1. Introduction
In the clone library screening, blood testing, multiple access communication, secure key distribution, and others, the group testing (or pooling design) is a mathematical tool to significantly reduce the number of tests (Du and Hwang, 2006). In the classical group testing, a test, which is also called a pool, is performed on an arbitrary subset of all clones with two possible outcomes: a negative outcome if all clones in the subset are negative and a positive outcome otherwise. The task is to identify all positive clones in all clones with group tests as few as possible.
In some applications, some special negative clones are called inhibitors whose effect is to cancel the effect of positive clones. Now, there exist many different inhibitor models that can be formulated by considering different canceled effects, such as 1-inhibitor model (Farach et al., 1997), 1-inhibitor model with error-tolerant version (Hwang and Liu, 2003), k-inhibitor model (Bonis and Vaccaro, 2003), and generalized inhibitor models (Chen et al., 2007). On the contrary, the decoding algorithms become even more difficult due to the experimental errors in practice. With the experimental errors, the test outcomes may consist of false positives or false negatives. In the former, a test yields a positive outcome when a test does not contain any positive clone. Likewise, in the latter, a test yields a negative outcome when a test contains at least one positive clone.
Damaschke (2006) proposed the threshold group testing, which is the generalization of the classic group testing. In threshold group testing problem, let l and u be non-negative integers with l < u, called the lower and upper threshold, respectively. Suppose that a group test for pool S says ‘Yes’ if S contains at least u positives and ‘No’ if at most l positives are present. If the number of positives in S is between l and u, the test can give an arbitrary answer, where l and u are constant and previously known. Furthermore, call g: = u−l−1 the (additive) gap between the thresholds. Obviously, the classical case of group testing is l = 0, u = 1.
Chen and Fu (2009) expanded threshold model to the error-tolerant version, where at most e erroneous outcomes are allowed. To construct an efficient nonadaptive algorithm for the threshold model, they started with the special case that g = 0 and found a relationship between the special case of the threshold model and graph search problems. By using a standard tool for graph search, they proposed an efficient nonadaptive algorithm for the threshold group testing problems.
He et al. (2012) generalized threshold group testing to the threshold model with k-inhibitor and at most e error-tolerant outcomes. In this testing problem, consider a set S of n items consisting of at most d positive items and at most q inhibitors with the other being negative items, where d + q < n and
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$$u + g < d \ll n$$
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, where g = u − l − 1 is the gap between the lower and upper thresholds. Let E be a group test for a subset of S and k be an integer with
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$$1 \le k \le q$$
\end{document}
. The group test E is positive if it contains at least u positive items and at most k−1 inhibitors, and E is negative if it contains at most l positive items or at least k inhibitors. When the number of positives in E is between l and u, and the number of inhibitors is at most k−1 inhibitors, the outcome of E is arbitrary. This testing model is denoted by (n, d, u, l, k, e)-threshold model. Moreover, threshold group testing is also related to guessing secrets (Chung et al., 2001), which is the special case l = 0 and u = p.
Motivated by the above works, in this article, we first advance a new group testing with multiple inhibitor sets and error-tolerant. By using a particular
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$$( d;z )$$
\end{document}
-disjunct matrix (Du and Hwang, 2006), we propose decoding algorithms to identify all positives for the new group testing model and analyze its decoding complexity. Then, we extend the new group testing model to the threshold group testing model and give threshold group testing with multiple inhibitor sets and error-tolerant. By using a particular
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$$( d , r;z ]$$
\end{document}
- disjunct matrix (Du and Hwang, 2006; Chen and Fu, 2009), we propose its decoding algorithms for gap g = 0 and g > 0, respectively. Finally, we point out that the new group testing is the natural generalization for the clone model.
2. Preliminaries
In this section, we briefly review some definitions and basic concepts concerning a group testing.
A group testing (or named pooling design) is usually represented by an incidence matrix M where the columns are the set of items, the rows are the set of tests, and cell (i, j) has a 1-entry if and only if item j is in the test i. For convenience, we view each column as the set of row indices where the column has 1-entry. Thus, we can talk about the union and the intersection of columns. We say that a set X of columns appears (or is contained in) a row if all columns in X have a 1-entry in the row. A test with a positive (negative) outcome is called a positive (negative) test, respectively.
Definition 2.1 (Du and Hwang, 2000) A binary matrix M is said to be (d, r]-disjunct if for any d + r column
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$${C_1} , {C_2} , \cdots , {C_{d + r}}$$
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,
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\begin{align*}
\vert \cap _{i = 1}^r{C_i} \backslash \cup _{i = r + 1}^{d + r}{C_i} \vert \ge 1.
