最新2024医学博士英语考试真题及答案全文共3篇示例，供读者参考篇12024 Medical Doctor English ExamIntroductionThe 2024 Medical Doctor English Exam is a standardized test conducted for medical students who are pursuing a career in medicine. The exam tests the students' proficiency in English language and their ability to comprehend and analyze medical texts, as well as their critical thinking and problem-solving skills.Section 1: Reading Comprehension1. According to the passage, what is the main function of the kidneys in the human body?A. Filtration of bloodB. Regulation of blood pressureC. Production of red blood cellsD. Digestion of foodAnswer: A. Filtration of blood2. Which of the following statements is true according to the passage?A. Insulin is produced by the pancreas.B. The liver is responsible for filtering waste products from the blood.C. The thyroid gland is located in the chest.D. The spleen is part of the digestive system.Answer: A. Insulin is produced by the pancreas.3. What does the phrase "immune response" refer to in the passage?A. The body's ability to fight off infectionsB. The process of digestionC. The function of the respiratory systemD. The production of hormonesAnswer: A. The body's ability to fight off infectionsSection 2: Listening ComprehensionListen to the following audio clip and answer the questions:1. What is the main topic of the conversation?A. The symptoms of a common coldB. The treatment for a broken boneC. The causes of diabetesD. The importance of physical exerciseAnswer: A. The symptoms of a common cold2. According to the speaker, what are the symptoms of a common cold?A. Fever and chillsB. Cough and sore throatC. Vomiting and diarrheaD. Muscle aches and joint painAnswer: B. Cough and sore throatSection 3: WritingWrite an essay on the following topic:"Discuss the impact of technology on modern healthcare."In your essay, you should address the following points:- How technology has revolutionized medical diagnosis and treatment.- The benefits and drawbacks of electronic health records.- The role of telemedicine in improving access to healthcare services.- The ethical considerations of using artificial intelligence in healthcare.ConclusionThe 2024 Medical Doctor English Exam is a comprehensive test that assesses students' knowledge and skills in the field of medicine. By preparing thoroughly for the exam and practicing with past papers, students can improve their chances of success and demonstrate their readiness to enter the medical profession.篇22024 Medical Doctor English Exam Questions and AnswersPart A: Reading ComprehensionRead the following passage and answer the questions below.Passage:The field of medicine is constantly evolving, with new technologies and treatments being developed every day. As a medical doctor, it is crucial to stay updated on the latestadvancements in order to provide the best care for your patients. One area that has seen significant growth in recent years is personalized medicine, which involves tailoring treatments to individual patients based on their genetic makeup.Question 1: What is personalized medicine?Answer: Personalized medicine involves tailoring treatments to individual patients based on their genetic makeup.Question 2: Why is it important for medical doctors to stay updated on the latest advancements in medicine?Answer: It is crucial for medical doctors to stay updated on the latest advancements in medicine in order to provide the best care for their patients.Question 3: Give an example of a recent advancement in the field of medicine.Answer: Personalized medicine is a recent advancement in the field of medicine.Question 4: How can personalized medicine improve patient care?Answer: Personalized medicine can improve patient care by tailoring treatments to individual patients based on their genetic makeup.Part B: Vocabulary and GrammarChoose the correct word or phrase to complete each sentence.1. The (affect/effect) of the new treatment on patients will be evaluated in a clinical trial.2. The doctor ordered a(n) (X-ray/ex-ray) to determine the cause of the patient's symptoms.3. It is important for medical professionals to have (comprehensive/comprehensible) knowledge of the human body.4. The patient's condition (improved/implored) after receiving the new medication.5. The medical team worked (collectively/collectably) to develop a treatment plan for the patient.Part C: WritingWrite a short essay (150-200 words) on the following topic:"Discuss the importance of communication skills for medical doctors."Communication skills are essential for medical doctors to effectively interact with patients, their families, and other healthcare professionals. Strong communication skills not only help doctors build rapport with patients but also ensure that important medical information is effectively conveyed. Patients rely on doctors to explain their diagnosis, treatment options, and prognosis in a clear and compassionate manner. Additionally, good communication skills enable doctors to listen attentively to patients' concerns, address any questions or fears they may have, and provide emotional support when needed.Furthermore, effective communication among healthcare professionals is crucial for coordinating patient care and ensuring that all members of the medical team are on the same page. Doctors must be able to communicate clearly with nurses, therapists, and other specialists to ensure that patients receive comprehensive and coordinated care.In conclusion, communication skills are a vital aspect of being a successful medical doctor, as they play a significant role in patient care, teamwork, and overall patient outcomes.