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Practical Multilinear Maps over the Integers

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Practical Multilinear Maps over the IntegersJean-S´e bastien Coron1,Tancr`e de Lepoint2,3,and Mehdi Tibouchi41University of Luxembourgjean-sebastien.coron@uni.lu2CryptoExperts,France3´Ecole Normale Sup´e rieure,Francetancrede.lepoint@4NTT Secure Platform Laboratories,Japantibouchi.mehdi@lab.ntt.co.jpAbstract.Extending bilinear elliptic curve pairings to multilinear maps is a long-standing openproblem.Thefirst plausible construction of such multilinear maps has recently been described by Garg,Gentry and Halevi,based on ideal lattices.In this paper we describe a different construction thatworks over the integers instead of ideal lattices,similar to the DGHV fully homomorphic encryptionscheme.We also describe a different technique for proving the full randomization of encodings:insteadof Gaussian linear sums,we apply the classical leftover hash lemma over a quotient lattice.We showthat our construction is relatively practical:for reasonable security parameters a one-round7-partyDiffie-Hellman key exchange requires less than40seconds per party.Moreover,in contrast with previouswork,multilinear analogues of useful,base group assumptions like DLIN appear to hold in our setting.1IntroductionMultilinear maps.Extending bilinear elliptic curve pairings to multilinear maps is a long-standing open problem.In2003Boneh and Silverberg showed two interesting applications of multilinear maps[BS03],namely multipartite Diffie-Hellman and very efficient broadcast encryption;however they were pessimistic about the existence of such maps from the realm of algebraic geometry.Thefirst plausible construction of multilinear maps has recently been described by Garg,Gentry and Halevi,based on ideal lattices[GGH13].The main difference with bilinear pairings is that the encoding a·g of an element a is randomized(with some noise)instead of deterministic;only the computed multilinear map e(a1·g,...,aκ·g)is a deterministic function of the a i’s only.The construction has bounded degree with a maximum degreeκat most polynomial in the security parameter.Indeed,the encoding noise grows linearly with the degree,and when the noise reaches a certain threshold the encoding can become incorrect,as for ciphertexts in a somewhat homomorphic encryption scheme.The security of the construction relies on new hardness assumptions which are natural extensions of the Decisional Diffie-Hellman(DDH)assumption.To gain more confidence in their scheme the authors provide an extensive cryptanalytic survey.The authors focus on one application:the multipartite Diffie-Hellman key exchange.The construction from[GGH13]works in the polynomial ring R=Z[X]/(X n+1),where n is large enough to ensure security.One generates a secret short ring element g∈R,generating a principal ideal I= g ⊂R.One also generates an integer parameter q and another random secret z∈R/qR.One encodes elements of the quotient ring R/I,namely elements of the form e+I for some e,as follows:a level-i encoding of the coset e+I is an element of the form u k=[c/z i]q, where c∈e+I is short.Such encodings can be both added and multiplied,as long as the norm of the numerators remain shorter than q;in particular the product ofκencodings at level1gives an encoding at levelκ.For such level-κencodings one can then define a zero-testing parameter p zt=[hzκ/g]q,for some small h∈R.Then given a level-κencoding u=[c/zκ]one can compute [p zt·u]q=[hc/g]q.When c is an encoding of zero we have c/g∈R,which implies that hc/g is small in R,and therefore[hc/g]q is small;this provides a way to test whether a level-κencoding c is an encoding of0.For the same reason the high-order bits of[p zt·u]q=[hc/g]q only depend on thecoset e +I and not on the particular c ∈e +I ;this makes it possible to extract a representation of cosets encoded at level κ,and eventually defines a degree-κmultilinear map for level-1encodings.Our contributions.Our main contribution is to describe a different construction that works over the integers instead of ideal lattices,similar to the DGHV fully homomorphic encryption scheme[DGHV10]and its batch variant [CCK +13].Our construction offers the same flexibility as the original from [GGH13];in particular it can be modified to support the analogue of asymmetric maps and composite-order maps.Moreover,it does not seem vulnerable to the “zeroizing”attack that breaks base group hardness assumptions like the analogues of DLIN and subgroup membership for the multilinear maps of [GGH13].Since those assumptions are believed necessary to adapt constructions of primitives like adaptively secure functional encryption and NIZK,our construction seems even more promising for applications than [GGH13].In Appendix C,we provide a comparison with the original scheme,to better highlight the similarities.As in [GGH13],the security of our construction relies on new assumptions;it cannot be derived from “classical”assumptions such as the Approximate-GCD assumption used in [DGHV10].We describe various possible attacks against our scheme;this enables us to derive parameters for which our scheme remains secure against these attacks.Our new construction works as follows:one first generates n secret primes p i and publishes x 0= n i =1p i (where n is large enough to ensure correctness and security);one also generates n small secret primes g i ,and a random secret integer z modulo x 0.A level-k encoding of a vector m =(m i )∈Z n is then an integer c such that for all 1 i n :c ≡r i ·g i +m iz k (mod p i )(1)for some small random integers r i ;the integer c is therefore defined modulo x 0by CRT.It is clear that such encodings can be both added and multiplied modulo x 0,as long as the numerators remain smaller than the p i ’s.In particular the product of κencodings c j at level 1gives an encoding at level κwhere the corresponding vectors m j are multiplied componentwise.For such level-κencodings one defines a zero-testing parameter p zt with:p zt =ni =1h i · z κ·g −1i mod p i · i =ip i mod x 0for some small integers h i .Given a level-κencoding c as in (1),one can compute ω=p zt ·c mod x 0,which gives:ω=n i =1h i · r i +m i ·(g −1i mod p i ) · i =ip i mod x 0.Then if m i =0for all i ,since the r i ’s and h i ’s are small,we obtain that ωis small compared to x 0;this enables to test whether c is an encoding of 0or not.Moreover for non-zero encodings the leading bits of ωonly depend on the m i ’s and not on the noise r i ;for level-κencodings this enables to extract a function of the m i ’s only,which eventually defines as in [GGH13]a degree-κmultilinear map.1Our second contribution is to describe a different technique for proving the full randomization of encodings.As in [GGH13]the randomization of encodings is obtained by adding a random subset-sum of encodings of 0from the public parameters.However as in [GGH13]the Leftover Hash Lemma (LHL)cannot be directly applied since the encodings live in some infinite ring instead of a finite group.The solution in [GGH13]consists in using linear sums with Gaussian coefficients;it is shown in [AGHS12]that the resulting sum has a Gaussian distribution (over some lattice).As noted by the authors,this can