“Algoritmlarni loyihalash” fani bo‘yicha 2-laboratoriya ishi variantlari va topshiriq
bandlari:
Berilgan jadvaldan n nomerdan boshlab 22 tа qiymat ko’chirib olinsin. bunda n
talabalarni guruh jurnalidagi tartib raqami. Jadvaldagi
variantga
deb olinadi
i=0,1,2,…,n ko’rinishida belgilanadi.
va variant jadvali
i
x
y
i
x
y
i
x
y
0
-3.0
84.0370
31
0.1
10.9361
61
3.1
-13.4719
1
-2.9
68.8241
32
0.2
12.7296
62
3.2
-14.6544
2
-2.8
55.2336
33
0.3
14.3601
63
3.3
-15.4959
3
-2.7
43.1361
34
0.4
15.8096
64
3.4
-15.9424
4
-2.6
32.4416
35
0.5
17.0625
65
3.5
-15.9375
5
-2.5
23.0625
36
0.6
18.1056
66
3.6
-15.4224
6
-2.4
14.9136
37
0.7
18.9281
67
3.7
-14.3359
7
-2.3
7.9121
38
0.8
19.5216
68
3.8
-12.6144
8
-2.2
1.9776
39
0.9
19.8801
69
3.9
-10.1919
9
-2.1
-2.9679
40
1.0
20.0370
70
4.0
-6.9630
10
-2.0
-6.9630
41
1.1
19.8801
71
4.1
-2.9679
11
-1.9
-10.1919
42
1.2
19.5216
72
4.2
1.9776
12
-1.8
-12.6144
43
1.3
18.9281
73
4.3
7.9121
13
-1.7
-14.3359
44
1.4
18.1056
74
4.4
14.9136
14
-1.6
-15.4224
45
1.5
17.0625
75
4.5
23.0625
15
-1.5
-15.9375
46
1.6
15.8096
76
4.6
32.4416
16
-1.4
-15.9424
47
1.7
14.3601
77
4.7
43.1361
17
-1.3
-15.4959
48
1.8
12.7296
78
4.8
55.2336
18
-1.2
-14.6544
49
1.9
10.9361
79
4.9
68.8241
19
-1.1
-13.4719
50
2.0
9.0370
80
5.0
84.0370
20
-1.0
-11.9630
51
2.1
6.9441
81
5.1
100.8561
21
-0.9
-10.2879
52
2.2
4.7936
82
5.2
119.4897
22
-0.8
-8.3824
53
2.3
2.5761
83
5.3
140.0001
23
-0.7
-6.3279
54
2.4
0.3216
84
5.4
162.4896
24
-0.6
-4.1664
55
2.5
-1.9375
85
5.5
187.0625
25
-0.5
-1.9375
56
2.6
-4.1664
86
5.6
213.8256
26
-0.4
0.3216
57
2.7
-6.3279
87
5.7
242.8882
27
-0.3
2.5761
58
2.8
-8.3824
88
5.8
274.3617
28
-0.2
4.7936
59
2.9
-10.2879
89
5.9
308.3601
29
-0.1
6.9441
60
3.0
-11.9630
90
6.0
345.0370
30
0.0
9.0370
1.Tanlangan jadval asosida chiziqli model tuzilsin
2.Tanlangan jadval asosida kvadratik model tuzilsin.
2-misol: N talabaning tartib raqami. n1=N%2, ya’ni 2ga bo’lgandagi qoldig’i.
f ( x)  ( N  2) x n1 1  (1) N (5N  3) ,
  ; 
Funksiyani furye qatoriga yoying. Masalan: N=5 bo’lsa,
f(x)=7x2-22,
  ;  
3-misol
1. Variantda berilgan graf qirralari narxlari matritsasiga ko‘ra planar graf chizing.
Izoh: C=(Cij) matritsa Cij elementi grafning i-va j-uchlarini tutashtiruvchi qirra
bo‘yicha harakat narxi. Cij=0 bo‘lgan hol i-, j- uchlarini tutashtiruvchi qirra
yo‘qligini bildiradi.
Variantlarni yechimiga namuna narxlar matritsasi bo‘lgan holni ko‘ramiz
С=
Grafni qurishda birinchi uchi sifatida eng yuqori karrali uchlaridan birini olgan
ma’qul. So‘ngra shu uchidan chiqqan qirralarini quramiz. Hosil bo‘lgan
qirralar uchlaridan esa navbatdagi qirralarini tuzishni toki barcha qirralari
shaklanguncha davom ettiriladi. Xususan, berilgan variantdagi graf
matritsasiga ko‘ra uch karrali uchlaridan biri V6 ni tanlaymiz va undan
chiquvchi
V6 V1, V6 V3, V6 V4 qirralarini chizamiz. So‘ngra V1 uchidan chiquvchi
1.
2. V1 V2, V1 V5 qirralarni chizamiz. Shundan so‘ng keyingi qirralarni
3. V3 V2, V3 V4 larni chizamiz
4. oxirida V4 V5 qo‘shiladi
Вариант № 1
0

