Báo cáo cuối kì môn cấu trúc rời rạc final report discrete structures

TỔNG LIÊN ĐOÀN LAO ĐỘNG VIỆT NAM TRƯỜNG ĐẠI HỌC TÔN ĐỨC THẮNG KHOA CÔNG NGHỆ THÔNG TIN BÁO CÁO CUỐI KÌ MƠN CẤU TRÚC RỜI RẠC FINAL REPORT Discrete Structures Người hướng dẫn: THẦY NGUYỄN QUỐC BÌNH Người thực hiện: ĐỒ Lớp : Khố THÀNH PHỐ HỒ CHÍ MINH, NĂM 2022 TỔNG LIÊN ĐOÀN LAO ĐỘNG VIỆT NAM TRƯỜNG ĐẠI HỌC TÔN ĐỨC THẮNG 0 : 24 KHOA CƠNG NGHỆ THƠNG TIN BÁO CÁO CUỐI KÌ MƠN CẤU TRÚC RỜI RẠC FINAL REPORT Discrete Structures Người hướng dẫn: THẦY NGUYỄN QUỐC BÌNH Người thực hiện: ĐỒN PHƯƠNG NAM Lớp : Khố : 24 THÀNH PHỐ HỒ CHÍ MINH, NĂM 2022 0 ii LỜI CẢM ƠN Đây phần tác giả tự viết ngắn gọn, thể biết ơn người giúp hồn thành Luận văn/Luận án/Báo cáo Khơng nên chép theo mẫu “lời cảm ơn” có Tơi xin chân thành cảm ơn Thầy môn cấu trúc rời rạc – ngành CNTT: Nguyễn Quốc Bình Các bạn bè khoa môn học hỗ trợ giúp đỡ Dù cố gắng nhiều suốt q trình làm báo cáo này, có nhiều hạn chế, hay thiếu sót mong đóng góp ý kiến thầy TP Hồ Chí Minh, ngày tháng năm 2022 Tác giả (Ký tên ghi rõ họ tên) Đoàn Phương Nam 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures iii ĐỒ ÁN / BÁO CÁO ĐƯỢC HỒN THÀNH TẠI TRƯỜNG ĐẠI HỌC TƠN ĐỨC THẮNG Tơi xin cam đoan cơng trình nghiên cứu riêng hướng dẫn khoa học Thầy Nguyễn Quốc Bình Các nội dung nghiên cứu, kết đề tài trung thực chưa cơng bố hình thức trước Những số liệu bảng biểu phục vụ cho việc phân tích, nhận xét, đánh giá tác giả thu thập từ nguồn khác có ghi rõ phần tài liệu tham khảo Ngoài ra, Khóa luận/Đồ án tốt nghiệp cịn sử dụng số nhận xét, đánh số liệu tác giả khác, quan tổ chức khác có trích dẫn thích nguồn gốc Nếu phát có gian lận tơi xin hồn tồn chịu trách nhiệm nội dung Khóa luận/Đồ án tốt nghiệp Trường Đại học Tơn Đức Thắng không liên quan đến vi phạm tác quyền, quyền tơi gây q trình thực (nếu có) TP Hồ Chí Minh, ngày tháng năm 2022 Tác giả (Ký tên ghi rõ họ tên) Đồn Phương Nam 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures iv TĨM TẮT QUESTION 1: HOÀN THÀNH Áp dụng kiến thức slide numberTheory để giải vấn đề QUESTION 2: HOÀN THÀNH Áp dụng kiến thức slide Sequences_and_Recursion để giải vấn đề QUESTION 3: HOÀN THÀNH Áp dụng kiến thức slide Set để giải vấn đề QUESTION 4: HOÀN THÀNH Vận dụng kiến thức slide, thử trường hợp điều kiện, từ đưa kết luận thuộc trường hợp QUESTION 5: HỒN THÀNH Tìm tịi tra tài liệu để đưa khái niệm xác hướng giải vấn đề QUESTION 6: CHƯA HỒN THÀNH QUESTION 7: HỒN THÀNH Tìm tài liệu kiến thức này, từ áp dụng định lí xem video giải vấn đề để từ áp dụng vào làm QUESTION 8: HOÀN THÀNH Áp dụng kiến thức slide Graphs_and_Trees để giải vấn đề 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures v MỤC LỤC TÓM TẮT iv MỤC LỤC v QUESTION 1: Euclid’s algorithm and bezout’s identity 1.1 Using Euclid’s algorithm to calculate GCD and LCM 1.2 find integer solutions pairs (x,y) QUESTION 2: RECURRENCE RELATION QUESTION 3: SET 1.1 Create a set 1.2 Find union, intersect, non- symmetric difference and symmetric difference of T and ∆ QUESTION 4: RELATIONS Reflexive: Symetric: Anti-symetric: Transitive: QUESTION 5: MUTIPLICATIVE INVERSION 1.1 Multiplicative inverses by using Extended Euclidean algorithm a Multiplicative inverse mean? b Phương pháp extended euclidean algorithm? 