Computer Science an Overview 12th Edition Brookshear Solutions Manual full download: people also search: computer science an overview 12th edition free download computer science an overview 12th edition answers computer science an overview 12th edition solutions pdf computer science an overview 12th edition pdf download free computer science an overview 12th edition ebook computer science an overview 11th edition solutions pdf computer science: an overview: global edition pdf computer science an overview 10th edition pdf free download. Computer science an overview 12th edition brookshear solutions manual. 1.

3 Computer Science An Overview 12th Edition Brookshear SOLUTIONS MANUAL Full download: edition-brookshear-solutions-manual/ Computer Science An Overview 12th Edition Brookshear TEST BANK Full download: edition-brookshear-test-bank/ Cha pt er One DATA S TOR AGE Chapter Summary This c ha pte r pr ese n ts t he rud imen ts of d a ta stora ge wit hin d igi ta l co mpute rs. It in troduce s the ba sics of d ig ita l circ ui try a nd ho w a simp le f lip - f lo p ca n be used to store a sing le bit. It t hen d iscusse s a d d re ssa b le me mory ce l ls a nd ma s s st ora ge syste ms ( ma g netic d i sk, compa ct d is ks, a nd f la sh me mor y ). Ha vin g es t a blished t hi s ba ckgr ou nd, the cha pte r d iscu sses ho w inf or ma tio n ( tex t, numer ic va lues, ima ges, a nd s ound ) a re enc od ed a s bit pa tter n s. The re a re t wo opti ona l secti on s, f irst in a br ief introduc tio n to Pyt ho n hig hli gh tin g th e mea ns by w hich a pr ogr a mmi ng la ng ua ges hide s the d eta ils of me mor y re pr ese ntio n. The secon d d elve s more d ee ply into d a ta stora ge topics by pr ese nt in g the pr oble ms of ove rf lo w er ror s, tr u nca tio n er rors, er r or d ete c tion a nd corr ec t io n tec hni que s, a nd d a ta compr essio n. Per ha ps the mo st i mport a nt com ment I ca n ma ke a bout t his cha pte r ( a nd the nex t one a s we ll) is to ex pla in i ts role i n the cha pte rs tha t f ol lo w.

This invol ve s t he d i stinc tio n be twee n ex po si ng stude nt s to a s ubjec t a nd re quiri ng t hem to ma ster the ma ter ia l —a d isti ncti on tha t is a t the hea rt of the spirit in w hic h the enti re tex t wa s written. The int entio n of this cha pte r is to pr ovide a re a listic ex posure t o a ve r y imp orta nt a re a of compu ter science. It is n ot nec es sa ry f or t he stude nt s to ma ster the ma ter ia l. A ll tha t is nee d ed f rom this cha pte r i n th e re ma inin g pa rt of the boo k a re the re mna nts tha t re ma i n f rom a br ief ex posure t o t he is sues of d a ta stora ge. E ve n if the cour se you tea c h re qu ire s a ma ster y of t hese d eta ils or the d ev el opme nt of m a nipula ti on s kil ls, I enc ou ra ge you t o a void empha siz i ng bit ma ni pula t ion s a nd re pr ese nta tion co nve rsio ns.

In pa rticula r, I urge you to a void be comi ng bo gged d ow n in the d e ta ils of conve rting b etwee n ba se ten a nd bina ry nota ti on. I ca n’t thin k of a nyt hi ng t ha t wo ul d be more bori ng f or t he st ude nts. ( I a polo giz e f or sta t ing my op ini on. The “ re quire d ” sect io ns i n th is c ha pte r cover the c o mpos iti on of ma in mem ory ( a s a ba ck grou nd f or ma chine a rc hitectur e i n cha pte r 2 a nd d a ta struc ture s in cha pte r 8 ), the phys ica l issues of ex ter na l d a ta stora ge sys te ms ( i n pr ep a ra tio n f or the s ubjec ts of f ile a nd d a ta ba se syste ms in c ha pte r 9 ), a nd the rud iments of d a ta encodin g ( tha t ser ve s a s a ba ckground f or t he sub jec t of d a ta types a nd hig h- leve l la ngua ge d ec la ra tion sta teme nts in cha pte r 6). The op tio na l sect io ns e xp lore t he i ssues of er ror ha nd lin g, includi ng tra nsmi ssi on er ror d ete ctio n a nd corr ec tion a s well a s the pr oblem of trunca tio n a nd ove rf l ow e r rors r esult ing f ro m n umer i c coding s yste ms. A s ment io ned in the pr ef a ce of the tex t, ther e a re se ve ra l theme s tha t run thr ou gho ut t he tex t, one of wh ich i s t he role of a bstr a ction. I l ike t o i nclude t his theme i n my lec t ure in wh i ch I i ntroduce f l ip - f lops.

