On The Longest Common Subsequence Problem-
General And Variants
A Thesis Submitted
in Partial Ful?llment of the Requirements
for the Degree of
Master of Technology
by
Sudheer Kumar Vootla
to the
DEPARTMENT OF COMPUTER SCIENCE&
ENGINEERING
INDIAN INSTITUTE OF TECHNOLOGY,KANPUR
July2006
Abstract
Let X and Y be any two sequences over an alphabetΣ,where each pair of elements inΣis comparable.The longest common increasing subse-quence(LCIS)problem is to?nd an increasing subsequence S common to both X and Y of greatest length.
For any two sequences X and Y over an alphabetΣ,the longest common subsequence problem of X and Y is to?nd a subsequence S common to both X and Y with greatest length.Unlike the LCIS problem where every pair of input alphabet is comparable,there is no such restriction in the LCS problem.Further the greatest common subsequence found in the LCIS must be strictly increasing,which need not be the case for LCS.
Let X,Y and P be any three sequences over an alphabetΣ.The con-strained longest common subsequence(CLCS)problem of X and Y with re-spect to P is to?nd a longest common subsequence S of X and Y,such that P is a subsequence of S.
In this thesis we propose an algorithm for solving the LCIS problem. Our algorithm takes O((n +m)log log m)time and O(n +mσ)space. Here is the length of the LCIS andσis the size of the alphabet.
We also describe an algorithm for retrieving the longest common sub-sequence(LCS).Our algorithm takes O(m log n+d)time and O(n+|Σ|) space.
Finally,we propose an algorithm for the CLCS problem.Our algorithm takes O(r(m log n+d))time and O(n+α+|Σ|)space to retrieve the CLCS.
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Acknowledgment
My foremost thanks go to Dr.Sanjeev Saxena for his invaluable supervision, guidance and support.He constantly encouraged me and gave many useful ideas throughout my Thesis work.Thank you sir,your guidance is of great help to me for completing my Thesis successfully.I would also like to thank Sitaram,Kiran and Siva Kumar for their help during many restless nights.
I would also like to take this opportunity to thank Lakshman,Swami along with my other lab mates Ravi Kumar,Sunil,Chandan,Abhijit,Sudeepa and Anjali for useful discussions and valuable suggestions.I must also thank our Department of Computer Science and IIT Kanpur for providing me world-class infrastructure with all the facilities and stimulating environment.
My special thanks to my parents who showered immense love on me and brothers for their a?ection and continuous support without which this work would not have been possible.
Sudheer Kumar Vootla
July2006
Indian Institute of Technology Kanpur
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Contents
1Introduction1
1.1Preliminaries (1)
1.2Organization of the rest of the report (2)
2Previous work3
2.1The LCIS problem (3)
2.2The LCS problem (4)
2.3The CLCS problem (5)
3The LCIS Problem6
3.1Preliminaries (6)
3.2An algorithm to?nd the LCIS[4] (8)
3.2.1Algorithm (9)
3.2.2Analysis of the algorithm (9)
3.3Our approach (10)
3.4Data Structure (12)
3.5Algorithm (14)
3.5.1Preprocessing (14)
3.5.2Finding LCIS (15)
3.6Analysis of the algorithm (17)
3.6.1Time Complexity (17)
3.6.2Space Complexity (19)
4The LCS Problem20
4.1Introduction (20)
iii
4.2Hirschberg’s Method (21)
4.3Sebastian’s Method (22)
4.3.1Algorithm (23)
4.3.2Complexity (24)
4.4Our approach (25)
4.4.1Overview (25)