1.2 Áp dụng phương pháp Extended Euclidean algorithm vào toán QUESTION 6: KRUSKAL’S ALGORITHM 11 QUESTION 7: EULERIAN CIRCUIT 12 a Does the following graph have an Eulerian circuit or Eulerian path? Why? 13 b Study and present your knowledge about Hierholzer’s algorithm to find an Eulerian circuit 13 c tìm Eulerian circuit phương pháp Hierholzer’s 13 QUESTION 8: MAP COLORING 15 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures vi 8.1 Modeling this map by a graph 16 8.2 Color the map with a minimum number of colors 17 TÀI LIỆU THAM KHẢO 22 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures QUESTION 1: Euclid’s algorithm and bezout’s identity 1.1 Using Euclid’s algorithm to calculate GCD and LCM a Calculate GCD gcd(2021,1000 + m) with m = 895 => gcd(2021,1895) B1: 2021 = 1895*1 + 126 126 = 2021 – 1895*1 B2: 1895 = 126*15 + 5 = 1895 – 126*15 B3: 126 = 5*25 + 1 = 126 – 5*25 B4: = 5*1 + Vậy: gcd(2021,1895) = b Calculate LCM lcm(2021,1000 + m) with m = 895 => lcm(2021,1895) ta có: gcd(2021,1895) = (a) Mà lcm(a,b) = (𝐚)(𝐛) 𝐠𝐜𝐝(𝐚,𝐛) (𝟐𝟎𝟐𝟏).(𝟏𝟖𝟗𝟓) => lcm(2021,1895) = 𝐠𝐜𝐝(𝟐𝟎𝟐𝟏,𝟏𝟖𝟗𝟓) = 0 𝟑𝟖𝟐𝟗𝟕𝟗𝟓 𝟏 = 𝟑𝟖𝟐𝟗𝟕𝟗𝟓 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 1.2 find integer solutions pairs (x,y) 𝟐𝟎𝟐𝟏𝐱 + (𝟏𝟎𝟎𝟎 + 𝐦)𝐲 = 𝐠𝐜𝐝(𝟐𝟎𝟐𝟏, 𝟏𝟎𝟎𝟎 + 𝐦) m = 895 𝟐𝟎𝟐𝟏𝐱 + 𝟏𝟖𝟗𝟓𝐲 = 𝐠𝐜𝐝(𝟐𝟎𝟐𝟏, 𝟏𝟖𝟗𝟓) Giải: 2021x + 1895y = gcd(2021,1895) = 126 - 25*5 = 1*126 + (-25)*5 = 1*126 + (-25)*(1895 - 126*15) = 1*126 + (-25)*1895 + (-15)*(-25)*126 = 1*126 + (-25)*1895 + 375*126 = (376)*126 + (-25)*1895 = (376)*(2021 - 1895*1) + (-25)*1895 = (376)*2021 - (376)*1895 + (-25)*1895 = (376)*2021 – (401)*1895 = (376)*2021 + (-401)*1895 One sol: x0 = 376 , y0 = -401 2021 = 1*2021 1895 = 1*1895 lcm (2021,1895) = 1*2021*1895 All sol: 2021*(376 + 1895m) + 1895*(-401 -2021m) = So, x = 376 + 1895m y = -401 -2021m 𝐟𝐨𝐫𝐦𝛜𝐙 Như vậy: +với m = ta có x = 376 + 1895*1, y = -401 -2021*1 => (x,y) = (2271, -2422) +với m = ta có x = 376 + 1895*2, y = -401 -2021*2 => (x,y) = (4166, -4443) +với m = ta có x = 376 + 1895*3, y = -401 -2021*3 => (x,y) = (6061, -6464) 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures +với m = ta có x = 376 + 1895*4, y = -401 -2021*4 => (x,y) = (7956, -8485) +với m = ta có x = 376 + 1895*5, y = -401 -2021*5 => (x,y) = (9851, -10506) 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Giải: tìm x cho: 96x mod 101=  96x = 101k +  96x + 101y = ri = Axi +Byi ri = 96xi + 101yi Sử dụng Extended Euclidean algorithm để tìm gcd(96,101) 101 = 96*1 + = 101 – 96*1  = 101 + (-1)*96 96 = 5*19 + = 96 – 5*19  = 96 + (-5)*19 = 1*5 + Giải: 96x + 101y = 1 = 96 + (-5)*19 = 96 + (-(101 + (-1)*96)*19) = 96 + (-19)*(101 + (-1)*96) = 96 + (-19)*101 + (19)*96 = 96*20 + 101*(-19) x = 20 ri = ri-2 mod ri – qi = [ri – / ri - 1] xi = xi – - qixi – yi = yi – - qiyi – 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 10 giá trị đầu xi,yi cho 0-1 1-0 i ri -1 qi xi yi 101 96 1 -1 19 20 -19 101 96 => 96-1 mod 101 = 20 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 11 QUESTION 6: KRUSKAL’S ALGORITHM 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 12 QUESTION 7: EULERIAN CIRCUIT a Does the following graph have an Eulerian circuit or Eulerian path? Why? b Study and present your knowledge about Hierholzer’s algorithm to find an Eulerian circuit c If the graph has an Eulerian circuit, use Hierholzer's algorithm to find an Eulerian circuit of that graph when the initial circuit R1 is:  % = then R1 is EINME i If 𝑎𝑏𝑐𝑑  % = then R1 is abhga ii If 𝑎𝑏𝑐𝑑  % = then R1 is UVbaU iii If 𝑎𝑏𝑐𝑑  % = then R1 is XCdX iv If 𝑎𝑏𝑐𝑑  is the 4-digit number combined by the last digits in your StudentID Where 𝒂𝒃𝒄𝒅  = 1234 For example, Student ID 520H1234 has 𝒂𝒃𝒄𝒅 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 13 Giải: a Does the following graph have an Eulerian circuit or Eulerian path? Why? Nó Eulerian circuit Vì: Ở đỉnh hình, chúng có cạch kề tạo nên chu trình khơng có cạch lẻ  đỉnh = 2.k (với k > 0) Khi bắt đầu đỉnh đó, chúng qua cạnh lần khơng bị lặp lại trở đỉnh b Study and present your knowledge about Hierholzer’s algorithm to find an Eulerian circuit Hiểu cách đơn giải thuật toán Hierholzer’s phải phác thảo Eulerian cycle cách kết nối cách vịng lại với Nó bắt đầu với nút(đỉnh) ngẫu nhiên sau chạy theo cạnh để qua đỉnh khác ( cịn gọi qua nhà hàng xóm) Bước lặp lặp lại quay trở nút(đỉnh) ban đầu Điều tạo nên vòng trịn đồ thị(graph) c tìm Eulerian circuit phương pháp Hierholzer’s  𝒂𝒃𝒄𝒅 = R1 is XCdX R1 is X WQRX CdX R1 is X WcbVW QRX CdX R1 is X WcbVW QRX CjhC dX R1 is X WcbVW QRX CjhC djicd X R1 is X WcbVW QRX CjhC djicd X R1 is X WcbVW QRX CjHDnH C djicd X R1 is X WcbVW QRX CjHDnH C djinmhi cd X R1 is X WcbVW QRX CjHDnH C djinmhi cd X R1 is X WcbVW QRX CjHDlmD nH C djinmhi cd X 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures QUESTION 4: RELATIONS Let ℜ be a binary relation defined on integers as follow: ∀a,b∈N(aRb↔m|(a.b)) where is the last digits of your student ID For example, if your student ID is 52000123 then the valid binary relation is ∀a,b∈N(aRb↔23|(a.b)) Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0transitive? Prove your answer Is R reflexive, symetric, anti-symetric, m = 95 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures ∀a,b∈N(aRb↔95|(a.b)) Giải: Reflexive: Not reflexisve when a = then 95 | (a.a)  95 | 2.2 = 95 | Symetric: ∀𝑎, 𝑏 ∈ 𝑁, 𝑖𝑓 95|𝑎 𝑏𝑡ℎ𝑒𝑛95|𝑏 𝑎 Therefore, R is symetric Anti-symetric: Not anti – symetric a= 1, b = if 95 | 1.5 and 95 | 5.1 but 𝑎 ≠ 𝑏  ≠ Transitive: ∀𝑎, 𝑏, 𝑐 ∈ 𝑁, 𝑖𝑓 95|𝑎 𝑏𝑎𝑛𝑑95|𝑏 𝑐 𝑡ℎ𝑒𝑛95|𝑎 𝑐 a=1 b=0 c=5 if 95 | 1.0 and 95| 0.5 then 95| 1.5 Therefore, R is transitive Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures QUESTION 5: MUTIPLICATIVE INVERSION a Study and present your knowledge about Extended Euclidean algorithm to compute