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ILMU KOM PUTER PRODI ILKOMP UGM

GP DALIYO. Daliyo. GP DALIYO. Daliyo. GP DALIYO. Daliyo. Daliyo. Daliyo. Daliyo. Daliyo. Daliyo. GP DALIYO. GP DALIYO. ILMU KOM PUTER PRODI ILKOMP UGM. Daliyo. Daliyo. GP DALIYO. Daliyo. Daliyo. GP DALIYO. p p p’ p’ q q’ q’ q. p p p’ p’

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ILMU KOM PUTER PRODI ILKOMP UGM

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  1. GP DALIYO Daliyo GP DALIYO Daliyo GP DALIYO Daliyo Daliyo Daliyo Daliyo Daliyo Daliyo GP DALIYO GP DALIYO ILMU KOMPUTER PRODIILKOMPUGM Daliyo Daliyo GP DALIYO Daliyo Daliyo GP DALIYO

  2. p p p’ p’ q q’ q’ q p p p’ p’ q q’ q’ q p p p’ p’ q q’ q’ q p p p’ p’ q q’ q’ q r r’ r r’ r r‘ r r‘ r diarsir (4 bjsk) p diarsir (4 bjsk) q diarsir (4bjsk) p’q p q p’q’ p q’ r r’ Perhatikan : 1 bjr-skr  3 literal 2 bjr-skr bersanding  2 literal 4 bjr-skr bersanding  1 literal Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Daliyo Kasus 3 variabel Peta Karnaugh yang berkaitan dengan formula F(p,q,r) di gambarkan sebagai berikut : Terdapat 8 konjungan/hasil-kali foundamental yaitu : p.q.r ; p.q.r’ ; p.q’.r ; p.q’.r’ ; p’.q.r ; p’.q.r’ ; p’.q’.r ; p’.q’.r’

  3. x y x’ y’ x y’ x’ y   z z’   Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Daliyo Kasus 3 variabel Contoh : E = xyz +xyz’ + x’yz’ +x’y’z F = pqr + pqr’ + pq’r + p’qr + p’q’r G = uvw + uvw’ + u’vw’ +u’v’w’ + u’v’w Fungsi F(p,q,r) E = xy + yz’ + x’y’z p q p’ q’ p q’ p’ q     u v u’ v’ u v’ u’ v r r’    w w’    F = p.q + r G = uv + u’v’ + u’w’ = uv + u’v’ + v w’

  4. p 1 1 1 1 0 0 0 0 q 1 1 0 0 1 1 0 0 r 1 0 1 0 1 0 1 0 F 1 1 0 0 1 0 1 0 p q 1 0 0 0 0 1 1 1 1 0 1 1 1 0 1 0 0 0 r Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Kasus 3 variabel Daliyo Daliyo Bagaiman jika fungsinya ditentukan dalam bentuk tabel kebenaran ? Misalnya : Digambarkan sebagai berikut : Daliyo Fungsi F(p,q,r) Daliyo Daliyo Jadi fungsinya F = p.q + p’.r Daliyo

  5. p 1 1 1 1 0 0 0 0 q 1 1 0 0 1 1 0 0 r 1 0 1 0 1 0 1 0 F 1 1 0 0 1 1 1 0 Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Kasus 3 variabel Daliyo Daliyo Digambarkan sebagai berikut : p q 1 0 0 0 0 1 1 1 1 0 1 1 1 0 1 0 0 1 r Fungsi F(p,q,r) Daliyo Daliyo Daliyo Jadi fungsinya F = q + p’.r Daliyo Daliyo Daliyo

  6. p p p’ p’ q q’ q’ q 1 1 0 0 p 1 0 0 1 q r s 1 1 1 0 0 0 0 1 r s r s’ r’s’ r’s Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Kasus 4 variabel Peta Karnaugh untuk 4 variabel lihat gb. Terdapat 16 bjr- skr yg ma sing-masing berkaitan dengan bndp (hasil-kali foundamental) : p .q .r .s , p .q .r .s’, p .q .r’.s , p .q .r’.s’, p .q’.r .s , p .q’.r .s’, p .q’.r’.s , p .q’.r’.s’, p’.q .r .s , p’.q .r .s’, p’.q .r’.s , p’.q.r’.s’, p’.q’.r .s , p’.q’.r .s’, p’.q’.r’.s , p’.q’.r’.s’ Daliyo