篇3Sorry, I can't provide the specific content of the latest 2024 Medical Doctor English Exam questions and answers as they are copyrighted materials. However, I can provide some general information and tips on how to prepare for the Medical Doctor English Exam.The Medical Doctor English Exam is designed to assess candidates' proficiency in English language skills, including reading comprehension, listening, writing, and speaking. It may also include medical terminology and scenarios to test their knowledge and communication abilities in a medical context.To prepare for the exam, candidates should focus on improving their English language skills by practicing reading medical journals, listening to medical podcasts or lectures, and writing essays on medical topics. They can also benefit from taking practice exams to familiarize themselves with the format and types of questions that may appear on the actual exam.In addition, candidates should pay attention to medical terminology and consider taking additional courses or workshops to enhance their knowledge in this area. They shouldalso practice speaking English in a medical setting to improve their communication skills and confidence.Overall, successful preparation for the Medical Doctor English Exam requires dedication, practice, and a comprehensive understanding of both English language skills and medical knowledge. Good luck to all candidates preparing for the exam!。
Efficient Lattice(H)IBE in the Standard ModelShweta Agrawal1,Dan Boneh2 ,and Xavier Boyen31University of Texas,Austin2Stanford University3Universit´e de Li`e ge,BelgiumAbstract.We construct an efficient identity based encryption systembased on the standard learning with errors(LWE)problem.Our securityproof holds in the standard model.The key step in the constructionis a family of lattices for which there are two distinct trapdoors forfinding short vectors.One trapdoor enables the real system to generateshort vectors in all lattices in the family.The other trapdoor enables thesimulator to generate short vectors for all lattices in the family exceptfor one.We extend this basic technique to an adaptively-secure IBE anda Hierarchical IBE.1IntroductionIdentity-Based Encryption(IBE)provides a public-key encryption mechanism where a public key is an arbitrary string such as an email address or a telephone number.The corresponding private key can only be generated by a Private-Key Generator(PKG)who has knowledge of a master secret.Identity-based encryption wasfirst proposed by Shamir[28],however,it is only recently that practical implementations were proposed.Boneh and Franklin[8]define a secu-rity model for identity-based encryption and give a construction based on the Bilinear Diffie-Hellman(BDH)problem.Cocks[13]describes a construction us-ing quadratic residues modulo a composite(see also[9])and Gentry et al.[16] give a construction using lattices.The security of all these systems requires cryp-tographic hash functions that are modeled as random oracles.For pairing-based systems,the structure of pairing groups enabled several secure IBE systems in the standard model[11,6,7,31,17,32].For systems based on quadratic residuosity it is still not known how to build a secure IBE in the standard model.In this paper we focus on lattice-based IBE.Cash et al.[12],Peikert[24]and Agrawal et al.[3]recently showed how to construct secure IBE in the standard model from the learning with errors(LWE)problem[27].Their constructions view an identity as a sequence of bits and then assign a matrix to each bit.The resulting systems,while quite elegant,are considerably less efficient than the underlying random-oracle system of[16]on which they are built.A full version of this paper is available at[1].Supported by NSF and the Packard Foundation.1.1Our ResultsWe construct a lattice-based IBE in the standard model whose performance is comparable to the performance of the random-oracle system from[16].In particular,we process identities as one chunk rather than bit-by-bit resulting in lattices whose dimension is similar to those in the random oracle system.Lattices in our system are built from two parts called“right”and“left”lattices.A trapdoor for the left lattice is used as the master secret in the real system and enables one to generate private keys for all identities.A trapdoor for the right lattice is only used in the proof of selective security and enables the simulator to generate private keys for all identities except for one.We use a“low norm”randomization matrix R to ensure that an attacker cannot distinguish between the real world and a simulation.In pairing-based IBE systems one uses large groups G and therefore identities can be encoded as integers in the range1...|G|.In contrast,lattice systems are typically defined over a relatively smallfield Z q and consequently encoding identities as integers in1...q would result in too few identities for the system.Instead,we represent identities as matrices in Z n×nq for some n.More precisely,we represent identities as elements in Z n q(for a total of q n identities)and thenuse an encoding function H:Z n q→Z n×nq to map identities to matrices.Oursecurity proof requires that for all id1=id2the matrix H(id1)−H(id2)∈Z n×nqis invertible.We present an encoding function H that has this property and expect this encoding to be useful in other lattice-based constructions.A similar function H was developed by Cramer and Damgard[14]in an entirely different context.Full IBE.In the full version of the paper[1]we show that our base construction extends to an adaptively-secure IBE using a lattice analog of the Waters IBE[31]. Our base construction requires that the underlyingfield Z q satisfy q>Q where Q is the number of private key queries issued by the adversary.This requirement can be relaxed using the framework of Boyen[10].Hierarchical IBE(HIBE).In the full version of the paper[1]we show how to extend our base IBE to an HIBE using the basis delegation technique from[12, 24].The construction assigns a matrix to each level of the hierarchy and the resulting lattice dimension is linear in the recipient identity’s depth.Since we do not process identities bit-by-bit we obtain an efficient HIBE where the lattice dimension is much smaller than