\end{align*}
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Definition 2.2 (Stinson and Wei, 2004; Du and Hwang, 2006) A binary matrix M is said to be (d, r, z]-disjunct if for any d + r column
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$${C_1} , {C_2} , \cdots , {C_{d + r}}$$
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,
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\begin{align*}
\vert \cap _{i = 1}^r{C_i} \backslash \cup _{i = r + 1}^{d + r}{C_i} \vert \ge z
\end{align*}
\end{document}
This means for any d + r column there exist at least z rows where each of the r designated columns has 1-entry and each of the other d columns has 0-entry. It is obvious that a (d, r]-disjunct matrix is (d, r; 1]-disjunct.
Let t(n, d, r; z] denote the minimum number of rows among all (d, r; z]-disjunct matrices with n columns. Engel (1996) first observed the equivalence between a (d, r; 1]-disjunct matrix and a cover of a properly defined hypergraph. Stinson and Wei (2004) eneralized the equivalence to (d, r; 1]-disjunct matrices for general z, but used the equivalence only to derive a lower bound of t(n, d, r; z]. Chen et al. (2008) gave an upper bound of the number of tests required in a pooling design for the complex model, that is,
Lemma 2.1 Chen et al. (2008) for any positive integer, d, r, z, and n, with
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$$d + r \le n$$
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, let
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$$t ( n , d , r;z ]$$
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denote the minimum number of rows among all (d, r; z]-disjunct matrices with n columns, then
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$$t ( n , d , r;z ] < z{ ( w / r ) ^r}{ ( w / d ) ^d} [ 1 + w ( 1 + \log ( n / w + 1 ) ) ]$$
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, where
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$$w = d + r$$
\end{document}
.
3. A New Group Testing Model and Its Decoding Algorithms
In this section, we advance a group testing with multiple inhibitor sets and error-tolerant and give its decoding algorithms.
3.1. A group testing with multiple inhibitor sets and error-tolerant
Recall an inhibitor in the Introduction is a special negative clone whose effect is to cancel the effect of positive clones. In this subsection, we advance a new group testing with multiple inhibitor sets and error-tolerant.
Definition 3.1 Let S be a set of n clones which consist of at most d positive clones and at most q inhibitors with the other being negative clones, where
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$$d + q \ll n$$
\end{document}
. Let P be the set of all positive clones, I be the set of all inhibitors, and N be the set of all negative clones. Moreover, I is composed of
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$$t ( t \ge 1 )$$
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kinds of different inhibitors, that is,
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$$I = \bigcup \nolimits_{j = 1}^t {I_j}$$
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and
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$${I_i} \cap {I_j} = \phi ( i \ne j$$
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and
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$$i , j \in \{ 1 , 2 , \ldots , t \} )$$
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, where
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$${I_j} ( j \in \{ 1 , 2 , \ldots , t \} )$$
\end{document}
is the nonempty subset of I and all inhibitors in Ij have the same property. For every Ij, there exists an integer kj satisfying inequality
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$$1 \le {k_j} \le \vert {I_j} \vert$$
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. Let E be a group test for a subset of S. The test E is positive if it contains at least a positive clone and at most kj−1 inhibitors in Ij for any
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$$j \in \{ 1 , 2 , \ldots , t \} $$
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. Otherwise, E is negative. If we let the function
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$$Outcome ( E )$$
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denote the outcome of test E, then the formulated description of the above group testing with multiple inhibitor sets is as follows.
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\begin{align*}
Outcome ( E ) = \begin{cases} {1 ( positive ) }
\quad& {if \vert E \cap P \vert \ge 1 \ and \ \forall j \in \{ 1
, 2 , \ldots , t \} , \vert E \cap {I_j} \vert \le {k_j} - 1 , }
\\ {0 ( negative ) } & {otherwise.} \\ \end{cases}
\end{align*}
\end{document}
Furthermore, if there are at most e erroneous outcomes in the above group testing, we obtain the group testing model with multiple inhibitor sets and at most e-erroneous outcomes. We use
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
to denote this group testing model.
Remark 3.1 If t = 1 in Definition 3.1, then the
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
is the classical group testing with error-tolerant and inhibitor (Hwang and Liu, 2003).
Definition 3.2 Let x be any element of S. We use
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$$T ( \{ x \} )$$
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to denote the number of negative tests in which the column x appears, and
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$${T^*} ( \{ x \} )$$
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to denote the number of negative tests in which the column x appears and each test contains at least one set R, where
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$$R \subseteq S \backslash \{ x \} $$
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and
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$$\vert R \vert \in \{ {k_1} , {k_2} , \ldots , {k_t} \} $$
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, Then, we define function
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$$f ( \{ x \} ) = T ( \{ x \} ) - {T^*} ( \{ x \} )$$
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. It is clear that
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$$f ( \{ x \} ) \ge 0$$
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by
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$$T ( \{ x \} ) \ge {T^*} ( \{ x \} )$$
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.
