Download - 6.3 & 7.3 NESTED LOOP
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1#33
6.3 & 7.3NESTED LOOP
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QUEUE( ANTRIAN )
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2. 01
A Queue is an ordered collection of items into which new items may be inserted at one end (called the rear of the queue) and from which items may be deleted at one end (called the front of the queue) .( Yedidyah L, Moshe J. A.,
and Aaron M. Tenenbaum; Data Structures Using C and C++)
InsertDeleteREARFRONT
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2. 01
Ada 3 macam struktur QUEUE (Antrian)yang menggunakan Array Satu Dimensi yangakan dibahas:
I.
II.
III.
LINEAR QUEUE
CIRCULAR QUEUE
DOUBLE ENDED QUEUE (DEQUE)
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2. 01
1. LINEAR QUEUE (Antrian Lurus).
F = FrontR = Rear
2. CIRCULAR QUEUE (Antrian Melingkar).
3. DOUBLE ENDED QUEUE (Antrian Dengan Ujung Ganda ).
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
L = LEFTR = RIGHT
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
n-10 1 2 3 4 5 6 7 8 9
FR
X X X X X X
n-10 1 2 3 4 5 6 7 8 9
L R
X X X X
ILUSTRASI
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2. 02
Front = depanRear = Belakang
front Rear
KELUAR MASUK
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2. 02
Front = depanRear = Belakang
front Rear
KELUAR MASUKA B C D
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8#33
2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
A B C D
A B C D
n-10 1 2 3 4 5 6 7 8 9
0 1 2 3 4
F R
FRONT REAR
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2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
A B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
A keluar
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11#33
2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
Geser B
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12#33
2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
Geser C
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13#33
2. 02Ilustrasi Linear Queue menggunakan array Satu dimensi
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
Geser D
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2. 02AlgoritmaPenggeseran isi antrian
A B C D
n-10 1 2 3 4 5 6 7 8 9
B C D D
n-10 1 2 3 4 5 6 7 8 9
Dari :
Menjadi :
F R
F R Algoritma :
for(I=F; I<=R-1; I++) { Q[I] = Q[I+1]; }R--;
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2. 02AlgoritmaPenggeseran isi antrian
A B C D
n-10 1 2 3 4 5 6 7 8 9
B C D D
n-10 1 2 3 4 5 6 7 8 9
Dari :
Menjadi :
F R
F R Algoritma :
for(I=F; I<=R-1; I++) { Q[I] = Q[I+1]; }R--;Penggeseran
BUTUH WAKTU!
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2. 02Loket yang digeser
A B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
perhatikan posisi F terhadap Loket
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2. 02Loket yang digeser
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
B C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
C D
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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21#33
2. 02Loket yang digeser
D
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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22#33
2. 02Loket yang digeser
D
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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2. 02Loket yang digeser
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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2. 02Loket yang digeser
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
F > Ratau F = R + 1berarti : ANTRIAN KOSONG
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2. 02Loket yang digeser
E
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
E F
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
E F G
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
E F G H
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
E F G H I
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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30#33
2. 02Loket yang digeser
E F G H I J
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
F G H I J
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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32#33
2. 02Loket yang digeser
E F G H I J
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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33#33
2. 02Loket yang digeser
F G H I J
n-10 1 2 3 4 5 6 7 8 9LOKET
F R
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2. 02Loket yang digeser
J
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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35#33
2. 02Loket yang digeser
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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2. 02Semua antriansudah keluar (dilayani)
n0 1 2 3 4 5 6 7 8 9LOKET
FRBila semua antrian sudah dilayanimaka keadaan perlu dikembalikan ke keadaan semula(perlu diRESET)
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2. 02Kembali kekeadaan awal
n-10 1 2 3 4 5 6 7 8 9LOKET
FR
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2. 02
1.1. ILUSTRASI ANTRIAN LURUS.
F = Front (depan)R = Rear (Belakang)
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
Representasi Linear Queue dalam array satu dimensi
#define n 10----int Q[n];int F, R, X;
misal dibuat dengan :
Q[ ]
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2. 03
PRINSIP / KONSEP PROSES :
FIFO (First In First Out)
atau FIFS (First In First Serve)
1.2.