  7. 1 1 0 0 p 1 0 0 1 q r s 1 1 1 0 0 0 0 1 Daliyo Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Kasus 4 variabel Daliyo p p p’ p’ q q’ q’ q pqrs pq’rs p’q’rs r s r s’ r’s’ r’s pqrs’ p’q’r’s’ p’qr’s Bagaimana dengan cara diatas ? Kerjakan sendiri !!!!! pr’ q’rs’ p’q Daliyo

  8. p p p’ p’ q q’ q’ q p p p’ p’ q q’ q’ q   r s r s’ r’s’ r’s r s r s’ r’s’ r’s      Diberikan peta Karnaugh, bagaimana fungsinya ??? E = q’r + pqr’ + qr’s’ Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Kasus 4 variabel Contoh (Fungsi : F(p,q,r,s)) Daliyo Daliyo Daliyo Diberikan : (1) E = pqr’s’ + pqr’s + pq’rs + pq’rs’ + p’q’rs + p’q’rs’ + p’qr’s’ (1) Fungsi F(p,q,r) ??

  9. Kasus 4 variabel Contoh (Tabel-Kebenaran) 1 1 0 0 p 1 0 0 1 q p 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 q 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 r 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 s 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 F(p,q,r,s) 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0   1 1 1 0 0 0 0 1 r s      F(p,q,r,s) = p q’r s + p’q’r s’ + q r’s’ + p’r’s + p’q s Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Daliyo

  10. Logika Proposisional[Aplikasi Peta Karnaugh] No Br 0 1 2 3 4 5 6 7 p 0 0 0 0 1 1 1 1 q 0 0 1 1 0 0 1 1 r 0 1 0 1 0 1 0 1 F 1 0 0 0 1 1 0 1 Daliyo Daliyo Daliyo Daliyo Penandaan Minterm dan Maxterm dp Fungsi Daliyo Dalam fungsi cetetan/switching fungsi didefinisikan dgn tabel kebe narannya atau suatu daftar drpd nilai fungsi untuk semua kombinasi yg mungkin daripada masukkannya. Pada tabel kebenaran disamping ini se tiap baris diberi nomor sesuai dengan kombinasi daripada masukkannya; mis. p=0, q=1, dan r=1 maka diinterpre tasikan dng 011]2 = 3]10 , maka didapat penandaan fungsi sebagi berikut : F(p,q,r) =  m(0,4,5,7) m – berarti minterm 0,4,5,7 – baris fungsi bernilai 1 m(0) = p’q’r’ ; m(4) = pq’r’ ; m(5) = pq’r ; m(7) = pqr f(p,q,r) = p’q’r’ + pq’r’ + pq’r + pqr (BNDP)

  11. Logika Proposisional[Aplikasi Peta Karnaugh] No Br 0 1 2 3 4 5 6 7 p 0 0 0 0 1 1 1 1 q 0 0 1 1 0 0 1 1 r 0 1 0 1 0 1 0 1 F 1 0 0 0 1 1 0 1 Daliyo Daliyo Daliyo Daliyo Penandaan Minterm dan Maxterm dp Fungsi Daliyo Dalam fungsi cetetan/switching fungsi didefinisikan dgn tabel kebe narannya atau suatu daftar drpd nilai fungsi untuk semua kombinasi yg mungkin daripada masukkannya. Pada tabel kebenaran disamping ini se tiap baris diberi nomor sesuai dengan kombinasi daripada masukkannya; mis. p=0, q=1, dan r=1 maka diinterpre tasikan dng 011]2 = 3]10 , maka didapat penandaan fungsi sebagi berikut : F(p,q,r) = Π M(1,2,3,6) M – berarti Maxterm 1,2,3,6 – baris fungsi bernilai 0 M(1) = p+q+r’ ; M(2) = p+q’+r ; M(3) = p+q’+r’ ; M(6) = p’+q’+r F(p,q,r) = (p+q+r’).(p+q’+r).(p+q’+r’).(p’+q’+r) (BNKP)

  12. Logika Proposisional[Aplikasi Peta Karnaugh] No Br 0 1 2 3 4 5 6 7 p 0 0 0 0 1 1 1 1 q 0 0 1 1 0 0 1 1 r 0 1 0 1 0 1 0 1 Minterm p’q’r’ = m0 p’q’r = m1 p’q r’ = m2 p’q r = m3 p q’r’ = m4 p q’r = m5 p q r’ = m6 p q r = m7 Maxterm p + q + r = M0 p + q + r’= M0 p + q’+ r = M0 p + q’+ r’= M0 p’+ q + r = M0 p’+ q + r’= M0 p’+ q’+ r = M0 p’+ q’+ r’= M0 Daliyo Daliyo Daliyo Daliyo Daliyo Sumari, kita kaitkan setiap minterm dengan kombinasi masukan di mana ia menghasilkan 1 dan setiap maxterm dng kombinasi masukan yang menghasilkan 0.