3.2. Decoding algorithms for
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
In this subsection, we give an efficient decoding algorithm for
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$$GTMIET$$
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by using the similar method of Thai and Znati (2009) and analyze the algorithm's computational complexity simultaneously.
The
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$$( d , r;z ]$$
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-disjunct matrix has been applied to a variety of problems such as graph search, DNA complexity screening, superimposed codes, and secure key distribution (Du and Hwang, 2006). It has become a major tool to construct decoding algorithms. In this subsection, we use
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$$( d + k + 1;2e + 1 )$$
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-disjunct matrix
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$$( k = \sum \nolimits_{j = 1}^t {k_j} - t + 1 )$$
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and function
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$$f ( \{ x \} )$$
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to identify all positive clones for
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
. Now, we first give an efficient decoding algorithm to identify all positives by using the result and analyze its computational complexity.
Theorem 3.2 Let M be a
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$$( d + k + 1;2e + 1 )$$
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-disjunct matrix of n columns,
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
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. For the
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
model, we have
(1)
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$$f ( \{ x \} ) \le e$$
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, for every
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$$x \in P$$
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; (2)
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$$f ( \{ x \} ) \ge e + 1$$
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, for every
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$$x \;\notin\, P$$
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.
Proof. (1) For any
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$$x \in P$$
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, the outcome of test containing {x} is either negative or positive. Obviously, the positive test is ineffective to compute
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$$T ( \{ x \} )$$
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and
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$${T^*} ( \{ x \} )$$
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. Therefore, we only consider the negative test, which contains {x}. By definition of
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
, it is clear that
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$$T ( \{ x \} ) = {T^*} ( \{ x \} )$$
\end{document}
,
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. For any negative test by the Drawer Principle, each negative test containing {x} must contain at least one of
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$${R_i} \subset {I_i} \ ( 1 \le i \le t )$$
\end{document}
, where
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$$\vert {R_i} \vert = {k_i}.$$
\end{document}
It implies that
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$$f ( \{ x \} ) = 0$$
\end{document}
by Definition 3.2. So, we have
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$$f ( \{ x \} ) \le 0$$
\end{document}
for every
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$$x \in P$$
\end{document}
when no erroneous outcome occurs. Hence, even for the worst case that the e tests are erroneous, we still have
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$$f ( \{ x \} ) \le e$$
\end{document}
for every
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$$x \in P$$
\end{document}
.
(2) When
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$$x \;\notin\, P$$
\end{document}
, by the property of
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$$( d + k + 1;2e + 1 )$$
\end{document}
-disjunctives, there must exist at least 2e + 1 rows containing {x} and the cell in these rows containing {x} column is 1-entry. It implies that
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$$T ( \{ x \} ) \ge 2e + 1$$
\end{document}
and
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$${T^*} ( \{ x \} ) = 0$$
\end{document}
. Hence,
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$$f ( \{ x \} ) = T ( \{ x \} ) - {T^*} ( \{ x \} ) \ge 2e + 1$$
\end{document}
. Even for the worst case that the
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$$e$$
\end{document}
tests are erroneous, we still have
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$$f ( \{ x \} ) \ge e + 1$$
\end{document}
for every \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$x \;\notin\, P$$
\end{document} ■.
The following decoding algorithm is the immediate consequence of Theorem 3.2.
Algorithm 3.1 Identify P under the
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
. Tool: Use a
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$$( d + k + 1; 2e + 1 )$$
\end{document}
-disjunct matrix of n columns, and
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
.
1.
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$$P \leftarrow \emptyset$$
\end{document}
2. for each
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$$x \in S$$
\end{document}
do
3. for any
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$$R \subseteq S \backslash\{ x \} $$
\end{document}
and
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$$\vert R \vert \in \{ {k_1} , {k_2} , \ldots , {k_t} \} $$
\end{document}
do
4. if
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$$f ( \{ x \} ) \le e$$
\end{document}
then
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$$P \leftarrow P \cup \{ x \} $$
\end{document}
5. endfor
6. endfor
7. return P.
Theorem 3.3 For
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
, the decoding complexity of Algorithm 3.1 is
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$$O ( {n^m}logn )$$
\end{document}
,
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$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
.
Proof. Because
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$$f ( \{ x \} )$$
\end{document}
is only dependent on the negative test, let t0 be the number of negative tests. To obtain the set P, it suffices to verify the condition
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$$f ( \{ x \} ) \le e$$
\end{document}
for each element in S and any subset
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$${R_1} , {R_2} , \ldots , {R_t} \subseteq S \backslash \{ x \} $$
\end{document}
, which satisfy
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$$\vert {R_i} \vert = {k_i} ( 1 \le i \le t )$$
\end{document}
, respectively. Without loss of generality, we let
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$${k_1} \le {k_2} \le \ldots \le {k_t}$$
\end{document}
. So the decoding complexity is
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\begin{align*}
{t_0}{n - 1 \choose {k_1}} {{n - 1} \choose {k_2}} \ldots {{n - 1} \choose {k_t}} \le {t_0}{n^{ \sum \nolimits_{j = 1}^t {{k_j}} }} = {t_0}{n^m} , m = \mathop\sum \limits_{j = 1}^t {k_j}.