in[12,24].We note that a recent result of[2]uses a different basis delegation mechanism to construct an improved HIBE where the lattice dimension isfixed for the entire hierarchy.2PreliminariesNotation.Throughout the paper we say that a function :R≥0→R≥0is neg-ligible if (n)is smaller than all polynomial fractions for sufficiently large n. We say that an event happens with overwhelming probability if it happens with2probability at least1− (n)for some negligible function .We say that integer vectors v1,...,v n∈Z m are Z q-linearly independent if they are linearly indepen-dent when reduced modulo q.2.1IBE and Hierarchical IBERecall that an Identity-Based Encryption system(IBE)consists of four algo-rithms[28,8]:Setup,Extract,Encrypt,Decrypt.The Setup algorithm generates system parameters,denoted by PP,and a master key MK.The Extract algorithm uses the master key to extract a private key corresponding to a given identity. The encryption algorithm encrypts messages for a given identity(using the sys-tem parameters)and the decryption algorithm decrypts ciphertexts using the private key.In a Hierarchical IBE[20,18],identities are vectors,and there is afifth algo-rithm called Derive.A vector of dimension represents an identity at depth . Algorithm Derive takes as input an identity id=(I1,...,I )at depth and the private key SK id| −1of the parent identity id| −1=(I1,...,I −1)at depth −1≥0.It outputs the private key SK id for identity id.We sometimes refer to the master key as the private key at depth0,given which the algorithm Derive performs the same function as Extract.The Setup algorithm in an HIBE scheme takes the maximum depth of the hierarchy as input.Selective and Adaptive ID Security.The standard IBE security model of[8]de-fines the indistinguishability of ciphertexts under an adaptive chosen-ciphertext and chosen-identity attack(IND-ID-CCA2).A weaker notion of IBE security given by Canetti,Halevi,and Katz[11]forces the adversary to announce ahead of time the public key it will target,which is known as a selective-identity attack (IND-sID-CCA2).As with regular public-key encryption,we can deny the adversary the ability to ask decryption queries(for the target identity),which leads to the weaker notions of indistinguishability of ciphertexts under an adaptive chosen-identity chosen-plaintext attack(IND-ID-CPA)and under a selective-identity chosen-plaintext attack(IND-sID-CPA)respectively.Security Game.We define IBE and HIBE selective security using a game that captures a strong privacy property called indistinguishable from random which means that the challenge ciphertext is indistinguishable from a random element in the ciphertext space.This property implies both semantic security and recip-ient anonymity,and also implies that the ciphertext hides the public parame-ters(PP)used to create it.This can make the IBE more resistant to subpoenas since an observer cannot tell from the ciphertext which authority holds the cor-responding master secret.For a security parameterλ,we let Mλdenote the message space and let Cλdenote the ciphertext space.The game,for a hierarchy of maximum depth d,proceeds as follows.Init:The adversary is given the maximum depth of the hierarchy d and outputs),k≤d.a target identity id∗=(I∗1, (I)k3Setup:The challenger runs Setup(1λ,1d)(where d=1for IBE)and gives the adversary the resulting system parameters PP.It keeps the master key MK to itself.Phase1:The adversary issues queries q1,...,q m where the i-th query q i is a query on id i,where id i=(I1,...,I u)for some u≤d.We require that id i is not a prefix of id∗,(i.e.,it is not the case that u≤k and I i=I∗i for all i=1,...,u).The challenger responds by running algorithm Extract to obtain a private key d i for the public key id i.It sends d i to the adversary.All queries may be made adaptively,that is,the adversary may ask q i with knowledge of the challenger’s responses to q1,...,q i−1.Challenge:Once the adversary decides that Phase1is over it outputs a plain-text M∈Mλon which it wishes to be challenged.The challenger picks a random bit r∈{0,1}and a random ciphertext C∈Cλ.If r=0it sets the challenge ciphertext to C∗:=Encrypt(PP,id∗,M).If r=1it sets the chal-lenge ciphertext to C∗:=C.It sends C∗as the challenge to the adversary. Phase2:The adversary issues additional adaptive queries q m+1,...,q n where q i is a private-key extraction query on id i,where id i is not a prefix of id∗.The challenger responds as in Phase1.Guess:Finally,the adversary outputs a guess r ∈{0,1}and wins if r=r . We refer to such an adversary A as an INDr–sID-CPA adversary.We define the advantage of the adversary A in attacking an IBE or HIBE scheme E asAdv d,E,A(λ)=Pr[r=r ]−1/2The probability is over the random bits used by the challenger and the adversary. Definition1.We say that an IBE or a depth d HIBE system E is selective-identity,indistinguishable from random if for all INDr–sID-CPA PPT adversaries A we have that Adv d,E,A(λ)is a negligible function.We abbreviate this by saying that E is INDr–sID-CPA secure for depth d.2.2Statistical distanceLet X and Y be two random variables taking values in somefinite setΩ.Define the statistical distance,denoted∆(X;Y),as∆(X;Y):=12s∈ΩPr[X=s]−Pr[Y=s]We say that X isδ-uniform overΩif∆(X;UΩ)≤δwhere UΩis a uniform random variable overΩ.Let X(λ)and Y(λ)be ensembles of random variables.We say that X and Y are statistically close if d(λ):=∆(X(λ);Y(λ))is a negligible function ofλ.42.3Integer LatticesLet B=b1...b m∈R m×m be an m×m matrix whose columns are linearlyindependent vectors b1,...,b m∈R m.The m-dimensional full-rank latticeΛgenerated by B is the set,Λ=L(B)=y∈R m s.t.∃s∈Z m,y=B s=mi=1s i b iHere,we are interested in integer lattices,i.e,when L is contained in Z m.Definition2.For q prime,A∈Z n×mqand u∈Z n q,define:Λq(A):=e∈Z m s.t.