SOAL :
Terdapat 1000 mahasiswa baru BSI, jika ingin dilakukan pencarian dengan nomer urut 212 dengan menggunnakan teknik
1. Linear search
2. Binary search
Maka ada berapa langkah penyelesaiannya…???
JAWAB :
1. Linear Search
212 langkah
2. Binary Search
Diket : L = 1
U =1000
Ditanya : X data = 212
Jawab : m = ( L + u ) / 2
= ( 1 + 1000 ) / 2
= 1001 /2
= 500,5
1) X data (m)
212 < 500 m = ( L + u ) /2
u = m - 1 = ( 1 + 499 ) /2
= 500 - 1 = 500 / 2
= 499 = 250
2) X data (m)
212 < 250 m = ( L + u ) /2
u = m - 1 = ( 1 + 249 ) /2
= 250 - 1 = 250 / 2
= 249 = 125
3) X data (m)
212 > 125 m = ( L + u ) /2
L = m + 1 = ( 126 + 249 ) /2
= 125 + 1 = 375 / 2
= 126 = 187,5
4) X data (m)
212 > 187 m = ( L + u ) /2
L = m + 1 = ( 188 + 249 ) /2
= 187 + 1 = 437 / 2
= 188 = 218,5
5) X data (m)
212 < 218 m = ( L + u ) /2
u = m - 1 = ( 188 + 217 ) /2
= 218 - 1 = 405 / 2
= 217 = 202,5
6) X data (m)
212 > 202 m = ( L + u ) /2
L = m + 1 = ( 203 + 217 ) /2
= 202 + 1 = 420 / 2
= 203 = 210
7) X data (m)
212 > 210 m = ( L + u ) /2
L = m + 1 = (211 + 217 ) /2
= 210 + 1 = 428 / 2
= 211 = 214
8) X data (m)
212 < 214 m = ( L + u ) /2
u = m - 1 = ( 211 + 213 ) /2 2
= 214 - 1 = 424 / 2
= 213 = 212
9) X data (m)
212 = 212
