arquitetura- 6/9/19

	ABCD	S
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001
10	1010
11	1011
12	1100
13	1101
14	1110
15	1111

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contagem só pode mudar um bit por vez -> de 01 para 11, não 10

AB\CD	00	01	11	10
00		0	1	1	1
01		0	1	1	0
11		0	1	0	0
10		0	0	1	0
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lista geral 6
b)  f(A,B,C,D) = {0,1,2,4,8,10,11,12}

	ABCD	S
0	0000	1
1	0001	1
2	0010	1
3	0011	0
4	0100	1
5	0101	0
6	0110	0
7	0111	0
8	1000	1
9	1001	0
10	1010	1
11	1011	1
12	1100	1
13	1101	0
14	1110	0
15	1111	0


AB\CD	00	01	11	10
00		1	1	0	1
01		1	0	0	0
11		1	0	0	0
10		1	0	1	1


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a) f(A,B,C,D) = {0,1,11,13,14,15}

	ABCD	S
0	0000	1
1	0001	1
2	0010	0
3	0011	0
4	0100	0
5	0101	0
6	0110	0
7	0111	0
8	1000	0
9	1001	0
10	1010	0
11	1011	1
12	1100	0
13	1101	1
14	1110	1
15	1111	1


AB\CD	00	01	11	10
00		1	1	0	0
01		0	0	0	0
11		0	1	1	1
10		0	0	1	0
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8)
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A	{0,2,3,5,6,7,8,9,10,12,14,15}

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B	{0,1,2,3,4,7,8,9,10,13}

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C	{0,1,3,4,5,6,7,8,9,10,11,13}

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D	{0,2,3,5,6,8,11,12,13,14}

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E	{0,2,6,8,10,11,12,13,14,15}

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F	{0,4,5,6,8,9,10,11,12,14,15}

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G	{2,3,4,5,6,8,9,10,11,13,14,15}

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