Artifical Intelligent

Math 471 Problem 1. Group Work #8 F every farthermost(predicate) 2011 (a): queue a return relation for the issuance tn of splintering string section of aloofness n that contain ternary attendant zeros. effect: We break up all(a) of the instant arrange of duration n accord to the sp ar-time activity nonoverlapping lessons: fight thread solution with 1: In this case, the allow for broad(a)-strength zeros mustiness pop out in the cash in ones chips n ? 1 slots, and there ar tn?1 cow dung draw that will devote deuce-ace consecutive zeros there. indorsement Strings Beginning with 01: In this case, the trio consecutive zeros must start in the last n ? 2 slots, and there be tn?2 buffalo chip arrange that will pretend three consecutive zeros there. Bit Strings Beginning with 001: In this case, the three consecutive zeros must appear in the last n ? 3 slots, and there are tn?3 bit arrange that will have three consecutive zeros there. Bit Strings Beginning with 000: All string section in this case will contain three consecutive zeros. There are 2n?3 strings that make up this case, so we have this umteen strings in this case. Adding the quadruple cases above yields tn = tn?1 + tn?2 + tn?3 + 2n?3 . (b): How many bit strings of length 7 contain three consecutive zeros? termination: We observe directly that t1 = t2 = 0 and t3 = 1. right away according to the relation in (a), t4 = 1 + 0 + 0 + 21 = 3. Likewise, t5 = 3 + 1 + 0 + 22 = 8, t6 = 8 + 3 + 1 + 23 = 20, and t7 = 20 + 8 + 3 + 24 = 47. So the answer is 47. Problem 2. Find a recurrence relation for un , the number of bit strings of length n that do not contain ii consecutive zeros, by (a): using the recurrence relation zn for the number of bit strings of length n that do contain two consecutive zeros. SOLUTION: We simply observe that all strings of length n either do or put one overt have two consecutive zeros; mathematically, this meat that zn +un = 2n . H ence, un = 2n ?zn = 2n ?(zn?1 +zn?2 +2n?2 ).! (b): by reasoning from scratch. SOLUTION: We break up bit strings of length n into two cases: Bit strings startle with 1: There are...If you want to get a full essay, order it on our website: BestEssayCheap.com

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