film reviews, films seen at home, guy green, john howard davies, kay walsh, london, robert newton, uk