4.4.2Detailed Description (25)
4.4.3Algorithm (26)
4.5Preprocessing (28)
4.6Complexity (28)
4.6.1Time Complexity (28)
4.6.2Space complexity (29)
4.7Proof of correctness (30)
5The CLCS Problem31
5.1Introduction (31)
5.2Chin’s work (31)
5.3He and Arslan’s work (35)
5.4Our approach (37)
5.5Algorithm (39)
5.6Complexity (40)
5.6.1Time complexity (40)
5.6.2Space complexity (40)
6Conclusions41
iv
List of Tables
2.1The LCIS problem (3)
2.2To?nd the length of the LCS (5)
2.3To retrieve the LCS (5)
2.4The CLCS problem (5)
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Chapter1
Introduction
1.1Preliminaries
De?nition1.An alphabet is a?nite non-empty set of symbols represented byΣ.Size of the alphabet is the number of symbols in the set.If each pair of elements in the alphabet is comparable,then the set forms a total order. 2
De?nition2.Sequence is a mapping from set of natural numbers Z+to the setΣof alphabets.2
De?nition3.Let S=a1a2...a n be a sequence over an alphabetΣ.A
sequence a i
1a i2...a i
is said to be a subsequence of S,if1≤i1 i ≤n.For a sequence of length n,total number of possible subsequences are2n.2 De?nition4.Let S=a1a2...a n be a sequence over an alphabetΣ,where each pair of elements inΣis comparable.S is said to be an increasing sequence only if,a1 De?nition5.Let X=a1a2...a m and Y=b1b2...b n,m≥n be two se-quences over an alphabetΣ,where each pair of elements inΣis comparable. A subsequence S is said to be a common increasing subsequence of X and Y,if S is increasing and is also a subsequence for both X and Y. 1 The longest common increasing subsequence(LCIS)problem is to?nd a common increasing subsequence with greatest length.2 De?nition6.Let X=a1a2...a m and Y=b1b2...b n be two sequences over an alphabetΣ.The longest common subsequence problem of X and Y is to?nd a subsequence common to both X and Y of greatest length.2 De?nition7.Let X=a1a2...a m,Y=b1b2...b n and P=p1p2...p r be three sequences of length m,n and r(m≥n≥r)respectively over an alphabetΣ.The constrained longest common subsequence problem of X and Y with respect to P is to?nd a longest common subsequence S of X and Y,such that P is a subsequence of S.The sequences X and Y are said to be input sequences and P the constraint sequence.2 1.2Organization of the rest of the report The report is organized as follows.In chapter2we talk about previous re-sults on the LCIS,LCS and CLCS problems.In chapter3,detailed descrip-tion of LCIS problem,our approach towards the solution and the proposed algorithm are given.In chapter4a linear space algorithm[9]to retrieve the LCS and an algorithm to?nd length of the LCS[14]are discussed.Our approach to the LCS problem,the time,space complexity of our algorithm and the proof of correctness are also given.In chapter5we will discuss the constrained longest common subsequence problem,a dynamic programming algorithm[5]and a linear space algorithm[7].In addition,we discuss our approach,time and space complexity of our algorithm.Finally we conclude in chapter6with conclusions and scope for future work. 2 Chapter2 Previous work 2.1The LCIS problem The longest common increasing subsequence(LCIS)problem has originated from two well known subsequence problems,the longest common subse-quence(LCS)problem and the longest increasing subsequence(LIS)prob-lem. The longest increasing subsequence problem can be solved in O( log m) time,where is the length of LIS(see Knuth’s paper cited in[4]).If the sequence is a permutation of numbers from{1,2,...,m},Bespamyatnikh and Michael[2]give an O(m log log m)time algorithm.In the data streaming model of the LIS problem Liben-Nowell,Vee,and Zhu et al.