\end{align*}
\end{document}
Moreover, note that
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$${t_0} \le t ( n , d + k + 1 , 1;2e + 1 ]$$
\end{document}
. By Lemma 2.1, the decoding complexity is
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$$O ( {n^m}logn )$$
\end{document}
,
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$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
. ■
Corollary 3.4 For
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
, there exists efficient decoding algorithm that successfully identifies all positive clones, using no more than
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$$( 2e + 1 ) ( d + k + 2 ) { \left( { { \frac { d + k + 2 } { d + k + 1 } } } \right)^ { d + k + 1 } } \cdot \ \left[ { 1 + ( d + k + 2 ) ( 1 + ln ( \frac { n } { { d + k + 2 } } + 1 ) ) } \right]$$
\end{document}
tests, where
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. Moreover, the decoding complexity is
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$$O ( {n^m}logn )$$
\end{document}
, where
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$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
.
Proof. The theorem is obtained by applying Lemma 2.1 and by the above discussion. ■
4. A New Threshold Group Testing Model and Its Decoding Algorithms
In this section, we first extend the
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
to the threshold model (Damaschke, 2006) and give a threshold group testing with multiple inhibitor sets error-tolerant (for short
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$$ThGTMIET$$
\end{document}
). Then, we give the decoding algorithm to identify all positives for the new threshold model without gap and some set
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$$P^{\prime}$$
\end{document}
with
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$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
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$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
when gap g > 0, respectively, and analyze the computational complexity of these algorithms.
4.1. Threshold group testing with multiple inhibitor sets and error-tolerant
The description of threshold group testing with multiple inhibitor sets and error-tolerant is as follows.
Definition 4.1 Based on
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$$GTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , e )$$
\end{document}
model, let l and u be integers with
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$$0 \le l < u \le d$$
\end{document}
and
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$$d + u + q \ll n$$
\end{document}
that called the lower and upper threshold, respectively. We always suppose that l and u are known constants. Let g = u − l − 1 denote the gap between the lower and upper thresholds. Let E be a group test for a subset of S. The test E is positive if it contains at least u positive clones and at most kj − 1 inhibitors in Ij for any
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$$j \in \{ 1 , 2 , \ldots , t \} $$
\end{document}
; E is negative if it contains at most l positive clones, or it contains at least kj inhibitors in Ij for certain
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$$j ( j \in 1 , 2 , \ldots , t )$$
\end{document}
; when the number of positives in E is between
l
and u, and the number of inhibitors is at most kj − 1 inhibitors in Ij for any
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$$j \in \{ 1 , 2 , \ldots , t \} $$
\end{document}
, the outcome of E is arbitrary, that is, negative or positive. If we let the function Outcome(E) denote the outcome of test E, then the formulated description of the threshold model with different kinds of inhibitors is as follows:
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\begin{align*}
Outcome ( E ) = \begin{cases} {1 ( positive ) } \quad\quad {if
\,\vert E \cap P \vert \ge u \ and \ \forall j \in \{ 1 , 2 ,
\ldots , t \} , \vert E \cap {I_j} \vert \le {k_j} - 1} \\ {0 (
negative ) } \quad\quad {if \,\vert E \cap P \vert \le l \ or \
\exists j \in \{ 1 , 2 , \ldots , t \} , \vert E \cap {I_j}
\vert \ge {k_j}} \\ {0 / 1 ( arbitrary ) } \quad {if \ l < \vert E
\cap P \vert < u \ and \ \forall j \in \{ 1 , 2 , \ldots , t \}
, \vert E \cap {I_j} \vert \le {k_j} - 1} \end{cases}
\end{align*}
\end{document}
Furthermore, if there are at most e erroneous outcomes in above threshold model, we can obtain the threshold model with t kinds of inhibitors and at most e-erroneous outcomes. We use the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
to denote the new threshold group testing. By the k-subset of S, we mean the subset with size k in S.
Remark 4.1 When u = 1 and l = 0, the Definition 4.1 is the Definition 3.1.
Definition 4.2 Let X be k-subset of S. We use T(X) to denote the number of negative tests in which all columns in X appear, and T#(X) to denote the number of negative tests in which all columns in X appear, and each test contains at least one set R, where
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$$R \subseteq S \backslash X$$
\end{document}
and
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$$\vert R \vert \in \{ {k_1} , {k_2} , \ldots , {k_t} \} $$
\end{document}
. Then, we define function
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$$f ( X ) = T ( X ) - {T^ \# } ( X )$$
\end{document}
. It is obvious that
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$$f ( X ) \ge 0$$
\end{document}
since
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$$T ( X ) \ge {T^ \# } ( X )$$
\end{document}
.
