LRR and LRS algorithm

종료 등록 시간: 1년 전 착불
종료 착불

R-programming in cryptography

R 프로그래밍 언어 알고리즘 암호 해독 Encryption

프로젝트 ID: #34349862

프로젝트 소개

9 건(제안서) 재택 근무형 프로젝트 서비스 이용 중: 1년 전

이 일자리에 대한 프리랜서 9 명의 평균 입찰가: $170

engrsaad03125

Dear Client! I am a Data science enthusiast and have an experience in R programming for more than 3 years. I can help you with Linear Recurring Sequence Algorithms in Cryptography application in R Studio. Looking forwa 기타

$250 AUD (15일 이내)
(5 리뷰)
2.6
marilatos12

Dear Sir, I've read your project description and I'm so much interested in your project. I am an innovative and strategic thinking professional with a proven track record of consistently going above and beyond in mee 기타

$140 AUD (7일 이내)
(0 리뷰)
0.0
josephwriter1996

Hi, Greetings and hoping you are doing well, i welcome you to my profile where quality and client satisfaction is the Priority. I am Expert Joseph and i hope to cooperate with you on your project . CERTIFIED EXPERT I 기타

$250 AUD (2일 이내)
(0 리뷰)
0.0
topgradeclubltd

Hi, Greetings and hoping you are doing well, i welcome you to my profile where quality and client satisfaction is the Priority. I am Expert DOMINIC and i hope to cooperate with you on your project. CERTIFIED EXPERT IN 기타

$250 AUD (5일 이내)
(0 리뷰)
0.0
anenkovakateryna

Hi I've read the project description carefully. I'm an expert in R programming with LRR algorithm for encryption. It would be a great pleasure for me to have the opportunity working with you. ✓ Looking forward to hear 기타

$100 AUD (2일 이내)
(0 리뷰)
0.0
rowangr2044

Hi, there. Before 2 months I have finished the similar project as you post . I read your requirements carefully and understood very well about the project scope and start working accordingly in stages. I am having more 기타

$140 AUD (7일 이내)
(0 리뷰)
0.0
danylovas

Hello, I have rich experience in cryptography and I can implement LRR and LRS algorithm with R programming. Let's discuss further details via chat. Looking forward to hearing from you. Thank you. Danylo

$200 AUD (7일 이내)
(0 리뷰)
0.0