Moslashuvchan - Research and education

Moslashuvchan (adaptiv) tasodifiy qidiruv usuli yordamida quyida keltirilgan
3ta test ko‘p o‘zgaruvchili optimallashtirish masalalarning yechimlarini topishda foydalanish mumkin.
”волна” funksiyasini hisoblash.
Masalaning berilishi:
𝑓(𝑥) = −2𝑥𝑒𝑖𝑥2 − 2 sin(200𝑥) → 𝑚𝑎𝑥
𝑥∗ = 0 𝑓∗ = −0.7 [−𝛼; 𝛼]| 𝛼 = 2
Masalaning yechimi:
-1.6285761637234164
-1.0108357120851046
-0.7
-1.0108357120851046
-1.6285761637234164
Ikki o‘lchamli funksiya. Levy funksiyasini hisoblash.
Masalaning berilishi:
𝑥 − 1
𝑥 − 1 2
𝑥 − 1
9
𝑓(𝑥) = 𝑠𝑖𝑛2 (𝜋 (1 + 1 )) + ∑ ( 𝑖−1 ) [1 + 10𝑠𝑖𝑛2 (𝜋 (1 + 1 ))]
4
2
+ (𝑥10 − 1)
4
→ 𝑚𝑖𝑛
4 4
𝑖=2
𝑥∗ = (0; 0; … . ; 0) 𝑓∗ = 0 [−𝛼; 𝛼]𝑥 𝛼 = 10
Masalaning yechimi:
[[50. 50.94739946 55.05140639 ... 55.05140639 50.94739946
50. ]
[50.94739946 51.89479891 55.99880584 ... 55.99880584 51.89479891
50.94739946]
[55.05140639 55.99880584 60.10281278 ..0... 60.10281278 55.99880584
55.05140639]
[55.05140639 55.99880584 60.10281278 ... 60.10281278 55.99880584
55.05140639]
[50.94739946 51.89479891 55.99880584 ... 55.99880584 51.89479891
50.94739946]
[50. 50.94739946 55.05140639 ... 55.05140639 50.94739946
50. ]]
Rastrigin funksiyasini hisoblash. Masalaning berilishi:
|𝑥|
𝑖
𝑓(𝑥) = ∑(𝑧2 − 10 cos(2𝜋𝑧𝑖) + 10) + 𝑓0 → 𝑚𝑖𝑛
𝑖=1
𝑥∗ = 𝑥0; … . . ; 𝑥0 𝑓∗ = 𝑓0 = −330
1 |𝑥|
|
[−𝛼; 𝛼]|𝑥|
|
𝛼 = 5
|
-300
|
|
|
-285
|
|
|
-242.0
|
|
|
-200.0
|
|
|
-162.0
|
|
|
-128.0
|
|
|
-98.0
|
|
|
-72.0
|
|
|
-50.0
|
|
|
-32.0
|
|
|
-18.0
|
|
|
-8.0
|
|
|
-2.0
|
|
|
1.3497838043956716e-31
|
|
|
2.0
|
|
|
8.0
|
|
|
18.0
|
|
|
32.0
|
|
|
50.0
72.0
98.0
128.0
162.0
Endi Logistik regressiya asosida klassifikatsiyalash algoritmi asosida hosil bo‘lgan ko‘p o‘zgaruvchili optimallashtirish masalasining yechimlarini topishda moslashuvchan (adaptiv) tasodifiy qidiruv usuli yordamida yaratilgan dastur orqali elektron pochtadagi xabarlarni toifalarga ajratish masalasining yechimini ko‘ramiz.
Jami 5575 ta spam va ham xabarlar olingan birinchi navbata 3 ustunga bo‘lib olingan ichidan 10 xabar tanlab olinadi.
ID
|
Turi
|
Xabar
|
1901
|
spam
|
Sorry, I’ll call later
|
5528
|
ham
|
Its just the effect of irritation. Just ignore it
|
3581
|
spam
|
You are right. Meanwhile how’s project twins c...
|
3960
|
spam
|
Your dad is back in ph?
|
403
|
ham
|
None of that’s happening til you get here though
|
2196
|
spam
|
Not much, just some textin’. How bout you? Website https://mov.com
|
917
|
ham
|
Not much, just some textin’. How bout you?
|
2825
|
ham
|
When people see my msgs, They think Iam addict...
|
2662
|
spam
|
Hello darling how are you today? I would love ...
|
1274
|
ham
|
Let me know how to contact you. I’ve you settl...
|
Shundan so‘ng dasturning o‘zi bu xabarlarni ikkita guruhga ajratib oladi yani spam va ham gurihiga va ularni fozi nisbatini chiqarib beradi bizning dasturimizda u quyidagi nisbatda bo‘ladi
Shundan so‘ng sapam va ham habarlani guruhlarga ajratilgandan so‘ng ularni 8 parametrlar bo‘yicha tekshirib chiqamiz
Natija quyidagicha chiqqanligini ko‘rishimiz mumkin.
