Question

1. ) Find a recurrence relation for the number of ternary strings of length n≥1 that do not contain two or more consecutive 2s. (Hint: A ternary string consists of 0s, 1s, and 2s.)

Answer 1

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Solution: Suppose an denote number of ternery shing of length nol that do not contain two or more consecutive 2s. case ( If string start with o then we just need to avoid consecutive as in later n-1 positions. Thus in this case number of such a strings an-1 case If string start with I then again we just need to avoid consecutive 2s in later n- in later n-t positions. Thus in this case number of such a strings = an- Case @ If string start with 2 then it may happen that later part of string can not contains consecutive 25 but together with first position contal will contain consecuhive 2s (for example 2210201---12). For this can take o or I in second position and latter part of string avoid 2 or more consective two's . Thus in this case number of such a shings We on 2 an.z recurrence From case 0, case © and case required relation is an= anyt any + 2an-2 1-2 an= 2an- + 2an-2 with initial condition 41 = 3, 92=8 Thank you.