be seen as a “leftover hash lemma over lattices”.In this paper wedescribe a different technique that does not use Gaussian coefficients;instead it consists in working modulo some lattice L ⊂Z n and applying the leftover hash lemma over the quotient Z n /L ,which is still a finite group.Such technique was already used to prove the security of the batch variant of the DGHV fully homomorphic encryption scheme [CCK +13,CLT13].Here we provide a more formal description:we clearly state our “LHL over lattices”so that it can later be applied as a black-box (as the corresponding Theorem 3in [AGHS12]).Our quotient lattice technique can independently be applied to the original encoding scheme from [GGH13],while the Gaussian sum technique from[AGHS12]is also applicable to ours.Our third contribution is to describe the first implementation of multilinear maps.It appears that the basic versions of both [GGH13]and our scheme are rather unpractical,because of the huge public parameter size required to randomize the encodings.Therefore we use a simple optimization that consists in storing only a small subset of the public elements and combining them pairwise to generate the full public-key.Such optimization was originally described in [GH11]for reducing the size of the encryption of the secret-key bits in the implementation of Gentry’s FHE scheme [Gen09].It was also used in [CMNT11]to reduce the public-key size of the DGHV scheme;however,as opposed to the latter work our randomization of encodings is heuristic only,whereas in [CMNT11]the semantic security was still guaranteed.Thanks to this optimization our construction becomes relatively practical:for reasonable security parameters a multipartite Diffie-Hellman computation with 7users requires less than 40seconds,with a public parameter size of roughly 2.6GBytes;a proof-of-concept implementation is available at [Lep].2Definition of Randomized Encodings and Multilinear MapsIn this section we recall the setting introduced in [GGH13]for the notion of randomized encodings and multilinear maps,which they call graded encoding schemes .There are essentially two main differences with classical bilinear pairings (and their generalization to cryptographic multilinear maps as considered in [BS03]):1.In bilinear pairings (and more generally cryptographic multilinear maps)we have a map e :G κ→G T that is linear with respect to all its κinputs:e (a 1·g,...,a κ·g )= κ i =1a i ·e (g,...,g ).(2)One can view a ·g as an “encoding”of the integer a ∈Z p over the group G of order p generated by g .The main difference in our setting is that encodings are now randomized.This means that an element a ∈R (where R is a ring that plays the role of the exponent space Z p )has many possible encodings;only the final multilinear map e (a 1·g,...,a κ·g )is a deterministic function of the a i ’s only,and not on the randomness used to encode a i into a i ·g .2.The second main difference is that to every encoding is now associated a level .At level 0we have the “plaintext”ring elements a ∈R ,at level 1we have the encoding a ·g ,and by combining k such encodings a i ·g at level 1one obtains a level-k encoding where the underlying elements a i are homomorphically multiplied in R .The difference with “classical”cryptographic multilinear maps is that we can now multiply any (bounded)subset of encodings,instead of strictly κat a time as with (2).For encodings at level κwe have a special zero-testing parameter p zt that can extract a deterministic function of the underlying ring elements.This enables to define a degree-κmultilinear map for encodings at level 1.2.1Graded Encoding SystemWe recall the formal definition of a κ-Graded Encoding System from [GGH13].For simplicity we only consider the symmetric case throughout the paper;we refer to [GGH13]for a more general framework that can handle the asymmetric case.Definition1(κ-Graded Encoding System[GGH13]).Aκ-Graded Encoding System for a ring R is a system of sets S={S(α)v∈{0,1}∗:v∈N,α∈R},with the following properties:1.For every v∈N,the sets{S(α)v:α∈R}are disjoint.2.There are binary operations+and−(on{0,1}∗)such that for everyα1,α2∈R,every v∈N,and every u1∈S(a1)v and u2∈S (a2)v,it holds that u1+u2∈S(α1+α2)v and u1−u2∈S(α1−α2)vwhereα1+α2andα1−α2are addition and subtraction in R.3.There is an associative binary operation×(on{0,1}∗)such that for everyα1,α2∈R,every v1,v2with0 v1+v2 κ,and every u1∈S(α1)v1and u2∈S (α2)v2,it holds that u1×u2∈S(α1·α2)v1+v2whereα1·α2is multiplication in R.2.2Multilinear map ProceduresWe also recall the definition of the procedures for manipulating encodings.As previously we consider only the symmetric case;we refer to[GGH13]for the general case.Instance Generation.The randomized InstGen(1λ,1κ)takes as inputs the parametersλandκ, and outputs(params,p zt),where params is a description of aκ-Graded Encoding System as above, and p zt is a zero-test parameter.Ring Sampler.The randomized samp(params)outputs a“level-zero encoding”a∈S(α)for anearly uniform elementα∈R R.Note that the encoding a does not need to be uniform in S(α).Encoding.The(possibly randomized)enc(params,a)takes as input a level-zero encoding a∈S(α)0 for someα∈R,and outputs the level-one encoding u∈S(α)1for the sameα.Re-Randomization.The randomized reRand(params,i,u)re-randomizes encodings relative to thesame level i.Specifically,given an encoding u∈S(α)v,it outputs another encoding u ∈S (α)v.Moreoverfor any two u1,u2∈S(α)v,the output distributions of reRand(params,i,u1)and reRand(params,i,u2) are nearly the same.Addition and negation.Given params and two encodings relative to the same level,u1∈S(α1)iand u2∈S(α2)i ,we have add(params,u1,u2)∈S(α1+α2)iand neg(params,u1)∈S(−α1)i.Below wewrite u1+u2and−u1as a shorthand for applying these procedures.Multiplication.For u1∈S(α1)i and u2∈S(α2)j,we have mul(params,u1,u2)=u1×u2∈S(α1·α2)i+j.Zero-test.The procedure isZero(params,p zt,u)outputs1if u∈S(0)κand0otherwise. Extraction.The procedure extracts a random function of ring elements from their level-κencoding. Namely ext(params,p zt,u)outputs s∈{0,1}λ,such that:1.For anyα∈R and u1,u2∈S(α)κ,ext(params,p zt,u1)=ext(params,p zt,u2).2.The distribution{ext(params,p zt,u):α∈R R,u∈S(α)κ}is nearly uniform over{0,1}λ.This concludes the definition of the procedures.In[GGH13]the authors consider a slightly relaxed definition of isZero and ext,where isZero can still output1even for some non-zero encoding u with negligible probability,and ext can extract different outputs when applied to encodings of the same elements,also with negligible probability;see[GGH13]for the corresponding definitions.2.3Hardness AssumptionsFinally we recall the hardness assumptions for multilinear maps from[GGH13];as previously we consider only the symmetric case and refer to[GGH13]for the general case.In this symmetric case given a set ofκ+1level-one encodings of random elements,it should be unfeasible to distinguish a level-κencoding of their product from random.Graded DDH (GDDH).Let GE be a graded encoding scheme consisting of all the routines above.For an adversary A and parameters λ,κwe consider the following process:1.