  13. No Brs 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 p 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 q 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 r 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 s 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Minterm p’q’r’s’= m0 p’q’r’s = m1 p’q’r s’= m2 p’q’r s = m3 p’q r’s’= m4 p’q r’s = m5 p’q r s’= m6 p’q r s = m7 p q’r’s’= m8 p q’r’s = m9 p q’r s’= m10 p q’r s = m11 p q r’s’= m12 p q r’s = m13 p q r s’= m14 p q r s = m15 Maxterm p + q + r + s = M0 p + q + r + s’ = M1 p + q + r ‘+ s = M2 p + q + r ‘+ s’ = M3 p + q’ + r + s = M4 p + q’ + r + s’ = M5 p + q’ + r ‘+ s = M6 p + q’ + r ‘+ s’ = M7 p’ + q + r + s = M8 p’ + q + r + s’ = M9 p’ + q + r ‘+ s = M10 p’ + q + r ‘+ s’ = M11 p’ + q’ + r + s = M12 p’ + q’ + r + s‘ = M13 p’ + q’ + r’ + s = M14 p’ + q’ + r’ + s’ = M15 Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo Daliyo Daliyo

  14. Contoh 1). Diberikan F(p,q,r,s) = p q r s + p q’r s + p’q r s + p q r s’ + p q’r s’ + p q r’s’ + p’q r s’ + p’q r’s’ sajikan ke dalam bentuk daftar minterm . Jawab : F(p,q,r,s) = pqrs+pq’rs+p’qrs+pqrs’+pq’rs’+pqr’s’+p’qrs’+p’qr’s’ 1111 10 11 0 111 1110 10 10 110 0 0 110 0 100 (15) (11) (7) (14) (10) (12) (6) (4) F(p,q,r,s) =  m(4,6,7,10,11,12,14,15) Daliyo Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Daliyo

  15. Daliyo Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Contoh 1). Diberikan G(p,q,r,s) = (p + q’+ r + s’).(p’+q + r + s).(p + q + r’+ s’). (p’+ q + r + s’) sajikan ke dalam bentuk daftar maxterm . Jawab : G(p,q,r,s) = (p+q’+r+s’).(p’+q+r+s).(p+q+r’+s’).(p’+q+r+s’) 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 1 (5) (8) (3) 9) Didapat : G(p,q,r,s) =  M(3,5,8,9)

  16. p 0 0 1 1 q 0 1 1 0 p+q 0 1 1 1 p 0 0 1 1 q 0 1 1 0 p.q 0 0 1 0 Tabel Kebenaran OR (+) Tabel Kebenaran AND ( . ) pq p 0 1 (AND) q 01 11 10 00 0 0 0 1 p . q 1 p . q p 0 1 0 1 (AND) pq q (OR) 0 1 01 11 10 00 0 1 p + q p + q 1 1 1 1 1 (OR) Logika Proposisional[Aplikasi Peta Karnaugh] Daliyo Daliyo Tabel Kebenaran dalam sajian lain 2 variabel Daliyo Daliyo

  17. Logika Proposisional[Aplikasi Peta Karnaugh] Pendekatan dengan diagram Venn p’ p p’ p q’ q Universal set Membangun peta Karnaugh (peta-K) dengan pendekatan diagram Venn p’ p p’ p Peta-K daripada AND (p.q) Bentuk Venn Peta-K daripada OR (p+q) Bentuk Venn q’ q q’ q

  18. Logika Proposisional[Aplikasi Peta Karnaugh] p 0 0 0 0 1 1 1 1 q 0 0 1 1 0 0 1 1 r 0 1 0 1 0 1 0 1 p.q.r 0 0 0 0 0 0 0 1 pq 00 01 11 10 r 0 0 0 1 1 0 0 0 0 0 p.q.r p p p p Daliyo 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1 1 OR r r r 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 r q q q q Daliyo p + q + r = p + q + r Daliyo Tabel Kebenaran dalam sajian lain 3 variabel AND Daliyo

