ics141-lecture18-Matrices

ics141-lecture18-Matrices - 18-1ICS 141: Discrete...

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Unformatted text preview: 18-1ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiICS141: Discrete Mathematics for Computer Science IDepartment of Information and Computer SciencesUniversity of HawaiiStephen Y. Itoga18-2ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiLecture 18Chapter 3. The Fundamentals3.7 Applications of Number Theory3.8 Matrices18-3ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiSome material in these slides were taken/adapted from the slides made by Prof. Michael P. Frank and Prof. Jonathan L. Gross which are provided through the publisher of “Discrete Mathematics and Its Applications” written by Kenneth H. Rosen.Some slides were done by Prof. Baek18-4ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiChinese Remainder TheoremTheorem:(Chinese remainder theorem.) Let m1,…,mnbe pairwise relatively prime and ai ,…,anarbitrary integers. Then the system of equations x≡ ai(mod mi)(for i=1 to n) has a unique solution modulo m= m1 m2···mn.Proof:18-5ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiComputer Arithmetic w. Large IntsBy Chinese Remainder Theorem, an integer awhere 0≤a<m=∏mi, gcd(mi,mj≠i)=1, can be represented by a’s residues mod mi:(amod m1, amod m2, …, amod mn) To perform arithmetic upon large integers represented in this way, Simply perform ops on these separate residues!Each of these might be done in a single machine op.The ops may be easily parallelized on a vector machine.18-6ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiComputer Arithmetic ExampleFor example, the following numbers are relatively prime:m1= 225−1 = 33,554,431 = 31 · 601 · 1,801m2= 227−1 = 134,217,727 = 7 · 73 · 262,657m3= 228−1 = 268,435,455 = 3 · 5 · 29 · 43 · 113 · 127m4= 229−1 = 536,870,911 = 233 · 1,103 · 2,089m5= 231−1 = 2,147,483,647 (prime)Thus, we can uniquely represent all numbers up to m= ∏mi≈ 1.4×1042≈ 2139.5by their residues rimodulo these five mi.E.g.,1030= (r1= 20,900,945; r2= 18,304,504; r3= 65,829,085; r4= 516,865,185; r5= 1,234,980,730)To add two such numbers in this representation, Just add their corresponding residues using machine-native 32-bit integers. Take the result mod 2k−1: If result is ≥ the appropriate 2k−1 value, subtract out 2k−1or just take the low kbits and add 1.Note: No carries are needed between the different pieces!18-7ICS 141: Discrete Mathematics I (Spr 2008)University of HawaiiPseudoprimesAncient Chinese mathematicians noticed that whenever nis prime, 2n−1≡1 (mod n).Some also claimed that the converse was true.However, it turns out that the converse is not true!...
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This note was uploaded on 09/12/2008 for the course ICS 141 taught by Professor Idk during the Fall '08 term at Hawaii.

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ics141-lecture18-Matrices - 18-1ICS 141: Discrete...

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