1. 변 우 성변 우 성 1/10 2001 년 6 월 2 일 변 우 성 연세대학교 전기전자공학과 부호 및 정보이론 연구실 Introduction and Explanation of Exercise #5 & #6

2. 변 우 성변 우 성 2/10 Table of Contents  Exercise #5 –Introduction and related Theorem –Explanation of exercise #5 –Discussion  Exercise #6 –Introduction and related Theorem –Explanation of exercise #6 –Discussion

3. 변 우 성변 우 성 3/10 Exercise #5  Introduction  Suppose a binary sequence of length 31 is selected at random. Calculate the expected number of 0-runs and 1-runs of lengths 1, 2, 3, 4, and 5. Compare your result to Theorem 10.2.  Theorem 10.2 Run-length property  Approach  Make a binary random sequence of length 31 by C programming  Find and Calculate the number of 0-runs and 1-runs of each length  Compare

4. 변 우 성변 우 성 4/10 Exercise #5  Explanation(1/2)  Binary random sequence (made by C) –“0000011100001100011111010101101” –Result

5. 변 우 성변 우 성 5/10 Exercise #5  Explanation(2/2)  Result to Theorem 10.2

6. 변 우 성변 우 성 6/10 Exercise #5  Discussion  The number of 0-runs and 1-runs different between m- sequence and arbitrary random sequence  In conclusion, Run-length property is exactly applied to m-sequence

7. 변 우 성변 우 성 7/10 Exercise #6  Introduction  According to Theorem 10.1, each of the 31 different nonzero 5-grams appears once in the m-sequence of length 31 given in Example 10.1. In this problem, we ask you to consider the 6-grams of that sequence. A. How many of the 64 possible 6-grams appear in the length 31 m-sequence of Example 10.1? B. Show that there exists a fixed length 6 vector a with the property that the 6-gram x appears in the m-sequence if and only if x is not zero and x ㆍ a=0. Find a explicitly.  Theorem 10.1 Window property  Approach  This problem is related to generate the m-sequence by the linear feedback shift registers

8. 변 우 성변 우 성 8/10 Exercise #6  Explanation of A  If a repeated 6-gram exists, the period of sequence is sooner than 31, because of recurrence relation  In conclusion, the number of 6-gram is 31  Explanation of B(1/2)  Vector a is a fixed length 6 vector and satisfies x ㆍ a=0  Here vector x is a 6-gram vector x = (x 1, x 2, x 3, x 4, x 5, x 6 )  By the way, x ㆍ a = a 1 x 1 +a 2 x 2 +a 3 x 3 +a 4 x 4 +a 5 x 5 +a 6 x 6 and this is the recursion

9. 변 우 성변 우 성 9/10 Exercise #6  Explanation of B(2/2)  Under circuit to implement the recursion x 6 = x 3 + x 1, whose characteristic polynomial is the primitive polynomial x 5 +x 2 +1  x 6 = x 3 + x 1 x 6 + x 3 + x 1 =0  In conclusion, a 6 = 1, a 3 = 1, a 1 = 1 a = (1, 0, 1, 0, 0, 1) m-sequence output(m=5) x6x6 x5x5 x1x1 x2x2 x3x3 x4x4

10. 변 우 성변 우 성 10/10 Exercise #6  Discussion  Check the result except all zero Total number of 6-grams  If a fixed length m+1 vector a of m-sequence satisfies x ㆍ a=0, vector a is the set of coefficient of recursion