(params ,p zt )←InstGen (1λ,1κ)2.Choose a j ←samp (params )for all 1 j κ+13.Set u j ←reRand (params ,1,enc (params ,1,a j ))for all 1 j κ+1//encodings at level 14.Choose b ←samp (params )//encoding at level 05.Set ˜u =a κ+1× κi =1u i //encoding of the right product at level κ6.Set ˆu =b × κi =1u i //encoding of a random product at level κThe GDDH distinguisher is given as input the κ+1level-one encodings u j and either ˜u (encoding the right product)or ˆu (encoding a random product),and must decide which is the case.The GDDH assumption states that the advantage of any efficient adversary is negligible in the security parameter.3Our new Encoding SchemeSystem parameters.The main parameters are the security parameter λand the required multilinearity level κ poly (λ).Based on λand κ,we choose the vector dimension n ,the bit-size ηof the primes p i ,the bit-size αof the primes g i ,the maximum bit-size ρof the randomness used in encodings,and various other parameters that will be specified later;the constraints that these parameters must satisfy are described in the next section.For integers z ,p we denote the reduction of z modulo p by (z mod p )or [z ]p with −p/2<[z ]p p/2.In our scheme a level-k encoding of a vector m =(m i )∈Z n is an integer c such that for all 1 i n :c ≡r i ·g i +m i z k(mod p i )(3)where the r i ’s are ρ-bit random integers (specific to the encoding c ),with the following secret parameters:the p i ’s are η-bit prime integers,the g i ’s are α-bit primes,and the denominator z is a random (invertible)integer modulo x 0= n i =1p i .The integer c is therefore defined by CRT modulo x 0,where x 0is made public.Since the p i ’s must remain secret,the user cannot encode the vectors m ∈Z n by CRT directly from (3);instead one includes in the public parameters a set of level-0encodings x j of random vectors a j ∈Z n ,and the user can generate a random level-0encoding bycomputing a random subset-sum of those x j ’s.Remark 1.From (3)each integer m i is actually defined modulo g i .Therefore our scheme encodes vectors m from the ring R =Z g 1×···×Z g n .Instance generation:(params ,p zt )←InstGen (1λ,1κ).We generate n secret random η-bit primes p i and publish x 0= n i =1p i .We generate a random invertible integer z modulo x 0.We generate nrandom α-bit prime integers g i and a secret matrix A =(a ij )∈Z n × ,where each component a ij israndomly generated in [0,g i )∩Z .We generate an integer y ,three sets of integers {x j }τj =1,{x j } j =1and {Πj }n j =1,a zero-testing vector p zt ,and a seed s for a strong randomness extractor,as described later.We publish the parameters params = n,η,α,ρ,β,τ, ,y,{x j }τj =1,{x j } j =1,{Πj }n j =1,s andp zt .Sampling level-zero encodings:c ←samp (params ).We publish as part as our instancegeneration a set of integers x j ,where each x j encodes at level-0the column vector a j ∈Z n of thesecret matrix A =(a ij )∈Z n × .More precisely,using the CRT modulo x 0we generate integers x j such that:1 j ,x j ≡r ij ·g i +a ij (mod p i )(4)where the r ij’s are randomly generated in (−2ρ,2ρ)∩Z .Our randomized sampling algorithm samp(params)works as follows:we generate a random binary vector b=(b j)∈{0,1} and output the level-0encodingc=j=1b j·x j mod x0.From Equation(4),this gives c≡j=1r ij b j·g i+j=1a ijb j(mod p i);as required the outputc is a level-0encoding:c≡r i·g i+m i(mod p i)(5) of some vector m=A·b∈Z n which is a random subset-sum of the column vectors a j.We note that for such level-0encodings we get|r i·g i+m i| ·2ρ+αfor all i.The following Lemma states that,as required,the distribution of m can be made statisticallyclose to uniform over R=Z g1×···×Z gn;the proof is based on applying the LHL over the set R(see Appendix A).Lemma1.Let c←samp(params)and write c≡r i·g i+m i(mod p i).Assume n·α+2λ.The distribution of(params,m)is statistically close to the distribution of(params,m )where m ←R. Encodings at higher levels:c k←enc(params,k,c).To allow encoding at higher levels,we publish as part of our instance-generation a level-one random encoding of1,namely an integer ysuch that:y≡r i·g i+1z(mod p i)for random integers r i∈(−2ρ,2ρ)∩Z;as previously the integer y is computed by CRT modulo x0.Given a level-0encoding c of m∈Z n as given by(5),we can then compute a level-1encoding of the same m by computing c1=c·y mod ly we obtain as required:c1≡r(1)i·g i+m iz(mod p i)(6)for some integers r(1)i ,and we get|r(1)i·g i+m i| ·22(ρ+α)for all i.More generally to generate alevel-k encoding we compute c k=c0·y k mod x0.In multipartite Diffie-Hellman key-exchange every user keeps a private level-0encoding c and publishes a level-1encoding of the same underlying(unknown)m;however one cannot publish c1=c·y mod x0directly since the private level-0encoding c could be recovered immediately from c=c1/y mod x0.Instead the level-1encoding c1mustfirst be re-randomized into a new level-1 encoding c 1whose distribution does not depend on the original c as long as it encodes the same m. Re-randomization:c ←reRand(params,i,c).To allow re-randomization of encodings at level one,we publish as part of our instance-generation a set of n integersΠj which are all level-1randomencodings of zero:1 j n,Πj≡ ij·g iz(mod p i).The matrixΠ=( ij)∈Z n×n is a diagonally dominant matrix generated as follows:the non-diagonal entries are randomly and independently generated in(−2ρ,2ρ)∩Z,while the diagonal entries are randomly generated in(n2ρ,n2ρ+2ρ)∩Z.We also publish as part of our instance-generation a set ofτintegers x j,where each x j is a level-1random encoding of zero:r ij·g iand where the column vectors of the matrix (r ij )∈Z n ×τare randomly and independently generated in the half-open parallelepiped spanned by the columns of the previous matrix Π;see Appendix E for a concrete algorithm.Given as input a level-1encoding c 1as given by (6),we randomize c 1with a random subset-sum of the x j ’s and a linear combination of the Πj ’s:c 1=c 1+τj =1b j ·x j +n j =1b j ·Πj mod x 0(7)where b j ←{0,1},and b j ←[0,2µ)∩Z .The following Lemma shows that as required the distributionof c 1is nearly independent of the input (as long as it encodes the same m ).This follows essentially from our “leftover hash lemma over lattices”;see Section 4.Lemma 2.Let c ←samp (params ),c 1←enc (params ,1,c ),and c 1←reRand (params ,1,c 1).Write c 1≡(r i ·g i +m i )/z (mod p i )for all 1 i n ,and r =(r 1,...,r n )T .Let τ n ·(ρ+log 2(2n ))+2λand µ ρ+α+λ.The distribution of (params ,r )is statistically close to that of (params ,r ),where r ∈Z n is randomly generated in the half-open parallelepiped spanned by the column vectors of 2µΠ.Writing c 1≡(r i ·g i +m i )/z (mod p i ),and using |r ij ·g i | 2n 2ρ+αfor all i ,j ,we obtain |r i ·g i +m i | 22(ρ+α)+τ·2n 2ρ+α+n ·2n 2µ+ρ+α.Using µ ρ+α+λthis gives |r i ·g i +m i | 4n 22µ+ρ+α.Adding and Multiplying Encodings.It is clear that one can homomorphically add encodings.Moreover the product of κlevel-1encodings u i can be interpreted as an encoding of the ly,given level-one encodings u j of vectors m j ∈Z n for 1 j κ,with u j ≡(r ij ·g i +m ij )/z (mod p i ),we simply let:u =κ j =1u j mod x 0.This gives:u ≡κ j =1(r ij ·g i +m ij )z κ≡r i ·g i + κ j =1m ij mod g i z κ(mod p i )for some r i ∈Z .This is a level-κencoding of the vector m obtained by componentwise product of the vectors m j ,as long as κj =1(r ij ·g i +m ij )<p i for all i .When computing the product of κlevel-1encodings from reRand and one level-0encoding from samp (as in multipartite Diffie-Hellman key exchange),we obtain from previous bounds |r i | (4n 22µ+ρ+α)κ· ·2ρ+1for all i .Zero Testing.isZero (params ,p zt ,u κ)?