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Articles written in Bulletin of Materials Science

• Molten salt synthesis, structural and photoluminescence properties of SrBi$_{0.95}$Sm$_{0.05}$Ta$_{2-x}$V$_x$O$_9$ (0${\le}$x${\le}$0.2)

SrBi$_{0.95}$Sm$_{0.05}$Ta$_{2-x}$V$_x$O$_9$ (x = 0, 0.05, 0.1, 0.2) compounds, which have been studied through X-ray diffraction Rietveld refinement, Fourier transform infrared (FTIR), absorption and emission techniques, were prepared through the molten salt method. All samples were refined in the orthorhombic system without any secondary phase detected. The Ta–O bond strength in the compounds is revealed by FTIR and Rietveld measurements. Along with scanning electron microscopic analysis, ceramics also consist of plate-like grains that decreased with the V content increasing. The difference in absorption spectra is attributed to localized tail stats within the bandgap. According to the impedance and AC conductivity results, after introducing vanadium into the structure, the V-doped SrBi$_{0.95}$Sm$_{0.05}$Ta$_2$O$_9$ ceramics exhibit higher conductivity. The reason that x = 0.1 ceramic has the smallest semicircle radius referring to ahigher efficient charge carrier transfer is that this ceramic sample is mainly composed of large particles. Samarium emits orange-red colour at wavelength 599 nm, which is attributed to ${}^4$G$_{5/2}$ ${\rightarrow}$ ${}^6$H$_{7/2}$ transition. Regarding photoluminescence intensity, the more vanadium ions are introduced into the lattice, the fewer emission peaks are obtained. The latter has been interpreted in terms of the inhibition of the transfer of energy from one ion to another. In essence, The red photoluminescence emission intensity of Sm$^{3+}$ ions were successfully enhanced in SrBi$_{0.95}$Sm$_{0.05}$Ta$_{2-x}$V$_x$O$_9$ (x = 0.05).

• Bulletin of Materials Science

Volume 46, 2023
All articles
Continuous Article Publishing mode

• Dr Shanti Swarup Bhatnagar for Science and Technology

Posted on October 12, 2020

Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru
Chemical Sciences 2020

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# Please check my work (friction on an incline plane)

two blocks of unequal mass are connected by a string over a smooth pulley. one mass of mass 2m rests on the incline while the second mass of mass m is hung over a pulley over the incline if the coefficient of friction is μk, what angle of the incline allows the masses to move at a constant speed?

|F1| = |F2|

F1= weight of mass hanging over the pulley

F2=Ʃforce acting on mass (2m) ie.. the x component of force along the direction of the incline + friction

mg=(2m)gsinθ+(μk)(2m)gcosθ

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

but since i know that.. μk=tanθ,

1/2=sinθ+tanθcosθ = sinθ+sinθ= 2sinθ

sinθ=(1/4)

θ=14.4775°

is the way i substituted tanθ for the coefficient valid? if not, why is it not valid?

Last edited:

Related Introductory Physics Homework Help News on Phys.org
mfb
Mentor
Why should the force from the hanging mass be equal to the horizontal component of the force on the other mass? I would assume the string goes along the incline.

mg=(2m)gsinθ+(μk)(2m)gcosθ
This looks like the mass m is hanging now, instead of the mass 2m. How did you get that formula?

The result will certainly depend on μk, a specific number cannot be right as result.

Please add explanations to your formulas, otherwise it is hard to understand what you did and what was wrong.

i fixed the m and 2m error.

i set mg equal to the rest of that because the problem wanted me to find the theta when the net force is zero.

so starting here,

mg=(2m)gsinθ+(μk)(2m)gcosθ

factoring mg..

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

now i'm supposed to solve for theta but i don't know how.. :(

mfb
Mentor
You have to start earlier. Did you draw a sketch? Did you identify all forces acting on the system?
Which forces have to be in equilibrium?

(sinθ)^2+(cosθ)^2=1 might be handy somewhere in the solving process, but first you need correct equations for the forces.

Doc Al
Mentor
here is the free body diagram with the arbitrary direction-sign specification (although that part wasn't needed)

http://i.imgur.com/BCmluRZ.png
You left out the tension force and the friction.

Hint: If the speed is constant, what must be the tension?

Last edited:
i essentially set the two tensions equal to one another. where..

T1=mg
T2=(2m)gsinθ+(μk)(2m)gcosθ

does this depend on the direction of the blocks? since if the blocks move to the right..

mg=(2m)gsinθ+(μk)(2m)gcosθ

and if the blocks move to the left..

mg=(2m)gsinθ-(μk)(2m)gcosθ

so this is actually $\hat{v}$ dependent?

Doc Al
Mentor
Solve it both ways and see if you get reasonable answers.


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# A Secretive Coded Caching for Shared Cache Systems using PDAs

This paper considers the secretive coded caching problem with shared caches in which no user must have access to the files that it did not demand. In a shared cache network, the users are served by a smaller number of helper caches and each user is connected to exactly one helper cache. To ensure the secrecy constraint in shared cache networks, each user is required to have an individual cache of at least unit file size. For this setting, a secretive coded caching scheme was proposed recently in the literature (Secretive Coded Caching with Shared Caches, in IEEE Communications Letters, 2021), and it requires a subpacketization level which is in the exponential order of the number of helper caches. By utilizing the PDA constructions, we propose a procedure to obtain new secretive coded caching schemes for shared caches with reduced subpacketization levels. We also show that the existing secretive coded caching scheme for shared caches can be recovered using our procedure. Furthermore, we derive a lower bound on the secretive transmission rate using cut-set arguments and demonstrate the order-optimality of the proposed secretive coded caching scheme.

## Authors

• 3 publications
• 5 publications
• 42 publications
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### A caching system with object sharing

We consider a public content caching system that is shared by a number o...
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### On the Effective Throughput of Coded Caching: A Game Theoretic Perspective

Recently coded caching has emerged as a promising means to handle contin...
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### Coded Caching with Uneven Channels: A Quality of Experience Approach

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We present a novel caching and coded delivery scheme for a multi-access ...
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## I Introduction

### I-a Contributions

In this work, we study the secretive coded caching with shared cache networks. Our contributions are summarized below:

• A procedure is proposed to obtain new secretive coded caching schemes for shared caches using PDAs. The advantage of our procedure is that it results in schemes with lesser subpacketization level than the corresponding scheme in [10] (Section V).

• A lower bound on the optimal rate-memory trade-off of secretive coded caching with shared caches is derived using cut-set based arguments and characterizes the multiplicative gap between the achievable rate and the lower bound (Section VI).

• We also show that the secretive coded caching scheme in [10] can be recovered using a PDA. Hence, our procedure subsumes the scheme in [10] as a special case (Section V).

The rest of the paper is organized as follows. In Section II, we briefly discuss some of the topics that are relevant for our scheme description. We describe the problem setup and present the main results in Sections III and IV, respectively. In Section V, we describe the proposed scheme and in Section VI, the lower bound and the order optimality of the scheme are presented. Section VII summarizes our results. Notations: For a positive integer , denotes the set . For any set , denotes the cardinality of . Binomial coefficients are denoted by , where and is zero for

. Bold uppercase and lowercase letters denote matrices and vectors, respectively. For a vector

, denotes the vector consisting the elements in at positions specified by the elements in the set . The columns of an matrix is denoted by ,

. An identity matrix of size

is denoted as . The finite field with elements is denoted by .

## Ii Preliminaries

In this section, we briefly review PDAs and secret sharing schemes which are required for describing our scheme.

### Ii-a Placement Delivery Array (PDA)

###### Definition 1.

([16]) For positive integers and , an array , and , composed of a specific symbol and positive integers , is called a placement delivery array (PDA) if it satisfies the following three conditions:
C1. The symbol appears times in each column.
C2. Each integer occurs at least once in the array.
C3. For any two distinct entries and , is an integer only if

1. [label=()]

2. , , i.e., they lie in distinct rows and distinct columns, and

3. , i.e., the corresponding sub-array formed by rows and columns must be of the following form:

or

Every PDA corresponds to a coded caching scheme for dedicated cache network with parameters and as in Lemma 1.

###### Lemma 1.

([16]) For a given PDA , a coded caching scheme can be obtained with subpacketization level and using Algorithm 1. For any distinct demand vector , the demands of all the users are met with a rate, .

In a PDA , the rows represent packets and the columns represent users. For any if , then it implies that the user has access to the packet of all the files. The contents placed in the user’s cache is denoted by in Algorithm 1. If is an integer, then it means that the user does not have access to the packet of any of the files. Condition guarantees that all users have access to some packets of all the files. According to the delivery procedure in Algorithm , the server sends a linear combination of the requested packets indicated by the integer in the PDA. Therefore, condition implies that the number of messages transmitted by the server is exactly , and the rate achieved is . Condition ensures the decodability.

### Ii-B Secret Sharing Schemes

The secretive coded caching schemes proposed so far in the literature rely on non-perfect secret sharing schemes. We also utilize the same in our scheme. The primary idea behind non-perfect sharing scheme is to encode the secret in such a way that accessing a subset of shares does not reveal any information about the secret and only accessing all the shares enable to recover the secret completely. The formal definition of the non-perfect secret sharing scheme is given below.

###### Definition 2.

([14]) For a secret with size bits and , an non-perfect secret sharing scheme generates equal-sized shares such that accessing any shares does not reveal any information about the secret and can be completely reconstructed from all the shares. i.e,

 I(W;S)=0 ∀S⊆{S1,S2,…,Sn} s.t |S|≤m, (1a) H(W|S1,S2,…,Sn)=0. (1b)

In the non-perfect secret sharing scheme, the size of each share should be at least bits [7]. For large enough , there exists non-perfect secret sharing schemes with size of each share being equal to bits.

## Iii Problem Setup

We consider a shared cache network as illustrated in Fig 1. There is a central server with a library of independent files , each of size

bits and is uniformly distributed over

. The server is connected to users through an error-free broadcast link, and there are helper caches, each of size equal to files. Each user gets connected to one helper cache and there is no limit on the number of users served by each helper cache. Further, each user has a dedicated cache of size files. The network operates in four phases as in [10]:

1. [label=()]

2. Helper Cache Placement phase: Let the contents stored in the helper cache be denoted as , where . The server fills each of the helper caches with functions of the library files and some randomness of appropriate size, such that

 I(W[1:N];Zλ)=0, ∀λ∈[Λ]. (2)

Equation (2) implies that no user is able to retrieve any information regarding any of the files from the cache contents that it gets access to. The placement is carried out without knowing the future demands of the users and their association to the caches and also, it satisfies the memory constraint at each helper cache. Let denote the contents stored in all the helper caches.