5
0
C 
0
0

8
Вариант № 2
5
0
0
0
0
7
0
9
7 0 10 6
0 10 0 0
9
6
0
0
0
0
12 4
8

0
0

12 
4

0 
 0 10 0 0 9 7 


10 0 12 0 0 6 
 0 12 0 13 0 8 
C 

 0 0 13 0 11 0 
 9 0 0 11 0 0 


 7 6 8 0 0 0
Вариант № 3
 0 8 0 0 10 0 


 8 0 12 0 0 0 
 0 12 0 6 0 7 
C 

 0 0 6 0 11 9 
10 0 0 11 0 5 


 0 0 7 9 5 0
Вариант № 4
0 8 0

 8 0 11
 0 11 0
C 
0 0 5
12 0 0

0 4 7
Вариант № 5
 0 6 0 0 10 8 


 6 0 12 0 0 0 
 0 12 0 9 0 7 
C 

 0 0 9 0 11 6 
10 0 0 11 0 0 


 8 0 7 6 0 0
0 12 0 

0 0 4
5 0 7

0 6 0
6 0 9

0 9 0 
Вариант № 6
0

5
8
C 
0
0

7
7

0 6 0 0 0
6 0 9 10 0 

0 9 0 7 0
0 10 7 0 11

0 0 0 11 0 
5
8
0
0
Вариант № 7
Вариант № 8
 0 11 0 9 7 6 


11 0 12 0 0 8 
 0 12 0 11 0 0 
C 

 9 0 11 0 5 0 
 7 0 0 5 0 0


 6 8 0 0 0 0
 0 10 0 0 9 7 


10 0 7 11 0 0 
 0 7 0 8 0 0
C 

 0 11 8 0 12 0 
 9 0 0 12 0 6 


 7 0 0 0 6 0
Вариант № 9
Вариант № 10
 0 12 0 0 10 9 


12 0 5 8 0 7 
 0 5 0 6 0 0
C 

 0 8 6 0 13 0 
10 0 0 13 0 0 


 9 7 0 0 0 0
 0 8 0 0 7 10 


 8 0 12 0 0 11 
 0 12 0 11 0 0 
C 

 0 0 11 0 6 0 
7 0 0 6 0 9


10 11 0 0 9 0 
Вариант № 11
0

8
12
C 
0
10

 13
10 13 

0 9 0 0 0
9 0 10 0 8 

0 10 0 11 0 
0 0 11 0 5 

0 8 0 5 0 
8 12
0
Вариант № 12
0 9 0 0 0 6 


 9 0 10 13 0 0 
 0 10 0 8 0 5 
C 

 0 13 8 0 7 0 
 0 0 0 7 0 11


 6 0 5 0 11 0 
Вариант № 13
0

9
0
C 
12
10

 11
9 0 12 10 11

0 7 9 0 0
7 0 8 0 0

9 8 0 6 0
0 0 6 0 8

0 0 0 8 0 
Вариант № 14
0

9
 11
C 
0
0

10
Вариант № 15
 0 10 0 0 6 7 


10 0 12 0 0 0 
 0 12 0 8 4 0 
C 

 0 0 8 0 0 5
 6 0 4 0 0 9


 7 0 0 5 9 0
9 11 0
0
7
0
7
0
0
6
6
0
0
8
7
0 13 0
10 

0 0
8 13 

7 0
0 10 

10 0 
0
Вариант № 16
0

9
10
C 
0
8

6
8 6

0 7 0 0 0
7 0 11 0 0 

0 11 0 7 0 
0 0 7 0 5

0 0 0 5 0 
9 10
0
Вариант № 17
Вариант № 18