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2. 03
PROSES.
a. AWAL (Inisialisasi)
b. INSERT (Isi, Sisip, Masuk, Simpan, Tulis)
c. DELETE ( Hapus, Keluar, Ambil atau Dilayani, Baca) d. RESET (Kembali ke keadaan awal)
1.3.
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2. 03Algoritma dasar untukproses AWAL (Inisialisasi),ditulis dalam bahasa C.
Ilustrasi hasil proses awal :
a.
void AWAL(void){ F = 0; R = -1; }
n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
0 -1
F R X
F > RatauF = R + 1Antrian KOSONG
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2. 03
Ilustrasi hasil proses awal :
n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
0 -1
F R X
Hanya bisa Insert
F > RatauF = R + 1Antrian KOSONG
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2.03
Dalam struktur LINEAR QUEUE,
digunakan istilah :
- INSERT untuk : Simpan, atau Masuk,
atau
Isi atau Tulis.
- DELETE untuk : Ambil, atau Keluar,
atau
Baca, atau HapusPrinsip atau Konsep prosesnya :
F I F O ( First In First Out )
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2. 03
F = Front (depan)R = Rear (Belakang)
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
DELETE INSERT
Ilustrasi Linear Queue
Pertanyaan : dalam keadaan seperti sekarang
Kalau ada perintah INSERT ( Isi, Masuk, Simpan)
maka akan diisi di elemen nomor : ?dan R akan menunjuk elemen nomor : ? F akan menunjuk elemen nomor : ?
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2. 03 n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
DELETE INSERT
Pertanyaan : dalam keadaan seperti sekarang
Kalau ada perintah INSERT ( Isi, Masuk, Simpan)
maka akan diisi di elemen nomor : 7dan R akan menunjuk elemen nomor : 7 F akan menunjuk elemen nomor : tetap di 3
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2. 03 n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
DELETE INSERT
Pertanyaan : dalam keadaan seperti sekarang
Kalau ada perintah DELETE ( Ambil, Keluar, Hapus)maka yang akan diambil adalah isi di elemen
nomor : ?Setelah itu akan terlihat :
R menunjuk elemen nomor : ? dan F menunjuk elemen nomor : ?
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2. 03 n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
DELETE INSERT
Pertanyaan : dalam keadaan seperti sekarang
Kalau ada perintah DELETE ( Ambil, Keluar, Hapus)maka yang akan diambil adalah isi di elemen nomor :
3Setelah itu akan terlihat :
R menunjuk elemen nomor : Tetap menunjuk 6 dan F menunjuk elemen nomor : 4
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Algoritma dasaruntuk
INSERT
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n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
0 -1
F R
AWAL
X
Mau Insert
F > RatauF = R + 1Antrian KOSONG
ANTRIAN KOSONG
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50#33
n-10 1 2 3 4 5 6 7 8 9
F
R
Q[ ]
0 0
F R
R maju satu langkah---> ( R=R+1);
X
F
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51#33
n-10 1 2 3 4 5 6 7 8 9
F
R
a Q[ ]
0 0
F R
Isi ditempat yang ditunjuk oleh R---> Q[R] = X;
a
X
F
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n-10 1 2 3 4 5 6 7 8 9
F R
a Q[ ]
0 1
F R
a
X
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53#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b Q[ ]
0 1
F R
b
X
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54#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b Q[ ]
0 2
F R
b
X
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55#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c Q[ ]
0 2
F R
c
X
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56#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c Q[ ]
0 3
F R
c
X
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57#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c d Q[ ]
0 3
F R
d
X
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58#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c d Q[ ]
0 4
F R
d
X
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n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e Q[ ]
0 4
F R
e
X
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n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e Q[ ]
0 5
F R
e
X
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61#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e f Q[ ]
0 5
F R
f
X
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62#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e f Q[ ]
0 6
F R
f
X
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63#33
n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e f g Q[ ]
0 6
F R
g
X
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2. 02
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
Algoritma dasar untuk INSERTpada antrian Q[ ]
Pilih satu yang benar
a. R = R + 1;Q[R] = X;
b. Q[R] = X;R = R + 1;
d. X = Q[R];R = R + 1;
c. R = R + 1;X = Q[R];
?