3. User-to-cache association phase: In this phase, each user gets connected to one of the helper caches and the set of users assigned to cache is denoted as . The overall user-to-cache association is represented as . All these disjoint sets together form a partition of the set of users and this association of users to helper caches is independent of the cached contents and the subsequent demands. For any user-to-cache association , the association profile describes the number of users accessing each cache. Therefore, where and . Without loss of generality, assume that and each to be an ordered set. Each user in is indexed as . Several user-to-cache associations result in the same . Therefore, each represents a class of . Let the helper cache accessed by any user be denoted as . For any two users and , if then users and are accessing the same cache.

4. User Cache Placement phase: Once is known to the server, there is an additional phase where the server fills each of the user’s dedicated cache with random keys satisfying the memory constraint. The contents stored in the user’s cache is denoted as and denotes the set of all users’ dedicated cache contents. User having access to and should not get any information about . That is,

 I(W[1:N];Zλk,Zk)=0, ∀k∈[K]. (3)
5. Delivery Phase: In this phase, each user demands one of the

files. The indices of the demanded files are denoted by random variables. Let

be a random variable denoting the user’s demand. Then, is a set of independent random variables, each uniformly distributed over the set . Let be a realization of . Upon receiving the demand vector , the server makes a transmission of size bits over the shared link to the users, where is a function of the association profile , , and . Each user must be able to decode its demanded file using the transmission and its available cache contents and and should not obtain any information about the remaining files. That is,

 H(Wdk|X,Zλk,Zk) =0 ∀k∈[K], and (4) I(W[1:N]∖dk;X,Zλk,Zk) =0 ∀k∈[K]. (5)

For a given association profile , the worst-case rate corresponds to . We aim to minimize the worst-case rate under the decodability and secrecy conditions mentioned in (4) and (5), respectively.

###### Definition 3.

For the above shared cache setting, a memory-rate pair is said to be secretively achievable if there exists a scheme for the memory point that satisfies the decodability condition in (4) and the secrecy condition in (5) with a rate less than or equal to for every possible realization of . The optimal rate-memory trade-off under secrecy condition is defined as

 R∗s (M,MU)= inf{Rs:((M,MU),Rs){ is secretively achievable}% }.

## Iv Main Results

Before presenting the main results, we first discuss the relevance of the dedicated user cache in our setting and show that the user cache must have a size of at least one file to ensure secrecy in any achievable coded caching scheme for shared cache system. In a shared cache network, several users will be sharing the same cache contents, hence multicasting opportunities cannot be created amongst the users accessing the same helper cache. Consider user connected to the helper cache . The transmissions that are useful for the user can be decoded by the remaining users in . Therefore, to ensure secrecy for the file content that user has requested against the users in

, the transmissions need to be encrypted using one-time pads that are known to only user

and unknown to other users in . To store these random keys, each user needs a dedicated memory unit in addition to the helper cache that it is accessing. As mentioned earlier, each user cache has a capacity to store files, and the condition needs to be satisfied to achieve a secretive coded caching scheme for shared caches. The formal proof of it is given below. Consider a cache which has more than one user connected to, say . Let be a demand vector where only the users in demand a file, and let be the corresponding transmission made by the server. Choose a user . Then,

 L1B ≤H(Wd) ≤I(Wd;X,Z1,Z[1:L1])+H(Wd|X,Z1,Z[1:L1]) ≤I(Wd;X,Z1,Z[1:L1]) (6a) ≤I(Wd;X,Z1,Zk) + I(Wd;Z[1:L1]∖{k}|X,Z1,Zk) (6b) ≤I(Wd∖{dk};X,Z1,Zk) + I(Wdk;X,Z1,Zk|Wd∖{dk}) + I(Wd;Z[1:L1]∖{k}|X,Z1,Zk) (6c) ≤I(Wdk;X,Z1,Zk|Wd∖{dk}) + I(Wd;Z[1:L1]∖{k}|X,Z1,Zk) (6d) ≤H(Wdk)+H(Z[1:L1]∖{k}) ≤B+B(L1−1)MU,

where (6a) follows from (4), (6b) and (6c

) follow from the chain rule of mutual information and (

6d) follows from (5). Thus, we obtain . It is sufficient to consider as unity as users’ individual caches are used only for storing the random keys that are used to encrypt those transmissions in which the user is involved. Therefore, in our further discussion, we fix , as taken in [10] and the shared caching problem described in Section III is referred to as shared caching problem henceforth. The following theorem presents a secretive coded caching scheme for shared caches obtained using PDAs.

###### Theorem 1.

For a given shared caching problem with an association profile , a secretive coded caching scheme with sub-packetization level can be derived from a PDA satisfying . The secretively achievable worst-case rate is obtained as

 Rs=∑s∈[S]LτsF−Z (7)

where, , .

The scheme that achieves performance in (7) is presented in Section V. Note that the rate varies according to the association profile for a given shared caching problem. For a uniform profile, the rate

is minimum and as the profile becomes more and more skewed, the rate increases. This follows from the description of the scheme.

###### Corollary 1.

For a uniform association profile, , the worst-case rate becomes

 Rs=K.SΛ(F−Z). (8)
###### Proof.

When the profile is uniform, , . Thus, (7) reduces to the expression in (8) . ∎

The following theorem provides an information-theoretic lower bound on the rate achievable by any secretive coded caching scheme for shared cache networks.

###### Theorem 2.

For any , and , the achievable secretive rate for a shared cache system is lower bounded by

 R∗s(M,MU)≥ (9) maxs∈{1,2,…,min(N2,K)}s⌊Ns⌋−1−(λs−1)M−(s−1)MU⌊Ns⌋−1.

The proof is given in Section VI. The lower bound expression in (LABEL:eq:cutset) has a parameter , which indicates the number of users under consideration. The term corresponds to the cache to which the user is connected to. In our setting, is fixed as unity and we define as .

###### Corollary 2.

For any , and , the achievable secretive rate is lower bounded by

 R∗s(M)≥maxs∈{1,2,…,min(N2,K)}s−(λs−1)M⌊Ns⌋−1. (10)
###### Proof.

The lower bound in (10) follows directly from (LABEL:eq:cutset) after letting . ∎

When , the server generates a set of independent random keys , each uniformly distributed over . Then in the user cache placement phase, the key is placed in the user’s cache. That is, . In the delivery phase, the server transmits , to satisfy the demands of the users. Thus, we obtain . It is straightforward to see that the conditions in (4) and (5) are satisfied by the above transmissions. Therefore, for . The following theorem demonstrates the order-optimality of the obtained scheme.

###### Theorem 3.

For and , if , the rate achieved by the secretive coded caching scheme obtained from PDAs is within the optimal rate by a factor which is a system parameter. i.e.,

 1≤Rs(M)R∗s(M)≤Λ. (11)

The proof is given in Section VI.

## V Secretive Coded Caching Scheme for Shared caches using PDAs

In this section, we present a procedure to obtain secretive coded caching scheme for shared caches using PDAs. Consider a shared cache network shown in Fig. 1. For the given shared caching problem, choose a PDA such that . The four phases involved are described below.

### V-a Helper Cache Placement Phase

The server first splits each file in into non-overlapping subfiles such that each subfile is of size, bits. Then, each file is encoded using a non-perfect secret sharing scheme. The shares of the file are denoted by , where and , . Let denote the set of shares of all the files. In the PDA , the rows represent the shares and the columns represent the helper caches. The placement of the shares in the helper caches is defined by the symbol ‘’ in the corresponding column. That is,

 Zλ={Sn,j,∀n∈[N]:pj,λ=⋆,j∈[F]} (12)

By Condition in Definition 1, each helper cache stores some shares of all the files such that the memory constraint is satisfied.

### V-B User-to-cache Association Phase

In this phase, each one of the users gets connected to one of the helper caches. Once the user-to-cache association and the profile are known, construct an array, of size from as described in lines - in Algorithm . The array is a generalized placement delivery array defined in [18]. In

, the numerical entries are an ordered pair, which come from a subset of

, where . Each column in corresponds to a user and the symbol ‘’ represents the shares that are available to each user. Thus, each user gets access to some shares of all the files but no information is gained about any of the files from these shares. This follows from the secret sharing scheme that we employ.

### V-C User Cache Placement Phase

To ensure the secrecy constraint while retrieving the demanded file, some keys need to be stored privately in each user’s cache which are essential for encrypting the transmissions. Since each user wants to decode the remaining shares of its demanded file, it needs to store independently and uniformly generated random keys of size . Thus, bits, which is in accordance with the memory constraint assumed for the user cache. The key which is used to encrypt a particular transmission will be stored in all those users’ caches which are involved in that transmission. Hence, the number of distinct keys that the server generates is same as the number of distinct ordered pairs present in and each key is indexed by an ordered pair (line of Algorithm ). As mentioned above, each user stores keys which are indexed by the ordered pair present in the column (described in lines - of Algorithm ).

### V-D Delivery Phase

In the delivery phase, users’ inform their demands to the server. Let be a demand vector. Consider the worst-case scenario where all the demands are distinct. The server transmits a message corresponding to every distinct ordered pair in . Let denote the number of times occurs in . Assume . Then, the sub-array formed by the rows and the columns is equivalent to a scaled identity matrix up to row or column permutations [18] as shown in (13).

 ⎛⎜ ⎜ ⎜ ⎜ ⎜⎝(s,i)⋆…⋆⋆(s,i)…⋆⋮⋮⋱⋮⋆⋆…(s,i)⎞⎟ ⎟ ⎟ ⎟ ⎟⎠m×m (13)

Each user , has the key and the set of shares wanted by other users. Hence, the server transmits a message of the form for every distinct in The delivery procedure is described in Algorithm 3.

### V-E Decoding

The decoding of the shares from the transmissions follows directly from (13). Each user has access to shares and obtains the remaining shares of its desired file from the transmissions, by using the helper cache contents and the keys that are stored privately in its cache. Hence, each user can retrieve its demanded file from its shares as mentioned in Definition 2.