  19. Peta-K baku (Peta Karnaugh dng Penomoran) pq p 00 01 11 10 0 1 r q 0 2 0 2 6 4 0 1 0 1 1 3 1 3 7 5 Tiga Variabel Dua Variabel p q no. 0 0 = 0 1 0 = 2 0 1 = 1 1 1 = 3 p q r no 1 1 0 = 6 1 1 1 = 7 1 0 0 = 4 1 0 1 = 5 p q r no 0 0 0 = 0 0 0 1 = 1 0 1 0 = 2 0 1 1 = 3 Logika Proposisional[Aplikasi Peta Karnaugh]

  20. p q r s No pq 00 01 11 10 rs 0 0 0 0 = 0 0 0 0 1 = 1 0 0 1 1 = 3 0 0 1 0 = 2 0 1 0 0 = 4 0 1 0 1 = 5 0 1 1 1 = 7 0 1 1 0 = 6 1 1 0 0 = 12 1 1 0 1 = 13 1 1 1 1 = 15 1 1 1 0 = 14 1 0 0 0 = 8 1 0 0 1 = 9 1 0 1 1 = 11 1 0 1 0 = 10 0 4 12 8 00 01 1 5 13 9 3 7 15 11 11 10 2 6 14 10 empat Variabel Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku

  21. p=0 p=1 qr qr 00 01 11 10 00 01 11 10 st st 0 4 16 20 12 8 28 24 00 01 00 01 1 5 17 21 13 9 29 25 3 7 19 23 15 11 31 27 11 10 11 10 2 6 18 22 14 10 30 26 Lima Variabel Daliyo Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku p q r s t 0 0 0 0 0 = 0 0 0 0 0 1 = 1 0 1 0 1 0 = 10 p q r s t 1 0 0 0 0 = 16 1 0 0 0 1 = 17 1 1 1 0 1 = 29

  22. Logika Proposisional[Aplikasi Peta Karnaugh] q=0 q=1 rs rs 00 01 11 10 00 01 11 10 tu tu 16 20 0 4 28 24 12 8 00 01 11 10 00 01 11 10 17 21 1 5 13 9 29 25 p=0 19 23 3 7 31 27 15 11 18 22 2 6 30 26 14 10 Daliyo 32 36 48 52 60 56 44 40 00 01 11 10 00 01 11 10 49 53 33 37 45 41 61 57 35 39 51 55 47 43 63 59 p=1 50 54 34 38 62 58 46 42 00 01 11 10 00 01 11 10 Peta-K baku enam Variabel Daliyo

  23. p p’ p’q’ p’q pq pq’ p’ p 0 2 0 2 6 4 r’ r q’ q 1 3 1 3 7 5 q Dua Variabel Tiga Variabel Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku Alternatif

  24. Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku Alternatif p p’q’ p’q pq pq’ 0 4 12 8 r’s’ r’s 1 5 13 9 r 3 7 15 11 rs rs’ s 2 6 14 10 q empat Variabel

  25. No p q r s f 0 0 0 0 0 0 1 0 0 0 1 0 2 0 0 1 0 0 3 0 0 1 1 0 4 0 1 0 0 0 pq rs 0 0 0 1 1 1 1 0 5 0 1 0 1 1 0 0 d 1 6 0 1 1 0 1 7 0 1 1 1 1 0 1 1 d 1 p 8 1 0 0 0 1 1 1 1 d d 9 1 0 0 1 1 qs 1 0 1 d d qr Logika Proposisional[Aplikasi Peta Karnaugh] Don’t care Contoh. Jika diketahui tabel kebenaran daripada fungsi f sbb : Dari tabel kebenaran, kita tahu bahwa f(5) = f(6) = f(7) = f(8) = f(8) = f(9) = 1 dan f(0) = f(1) = f(2) = f(3) = f(4) = 0 , sedangkan f(10) = f(11) = f(12) = f(13) = f(14) = f(15) = don’t care , karena tidak didefinisikan pada tabel sehingga didapat peta Karnaugh :

  26. p=0 p=1 qr qr 00 01 11 10 00 01 11 10 st st 0 4 16 20 12 8 28 24 00 01 00 01 1 5 17 21 13 9 29 25 3 7 19 23 31 27 15 11 11 10 11 10 2 6 18 22 14 10 30 26 Lima Variabel Daliyo Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku Alternatif