 0 12 10 0 13 7 


12 0 11 0 0 0 
10 11 0 9 0 0 
C 

 0 0 9 0 11 0 
 13 0 0 11 0 9 


 7 0 0 0 9 0
 0 13 0 0 12 8 


 13 0 11 0 0 0 
 0 11 0 9 0 0 
C 

 0 0 9 0 10 7 
12 0 0 10 0 6 


 8 0 0 7 6 0
Вариант № 19
Вариант № 20
 0 9 0 0 7 12 


 9 0 11 0 6 0 
 0 11 0 10 0 0 
C 

 0 0 10 0 8 6 
7 6 0 8 0 0


12 0 0 6 0 0 
0 8 0 0 5

 8 0 12 0 0
 0 12 0 8 0
C 
 0 0 8 0 11
 5 0 0 11 0

0 9 0 4 7
0

9
0

4
7

0 
Вариант № 21
Вариант № 22
 0 6 0 0 12 7 


 6 0 11 0 0 9 
 0 11 0 10 0 5 
C 

 0 0 10 0 8 0 
12 0 0 8 0 0 


 7 9 5 0 0 0
 0 11 0 0 0 12 


 11 0 6 0 0 0 
0 6 0 7 9 0
C 

 0 0 7 0 10 8 
 0 0 9 10 0 11 


12 0 0 8 11 0 
Вариант № 23
Вариант № 24
0

5
8
C 
0
7

12
5 8
0 6
6 0
0 9
0 0
0 0
7 12 

0 0 0
9 0 0

0 0 10 
0 0 9

10 9 0 
0
0

7
 11
C 
0
0

10
0 10 

0 8 0 0 0
8 0 12 0 0 

0 12 0 6 9 
0 0 6 0 7

0 0 9 7 0 
7 11
0
Вариант № 25
Вариант № 26
 0 11 0 0 12 0 


 11 0 5 9 0 0 
 0 5 0 7 0 0
C 

 0 9 7 0 11 6 
12 0 0 11 0 8 


 0 0 0 6 8 0
0 8 0 0 0 7


 8 0 6 0 12 9 
 0 6 0 8 10 0 
C 

 0 0 8 0 11 0 
 0 12 10 11 0 9 


7 9 0 0 9 0
Вариант № 27
Вариант № 28
0

6
8
C 
13
 11

7
13 11 7 

0 7 0 0 0
7 0 10 0 0 

0 10 0 10 0 
0 0 10 0 9 

0 0 0 9 0 
6
8
 0 11 0 0 10 0 


 11 0 10 0 0 5 
 0 10 0 8 12 13 
C 

0 0 8 0 9 0
10 0 12 9 0 8 


 0 5 13 0 8 0 
Вариант № 29
Вариант № 30
 0 7 0 0 13 8 


 7 0 11 0 0 0 
 0 11 0 6 0 5 
C 

0 0 6 0 9 0
13 0 0 9 0 10 


 8 0 5 0 10 0 
 0 9 0 0 10 11 


 9 0 11 0 0 0 
 0 11 0 9 13 0 
C 

0 0 9 0 7 0
10 0 13 7 0 12 


 11 0 0 0 12 0 
Вариант № 31
Вариант № 32
0

9
0
C 
0
9

7
7

0 8 0 0 6
8 0 13 0 8 

0 13 0 11 0 
0 0 11 0 0 

6 8 0 0 0 
9
0
0
9
0

5
0
C 
0
0

8
5
0
0
0
0
7
0
9
7 0 10 6
0 10 0 0
9
6
0
0
0
0
12 4
8

0
0

12 
4

0 