∃s∈Z n q where A s=e(mod q)Λ⊥q(A):=e∈Z m s.t.A e=0(mod q)Λu q(A):=e∈Z m s.t.A e=u(mod q)Observe that if t∈Λu q(A)thenΛu q(A)=Λ⊥q(A)+t and henceΛu q(A)is a shift ofΛ⊥q(A).2.4The Gram-Schmidt Norm of a BasisLet S be a set of vectors S={s1,...,s k}in R m.We use the following notation:– S denotes the L2length of the longest vector in S,i.e. S :=max i s i for1≤i≤k.–˜S:={˜s1,...,˜s k}⊂R m denotes the Gram-Schmidt orthogonalization of the vectors s1,...,s k taken in that order.We refer to S as the Gram-Schmidt norm of S.Micciancio and Goldwassser[22]showed that a full-rank set S in a latticeΛcan be converted into a basis T forΛwith an equally low Gram-Schmidt norm. Lemma1([22,Lemma7.1]).LetΛbe an m-dimensional lattice.There is a deterministic polynomial-time algorithm that,given an arbitrary basis ofΛand a full-rank set S={s1,...,s m}inΛ,returns a basis T ofΛsatisfyingT ≤ S and T ≤ S √m/2Ajtai[4]showed how to sample an essentially uniform matrix A∈Z n×mq with an associated basis S A ofΛ⊥q(A)with low Gram-Schmidt norm.We use an improved sampling algorithm from Alwen and Peikert[5].The following follows from Theorem3.2of[5]takingδ:=1/3.5Theorem 1.Let q ≥3be odd and m := 6n log q .There is a probabilistic polynomial-time algorithm TrapGen (q,n )that outputs apair (A ∈Z n ×m q ,S ∈Z m ×m )such that A is statistically close to a uniform matrix in Z n ×m q and S is a basis for Λ⊥q (A )satisfyingS≤O ( n log q )and S ≤O (n log q )with all but negligible probability in n .We will also need the following simple lemma about the effect of matrix multiplication on the Gram-Schmidt norm.Lemma 2.Let R be a matrix in R ×m and S ={s 1,...,s k }⊂R m a linearly independent set.Let S R :={Rs 1,...,Rs k }.ThenS R ≤max 1≤i ≤k R ˜s i Proof.We show that for all i =1,...,k the i -th Gram-Schmidt vector of S R has L 2norm less than R ˜s i .This will prove the lemma.For i ∈{1,...,k }let V :=span R (Rs 1,...,Rs i −1).Set v :=s i −˜s i .Then v ∈span R (s 1,...,s i −1)and therefore Rv ∈V .Let u be the projection of R ˜s i on V and let z :=R ˜s i −u .Then z is orthogonal to V andRs i =Rv +R ˜s i =Rv +u +z =(Rv +u )+z .By construction,Rv +u ∈V and hence,since z is orthogonal to V ,this z must be the i -th Gram-Schmidt vector of S R .Since z is the projection of R ˜s i on V ⊥we obtain that z ≤ R ˜s i .Hence,for all i =1,...,k the i -th Gram-Schmidt vector of S R has L 2norm less than R ˜s i which proves the lemma.2.5Discrete GaussiansLet L be a subset of Z m .For any vector c ∈R m and any positive parameter σ∈R >0,define:ρσ,c (x )=exp −π x −c 2σ2 :a Gaussian-shaped function on R m with center cand parameter σ,ρσ,c (L )= x ∈L ρσ,c (x ):the (always converging)sum of ρσ,c over L ,D L,σ,c :the discrete Gaussian distribution over L with parameters σand c ,∀y ∈L ,D L,σ,c (y )=ρσ,c (y )ρσ,c (L )We abbreviate ρσ,0and D L,σ,0as ρσand D L,σ.We write ρto denote ρ1.The distribution D L,σ,c will most often be defined over the lattice L =Λ⊥q (A )for amatrix A ∈Z n ×m q or over a coset L =t +Λ⊥q (A )where t ∈Z m .6Properties.The following lemma from [24]captures standard properties of these distributions.The first two properties follow from Lemma 4.4of [23]and Corol-lary 3.16of [27]respectively (using Lemma 3.1from [16]to bound the smoothing parameter).We state in property (2)a stronger version of Regev’s Corollary 3.16found in [2].The last two properties are algorithms from [16].Lemma 3.Let q ≥2and let A be a matrix in Z n ×m q with m >n .Let T A be a basis for Λ⊥q (A )and σ≥ T A ω(√log m ).Then for c ∈Rm and u ∈Z n q :1.Pr x ∼D Λ⊥q (A ),σ: x >√m σ ≤negl(n ).2.A set of O (m log m )samples from D Λ⊥q (A ),σcontains a full rank set in Z m ,except with negligible probability.3.There is a PPT algorithm SampleGaussian (A,T A ,σ,c )that returns x ∈Λ⊥q (A )drawn from a distribution statistically close to D Λ,σ,c .4.There is a PPT algorithm SamplePre (A,T A ,u,σ)that returns x ∈Λu q (A )sampled from a distribution statistically close to D Λu q (A ),σ.Recall that if Λu q (A )is not empty then Λu q (A )=t +Λ⊥q (A )for some t ∈Λu q (A ).Algorithm SamplePre (A,T A ,u,σ)works by calling SampleGaussian (A,T A ,σ,t )and subtracts t from the result.2.6The L WE hardness assumptionSecurity of all our constructions reduces to the LWE (learning with errors)prob-lem,a classic hard problem on lattices defined by Regev [27].Definition 3.Consider a prime q ,a positive integer n ,and a distribution χover Z q ,all public.An (Z q ,n,χ)-LWE problem instance consists of access to an unspecified challenge oracle O ,being,either,a noisy pseudo-random sampler O s carrying some constant random secret key s ∈Z n q ,or,a truly random sampler O $,whose behaviors are respectively as follows:O s :outputs samples of the form (u i ,v i )= u i ,u T i s +x i ∈Z n q ×Z q ,where,s ∈Z n q is a uniformly distributed persistent value invariant across invocations,x i ∈Z q is a fresh sample from χ,and u i is uniform in Z n q .O $:outputs truly uniform random samples from Z n q ×Z q .The (Z q ,n,χ)-LWE problem allows repeated queries to the challenge oracleO .We say that an algorithm A decides the (Z q ,n,χ)-LWE problem if Pr[A O s =1]−Pr[A O $=1] is non-negligible for a random s ∈Z n q .Regev [27]shows that for certain noise distributions χ,denoted Ψα,the LWE problem is as hard as the worst-case SIVP and GapSVP under a quantum reduction (see also [25]).Definition 4.Consider a real parameter α=α(n )∈(0,1)and a prime q .Denote by T =R /Z the group of reals [0,1)with addition modulo 1.Denote7by Ψαthe distribution over T of a normal variable with mean 0and standard deviation α/√2πthen reduced modulo 1.Denote by x = x +12 the nearest integer to the real x ∈R .We denote by Ψαthe discrete distribution over Z q of the random variable q X mod q where the random variable X ∈T has distribution Ψα.Theorem 2([27]).If there exists an efficient,possibly quantum,algorithm for deciding the (Z q ,n,Ψα)-LWE problem for q >2√n/αthen there exists an effi-cient quantum algorithm for approximating the SIVP and GapSVP problems,to within ˜O(n/α)factors in the 2norm,in the worst case.If we assume the hardness of approximating the SIVP or GapSVP problems in lattices of dimension n to within approximation factors that are polynomial in n ,then it follows from Lemma 2that deciding the LWE problem is hard when n/αis polynomial in n .3Randomness ExtractionWe will need the following lemma which follows directly from a generalization of the left over hash lemma due to Dodis et al.