欢迎来到麦多课文档分享! | 帮助中心 海量文档,免费浏览,给你所需,享你所想!
麦多课文档分享
全部分类
  • 标准规范>
  • 教学课件>
  • 考试资料>
  • 办公文档>
  • 学术论文>
  • 行业资料>
  • 易语言源码>
  • ImageVerifierCode 换一换
    首页 麦多课文档分享 > 资源分类 > PPT文档下载
    分享到微信 分享到微博 分享到QQ空间

    Asynchronous Pattern Matching -Metrics.ppt

    • 资源ID:378677       资源大小:896KB        全文页数:34页
    • 资源格式: PPT        下载积分:2000积分
    快捷下载 游客一键下载
    账号登录下载
    微信登录下载
    二维码
    微信扫一扫登录
    下载资源需要2000积分(如需开发票,请勿充值!)
    邮箱/手机:
    温馨提示:
    如需开发票,请勿充值!快捷下载时,用户名和密码都是您填写的邮箱或者手机号,方便查询和重复下载(系统自动生成)。
    如需开发票,请勿充值!如填写123,账号就是123,密码也是123。
    支付方式: 支付宝扫码支付    微信扫码支付   
    验证码:   换一换

    加入VIP,交流精品资源
     
    账号:
    密码:
    验证码:   换一换
      忘记密码?
        
    友情提示
    2、PDF文件下载后,可能会被浏览器默认打开,此种情况可以点击浏览器菜单,保存网页到桌面,就可以正常下载了。
    3、本站不支持迅雷下载,请使用电脑自带的IE浏览器,或者360浏览器、谷歌浏览器下载即可。
    4、本站资源下载后的文档和图纸-无水印,预览文档经过压缩,下载后原文更清晰。
    5、试题试卷类文档,如果标题没有明确说明有答案则都视为没有答案,请知晓。

    Asynchronous Pattern Matching -Metrics.ppt

    1、Asynchronous Pattern Matching - Metrics,Amihood Amir CPM 2006,Motivation,Motivation,In the “old” days: Pattern and text are given in correct sequential order. It is possible that the content is erroneous. New paradigm: Content is exact, but the order of the pattern symbols may be scrambled. Why? Tra

    2、nsmitted asynchronously? The nature of the application?,Example: Swaps,Tehse knids of typing mistakes are very common So when searching for pattern These we are seeking the symbols of the pattern but with an order changed by swaps. Surprisingly, pattern matching with swaps is easier than pattern mat

    3、ching with mismatches (ACHLP:01),Example: Reversals,AAAGGCCCTTTGAGCCC AAAGAGTTTCCCGGCCC Given a DNA substring, a piece of it can detach and reverse. This process still computationally tough. Question: What is the minimum number of reversals necessary to sort a permutation of 1,n,Global Rearrangement

    4、s?,Berman & Hannenhalli (1996) called this Global Rearrangement as opposed to Local Rearrangement (edit distance). Showed it is NP-hard.Our Thesis: This is a special case of errors in the address rather than content.,Example: Transpositions,AAAGGCCCTTTGAGCCC AATTTGAGGCCCAGCCCGiven a DNA substring, a

    5、 piece of it can be transposed to another area. Question: What is the minimum number of transpositions necessary to sort a permutation of 1,n ?,Complexity?,Bafna & Pevzner (1998), Christie (1998), Hartman (2001): 1.5 Polynomial Approximation.Not known whether efficiently computable.This is another s

    6、pecial case of errors in the address rather than content.,Example: Block Interchanges,AAAGGCCCTTTGAGCCC AAGTTTAGGCCCAGCCC Given a DNA substring, two non-empty subsequences can be interchanged. Question: What is the minimum number of block interchanges necessary to sort a permutation of 1,n ? Christi

    7、e (1996): O(n ),2,A General-Purpose Metric,Options: 1. count interchanges,interchange,interchange matches,S1=bbaca S2=bbaac,2. L1 , L2 ,or any other metric on the address. Example: AGGTTCCAATC1 22 1 12 215 11 GTAGCAACTCT,In This Talk:,We concentrate on counting the interchanges As a metric. (we also

    8、 have results on the L2 metric, partial results on L1, and Address register errors)We have a pedagogical reason for this,Summary,Biology: sorting permutations Reversals (Berman & Hannenhalli, 1996) Transpositions (Bafna & Pevzner, 1998),Pattern Matching: Swaps (Amir, Lewenstein & Porat, 2002),NP-har

    9、d ?,Block interchanges O(n2) (Christie, 1996),O(n log m),Note: A swap is a block interchange simplification,1. Block size,2. Only once,3. Adjacent,Edit operations map,Reversal, Transposition, Block interchange: 1. arbitrary block size 2. not once 3. non adjacent4. permutation 5. optimization Interch

    10、ange: 1. block of size 1 2. not once 3. non adjacent4. permutation 5. optimization Generalized-swap: 1. block of size 1 2. once 3. non adjacent4. repetitions 5. optimization/decision Swap: 1. block of size 1 2. once 3. adjacent4. repetitions 5. optimization/decision,interchange,interchange matches,S

    11、1=bbaca S2=bbaac,generalized-swapmatches,S1=bbaca S2=bcaba,Definitions,Generalized Swap Matching,INPUT: text T0n, pattern P0m OUTPUT: all i s.t. P generalized-swap matches Tii+m,Reminder: Convolution The convolution of the strings t1n and p1m is the string t*p such that:,Fact: The convolution of n-l

    12、ength text and m-length pattern can be done in O(n log m) time using FFT.,In Pattern Matching,Convolutions:,O(n log m) using FFT,b0 b1 b2,b0 b1 b2,b0 b1 b2,Problem: O(n log m) only in algebraically closed fields, e.g. C.,Solution: Reduce problem to (Boolean/integer/real) multiplication. S,This reduc

