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Bilangan Prima < 200 adalah : 46 buah 2.3.5.7.11.13.17.19.23.29 31.37.41.43.47.53.59.61.67.71 73.79.83.89.97.101.103.107.109.113 127.131.137.139.149.151.157.163.167.173 179.181.191.193.197.199 1. Masukkan ke dalam hash dengan ukuran table hash = 47; h(x) = x mod 47, Gunakan resolusi linear. Hitung waktu akses rata
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
47
191.(6) 2
3
97.(2 )
5
53
7
101.(2 )
103
149.(2 )
11
59
13
61
22
2 3
2 4
25
2 6
27
28
29
2 3
7 1
163.(3 )
7 3
16 7
193.(33 )
29
15
16
1 7
18
1 9
2 0
21
107.(3 )
109.(2 )
1 7
151.(8 )
1 9
6 7
113.(3) 157.(6 )
30
3 1
3 2
33
34
35
3 6
3 7
38
39
197.(29 )
3 1
7 9
12 7
173.(2 )
199.(32 )
8 3
3 7
131.(2 )
17 9
4 0
4 1
4 2
4 3
44
45
46
4 1
8 9
4 3
13 7
181.(5 )
139
Waktu akses Rata – rata : ( 6 + 2 + 2+2+3+2+8+3+6+3+33+29+2+32+2+5)/47 = 140/47 = 2.978
2. tabel Utama ukuran 29 ( h(x) = x mod 29) 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
29
59
2
3
89
61
149
7
37
181
11
41
13
43
15
16
17
18
19
20
21
22
23
24
25
71
101
17
47
19
163
193
107
23
53
139
26
27
28
83
113
Overflow1 ukuran 17 ( h(x) = x mod 17) 0
1
2
10 3
13 7
3
4
5
6
73
7
8
9
10
11
12
10 9
59
19 7
17 9
79
97
Overflow2 ukuran 11 ( h(x) = x mod 11) 0 1 2 3 4
5
191
127
173
157
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(173) = ( 173 + 4 mod 11 = 1 ) Waktu akses rata – rata : 19 / 46 = 0.412
3. a). Ukuran Index ( terbuka) = 7 Ukuran Index ( tertutup) = 7
6
13
14
15
16
31
16 7
67
7
8
9
151
199
131
10
0
1
2
3
4
5
6
0
7
79
107
149
163
181
191
1
29
43
71
101
157
199
2
151
179
2
23
37
109
137
3
73
89
103
3
17
31
59
4
11
53
67
83
97
139
167
5
47
61
193
5
19
6
13
41
113
173
197
127
131
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(131) = ( h(131) +1 mod 7 = 6 H(173) = (h(173)+1 mod 7 = 6 H(181) = (h(181)+1 mod 7 = 0 H(193) = ( h(193) + 1 mod 7 = 5 Waktu rata – rata = (2 + 3+4+5+6+7+2+3+4+5+6+6+7+2+2+3+4+5+5+6+7+2+3+4+2+3+4+5+6+7+3+4+5+2+2+3+4 +5+6+7)/46 = 288/46 = 6,26
B). Beda Rumus a).H1(x)= (x+1) mod 7
0
1
2
3
4
5
6
0
13
71
79
107
149
163
191
1
7
43
101
157
199
2
151
2
29
37
73
109
137
3
67
89
103
3
23
31
59
4
17
53
83
97
139
167
181
5
47
61
179
193
5
11
6
19
41
113
127
173
197
131
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(131) = ( h (131) + 1 mod 7 = 6 H( 173 )= ( h ( 173 ) + 1 mod 7 = 6 H( 179 ) = ( h(179) + 1 mod 7= 5 Waktu rata – rata = ( 169 / 46 = 3,673 )
b).h(x) = (x+2) mod 7
0
1
2
3
4
5
6
0
17
47.(2)
61.(3)
89.(4)
103.(5)
139.(6)
167(7)
1
11
41.(2)
83.(3)
97.(4)
179(5)
181(6)
193(7)
2
7.(2)
19.(3)
197(4)
2 3
71.(5)
113.(6)
127.(7)
3
13.(2)
29.(3)
43.(4)
4
107.(4)
157(5)
191(6)
199(7)
