Berikut ini adalah pertanyaan dari netanyasembiring11 pada mata pelajaran TI untuk jenjang Sekolah Menengah Atas
Jawaban dan Penjelasan
Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.
Jawaban:
1. buatlah algoritma pengurutan secara ascending (dari angka terkecil ke yang terbesar) dari data 5,11,7,3,9,2,5,13,23,20 dengan metode insertion short!
code in python:
def insertionSort(arr):
for i in range(1, len(arr)):
element = arr[i]
j = i-1
while j >= 0 and element < arr[j] :
arr[j + 1] = arr[j]
j -= 1
print('pergeseran iterasi ke', i - 1, arr)
arr[j + 1] = element
print('iterasi', i-1,arr)
arr = [5,11,7,3,9,2,5,13,23,20]
insertionSort(arr)
Output code python:
iterasi 0 [5, 11, 7, 3, 9, 2, 5, 13, 23, 20]
pergeseran iterasi ke 1 [5, 11, 11, 3, 9, 2, 5, 13, 23, 20]
iterasi 1 [5, 7, 11, 3, 9, 2, 5, 13, 23, 20]
pergeseran iterasi ke 2 [5, 7, 11, 11, 9, 2, 5, 13, 23, 20]
pergeseran iterasi ke 2 [5, 7, 7, 11, 9, 2, 5, 13, 23, 20]
pergeseran iterasi ke 2 [5, 5, 7, 11, 9, 2, 5, 13, 23, 20]
iterasi 2 [3, 5, 7, 11, 9, 2, 5, 13, 23, 20]
pergeseran iterasi ke 3 [3, 5, 7, 11, 11, 2, 5, 13, 23, 20]
iterasi 3 [3, 5, 7, 9, 11, 2, 5, 13, 23, 20]
pergeseran iterasi ke 4 [3, 5, 7, 9, 11, 11, 5, 13, 23, 20]
pergeseran iterasi ke 4 [3, 5, 7, 9, 9, 11, 5, 13, 23, 20]
pergeseran iterasi ke 4 [3, 5, 7, 7, 9, 11, 5, 13, 23, 20]
pergeseran iterasi ke 4 [3, 5, 5, 7, 9, 11, 5, 13, 23, 20]
pergeseran iterasi ke 4 [3, 3, 5, 7, 9, 11, 5, 13, 23, 20]
iterasi 4 [2, 3, 5, 7, 9, 11, 5, 13, 23, 20]
pergeseran iterasi ke 5 [2, 3, 5, 7, 9, 11, 11, 13, 23, 20]
pergeseran iterasi ke 5 [2, 3, 5, 7, 9, 9, 11, 13, 23, 20]
pergeseran iterasi ke 5 [2, 3, 5, 7, 7, 9, 11, 13, 23, 20]
iterasi 5 [2, 3, 5, 5, 7, 9, 11, 13, 23, 20]
iterasi 6 [2, 3, 5, 5, 7, 9, 11, 13, 23, 20]
iterasi 7 [2, 3, 5, 5, 7, 9, 11, 13, 23, 20]
pergeseran iterasi ke 8 [2, 3, 5, 5, 7, 9, 11, 13, 23, 23]
iterasi 8 [2, 3, 5, 5, 7, 9, 11, 13, 20, 23]
2. buatlah algoritma pengurutan secara ascending data 17,9,12,8,10,6,13,20,11,21 dengan metode selection short!
code in python:
def Selection_Sort(A):
for i in range(len(A)):
print('iterasi', i, A)
min_idx = i
for j in range(i+1, len(A)):
if A[min_idx] > A[j]:
min_idx=j
A[i], A [min_idx] = A[min_idx], A[i] # proses pertukaran di python
A = [17,9,12,8,10,6,13,20,11,21]
Selection_Sort(A)
Output code python:
iterasi 0 [17, 9, 12, 8, 10, 6, 13, 20, 11, 21]
iterasi 1 [6, 9, 12, 8, 10, 17, 13, 20, 11, 21]
iterasi 2 [6, 8, 12, 9, 10, 17, 13, 20, 11, 21]
iterasi 3 [6, 8, 9, 12, 10, 17, 13, 20, 11, 21]
iterasi 4 [6, 8, 9, 10, 12, 17, 13, 20, 11, 21]
iterasi 5 [6, 8, 9, 10, 11, 17, 13, 20, 12, 21]
iterasi 6 [6, 8, 9, 10, 11, 12, 13, 20, 17, 21]
iterasi 7 [6, 8, 9, 10, 11, 12, 13, 20, 17, 21]
iterasi 8 [6, 8, 9, 10, 11, 12, 13, 17, 20, 21]
iterasi 9 [6, 8, 9, 10, 11, 12, 13, 17, 20, 21]
maaf klo salah
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Last Update: Tue, 11 Jul 23