=0/1.We can test equality between encodings by subtracting them and testing for zero.To enable zero testing we randomly generate an integer matrix H =(h ij )∈Z n ×n such that H is invertible in Z and both H T ∞ 2βand (H −1)T ∞ 2β,for some parameter βspecified later;here · ∞is the operator norm on n ×n matrices with respect to the ∞norm on R n .A technique for generating such H is discussed in Appendix F.We then publish as part of our instance generation the following zero-testing vector p zt ∈Z n :(p zt )j =ni =1h ij · z κ·g −1i mod p i · i =i p i mod x 0.(8)To determine whether a level-κencoding c is an encoding of zero or not,we compute the vector ω=c ·p zt mod x 0and test whether ω ∞is small:isZero (params ,p zt ,c )= 1if c ·p zt mod x 0 ∞<x 0·2−ν0otherwisefor some parameterνspecified later.Namely for a level-κciphertext c we have c≡(r i·g i+m i)/zκ(mod p i)for some r i∈Z; therefore for all1 i n we can write:c=q i·p i+(r i·g i+m i)·z−κmod p i(9)for some q i∈Z.Therefore combining(8)and(9),we get:(ω)j=(c·p zt mod x0)j=ni=1h ij·r i+m i·(g−1imod p i)·i =ip i mod x0.(10)Therefore if m i=0for all1 i n,then ω ∞= c·p zt mod x0 ∞is small compared to x0 when the r i’s are small enough,i.e.a limited number of additions/multiplications on encodings has been performed.Conversely if m i=0for some i we show that ω ∞must be large.More precisely we prove the following lemma in Appendix B.Lemma3.Let n,η,αandβbe as in our parameter setting.Letρf be such thatρf+λ+α+2β η−8,and letν=η−β−ρf−λ−3 α+β+5.Let c be such that c≡(r i·g i+m i)/zκ(mod p i) for all1 i n,where0 m i<g i for all i.Let r=(r i)1 i n and assume that r ∞<2ρf.If m=0then ω ∞<x0·2−ν−λ−2.Conversely if m=0then ω ∞ x0·2−ν+2.Extraction.sk←ext(params,p zt,uκ).This part is essentially the same as in[GGH13].To extract a random function of the vector m encoded in a level-κencoding c,we multiply c by the zero-testing parameter p zt modulo x0,collect theνmost significant bits of each of the n components of the resulting vector,and apply a strong randomness extractor(using the seed s from params):ext(params,p zt,c)=Extract smsbsν(c·p zt mod x0)where msbsνextracts theνmost significant bits of the ly if two encodings c and c encode the same m∈Z n then from Lemma3we have (c−c )·p zt mod x0 ∞<x0·2−ν−λ−2, and therefore we expect thatω=c·p zt mod x0andω =c ·p zt mod x0agree on theirνmost significant bits,and therefore extract to the same value.2Conversely if c and c encode different vectors then by Lemma3we must have (c−c )·p zt mod x0 ∞>x0·2−ν+2,and therefore theνmost significant bits of the correspondingωandω must be different.This implies that for random m∈R=Z g1×···×Z gnthe min-entropy ofmsbsν(c·p zt mod x0)when c encodes m is at least log2|R| n(α−1).Therefore we can use a strong randomness extractor to extract a nearly uniform bit-string of length log2|R| −λ.This concludes the description of our multilinear encoding scheme.In Appendix C we provide a comparison with the original scheme from[GGH13].3.1Setting the ParametersThe system parameters must satisfy the following constraints:•The bit-sizeρof the randomness used for encodings must satisfyρ=Ω(λ)to avoid brute force attack on the noise,including the improved attack from[CN12]with complexity˜O(2ρ/2).2Two coefficientsωandω fromωandω could still be on the opposite sides of a boundary,with ω/2k =v and ω /2k =v+1,so thatωandω would extract to different MSBs v and v+1.Heuristically this happens with probability O(2−λ).The argument can be made rigorous by generating a public random integer W modulo x0in the parameters,and extracting the MSBs ofω+W mod x0instead ofωmod x0for all coefficientsωof the vector ω.•The bit-size αof the primes g i must be large enough so that the order of the group R =Z g 1×···×Z g n does not contain small prime factors;this is required to prove the security of the multipartite Diffie-Hellman Key Exchange (see Appendix D).One can take α=λ.•The parameter n must be large enough to thwart lattice-based attacks on the encodings,namely n =ω(ηlog λ);see Section 5.1.•The number of level-0encodings x j for samp must satisfy n ·α+2λin order to apply the leftover hash lemma;see Lemma 1.•The number τof level-1encodings x j must satisfy τ n ·(ρ+log 2(2n ))+2λin order to apply our “leftover hash lemma over lattices”.For the same reason the bit-size µof the linear sum coefficients must satisfy µ α+ρ+λ;see Lemma 2.•The bitsize βof the matrix H entries must satisfy β=Ω(λ)in order to avoid the GCD attack from Section 5.2.One can take β=λ.•The bit-size ηof the primes p i must satisfy η ρf +α+2β+λ+8,where ρf is the maximum bit size of the randoms r i a level-κencoding (see Lemma 3).When computing the product of κlevel-1encodings and an additional level-0encoding (as in a multipartite Diffie-Hellman key exchange with N =κ+1users),one obtains ρf =κ·(µ+ρ+α+2log 2n +2)+ρ+log 2 +1(see previous Section).•The number νof most significant bits to extract can then be set to ν=η−β−ρf −λ−3(see Lemma 3).3.2Security of our Construction As in [GGH13]the security of our construction relies on new assumptions that do not seem to be reducible to more classical ly,as in [GGH13]one can make the assumption that solving the Graded DDH problem (GDDH)recalled in Section 2.3is hard in our scheme.This enables to prove the security of the one-round N -way Diffie-Hellman key exchange protocol [GGH13].Ideally one would like to reduce such assumption to a more classical assumption,such as the Approximate-GCD assumption,but that does not seem feasible.Therefore to gain more confidence in our scheme we describe various attacks in Section 5.4Another Leftover Hash Lemma over LatticesAs mentioned in the introduction,to prove the full randomization of encodings (Lemma 2)one cannot apply the classical Leftover Hash Lemma (LHL)because the noise in the encodings live in some infinite ring instead of a finite group.In [GGH13]the issue was solved by using linear sums with Gaussian coeffily the analysis in [AGHS12]shows that the resulting sum has a Gaussian distribution (over some lattice).As noted by the authors this technique can be seen as a “leftover hash lemma over lattices”.Such a technique would be applicable to our scheme as well.In this section we describe an alternative technique (without Gaussian coefficients)that can also be seen as a “leftover hash lemma over lattices”.It consists in working modulo a lattice L ⊂Z n and applying the classical leftover hash lemma over the finite group Z n /L .This technique was already used in [CCK +13,CLT13]to prove the security of a batch extension of the DGHV scheme.In this paper we provide a more formal description;namely we clearly state our “LHL over lattices”so that it can later be applied as a black-box (as the corresponding Theorem 3in [AGHS12]).Namely our quotient lattice technique can independently be applied to the original encoding scheme from [GGH13].4.1Classical Leftover Hash LemmaWe first recall the classical Leftover Hash Lemma.We say that the distributions D 1,D 2over a finite domain X are ε-statistically close if the statistical distance ∆(D 1,D 2)=12x ∈X |D 1(x )−D 2(x )|is。

