TUGAS AKHIR MODUL 1 PROFESIONAL SOAL NO 1 1a. Adapun langkah-langkah yang harus dilakukan untuk menyelesaikan permasalahan permasalahan ( p q) (r q) ( p r ) q adalah sebagai berikut Langkah ke-1 susunan nilai kebenaran untuk tiap-tiap pernyataan pernyataan tunggal tunggal p, q, dan r sesuai sesuai dengan susunan nilai kebenaran pernyataan tunggal p, q, dan r dalam kolom pertama Langkah ke-2
susunan nilai kebenaran “p r” dari hasil implikasi nilai kebenaran p dan r pada langkah ke-1
Langkah ke-3 susunan nilai kebenaran “ r Langkah ke-4 susunan nilai kebenaran “ p
q ” dari hasil implikasi nilai kebenaran r dan s pada langkah ke-1
r ” dari hasil implikasi nilai kebenaran p dan s pada langkah ke-1
Langkah ke-5 susunan nilai kebenaran konjungsi dari hasil langkah ke -2 -2 dan langkah ke-3, yaitu susunan nilai kebenaran dari (p r) r) (r q) q) Langkah ke-6 susunan nilai kebenaran implikasi dari hasil langkah ke -4 dan dan pengikut hasil langkah ke-2 , yaitu susunan nilai kebenaran dari ( p r ) q Langkah ke-7 susunan nilai kebenaran implikasi dari hasil langkah ke -5 dan pengikut hasil langkah ke-6 , yaitu susunan nilai kebenaran dari ( p q) (r q) ( p r ) q
1|Page
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Berdasarkan tabel di bawah : pernyataan ( p q) (r q) ( p r ) q bukan suatu tautology atau kontradiksi. Tabel kebenaran yang sesuai dengan langkah-langkah diatas adalah sebagai berikut Langkah Ke-1
Ke-1
Ke-1
Ke-2
Ke-3
p
q
r
p q
B
B
B
B
B
B
B
B
B
B
B
S
B
B
B
S
B
B
B
S
B
S
S
S
B
S
B
B
S
S
S
B
S
S
B
B
S
B
B
B
B
B
B
B
B
S
B
S
B
B
B
B
B
B
S
S
B
B
S
S
B
B
B
S
S
S
B
B
B
B
S
S
r
q
Ke -5
Ke -4
Ke-6
Ke -7
( p q) ( r q)
p r
( p r ) q
( p q) (r q) ( p r ) q
1b. Adapun langkah-langkah yang harus dilakukan untuk menyelesaikan permasalahan
p
~
pq
adalah sebagai
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Langkah ke-4 susunan nilai kebenaran konjungsi dari hasil langkah langkah ke -1 dan pengikut hasil langkah langkah ke-3, yaitu susunan nilai kebenaran dari
p
~
pq
Berdasarkan tabel di bawah : pernyataan
p
~
pq
adalah
kontradiksi.
Tabel kebenaran yang sesuai dengan langkah-langkah diatas adalah sebagai berikut langkah
Ke-1
Ke-1
Ke-2
Ke-3
p
q
~p
~ p q
B
B
S
S
S
B
S
S
S
S
S
B
B
B
S
S
S
B
S
S
Ke-4 p
~
p q
SOAL NO 2 CARA I
( p q ) (r s)
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Kesimpulan Langkah 2
Premis 2 :
~
~
p
r s
~
q
Dengan menggunakan hukum De Morgan ~
Langkah 3
Premis 1
:
Premis 2
:
p q (r s) ~
~
Langkah 4
Kesimpulan Kesimpulan :
p
~ r ~ s ~ p ~ q
b
~
a
~
b
q
r s
~
p
~
q
Jadi argumen tersebut sah/valid (terbukti).
( p q ) (r s)
Menggunakan modus Tollens
Dengan menggunakan hukum De Morgan ~
Cara II Bukti keabsahan
a
a
b
~
a
~
b
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SOAL NO 3 Diketahui : Persamaan 1 + 2 + 3 = 20 dengan syarat 1 ≥ 2, 0 ≤ Ditanya : Banyaknya solusi dari persamaan 1 + 2 + 3 = 20 Jawab : Fungsi pembangkit untuk kemungkinan terambilnya objek Fungsi pembangkit untuk kemungkinan terambilnya objek Fungsi pembangkit untuk kemungkinan terambilnya objek Banyak solusi dari G ( x)
x x
2
2
1
x x 2 3
20 dikaitkan
3
x
x x
4
2
x x
5
3
1 ...1
...
x x
x x
2
2
x x
1
3
3
x
x
3
3
x
4
x
x x
2
5
4 3 1 x 1 x 1 x 5 1 x 1 x 1 x
1
1
3
x
x
5
1
x
4
1
x
3
1
x x
adalah
2
3
≤ 3, dan 3 ≤
adalah adalah
A( x) B( x)
C ( x)
x
2
3
x
3
≤5
x
1 x x
x
3
x
4
2
4
x
x
x
5
....
3
5
dengan fungsi pembangkit adalah menentukan koefesien dari
A( x) B ( x) C ( x)
x
1
x
x
2
……. (*)
x
20
dalam ekspansi :
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G ( x )
1
1
x
3
x
5
1
x
4
1
x
3
2 3 4 5 6 7 x 0 x1 x 2 x 3 x 4 x 5 .... x 5 1 x 4 1 x 3 1 2 3 4 5 0 2 3 4 5 6 7 x x 2 x 3 x 4 x 5 .... x 5 1 x 3 x 4 x 7 3 4 5 0 1 2 2 3 4 5 6 7 x x 2 x 3 x 4 x 5 .... x 5 x 8 x 9 x12 3 4 5 0 1 2 Koefesien dari
x
20
dapat ditentukan dengan memperhatikan beberapa bagian suku pada penjabaran di atas yaitu
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SOAL NO 4 Diketahui : Suatu graf yang digambarkan sebagai berikut !
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Berdasarkan definis graf bipartisi lengkap lengkap maka graf G bukan merupakan graf bipartisi lengkap karenaa ada anggota di V1 yang tidak berpasangan dengan salah satu anggota di V2. Contohnya titik a pada V1 tidak berpasangan dengan titik g di V2, titik a hanya berpasangan dengan titik b, titik d, dan titik e. Begitu pula dengan titik c, titik f dan titik h di V1 tidak semuanya dipasangkan dengan semua titik di V2. Jadi, graf G adalah graf bipartisi dan bukan graf bipartisi lengkap. SOAL NO 5 Diketahui : Suatu graf yang yang yang verteknya akan di di warnai dengan 6 warna yang yang disediakan Ditanya : banyak : banyak cara mewarnai vertek graf dengan syarat vertek yang bertetangga tidak boleh memiliki warna yang sama.
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Gunakan Algoritma Welch Powell untuk pewarnaan graf G Langkah-langkahnya sebagai berikut : 1. Jumlah titik Graf G adalah 6 buah buah dan urutan titik dari derajat tertinggi hingga terendah seperti tabel tabel 1 2. Pilih titik pada graf yang berderajar tertinggi. Berdasarkan tabel 1 ada dua titik yang berderat tertinggi yaitu 4 yaitu titik E dan titik B. Ambil salah satu titik misalnya titik E dan diwarnai dengan warna merah. Ada satu titik yang tidak bertetangga dengan E yaitu titik C dapat diwarnai diwarnai dengan warna merah juga. Dapat ditunjukkan seperti seperti gambar di bawah
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