[15].Lemma 4.Suppose that m >(n +1)log 2q +ω(log n )and that q is prime.LetA,B be matrices chosen uniformly in Z n ×m q and let R be an m ×m matrix chosen uniformly in {1,−1}m ×m mod q .Then,for all vectors w in Z m q ,the distribution(A,AR,R w )is statistically close to the distribution (A,B,R w ).To prove the lemma recall that for a prime q the family of hash functionsh A :Z m q →Z n q for A ∈Z n ×m q defined by h A (x )=A x is universal.Therefore,when the columns of R are sampled independently and have sufficient entropy,the left over hash lemma (e.g.as stated in [29,Theorem 8.38])shows that the distributions (A,AR )and (A,B )are statistically close.A generalization by Dodis et al.[15](Lemma 2.2b and 2.4)shows that the same holds even if some small amount of information about R is leaked.In our case R w is leaked which is precisely the settings of Dodis et al.We provide the complete proof of Lemma 4in the full version of the paper [1].3.1Random Subset SumsWe will also need the following simple lemma.Lemma 5.Let R be an m ×m matrix chosen at random from {−1,1}m ×m .Then for all vectors u ∈R m we havePr Ru > u √m ·ω( log m ) <negl(m ).8Proof.Let r ∈{−1,1}m be a row vector of the matrix R .Then r ·u can be written as r u = m i =1x i where x i =r i u i .We know that E [x i ]=0and that x i ∈[−u i ,u i ]for all i =1,...,m .Then,by the Hoeffding bound [19,Theorem 2]we obtain thatPr |r ·u |> u ω( ) <negl(m )The lemma now follows since an m -vector whose entries are less than some bound B has L 2norm less than √mB . 4Sampling AlgorithmsLet A and B be matrices in Z n ×m q and let R be a matrix in {−1,1}m ×m .Our construction makes use of matrices of the form F =(A |AR +B )∈Z n ×2m q and we will need to sample short vectors in Λu q (F )for some u in Z n q .We show thatthis can be done using either a trapdoor for Λ⊥q (A )or a trapdoor Λ⊥q (B ).Moreprecisely,we define two algorithms:1.SampleLeft takes a basis for Λ⊥q (A )(the left side of F )and outputs a shortvector e ∈Λu q (F ).2.SampleRight takes a basis for Λ⊥q (B )(the right side of F )and outputs ashort vector e ∈Λu q (F ).We will show that,with appropriate parameters,the distributions on e produced by these two algorithms are statistically indistinguishable.4.1Algorithm SampleLeftAlgorithm SampleLeft (A,M 1,T A ,u,σ):Inputs:a rank n matrix A in Z n ×m q and a matrix M 1in Z n ×m 1q ,a “short”basis T A of Λ⊥q (A )and a vector u ∈Z n q ,a gaussian parameter σ> T A ·ω( log(m +m 1)).(1)Output:Let F 1:=(A |M 1).The algorithm outputs a vector e ∈Z m +m 1sampled from a distribution statistically close to D Λu q (F 1),σ.In particular,e ∈Λu q (F 1).The algorithm appears in Theorem 3.4in [12]and also in the signing algo-rithm in [24].For completeness,we briefly review the algorithm.1.sample a random vector e 2∈Z m 1distributed statistically close to D Z m 1,σ,2.run e 1R ←SamplePre (A,T A ,y,σ)where y =u −(M 1·e 2)∈Z n q ,note that Λy q (A )is not empty since A is rank n ,3.output e ←(e 1,e 2)∈Z m +m 19Clearly (A |M 1)·e =u mod q and hence e ∈Λu q (F 1).Theorem 3.4in [12]shows that the vector e is sampled from a distribution statistically close to D Λu q (F 1),σ.Peikert’s basis extension method [24]gives an alternate way to view this.Given the basis T A of Λ⊥q (A )Peikert shows how to build a basis T F 1of Λ⊥q (F 1)with the same Gram-Schmidt norm as T A .Then calling SamplePre (F 1,T F 1,u,σ)generates a vector e sampled from a distribution close to D Λu q (F 1),σ.We summa-rize this in the following theorem.Theorem 3.Let q >2,m >2n log q and σ> T A·ω( 1.Then Algorithm SampleLeft (A,M 1,T A ,u,σ)taking inputs as in (1),outputs a vector e ∈Z m +m 1distributed statistically close to D Λu q (F 1),σwhere F 1:=(A |M 1).4.2Algorithm SampleRightAlgorithm SampleRight (A,B,R,T B ,u,σ).Inputs:matrices A,B in Z n ×m q where B is rank n ,a uniform random matrix R ∈{−1,1}m ×m ,a basis T B of Λ⊥q (B )and a vector u ∈Z n q ,a parameter σ> T B ·√m ·ω(log m ).(2)Output:Let F 2:=(A |AR +B ).The algorithm outputs a vector e ∈Z 2m sampled from a distribution statistically close to D Λu q (F 2),σ.In particular,e ∈Λu q (F 2).The algorithm uses the basis growth method of Peikert [24,Sec.3.3]and works in three steps:1.First,it constructs a set T F 2of 2m linearly independent vectors in Λ⊥q (F 2)such that T F 2 < T B·√m ·ω( log m )<σ/ω( log m )with overwhelming probability over the choice of R .2.Next,if needed it uses Lemma 1to convert T F 2into a basis T F 2of Λ⊥q (F 2)with the same Gram-Schmidt norm as T F 2.3.Finally,it invokes SamplePre (F 2,T F 2,u,σ)to generate a vector e ∈Λu q (F 2).Since σ> T F 2 ω(√log m )w.h.p,this e is distributed as D Λu q (F 2),σ,as re-quired.Step 1is the only step that needs explaining.Let T B ={b 1,...,b m }∈Z m ×mbe the given basis of Λ⊥q (B ).We construct the 2m vectors in Λ⊥q (F 2)as follows:1.for i =1,...,m set t i :=(−Rb i |b i )∈Z 2m and view it as a column vector;then clearly F 2·t i =B b i =0mod q and therefore t i is in Λ⊥q (F 2).2.for i =1,...,m let w i be the i -th column of the identity matrix I m .Let u i be an arbitrary vector in Z m satisfying Aw i +Bu i =0mod q .This u i exists since B is rank n .Set t i +m to be t i +m := w i −Ru i u i∈Z 2mThen F 2·t i +m =Aw i +Bu i =0mod q and hence,t i +m ∈Λ⊥q (F 2).10We show that T F2:={t1,...,t2m}are linearly independent in Z2m.First,ob-serve that thefirst m vectors are linearly independent and span the linear space V of vectors of the form(−Rx|x)where x∈Z m q.For all i>m,the vector t i isthe sum of the unit vector(w i|0m)plus a vector in V.It follows that T F2is alinearly independent set.This also means that for i>m the i-th