[13]obtained one pass algorithm that uses O( log m)space and with update time of O(log ) or O(log log m). Paper Time Space [11]O(mn)O(mn) [12]O(n log n+m log m)O(n +m) [6]O(min(r log ,n +r)log log m+m log m)O(r) [3]O((n +m)log logσ+sort(m))-Expected O(n ) Table2.1:The LCIS problem. 3 The LCIS problem was?rst introduced and solved by Yang and Huang et al.[18].They solved the LCIS problem in O(mn)time and O(mn)space, using dynamic programming approach.For applications where the length of the LCIS is known to be small,Katriel and Kut et al.[12]gave an algorithm,which solves the LCIS problem in O(n log n+m log m)time and O(n +m)https://www.doczj.com/doc/f29424295.html,ter Chan and Yong et al.[4]gave an O(min(r log ,n + r)log log m+m log m)time and O(r)space algorithm.In[3]Katriel and Kutz gave an algorithm which solves the problem in O((n +m)log logσ+ sort(m))expected time and O(n )space.See Table2.1for summary of these results. 2.2The LCS problem For two sequences of length m and n(m≥n),the LCS problem can be solved in O(mn)time and space using simple dynamic programming approach[17]. For some special cases Hunt and Szymanski et al.[11]obtained O((d+ n)log n)time algorithm,where d is the total number of matches.This algorithm has better running time when most positions of a sequence match relatively few positions in the other sequence.Hirschberg[10]solved the problem in O(n +n log n)time.Hirschberg[9]give another algorithm using divide and conquer approach to solve the LCS problem with O(mn)time O(m+n)space.Recently Guo and Hwang[6]solved the LCS problem in O(n +m|Σ|)time and O(m|Σ|)space,where is the length of LCS.For various approaches to solve the LCS problem refer to Bergroth’s et al.[1] survey paper. If only length of the LCS is required,Sebastian[14]give an algorithm with time complexity O(m +d).See Table2.2for a summary for?nding length of the LCS and Table2.3for actually?nding a longest common subsequence. 4 Method Time Space [17]O(mn)O(mn) [9]O(mn)O(n) [14]O(m +d)O( ) Table2.2:To?nd the length of the LCS. Method Time Space [17]O(mn)O(mn) [9]O(mn)O(m+n) Table2.3:To retrieve the LCS. 2.3The CLCS problem The constrained longest common subsequence problem was?rst introduced by Tsai in[15].Tsai proposed a dynamic programming algorithm,which solves the the constrained longest common subsequence problem in O(n2m2r) time and O(nmr)space. Chin,Santis,Ferrara,Ho and Kim et al.[5]proposed a new Dynamic Pro-gramming algorithm which solves the problem in O(nmr)time and space. Chin et al.[5]also showed that,the CLCS problem is a special case of the constrained pairwise sequence alignment(CPSA)problem which can also be solved in O(nmr)time and space.By applying Hirschberg’s divide and conquer approach[9]the CLCS problem can be solved in O(nr)space and O(nmr)time. Sebastian[14]solved the CLCS problem in O(r(m +d)+m)time and O(dr)space,where d is the total number of matches between X,Y and is the length of the LCS(X,Y).Table2.4is a summary of these results. Paper Time Space [15]O(n2m2r)O(nmr) [5]O(nmr)O(nr) [14]O(r(m +d)+m)O(dr) Table2.4:The CLCS problem. 