### V-F Proof of Secrecy

Consider a user , and its accessible cache contents and . According to the placement procedure described in Section V-A and Section V-C, the helper cache contents consist of some shares of all the files and is constituted by independently and uniformly generated random keys which are used for one-time padding. By virtue of the secret sharing scheme that is used, the shares of a file do not reveal any information about it and the shares of one file are independent of the other. Therefore, having access to all the shares of one file do not convey any information about other files as well. Thus, we obtain:

 I(W[1:N];Zλk,Zk)=0, (14) I(W[1:N]∖dk;(Zλk∪Sdk),Zk)=0. (15)

### V-G Calculation of Rate

Now, we calculate the required transmission rate in the worst-case scenario. According to the delivery procedure summarized in Algorithm , there is a transmission corresponding to every distinct in and each transmission is of size bits. For each , the value that takes is different as it depends on the association profile . Assume appears times in the PDA and let be the set of column indices in which occurs. Then, the number of transmissions in which is involved depends on the number of users connected to the most populated cache amongst the above set of helper caches. The most populated cache in the above set corresponds to the minimum of as it is assumed that the helper caches are labelled in a non-increasing order of the number of users accessing it. Thus, we obtain the normalized rate as

 Rs=∑s∈[S]LτsF−Z

where, is defined as the minimum column index in which appears in the PDA . This concludes the proof of Theorem 1.

### V-H Example: (Λ=6,K=21,M=21,N=21) shared caching problem with L=(6,5,4,3,2,1).

Consider a setting with a server having access to files , each of size bits. The server is connected to users, each possessing a cache of size file. There are helper caches, each of size equal to files. For this setting, we start with a PDA given in (16), which satisfies the condition .

 P=⎛⎜ ⎜ ⎜⎝⋆⋆⋆123⋆12⋆⋆41⋆3⋆4⋆23⋆4⋆⋆⎞⎟ ⎟ ⎟⎠ (16)

Each file gets splits into subfiles , each of size bits. Then, each file is encoded using a non-perfect secret sharing scheme. To generate the shares of a file, first form an column vector comprised of subfiles and independent random keys , each uniformly distributed over . Then, pre-multiply it with the parity check matrix of a MDS code over , where is sufficiently large such that the MDS code exists. In this example, we consider a Cauchy matrix over . Thus, the four shares of the file , are obtained as follows:

 ⎡⎢ ⎢ ⎢ ⎢⎣Sn,1Sn,2Sn,3Sn,4⎤⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣1624614224164216⎤⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢ ⎢⎣Wn,1Wn,2Vn,1Vn,2⎤⎥ ⎥ ⎥ ⎥⎦

The contents stored in each helper cache are:

 Z1 ={Sn,1,Sn,2:n∈[21]} Z2 ={Sn,1,Sn,3:n∈[21]} Z3 ={Sn,1,Sn,4:n∈[21]} Z4 ={Sn,2,Sn,3:n∈[21]} Z5 ={Sn,2,Sn,4:n∈[21]} Z6 ={Sn,3,Sn,4:n∈[21]}

Let the user-to cache association be with . The generalized PDA is obtained as given in (17). Each user has access to shares of each file, but the helper cache contents do not reveal any information about due to the non-perfect secret sharing encoding.

Corresponding to each distinct ordered pair in , the server generates independent random keys, each uniformly distributed over and is indexed by an ordered pair. The keys stored in each user’s cache are:

 Z1 ={K(1,1),K(2,1)}, Z2={K(1,2),K(2,2)} Z3 ={K(1,3),K(2,3)}, Z4={K(1,4),K(2,4)} Z5 ={K(1,5),K(2,5)}, Z6={K(1,6),K(2,6)} Z7 ={K(1,1),K(3,1)}, Z8={K(1,2),K(3,2)} Z9 ={K(1,3),K(3,3)}, Z10={K(1,4),K(3,4)} Z11 ={K(1,5),K(3,5)}, Z12={K(2,1),K(3,1)} Z13 ={K(2,2),K(3,2)}, Z14={K(2,3),K(3,3)} Z15 ={K(2,4),K(3,4)}, Z16={K(1,1),K(4,1)} Z17 ={K(1,2),K(4,2)}, Z18={K(1,3),K(4,3)} Z19 ={K(2,1),K(4,1)}, Z20={K(2,2),K(4,2)} Z21 ={K(3,1),K(4,1)}.

In the delivery phase, each user requests a file from the server. Then, the messages transmitted are as follows:

 X(1,1) =S1,3⊕S7,2⊕S16,1⊕K(1,1) X(1,2) =S2,3⊕S8,2⊕S17,1⊕K(1,2) X(1,3) =S3,3⊕S9,2⊕S18,1⊕K(1,3) X(1,4) =S4,3⊕S10,2⊕K(1,4) X(1,5) =S5,3⊕S11,2⊕K(1,5) X(1,6) =S6,3⊕K(1,6) X(2,1) =S1,4⊕S12,2⊕S19,1⊕K(2,1) X(2,2) =S2,4⊕S13,2⊕S20,1⊕K(2,2) X(2,3) =S


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BattleField 2 The following people contributed to this publication: Björn Christoph, Ken Creedon, Stuart Dunsmore, Daniel Ferguson, Tony Ray, Jason Roman, Paul Seale, and Hendrik Thole. This publication is © Copyright 2003-2010 by Even Balance, Inc. All Rights Reserved. PunkBuster™ Software and the PunkBuster Logo are trademarks of Even Balance, Inc. Other trademarks referenced by this publication are the property of their respective owners.

 Introduction What PunkBuster Is All About Quick Start Guide The Basics Enabling and Disabling PunkBuster Finding PunkBuster Enabled Servers Updating PunkBuster Commands and Settings Using PunkBuster to the Fullest Screen Capture Facility PunkBuster Player Power The Name Registry Remote Logging Facility Violations Frequently Asked Questions

 This publication is for Players of online games who are interested in using or finding out more about the PunkBuster Anti-Cheat system.  This publication is meant to document PunkBuster for the following games: Battlefield 2 by EA Games.  PunkBuster editions for other games may vary from the edition described herein. A great effort has been made to make this publication useful for those new to PunkBuster and also for advanced users of PunkBuster.  Any comments or suggestions will be gladly received via email to docs@evenbalance.com. If you wish to learn about using the PunkBuster Server software for running a game server, please see our related publication entitled 'PunkBuster for Server Administrators'.

 From the main game menu, choose Multiplayer then Join Internet to get to the In-Game Server Browser.  On the In-Game Server Browser screen, make sure the 'PunkBuster' button is checked. If not, check it to enable PunkBuster. The In-Game Server Browser You are now ready to join any servers that are running PunkBuster. There is a small PunkBuster logo to the left of the servers that are PunkBuster Enabled.  When PunkBuster needs to give you information, it will appear at the top of your playing screen briefly and then scroll off he top of the screen.  If you wish to review previous messages, press the tilde (~) key to bring down the game console.  From inside the game console, use the 'PageUp' and 'PageDown' keys to scroll the console text up and down if necessary.

 From the game's main screen, clicking 'Multiplayer' moves to the in-game Server Browser and Selection Screen.  In the bottom portion of this screen will be a check box.  Click this button to toggle the current status of PunkBuster.  There is no additional cost involved for using PunkBuster in a personal, non-commercial environment.  If you are playing in a prize tournament or any event (whether over the Internet or in a LAN environment) that charges an entry fee and/or awards prizes to participants, the organizers of that event are required to obtain a commercial license with Even Balance, Inc. if they wish to utilize PunkBuster Software during the event.  More information about this requirement can be obtained by emailing license@evenbalance.com NOTICE: PunkBuster cannot be Disabled immediately during the middle of gameplay.  Choosing to Disable PunkBuster by any methods described above will set PunkBuster into Disabled mode but that does not take effect until the entire game is restarted.  Neither Players nor Server Admins can simply turn PunkBuster off and on during the middle of gameplay.  Once PunkBuster Client and/or Server has been Enabled for a gaming session, it stays Enabled regardless of how many times the Enabled/Disabled mode is toggled.  When the game is exited, the current PunkBuster status is saved and the next time the game is started, PunkBuster will initialize to the mode in which it was last set, either Enabled or Disabled.  That means players can Enable PunkBuster once and then never worry about it again if desired, as it will continue to be enabled from that point on even when the Game is exited unless specifically Disabled at some later point in time.

 Most Players use some type of Game Browser to find Game Servers to join.  The Game itself has a very nice integrated Game Browser built in so that Players can usually find what they are looking for very easily.  Some Players prefer to use external Game Browsers such as the ones produced by GameSpy, Pingtool and Kali. The In-Game Server Browser For Players who use the in-game browser, a column is displayed that specifically shows whether or not each server listed currently has PunkBuster Enabled. For external Game Browsers, some configuration may be required to display and/or filter servers based on whether or not they currently have PunkBuster Enabled.  First, a little background: while a Game Server is running, it sends a "heartbeat" packet of information every few seconds to several Master Game Servers located around the world.  This heartbeat packet tells the Master Game Servers that the Game Server is running, how many players are currently connected, and the current status of all sorts of things such as the Name of the server, the maximum number of players and what is commonly referred to as the "server rules".  One of the server rules returned by Game Servers is called sv_punkbuster.  If sv_punkbuster is set to "1" for a particular Server, then PunkBuster is Enabled on that Server.  Consult the documentation for your chosen external Game Browser to determine how to display and/or filter servers based on the value of sv_punkbuster.

 One of the core aspects of the PunkBuster system is the automatic update feature.  As new versions of PunkBuster files are made available for download from our Internet-based Master PunkBuster Servers, the PunkBuster software running on Players' and Admin's computers will attempt to retrieve those new versions and perform an automatic "self-update" function in the background without interrupting gameplay.  For most users, this occurs automatically and seamlessly.  Since PunkBuster Servers require all connected Players to be running identical versions of all PunkBuster files in order to be authenticated for "clean" online play, the automatic update facility inside PunkBuster is an important component of the overall system.  PunkBuster will retain older versions in case there is ever a need to connect to a Server that is still running old versions of the PunkBuster software. All PunkBuster update files are stored in special compressed HTM files and can be opened with any Internet Web Browser.  We have designed and implemented a robust method for securely delivering updates from our Master PunkBuster Servers to Regular PunkBuster Servers and then on to computers that have Enabled the PunkBuster Client software.  Great care has been taken to design and develop the system such that only authentic PunkBuster files will ever be accepted and used by the PunkBuster System during the auto-update process. There may be times, however, when it is necessary to manually update a PunkBuster System.  For instance, if there is no Internet Access available and multiplayer gameplay is occurring in a LAN environment.  Or perhaps, there is a temporary routing problem causing locally running PunkBuster software to not be able to "find" any Master PunkBuster Servers whereby an update can be obtained.  There may also be times that the PunkBuster Team offers pre-release beta versions of PunkBuster software for manual download so that it can be tested before broad distribution.  For whatever reason that may come up that requires a manual update, we have prepared a separate section, called 'Updating PunkBuster' on our website at Even Balance.