4.2. Decoding algorithm for
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap
For
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap (that is g = 0), we propose an efficient decoding algorithm to identify all positive clones in S by using
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$$( d + k - u + 1 , u;2e + 1 ]$$
\end{document}
-disjunct matrix, where
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
, and function f (X). Now, we first give the decoding algorithm to identify all positives by using the result and analyze its computational complexity.
Theorem 4.2 Let M be a
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$$( d + k - u + 1 , u;2e + 1 ]$$
\end{document}
-disjunct matrix of n columns, where
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. For the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap, we have
(1)
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$$f ( X ) \le e$$
\end{document}
, for every u-subset
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$$X \subseteq P$$
\end{document}
;
(2)
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$$f ( X ) \ge e + 1$$
\end{document}
, for every u-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq S$$
\end{document}
and
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$$\vert X \backslash P \vert \ge 1$$
\end{document}
.
Proof. (1) For any u-subset
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$$X \subseteq P$$
\end{document}
, the outcome of test containing X is either negative or positive. Obviously, the positive test is ineffective to compute f (X). Therefore, we only consider the negative test, which contains X. By Definition 4.2, for any
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$$t$$
\end{document}
subset
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$${R_1} \subseteq {I_1} , {R_2} \subseteq {I_2} , \ldots , {R_t} \subseteq {I_t}$$
\end{document}
and
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$$\vert {R_j} \vert = {k_j} , 1 \le j \le t$$
\end{document}
, it is clear that
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$$T ( X ) = {T^ \# } ( X )$$
\end{document}
because
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. By Drawer Principle, every negative test containing X must contain at least one set in
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$${R_1} , {R_2} , \ldots , {R_j}$$
\end{document}
. It implies that f (X) = 0 by definition of
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
. So, we have
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$$f ( X ) \le 0$$
\end{document}
for every u-subset
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$$X \subseteq P$$
\end{document}
when no erroneous outcome occurs. Hence, even for the worst case, the e tests are erroneous, we still have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$f ( X ) \le e$$
\end{document}
for every
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$$u$$
\end{document}
-subset
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$$X \subseteq P$$
\end{document}
.
(2) For each u-subset
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$$X \subseteq S$$
\end{document}
and
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$$\vert X \backslash P \vert \ge 1$$
\end{document}
, X contains at least one clone not in P. Note that |X| = u and g = u − l − 1 = 0. Thus
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$$\vert X \cap P \vert \le u - 1 = l$$
\end{document}
. First, we can choose any t subset
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$${R_1} , {R_2} , \ldots , {R_t}$$
\end{document}
of
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$S \backslash X$$
\end{document}
, which satisfies
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$${R_i} \cap {R_j} = \emptyset , i \ne j$$
\end{document}
and
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$$1 \le i , j \le t , \sum \nolimits_{j = 1}^t \vert {R_i} \vert = k$$
\end{document}
. Let
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$$R = \bigcup \nolimits_{i = 1}^t {R_i}$$
\end{document}
, for such X and
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$${R_1} , {R_2} , \ldots , {R_t}$$
\end{document}
, we can choose a
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$$( d - u + 1 )$$
\end{document}
-subset
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$$Y \subseteq S$$
\end{document}
as follows:
If
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$$\vert P \backslash ( X \cup R ) \vert \ge d - u + 1$$
\end{document}
, then we choose Y such that
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$$Y \subseteq P \backslash ( X \cup R )$$
\end{document}
. Otherwise, we can choose Y such that
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$$P \backslash ( X \cup R ) \subseteq Y \subseteq S \backslash ( X \cup R )$$
\end{document}
.
By the property of
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$$( d + k - u + 1 , u;2e + 1 ]$$
\end{document}
-disjunctions, there must exist at least
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$$2e + 1$$
\end{document}
rows containing X but none of
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$$Y \cup R$$
\end{document}
, and the number of positives in these rows is at most l. In fact, when
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$$P \backslash ( X \cup R ) \subseteq Y$$
\end{document}
, it is clear that these rows contain no more than l positive clones since
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$$\vert X \cap P \vert \le l$$
\end{document}
. In contrast, when
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash ( X \cup R ) \vert \ge d - u + 1$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$Y \subseteq P \backslash ( X \cup R )$$
\end{document}
, we have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash ( Y \cup R ) \vert \le \vert P \backslash Y \vert = \vert P \vert - \vert Y \vert \le d - ( d - u + 1 ) = u - 1 = l$$
\end{document}
. So the number of positives in these rows is at most l too. The test corresponding to these rows should be negative since they contain at most l positive clones. It implies that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$T ( X ) \ge 2e + 1$$
\end{document}
and
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$${T^ \# } ( X ) = 0$$
\end{document}
. Hence
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$$f ( X ) = T ( X ) - {T^ \# } ( X ) \ge 2e + 1$$
\end{document}
. Since
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$${R_1} , {R_2} , \ldots , {R_t}$$
\end{document}
are arbitrary, we have
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$$f ( X ) \ge 2e + 1$$
\end{document}
when no erroneous outcome occurs. Hence, even for the worst case, the e tests are erroneous, we still have
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$$f ( X ) \ge e + 1$$
\end{document}
for every u-subset
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$$X \subseteq S$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \backslash P \vert \ge 1$$
\end{document}
. ■
The following decoding algorithm is the immediate consequence of Theorem 4.2.