I end up with both f lip - f lop d ia gra ms f rom the tex t on the boa rd, a nd I empha siz e tha t they re pr ese nt tw o wa ys of a cc ompl ish in g the sa me ta s k. I then d ra w a re cta ngle a round ea ch d ia gra m a nd er a se the circ ui ts wi thin the re cta n gles lea vin g on ly t he i nputs, outp u ts, a nd re cta ngle s sho wi ng. A t this po int the two lo ok ide ntica l.

I thi n k tha t this cr ea tes a strong visua l ima ge tha t d rive s h ome t he d ist inct io n be twee n a n a bstra ct t oo l’s i nter f a ce wit h the out si d e world a nd t he inter na l d e ta il s of the t ool. This is a spec ific ex a mple o f tea ching seve ra l topics a t the sa me time — in th is ca se, the conce pts of a bstra ction a nd e nca psul a tion a re ta ught in t he cont ex t of tea ching d i gita l cir cu its.

Don’t f orge t a bout t he circ uits i n A ppe nd ix B. I used to ha ve stude n ts w ho co nti nued to re cord a n ex tra bit in the a nswer to a two’s co mpleme nt a d d ition pr ob lem w hen a ca rry occ urr ed —ev en thou gh I ha d ex pla i ned th a t a ll va lues in a two’ s co mpleme nt sy stem were re pr ese nted wit h the sa me n umbe r of bit s. O nc e I sta rte d pr ese nti ng t he a d d ition c ircuit in A ppe nd ix B, thi s pr oblem d isa ppe a re d. It ga ve the stude nts a concr ete u nd er s ta nd ing t ha t the ca rr y is t hrow n a wa y.

( Of course, in a la ter course co mputi ng st ude nt s will lea r n tha t it re a lly is n’t thr ow n a w a y but sa ve d a s the 'c a rr y bit' f or po tentia l use in the f u ture, but f or n o w I i gn ore th is. ) I ha ve a ls o f ound t ha t a g ood ex er cise is to a sk stude nts to ex tend the circ uit in Figu re B. 3 so tha t it pr od uce s an a d d itiona l outp ut tha t ind ica tes whe ther a n ov er f low ha s occ urr ed.

For ex a mple, the output cou ld be 1 in the ca se of a n ove rf low a nd 0 ot her wi s e. For most stude nts, see i n g the re a lity of the things t hey a re told is a mea ningf ul ex pe rie nce.

For this re a so n I of te n f ind it a d va nta geous to d e mo ns tra te the d isti ncti on be t w ee n nu mer ic a nd cha ra cte r d a ta using a spre a d shee t. I like to sho w the m ho w the ma nip ula ti on of la rge numbe rs ca n lea d to er rors. I ha ve f ound tha t stude nts re sp ond wel l to hea ri n g a bout CD a nd DV D st or a ge system s, ho w sou nd is enc od ed, a nd ima ge re pr ese nta tio n sy stem s s uch a s GIF a nd J PE G. I ha v e of ten used the se topics a s a wa y of get tin g n on - ma j ors i nter ested i n tec hnica l i ssues. For stude nts n ot ma j ori ng in co mpute r sc ience, to pics such a s tw o's c omp lement a nd f loa ti ng - poin t nota ti on ca n get a bit d ry. The ma in point f or them to und er sta nd is tha t w hen inf or ma tio n is encode d, so me inf or ma tio n usua ll y gets lo st.

Thi s po int ca n be ma d e just a s we ll usi ng a udio a nd vide o, wh ich a re contex ts t ha t see m to be more i nter es ting t o the no n- ma j ors. Answers to Chap ter R ev iew Pro blems 1. W ith a 1 o n the up p er inp u t a nd a 0 o n t he lo wer i np ut, al l c ir cuit s wil l p r o d uce an o utp ut 0.

I f in stead a 0 i s o n the up p er inp ut and 1 is o n the lo wer i np ut, cir c uits b an d c will p r o d uce an o utp u t 1, and cir cuit a will still p r o d uce a 0. T he entire cir cuit i s eq ui val ent to a sin gle AN D gate. T he entire cir cuit is eq ui val ent to an E xc lu sive O R gate. After the t hir d p ulse, t hi s cir cuit wil l p r o d uce an o utp ut of 1 and 1. After the fo ur t h p ulse, b o th f lip - flo p s ar e flip p ed b ac k to a 0 state so the cir cuit will aga in p r o d uce an o utp ut of 0 and 0. I t is inter estin g to no te that thi s cir cuitr y fo r ms a b inar y co u nter that will r ep ea ted l y co unt fr o m 0 0 to 1 1. T hus, th is cir cuit fo r ms a n ab str ac t to o l t hat ca n b e us ed as a b uild i n g b lo ck i n o t he r cir cuits.

