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编译原理第三章习题答案

时间:2024-02-19 来源:飒榕旅游知识分享网
P64–7 (1)

1(0|1)101 X Y

0 1   1 0 1 X 1 2 3 4 5 Y 1 确定化: 0 1 {X} φ {1,2,3} φ φ φ {1,2,3} {2,3} {2,3,4} {2,3} {2,3} {2,3,4} {2,3,4} {2,3,5} {2,3,4} {2,3,5} {2,3} {2,3,4,Y} {2,3,4,Y} {2,3,5} {2,3,4,} 0

1 0 0 2 3

0 0 1 1 0 0 1 1 4 5 0 6 1 1 1 最小化: *{0,1,2,3,4,5},{6}{0,1,2,3,4,5}{1,3,5} {0,1,2,3,4,5}{1,2,4,6}{0,1,2,3,4},{5},{6}{0,1,2,3,4}{1,3,5}00101{0,1,2,3},{4},{5},{6}{0,1,2,3}{1,3} {0,1,2,3}{1,2,4}{0,1},{2,3}{4},{5},{6}{0,1}{1} {0,1}{1,2}{2,3}01

{0},{1},{2,3},{4},{5},{6} 0

1 2 0

0 0 1 0 0 1 1 3 4 0 1 1 1

0{3} {2,3}{4}15 P64–8 (1)

(1|0)01

(2)

*(1|2|3|4|5|6|7|8|9)(0|1|2|3|4|5|6|7|8|9)(0|5)|(0|5)

*(3)

01(0|101)|10(0|101)

P64–12 (a)

a

a,b

1 0 a

确定化: a {0} {0,1} {0,1} {0,1} {1} {0} φ φ 给状态编号: a 0 1 1 1 2 0 3 3 a

a

0 1

a b b b

b 2 3

a

最小化:

******b {1} {1} φ φ b 2 2 3 3 {0,1},{2,3}{0,1}{1} {0,1}{2,3}

a{0,1},{2},{3}a{0,3} {2,3}b{2}b{3}

a a

b b

1 2 0 a b (b)

b b a

2 3 0 a b a a b b a 4 5 a a 1

已经确定化了,进行最小化 最小化:

{{0,1}, {2,3,4,5}}{0,1}a{1} {0,1}b{2,4}{2,3,4,5}a{1,3,0,5} {2,3,4,5}b{2,3,4,5}{2,4}a{1,0} {2,4}b{3,5}{3,5}a{3,5} {3,5}b{2,4}{{0,1},{2,4},{3,5}}{0,1}a{1} {0,1}b{2,4}{2,4}a{1,0} {2,4}b{3,5}

b b a

1 2 0 a b

a

P64–14

(1) 0 1

0 0

(2):

(010|)

*{3,5}a{3,5} {3,5}b{2,4}1 X Y 2 0 1   1 Y X

0

确定化: 0 {X,1,Y} {1,Y} {1,Y} {1,Y} {2} {1,Y} φ φ 给状态编号: 0 0 1 1 1 2 1 3 3 0

0 1 0 1 0 1 1 1 2 3

0 最小化: 1 {2} {2} φ φ 1 2 2 3 3 0,1},{2,3}{0,1}0{1} {0,1}1{2}{2,3}0{1,3} {2,3}1{3}{0,1},{2},{3}

0 1 1 1

1 3 0 0 0

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