 The use of PunkBuster commands and settings is optional.  This information is provided for advanced users who want to get more out of the PunkBuster software or who want to customize it.  A brief overview of using commands and settings is given, as well as a reference list of all available commands and settings along with explanations of their correct use. The most common way to enter a command or to check / change a setting in PunkBuster is to bring down the Game Console and type away.  However, since PunkBuster is directly integrated into the game, PunkBuster commands and settings can also be treated just like any other game command or setting in that they can be placed into game script files, bound to keystrokes and used in aliases. The easiest way to quickly test that your PunkBuster system is properly installed, Enabled, and working is to type "PB_VER" into the game console.  If PunkBuster is working properly, it will respond by displaying the version number that is currently installed and running.  If there is no response, then PunkBuster is either currently Disabled or there is a problem that may require some troubleshooting.  Consult the FAQ (Frequently Asked Questions) Appendix later in this publication for suggestions on ways to troubleshoot this type of problem. Some commands are standalone and others may require or accept additional parameters.  For example, the command "PB_VER" is standalone, typing in "PB_VER" tells PunkBuster that you would like to know the currently running Version number.  The command "PB_LOAD" takes one parameter (a filename), typing in "PB_LOAD ABC.CFG" tells PunkBuster to load the configuration file called "ABC.CFG". PunkBuster settings, sometimes also called variables, hold numbers or textual information that PunkBuster uses while it is operating.  Changing PunkBuster settings will affect the way PunkBuster runs in specific ways.  All PunkBuster settings start out with default values that are the recommended setting values for most users.  To find out the current value of a setting, simply type in the name of the setting all by itself.  For example, typing "PB_LOGTOFILE" will cause PunkBuster to display the current value.  Also displayed is the allowable range that the value can contain.  This particular setting can be set either to 1 or 0 and 0 is the default.  To set it to 1, type in "PB_LOGTOFILE 1", PunkBuster will then respond by showing the setting name along with its new value.  Trying to set a PunkBuster setting outside of its allowable range will cause PunkBuster to give the setting the closest allowable value to what was originally specified. Listed below (in alphabetical order) is a list of PunkBuster Commands and Settings along with a general description and usage instructions where necessary. PB_GetSs Captures and saves a screen shot based on the parameters found in the PB_SSOPTIONS setting listed below; described more fully in the Screen Capture Facility section PB_Kick [slot #] Requests that a specific player be removed from the game; this command is part of the PB Player Power facility PB_Load [File Name] Loads the specified PunkBuster config file which can contain commands and/or setting changes PB_MyGuid Displays your PunkBuster GUID and your Game CDKey (Note: Cdkey is not shown if currently connected to a server); The GUID is used by PunkBuster to identify players and it is based on the CDKey entered into the game PB_PList Requests that the connected PunkBuster Server display a list of players and their current status to the console PB_Power Requests that the connected PunkBuster Server display a list of players and their current PB Player Power status to the console PB_SsOptions Sets and/or displays the parameters used for taking local PunkBuster Screenshots; used by the Screen Capture Facility PB_Ver Displays the currently running version of the PunkBuster Client software PB_WriteCfg Writes the current values of the PunkBuster settings to the local hard drive (creating or overwriting files called pbcl.cfg and pbcllog.cfg) in such a way that they will be loaded automatically the next time PunkBuster starts; if you make changes to your PunkBuster settings and you want the changes to be permanent, then enter this command after making the changes PB_AutoExec [Filename] Holds the name of a game script file that is to be executed each time a server is joined; This is useful if you need to reset game settings each time you join a server PB_Lan [0=no, 1=yes] Defaults to 0; when set to 1, PB will behave as though it has no Internet access PB_LogAddr [Address / Internet Hostname] Holds the address of the machine to which remote logging is sent (default is empty meaning the feature is not used) PB_LogMD5Tool [0=no, 1=yes] Setting defaults to 0; setting to 1 will cause PB to log all MD5Tool file check query replies sent to connected PB Servers during gameplay PB_LogPort [Port #] Holds the listen port of the machine to which remote logging is sent (default is 0 meaning the feature is not used) PB_LogPw [Password] Holds the password required to send logging output to a listening remote logging machine PB_LogToFile [0=no, 1=yes] When set to 1, PunkBuster will log activity to the pbcl.log file; the default is 0 which meansonly errors are logged PB_LogUser [Username] Holds the username required to send logging output to a listening remote logging machine PB_MsgPrefix [New Prefix] Holds the text that PunkBuster displays in front of every output line displayed; the default is "PunkBuster Client". PB_Security [0=no, 1=yes] When set to 0, PunkBuster will accept all "update" files without verifying that they are authentic; the default is 1 which should not normally be changed except for LAN environments where there is no Internet Access and/or you fully trust the Servers you are connecting to PB_Sleep [Period] Holds the period of time (milliseconds) that PunkBuster "sleeps" between processing cycles; the default is 60; lower numbers will cause PunkBuster to process events more times each second which also has the effect of increasing the bandwidth used by PunkBuster; Players with a modem connection will probably want to set this as high as possible PB_SsLog [0=no, 1=yes] When set to 1 (which is the default), PunkBuster will create an entry in the pbsslog.htm log file every time a PunkBuster Server requests a screenshot to be taken and delivered; this log file is designed to be viewed with a web browser and contains information about the size of the image, when it was taken, and also a digital signature of the image itself PB_SsSave [0=no, 1=yes] When set to 1 (0 is the default), PunkBuster will save all Server-requested Screenshots to the local hard drive so that Players can see the actual screenshots that have been delivered to Game Servers that they were connected to; if this setting and the PB_SsLog setting are both set to 1, then the pbsslog.htm log file will automatically have links created for local viewing directly from the log file PB_System [0/1] PB can conflict with other software on some systems due to vigorous memory scanning; Changing this setting may reduce or eliminate conflicts if you have problems on your system; this setting can only be changed before connecting to a server by editing the pbsys.cfg file in your PB folder. This file is created the first time you run PB.

 PunkBuster issues several types of messages during gameplay.  These will appear at the top of the playing screen for a few seconds before scrolling off.  All messages also are logged to the console and old messages can be reviewed by bringing down the game console.  The game console can be scrolled up or down by using the PageUp or PageDown keys respectively.  During almost every game, there will be periods of inactivity where players are not actively playing the game.  This is a good time to review previous PunkBuster messages to see if you missed any important messages while playing. The most useful PunkBuster command to get acquainted with is the pb_plist command.  By occasionally issuing a pb_plist command, you can view the PunkBuster status of all players in the game.  The status of each player can either be INIT, UPDT or OK.  The INIT status means that the PunkBuster Server is still initializing that player’s connection.  The UPDT status indicates that the PunkBuster Server is updating that player’s PunkBuster installation.  Other information is displayed by the pb_plist command such as the Auth Rate and RecentSS columns.  The Auth Rate of a player is how many times per minute the player’s PunkBuster has successfully authenticated with the PunkBuster Server.  Once each player has reached OK status, then the Auth Rate should never fall below 2.0.  If it does, that would indicate that PunkBuster is not functioning normally for some reason.  The RecentSS column indicates how many screenshots the PunkBuster Server has requested from each player during the past few minutes. After each map ends and a few seconds after the next one is starting, the PunkBuster Server will report some information to connected players about the Server log files.  This information gives players the filenames of the log files and also the Digital Signatures of the log files saved at the Server.  This is useful mostly in organized matches where Server Admins provide logs to administrators after each match.  The Digital Signatures that are reported to players keep the Server Admin or anyone else from being able to change or edit the log files without altering the reported Digital Signatures.  If a log file is changed at all, then its Digital Signature will no longer match what was broadcast to all players in the game. We have for download, on our website at evenbalance.com, a free utility called pbmd5 that will compute the Digital Signature (also known as MD5) of any file.  If you would like to "save" the Digital Signature of log files reported by a Server, we recommend bringing down the game console and then taking a screenshot of your playing screen while that information is displayed.  If desired, the PB_GETSS command can be used to save the screenshot.