Masalani nazorat tanlanmasi k=5575ta obyektlar uchun hisoblash tajribasini о‘tqazamiz. Bunda nazorat tanlanmasidagi obyektlar(xabarlar)ni ikkita sinf(ham yoki spam)ga toifalash maqsadida quyidagi produksion model yordamida ifodalangan qoidadan foydalanamiz:
6
spam, 𝑎𝑔𝑎𝑟 ∑ 𝑥 𝑗 ≥ 𝑏
𝑦𝑜𝑘𝑖 (𝑥7 > 0.06 𝑦𝑜𝑘𝑖 𝑥8 > 0.03 )
𝑥𝑖 = {
𝑖 𝑞 𝑖
𝑗=1
𝑖 }
ham, 𝑏𝑜𝑠ℎ𝑞𝑎 ℎ𝑜𝑙𝑙𝑎𝑟𝑑𝑎
i=1,2, . . . , k. bq - belgilar yig‘indisi uchun bo‘sag‘aviy parametr, ko‘rilayotgan masala
uchun xususiy xolda
bq = 2 olish tavsiya etiladi. Ushbu k=5575 nazorat tanlanmasi
uchun to‘g‘ri toifalash ko‘rsatkichlari “spam” sinfi obektlari uchun 4828ta obektdan 4791tani, ya’ni 99,23%ni, “ham” sinfi obektlari uchun 747ta obektdan 738tani, ya’ni 98,8%ni tashkil etdi.
Ikkinchi turdagi xatolik bo‘yicha 10 ta “spam” xabarni 0.2% va 5 ta “ham” xabarni 0.66% noaniqlik bilan topdi. Umumiy natija alohida “spam” xabarlar uchun 99% va ham xabarlar uchun 97.99%ni hamda birgalikdagi “spam” va “ham” xabarlar uchun 98.9%ni tashkil etdi.
1-jadval. Testlash natijalari.
obekt
|
Nazorat tanlanmasi ( summ - sinov xabarlar soni)
|
Birinchi turdagi xatolik (FRR - False Rejection Rate)
- noto‘g‘ri rad etishlar soni (% yolg‘on rad
etish)
|
To‘g‘ri toifalash natijasi, % da
|
Ikkinchi turdagi xatolik (FAR - False Acceptance Rate) - noto‘g‘ri aniqlashlar soni (noto‘g‘ri aniqlash%)
|
Identifika tsiya (umumiy) natijasi % da
|
spam
|
4828
|
37 (0,8%)
|
99,23%
|
10 (0.2%)
|
99%
|
ham
|
747
|
9 (1,2%)
|
98,8%
|
5 (0.66%)
|
97,99%
|
Jami (e mail = spam +
ham )
|
5575
|
46 (0.83%)
|
99.17%
|
15 (0,6%)
|
98,9%
| XULOSA
Yuqoridagi masalalardan shuni ko‘rishimiz mumkinki logistik regressiya toifalashtirish usullari orasida eng yaxshilaridan biri hisoblanadi. Bu usulni takomillashtirish orqali axborot xavfsizligini turli masalalarini yechishda yuqori samaradorlikka erishish mumkin.
ADABIYOTLAR RO‘YXATI
Л.А.Растригин. Современние принципи управления сложними объектами.
М.: Советское радио. 1980. - 230 с.
Abdulhamit Subasi, in Practical Machine Learning for Data Analysis Using Python, 2020
А.П. Карпенко.Современние алгоритми поисковий оптимизасии. М.:Издателство МГТУ им. Н.Э.Баумана, 2014. - 446 с.
Thomas W. Edgar, David O. Manz, in Research Methods for Cyber Security, 2017
Бурков Андрей, Машинное обучение без лишних слов. — СПб.: Питер, 2020. — 192 с.
Khamdamov R.Kh., Khaydarov E.D. “Pre-processing of primary spam classification data from email messages” Современное состояние и перспективи развития сифрових технологий и искусственного интеллекта Сборник докладов республиканской научно-технической конференции Самарканд, 26-27 октябрья 2022 г.
https://t.me/ResearchEdu_Journal
Multidisciplinary Scientific Journal
December, 2022
|