5 Chapter3 The LCIS Problem Let X=a1a2...a n and Y=b1b2...b m,m≥n be two sequences over an alphabetΣ.Letσbe the size(number of di?erent symbols)ofΣ.We assume each pair of elements in the alphabet is comparable. The LCIS of X and Y is to?nd a common increasing subsequence S of greatest length that occurs both in X and Y.Next we explain some concepts as in et al.[4] 3.1Preliminaries De?nition8.Pre?x of X,denoted by X i,for0≤i≤n is de?ned as,?rst i symbols in the sequence X.Mathematically X i can be represented as X i=a1a2...a i. X0is a special case and represents the empty pre?x.2 We let a pre?x pair of X and Y be a point on a plane.The pre?x pair [X i,Y j]is mapped to point(i,j).Then all pre?x pairs[X i,Y j](1≤i≤n and 1≤j≤m)forms an(n+1)×(m+1)size square grid of points(including the empty pre?xes).With this representation we have following de?nitions, 6 De?nition9.A match of X and Y is an ordered pair(i,j),1≤i≤n and 1≤j≤m such that a i=b j.The ordered pair(0,0)is also considered as a match for convenience.2 The set of all matches between X and Y is represented by M and is de?ned as M={(i,j)|a i=b j,1≤i≤n and1≤j≤m} From here on we call the?rst co-ordinate of a match as the i-value and the second co-ordinate as the j-value. De?nition10.For a match(i,j)of X and Y,a i(=b j)is called as‘match value’or simply‘value’.The match value of the match(0,0)is considered as the lower of all the match values.2 De?nition11.A match m=(i,j)of X and Y,is said to dominate(or is a dominating match for)another match m =(i ,j )if,i De?nition12.Rank of a match m=(i,j)denoted as R(i,j)is k,if the length of the LCIS ending at the match m is k.Rank of the match(0,0)is considered as0.2 If a match(i,j)is having rank k≥1,then there must be a match(i ,j ) of rank(k?1)such that i From the above discussion the rank function of matches can be expressed as a recurrence relation as follows, R(i,j)=? ?? ?? 0if(i,j)is not a match, 1+k otherwise. where k =max{R(i ,j )|(i ,j )is a dominating match for(i,j)}.(3.1) 7 De?nition13.A chain is de?ned as a sequence of matches that are in-creasing in their i-value,j-value and the match value.2 The dominance relation de?nes a partial order on the set of all matches. With this setup,the the longest common increasing subsequence(LCIS) problem can be now viewed as,the problem to?nd the longest chain in the poset of matches under the dominance relation. 3.2An algorithm to?nd the LCIS[4] From the recurrence relation for the rank of a match,it is su?cient to?nd ranks only for the match points in(n+1)×(m+1)grid of points.The length of the longest common increasing subsequence(LCIS)will be the maximum of all these rank values. To?nd all the matches e?ciently,sort the two sequences in non-decreasing order of the symbol values.Now by making sequential search from begin-ning(like in merge sort sorted sequence)on both of the sequence record all matches between X and Y.The matches will be in the non-decreasing order of their match values.For each match(i,j)in this order?nd ranks.The non-decreasing order of match values ensures that,for a match(i,j)all the dominating matches of(i,j)are processed prior to it.To?nd the dominat-ing matches for(i,j),do sequential search on the matches for which ranks have been calculated so far.Keep a variable to record the largest rank of all matches whose ranks have been calculated so far and record the match t(say)which is having the largest rank.To?nd the actual longest common increasing subsequence,for every match(i,j),record the dominating match (i ,j )from all the dominating matches of(i,j),which has the largest rank. After?nding rank for the last match,length of the LCIS will be recorded in and the match with largest rank will be in t.The actual longest common increasing subsequence can be retrieved by back tracking from the recorded dominating matches. 