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2. 02
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
Algoritma dasar untuk INSERTpada antrian Q[ ]
Algoritma Insert yang benar
a. R = R + 1;Q[R] = X;
b. Q[R] = X;R = R + 1;
d. X = Q[R];R = R + 1;
c. R = R + 1;X = Q[R];
![Page 66: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/66.jpg)
66#33
2. 02Algoritma dasar untuk INSERTpada antrian Q[ ]
R = R + 1;Q[R] = X;
R+=1;Q[R] = X;
R++;Q[R] = X;
Q[++R] = X;
![Page 67: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/67.jpg)
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n-10 1 2 3 4 5 6 7 8 9
F R
a b c d e f g Q[ ]
0 6
F R
Dalam Keadaan seperti iniAkan di DELETE
g
X
Akan diDELETE
Isi elemen no : ?
![Page 68: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/68.jpg)
68#33
n-10 1 2 3 4 5 6 7 8 9
F R
b c d e f g Q[ ]
0 6
F R
Ambil isi yang ditunjuk oleh F ---> X = Q[F];
a
X
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n-10 1 2 3 4 5 6 7 8 9
F R
b c d e f g Q[ ]
1 6
F R
Kemudian F maju satu langkah ---> F = F + 1;
a
X
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n-10 1 2 3 4 5 6 7 8 9
F R
c d e f g Q[ ]
1 6
F R
b
X
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n-10 1 2 3 4 5 6 7 8 9
F R
c d e f g Q[ ]
2 6
F R
b
X
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n-10 1 2 3 4 5 6 7 8 9
F R
d e f g Q[ ]
2 6
F R
c
X
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n-10 1 2 3 4 5 6 7 8 9
F R
d e f g Q[ ]
3 6
F R
c
X
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n-10 1 2 3 4 5 6 7 8 9
F R
X X X X Q[ ]
3 6
F R
Ini yang digunakan untuk :ILUSTRASI LINEAR QUEUE
X
![Page 75: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/75.jpg)
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n-10 1 2 3 4 5 6 7 8 9
F R
d e f g Q[ ]
3 6
F R
DELETE LAGI SAMPAI KOSONG
c
X
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n-10 1 2 3 4 5 6 7 8 9
F R
e f g Q[ ]
3 6
F R
d
X
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n-10 1 2 3 4 5 6 7 8 9
F R
e f g Q[ ]
4 6
F R
d
X
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n-10 1 2 3 4 5 6 7 8 9
F R
f g Q[ ]
4 6
F R
e
X
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n-10 1 2 3 4 5 6 7 8 9
F R
f g Q[ ]
5 6
F R
e
X
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n-10 1 2 3 4 5 6 7 8 9
F R
g Q[ ]
5 6
F R
f
X
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n-10 1 2 3 4 5 6 7 8 9
F R
g Q[ ]
6 6
F R
f
X
F = R
Isi antrian tinggalSatu elemen
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n-10 1 2 3 4 5 6 7 8 9
F R
Q[ ]
6 6
F R
g
X
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n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
7 6
F R
g
X
F > RatauF = R + 1Antrian Kosong
ANTRIAN KOSONG
![Page 84: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/84.jpg)
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n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
7 6
F R
g
X
ANTRIAN KOSONG SEKARANG MAU DIISI LAGI
Disi di kolom : ?
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n-10 1 2 3 4 5 6 7 8 9
F R
Q[ ]
7 7
F R
g
X
Misal akan diisi sampai penuh
R maju satu langkah
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n-10 1 2 3 4 5 6 7 8 9
F R
h Q[ ]
7 7
F R
h
XIsi ditempat yangditunjuk oleh R
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n-10 1 2 3 4 5 6 7 8 9
F R
h Q[ ]
7 7
F R
h
X
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n-10 1 2 3 4 5 6 7 8 9
F R
h i Q[ ]
7 8
F R
i
X
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n-10 1 2 3 4 5 6 7 8 9
F R
h i Q[ ]
7 9
F R
i
X
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n-10 1 2 3 4 5 6 7 8 9
F R
h i jQ[ ]
7 9
F R
j
XR = n
Antrian PENUH
ANTRIAN PENUH
![Page 91: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/91.jpg)
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n-10 1 2 3 4 5 6 7 8 9
F R
h i jQ[ ]
7 9
F R
j
X
ANTRIAN PENUH Sekarang mau di DELETEDelete kolom berapa : ?