 The PunkBuster Screen Capture Facility allows Server Admins to request actual screenshots from players' screens while they are playing.  The screenshots are transferred over the network and saved by the Server Admins either for private use or possibly for publication to a website.  PunkBuster limits the filesize of the screenshots by setting a maximum number of pixels that can be captured.  PunkBuster also limits how many times a PunkBuster Server can request a screenshot within a few minutes to prevent Server Admins from taking screenshots one after another continuously.  All PunkBuster screenshots are saved in the PNG image format and therefore have the ".png" extension.  This format is similar to the JPEG format but does not reduce image quality during compression.  Most newer Web browsers can view PNG files directly.  Programs that convert PNG format images to and from other formats are widely available.  More information about the PNG format can be found at the PNG-Homepage. There are two PunkBuster settings that can be used by players to track the activity of PunkBuster Servers that request screenshots from their system.  The settings are called PB_SSLOG and PB_SSSAVE.  If PB_SSLOG is set to 1 (which is the default), then PunkBuster will keep a log file on your hard drive in the "pb" folder called "pbsslog.htm".  This log can be viewed using any web browser such as Netscape or Internet Explorer.  The log contains information about each screenshot that was requested by a PunkBuster Server.  If PB_SSSAVE is set to 1 (note that 0 is the default), then PunkBuster will save the actual screenshot to your hard drive after the screenshot image has been transmitted to a PunkBuster Server.  Note that screenshot images vary in size usually between 40 and 80 thousand bytes.  If you play frequently, saving Screenshots to your hard drive can consume a lot of disk space after only a few days.  We don't recommend setting PB_SSSAVE to 1 unless you are sure that you will routinely view the screenshot images and remove them if they start to take up too much disk space for your system. If PB_SSSAVE is set to 1, PunkBuster saves the screenshots into the folder called "scrnshot" inside the "pb" folder.  The filename used is based on the date that the screenshot was taken.  For example, the first screenshot taken on March the 15th will have the filename "0315_000.png".  The next one would have "0315_001.png", etc.  If you exit the game and play again later the same day, PunkBuster will start over at 000 and overwrite the images taken in earlier sessions.  Please keep this in mind if you intend to keep all of your screenshot images. When PunkBuster writes log lines to the "pbsslog.htm" file due to a screenshot request from a connected PunkBuster Server, the log lines will have the following format: "

 This facility is not associated with the votekick system that is built into the game itself. Both systems work independently of each other and can be used together or be enabled/disabled separately depending on the desires of each PB Server Admin. PB Player Power allows PB Server Admins to empower their trusted regular players with more weight in a new elective process used to remove an undesirable player from the game server. Frequently, troublemakers who attack teammates, block doorways, or otherwise ruin the gameplay are able to get around other methods of keeping them off of the server. This new facility aims to tackle this issue so that gameplay on good public servers doesn't suffer when troublemaking punks show up. The use of this facility by players requires being comfortable with issuing PB Commands. Please refer to the Commands section of this publication for general information on how to access the game console and enter PunkBuster commands. When connected to a PB Server, use the pb_power command to determine if PB Player Power is enabled or disabled on that server. The first line of output shows this indication. Below that line, a list of all connected players is displayed showing each player's PB slot #, power rating and also the number of power points that have been applied against each player (if any). Once the number of power points applied against a player has reached the limit for the server, then that player is removed from the game and won't be allowed to rejoin until several minutes have passed (the exact amount of time is set by each PB Server Admin). Notice that if any players are shown to have 100 power points, then those players are considered deputies to the PB Server Admin. If one or more deputies are connected, then only deputies are able to use the PB Player Power facility. When no deputies are present, all players on the server may each apply their power points against other players. To apply your power points against a player that you want removed from the game, use the pb_kick command. You must specify the slot # of the player who you wish to have removed. The slot # of each player is shown in the list displayed by the pb_power command described above. For example, if you wish to remove the player in slot #5, then press your tilde key (~) to bring down the game console, then enter "pb_kick 5" (without the quotes), press the Tab key, and then press the Enter key. The PB Server will ignore your pb_kick commands if there are one or more deputies connected or if you enter the command more frequently than once per minute. There is no duration for the pb_kick command. It stays in effect until you leave the server, or the target player leaves the server or until you specify a different player with the pb_kick command. That means that your power points can only be applied against any one other player at a time. Players should be aware that when power points are applied against another player, the event is reported to all players on the server by PunkBuster. The report includes the target player's slot # and name as well as the name of the player who issued the pb_kick command and the associated number of power points applied.

 Even Balance, Inc. offers a premium, optional Name Registry hosting service for Players, Clans, Guilds and other groups who wish to take advantage of our facility for online identity verification.  Registered names are password protected so that only persons knowing the correct password will be designated as *Registered while playing on PunkBuster-Enabled Game Servers.  Subscribers will be able to Register Names either with Partial or Full matching.  Registered Names for Partial matching are designed for groups such as clans, guilds, etc. whose members wear a "tag" (usually at the beginning or end of each member's playing name).  For Partial type Names, the Registered Name can appear anywhere in the player's name to attain *Registered status.  Full matching Registered Names will require an exact match of the playing name to the Name Registration record in order to attain *Registered status. Each Registered Name comes with a descriptive line of text (up to 80 characters) that is provided so that a website, email address, membership requirements, or other information describing the group or person represented by the Registered Name can be displayed to other players during gameplay.  The Name Registry is not exclusive with regards to unique Registered Names.  Different groups / players will be allowed to Register a Name that is already in the Registry.  All Registrations of a unique Name after the first will have a serial number automatically appended to the Name Registration so that different Registrations can be easily distinguished. More information is available on our website at evenbalance.com.

 Even Balance, Inc. will soon be offering a premium, optional hosting service for Players who wish to remotely log screenshot signatures in real time for automatic publication to a fully and openly accessible website.  We will offer a similar service for Server Admins.  Each subscriber gets a separate page and each days' activity is further separated into web pages.  The subscription service is password protected so that players who play on multiple machines can log screenshot signatures from their various playing machines.  When this Service is available, more information will be available on our website at evenbalance.com.


text:
# Daddy Was A Number Runner

SBN: 1558614427    EAN: 9781558614420

Publisher: Feminist Press     (View Publisher's Titles)

US SRP: \$16.95 US  -  (Discount: 15%)

Binding: Paperback

Pub Date: December 01, 2002


text:
# Dirac Delta function question

I cannot get the answer as from the solution manuel.

Please tell me what am I assuming wrong.
Thanks

Dick
Homework Helper
Just showing that the value of a function is '-infinity' at a point and zero otherwise doesn't tell you it's a delta function. Apply the divergence theorem to grad(1/r) on a sphere around the origin to figure out how much 'charge' is there.

Just showing that the value of a function is '-infinity' at a point and zero otherwise doesn't tell you it's a delta function. Apply the divergence theorem to grad(1/r) on a sphere around the origin to figure out how much 'charge' is there.

I understand that and I showed in my own answer. What I don't get is where the 4$$\pi$$ come from!!
Thanks

Dick


text:
§ Pavel: bridges, articulation points for UNDIRECTED graphs

• Two vertices are 2-edge connected if there are 2 paths between them. The two paths cannot share ANU edges.
• Every bridge must occur as a DFS tree edge, because DFS connects all components together.
• More generally, every spanning tree contains all bridge edges.
• Now we check if each edge is a bridge or not.
• To check, we see what happens when we remove the edge $(u, v)$. If the edge is not a bridge, then the subtree of $v$ must connect to the rest of the graph.
• Because we run DFS, the subtree rooted at $v$ must go upwards , it cannot go cross. On an undirected graph, DFS only gives us tree edges and back edges.
• This means that if the subtree rooted at $v$ is connected to the rest of the graph, it must have a backedge that is "above" $u$, and points to an ancestor of $u$.
• Instead of creating a set of back edges for each vertex $v$, we take the highest / topmost back edge, since it's a safe approximation to throw away the other back-edges if all we care about is to check whether there is a backedge that goes higher than $u$.
• To find the components, push every node into a list. When we find an edge that is a bridge, take the sublist from the vertex $v$ to the end of the list. This is going to be one connected component. We discover islands in "reverse order", where we find the farthest island from the root first and so on.

§ Vertex connectivity

• The problem is that vertex connectivity is not an equivalence relation on vertices!
• So we define it as an equivalence relation on edges .
• More subtle, we cannot "directly" condense. We need to build a bipartite graph, with components on one side and articultion points on the other side.


text:
## 前言

• 一层循环就是一层O(N), 循环套循环需要相乘
• 如果是树状结构, 比如说有分治(二叉搜索, 快排)的时候, 那么就是O(logN)
• 如果有递归:

• 如果是尾递归, 那么其实是和循环等价的
• 如果是树状递归, 比如说斐波拉契数列, 那么就是O(2^N), 确切的说是O(branches^depth)

## Example1

Suppose we had an algorithm that took in an array of strings, sorted each string, and then sorted the full array. What would the runtime be?

Let's define new terms-and use names that are logical.

• Let s be the length of the longest string.
• Let a be the length of the array.

Now we can work through this in parts:

• Sorting each string is O(s log s).
• We have to do this for every string (and there are a strings), so that's O(a*s log s).
• Now we have to sort all the strings. There are a strings, so you'll may be inclined to say that this takes O(a log a) time. This is what most candidates would say. You should also take into account that you need to compare the strings. Each string comparison takes O(s) time. There are O(a log a) comparisons, therefore this will take O(a*s log a) time.

If you add up these two parts, you get 0(a*s (log a + log s)).

## Example2

The following code prints all Fibonacci numbers from 0 to n. What is its time complexity?
void allFib(int n) {
for (int i= 0; i < n; i++) {
System.out.println(i + ": "+ fib(i));
}
}

int fib(int n) {
if (n <= 0) return 0;
else if (n == 1) return 1;
return fib(n - 1) + fib(n - 2);
}

Instead, let's walk through each call.

• fib(!) -> 2^1 steps
• fib(2) -> 2^2 steps
• fib(3) -> 2^3 steps
• fib(4) -> 2^4 steps
• fib(n) -> 2^n steps

Therefore, the total amount of work is: 2^1 + 2^2 + 2^3 + 2^4 + ... + 2^N = 2^(N+1)

## Reference

Cracking the Coding Interview 6th Edition


text:
# EMF induced in a rolling disk

1. May 3, 2016

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

O is the center of the disk

Part a) The emf induced across PS can be found by thinking as if a rod PS is rotating and translating .

The component of velocity of rod perpendicular to its length is vsin45° = v/√2 . The length PS = √2R . Hence emf induced would be B(v/√2)(√2R) = BvR.

Is that correct ?

Thanks

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2. May 4, 2016

### TSny

I believe that's right. I assume "due to translation" means to treat the disk as sliding along the surface with speed v with no rotation, while "due to rotation" means to keep the center of the disk at rest and rotate the disk about the center with angular speed ω.

3. May 4, 2016

### Vibhor

Ok.

But what if the question asked "What is the emf induced in PS due to rotation ?" . I found this a bit difficult .

For this I instead took the bottommost point as instantaneous axis and considered rod PS in pure rotation . Then I applied the result $\frac{1}{2}Bωl^2$ to this and found total induced emf = $\frac{1}{2}B( \frac{v}{R})(√2R)^2$ = BvR . On comparing this "total induced emf" with the previous result i.e induced emf due to translation , I concluded that emf induced in rod PS due to rotation is 0 .

Is this result correct ?

4. May 4, 2016

### TSny

You should be able to argue that the emf for PS is zero "due to rotation" if "due to rotation" means rotation about the center.

5. May 4, 2016

### Vibhor

Do you find my above reasoning inadequate ?
Are you suggesting that I should imagine the rod PS in pure rotation about the center O and see if it cuts any magnetic field lines ?