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地理信息系统导论课后习题英文版8

地理信息系统导论课后习题英文版8

Chapter 8 Review Questions1. Explain the difference between location errors and topological errors. Location errors such as missing polygons or distorted lines relate to the geometric inaccuracies of spatial features, whereas topological errors such as dangling lines and unclosed polygons relate to the logical inconsistencies between spatial features.2. What are the primary data sources for digitizing?Global positioning systems (GPS) and remote sensing imagery provide the primary data sources for digitizing. These data sources can bypass printed maps and the practice of various methods of map generalization.3. A digitized map from a secondary data source such as a USGS quadrangle map is subject to more location errors than a primary data source. Why?A secondary data source such as a USGS quadrangle map is subject to more location errors because the map has undergone simplification, generalization, and other practices during the mapmaking process.4. Although the U.S. National Map Accuracy Standard adopted in 1947 is still printed on USGS quadrangle maps, the standard is not really applicable to GIS data. Why?A GIS uses digital spatial data, which can be easily manipulated and output to any scale. The U.S. National Map Accuracy Standard, on the other hand, is scale dependent.5. According to the new National Standard for Spatial Data Accuracy, a geospatial data producer is encouraged to report a RMS statistic associated with a data set. In general terms, how does one interpret and use the RMS statistic? One can first multiply the RMS by 1.7308 to get the NSSDA (National Standard for Spatial Data Accuracy) statistic. This statistic represents the standard error of the mean at the 95 percent confidence level. In other words, one can be sure that, 95 percent of the time, the accuracy of a point or a line is within the NSSDA statistic.6. Suppose a point location is recorded as (575729.0, 5228382) in data set 1 and (575729.64, 5228382.11) in data set 2. Which data set has a higher data precision? In practical terms, what does the difference in data precision in this case mean?Data set 2 has a higher data precision than data set 1. If the measurement unit is meters, the recording of a point location in data set 2 is down to one hundredth of a meter and the recording in data set 1 is rounded off at meters.7. The ArcGIS Desktop Help has a poster illustrating topology rules in the geodatabase data model (ArcGIS Desktop Help > Editing in ArcMap > Editing Topology > Topology rules). View the poster. Can you think of an example (other than those on the poster) that can use the polygon rule of “Must be covered by feature class of”?[The poster illustrates the polygon rule with the example of “States are covered by counties.” By extension, counties must be covered by census tracts, census tracts by block groups, and block groups by blocks.]8. Give an example (other than those on the poster) that can use the polygon rule of “Must not overlap with.”[The poster illustrates the polygon rule with the example of “Lakes and land parcels from two different feature classes must not overlap.” The Census Administrative Boundaries Data Model poster that can be downloaded from the ESRI website has two other examples: the feature class of American Indian Reservation must not overlap with the feature class of Place:city, and the feature class of American Indian Reservation must not overlap with the feature class of Place:town.]9. Give an example (other than those on the poster) that can use the line rule of “Must not intersect or touch interior.”[The poster illustrates the line rule with the example of “Lot lines cannot intersect or overlap and must connect to one another only at the endpoint of each line feature.”Like lot lines, road center lines cannot intersect or overlap.]10. Use a diagram to illustrate how a large nodesnap for editing can alter the shape of line features.Node a is supposed to be snapped to node b. But a large nodesnap can snap node a to node c instead.11. Use a diagram to illustrate how a large cluster tolerance for editing can alter the shape of line features.A large cluster tolerance can incorrectly snap the two lines in the center of the diagram indicating a small stream channel.12. Explain the difference between a dangling node and a pseudo node.A dangling node is at the end of a dangling arc, whereas a pseudo node appears along a continuous line and divides the line unnecessarily into separate lines.13. What is a map topology?A map topology is a temporary set of topological relationships between the parts of features that are supposed to be coincident.14. Describe the three basic steps in using a topology rule.Step 1: create a new topology by defining the participating feature classes, the ranks for each feature class, the topology rule(s), and a cluster tolerance.Step 2: evaluate the topology rule and creates errors indicating those features that have violated the topology rule.Step 3: fix errors or accepts errors as exceptions.15. Some nontopological editing operations can create features from existing features. Give two examples of such operations.[Examples include merge features, buffer features, union features, and intersect features.]16. Edgematching requires a source layer and a target layer. Explain the difference between these two types of layers.Features, typically vertices, on the source layer are moved to match those on the target layer during the edgematching process.17. The Douglas-Peucker algorithm typically produces simplified lines with sharp angles. Why?The Douglas-Peucker algorithm connects trend lines to create simplified lines. Because the trend lines are straight lines, they form sharp angles when connected.。

