1 / 40

Advanced Computer Arithmetic RNS in the Complex Domain Week 13

CENG536 Computer Engineering Department Ç ankaya University. Advanced Computer Arithmetic RNS in the Complex Domain Week 13. Introduction.

vinny
Download Presentation

Advanced Computer Arithmetic RNS in the Complex Domain Week 13

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CENG536 Computer Engineering Department Çankaya University Advanced Computer ArithmeticRNS in the Complex DomainWeek 13

  2. Introduction CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  3. In famous “DisquisitionesArithmeticae” of German mathematician Carl Friedrich Gauss there were introduced integer complex numbers a + bi, where a and b are integer real numbers and , on the base of which was built the theory of congruences of the integer complex numbers. Hereinafter the “complex number” we mean “integer complex number”, unless otherwise specified. Let there is , then numbers obtained by multiplying of by – 1, i, and – i are numbers associated with number . Number obtained from by replacing of i by – iis conjugated number. Value is the norm of complex number. Introduction CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  4. Simple complex number is the complex number that can not be represented as a product of two complex numbers distinct of 1. In other case this number is a composite complex number. From this follows, that composite real number is at the same time a composite complex number. The reverse is not always true, for example By the same way, any prime number of form is the composite complex number, for example If the prime numbers are represented as they can not be represented by this way and such numbers are the simple complex numbers. Introduction CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  5. Complex numbers poses important property of parity. As a complex “two” here is used number 1 + i . Complex number a + bi is odd if its not divisible by 1 + i. Complex number a + bi is even, if a and b are even. Such a numbers are always divisible by 1 + i. But there is another class of complex numbers, that are divisible by 1 + i, but they have odd values of a and b. This class of numbers is referred to as half-even complex numbers. Introduction CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  6. Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  7. Complex number is multiple of complex number (or will be divisor of number ) if quotient of division is integer complex number. Result of this division will be an integer complex number only if The number is not divisible by if this condition is not satisfied. Let is such that is divisible of , so we can write or is residue of number by modulo . Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  8. Example: • Determine divisibility of numbers and • Here we have • Necessary conditions are satisfied as • . Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  9. Theorem 27. Let there are numbers and let congruences are satisfied (*) In such case a congruence will take place. To proof we divide by that gives Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  10. To have after division integer complex number must be satisfied congruences that are equivalent to (*). From this, if condition (*) is satisfied, the number is the residue of number by modulo Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  11. For complex numbers we cant say “greater” or “less” but may be introduced the least residue. If expressions xp + yq and yp – xqwill be the least residui by real modulo p2 + q2, there may be determined complex number x + iy, that may be named the least residue of by modulo In other words, must be satisfied Here should be distinguished least residues and absolute least residues. For first case we suppose that xp + yqand yp – xqare integer positive numbers that less then p2 + q2 – 1 and for second case these values may be both positive and negative but by absolute value less than (p2 + q2) / 2. Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  12. If there are determined the least residues of expressions ap + bq and bp – aq of form the least residue of number by modulo is Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  13. Example 1: • Determine least residue of number by modulo • Lets write system of congruences • where • ap + bq= 153 + 22 = 49 , • bp – aq= 23 – 152 = – 24 . • From this we have equations • Solving them gives i.e. 2 + 2i is the least residue. Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  14. Example 2: • Determine absolute least residue of number by modulo • Here we have • 49 (mod 13) = 10 (mod 13) = –3, • – 24 (mod 13) = – 11 (mod 13) = 2. • that gives system of equations • Solving them gives i.e. – 1 is the absolute least residue. Congruence of Complex Numbers CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  15. Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  16. If two sets of numbers may be mutually reflected one to another without distortion of correlation between them, i.e. if each element a from set will have connection to the only element a from set X, so that correlation between any element of X will take place between appropriate elements of X and vice versa, these sets are referred to as isomorphic. As it was state before, the least residue of any complex number a + bi by complex modulo p + qi may be determined solving system of real congruences (*) where r and r– the least residui by real modulo N = p2 + q2. Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  17. For rand rare possible values from set and there is correlation between them. Lets determine this connection between r and r. Multiplying first equation of (*) by p and second by q and then subtracting second from first we get or If p and q are coprime (relatively prime) numbers, the congruence has the only solution where and z is such that t is integer and less than p2 + q2. Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  18. Example: • Determine all possible pairs of r and r for modulo • Because 3 and 4 are coprime numbers , we have • and congruence has solution • . • This gives next pairs of r and r: • (Here is applied technique of enumeration). Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  19. Theorem 28. (The fundamental Gauss Theorem 1) For given complex modulo that has norm and p and q are coprime, every integer complex number is congruent to the only residue from the set From the number theory we know, that for pair of coprime numbers p and q there is possible to find such integer numbers u and v that gives Lets write an identity (its easy to check) Let there is a complex number Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  20. We rewrite it by changing representation of i by introduced value. Lets denote by h the least positive residue of number by modulo N and determine In such case will be satisfied equality or in form of congruence Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  21. This is the proof that a + bi is congruent to one of numbers by modulo Now lets proof, that this number is the only number of that set. Let there are two congruences According property of congruences numbers h1 and h2 are congruent by modulo i.e. or and this congruence may be replaced by Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  22. From last equality follows, that will be true (this is multiplication of both sides by conjugated modulo) which is equivalent next two real equalities Multiplying first by u, second by v and adding them gives From which, taking in account , follows or (*) By hypothesis and from which follows that for (*) be true we need Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  23. This shows impossibility of presence of two numbers h1 and h2less than N, which will be congruent to a + bi by modulo There is the only number that can be determined from congruence or This theorem states isomorphism between integer complex numbers and real residui, determined by method shown. Definition: Expression uq– vpset up connection between complex and real residui by modulo p + qi and is referred to as isomorphism coefficient that will be denoted by . Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  24. Example: • Solve congruence 16 + 7i h (mod 5 + 2i). • Here p = 5, q = 2 and N = 25 + 4 = 29. • We need up + vq = 1. Substituting gives 5u + 2v = 1. • Selecting u = 1 and v = –2 produce • 5u + 2v = 51 – 22 = 1 • Isomorphism coefficient of modulo 5 + 2i is • = uq – vp = 12 – (–2) 5 = 2 + 10 = 12. • Now, a + b  h (mod N) • 16 + 712  h (mod 29) or h  13 (mod 29) • So, result is • 16 + 7i 13 (mod 5 + 2i). Fundamental Gauss Theorem CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  25. Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  26. Lets check validity of the next statement. Let there are two numbers and . Let for this numbers exist such and that then, there will be correct where N – is norm of Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  27. We start from congruence introduced before (*) Before this moment there were analyzed Gauss theorem and solution of this congruence having that p and q are coprime numbers. Now lets consider case of p , q and N = p2 + q2 having common factor d. Now we have p = ed, q = fd and N = (e2 + f2)d2. In concordance with (*) in this case there will be d solutions of form Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  28. where satisfies congruence , (**) in which is satisfied condition of coprime property for e and f and solution of which is obtained by general way. Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  29. Example: • Determine all possible pairs of r and r for p + qi = 3 + 6i • For this example we have that congruence • will be of form • Here common factor is d = 3, and now p = 13, q = 23 • that give for • result of form • Solution of this congruence is Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  30. This congruence has three groups of solutions • Here is the table for all pairs of • Group 1. • Group 2. • Group 3. Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  31. Gauss isomorphism theorem is established, for the modulo with relatively prime (coprime) components, that implementation of rational operations on a complex residui can be replaced by implementation of these operations with the corresponding real residui by modulo that equal to the norm of the complex modulo. These operations may be executed by using arithmetic units of traditional type and on the base of tables. • For the complex modulo of which components have common factor, that means absence of isomorphism, mathematic operations may be realized by using real equivalents of complex residui only on the base of tables. Arithmetic in Complex Domain CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  32. Geometric Interpretation CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  33. Complex numbers are represented by points on a plane. Let we apply rectangular system of coordinates (introduced by René Descartes) that has axes X and Y and point of origin O. Abscissa axis will be used for representation of real parts of complex numbers and ordinate axis – for imaginary parts. Point M with coordinates (p, q) represents complex number p + qi, line OM is value N, and the norm N is square area built on the line OM (square ORLM). Here OR = OM = RL = LM = N. Geometric Interpretation CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  34. Y L R M O X Geometric Interpretation CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  35. Let is given modulo and x + yileast residue by this modulo. Determining the boundaries should be satisfied conditions • or, in another form • For absolute least residues this equations will be of form • or The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  36. Geometric representation of full system of least residues by modulo The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  37. Example: • Determine full system of absolute least residues by modulo • System of inequalities will be of form • For there will possible values – 12, – 11, . . . , 0, 1, . . . , 12. From this system we get an equation • Value of x will get the largest positive value for • i.e. from which • or The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  38. The smallest negative value x will have at • that gives • So for x possible values are – 3, – 2, – 1, 0, 1, 2, 3. • Lets compute corresponding values for y. • 1. • Checking inequalities we state, that only y = 0 satisfies them. • 2. • Here we get The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  39. 3. • Values of y are • 4. • For y we have • 5. • That gives • 6. • Here values of y are The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

  40. 7. Where we get Totally we have N = 25 absolute least residues. All of them are inside the square, shown below. O x The method of determining of boundaries CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

More Related