6. May 4, 2016

### TSny

No, your reasoning is correct. I was just saying that you can see that the emf due to rotation about the center is zero more directly.

Yes
More specifically, consider the emf for small sections of PS. You shouldn't need to actually carry out an integration to deduce the total emf.

7. May 4, 2016

### Vibhor

Ok .

So , if a rod translates and rotates then Is induced emf across the rod = emf induced due to translation + emf induced due to rotation ??

Can I think of it as a general result ?

8. May 4, 2016

### TSny

Yes. The velocity of any element of the rod is a sum of the velocity due to translation and the velocity due to rotation.

( til tomorrow)

9. May 4, 2016

### Vibhor

Ok .

10. May 4, 2016

### Vibhor

The line PS divides the disk into two unequal sized circular segments ,say I and II .

When we say induced emf across PS is BvR, we are basically dealing with two parallel EMF's ,each of magnitude BvR , between points P and S . These emfs are due to portion I and II of the disk . The equivalent of two parallel EMF's of magnitude BvR , across PS , is BvR .

Makes sense ?

11. May 4, 2016

### Vibhor

This is how I tried to reason it out .

There would be only one point on the rod (call it M) which would be moving at right angles to the line joining the point to the center O . 'M' is also the mid point of the rod . In other words this point would have velocity directed along the length of the rod . Except this point , every point on PS will have a velocity component perpendicular to the length of the rod .

Now , consider a tiny length AB in the portion PM of the rod .This will have a component of velocity vP perpendicular to the length PS , but directed away from center O . Corresponding to this AB there will be a tiny length CD in MS part of the rod .This will also have a component of velocity vP perpendicular to the length PS , but directed towards center O .

The two tiny EMF's in AB and CD wil cancel each other as far as induced emf across PS is concerned . Similarly EMF in each part of PM will cancel that generated in MS . The result is that emf induced in PS due to rotation is 0 .

Is this reasoning fine ?

12. May 4, 2016

### TSny

Yes, very nice.

13. May 4, 2016

### Vibhor

Ok . Is post#10 correct ??

14. May 4, 2016

### TSny

Yes, I believe it's ok to think of it that way. It's like the picture shown below where you have rods moving through a B field into the page (not shown). In the figure on the left, you can think of the emf from A to B of the wide rod as due to two parallel emf's by dividing the rod into two halves as shown. Likewise, in the figure on the right, the emf of the long rod from A to C can be thought of as two emf's in series (A to B plus B to C).

#### Attached Files:

• ###### emfs.png
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15. May 4, 2016

### Vibhor

Thanks a lot .


text:
English Információ A lap Pontverseny Cikkek Hírek Fórum

Rendelje meg a KöMaL-t!

VersenyVizsga portál

Kísérletek.hu

Matematika oktatási portál

I. 346. The present exercise is similar to our earlier Problem I. 343. Now we again have an $\displaystyle N\times N$ ($\displaystyle 4\le N\le 50$) board, and each cell of the board is either empty or has some color. When the board is in its vertical position, the colored squares try to occupy the lowest possible empty positions due to the gravitational force. This time however the board also contains some preset and fixed obstacles that block the way of the sliding colored squares. Empty squares are denoted by a . character, the colored squares can be either red (denoted by P), blue (K) or green (Z), while a fixed obstacle is denoted by A. The board is surrounded by a border preventing the colored squares from falling out. In this exercise the board can be rotated only to the right. After the rotation, the colored squares will again slide to the lowest free positions, and they will stay on top of each other. We are now interested in the mixing effect due to the presence of the obstacles and the rotations. In the example, indulás'' is the starting position, and 1. forgatás után'' is the state of the board after the first right turn.

In the starting position and after each rotation you should determine the number of colored squares having at least two neighbors of the same color. The figure shows that a square $\displaystyle (i,j)$ has by definition 4 neighbors.

Create your program i346 to simulate the rotation of the board and the mixing of the colors.

The first command line argument to your program is the name of the data file describing the board. The first line of the file contains the board size $\displaystyle N$ ($\displaystyle 4\le N\le 50$) and the number of rotations $\displaystyle K$ ($\displaystyle 1\le K\le 100$). The following $\displaystyle N$ lines contain the initial cell content of the board.

The second command line argument to your program is the name of the output file. The first line of this file should contain the number of colored squares having at least two neighbors of the same color after each rotation; these numbers should be separated by a space. The remaining $\displaystyle N$ lines of the output file should describe the content of each cell of the board after the last rotation; each line of the file corresponds to a row of the board. In the example, Bemenet'' is an input, while Kimenet'' is the corresponding output.

The source code (i346.pas, i346.cpp, ...) of your program and a brief documentation (i346.txt, i346.pdf, ...) of your solution should be submitted, also specifying the name of the developer environment to use for compiling your source.

(10 points)

Deadline expired on 12 May 2014.

Google Translation (Sorry, the solution is published in Hungarian only.)

A kimeneti fájl első sorába a forgatások számánál eggyel több számot ír a program. Az első szám a kezdeti állapotot mutatja. Sajnos egy kis elírás történt a lapban és a minta bemenetre ez az első érték nem 5 a kimenetben, hanem 6.

Tesztállományok: tesztbe.zip teszteredmenyek.ki

Minta megoldás:

Paulovics Zoltán 11. osztályos tanuló (Zrínyi Miklós Gimnázium, Zalaegerszeg) megoldása: i346.vb

Statistics on problem I. 346.
 12 students sent a solution. 10 points: Csernák Tamás, Fényes Balázs, Kovács 246 Benedek, Kovács Balázs Marcell, Paulovics Zoltán, Radnai Bálint, Uzonyi 000 Ákos. 8 points: 2 students. 5 points: 2 students. Unfair, not evaluated: 1 solution.

• Problems in Information Technology of KöMaL, April 2014

•  Támogatóink: Morgan Stanley


text: Nice finding today! Just when I thought I would need to write a full macOS native app for viewing Fasta files or MAFFT-aligned sequences, I found it already exists, and it’s so much faster and prettier than Jalview. Thank you so much Mathieu Fourment! #bioinfo


text:
# TÀI LIỆU TOÁN - WWW.MOLYMPIAD.NET - ĐỀ THI TOÁN

## $show=home 1. Find all triple of natural numbers$a, b, c$less than$20$such that $$a(a+1)+b(b+1)=c(c+1)$$ where$a$is a prime number and$b$is a multiple of$3 .$2. Let$f(n)=\left(n^{2}+n+1\right)^{2}+1,$where$n$is a positive integer and let $$P_{n}=\frac{f(1) \cdot f(3) \cdot f(5) \ldots f(2 n-1)}{f(2) \cdot f(4) \cdot f(6) \ldots f(2 n)}.$$ Prove the inequality $$P_{1}+P_{2}+\ldots+P_{n}<\frac{1}{2}.$$ 3. Find the maximum value of the expression $$T=\frac{(y+z)^{2}}{y^{2}+z^{2}}-\frac{(x+z)^{2}}{x^{2}+z^{2}},$$ where$x$,$y$,$z$are real numbers such that$x>y$,$z>0$and$z^{2} \geq x y$. 4. Solve for$x$$$\sqrt[3]{14-x^{3}}+x=2\left(1+\sqrt{x^{2}-2 x-1}\right).$$ 5. An acute triangle$A B C$is inscribed in a fixed circle with center at$O$. Let$A I$,$B D$and$C E$denote the altitudes through$A$,$B$and$C$respectively. Prove that the perimeter of the triangle$I D E$does not change when$A, B,$and$C$move on the circle$(O)$such that the area of the triangle$A B C$is always equal to$a^{2}$. 6. Solve the system of equations $$\begin{cases} \sqrt{x^{2}+91} &=\sqrt{y-2}+y^{2} \\ \sqrt{y^{2}+91} &=\sqrt{x-2}+x^{2}\end{cases}.$$ 7. Let$a, b, c$be real numbers such that$4(a+b+c)-9=0 .$Find the maximum value of the sum $$S=\left(a+\sqrt{a^{2}+1}\right)^{b}\left(b+\sqrt{b^{2}+1}\right)^{c}\left(c+\sqrt{c^{2}+1}\right)^{a}.$$ 8. Let$I$and$O$denote respectively the incenter and the circumcenter of a triangle$A B C .$Given that$\widehat{A I O}=90^{\circ},$prove that the area of the triangle$A B C$is less than$\dfrac{3 \sqrt{3}}{4} A I^{2}$. 9. Let$a, b, n$be positive integers,$b>1$and$a$is a multiple of$b^{n}-1 .$Rewritten$a$to the base$b,$prove that the resulting contains at least$n$non-zero digits. 10. Let$x, y, z$be real numbers such that$0<z \leq y \leq x \leq 8$and $$3 x+4 y \geq \max \left\{x y ; \frac{1}{2} x y z-8 z\right\}.$$ Find the maximum value of $$A=x^{5}+y^{5}+z^{5}.$$ 11. Find all functions$f: \mathbb{R} \rightarrow \mathbb{R}$such that $$f\left(f(x)+y^{2}\right)=f^{2}(x)-f(x) f(y)+x y+x.$$ 12. Let$K$denote the intersection of the two diagonals of a quadrilateral$A B C D$where$\widehat{A B C}=\widehat{A D C}=90^{\circ} ;$and$A C=A B+A D .$Prove that the radii of the inscribed of the triangles$A B K$and$A D K$are equal. ##$type=three$c=3$source=random$title=oot$p=1$h=1$m=hide$rm=hide ## Anniversary_$type=three$c=12$title=oot$h=1$m=hide\$rm=hide

Name

2006,1,2007,12,2008,12,2009,12,2010,12,2011,12,2012,12,2013,12,2014,12,2015,12,2016,12,2017,12,2018,11,2019,12,2020,12,2021,6,Anniversary,4,
ltr
item
Mathematics & Youth: 2008 Issue 374
2008 Issue 374
Mathematics & Youth
https://www.molympiad.org/2020/09/2008-issue-374.html
https://www.molympiad.org/
https://www.molympiad.org/
https://www.molympiad.org/2020/09/2008-issue-374.html
true
8958236740350800740
UTF-8


text:
• # question_answer AB, ${{A}_{2}}$ and ${{B}_{2}}$are diatomic molecules. If the bond enthalpies of ${{A}_{2}},$AB and ${{B}_{2}}$are in the ratio $1:1\text{ }:0.5$ and enthalpy of formation of AB from ${{A}_{2}}$ and ${{B}_{2}}$is $-100kJ\,mo{{l}^{-1}}$. What is the bond energy of ${{A}_{2}}:$ : A)  $200\,kJ\,mo{{l}^{-1}}$                    B)  $100\,kJ\,mo{{l}^{-1}}$C)  $300\,kJ\,mo{{l}^{-1}}$       D)  $400\,kJ\,mo{{l}^{-1}}$

Let bond energy of ${{A}_{2}}$ be x then bond energy of AB is also x and bond energy of ${{B}_{2}}$ is $x/2$. Enthalpy of formation of AB is$~-100\text{ }KJ/mole$: ${{A}_{2}}+{{B}_{2}}\to 2AB;$      $\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\,\Delta =-100KJ$ or$-100=\left( \frac{x}{2}+\frac{x}{4} \right)-x$$\therefore$                               $-100=\frac{2x+x-4x}{4}\therefore x=400KJ$


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Definition of Variable | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.