8 3.2.1Algorithm Algorithm 1 1:procedure LCIS (X,Y ) 2:Sort the sequence X in non-decreasing order of symbol values.3:Sort the sequence Y in non-decreasing order of symbol values.4:Find the set of all matches M between X and Y . 5:Initialize set P with (0,0)and set it’s rank as zero. 6:/*P contains the set of matches whose ranks have been computed so far.*/ 7:for all (i,j )∈M do 8:R (i,j )=1+max {R (i ,j )|(i ,j )∈P and dominates (i,j )}9:/*To ?nd all such (i ,j )make sequential search on set P .*/10:P ←P {(i,j )} 11:end for 12:Output.max {R (i,j )|(i,j )∈P } 13:end procedure 3.2.2Analysis of the algorithm Sorting of the sequences X and Y (line 2-3)in Algorithm 1,can be done in O (n +m )time and space using integer sorting.If the total number of matches between X and Y are d ,then to ?nd all the matches(line 4)takes O (d )time and O (d )space.For a match (i,j )∈M ,to ?nd the rank of (i,j )we are making a sequential search on the set P (line 8)which costs O (d )time.It takes O (d 2)time to compute ranks for the all matches in M .Finding the match with largest rank (line 12)takes O (1)time.Thus the total running time of the algorithm is O (n +m +d +d 2)=O (d 2+n +m ).Space required for this algorithm is O (d ). If the number of matches between X and Y is O (mn ),the algorithm will take O (m 2n 2)time to ?nd LCIS(X,Y ).The critical and most time consum-ing step in the above algorithm is ?nding a dominating match with the great-est rank.Chan and Yong et al.[6]have given two implementations to ?nd LCIS,which take O (d log log log m +m log m )and O ((n +d )log log m +m log m )time respectively.Thus they have an O (min(d log ,n +d )log log m + 9 m log m)time algorithm to?nd LCIS(X,Y). 3.3Our approach De?nition14.Rank of a symbol denoted by R(α),whereα∈Σ,is the number of symbols less than or equal toαin the setΣ.2 LetΣ={α1,α2,...ασ}be the input alphabet andα1<α2<...<ασbe the total order. As in LCIS all symbols beforeαx must be smaller thanαx, Observation1.All matches(i,j)with match valueαx(1≤x≤σ)will have R(i,j)≤R(αx).2 Observation2.All matches with match valueαx(1≤x≤σ)can not have dominating matches with match values greater than or equal toαx.2 De?nition15.[4]x-partition of matches denoted as P x is de?ned as,the set of all matches whose rank is x. P x={(i,j)|R(i,j)=x}.(3.2) 2 From the above de?nition it is easy to see that the set of all matches M is partitioned into -disjoint partitions P1,P2,...P ,where is the length of the LCIS(X,Y). M=P1 P2 ... P .(3.3) To?nd the rank of a match(i,j)of value v,we must compute ranks of all the dominating matches.As from Observation2all dominating matches of (i,j)will have match values less than v;it is su?cient to?nd the ranks of matches in non decreasing order of match values.Let P be the set of all matches with match values no more than v and ranks calculated so far in non decreasing order of match values. 