  27. q’ (0) q (1) r (1) r (1) 0 4 16 20 28 24 12 8 17 21 1 5 13 9 29 25 u (1) p’(0) 3 7 19 23 15 11 31 27 t(1) 18 22 2 6 30 26 14 10 Daliyo 32 36 48 52 60 56 44 40 49 53 33 37 45 41 61 57 u (1) p(1) 35 39 51 55 47 43 63 59 t(1) 50 54 34 38 62 58 46 42 Logika Proposisional[Aplikasi Peta Karnaugh] Peta-K baku Alternatif s (1) s(1)

  28. v=0 v=1 wx wx 00 01 11 10 00 01 11 10 yz yz 0 4 12 8 16 20 28 24 00 01 00 01 1 5 13 9 17 21 29 25 3 7 15 11 19 23 31 27 11 10 11 10 2 6 14 10 18 22 30 26 Logika Proposisional[Aplikasi Peta Karnaugh] Contoh Cari peta-K untuk fungsi sbb : f(v,w,x,y,z) =  m(9,20,21,29,30,31) Daliyo

  29. a’ (1) wx 00 01 11 10 yz 16 20 28* 24* 00 01 d’ 17 21 29* 25* 19+& 23& 31*& 27*& 11 10 d 18+ 22 30* 26* Logika Proposisional[Aplikasi Peta Karnaugh] Cari peta-K untuk fungsi sbb : f(a,b,c,d,e) = ab +c’d +de • * = a.b ; + = c’d • & = de Didapat : f(a,b,c,d,e) =  m(2, 3, 7,10,11,15,18,19, 23, 24, 25, 26, 27, 28, 29, 30, 32 a’ (0) bc 00 01 11 10 de 0 4 12 8 00 01 1 5 13 9 e 3+& 7& 15& 11+& 11 10 2+ 6 14 10+

  30. c d a b’ . c d’ a + f b a + b c c d’ . a b’ c’ d f d a c d’ a b c’ d a c’ d a’ f b c b c d’ d’ a’ b b c’ d c’ d Logika Proposisional[Aplikasi Peta Karnaugh] Diberikan : f(a,b,c,d) = (a+b)(cd’+c’d) = acd’+ac’d+bcd’+bc’d =  m(5,6,9,10,13,14) (tunjukan !!!) III I 3 level II 2 level 2 level

  31. a b 4 1 12 1 5 1 `3 1 15 1 11 1 c c 14 1 10 1 ac bc’ b 0 1 8 1 1 1 9 1 b’c’ c Logika Proposisional[Aplikasi Peta Karnaugh] Set of four on K-map

  32. Set of four on K-map b b 4 1 12 1 0 1 8 1 ` d d 6 1 14 1 2 1 10 1 bd’ b’d’ b 1 1 9 1 b’d d 3 1 11 1 Logika Proposisional[Aplikasi Peta Karnaugh]

  33. Set of eight on K-map a a’ a 12 1 8 1 4 1 12 1 0 1 8 1 13 1 9 1 5 1 `13 1 1 1 9 1 d 15 1 11 1 c 14 1 10 1 a c’ a a’ 0 1 4 1 12 1 8 1 d’ d 2 1 6 1 14 1 10 1 Logika Proposisional[Aplikasi Peta Karnaugh]

  34. a a’ b’ b b’ b 1 1 1 1 d’ 1 1 e 1 1 1 1 d 1 1 c c Kelompok 4 = acd’ Kelompok 1 = a’c’d’e’ Kelompok 2 = b’c’de’ Kelompok 3 = bde Logika Proposisional [Aplikasi Peta Karnaugh] Sets on a 5-variable map

  35. a a’ b’ b b’ b d’ e d c c Logika Proposisional [Aplikasi Peta Karnaugh] Sets on a 5-variable map , example Diberikan : f(a,b,c,d,e) =  m(0,1,3,4,5,7,8,9,10,12,13,21,24,25,26,28,29) 0 1 4 1 12 1 8 1 28 1 24 1 1 1 5 1 13 1 9 1 21 1 29 1 25 1 3 1 7 1 10 1 26 1 f = cd’e  m(0,1,4,5,8,9,12,13) + a’b’e  m(1,3,5,7) + a’d’  m(5,13,21,29) + bc’e’  m(8,10,24,26) + bd’  m(8,9,12,13,24,25,28,29)

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