# 1.1: Definition of Variable

Difficulty Level: At Grade Created by: CK-12
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Practice Definition of Variable
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What if you had a jar filled with dimes and quarters? You know that the total of the coins in the jar is $8.60. How could you write an equation to represent this situation? After completing this Concept, you'll be able to use variables to write equations like this one with unknown quantities. ### Watch This ### Guidance No one likes doing the same problem over and over again—that’s why mathematicians invented algebra. Algebra takes the basic principles of math and makes them more general, so we can solve a problem once and then use that solution to solve a group of similar problems. In arithmetic, you’ve dealt with numbers and their arithmetical operations (such as $+, \ -, \ \times, \ \div$ ). In algebra, we use symbols called variables (which are usually letters, such as $x, \ y, \ a, \ b, \ c, \ \ldots$ ) to represent numbers and sometimes processes. For example, we might use the letter $x$ to represent some number we don’t know yet, which we might need to figure out in the course of a problem. Or we might use two letters, like $x$ and $y$ , to show a relationship between two numbers without needing to know what the actual numbers are. The same letters can represent a wide range of possible numbers, and the same letter may represent completely different numbers when used in two different problems. Using variables offers advantages over solving each problem “from scratch.” With variables, we can: • Formulate arithmetical laws such as $a + b = b + a$ for all real numbers $a$ and $b$ . • Refer to “unknown” numbers. For instance: find a number $x$ such that $3x + 1 = 10$ . • Write more compactly about functional relationships such as, “If you sell $x$ tickets, then your profit will be $3x - 10$ dollars, or “ $f(x) = 3x - 10$ ,” where “ $f$ ” is the profit function, and $x$ is the input (i.e. how many tickets you sell). #### Example A Write an algebraic equation for the perimeter and area of the rectangle below. To find the perimeter, we add the lengths of all 4 sides. We can still do this even if we don’t know the side lengths in numbers, because we can use variables like $l$ and $w$ to represent the unknown length and width. If we start at the top left and work clockwise, and if we use the letter $P$ to represent the perimeter, then we can say: $P = l + w + l + w$ We are adding $2 \ l$ ’s and $2 \ w$ 's, so we can say that: $P = 2 \cdot l + 2 \cdot w$ It's customary in algebra to omit multiplication symbols whenever possible. For example, $11x$ means the same thing as $11 \cdot x$ or $11 \times x$ . We can therefore also write: $P = 2l + 2w$ Area is length multiplied by width . In algebraic terms we get: $A = l \times w \to A = l \cdot w \to A = lw$ Note: $2l + 2w$ by itself is an example of a variable expression ; $P = 2l + 2w$ is an example of an equation . The main difference between expressions and equa tions is the presence of an equa ls sign (=). In the above example, we found the simplest possible ways to express the perimeter and area of a rectangle when we don’t yet know what its length and width actually are. Now, when we encounter a rectangle whose dimensions we do know, we can simply substitute (or plug in ) those values in the above equations. In this chapter, we will encounter many expressions that we can evaluate by plugging in values for the variables involved. #### Example B Eric has some money in his savings account. How much more money does he need in order to buy a game that costs$98?

Solution:

Let $M$ be the money that Eric still needs and let $S$ be the money that Eric has in his savings account. Then, by subtracting the money he already has from the total money needed, we can figure out how much money he still needs:

$M=98-S.$

#### Example C

Write an equation for the sum of 3 times some number and 5.

Solution:

Let $S$ be the total sum. Let $n$ be some number. Then 3 times some number is $3\cdot n$ and then the sum of that and 5 is:

$S=3n+5.$

Watch this video for help with the Examples above.

### Vocabulary

• We use symbols called variables (which are usually letters, such as $x, \ y, \ a, \ b, \ c, \ \ldots$ ) to represent numbers and sometimes processes.
• $2l + 2w$ by itself is an example of a variable expression ; $P = 2l + 2w$ is an example of an equation . The main difference between expressions and equa tions is the presence of an equa ls sign (=).

### Guided Practice

Alex has a certain amount of nickels and dimes in a jar. Write an algebraic equation for how much money she has, in terms of how many nickles and dimes she has.

Solution:

Let $n$ be the number of nickels and $d$ be the number of dimes that Alex has in the jar. Since each nickel is worth $0.05, the amount of money she has in nickels will be: $0.05\cdot n$ Since each dime is worth$0.10, the amount of money she has in dimes will be:

$0.10\cdot d$

This means that the total amount of money $M$ that Alex has will be:

$M=0.05 \cdot n +0.10 \cdot d.$

Simplifying the expressions, we get:

$M=0.05n +0.10d.$

### Practice

For 1-4, write the following in a more condensed form by leaving out a multiplication symbol.

1. $2 \times 11x$
2. $1.35 \cdot y$
3. $3 \times \frac { 1 } { 4 }$
4. $\frac { 1 } { 4 } \cdot z$

For 5-10, write an equation for the following situations.

1. The amount of money Andrea has in a jar full of quarters and dimes.
2. The amount of money Michelle has in her coin purse if it only contains quarters, dimes and pennies.
3. The sum of 7 and 6 times some number.
4. 4 less than 20 times some number.
5. The amount of money you will earn if you are paid $10.25 an hour and spend$4.00 round trip to get too and from work.
6. A father earns a \$2000 dividend from an oil investment and distributes it equally amongst his children.

## Date Created:

Aug 13, 2012

Nov 25, 2014
Files can only be attached to the latest version of Modality


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# 8.12.3.3 IfcColourRgbList

AbRV ⓘ
RV ⓘ

## 8.12.3.3.1 Semantic definition

The IfcColourRgbList defines an ordered collection of RGB colour values. Each colour value is a fixed list of three colour components (red, green, blue). The attribute ColourList is a two-dimensional list, where:

• first dimension is an unbounded list representing each colour value;
• second dimension is a fixed list of three list members, where [1] is the red component, [2] the green component, and [3] the blue component.

## 8.12.3.3.5 Formal representation

ENTITY IfcColourRgbList
SUBTYPE OF (IfcPresentationItem);
ColourList : LIST [1:?] OF LIST [3:3] OF IfcNormalisedRatioMeasure;
END_ENTITY;


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### LOW-FREQUENCY SPECTRUM ASYMPTOTICS AS A MEASURE OF NONSTATIONARITY OF SOME GEOPHYSICAL PROCESSES

V.B.Lysenko and V. F. Pisarenko

Abstract

We present a method of statistical estimation of power $\alpha$ for low-frequency spectrum asymptotics $f(\lambda) \sim c/\lambda^{\alpha}$ for Gaussian and point processes. The method is applied to several geophysical processes: Earth's nutation, atmospheric pressure, and regional seismicity.

Back to
Computational Seismology, Vol. 3.


text:
# On perfectness of dot product graph of a commutative ring

Document Type : Research Paper

Authors

Department of Mathematics, Islamic Azad University, Central Tehran Branch, P. O. Box 14168-94351, Iran

10.29252/as.2019.1399

Abstract

Let $A$ be a commutative ring with nonzero identity, and $1\leq n<\infty$ be an integer, and
$R=A\times A\times\cdots\times A$ ($n$ times). The total dot product graph of $R$ is the (undirected) graph $TD(R)$ with vertices $R^*=R\setminus \{(0,0,\dots,0)\}$, and two distinct vertices $x$ and $y$ are adjacent if and only if $x\cdot y=0\in A$ (where $x\cdot y$ denote the normal dot product of $x$ and $y$).
Let $Z(R)$ denote the set of all zero-divisors of $R$.  Then the zero-divisor dot product graph of $R$ is the induced subgraph $ZD(R)$ of $TD(R)$ with vertices  $Z(R)^*=Z(R)\setminus \{(0,0,\dots,0)\}$. It follows that if  $\Gamma(A)$ is not  perfect, then  $ZD(R)$ (and hence $TD(R)$) is not  perfect.
In this paper we investigate perfectness of the graphs $TD(R)$ and $ZD(R)$.

Keywords

#### References

[1] D. F. Anderson, A. Badawi, On the total graph of a commutative ring without the zero element, J. Algebra Appl. 11, 1250074 (2012) [18 pages].
[2] D. F. Anderson, P. S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra, 217 (1999), 434-447.
[3] D. F. Anderson, On the diameter and girth of a zero-divisor graph, II. Houston J. Math, (2008) 34:361-371.
[4] A. Badawi, On the Annihilator Graph of a commutative ring, Communications in Algebra, 42 (2014) 108-121.
[5] A. Badawi, On the dot product graph of a commutative ring, Communications in Algebra, 43 (2015) 43-50.
[6] R. Diestel, Graph Theory NY, USA: Springer-Verlag, (2000).
[7] D.B. West, Introduction to Graph Theory, 2nd ed., Prentice Hall, Upper Saddle River (2001).


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# Scaling property of Fourier Transform

Problem 4.6(b) from Oppenheim, Wilsky & Nawab (2nd ed) reads:

Given that $x(t)$ has the Fourier transform $X(j\omega)$, express the Fourier transform of $x(3t - 6)$ in terms of $X(j\omega)$.

