{"title":"“迹线重建的下限”勘误表","authors":"N. Holden, R. Lyons","doi":"10.1214/22-aap1827","DOIUrl":null,"url":null,"abstract":"Lemma 3.1 asserts that Eyn[Z(̃yn)] − Exn[Z(̃xn)] = (n−1/2) and Eyn[Z(̃yn)] > Exn[Z(̃xn)] for all sufficiently large n. Our proof was not correct: As Benjamin Gunby and Xiaoyu He pointed out to us, we missed four terms in the computation of equation (3.3). Those terms contribute a negative amount, so the proof is more delicate. Here is a correct proof. The intuition behind the result is that a string with a defect of the type we consider, namely, a 10 in a string of 01’s, is likely to cause more 11’s in the trace than a string without the defect. Since the defect in yn is shifted to the right as compared to the defect in xn, the defect of yn is slightly more likely to “fall into” the test window { 2np + 1 , . . . , 2np + √npq } of the trace than is the defect of xn. More precisely, the difference in probability is of order n−1/2. In the proof below, we make this intuition rigorous. PROOF. We assume throughout the proof that k ∈ { 2np + 1 , . . . , 2np + √npq }. Let E(m,k) denote the event that bit m in the input string is copied to position k in the trace. First observe that","PeriodicalId":1,"journal":{"name":"Accounts of Chemical Research","volume":null,"pages":null},"PeriodicalIF":16.4000,"publicationDate":"2022-08-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Erratum to “Lower bounds for trace reconstruction”\",\"authors\":\"N. Holden, R. Lyons\",\"doi\":\"10.1214/22-aap1827\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"Lemma 3.1 asserts that Eyn[Z(̃yn)] − Exn[Z(̃xn)] = (n−1/2) and Eyn[Z(̃yn)] > Exn[Z(̃xn)] for all sufficiently large n. Our proof was not correct: As Benjamin Gunby and Xiaoyu He pointed out to us, we missed four terms in the computation of equation (3.3). Those terms contribute a negative amount, so the proof is more delicate. Here is a correct proof. The intuition behind the result is that a string with a defect of the type we consider, namely, a 10 in a string of 01’s, is likely to cause more 11’s in the trace than a string without the defect. Since the defect in yn is shifted to the right as compared to the defect in xn, the defect of yn is slightly more likely to “fall into” the test window { 2np + 1 , . . . , 2np + √npq } of the trace than is the defect of xn. More precisely, the difference in probability is of order n−1/2. In the proof below, we make this intuition rigorous. PROOF. We assume throughout the proof that k ∈ { 2np + 1 , . . . , 2np + √npq }. Let E(m,k) denote the event that bit m in the input string is copied to position k in the trace. First observe that\",\"PeriodicalId\":1,\"journal\":{\"name\":\"Accounts of Chemical Research\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":16.4000,\"publicationDate\":\"2022-08-01\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Accounts of Chemical Research\",\"FirstCategoryId\":\"100\",\"ListUrlMain\":\"https://doi.org/10.1214/22-aap1827\",\"RegionNum\":1,\"RegionCategory\":\"化学\",\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q1\",\"JCRName\":\"CHEMISTRY, MULTIDISCIPLINARY\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Accounts of Chemical Research","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.1214/22-aap1827","RegionNum":1,"RegionCategory":"化学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q1","JCRName":"CHEMISTRY, MULTIDISCIPLINARY","Score":null,"Total":0}
Erratum to “Lower bounds for trace reconstruction”
Lemma 3.1 asserts that Eyn[Z(̃yn)] − Exn[Z(̃xn)] = (n−1/2) and Eyn[Z(̃yn)] > Exn[Z(̃xn)] for all sufficiently large n. Our proof was not correct: As Benjamin Gunby and Xiaoyu He pointed out to us, we missed four terms in the computation of equation (3.3). Those terms contribute a negative amount, so the proof is more delicate. Here is a correct proof. The intuition behind the result is that a string with a defect of the type we consider, namely, a 10 in a string of 01’s, is likely to cause more 11’s in the trace than a string without the defect. Since the defect in yn is shifted to the right as compared to the defect in xn, the defect of yn is slightly more likely to “fall into” the test window { 2np + 1 , . . . , 2np + √npq } of the trace than is the defect of xn. More precisely, the difference in probability is of order n−1/2. In the proof below, we make this intuition rigorous. PROOF. We assume throughout the proof that k ∈ { 2np + 1 , . . . , 2np + √npq }. Let E(m,k) denote the event that bit m in the input string is copied to position k in the trace. First observe that
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Accounts of Chemical Research presents short, concise and critical articles offering easy-to-read overviews of basic research and applications in all areas of chemistry and biochemistry. These short reviews focus on research from the author’s own laboratory and are designed to teach the reader about a research project. In addition, Accounts of Chemical Research publishes commentaries that give an informed opinion on a current research problem. Special Issues online are devoted to a single topic of unusual activity and significance.
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