10 To?nd the rank of a match(i,j)with value v+1,?nd x(0≤x≤ ?1)such that,the partition P x has a dominating match and P x+1has no dominating match.Then the match is inserted into the partition P x+1. To?nd such x,we make a sequential search on the partitions from P1to P .There may be more than one match in a partition,but not all these matches are necessary to?nd a dominating match for the match(i,j)in that partition.To show that some matches are redundant in a partition,we need the following de?nition of necessary matches in a partition. De?nition16.[4]In a non empty partition of P,a match(i,j)is called a critical(necessary)match if there is no match(i ,j )with i ≤i and j ≤j. 2 In a partition P x let the set of all critical matches and the set of non critical matches be P C x and P N x respectively.Now we prove,that the set P C x is su?cient to?nd a dominating match in the partition P x. Lemma1.[4]Let P contains all matches of values no more than v.Given a match m=(i,j)of value more than v,for any partition P x,if there is a match in P x dominating m,then there is also a critical match in P x domi-nating m. Proof.[4]Suppose that there is a match(i ,j )in P x dominating m and is not a critical match in P x.By the de?nition of dominating match we have, i As(i ,j )is not a critical match in P x,there must be a critical match(i ,j ) in P x such that,i ≤i and j ≤j ,with any relation between val(i ,j ) and val(i ,j ). From the above,it implies that i From the above discussion to?nd LCIS(X,Y),it is su?cient to?nd ranks only for the critical matches in the partitions.Let P k(0≤k≤ )be any non empty partition of P contains a single match,then it is easy to see 11 that the match is a critical match in P k. De?nition17.Min(k)(x)is de?ned as the minimum j-value of all matches having rank k and i-values are less than x if such matches exists,and+∞otherwise.2 Lemma2.Let P contain all the matches of values no more than v.For an index‘i’in the sequence X,let(i,j)be a match with value greater than v, (i,j)will be a critical match in P k only if Min(k?1)(i) Proof.To become a critical match in P k,for a match(i,j)we need to prove Min(k?1)(i) As P contains matches with values up to v,j=Min(k)(i).Now we prove j To prove minimality,let(i,j1),(i,j2),...(i,j k)be matches with the same i-value and j1 3.4Data Structure To store matches in the partitions we are using Van Emde Boas trees[16]. A Van Emde Boas(VEB)tree is a data structure,which stores a subset of elements from a?xed total ordered set of elements.If size of the?xed set is m,then a VEB-tree supports various operations in O(log log m)time and O(m)space.The operations used in our algorithm are 12 ?Insert(T,x):Inserts an element x into the VEB-tree T. ?Delete(T,x):Delete the element x from VEB-tree T. ?Predecessor(T,x):Gives the largest element t ?Successor(T,x):Gives the least element t>x in VEB-tree T. A Van Emde Boas tree supports all the above operations on a to-tal(linear)order data.But our algorithm needs to support operations such as Min(k)(i),in which no such ordering exists between matches.However in a partition if we order all the critical matches in increasing order of their i-values,then their j-values must be in decreasing order(see lemma3).We store matches as ordered pairs with i-value as the key to insert into VEB-tree and j-value as satellite data.With this structure all four operations have the form Operation(T,(i,j)). Result1.With above structure Min(k)(i)can be obtained as Min(k)(i)=j?value(P redecessor(P k,(i,j d))). Here j d is any dummy value.Predecessor(P k,(i,j d))operation returns a match(i ,j )∈P k with greatest i Lemma3.Suppose that P contains only critical matches in the disjoint partitions.In any partition P x,1≤x≤ ,if we order the matches in ascending order of their i-values,then j-values will be in descending order. Proof.We prove the lemma by contradiction.Let the disjoint partition P x contains only critical matches of rank x.Let the order of matches after sorting them in ascending order of their i-values be(i1,j1),(i2,j2)...