1. 2001 년 6 월 진석용 연세대학교 기계전자공학부 The Theory of m-sequence The Run Property of m-sequence

2. Introduction  One of the most remarkable of pseudo- randomness properties of m-sequences is their run-distribution properties.  We investigate an example and prove general result, Theorem. 10.2

3. Definition and Simple Example Def (run of length r) A run of length r in an binary sequence is a subsequence of exactly r consecutive(or) Example (a) The Sequence 01101100 of length 8 has (when viewed cyclically) one 0-run of length 1, no 1-run of length 1 no 0-run of length 2, two 1-run of length 2 one 0-run of length 3, no 1-run of length 3

4. Distribution of run-lengths of m-sequence by example(direct counting) Example (b) Then Consider only the case whenin Example 10.1 of the text.,,,,() We gather some data by counting the number of runs in the m-sequence of length 31 given above by just counting the number of runs. A histogram of the runs is as follows(we view the sequence cyclically).

5. Example(b) — continued length0-runs Total runs 1448 2224 3112 4101 5011 Totals8816 Observation  There are total 16 runs, half have length 1, one-fourth have length 2, one-eighth have length 3.  The number of 0-runs and 1-runs of the same length are equal except for length of 4 & 5. This is a very regular and predictable distribution in a completely random sequence of 0’s and 1’s

6. Example (b) — logical counting Q: Is there a logical method to predict the distribution of run lengths of any m-sequence? A: Yes! (Theorem 10.2) Before general theorem, we observe above example again. Count the runs of various lengths in an m-sequence by distinguishing three cases: length 5, length 4, length 3 1) Length 5 No 0-runs and (just) one 1-runs, by Theorem 10.1 2) Length 4 No 1-runs and (just) one 0-runs (Why? -continued )

7. Logical counting(continued) a. No 1-runs of length 4. 5-gram 11111 appears exactly once by Theorem 10.1, and moreover, 11111 must be sandwiched between. (If not, there would be one more all 5-gram;, Contradiction!) Thus consider subsequence 0111110 of length 7 which appears somewhere in the m-sequence. Within this subsequence two 5-gram s 01111 & 11110 appears. If there were a separate 1-run of length 4, i.e, 1111 somewhere else in the given m- sequence, it too would be sandwiched between, leading to the subsequence 011110, and this would lead to another appearance of the 5- gram s 01111 & 11110. Since by Theorem 10.1, this is impossible, there can be no separate 1-run of length 4.

8. Logical counting(continued) b. One 0-runs of length 4. Consider 5-gram 10000. The next digit of the last 0 must be 1. (Otherwise, there would be subsequence 100000 and consequently 0-run of length 5 occurs in given m-sequence. This is impossible by part a.) Thus 5-gram 10000 and 00001 must occur together as 100001. Obviously, there can not be separate subsequence 0000 somewhere in given m-sequence. 3) Length (will be continued)

9. Logical counting(continued) 3) Length a. Each 1-run of length 3 r=3: 01110 just one 1-run of length 3 r=2: ortwo 1-run of length 3 (Be careful no to count 5-gram of the form.) r=1:four 1-run of length 3 b. The number of each 0-run of length r for is equal to that of 1-run by same argument.

10. Theorem 10.2 Now, we state general result. Theorem 10.2 The run distribution for any m-sequence of length is as given in the table below. length12rm-2m-1mTotals 0-runs110 1-runs101

11. Proof of Theorem. 10.2 Proof: Let ( ), denote the m-sequence in question. Using Theorem 10.1, we can count the runs of various lengths in an m-sequence by distinguishing three cases: length m, length m-1, length If we put m=5, then the proof is exactly same as the argument given in the preceding example. ( 자세한 내용은 또다른 유인물 3, 4 페이지 참조 )