multiplicative inverses in modular structures b Apply the algorithm to find (m+1) (mod 101) where m is the last digits of your student ID For example, if your student ID is 52000123 then m = 23 and you need to find 24 (mod 101) -1 -1 1.1 Multiplicative inverses by using Extended Euclidean algorithm a Multiplicative inverse mean? - Là số nhân với số đão ngược cho kết 1 ( 𝐴 × 𝐴 = ) Ex: x 5-1 =  × =1 - Khi có số mà nhân với nghịch đão số chia n mà cho kết 1, ta gọi Multiplicative inverse under modular arithmetic operations 𝐴 × 𝐴−1 ≡ 1𝑚𝑜𝑑 𝑛 Ex: ×? ≡ 1𝑚𝑜𝑑5 Để tìm số nhân cho số mà chia dư Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Sẽ thử từ đến tăng dần × = 3𝑚𝑜𝑑5 = ≠ 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures × = 6𝑚𝑜𝑑5 = = So, × = 1𝑚𝑜𝑑 - Lưu ý: +Số (A) số (n) không số nguyên tố GCD số khác mà khơng phải chúng không số nguyên tố nha nên có multiplicative inverse Vậy, để có multiplicative inverse số (A) số (n) phải số nguyên tố + Multiplicative inverse phụ thuộc vào số mod (n) + Nếu số mod nhỏ bắt đầu nhẩm từ 1,2,3… Cịn số mod lớn prime number lớn phải cần số algorithms để giúp Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures tìm multiplicative inverse cách dễ dàng để tìm ta dùng phương pháp extended euclidean algorithm b Phương pháp extended euclidean algorithm? - Mình tìm gcd số tìm multiplicative inverse - Nếu số dương x < n thỏa x*a mod n = x nghịch đão a mod n, kí hiệu a-1 mod n Ax + By = Tìm x,y thỏa Ax + By = - Sử dụng Euclidean algorithm để tìm gcd - Mở rộng để tìm x,y ri = Axi +Byi ri = ri-2 mod ri – qi = [ri – / ri - 1] xi = xi – - qixi – yi = yi – - qiyi – i ri qi xi yi 1.2 Áp dụng phương pháp Extended Euclidean algorithm vào toán Apply the algorithm to find (m+1)-1 (mod 101) where m is the last digits of your student ID For example, if your student ID is 52000123 then m = 23 and you need to find 24-1 (mod 101) m = 95 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 96-1 (mod 101) 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Giải: Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures tìm x cho: 96 d 101 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 96x mod 101=  96x = 101k +  96x + 101y = ri = Axi +Byi ri = 96xi + 101yi Sử dụng Extended Euclidean algorithm để tìm gcd(96,101) 101 = 96*1 + = 101 – 96*1  = 101 + (-1)*96 96 = 5*19 + = 96 – 5*19  = 96 + (-5)*19 = 1*5 + Giải: 96x + 101y = 1 = 96 + (-5)*19 = 96 + (-(101 + (-1)*96)*19) = 96 + (-19)*(101 + (-1)*96) = 96 + (-19)*101 + (19)*96 = 96*20 + 101*(-19) x = 20 ri = ri-2 mod ri – qi = [ri – / ri - 1] xi = xi – - qixi – yi = yi – - qiyi – Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 10 giá trị đầu xi,yi cho 0-1 1-0 i ri -1 qi xi yi 101 96 1 -1 19 20 -19 101 96 => 96-1 mod 101 = 20 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures 0 Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures Bao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structuresBao.cao.cuoi.ki.mon.cau.truc.roi.rac.final.report.discrete.structures