Algorithm 4.1 Identify P under the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap.
Tool: Use a
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$$( d + k - u + 1 , u;2e + 1 ]$$
\end{document}
-disjunct matrix of n columns,
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
.
1.
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$$P \leftarrow \emptyset$$
\end{document}
2. for each
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$$u$$
\end{document}
-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq S$$
\end{document}
do
3. for any
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${R_1} , {R_2} , \ldots , {R_t} \subseteq S \backslash X$$
\end{document}
and
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$${R_i} \cap {R_j} = \emptyset$$
\end{document}
, and
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$$\vert {R_j} \vert = {k_j} ( i \ne j , 1 \le i , j \le t )$$
\end{document}
do
4. if
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$$f ( X ) \le e$$
\end{document}
then
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P \leftarrow P \cup X$$
\end{document}
5. endfor
6. endfor
7. return P
Clearly, we easily obtain the following theorem.
Theorem 4.3 For
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap, the decoding complexity of Algorithm 4.1 is
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$$O ( {n^{u + m}}logn )$$
\end{document}
, where
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$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
.
Proof. Because f (X) is only dependent on the negative test, let t0 be the number of negative tests. To obtain the set P, it suffices to verify the condition
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$$f ( X ) \le e$$
\end{document}
for each u-subset X and any t subset
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$${R_1} , {R_2} , \ldots , {R_t} \subseteq S \backslash X$$
\end{document}
, which satisfy
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$${R_i} \cap {R_j} = \emptyset ( i \ne j , 1 \le i , j \le t )$$
\end{document}
and
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$$\bigcup \nolimits_{j = 1}^t \vert {R_j} \vert = k$$
\end{document}
, separately. So the complexity is
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\begin{align*}
{t_0} ( u + k ) { n \choose u} {{n - u} \choose {k_1}} {{n - u - {k_1}} \choose {k_2}} \ldots { {n - u - \sum \nolimits_{j = 1}^{t - 1} {{k_j}} } \choose {k_t}} \le {t_0} ( u + k ) {n^u}{n^{ \sum \nolimits_{j = 1}^{t - 1} {{k_j}} }}.
\end{align*}
\end{document}
Moreover, note that
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$${t_0} \le t ( n , d + k - u + 1 , u;2e + 1 ]$$
\end{document}
by Lemma 2.1, the decoding complexity is
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$$O ( {n^{u + m}}logn )$$
\end{document}
, where
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$$m = \sum \nolimits_{j = 1}^{t - 1} {k_j}$$
\end{document}
. ■
Corollary 4.4 For
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
without gap, there exists efficient decoding algorithm that successfully identifies all positive clones, using no more than
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$$( 2e + 1 ) { \left( { \frac { { d + k + 1 } } { u } } \right)^u } { \left( { { \frac { d + k + 1 } { d + k - u + 1 } } } \right)^ { d + k - u + 1 } } \ \left[ { 1 + ( d + k + 1 ) ( 1 + ln ( \frac { n } { { d + k - 1 } } + 1 ) ) } \right]$$
\end{document}
tests, where
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. Moreover, the decoding complexity is
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$$O ( {n^{u + m}}logn )$$
\end{document}
, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$m = \sum \nolimits_{j = 1}^{t - 1} {k_j}$$
\end{document}
.
Proof. The theorem is obtained by applying Lemma 2.1 and by the above discussion. ■
4.3. Decoding algorithm for
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0
In this subsection, we consider the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0. Things will become more complicated if there is a gap between the upper and lower threshold. Damaschke (2006) showed that to find some set
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$$P^{\prime}$$
\end{document}
with
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$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
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$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
is the best result even for threshold model in which there are no inhibitors. In view of this, for the
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$$ThGTMIET ( n , d , q ( {k_1} , \ {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0, we attempt to find some set
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$$P^{\prime}$$
\end{document}
with the properties mentioned above by using
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$$( d + k - l , u;2e + 1 ]$$
\end{document}
-disjunct matrix, where
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
.