《11规则》题库第二章航海仪器测试题

《11规则》题库第二章航海仪器测试题

《11规则》题库第二章航海仪器测试题基本信息：[矩阵文本题] *1. ______ is not required by bridge-to-bridge communications. [单选题] *VHFSARTNAVTEXLES(正确答案)2. As for Racon, which one is incorrect in the following? [单选题] *Racon is a radar transponderRacon is often installed on major waypointsWhen activated by a radar signal, the Racon sends it back with an identifier Racon can be used to take place of radar(正确答案)3. PAN-PAN repeated three times over the radiotelephone indicates which type of message will follow? [单选题] *DistressSafetyAll clearUrgency(正确答案)4. Automatic identification systems (AIS) are expected to broadcast all of the following information except ______. [单选题] *port of origin(正确答案)name of vesselcourse and speed over grounddraft of vessel5. What does a navigator who uses an ECDIS require? [单选题] *Good navigational knowledge and a professional job attitude(正确答案)Good computer knowledgeGood system knowledgeNothing special6. The terrestrial system of GMDSS consists of ____________. [单选题] *radio-telephony, DSC, Direct Printing Telegraphy, Navtex and SART(正确答案) INMARSAT, DSC, Direct Printing Telegraphy, Navtex and SARTradio-telephony, DSC, EPIRB, Navtex and SARTStatus Recording system, DSC, Direct Printing Telegraphy, Navtex and SART7. The ______ provides the link between the Space Segment and the land-based National/International fixed communications networks. [单选题] *VHFSARTNAVTEXLES(正确答案)8. If the electronic chart is part of an ECDIS, it must display the minimum data required by IMO/IHO, to include all the following EXCEPT ______. [单选题] *hydrographyaids to navigationtidal current(正确答案)regulatory boundaries9. When a call is complete，and subsequently during an exchange of messages，a station invites a reply by saying ___. [单选题] *“Over”(正确答案)“Out”“OK”“Roger”10. What is the correct speed input to an ARPA used for traffic surveillance? [单选题] *Ground speedSpeed through water(正确答案)Speed from GPSSpeed from Doppler11. All echo-sounders can measure the ______. [单选题] *actual depth of wateractual depth of water below keel(正确答案)average depth from waterline to hard bottomaverage depth of water to soft bottom12. A satellite navigation system with global coverage may be termed as global navigation satellite system or _____. [单选题] *GNSS(正确答案)EPIRBSSASECDIS13. The VDR system is designed to operate_____ once it is set up correctly. There is no user interaction . [单选题] *automatically(正确答案)accuratelypromptlyConspicuously14. What is the MOST important thing you should do before transmitting on a marine radio? [单选题] *Ask for permissionRecord the time in your radio logPress the push to talk button three timeMonitor the channel to ensure that it is clear(正确答案)15. While underway, a vessel over 100,000 gross tons with an automatic identification system (AIS) is expected to broadcast all of the following information every 1 to 10 seconds EXCEPT ______. [单选题] *rate of turnname of the vessel(正确答案)navigational statusship’s heading16. The GPS system was designed for ______ satellites. [单选题] *3121824(正确答案)17. What is important to check when transferring a position from GPS to a chart? [单选题] *Reading the position correctlyPlotting the position correctlyMake sure that the chart and the GPS use same datum(正确答案)Make sure the map is updated18. You are using a radar in which your own ship is shown at the center, and the heading flash always points to 0 degree, If bearings are measured in relation to the flash, what type of bearings are produced? ______. [单选题] *Relative(正确答案)TrueCompassMagnetic19. The compass error of a magnetic compass that has no deviation is ______. [单选题] *zeroequal to variation(正确答案)eliminated by adjusting the compassconstant at any geographical location20. A radar range to a small，charted object such as a light will provide a line of position in the form of ______. [单选题] *straight linearc(正确答案)parabolahyperbola21. You have another ship overtaking you close aboard to starboard. You have 3 radar targets bearing 090，0.5 mile, 1 mile，and 1.5 miles. In this case，the unwanted echoes are called ______. [单选题] *Multiple echoes(正确答案)SpokingIndirect echoesSide-lobe echoes22. What is the main purpose of DGPS? [单选题] *Improve positioning accuracy(正确答案)Decrease positioning accuracyReduce operational costI don't know23. When it is accepted to remain on the frequency indicated，you should say ______. [单选题] *standing by on VHF frequency(正确答案)coming to VHF frequencyVHF frequency is the best place for you to stay byremaining in frequency and do not change24. The satellite systems of GMDSS comprise ______. [单选题] *NAVTEX, COSPAS/SARSAT, EPIRBs and Status Recording-systemInmarsat, COSPAS/SARSAT, EPIRBs and Status Recording-system(正确答案) NAVTEX, Inmarsat, EPIRBs and SARTInmarsat, Direct Printing Telegraphy, and DSC25. When another calling channel/frequency is available，do not use ______ or other safety frequencies to make a transmission. [单选题] *2182 kHz or VHF Channel 16(正确答案)VHF Channel 76FAXCABLES26. How to reduce beam width distortion? [单选题] *Adjust centre positionAdjust brillianceAdjust heading markerReduce gain(正确答案)27. ______ is on the panel of DF. [单选题] *ZERO CLEARING(正确答案)ANTI-CLUTTER-RAINDIMMERCURSOR28. In radiotelephony，the spoken word for distress is ______. [单选题] *SecuriteSafetyEmergencyMAYDAY(正确答案)29. Which one of the followings is incorrect about magnetic compass? [单选题] *The magnetic compass is compulsory on my vesselThe magnetic compass is prone to errorThe magnetic compass is always placed inside steel constructions(正确答案)The magnetic compass needs to be calibrated to compensate for local magnetic distortion 30. Radar reflectors are required for ______. [单选题] *all fishing vessels over 39 feet in lengthsail-propelled fishing vesselsall fishing vessels of less than 200 GTwooden hull fishing vessels with a poor radar echo(正确答案)31. If there were suddenly an immediate danger for both the vessel and its crew, would you send a MAYDAY on__ [单选题] *Ch06 with dual watchCh06 with 25 watts outputCh16 with 1 watt outputCh16 with 25 watts output(正确答案)32. The line connecting the Loran-C master station with a secondary station is called the ______. [单选题] *focus linebase line(正确答案)side linecenter line33. If a GMDSS radio operator initiates a DSC distress transmission but does not insert a message，what happens? [单选题] *The transmission is aborted and an alarm sounds to indicate this data must be provided by the operatorThe transmission is not initiated and “ERROR” is indicated on the display readoutThe transmission will be made with “default” information provided automatically(正确答案)The receiving station will poll the DSC unit of the vessel in distress to download the necessary information34. Ship to ship communications during SAR should be executed by _____. [单选题] * VHFMFSatelliteVHF and MF(正确答案)35. The indicated heading should not be relied on ________. [单选题] *until the gyrocompass has settled(正确答案)until the “Ready” lamp will be litwhen the alarm is generatedwhen the red “Power” lamp RCU illuminates36. When your vessel is proceeding to the area of traffic density，______ is used to determine the exact ranges of other ships or objects in the vicinity. [单选题] * Radar(正确答案)GPSDFSatellite Navigator37. Magnetic heading differs from compass heading by _______. [单选题] * compass errortrue headingvariationdeviation(正确答案)38. What is the primary equipment for receiving MSI ________. [单选题] *SARTEPIRBNAVTEX(正确答案)INMARSAT-B39. If the gyrocompass error is east，what describes the error and the correction to be made to gyrocompass headings to obtain true headings ________. [单选题] *The readings are too low（small numerically）and the amount of the error must be added to the compass to obtain true(正确答案)The readings are too low and the amount of the error must be subtracted from the compass to obtain trueThe readings are too high（large numerically）and the amount of the error must be added to the compass to obtain trueThe readings are too high and the amount of the error must be subtracted from the compass to obtain true40. If the magnetic heading is greater than the compass heading，the deviation is ______. [单选题] *east(正确答案)westnorthsouth41. Deviation which is maximum on intercardinal compass headings may be removed by the ______. [单选题] *Flinders barTransverse magnetsFore-and-aft magnetsSoft iron spheres on the sides of the compass(正确答案)42. The accuracy of the DGPS mainly depends on_______. [单选题] *the price of the equipmentthe relative angle between the satellitethe position of the observerthe age of the calculated correction(正确答案)43. Which statement concerning locating signals in the GMDSS is FALSE? [单选题] *Locating signals are transmitted by survival craft VHF transceivers.