11 Let <p> and <q> be a target class files, and <f> be a birthmark
\r
12 extraction method. Then, <f(p)> and <f(q)> be a extracted birthmarks
\r
13 which elements are <(e^p_1, e^p_2, ..., e^p_n)> and <(e^q_1, e^q_2,
\r
18 Let <L> be a number of matched elements of two birthmarks and same
\r
19 index. Then, the similarity of this method is calculated by
\r
25 |<f(p) cap f(q)>| / |<f(p)>| |<f(q)>|
\r
37 Using this comparison method, birthmarks must have name and its
\r
38 frequency. Threfore, elements of <f(p)> be a set of <(\{name_1,
\r
39 freq_1\}, \{name_2, freq_2\}, ..., \{name_n, freq_n\})>.
\r
41 Next, if <f(p)> have name <FOO> and <f(q)> do not have <FOO>, we add
\r
42 element <\{FOO, 0\}> to f(q). Both birthmarks makes to appearing all
\r
45 Then, the similarity of <f(p)> and <f(q)>is denoted by
\r
47 <norm1 = sqrt(freq^p_1 * freq^p_1 + freq^p_2 * freq^p_2 + ... + freq^p_n * freq^p_n)>
\r
49 <norm2 = sqrt(freq^q_1 * freq^q_1 + freq^q_2 * freq^q_2 + ... + freq^q_n * freq^q_n)>
\r
51 <product = freq^p_1 * freq^q_1 + freq^p_2 * freq^q_2 + ... + freq^p_n * freq^q_n>
\r
53 <similarity = product / (norm1 * norm2)>