23
37.(2)
79.(3)
5
59.(3)
73.(4)
101.(5)
137.(6)
151(7)
5
31.(2)
6
53
67.(2)
109.(3)
131.(4)
149(5)
163(6)
173(7)
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(149) = ( h(149) + 1 mod 7 = 6 H( 163) = ( h(163) + 1 mod 7 = 6 H( 179) = ( h(179) + 1 mod 7 = 1 H( 181) = ( h (181) + 2 mod 7 = 1 H( 191) = ( h ( 191) + 2 mod 7 = 4 H(193) = ( h (193)+1 mod 7 = 1
Waktu rata rata = 171 / 46 = 3,717
c). H(x) = (x+6)mod 7
0
1
2
3
4
5
6
0
29
37.(2)
101.(3)
113.(4)
127.(5)
149.(6)
151.(7)
1
23
31.(2)
59.(3)
73.(4)
109.(5)
137.(6)
173.(7)
2
131.(6)
167.(7)
2
17.(2)
53.(3)
67.(4)
103.(5)
3
83.(5)
97.(6)
139.(7)
3
11.(2)
47.(3)
61.(4)
4
19
41.(2)
179(3)
193(4)
197(5)
199(6)
5
163.(3)
181(4)
191(5)
6
7
17.(2)
43.(3)
71.(4)
Resolusi Linear : H(x) = h(x) + 6 mod 7 H(179) = H( 179 ) + 7 mod 7 = 4 H(181) = H ( 181 ) + 6 mod 7 = 2 H( 193) = H ( 193) + 7 mod 7 = 4 H( 199) = H ( 199) + 8 mod 7 = 4
Waktu rata – rata = 169 / 46 = 3,673
d). H(x) = ( x + 8 ) mod 7
79.(5)
5
13.(2)
107.(6)
157.(7)
0
1
2
3
4
5
6
0
13
41.(2)
113.(3)
127.(4)
149.(5)
157.(6)
199(7)
1
7
71.(2)
107.(3)
179.(4)
193(5)
197(6)
2
151.(6)
173.(7)
2
29.(2)
43.(3)
79.(4)
101.(5)
3
109.(5)
137.(6)
167.(7)
3
23.(2)
37.(3)
73.(4)
4
17
31.(2)
59.(3)
67.(4)
89.(5)
103.(6)
131.(7)
5
53.(3)
61.(4)
83.(5)
97.(6)
139.(7)
5
11.(2)
6
19
47.(2)
163.(3)
181(4)
191(5)
Overflow pada bilangan prima 181 H(x) = ( x + 8 ) mod 7 H(181) = ( 181 + 8 mod 7 = 6) H ( 197) = ( 197 + 8 mod 7 = 1) Waktu rata – rata = 169 / 46 = 3.673
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
47
191.(6) 2
3
97.(2 )
5
53
7
101.(2 )
103
149.(2 )
11
59
13
61
22
2 3
2 4
25
2 6
27
28
29
2 3
7 1
163.(3 )
7 3
16 7
193.(33 )
29
15
16
1 7
18
1 9
2 0
21
107.(3 )
109.(2 )
1 7
151.(8 )
1 9
6 7
113.(3) 157.(6 )
30
3 1
3 2
33
34
35
3 6
3 7
38
39
197.(29 )
3 1
7 9
12 7
173.(2 )
199.(32 )
8 3
3 7
131.(2 )
17 9
4 0
4 1
4 2
4 3
44
45
46
4 1
8 9
4 3
13 7
181.(5 )
139
Waktu akses Rata – rata : ( 6 + 2 + 2+2+3+2+8+3+6+3+33+29+2+32+2+5)/47 = 140/47 = 2.978
2. tabel Utama ukuran 29 ( h(x) = x mod 29) 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
29
59
2
3
89
61
149
7
37
181
11
41
13
43
15
16
17
18
19
20
21
22
23
24
25
71
101
17
47
19
163
193
107
23
53
139
26
27
28
83
113
Overflow1 ukuran 17 ( h(x) = x mod 17) 0
1
2
10 3
13 7
3
4
5
6
73
7
8
9
10
11
12
10 9
59
19 7
17 9
79
97
Overflow2 ukuran 11 ( h(x) = x mod 11) 0 1 2 3 4
5
191
127
173
157
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(173) = ( 173 + 4 mod 11 = 1 ) Waktu akses rata – rata : 19 / 46 = 0.412
3. a). Ukuran Index ( terbuka) = 7 Ukuran Index ( tertutup) = 7
6
13
14
15
16
31
16 7
67
7
8
9
151
199
131
10
0
1
2
3
4
5
6
0