(正确答案) Locating signals are transmitted by SARTS.Locating signals are intended to facilitate the finding of a distressed vessel or its survivorsLocating signals are not transmitted by auto alarm generators.44. Coral atolls, or a chain of islands at right angles to the radar beam, may show as a long line rather as an individual targets due to ______. [单选题] *the effects of beam width(正确答案)limitations on range resolutionthe pulse length of the radarthe multiple-target resolution factor45. Quadrantal error in a gyrocompass has its greatest effect ______. [单选题] *in high latitudesnear the equatoron north or south headingson intercardinal headings(正确答案)46. What is important to remember when using AIS for collision avoidance? [单选题] *AIS may not give a complete picture of the traffic situation(正确答案)AIS is more accurate than ARPAAIS is not as accurate as ARPAAIS is not allowed to be used for collision avoidance47. Integrated Bridge and Navigation Systems (IBS/INS) provide excellent performance and reliable navigation under all conditions by _____. [单选题] *integrating well-proven, harmonizing equipment(正确答案)integrating all equipment in the bridgean intelligent platform with ECDIS and Radarthe control center based on ECDIS48. Which statement concerning GPS is TRUE ________. [单选题] *It cannot be used in all parts of the worldThere are 12 functioning GPS satellites at presentIt may be suspended without warning(正确答案)Two position lines are used to give a 2D fix49. When hitting a solid object such as a ship or an airplane, the radar waves are reflected back ______ they came. [单选题] *in the waythe wayby the waythrough the path(正确答案)50. A Doppler log in the volume reverberation mode indicates ______. [单选题] *speed being made goodspeed through the water(正确答案)the set of the currentthe depth of the water51. ECDIS must be able to perform all of the following EXCEPT ______. [单选题] *determine true bearing and distance between two geographical pointsdetermine magnetic compass deviation(正确答案)transform a local datum to the WGS-84 datumconvert “graphical coordinates” to “display coordinates”52. According to SOLAS the breathing air apparatus that must be onboard shall have sufficient capacity for how many minutes of operation? [单选题] *2030(正确答案)456053. Which factor has the greatest effect on the amount of gain required to obtain a fathometer reading? [单选题] *Salinity of waterTemperature of water(正确答案)Atmosphere pressureType of bottom54. Your ARPA has two guard zones. What is the purpose of the inner guard zone? [单选题] *Alert the watch officer that a vessel is approaching the preset CPA limitWarn of small targets that are initially detected closer than the outer guard zone(正确答案)Guard against target loss during critical maneuvering situationsSound an alarm for targets first detected within the zone55. ETA／PILOT REVERTING means [单选题] *ETA pilot station has been givenETA pilot station will be given afterwards(正确答案)ETA pilot station was not givenETA pilot station is given56. ________is used for providing homing signals from survival craft for detection by 9 GHz radar. [单选题] *NAVETEX receiverSearch and Rescue Transponder(正确答案)Emergency Position Indicating Radio BeaconDigital selective calling57. ______ is used for safety of navigation ship-to-ship. [单选题] *Channel 6Channel 13(正确答案)Channel 16Channel 7058. Which action should you take after sending a false distress alert on VHF? [单选题] *Send a DSC cancellation message on Ch-70.Make a voice announcement to cancel the alert on Ch-16.(正确答案)Make a voice announcement to cancel the alert on Ch-13.Make a voice announcement to cancel the alert on Ch-22A.59. ______ typically extends from close as 0.1 nautical miles out to 32 NM. [单选题] *EBLVRMCRTtarget tracking range(正确答案)60. When own ship’s position input to ECDIS is wrong, what is the result? [单选题] *NothingECDIS will give warningECDIS will automatically be switched offPosition, range and bearing taken on the ECDIS will be wrong(正确答案)61. The radar control used to reduce sea return at close ranges is the ______. [单选题] *Gain controlSensitivity time control(正确答案)Fast time constantPulse length control62. The common way to obtain your ship’s position is ______. [单选题] *keeping a close watch and lookouttaking a radar range and bearing(正确答案)observing a radar target and listening to signalskeeping a well clear caution63. A radar display in which North is always at the top of the screen is a（n）______. [单选题] *Unstabilized displayStabilized display(正确答案)Composition displayRelative display64. ______ is a radio receiver with ability to sense direction of the incoming radio waves. [单选题] *The echo sounderThe radarThe course recorderThe direction finder(正确答案)65. The abbreviation RYC in a marine cable generally stands for ______. [单选题] *referring to your crewreference for your captainreturn to your cabinreceived your cable(正确答案)66. How long must the GMDSS radio log be retained on board ________. [单选题] *At least two years after the last entry(正确答案)At least one year after the last entryAt least 90 days after the last entryAt least 30 days after the last entry67. The angular difference between the true meridian（great circle connecting the geographic poles）and the magnetic meridian（direction of the lines of magnetic flux）is called ______. [单选题] *deviationvariation(正确答案)errordifference68. If you receive the signal over radiotelephone of Romeo Papa Tango while using the International Code of Signals, you should ______. [单选题] *report to the callerrepeat your last transmission(正确答案)continue since he received your last transmissionend the transmission69. Deviation changes with a change in ______. [单选题] *latitudeheading(正确答案)longitudesea conditions70. How does current and drift effect the relative motion, relative vector presentation? [单选题] *No effect(正确答案)Producing small errors in calculated aspectProducing large errors in calculated aspectProduce errors in calculated CAP/TCPA71. A single vertical magnet placed underneath the compass in the binnacle is used to compensate for ______. [单选题] *the horizontal component of the permanent magnetismdeviation caused by the vessel's inclination from the vertical(正确答案)induced magnetism in the horizontal soft ironinduced magnetism in the vertical soft iron72. The receiver uses ______ satellites to compute latitude，longitude，altitude，and velocity. [单选题] *onetwothreefour(正确答案)73. ______ is used for calling and replying, and for transmitting acknowledging and relaying distress alerts. [单选题] *NAVTEX receiverSearch and Rescue TransponderEmergency Position Indicating Radio BeaconDigital selective calling(正确答案)74. How is an uncoded racon displayed on the PPI? [单选题] *As a line(正确答案)As a dotAs a small circleAs a large circle75. Variation is a compass error that you ______. [单选题] *can correct by adjusting the compass cardcan correct by adjusting the compensation magnetscan correct by changing the vessel’s headingcannot correct(正确答案)76. Before sailing, mariners on duty shall check the headings of magnetic compass by comparison with _____. [单选题] *ship’s clockengine movement recorderrepeaters(正确答案)navigation lights77. ____ is the navigational timekeeper of the vessel. [单选题] *The marine sextantThe chronometer(正确答案)The magnetic compassThe gyrocompass78. A(n) ______ indicates that there is serious and immediate danger for vessel, crew and passengers. [单选题] *distress alert(正确答案)urgency messagesafety messageroutine message79. Which one of the followings is a standard phrase? [单选题] *WARNING. You are running into danger.(正确答案)You are possibly running into danger.You could be in the case of running into danger.You could , I think , be running into danger.80. What is the right way to use VHF CH16 and working channel? [单选题] *Ships can call other ships on Ch16 but should move to a working channel as soon as possible.(正确答案)Ship should use working channel to call other ships at any time.Ch16 is only used as the international distress and calling frequency.Ship can use Ch16 or working channel as they like81. The ship LRIT system shall transmit the ship’s LRIT information at ______ to an LRIT DATA center ? [单选题] *1-hour interval2-hour interval3-hour interval6-hour intervals(正确答案)82. In ________ within coverage of an Inmarsat geostationary satellite, continuous alerting is available. [单选题] *Sea Area A1Sea Area A2Sea Area A3(正确答案)Sea Area A483. How to report your ship call sign to a shore station? [单选题] *Use capital lettersUse phonetic alphabet(正确答案)Use letters and numbersUse flag signals84. Any piece of metal on becoming magnetized will develop regions of concentrated magnetism called ______. [单选题] *fluxpoles(正确答案)magnetsazimuth85. ____ is used for measuring horizontal and vertical angles. [单选题] *The marine sextant(正确答案)The chronometerThe magnetic compassThe gyrocompass86. Which of the following is defined as static information? [单选题] *Navigational statusSafety related messageMMSI(正确答案)Route plan87. The ______ transmits own ship data cyclically via two defined VHF channels and receives the same data of the other ships and objects that are equipped with AIS systems. [单选题] *AIS(正确答案)ECDISGPSVDR88. What are the only magnetic compass correctors that correct for both permanent and induced effects of magnetism？ [单选题] *Quadrantal spheresHeeling magnets(正确答案)Athwartships magnetsFore-and-aft magnets89. All VHF marine band radios operate in the simplex mode, which means that ______. [单选题] *only one person may talk at a time(正确答案)only two persons may talk at the same timethe radio only transmitsthe radio only receives90. What type of radar can activate a racon? [单选题] *X-band radar(正确答案)S-band radarC-band radarNo radar can91. Your radar is set on a true motion display. Which of the following will NOT appear to move across the PPI scope? [单选题] *Echoes from a buoy(正确答案)Own ship's markerEcho from a ship on the same course at the same speedEcho from a ship on a reciprocal course at the same speed92. What is the spoken emergency signal for a “man overboard” on the VHF radio? [单选题] *Man OverboardSecurityMaydayPan-Pan(正确答案)93. When the transmissions of a radio station have broken down, switched off or suspended, it is _______. [单选题] *unfunctionalbreak downoff poweroff air(正确答案)94. The maritime radio system consisting of a series of coast stations transmitting coastal warning is called______. [单选题] *NAVTEX(正确答案)HYDROLANT/HYDROPACNAVAREASAFESEA95. If a magnetic compass is not affected by any magnetic field other than the Earth's，which statement is TRUE? [单选题] *Compass error and variation are equal(正确答案)Compass north will be true northVariation will equal deviationThere will be no compass error96. Which device provides the main means in the GMDSS for locating ships in distress or their survival craft? [单选题] *Radio direction finderSatellite EPIRBs(正确答案)MF/HF DSCVHF homing device97. The description of RACON beside an illustration on a chart would near a ______. [单选题] *radar conspicuous beaconcircular radiobeaconradar transponder beacon(正确答案)radar calibration beacon98. A navigator fixing a vessel’s position by radar______. [单选题] *should never use radar bearingsshould only use radar bearings when the range exceeds the distance to the horizoncan use radar information from one object to fix the position(正确答案)must use information from targets forward of the beam99. Lines on a chart which connect points of equal magnetic variation are called ______. [单选题] *Magnetic latitudesMagnetic declinationsDipIsogonic lines(正确答案)100. AMVER is a system which provides ______. [单选题] *satellite communicationsnavigational informationweather informationposition reporting service(正确答案)。