Gram-Schmidt vector of T F2cannot be longer than(w i|0m)and therefore has norm at most1.Hence,to boundT F2 it suffices to bound the Gram-Schmidt norm of thefirstm vectors{t1,...,t m}.Let W∈Z2m×m be the matrix(−R |I m) .Then t i=W b i for i=1,...,m. Since R is uniform in{−1,1}m×m we know by Lemma5that for all vectors x∈R m we have w.h.pW x ≤ R x + x ≤ x √m·ω(log m)+ x ≤ x√m·ω(log m)Now,since t i=W b i for i=1,...,m,applying Lemma2to the matrix W gives a bound on the Gram-Schmidt norm of{t1,...,t m}(and hence also onT F2):T F2 ≤max1≤i≤mW˜b i ≤max1≤i≤m˜b i ·√m·ω(log m)≤ T B ·√m·ω()Thus,we built2m linearly independent vectors inΛ⊥q(F2)that w.h.p.have a short Gram-Schmidt norm as required for Step1.This completes the description of algorithm SampleRight.We summarize this in the following theorem. Theorem4.Let q>2,m>n andσ> T B ·√m·ω(log m).Then Algo-rithm,SampleRight(A,B,R,T B,u,σ)taking inputs as in(2),with R uniform in{1,−1}m×m,outputs a vector e∈Z2m distributed statistically close to DΛuq (F2),σwhere F2:=(A|AR+B).5Encoding Identities as MatricesOur construction uses an encoding function H:Z n q→Z n×nq to map identities inZ n q to matrices in Z n×nq .Our proof of security requires that the map H satisfya strong notion of injectivity,namely that,for any two distinct inputs id1and id2,the difference between the outputs H(id1)and H(id2)is never singular,i.e., det(H(id1)−H(id2))=0.Definition5.Let q be a prime and n a positive integer.We say that a function H:Z n q→Z n×nqis an encoding with full-rank differences(FRD)if:1.for all distinct u,v∈Z n q,the matrix H(u)−H(v)∈Z n×nq is full rank;and2.H is computable in polynomial time(in n log q).Clearly the function H must be injective since otherwise,if u=v satisfies H(u)=H(v),then H(u)−H(v)is not full-rank and hence H cannot be FRD.11The function H in Definition5has domain of size q n which is the largest possible for a function satisfying condition1of Definition5.Indeed,if H had domain larger than q n then its image is also larger than q n.But then,by pigeon-hole,there are two distinct inputs u,v such that the matrices H(u)and H(v) have the samefirst row and therefore H(u)−H(v)is not full rank.It follows that our definition of FRD,which has domain of size of q n,is the largest possible. An Explicit FRD Construction.We construct an injective FRD encoding for the exponential-size domain id∈Z n q.A similar construction is described in[14].Our strategy is to construct an additive subgroup G of Z n×nq of size q n suchthat all non-zero matrices in G are full-rank.Since for all distinct A,B∈G the difference A−B is also in G,it follows that A−B is full-rank.While our primary interest is thefinitefield Z q we describe the construction for an arbitraryfield F.For a polynomial g∈F[X]of degree less than n define coeffs(g)∈F n to be the n-vector of coefficients of g(written as a row-vector).If g is of degree less than n−1we pad the coefficients vector with zeroes on the right to make it an n-vector.For example,for n=6we have coeffs(x3+2x+3)= (3,2,0,1,0,0)∈F6.Let f be some polynomial of degree n in F[X]that is irreducible.Recall that for a polynomial g∈F[X]the polynomial g mod f has degree less than n and therefore coeffs(g mod f)is a vector in F n.Now,for an input u=(u0,u1,...,u n−1)∈F n define the polynomial g u(X)= n−1i=0u i x i∈F[X].Define H(u)asH(u):=coeffs(g u)coeffs(X·g u mod f)coeffs(X2·g u mod f)...coeffs(X n−1·g u mod f)∈F n×n(3)This completes the construction.Since for all primes q and integers n>1there are(many)irreducible polynomials in Z q[X]of degree n,the construction can accommodate any pair of q and n.The following theorem proves that the function H in(3)is an FRD.The proof,given in[14],is based on the observation that the matrix H(u) corre-sponds to multiplication by a constant in the numberfield K=F[X]/(f)and is therefore invertible when the matrix is non-zero.We note that similar matrix encodings of ring multiplication were previously used in[26,21].Theorem5.Let F be afield and f a polynomial in F[X].If f is irreducible in F[X]then the function H defined in(3)is an encoding with full-rank differences (or FRD encoding).12。
Conversation 1 (1)Lecture1 (2)Lecture 2 (4)答案： (6)Conversation 2 (6)Lecture3 (8)Lecture4 (10)答案： (12)Conversation 11.What do the speakers mainly discuss?O Methods that the professor uses to challenge her studentsO Reasons that the student turned in his paper a week late .O The two parts of an assignment for a writing class.O Seldom discussed aspects of a famous poet's work.2.What reason does the professor give for wanting to meet with the student?O She wants to compliment him on the work he has done so far.O She is concerned that the student is not reading a wide enough variety of poetry. O She wants to lend him a book of poetry.O She routinely meets individually with her students.3.What does the student like about Pablo Neruda's poems in the book Elemental Odes?O That Neruda challenged himself by limiting the theme of all the poems to fruits and vegetables O That the poems have been written in many different styles.O How Neruda focuses on color to make his poems more memorable.O How Neruda describes common objects in unexpected ways.4.Why does the student mention the meter called iambic pentameter?O To point out that his poem is longer than The Lemon.O To explain how he approached the composition of his poem.O To explain what he liked most about Neruda's poem The LemonO To distinguish Neruda's poetry from that of other poets.5.What does the professor mean when she says this:O Neruda was an extraordinary writer.O Neruda should not have won a Nobel Prize.O It is surprising that Neruda's poems are not more popularO It is unfortunate that Neruda did not win a Nobel Prize.Lecture16.What aspects of snowflakes does the professor mainly discuss? Click on 2 answers.O How they develop into complex structures.O How they are affected by the presence of ozone.O The challenges researchers face in studying them.O