![Page 92: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/92.jpg)
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n-10 1 2 3 4 5 6 7 8 9
F R
h i jQ[ ]
7 6
F R
j
X
Misal mau diDELETE sampai kosong(habis)
![Page 93: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/93.jpg)
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n-10 1 2 3 4 5 6 7 8 9
F R
i jQ[ ]
7 6
F R
h
X
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n-10 1 2 3 4 5 6 7 8 9
F R
i jQ[ ]
8 9
F R
h
X
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n-10 1 2 3 4 5 6 7 8 9
F R
jQ[ ]
8 9
F R
i
X
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n-10 1 2 3 4 5 6 7 8 9
F R
jQ[ ]
9 9
F R
i
XF = RIsi antrianTinggal satu elemen
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n-10 1 2 3 4 5 6 7 8 9
F R
Q[ ]
9 9
F R
j
X
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n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
10 9
F R
j
X
F = R+1Antrian KOSONG
tak ada isnya
Antrian KOSONG
F=n+1
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n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
10 9
F R
j
X
F = R+1Antrian KOSONG
R = n, antrian PENUHdalam arti antrian Tak Bisa Diisi
Antrian Kosong,juga Penuh,Antrian perlu diRESET
![Page 100: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/100.jpg)
100#33
n-10 1 2 3 4 5 6 7 8 9
FR
Q[ ]
0 -1
F R X
Antrian sudah diRESETF = 0,R = -1
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101#33
2. 02
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
Algoritma dasar untuk DELETEpada antrian Q[ ]
Pilih satu yang benar
a.
F = F + 1;Q[F] = X;
b.
Q[F] = X;F = F + 1;
d.
X = Q[F];F = F + 1;
c.
F = F + 1;X = Q[F];
?
![Page 102: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/102.jpg)
102#33
2. 02
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
3 6
F R X
Algoritma dasar untuk DELETEpada antrian Q[ ]
Pilih satu yang benar
a.
F = F + 1;Q[F] = X;
b.
Q[F] = X;F = F + 1;
d.
X = Q[F];F = F + 1;
c.
F = F + 1;X = Q[F];
?
![Page 103: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/103.jpg)
103#33
2. 02Algoritma Deletepada LINEAR QUEUE
X = Q[F];F = F + 1;
X = Q[F];F+=1;
X = Q[F];F++;
X = Q[F++];
atau :
atau :
atau :
![Page 104: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/104.jpg)
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1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
Gambar no : ?
![Page 105: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/105.jpg)
105#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
Ciri : ?
![Page 106: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/106.jpg)
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1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
F = R + 1
Gambar no : ?
1 75
![Page 107: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/107.jpg)
107#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
F = R + 1 Gambar no : ?
Ciri : ?
![Page 108: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/108.jpg)
108#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
F = R + 1
R = n-1
Gambar no : ?
![Page 109: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/109.jpg)
109#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
1 2 3
4 5
F = R + 1
R = n-1
Ciri : ?
![Page 110: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/110.jpg)
110#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
1 2 3
4 5
F = R + 1
R = n-1
R < n-1
Gambar no : ?
![Page 111: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/111.jpg)
111#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
2 3 4 6
1 2 3
4 5
F = R + 1
R = n-1
R < n-1
Ciri : ?
![Page 112: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/112.jpg)
112#33
1.4 KONDISI ANTRIAN n-1 0 1 2 3 4 5 6 7 8 9
2. 05
FR
X X X X X X X
0 1 2 3 4 5 6 7 8 9
X X X X
X
0 1 2 3 4 5 6 7 8 9
X X X
F R
F R
F R
FR
F R
FR
1
2
3
4
5
6
7
KOSONG :
PENUH :
BISA DIISI :
ADA ISINYA :
1 75
6 7
2 3 4 6
1 2 3
4 5
F = R + 1
R = n-1
R < n-1
F < R+1
![Page 113: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/113.jpg)
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1.5Algoritma yang lengkap
untuk
INSERT&
DELETE
![Page 114: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/114.jpg)
114#33
2. 07Algoritma yang lengkap untuk proses INSERT
void INSERT(void){ if ( R < n-1 ) { R = R + 1; Q[R] = X; } else printf( “Antrian Penuh”);}
Algoritma yang lengkap untuk proses INSERTdengan ‘logika terbalik’
void INSERT(void){ if ( R == n-1 ) printf(“Antrian Penuh’); else { R = R + 1; Q[R] = X; } }
a.
