336 FINITE STATE MACHINES AND TURING MACHINES [CHAP. 13
13.24. Letf be the functionf (n)=n−2 whenn>1 andf (n)=0 whenn=0 or 1. Show thatf is
computable.
13.25. Letfbe the functionf (x, y)=x. Show thatfis computable.
Answers to Supplementary Problems
13.10. (a)A=(a, b),S={s 0 ,s 1 ,s 2 },Z={x, y, z}, ands 0
is the initial state. (b) See Fig. 13-8(a).
(c)v=y^2 zyzxzxyz.
13.11. (a) See Fig. 13-8. (b) (i)v=xyz^2 x^2 zx^3 z^2 ,
(ii)v=xy^2 xz^3 xyx.
13.12. (a)xy^3 zxyzxz^2 ; (b)xyzxy^2 z^2 x^2 z^2 y^2.
13.13. (a)zyz^2 xy^2 xyzy; (b)zyxy^2 zx^2 zxy^2 xy.
13.14. (a)α=abs 2 baa; (b)α=aabs 3 b; (c)α=s 0 aaabbb;
(d)α=s 01111 B111.
13.15. (a)β=abbs 3 a; (b)β=as 3 baa; (c)β=abbs 2 a;
(d)β=as 3 bba; (e)αa is not changed byq;
(f)β=as 2 baa.
13.16. (a)β=bs 1 Bab; (b)β=s 3 BaBab;(c)β=bs 2 Bab;
(d)β=s 3 BbBab; (e)αis not changed byq;
(f)β=s 2 BaBab.
13.17. α 1 =s 0 ab,α 2 =bs 1 b,α 3 =s 2 bb,α 4 =as 3 b;
q 1 =s 0 abs 1 R,q 2 =s 1 bbs 2 L,q 3 =s 2 bas 3 R,
q 4 =s 3 bbs 0 L.
13.18. Yes.
13.19. No, sinceα→β→α→β→α→β→ ···
never ends.
13.20. q 1 =s 0 BBsNR(NO);q 2 =s 0 bbsNR(NO);
q 3 =s 0 aas 1 R;q 4 =s 1 BBsNR(NO);q 5 =s 1 aasNR
(NO);q 6 =s 1 bbsNR;q 7 =s 2 bbs 2 R;q 8 =s 2 aasNR
(NO);q 9 =s 2 BBsYR(YES).
13.21. q 1 =s 0 BBsNR(NO);q 2 =s 0 bbsNR(NO);
q 3 =s 0 aas 1 R;q 4 =s 1 BBsYR(YES);
q 5 =s 1 bbsNR(NO);q 6 =s 1 aas 2 R;q 7 =s 2 BBsYR
(YES);q 8 =s 2 aasNR(NO);q 9 =s 2 bbsNR(NO).
13.22. (a)〈 6 〉= 17 ; (b)〈m〉= 16 B 1 B 14 B 12 ;
(c)〈m〉= 1 B 1 B1; (d) not defined.
13.23. (a)[E]=7; (b)[E]=2; (c)[E]=14.
13.24. Strategy: Erase first three 1s
q 1 = S 01 Bs 1 R;q 2 = s 1 BBsHR(HALT);q 3 =
S 11 Bs 2 R;q 4 =s 2 BBsHR(HALT);q 5 =s 2 BBsHR
(HALT).
13.25. Strategy: Erase first 1 and then all 1s afterB.
q 1 = s 01 Bs 1 R;q 2 =S 111 s 1 R;q 3 =S 1 BBs 2 R;
q 4 =S 21 Bs 3 R;q 5 = S 31 Bs 3 R;q 6 =S 3 BBsHR
(HALT).
Fig. 13-8