Discrete Structures Lecture Integer Representation and Algorithms

Transcription

1 THEOREM 1 Let b be postve nteger greter thn 1. Then f n s postve nteger, t cn be expressed unquely n the form. where k s nonnegtve nteger,,, re nonnegtve ntegers less thn b, nd 0. Remrk Theorem 1 s employng forml mthemtcl termnology to express tht our number system s postonl number system. For exmple, 954 mens 954 bse 10 nd cn be expressed s The dgts of the number 954 re n postons 2, 1, nd 0 respectvely. Furthermore, the postons re exponents of the bse. EXAMPLE 1.1 Fnd the Decml equvlent of 20 Soluton: poston dgt decml vlue exponentted bse subtotl A E B Totl EXAMPLE 1.2 Fnd the decml equvlent of Soluton: 1. Frst, convert the bnry nteger to hexdecml by mkng groups of four bts strtng from the bnry pont Next convert ech group of four ech nbble to ts hexdecml equvlent. D A B 2 3. Use the method of exmple 1.1 to convert the hexdecml vlue to decml. poston dgt decml vlue exponentted bse subtotl 3 D A B Totl

2 The Rdx Dvde/Multply Method Objectve: Covert decml number to n equvlent number n nother rdx (bse). Soluton: Step Dscusson 1 Seprte the nteger nd frctonl portons of the decml number Integer porton converson lgorthm: Step Dscusson 0 Dvde the decml number by the rdx. The remnder s, =0,1,2,3 n 1. Quotent q becomes the dvdend n the next terton. r : rdx, dvsor d : dvdend n terton. d 0 s the ntl dvdend, the decml number to be converted to the foregn bse. q : quotent n terton : remnder produced n terton, th dgt of the number n rdx r. d mod r, d d r 1 n 1 Perform step 0 untl the dvdend equl zero. Exmple: Convert to bnry. Step Rdx Dvsor Dvdend Quotent Remnder stop! = = = = = = =

3 EXAMPLE 2.1 Fnd the bnry equvlent of (29) 10 Soluton: 1. Frst, convert the decml number to hexdecml usng the rdxdvde method shown bove. Rdx Dvdend Quotent Remnder Hexdecml Dvsor D stop! 2. Next, convert ech hexdecml dgt to bnry. 1 D = Lst, remove ledng zeroes

4 Purpose: complements smplfy bnry subtrcton Bnry subtrcton requres: 1. complement 2. fxed feld wdths for bnry numbers One s complement: nvert ll bts 1 0,0 1 Note: The only tme number s complemented s when the number s negtve. Exmple: consder , the one s complement s Thnk of the one s complement s the dfference between the ntl opernd nd number of equl length hvng one n every poston. Exmple: Two s complement s one more thn the one s complement Exmple: fnd the two s complement of Orgnl vlue One s complement Add one

5 Two s Complement Representton 1. Choose feld wdth. Common feld wdths re 4, 8, 16, 32, nd 64 bts. 2. A bnry number s postve f the most sgnfcnt dgt s zero (0), otherwse t s negtve. Exmple: Fnd the decml equvlent of the followng 16 bt two s complement number Mke the two s complement number postve. Fnd the mgntude of the two s complement Frst, fnd the one s complement by nvertng ll the bts Next, dd one (1) to fnd the two s complement Convert to decml Frst convert to hexdecml F B 2.2. Next, convert to decml. Hex Dec F B Remember tht the number ws negtve. Thus

6 r s complement n ( N) r N f N 0, 0 f N 0 where r, c Complements of Numbers n number of dgts n the nteger porton of N r rdx Exmple: Fnd the ten s complement of ( n 3, r 10,(147) ) 10,c 10, c Exmple: Fnd the ten s complement of (0.53) n 0, r 10,(0.53) , c 10, c Exmple: Fnd the two s complement of ) n 4, r 2,(1010) ( 2, c 2, c Exmple: Fnd the two s complement of ), n 4, r 2 ( 2, c

7 Subtrcton wth r s complement M nd S re two postve bse r numbers where r must be evenly dvsble by 2. Fnd the dfference D M S 1. Defne feld wdth nd dd M to the r' s complement of S. Dscrd ny dgts tht crry nto postons more sgnfcnt thn those defned by the feld wdth. Dscrd ny crry out dgts. Exmple Fnd the dfference ( ) Fnd the 2 s complement of S s complement of S 1000 Add 1 to fnd 2 s complement of S s complement of S 1001 Add M to the r' s complement of S. M s complement of S D M S Fnl result Determnng the sgn. r 2.1. If the most sgnfcnt dgt, d, then the number s negtve, otherwse t s 2 postve To fnd the mgntude of negtve number represented n r' s complement form, complement the number. Exmple Fnd the dfference ( ) Fnd the 2 s complement of S s complement of S 0101 Add 1 to fnd 2 s complement of S s complement of S 0110 Add M to the r' s complement of S. M s complement of S D M S s complement s complement

8 3. Determnng f overflow occurred Overflow occurs when ddng two postve numbers or when ddng two negtve numbers. When the mgntude of the sum exceeds the rnge of vlues tht cn be represented n the feld n overflow hs occurred. Exmple: Consder 10 s complement representton nd 2 dgt feld. A number, n, rnges over the ntervl, 50 n 49. When the sum s greter thn 49 or less thn 50 n overflow hs occurred Specl cse: Overflow n 2 s complement numbers. Overflow occurs when the crry nto the sgn bt s unequl to the crry out. Exmple: consder 4 bt 2 s complement number. A number, n, rnges over the ntervl, 8 n 7. We consder two possbltes. Decml Bnry 2 s Complement Decml Bnry 2 s Complement

9 Algorthms for Integer Opertons ALGORITHM 2 Addton of Integers procedure (,: postve ntegers) {The bnry expnsons of nd b re nd respectvely where nd b re both bnry numbers.} c:=0 for :=0 to n 1 begn 2 f 1 then else end {The bnry expnson of the sum s } EXAMPLE Add 1110 nd 1011 Soluton: Follow the procedure specfed n lgorthm 2 c:= f 1 then else f 1 then else f 1 then else f 1 then else Check Crry

10 ALGORITHM 5 Modulr Exponentton procedure (: nteger, : postve nteger,m: postve nteger ) 1 { } EXAMPLE 11 Use Algorthm 5 to fnd Soluton: Algorthm 5 ntlly sets 1 nd ; In the computton of 3 645, ths lgorthm determnes for 1,2,,9 by successvely squrng the nd reducng modulo 645. If 1 (where s the bt n the th poston n the bnry expnson of 644), t multples the current vlue of by nd reduces the result modulo 645. Here re the steps used: , , , ,