The function of their quasi-liquid layer.7.What does the professor say about the role of water vapor in snowflake formation?O Too much water vapor prevents the initial dinner plate from forming.O Water vapor's role in snowflake formation is not completely understood.O Water vapor molecules in snowflakes attract ice particles from the air.O Water vapor is necessary for snowflakes to be able to form branches.8.What factor helps explain why no two snowflakes are alike?O They all freeze at different rates.O They all form in slightly different air temperatures.O They all begin with a different number of water moleculesO They all follow different paths through clouds.9.How do molecules in the quasi-liquid layer differ from those in other parts of the snowflake?O They are not held in place as tightly as other molecules.O They react with ozone to keep the layer from completely freezing.O They prevent ice crystals from forming additional branches.O They are thinner than other molecules.10.What does the professor imply about ice crystals with a large number of branches? O They help block harmful radiation from the Sun.O They form as a result of complex reactions with ozone.O They contribute to a reduction in ground-level ozone.O They have a thinner quasi-liquid layer than ice crystals with fewer branches 11.What can be inferred about the professor when he says this:O He doubts that the students have understood his explanation.O He does not think that bricks are an ideal illustration of his point.O He is not sure that the information he has just given is accurate.O He thinks that the similarities between liquid and bricks are surprising Lecture 212.What is the lecture mainly about?O The discovery of a previously unknown trace metal.O The role trace metals play in carbon cycling.O Ways that living organisms rid themselves of trace metals.O Ways that zinc interacts with carbon dioxide.13.What does the professor imply about the conversation of carbon dioxide molecules in plants?O It is an unusually complex chemical process.O It only takes place in full sunlight.O It proceeds slowly when cadmium is present.O It is regulated by an enzyme that may contain zinc.14.According to the professor, why is it surprising that many marine plants areable to survive near the surface of oceans?O Weather conditions near the surface disrupt certain life processes.O The salt content of surface waters is constantly changing.O Surface waters contain low quantities of zinc.O Surface waters absorb large amounts of carbon dioxide.15.A ccording to the professor, what important function do diatoms serve?O They alter cadmium so it is less toxic to humans.O They help cycle zinc in places where it is scarce.O They distribute carbon throughout the ocean.O They remove cadium from the ocean floor.16.What point does the professor make when she talks about cadmium being poisonous to humans?O That cadmium and zinc can serve a similar function in plant enzymes.O That both cadmium and zinc are rare in plant enzymes.O That most trace metals are poisonous to humans.O That cadmium does not serve any biological purpose.17.The professor states that the discovery of an enzyme containing cadmium is important. What are two reasons that this discovery is important? Click on 2 answers.O It may lead to the discovery of new enzymes that use other trace metals.O It may explain the ocean's increased levels of carbon dioxide.O It may explain the scarcity of some elements in the ocean.O It may help scientists better understand global warming.答案：1.c2.D3. D4.B5. D6.AD7.A8.D9.A10.B11.C12.B13.D14.C15.C16.A17.ADConversation 21.Why does the woman go to talk to the man?O To find out how the store pays artists for their work.O To purchase some ceramic coffee mugs.O To find out if the store sells objects made by students.O To ask about the advantages and disadvantages of consignment sales2.What is the main reason that the woman cannot display her ceramic bowls in the campus store?O Her bowls are too expensive.O There is not enough room for her display case.O The store gets merchandise only from approved suppliers. O There is little demand on campus for ceramic bowls.3.According to the conversation, what is a reason that the woman wants to sell her bowls?O To earn enough money to buy a second display case. O To fulfill a requirement of one of her courses.O To impress her studio art professor.O To gain experience that could help in her future career.4.What is the woman's attitude toward selling items at the Emporium?O She is eager to display her work to the public there.O She is encouraged because the Emporium specializes in selling ceramics.O She is worried because she does not fully understand the consignment process.O She is worried that she might not make much money.5.What concerns does the woman initially express about selling items at the craft fair? Click on 2 answers.O Whether doing so would interfere with her studies.O Whether customers would appreciate her artistryO Whether she could afford the fee charged to sellers.O Whether she would be able to transport her items to the fair.Lecture36.What is the main purpose of the lecture?O To familiarize students with the Mayan civilization in the Classic Period.O To prepare students for an archaeology project about the Mayan civilization.O To provide evidence for a point made in a previous class about the Mayan civilization O To call into question a common view about the decline of ancient Mayan civilization.7.Why does the professor discuss Lamanai in detail?O To present findings about one Mayan settlement from the Postclassic period.O To describe the physical layout of the first Mayan settlements in Central America.O To criticize the excavation methods used there during the 1970s.O To note how the size of a typical Mayan settlement varied throughout its history.8.What is one of the features that gives Lamanai special archaeological significance?O It was the first Mayan site in Belize to