1404505038 putuwahyusaputra jarkom(a) kompresidata
DESCRIPTION
Tugas JarkomTRANSCRIPT
Nama: Putu Wahyu SaputraNIM: 1404505038Kelas: Jaringan Komputer dan Komunikasi (A)
Pemampatan file menggunakan algoritma Huffman
Data : PUTUWAHYUSAPUTRA1. Menghitung frekuensi kemunculan karaktera. H = 1b. R = 1c. S = 1d. W = 1e. Y = 1f. P = 2g. T = 2h. A = 3i. U = 4
2. Jika diubah ke dalam ASCII :P= 50= 01010000U = 55= 01010101T = 54= 01010100U= 55= 01010101W= 57= 01010111A= 41= 01000001H= 48= 01001000Y= 59= 01011001U= 55= 01010101S= 53= 01010011A= 41= 01000001P= 50= 01010000U= 55= 01010101T= 54= 01010100R= 52= 01010010A= 41= 01000001
Jadi, ukuran data yang dikirim yaitu (8 x 16) = 128 byte
3. Proses penyandian :a. Tahap 1H,1R,1S,1W,1Y,1P,2T,2A,3U,4
b. Tahap 2HR,2S,1W,1Y,1P,2T,2A,3U,4H,1R,1S,1W,1Y,1P,2T,2A,3U,4HR,2H,1R,1
c. Tahap 3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1
Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1
d. Tahap 4Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1YSW,3
YSW,3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1
e. Tahap 5Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1YSW,3HRP,4HRP,4Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1YSW,3
f. Tahap 6HRP,4TYSW,5YSW,3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1YSW,3TYSW,5HRP,4
g. Tahap 7TYSW,5YSW,3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1HRP,4AHRP,7AHRP,7HRP,4Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1TYSW,5YSW,3
h. Tahap 8UTYSW,9Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1TYSW,5YSW,3HRP,4AHRP,7
UTYSW,9TYSW,5YSW,3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1HRP,4AHRP,7
i. Tahap 9AHRPUTYSW,16Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1YSW,3HRP,4TYSW,5AHRP,7UTYSW,9
j. Tahap 10Setelah proses pohon Hoffman selesai, selanjutnya memberikan nilai biner 0 dan 1 di masing masing cabang.
0000000011111111UTYSW,9AHRP,7TYSW,5HRP,4YSW,3Y,1P,2T,2A,3U,4HR,2H,1R,1SW,2S,1W,1AHRPUTYSW,16
4. Proses pengkodean masing masing karakterP= 011U = 10T = 110U= 10W= 11111A= 00H= 0100Y= 1110U= 10S= 11110A= 00P= 011U= 10T= 110R= 0101A= 00
Jadi, ukuran data setelah dikompresi yaitu (2 x 7) + (3 x 4)+(4 x 3)+(5 x 2) = 48 byte. Penghematan file sebanyak (128 48) = 80 byte.