(i k,j k). In other words,for i1 We need to prove, j1>j2>j3>......>j k. 13 【摘要】黄自先生是我国杰出的音乐家，他以艺术歌曲的创作最为代表。而黄自先生特别强调了钢琴伴奏对于艺术歌曲组成的重要性。本文是以黄自先生创作的具有爱国主义和人道主义的艺术歌曲《天伦歌》为研究对象，通过对作品分析，归纳钢琴伴奏的弹奏方法与特点，并总结黄自先生的艺术成就与贡献。 【关键词】艺术歌曲；和声；伴奏织体；弹奏技巧 一、黄自艺术歌曲《天伦歌》的分析 （一）《天伦歌》的人文及创作背景。黄自的艺术歌曲《天伦歌》是一首具有教育意义和人道主义精神的作品。同时，它也具有民族性的特点。这首作品是根据联华公司的影片《天伦》而创作的主题曲，也是我国近代音乐史上第一首为电影谱写的艺术歌曲。作品创作于我国政治动荡、经济不稳定的30年代，这个时期，这种文化思潮冲击着我国各个领域，连音乐艺术领域也未幸免――以《毛毛雨》为代表的黄色歌曲流传广泛，对人民大众，尤其是青少年的不良影响极其深刻，黄自为此担忧，创作了大量艺术修养和文化水平较高的艺术歌曲。《天伦歌》就是在这样的历史背景下创作的，作品以孤儿失去亲人的苦痛为起点，发展到人民的发愤图强，最后升华到博爱、奋起的民族志向，对青少年的爱国主义教育有着重要的影响。 （二）《天伦歌》曲式与和声。《天伦歌》是并列三部曲式，为a+b+c，最后扩充并达到全曲的高潮。作品中引子和coda所使用的音乐材料相同，前后呼应，合头合尾。这首艺术歌曲结构规整，乐句进行的较为清晰，所使用的节拍韵律符合歌词的特点，如三连音紧密连接，为突出歌词中号召的力量等。 和声上，充分体现了中西方作曲技法融合的创作特性。使用了很多七和弦。其中，一部分是西方的和声，一部分是将我国传统的五声调式中的五个音纵向的结合，构成五声性和弦。与前两首作品相比，《天伦歌》的民族性因素增强，这也与它本身的歌词内容和要弘扬的爱国主义精神相对应。 （三）《天伦歌》的伴奏织体分析。《天伦歌》的前奏使用了a段进唱的旋律发展而来的，具有五声调性特点，增添了民族性的色彩。在作品的第10小节转调入近关系调，调性的转换使歌曲增添抒情的情绪。这时的伴奏加强和弦力度，采用切分节奏，节拍重音突出，与a段形成强弱的明显对比，突出悲壮情绪。 c段的伴奏采用进行曲的风格，右手以和弦为主，表现铿锵有力的进行。右手为上行进行，把全曲推向最高潮。左手仍以柱式和弦为主，保持节奏稳定。在作品的扩展乐段，左手的节拍低音上行与右手的八度和弦与音程对应，推动音乐朝向宏伟、壮丽的方向进行。coda 处，与引子材料相同，首尾呼应。 二、《天伦歌》实践研究 《天伦歌》是具有很强民族性因素的作品。所谓民族性，体现在所使用的五声性和声、传统歌词韵律以及歌曲段落发展等方面上。 作品的整个发展过程可以用伤感――悲壮――兴奋――宏达四个过程来表述。在钢琴伴奏弹奏的时候，要以演唱者的歌唱状态为中心，选择合适的伴奏音量、音色和音质来配合，做到对演唱者的演唱同步，并起到连接、补充、修饰等辅助作用。 作品分为三段，即a+b+c+扩充段落。第一段以五声音阶的进行为主，表现儿童失去父母的悲伤和痛苦，前奏进入时要弹奏的使用稍凄楚的音色，左手低音重复进行，在弹奏完第一个低音后，要迅速的找到下一个跨音区的音符；右手弹奏的要有棱角，在前奏结束的时候第四小节的t方向的延音处，要给演唱者留有准备。演唱者进入后，左手整体的踏板使用的要连贯。随着作品发展，伴奏与旋律声部出现轮唱的形式，要弹奏的流动性强，稍突出一些。后以mf力度出现的具有转调性质的琶音奏法，要弹奏的如流水般连贯。在重复段落，即“小 我国艺术歌曲钢琴伴奏-精 2020-12-12 【关键字】传统、作风、整体、现代、快速、统一、发展、建立、了解、研究、特点、突出、关键、内涵、情绪、力量、地位、需要、氛围、重点、需求、特色、作用、结构、关系、增强、塑造、借鉴、把握、形成、丰富、满足、帮助、发挥、提高、内心 【摘要】艺术歌曲中，伴奏、旋律、诗歌三者是不可分割的重 要因素，它们三个共同构成一个统一体，伴奏声部与声乐演唱处于 同样的重要地位。形成了人声与器乐的巧妙的结合，即钢琴和歌唱 的二重奏。钢琴部分的音乐使歌曲紧密的联系起来，组成形象变化 丰富而且不中断的套曲，把音乐表达的淋漓尽致。 【关键词】艺术歌曲；钢琴伴奏；中国艺术歌曲 艺术歌曲中，钢琴伴奏不是简单、辅助的衬托，而是根据音乐 作品的内容为表现音乐形象的需要来进行创作的重要部分。准确了 解钢琴伴奏与艺术歌曲之间的关系,深层次地了解其钢琴伴奏的风 格特点,能帮助我们更为准确地把握钢琴伴奏在艺术歌曲中的作用 和地位，从而在演奏实践中为歌曲的演唱起到更好的烘托作用。 一、中国艺术歌曲与钢琴伴奏 “中西结合”是中国艺术歌曲中钢琴伴奏的主要特征之一，作 曲家们将西洋作曲技法同中国的传统文化相结合，从开始的借鉴古 典乐派和浪漫主义时期的创作风格，到尝试接近民族乐派及印象主 义乐派的风格，在融入中国风格的钢琴伴奏写作，都是对中国艺术 歌曲中钢琴写作技法的进一步尝试和提高。也为后来的艺术歌曲写 作提供了更多宝贵的经验，在长期发展中，我国艺术歌曲的钢琴伴 奏也逐渐呈现出多姿多彩的音乐风格和特色。中国艺术歌曲的钢琴 写作中，不可忽略的是钢琴伴奏织体的作用，因此作曲家们通常都以丰富的伴奏织体来烘托歌曲的意境，铺垫音乐背景，增强音乐感染力。和声织体，复调织体都在许多作品中使用，较为常见的是综合织体。这些不同的伴奏织体的歌曲，极大限度的发挥了钢琴的艺术表现力，起到了渲染歌曲氛围，揭示内心情感，塑造歌曲背景的重要作用。钢琴伴奏成为整体乐思不可缺少的部分。优秀的钢琴伴奏织体，对发掘歌曲内涵，表现音乐形象，构架诗词与音乐之间的桥梁等方面具有很大的意义。在不断发展和探索中，也将许多伴奏织体使用得非常娴熟精确。 二、青主艺术歌曲《我住长江头》中钢琴伴奏的特点 《我住长江头》原词模仿民歌风格，抒写一个女子怀念其爱人的深情。青主以清新悠远的音乐体现了原词的意境，而又别有寄寓。歌调悠长，但有别于民间的山歌小曲；句尾经常出现下行或向上的拖腔，听起来更接近于吟哦古诗的意味，却又比吟诗更具激情。钢琴伴奏以江水般流动的音型贯穿全曲，衬托着气息宽广的歌唱，象征着绵绵不断的情思。由于运用了自然调式的旋律与和声，显得自由舒畅，富于浪漫气息，并具有民族风味。最有新意的是，歌曲突破了“卜算子”词牌双调上、下两阕一般应取平行反复结构的惯例，而把下阕单独反复了三次，并且一次比一次激动，最后在全曲的高音区以ff结束。这样的处理突出了思念之情的真切和执著，并具有单纯的情歌所没有的昂奋力量。这是因为作者当年是大革命的参加者，正被反动派通缉，才不得不以破格的音乐处理，假借古代的黄自艺术歌曲钢琴伴奏及艺术成就
我国艺术歌曲钢琴伴奏-精
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