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3. (20 points) Let alphabet 101 1 Alphabet contains all columns of 0s and is of height two. A string of symbols in gives two rows of Os and 1s. Consider each row to be an unsigned binary number and let language L-{w over | the reverse of the bottom row of w is two times the reverse of the top row) Examples, 11 11 [01 [o 0] [17 11 [11 011 [0 Please construct a minimum DFA for language L, that should read an input string from left to right (15 points). please briefly describe the strings corresponding to each state (5 points) Page 1 of 2 3 Question 3 For 1 vector: For 2 vector For 3 vector: 0 1 0 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 For 4 vectors(1 value), like 8-4*2, 4-2*2, 2-2*1: 0 0 1 0 [0 1 0 0 [1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 For 4 vectors(2 value start 8), like 8+4=(4+2)*2, 8+2=(4+1)*24+2=(2+1)*2 0 1 1 0 1 1 0 0 1 0 1 0 0 0 1 1 0 1 1 0 0 1 0 1 For 4 vectors(3 values),like 8+4+2 (4+2+1)*2 For 5 vectors(1 value),like 16-8*2,8-4*2, 4-2*1, 2-1*2 010 0 1 0 0][0 1 0001[1 0 0 0 0 0 0 10 0 0 1 0 0 01 0 0 0 1 0 0 0 For 5 vectors(2 value) ,like 16+8=(844)*2,844-(4+2)*2, 4t2-(2+1)*2 0 01 1 0 [0 1 1 0 0 [1 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 For 5 vectors(2 value),like 16+4 (8+2)*2,8+2 (4+2)*2 0 1 0 1 01 0 1 0 0 0 0 1 0 1 0 1 0 1 0 For 5 vectors(2 value),like 16+2-2*(8+1) 0 1 0 0 1 For 5 vectors (3 values),like 16+8+4-(8+4+2)*2, (8+4+2)*2-(4+2+1)*2 1 1 1 0 01 [0 1 1 1 0 0 1 1 1 00 0 1 1 1 For 5 vectors(3 values) (16+8+2 (8+4+1)*2): 1 1 0 1 0 For 5 vectors(4 values) 3 For 6 vectors(1 values), like 32-16 2, 16-8*2, 8-4*2, 4-2 2, 2-1*2 00001010 0 0 1 0 0][0 0 1 0 0 0][0 1 0 0 0 0][1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 00 For 6 vectors(2 values), like 32+16 (16+8)*2, 16+8-(8+4)*2, 8+4 (4+2)*2, 4+2 (2+1)*2 11000010 1 1 0 0 0][0 0 1 1 0010 0 0 1 1 0 0 1 1 0 0 0 0 01 1 0 0 0 0 0 11 0l0 0 0 0 1 1 For 6 vectors(2 values), like 32+8=(16+4)*2, 16+4=(842)*2, 8+2 (4+1)*2, 00101010 1 0 1 0 0]1 0 1 0 0 0 0 0 0 1 0 10 0 1 0 1 0 01 0 1 0 0 For 6 vectors(2 values), like 32+4 (16+2)*2, 16+2-(8+1)*2, 1 0 0 1 0 01 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 For 6 vectors(2 values), like 32+2 (16+1)*2, 1 0 0 0 1 0 0 1 0 0 0 1 For 6 vectors(3 values), like 2+4+8 (1+2+4)*2, 1 1 1 0 0 01 [0 1 1 1 0 01 [0 0 1 1 1 0 0 1 1 1 0 0 0 0 1 1 1 0 0 0 0 11 1 For 6 vectors(3 values), like 2+4+16 (1+2+8)*2, 1 1 0 1 0 0 [0 1 1 0 1 0 0 11 0 1 0 0 0 1 1 0 1 For 6 vectors (3 values), like 2+4+32-(1+2+16)*2, 0 1 1 0 0 1 For 6 vectors(4 values), like 2+4+8+161+2+4+8)*2,4+8+16+32 (2+4+8+16) 2 1 11 1 0 01 [0 1 1 1 1 0 0 1 1 1 1 0 001 1 1 1 For 6 vectors(4 values), like 2+4+8+32-1+2+4+16)*2 1 1 1 0 1 0 0 1 1 1 0 1 (0 1 80 S1 Figure 3: minimum DFA Figure 4: minimum NFA For the fourth part: 0 For the final part ntimesofo ntimesof0 So the graph is easily to get the figure 3