100句想学好英语的一定不要错过

100句想学好英语的一定不要错过

100句。

想学好英语的，一定不要错过。

1. Typical of the grassland dwellers of the continent is the American antelope, or pronghorn.1.美洲羚羊，或称叉角羚，是该大陆典型的草原动物。

2. Of the millions who saw Haley’s comet in 1986, how many people will live long enough to see it return in the twenty-first century.2. 1986年看见哈雷慧星的千百万人当中，有多少人能够长寿到足以目睹它在二十一世纪的回归呢？3. Anthropologists have discovered that fear, happiness, sadness, and surprise are universally reflected in facial expressions.3.人类学家们已经发现，恐惧，快乐，悲伤和惊奇都会行之于色，这在全人类是共通的。

4. Because of its irritating effect on humans, the use of phenol as a general antiseptic has been largely discontinued.4.由于苯酚对人体带有刺激性作用，它基本上已不再被当作常用的防腐剂了。

5. In group to remain in existence, a profit-making organization must, in the long run, produce something consumers consider useful or desirable.5.任何盈利组织若要生存，最终都必须生产出消费者可用或需要的产品。

好奇绘星图的英语作文

好奇绘星图的英语作文

Curiosity has always been a driving force in human exploration and discovery.The desire to understand the universe and our place within it has led to the creation of star mapsa visual representation of the celestial bodies that light up the night sky.In this essay, we will delve into the process of creating a star map,its importance in history,and the role it plays in modern astronomy.Firstly,the creation of a star map involves meticulous observation and recording of the positions of stars and other celestial objects.Astronomers use specialized tools such as telescopes and cameras to capture the night sky.They then analyze the data,often with the help of computer software,to determine the exact coordinates of each star.This process requires patience and precision,as even the slightest error can lead to inaccuracies in the final map.The history of star maps dates back to ancient civilizations.The Babylonians,for example,were known to have recorded the positions of stars on clay ter,the Greeks and the Chinese also developed their own star maps,which were used for navigation and religious purposes.The most famous of these ancient maps is the Farnese Atlas,a2ndcentury Roman copy of a Greek star map that includes constellations and their mythological stories.Star maps have played a crucial role in the advancement of astronomy.They have allowed scientists to track the movement of celestial bodies,predict celestial events such as eclipses and meteor showers,and even discover new planets and galaxies.The development of more accurate star maps has also contributed to the understanding of the universes structure and the laws that govern it.In the modern era,star maps have evolved with the advent of technology.Digital star maps,for instance,can be accessed through various online platforms and mobile applications.These digital maps offer interactive features,allowing users to explore the night sky in realtime and learn about the stars and constellations they see.Moreover,star maps are not just for professional astronomers.They have become a popular tool for amateur stargazers and educators alike.They provide a means to connect with the cosmos,fostering a sense of wonder and curiosity about the universe.Star maps can also be used as educational tools to teach students about astronomy,constellations, and the science behind celestial movements.In conclusion,the creation and use of star maps have been integral to our understanding of the universe.From ancient civilizations to the modern age,these maps have served as a bridge between Earth and the cosmos,sparking curiosity and inspiring exploration.Aswe continue to develop more advanced technology,the role of star maps in our quest for knowledge about the stars and beyond will only grow.。