    13、tion costs!,Example: Hamming distance.,Counting mismatches is equivalent to Counting matches,A B A B C A B B B A,Example:,Count all “hits” of 1 in pattern and 1 in text.,1 0 1,1 0 1,1 0 1,For,Define:,1 if a=b,0 o/w,Example:,For,Do:,+,+,Result: The number of times a in pattern matches a in text + the

    14、 number of times b in pattern matches b in text + the number of times c in pattern matches c in text.,Idea: assign natural numbers to alphabet symbols, and construct: T: replacing the number a by the pair a2,-a P: replacing the number b by the pair b, b2. Convolution of T and P gives at every locati

    15、on 2i:j=0mh(T2i+j,Pj) where h(a,b)=ab(a-b). 3-degree multivariate polynomial.,Generalized Swap Matching: a Randomized Algorithm,Generalized Swap Matching: a Randomized Algorithm,Since: h(a,a)=0 h(a,b)+h(b,a)=ab(b-a)+ba(a-b)=0, a generalized-swap match 0 polynomial.,Example: Text: ABCBAABBC Pattern:

    16、CCAABABBB,1 -1, 4 -2, 9 -3,4 -2,1 -1,1 -1,4 -2,4 -2,9 -3 3 9, 3 9, 1 1,1 1,2 4, 1 1,2 4, 2 4,2 4,3 -9,12 -18,9 -3,4 -2,2 -4,1 -1,8 -8,8 -8,18 -12,Problem: It is possible that coincidentally the result will be 0 even if no swap match.Example: for text ace and pattern bdf we get a multivariate degree

    17、3 polynomial:We have to make sure that the probability for such a possibility is quite small.,Generalized Swap Matching: a Randomized Algorithm,Generalized Swap Matching: a Randomized Algorithm,What can we say about the 0s of the polynomial?,By Schwartz-Zippel Lemma prob. of 0degree/|domain|.Conclud

    18、e:,Theorem: There exist an O(n log m) algorithm that reports all generalized-swap matches and reports false matches with prob.1/n.,Generalized Swap Matching: De-randomization?,Can we detect 0s thus de-randomize the algorithm?,Suggestion: Take h1,hk having no common root.,It wont work, k would have t

    19、o be too large !,Generalized Swap Matching: De-randomization?,Theorem: (m/log m) polynomial functions are required to guarantee a 0 convolution value is a 0 polynomial.,Proof: By a linear reduction from word equality. Given: m-bit words w1 w2 at processors P1 P2 Construct: T=w1,1,2,m P=1,2,m,w2. Now

    20、, T generalized-swap matches P iff w1=w2.,Communication Complexity: word equality requires exchanging (m) bits, We get: klog m= (m), so k must be (m/log m).,P1 computes: w1 * (1,2,m),log m bit result,P2 computes: (1,2,m) * w2,Interchange Distance Problem,INPUT: text T0n, pattern P0m OUTPUT: The mini

    21、mum number of interchanges s.t. Tii+m interchange matches P.,Reminder: permutation cycle The cycles (143) 3-cycle, (2) 1-cycle represent 3241. Fact: The representation of a permutation as a product of disjoint permutation cycles is unique.,Interchange Distance Problem,Lemma: Sorting a k-length permu

    22、tation cycle requires exactly k-1 interchanges. Proof: By induction on k.,Theorem: The interchange distance of an m-length permutation is m-c(), where c() is the number of permutation cycles in .,Result: An O(nm) algorithm to solve the interchange distance problem.,A connection between sorting by in

    23、terchanges and generalized-swap matching?,Cases: (1), (2 1), (3 1 2),Interchange Generation Distance Problem,INPUT: text T0n, pattern P0m OUTPUT: The minimum number of interchange- generations s.t. Tii+m interchange matches P.,Definition: Let S=S1,S2,Sk=F, Sl+1 derived from Sl via interchange Il. An

    24、 interchange-generation is a subsequence of I1,Ik-1 s.t. the interchanges have no index in common.,Note: Interchanges in a generation may occur in parallel.,Interchange Generation Distance Problem,Lemma: Let be a cycle of length k2. It is possible to sort in 2 generations and k-1 interchanges. Examp

    25、le: (1,2,3,4,5,6,7,8,0)generation 1:(1,8),(2,7),(3,6),(4,5)(8,7,6,5,4,3,2,1,0)generation 2:(0,8),(1,7),(2,6),(3,5)(0,1,2,3,4,5,6,7,8),Interchange Generation Distance Problem,Theorem: Let maxl() be the length of the longest permutation cycle in an m-length permutation . The interchange generation dis

    26、tance of is exactly: 0, if maxl()=1. 1, if maxl()=2. 2, if maxl()2.,Note: There is a generalized-swap match iff sorting by interchanges is done in 1 generation.,Open Problems,1. Interchange distance faster than O(nm)? 2. Asynchronous communication different errors in address bits. 3. Different error measures than interchange/block interchange/transposition/reversals for errors arising from address bit errors.,Note: The techniques employed in asynchronous pattern matching have so far proven new and different from traditional pattern matching.,The End,


    注意事项

    本文(Asynchronous Pattern Matching -Metrics.ppt)为本站会员(王申宇)主动上传,麦多课文档分享仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知麦多课文档分享(点击联系客服),我们立即给予删除!




    关于我们 - 网站声明 - 网站地图 - 资源地图 - 友情链接 - 网站客服 - 联系我们

    copyright@ 2008-2019 麦多课文库(www.mydoc123.com)网站版权所有
    备案/许可证编号:苏ICP备17064731号-1 

    收起
    展开