![Page 115: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/115.jpg)
115#33
Algoritma yang lengkap untuk proses DELETE
void DELETE(void){ if ( F < R+1 ) { X = Q[F] ; F = F + 1; if ( F == n ) { F = 0; R = -1; } } else printf( “Antrian Kosong”);}
2. 08
Algoritma yang lengkap untuk proses DELETE dengan ‘logika terbalik’
void DELETE(void){ if ( F == R+1 ) printf(“Antrian Kosong’); else { X = Q[R]; F = F + 1; if ( F == n ) { F = 0; R = 0; } } }
b.
![Page 116: 6.3 & 7.3 NESTED LOOP](https://reader036.vdocument.in/reader036/viewer/2022081415/56814dd5550346895dbb3954/html5/thumbnails/116.jpg)
116#33
2. 091.6. CONTOH PROGRAM LINEAR QUEUE.
#include<stdio.h>#include<conio.h>#define n 10
char Q[n];int F,R; void Insert(char X);void Delete(void);void Awal(void);void Reset(void);
int main(){ int I; char X, Y; clrscr(); Awal(); - scanf(“%i”, &X); Insert(X); - - Delete(); - -}
void Awal(void) { F = 0; R = -1; }
void Insert( char X) { if (R < n-1) { R++; Q[R] = X; } else printf(“\nAntrian Penuh “); }
void Delete( void) { char X; if( F < R+1 ) { X = Q[F]; printf(“%c”, X); F++; if(F == n) Reset( ); else printf(“Antrian Kosong”); }
void Reset(void) { F = 0; R = -1; }
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2. 101.7. SOAL SOAL LATIHAN MANDIRI
1. Sebutkan ciri bahwa Queue :
a.b.c.d.e.
f.g.h.
Kosong tak ada isinyaPenuh, tak bisa diisiBisa diisiAda isinyaAntrian tak bisa diisi lagi, tapi belum ada isi antrianyang sudah keluar atau sudah dilayani.Antrian perlu diReset.Hanya berisi 10 pengantri (yang belum keluar/dilayani)Sisa tempat yang masih bisa diisi hanya ada 10 tempat (elemen)
Soal untuk Linear Queue seperti yang diilustrasikan pada gambar diatas
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
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2. 101.7. SOAL SOAL LATIHAN MANDIRI
2. Tulis algoritma dasar untuk:
4.Tulis algoritma yang lengkap untuk mengisi antrian record per record sampai antrian penuh tak bisa diisi lagi.
a.b.c.d.
Inisialisasi.Insert sebuah recordDelete sebuah recordReset.
3. Tulis algoritma yang lengkap untuk :a.b,Insert sebuah recordDelete sebuah record
Soal untuk Linear Queue seperti yang diilustrasikan pada gambar diatas
n-10 1 2 3 4 5 6 7 8 9
F R
X X X X
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2. 101.7. SOAL SOAL LATIHAN MANDIRI
5.Tulis algoritma yang lengkap untuk mendelete isi antrian record per record sampai antrian kosong
7.
6.Tulis algoritma yang lengkap untuk mengisi antrian record per record sebanyak 10 recordselama antrian beleum penuh. Apabila antrian penuh, walaupun belum mengisi 10 record, proses pengisian dihentikan.Tulis algoritma yang lengkap untuk mendelete isi antrian record per record sebanyak 10 record selama antrian masih ada isinya. Bila antrian sudah kosong, walaupun belum mendelete sebanyak 10 record, maka proses delete dihentikan.
8. Tulis algoritma untuk menghitung dan mencetak jumlah tempat (elemen) yang ada isinya.
9. Tulis algoritma untuk menghitung dan mencetak jumlah tempat yang masih bisa diisi.
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6.3 & 7.3NESTED LOOP
bersambungke
Circular Queue