The way I proceeded, $$x(3t - 6) = x(3(t-2))$$

$$x(t) \longleftrightarrow X(j\omega)$$ $$\implies x(t-2) \longleftrightarrow X(j\omega)e^{-2j\omega}$$ $$\implies x(3(t-2)) \longleftrightarrow \frac{1}{3}X(\frac{j\omega}{3})e^{\frac{-2j\omega}{3}}$$ But the answer given at the back of the book is $$x(3(t-2)) \longleftrightarrow \frac{1}{3}X(\frac{j\omega}{3})e^{-2j\omega}$$

My question is, while applying the scaling property of Fourier Transform, why does scaling happen only to the $X(j\omega)$ term and not the exponential term, when even the latter forms part of the actual $X(j\omega)$ on RHS?

Ok you are mixing the order of the transforms. When you decompose the argument $3t-6$ into $3\cdot(t-2)$ , then you should first scale $x(t)$ by $3$ and then shift the scaled result by $2$, that's where you have chosen the wrong order. Let's show the correct order:

Let $x(t)$ be the original signal whose CTFT is $X(\omega)$, then define the following signals in two steps to get the final signal $z(t)= x(3t-6)$:

$$y(t) = x(3t) \tag{1}$$ and $$z(t) = y(t-2) \implies z(t) = x(3(t-2)) = x(3t-6) \tag{2}$$

Notice how we replaced $y(t-2)$ by using its definition from the eq. (1).

Now applying CTFT to the above rows yields:

$$Y(\omega) = \frac{1}{3} X(\omega/3) \tag{3}$$ and $$Z(\omega) = Y(\omega) e^{-j2\omega} \tag{4}$$

finally insert $Y(\omega)$ from eq. (3) into eq. (4) to get:

$$\boxed{Z(\omega) = Y(\omega) e^{-j2\omega} = \frac{1}{3} X(\omega/3) e^{-j2\omega} }$$

By the way, you could have also used the alternate approach: first shift $x(t)$ by $6$: $$y(t) = x(t-6) \implies Y(\omega) = X(\omega)e^{-j6\omega}$$ and then scale $y(t)$ by $3$ to obtain $$z(t) = y(3t) = x(3t-6) \implies Z(\omega) = \frac{1}{3}Y(\omega/3) = \frac{1}{3}X(\omega/3)e^{-j2\omega}$$

Fat32's answer is correct and that's the way you should do it. Here I just want to add an explanation of your error, and a way to solve the problem correctly continuing from where you went wrong.

The problem occurs in the $4^{th}$ line of your derivation. The expression on the right-hand side is the Fourier transform of $x(3t-2)$, and not of $x(3(t-2))$. Scaling is related to the independent variable, nothing else. So now you got the Fourier transform of $x(3t-2)$:

$$\mathcal{F}\{x(3t-2)\}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega/3}\tag{1}$$

Another shifting operation will give you the final result:

$$\mathcal{F}\{x(3t-6)\}=\mathcal{F}\{x(3(t-4/3)-2)\}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega/3}e^{-4j\omega/3}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega}\tag{2}$$

Note that I do not recommend this roundabout way, but this was just to make sure you understand your mistake, and you also understand how to arrive at the correct result from how you approached the solution.


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## Quantile Bandits for Best Arms Identification

### Mengyan Zhang, Cheng Soon Ong

[ Abstract ] [ Livestream: Visit Bandits 4 ] [ Paper ]

We consider a variant of the best arm identification task in stochastic multi-armed bandits. Motivated by risk-averse decision-making problems, our goal is to identify a set of $m$ arms with the highest $\tau$-quantile values within a fixed budget. We prove asymmetric two-sided concentration inequalities for order statistics and quantiles of random variables that have non-decreasing hazard rate, which may be of independent interest. With these inequalities, we analyse a quantile version of Successive Accepts and Rejects (Q-SAR). We derive an upper bound for the probability of arm misidentification, the first justification of a quantile based algorithm for fixed budget multiple best arms identification. We show illustrative experiments for best arm identification.


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# Bell basis

The Bell basis is a basis for the Hilbert space of a 2-qubit system where the basis vectors are defined in terms of the computational basis as :

$$\begin{cases} |\Psi^- \rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) \\ |\Psi^+ \rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle) \\ |\Phi^- \rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) \\ |\Phi^+ \rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) \end{cases}$$

The quantum states represented by these vectors are called Bell states and are maximally entangled. Density matrices which are diagonal in this basis are called Bell-diagonal.

### See also

Category:Quantum States

## Last modified:

Monday, October 26, 2015 - 17:56


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# Multiply two complex numbers

multiplication of two complex numbers - it's the same as multiplication of vectors.

From physics i know that's result of multiplication of two vectors - it's a number.

But when we multiply complex numbers - it's a vector. It's some kind of contradiction, isn't it?

• Multiplication of vectors can be defined in different ways. – Karl Aug 8 '15 at 13:36
• Do you mean Cross product for example? – Evgeny Semyonov Aug 8 '15 at 13:38
• @OP: you may be confusing the notion of multiplication of vectors (which give a vector, and can be defined in several ways), and dot product (which gives a scalar/number). – Clement C. Aug 8 '15 at 13:38
• This is a strange thing to say... a really confusing comparison. Unless this is in the context of Clifford algebra, it doesn't make much sense. – orion Aug 8 '15 at 13:40
• Short answer is that multiplication of complex numbers has nothing to do with vectors. – user21820 Aug 8 '15 at 13:58

If the complex numbers are written with real and imaginary parts, $a+ib$ and $c+id$, then: $$(a+ib)(c+id) = ac - bd + i(ad + bc)$$
If the complex numbers are written as a direction and a magnitude, $(\theta_a, a)$ and $(\theta_b, b)$, then their product is the complex number $(\theta_a + \theta_b, ab)$.


text:
Articles
Permutation - Permutation order

The order of a permutation \sigma is the smallest positive integer m so that. It is the least common multiple of its cycles lengths. For example, the order of is 2\cdot3 = 6.

Permutation - Permutation type

The cycles of a permutation partition the set S_n so the lengths of the cycles of a permutation \sigma forms a partition of n called the cycle type of \sigma. There is a "1" in the cycle type for every fixed point of σ, a "2" for every transposition, and so on. The cycle type of is (3,2,2,1) which is sometimes written in a more compact form as [1 1 2 2 3 1 ].

Pseudorandom permutation - Unpredictable permutation

An adversary for an unpredictable permutation is defined to be an algorithm that is given access to an oracle for both forward and inverse permutation operations. The adversary is given a challenge input k and is asked to predict the value of F k. It is allowed to make a series of queries to the oracle to help it make this prediction, but is not allowed to query the value of k itself.

Pseudorandom permutation - Pseudorandom permutation

A pseudorandom permutation family is a collection of pseudorandom permutations, where a specific permutation may be chosen using a key.

Pseudorandom permutation - Unpredictable permutation

A randomized algorithm for generating permutations generates an unpredictable permutation if its outputs are permutations on a set of items (described by length-n binary strings) that cannot be predicted with accuracy significantly better than random by an adversary that makes a polynomial (in n) number of queries to the oracle prior to the challenge round, whose running time is polynomial in n, and whose error probability is less than 1/2 for all instances. That is, it cannot be predicted in the complexity class PP, relativized by the oracle for the permutation.

Permutation - Parity of a permutation

Every permutation of a finite set can be expressed as the product of transpositions. Although many such expressions for a given permutation may exist, either they all contain an even or an odd number of transpositions. Thus all permutations can be classified as even or odd depending on this number.

Permutation group - Permutation isomorphic groups

If G and H are two permutation groups on sets X and Y with actions f 1 and f 2 respectively, then we say that G and H are permutation isomorphic (isomorphic as permutation groups) if there exists a bijective map λ : X → Y and a group isomorphism ψ : G → H such that: : λ(f 1 (g, x)) = f 2 (ψ(g), λ(x)) for all g in G and x in X.

Permutation representation - Abstract permutation representation

The image is a permutation group and the elements of G are represented as permutations of X. A permutation representation is equivalent to an action of G on the set X:

Layered permutation - Related permutation classes

Every layered permutation is an involution. They are exactly the 231-avoiding involutions, and they are also exactly the 312-avoiding involutions.

Permutation representation - Linear permutation representation

If G is a permutation group of degree n, then the permutation representation of G is the linear representation of G :which maps g\in G to the corresponding permutation matrix (here K is an arbitrary field). That is, G acts on K^n by permuting the standard basis vectors.

Permutation representation - Linear permutation representation

This notion of a permutation representation can, of course, be composed with the previous one to represent an arbitrary abstract group G as a group of permutation matrices. One first represents G as a permutation group and then maps each permutation to the corresponding matrix.

Permutation representation - Abstract permutation representation

A permutation representation of a group G on a set X is a homomorphism from G to the symmetric group of X:

Permutation - Other uses of the term permutation

The concept of a permutation as an ordered arrangement admits several generalizations that are not permutations but have been called permutations in the literature.

Permutation matrix - Permutation of rows

The permutation matrix P π corresponding to the permutation : is :

Permutation model - Construction of permutation models

If V is a model of ZFA with A the set of atoms, then an element of V is called symmetric if the subgroup fixing it is in F, and is called hereditarily symmetric if it and all elements of its transitive closure are symmetric. The permutation model consists of all hereditarily symmetric elements, and is a model of ZFA.

Permutation representation - Character of the permutation representation

Given a group G and a finite set X with G acting on the set X then the Character (mathematics) \chi of the permutation representation is exactly the number of fixed points of X under the action of \rho(g) on X. That is \chi(g)= the number of points of X fixed by \rho(g).

Permutation polynomial - Some classes of permutation polynomials

Beyond the above examples, the following list, while not exhaustive, contains almost all of the known major classes of permutation polynomials over finite fields.

Permutation representation - Character of the permutation representation

This follows since, if we represent the map \rho(g) with a matrix with basis defined by the elements of X we get a permutation matrix of X. Now the character of this representation is defined as the trace of this permutation matrix. An element on the diagonal of a permutation matrix is 1 if the point in X is fixed, and 0 otherwise. So we can conclude that the trace of the permutation matrix is exactly equal to the number of fixed points of X.

Permutation matrix - Permutation of rows and columns

Permutations of rows and columns are for example reflections (see below) and cyclic permutations (see cyclic permutation matrix).

Permutation polynomial - Some classes of permutation polynomials

. * Only a few other specific classes of permutation polynomials over