7
79
107
149
163
181
191
1
29
43
71
101
157
199
2
151
179
2
23
37
109
137
3
73
89
103
3
17
31
59
4
11
53
67
83
97
139
167
5
47
61
193
5
19
6
13
41
113
173
197
127
131
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(131) = ( h(131) +1 mod 7 = 6 H(173) = (h(173)+1 mod 7 = 6 H(181) = (h(181)+1 mod 7 = 0 H(193) = ( h(193) + 1 mod 7 = 5 Waktu rata – rata = (2 + 3+4+5+6+7+2+3+4+5+6+6+7+2+2+3+4+5+5+6+7+2+3+4+2+3+4+5+6+7+3+4+5+2+2+3+4 +5+6+7)/46 = 288/46 = 6,26
B). Beda Rumus a).H1(x)= (x+1) mod 7
0
1
2
3
4
5
6
0
13
71
79
107
149
163
191
1
7
43
101
157
199
2
151
2
29
37
73
109
137
3
67
89
103
3
23
31
59
4
17
53
83
97
139
167
181
5
47
61
179
193
5
11
6
19
41
113
127
173
197
131
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(131) = ( h (131) + 1 mod 7 = 6 H( 173 )= ( h ( 173 ) + 1 mod 7 = 6 H( 179 ) = ( h(179) + 1 mod 7= 5 Waktu rata – rata = ( 169 / 46 = 3,673 )
b).h(x) = (x+2) mod 7
0
1
2
3
4
5
6
0
17
47.(2)
61.(3)
89.(4)
103.(5)
139.(6)
167(7)
1
11
41.(2)
83.(3)
97.(4)
179(5)
181(6)
193(7)
2
7.(2)
19.(3)
197(4)
2 3
71.(5)
113.(6)
127.(7)
3
13.(2)
29.(3)
43.(4)
4
107.(4)
157(5)
191(6)
199(7)
23
37.(2)
79.(3)
5
59.(3)
73.(4)
101.(5)
137.(6)
151(7)
5
31.(2)
6
53
67.(2)
109.(3)
131.(4)
149(5)
163(6)
173(7)
Resolusi linear : H(x)= ( h(x) + 1 mod n ) H(149) = ( h(149) + 1 mod 7 = 6 H( 163) = ( h(163) + 1 mod 7 = 6 H( 179) = ( h(179) + 1 mod 7 = 1 H( 181) = ( h (181) + 2 mod 7 = 1 H( 191) = ( h ( 191) + 2 mod 7 = 4 H(193) = ( h (193)+1 mod 7 = 1
Waktu rata rata = 171 / 46 = 3,717
c). H(x) = (x+6)mod 7
0
1
2
3
4
5
6
0
29
37.(2)
101.(3)
113.(4)
127.(5)
149.(6)
151.(7)
1
23
31.(2)
59.(3)
73.(4)
109.(5)
137.(6)
173.(7)
2
131.(6)
167.(7)
2
17.(2)
53.(3)
67.(4)
103.(5)
3
83.(5)
97.(6)
139.(7)
3
11.(2)
47.(3)
61.(4)
4
19
41.(2)
179(3)
193(4)
197(5)
199(6)
5
163.(3)
181(4)
191(5)
6
7
17.(2)
43.(3)
71.(4)
Resolusi Linear : H(x) = h(x) + 6 mod 7 H(179) = H( 179 ) + 7 mod 7 = 4 H(181) = H ( 181 ) + 6 mod 7 = 2 H( 193) = H ( 193) + 7 mod 7 = 4 H( 199) = H ( 199) + 8 mod 7 = 4
Waktu rata – rata = 169 / 46 = 3,673
d). H(x) = ( x + 8 ) mod 7
79.(5)
5
13.(2)
107.(6)
157.(7)
0
1
2
3
4
5
6
0
13
41.(2)
113.(3)
127.(4)
149.(5)
157.(6)
199(7)
1
7
71.(2)
107.(3)
179.(4)
193(5)
197(6)
2
151.(6)
173.(7)
2
29.(2)
43.(3)
79.(4)
101.(5)
3
109.(5)
137.(6)
167.(7)
3
23.(2)
37.(3)
73.(4)
4
17
31.(2)
59.(3)
67.(4)
89.(5)
103.(6)
131.(7)
5
53.(3)
61.(4)
83.(5)
97.(6)
139.(7)
5
11.(2)
6
19
47.(2)
163.(3)
181(4)
191(5)
Overflow pada bilangan prima 181 H(x) = ( x + 8 ) mod 7 H(181) = ( 181 + 8 mod 7 = 6) H ( 197) = ( 197 + 8 mod 7 = 1) Waktu rata – rata = 169 / 46 = 3.673
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