看板 CSIE_ASM 關於我們 聯絡資訊
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) ◆ From: 140.112.25.78
gglk:ㄟ謝謝助教,所以USEFUL就是可以幫你算出要窮舉幾個這樣嗎? 10/05 22:30
gglk:另外如果真的要證COMPLETENESS是不是應該還要推到N個INPUT 10/05 22:31