We first give the algorithm of Damaschke's method for threshold model without inhibitors and prove a result using
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$$( d + k - l , u;2e + 1 ]$$
\end{document}
-disjunct matrix. Then, using these abovementioned, we propose an efficient decoding algorithm for the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0 and analyze its computational complexity.
Let
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$$U = \{ X \vert X$$
\end{document}
be a u-subset of S such that
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$$X \subseteq P$$
\end{document}
or
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$$\vert X \backslash P \vert \le g \} $$
\end{document}
. Let
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$$P^{\prime}$$
\end{document}
be the set with
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$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
what we want to obtain and
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$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
. The algorithm description of Damaschke's method (Damaschke, 2006) is as follows:
Algorithm 4.2 Damaschke's method for obtaining set
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$$P^{\prime}$$
\end{document}
with
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$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
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$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
Input: Arbitrarily choose a u-subset
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$$P^{\prime} \in U$$
\end{document}
.
1. while
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$$\vert P^{\prime} \vert < d$$
\end{document}
do
2. for each
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$$( g + 1 )$$
\end{document}
-subset
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$$A \subseteq S \backslash P^{\prime}$$
\end{document}
and
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$$g$$
\end{document}
-subset
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$$B \subseteq P^{\prime}$$
\end{document}
do
3. if exists u-subset
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$$Y \subseteq ( P^{\prime} \cup A ) \backslash B$$
\end{document}
satisfies
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$$Y \in U$$
\end{document}
then goto 6
4. else return
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$$P^{\prime}$$
\end{document}
5. endfor
6.
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$$P^{\prime} \leftarrow ( P^{\prime} \cup A ) \backslash B$$
\end{document}
and
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$$\vert P^{\prime} \vert \leftarrow \vert P^{\prime} \vert + 1$$
\end{document}
7. endwhile
8. return
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$$P^{\prime}$$
\end{document}
The proof of the correctness and computational complexity of Algorithm 4.2 refers to Damaschke (2006) and He et al. (2012). We only give these conclusions as the following lemma:
Lemma 4.5 When Algorithm 4.2 terminates, it outputs set
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$$P^{\prime}$$
\end{document}
with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
, and the computational complexity is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$O ( {n^{g + 1}} )$$
\end{document}
.
Theorem 4.6 Let
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$$M$$
\end{document}
be a
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$$( d + k - l , u;2e + 1 ]$$
\end{document}
-disjunct matrix of n columns,
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$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. For the
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$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0, we have
(1)
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$$f ( X ) \le e$$
\end{document}
, for every u-subset
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$$X \subseteq P$$
\end{document}
;
(2)
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$f ( X ) \ge e + 1$$
\end{document}
, for every u-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq S$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \backslash P \vert \ge g + 1$$
\end{document}
.
Proof. (1) Obviously, the proof is similar to that of the Theorem 4.2. So we omit it here.
(2) For each u-subset
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$$X \subseteq S$$
\end{document}
with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \backslash P \vert \ge g + 1$$
\end{document}
, X contains more than g items not in P. Note that |X| = u and g = u−l−1 > 0. We have
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$$\vert X \cap P \vert \le u - ( g + 1 ) = l$$
\end{document}
. First, we can choose any t subset
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$${R_1} , {R_2} , \ldots , {R_t}$$
\end{document}
of
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$S \backslash X$$
\end{document}
, which satisfies
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${R_i} \cap {R_j} = \emptyset , i \ne j$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$1 \le i , j \le t , \sum \nolimits_{j = 1}^t \vert {R_j} \vert = k$$
\end{document}
. Let
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$R = \bigcup \nolimits_{j = 1}^t {R_j}$$
\end{document}
, for such X and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${R_1} , {R_2} , \ldots , {R_t}$$
\end{document}
, we can choose a
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$( d - l )$$
\end{document}
-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$Y \subseteq S$$
\end{document}
as follows:
If
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash ( X \cup R ) \vert \ge d - l$$
\end{document}
, then we choose Y such that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$Y \subseteq P \backslash ( X \cup R )$$
\end{document}
. Otherwise, we can choose Y such that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P \backslash ( X \cup R ) \subseteq Y \subseteq S \backslash ( X \cup R )$$
\end{document}
.
