Re: [問題] DLD 投影片-6
給ㄧ個logic gate的set
要看它滿不滿足completeness
只要看這個set任意組合出來的電路
能夠符合每個function
就是COMPLETENESS
我舉個例
如有2個logic variable--A B
則FUNCTION有2的2的2次方 也就是16種
只要能用gate的set完成(在此我用and or not示範)
就符合completeness
(A,B)---->OUTPUT
FUNCTION1(A*NOT A)
(0,0)->0
(0,1)->0
(1,0)->0
(1,1)->0
FUCTION2(A+NOT A)
(0,0)->1
(0,1)->1
(1,0)->1
(1,1)->1
FUCTION3(NOT(A+B))
(0,0)->1
(0,1)->0
(1,0)->0
(1,1)->0
FUCTION4(NOT A*B)
(0,0)->0
(0,1)->1
(1,0)->0
(1,1)->0
FUCTION5(A*NOT B)
(0,0)->0
(0,1)->0
(1,0)->1
(1,1)->0
FUCTION6(A*B)
(0,0)->0
(0,1)->0
(1,0)->0
(1,1)->1
FUCTION7(NOT A)
(0,0)->1
(0,1)->1
(1,0)->0
(1,1)->0
FUCTION8(NOT B)
(0,0)->1
(0,1)->0
(1,0)->1
(1,1)->0
FUCTION9(A*B+NOT A*NOT B)
(0,0)->1
(0,1)->0
(1,0)->0
(1,1)->1
FUCTION10((NOT A+NOT B)*(A+B))
(0,0)->0
(0,1)->1
(1,0)->1
(1,1)->0
FUCTION11(B)
(0,0)->0
(0,1)->1
(1,0)->0
(1,1)->1
FUCTION12(A)
(0,0)->0
(0,1)->0
(1,0)->1
(1,1)->1
FUCTION13(NOT A+NOT B)
(0,0)->1
(0,1)->1
(1,0)->1
(1,1)->0
FUCTION14(A+NOT(A+B))
(0,0)->1
(0,1)->0
(1,0)->1
(1,1)->1
FUCTION15((NOT A*NOT B)+B)
(0,0)->1
(0,1)->1
(1,0)->0
(1,1)->1
FUCTION16(A+B)
(0,0)->0
(0,1)->1
(1,0)->1
(1,1)->1
※ 引述《gglk (錦州挖挖)》之銘言：
: 老師不好意思，
: 也許我問的有些問題您上課有講過，
: 或是您覺得很直觀，
: 不過還是請您指點一下愚昧的學生。
: 請問
: 2.最後一行，可以說明一下NUMBERS OF FUNCTIONS的定義對於證明COMPLETENESS
: 是怎樣USEFUL嗎?
: 謝謝!
--
※ 發信站: 批踢踢實業坊(ptt.cc)
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