be excavated in modern times.O It was occupied by two distinct cultural groups during the Classic period.O It was continuously occupied by the Maya longer than any other site.O It had an economic structure that was distinct from that of other Mayan cities9.Why does the professor say that it would require lots of funding to uncover the stone structures that she discusses?O Because there are so many of them.O Because very few archaeologists are given access to them. O Because they arelocated on so many different islands.O Because of difficult weather conditions in the region.10.According to the professor, what can be inferred from the ceramic artifactsfound on the island? O The island was the source of most of the pottery used at Lamanai.O Much of Lamanai's population relocated to the island during the Postclassicperiod. O The Mayan trading network remained strong during the Postclassicperiod.O The Maya developed new technological capabilities on the island.11.Why does the student say this:O To request that the professor repeat the point she just made.O To express his doubt about the period of time being discussed.OTo disagree with the professor's interpretation of the evidence about Lamanai. OTo find out if he correctly understands the professor's point.Lecture412.What is the main purpose of the lecture?O To explain how musicians can perform successfully in theaters and concert halls with poor acoustics.O To explain how the design of theaters and concert halls has changed over time.O To discuss design factors that affect sound in a room.O To discuss a method to measure the reverberation time of a room.13.According to the lecture, what were Sabine's contributions to architectural acoustics? Click on 2 a nswers.O He founded the field of architectural acoustics.O He developed an important formula for measuring a room’s reverberation time. O H e renewed architects ’ interest in ancient theaters.O He provided support for using established architectural principles in the design of concert halls.14.According to the professor, what is likely to happen if a room has a very long reverberation time?O Performers will have to make an effort to be louder.O Sound will not be scattered in all directions.O Older sounds will interfere with the perception of new sounds.O Only people in the center of the room will be able to hear clearly.15.Why does the professor mention a piano recital?O To illustrate that different kinds of performances require rooms with different reverberation times.O To demonstrate that the size of the instrument can affect its acoustic properties.O To cite a type of performance suitable for rectangular concert hall.O To exemplify that the reverberation time of a room is related to its size.16.According to the professor, what purpose do wall decorations in older concert halls serve?O They make sound in the hall reverberate longer.O They distribute the sound more evenly in the hall.O They make large halls look smaller and more intimate.O They distuise structural changes made to improve sound quality.17.Why does the professor say this:O To find out if students have understood his point.O To indicate that he will conclude the lecture soon.O To introduce a factor contradicting his previous statement. O To add emphasis to his previous statement.答案：1.C2.B3.B4.D5.AD6.D7.A8.C9.A10.C11D12.C13.AB14.C15.AD16.B17.D。
蛋白及多肽药物干粉吸入剂研究新进展周洁雨, 张兰, 毛世瑞*(沈阳药科大学药学院, 辽宁沈阳 110016)摘要: 为成功设计蛋白及多肽类药物物理混合型干粉吸入剂提供理论和实践依据, 本文综述和阐明了干粉吸入剂常用载体, 药物微粉化制备技术, 影响干粉吸入剂肺部沉积的处方工艺因素, 包括载体性质、药物载体比例、混合顺序、混合方法和混合时间、药物载体相互作用, 以及粉体学性质包括粒径大小和形态、密度、粉体流动性、带电性、分散性、吸湿性对肺部沉积率的影响。
依据上述讨论和粉末分散的机制, 提出了增加干粉吸入剂肺部沉积率的策略, 包括加入载体细粉、加入黏附力控制物质和对药物微粉再加工等。
因此, 设计肺部沉积率高的蛋白及多肽药物的物理混合型干粉吸入剂需系统地研究药物与载体相互作用, 阐明处方工艺及粉体学性质的影响。
关键词: 蛋白质及多肽类药物; 干粉吸入剂; 载体; 肺部沉积率中图分类号: R943 文献标识码:A 文章编号: 0513-4870 (2015) 07-0814-10Recent progress of dry powder inhalation of proteins and peptidesZHOU Jie-yu, ZHANG Lan, MAO Shi-rui*(School of Pharmacy, Shenyang Pharmaceutical University, Shenyang 110016, China)Abstract: To provide theoretical and practical basis for the successful formulation design of physically-mixed inhalation dry powder of proteins and peptides, related references were collected, analyzed and summarized. Inthis review drug micronization technology and commonly used carriers for inhalation dry powder preparationwere introduced. For proteins and peptides, supercritical fluid technology and spray-drying are more suitablebecause of their capabilities of keeping drug activity. Being approved by U. S. Food and Drug Administration,lactose has been extensively used as carriers in many inhalation products. Formulation and process factors influencing drug deposition in the lung, including carrier properties, drug-carrier ratio, blending order, mixingmethods, mixing time and the interaction between drug and carrier, were elucidated. The size, shape and surface properties of carries all influence the interaction between drug and carrier. Besides, influence of micromeritic properties of the dry powder, such as particle size, shape, density, flowability, charge, dispersibilityand hygroscopicity, on drug deposition in the lung was elaborated. Among these particle size plays the mostcrucial role in particle deposition in the lung. Moreover, based on the mechanisms of powder dispersity, somestrategies to improve drug lung deposition were put forward, such as adding carrier fines, adding adhesive- controlling materials and reprocessing micronized drug. In order to design physically-mixed inhalation drypowder for proteins and peptides with high lung deposition, it is essential to study drug-carriers interactions systematically and illustrate the potential influence of formulation, process parameters and micromeritic properties of the powder.Key words: protein and peptide; dry powder inhalation; carrier; lung deposition收稿日期: 2015-03-13; 修回日期: 2015-04-23.*通讯作者 Tel / Fax: 86-24-23986358, E-mail: maoshirui@因蛋白及多肽类大分子药物膜通透性差且对酶敏感, 所以一直以来这类药物在临床上以注射给药为主。