By the property of
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$( d + k - l , u;2e + 1 ]$$
\end{document}
-disjunctions, there must exist at least 2e + 1 rows containing X but none of
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$Y \cup R$$
\end{document}
, and the number of positives in these rows is at most l. In fact, when
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P \backslash ( X \cup R ) \subseteq Y$$
\end{document}
, clearly these rows contain no more than l positive clones since
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \cap P \vert \le l$$
\end{document}
. On the contrary, when
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash ( X \cup R ) \vert \ge d - l$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$Y \subseteq P \backslash ( X \cup R )$$
\end{document}
, we have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash ( Y \cup R ) \vert \le \vert P \backslash Y \vert = \vert P \vert - \vert Y \vert \le d - ( d - l ) = l$$
\end{document}
. So the number of positives in these rows is at most l too. The test corresponding to these rows should be negative since they contain at most l positive clones. It implies that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$T ( X ) \ge 2e + 1$$
\end{document}
,
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${T^ \# } ( X ) = 0$$
\end{document}
, and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$f ( X ) = T ( X ) - {T^ \# } ( X ) \ge 2e + 1$$
\end{document}
. Note that R is arbitrary. Hence, we have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$f ( X ) \ge 2e + 1$$
\end{document}
when no erroneous outcome occurs. Therefore, even for the worst case that the e tests are erroneous, we have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$f ( X ) \ge e + 1$$
\end{document}
for every
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$u$$
\end{document}
-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq S$$
\end{document}
with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \backslash P \vert \ge g + 1$$
\end{document}
. ■
By applying Theorem 4.6 and the Algorithm 4.2, we can obtain the following algorithm for identifying some set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
.
Algorithm 4.3 Identifies
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
under the
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap g > 0.
Tool: Use a
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$( d + k - l , u;2e + 1 ]$$
\end{document}
-disjunct matrix of n columns,
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
.
1.
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$$U \leftarrow \emptyset$$
\end{document}
2. for each u-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq S$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${R_1} , {R_2} , \ldots , {R_t} \subseteq S \backslash X$$
\end{document}
,
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$${R_i} \cap {R_j} = \emptyset$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert {R_j} \vert = {k_j} ( i \ne j , 1 \le i , j \le t )$$
\end{document}
do
3. if
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$${T^*} ( X ) \le e$$
\end{document}
then
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$U \leftarrow U \cup \{ X \} $$
\end{document}
4. endfor
5. arbitrarily choose an u-subset
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime} \in U$$
\end{document}
as the input of Algorithm 4.2
6. call Algorithm 4.2 and gain a new set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
7. return
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
Now, we prove Algorithm 4.3 and analyze its computational complexity.
Theorem 4.7 The set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
that gives output of Algorithm 4.3 satisfies
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
, and the complexity of Algorithm 4.3 is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$O ( {n^{u + m}}logn )$$
\end{document}
, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
, for fixed parameters
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
.
Proof. By Theorem 4.6, after performed Step 2 to Step 4 of Algorithm 4.3, it is clear that the set U is the set
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$$\{ X \vert X$$
\end{document}
is a
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$u$$
\end{document}
-subset of S such that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$X \subseteq P$$
\end{document}
or
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert X \backslash P \vert \le g \} $$
\end{document}
. So it satisfies the condition of Algorithm 4.2. Moreover, by Theorem 4.3 and Algorithm 4.2, the set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
,that is, output of Algorithm 4.3 must satisfy
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
.
The complexity of Algorithm 4.3 mainly depends on the Step 2 to Step 4 and Algorithm 4.2 (i.e., Step 6 in Algorithm 4.3). By the similar analysis of Theorem 4.3, it is clear that the complexity of Step 2 to Step 4 is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$O ( {n^{u + m}}logn )$$
\end{document}
, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
. By Lemma 4.5, the complexity of Algorithm 4.3 is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$O ( {n^{u + m}}logn ) + O ( {n^{g + 1}} ) = O ( {n^{u + m}}logn )$$
\end{document}
, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
. ■
Corollary 4.8 For the
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$ThGTMIET ( n , d , q ( {k_1} , {k_2} , \ldots , {k_t} ) , u , g , e )$$
\end{document}
with gap
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$g > 0$$
\end{document}
, there exists efficient decoding algorithm that successfully identifies some set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$P^{\prime}$$
\end{document}
with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P \backslash P^{\prime} \vert \le g$$
\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$\vert P^{\prime} \backslash P \vert \le g$$
\end{document}
, using no more than
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$( 2e + 1 ) { \left( { \frac { { d + u + k - l } } { u } } \right) ^u } { \left( { { \frac { d + u + k - l } { d + k - l } } } \right) ^ { d + k - l } } \ \left[ { 1 + ( d + u + k - l ) ( 1 + ln ( \frac { n } { { d + u + k - l } } + 1 ) ) } \right]$$
\end{document}
tests, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$k = \sum \nolimits_{j = 1}^t {k_j} - t + 1$$
\end{document}
. Moreover, the decoding complexity is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$O ( {n^{u + m}}logn )$$
\end{document}
, where
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$m = \sum \nolimits_{j = 1}^t {k_j}$$
\end{document}
.
Proof. The theorem is obtained by applying Lemma 2.1 and by the aforesaid discussion. ■