Ad d itio nal flip -flo p s ca n b e ad d ed to co unt thr o ug h a lar ger r an g e o f nu mb er s. A 1 will b e se nt o n O utp ut B o n t he 2 nd, 6 th, 1 0 th p ul ses o f t he clo ck. Li ke wise, a 1 will b e sent to O utp u t C o n the 3 r d, 7 th, 11 th p ulses o f the clo ck. A 1 will no t b e sent to any o utp ut o n the 4 th, 8 th, 1 2 th p ulse s o f t he clo ck. As we mo ve fo r war d into t he ne xt c hap ter, a cir cuit si milar to thi s ca n b e used to d r ive the mac hine c ycle ( co mp o sed o f f etch, d ec o d e, and exec ute). O utp u t A wo uld b e co nnec ted to th e inp ut t hat ac ti vate s the fetc h cir cuitr y.

Li ke wise Outp ut B and Outp ut C wo uld b e co nnec ted to the d ec o d e and exec ute cir cuit s r esp ec tivel y. This i s a f lip - f lop tha t i s trigger ed b y 0 s ra th er t ha n 1 s. Tha t is, te mpora rily c ha ngi ng t he up pe r input t o 0 will esta bli sh a n outpu t of 1, wher ea s tempora ril y cha ngi ng t he lo wer input t o 0 will esta blis. 6 h a n output of 0.

To obta in a n equ iva lent circ uit usi ng NA ND ga tes, si mply re pla ce ea ch A ND- NOT ga t e pa ir w ith a NA ND ga te. A d d re ss Conte nts 00 02 01 53 02 01 03 53 6. 2 56 using tw o hex a d ec i ma l d igit s ( 1 6 bits), 6 55 3 6 usin g f our hex a d ec i ma l d ig its ( 3 2 bits). 1 1 00 1 10 1 b. 0 11 0 01 1 1 c. 1 0 01 1 01 0 d. 11 1 11 1 11 e.

0 0 0 10 0 00 8. The ima ge co nsi sts of 1 0 2 4 x 1 02 4 = 1, 0 48, 5 76 pixe ls a nd ther ef ore 3 x 1, 0 4 8, 5 76 = 3, 1 4 5, 72 8 bytes, or a bout 3 M B. This mea ns tha t a bout 8 6 ima ges could be store d in the 2 5 6M B ca mer a stora ge syste m.

( B y compa r ing t hi s to a ctua l ca mer a stora ge ca pa cities, stude nts ca n ga in a n a ppr ec ia tion f or the be nef its of i ma ge compr es sio n tec h ni ques. U s ing t hem, a typica l 2 5 6 M B stora ge sy stem ca n hold a s ma ny a s 3 0 0 ima ges.) 1 1. ( E a ch pixe l wo uld re quire o ne me mor y ce l l.) 1 2.

Da ta re triev a l f rom ma in mem ory i s muc h f a ster tha n f rom d i sk st ora ge. A lso d a ta in ma i n mem ory ca n be re f er ence d in byte - siz ed units ra ther tha n in la rge block s. On th e other ha nd, d isk stora ge s yste ms ha ve a la rg er ca pa city tha n ma i n mem ory a nd the d a ta st ore d o n d isk i s les s v ola tile tha n tha t s tore d in ma in me mory. Ther e a re 70 GB of ma ter ia l on t he ha rd - d isk d rive. E a ch CD ca n h old n o m ore tha n 7 0 0 M B. Thus, it will re q uire a t lea st 1 00 CDs to st ore a ll the ma ter ia l. Tha t d oes not see m pr a ctica l to me.

On the other ha nd, D VDs ha ve ca pa cities of a bout 4. 7 GB, mea nin g tha t o nly a bo ut 1 5 DVDs wou ld be re quire d. Thi s ma y stil l be impr a ctica l, but i ts a big i mpr ove ment ove r CD s.

( The re a l poin t of t his pr oblem i s to ge t stude nts t o thi nk a bou t s t ora ge ca pa cities i n a mea nin gful wa y.) 1 4. Ther e would be a bout 5, 0 00 cha ra cte rs on the p a ge re quiri ng t wo by tes f or ea ch tw o byte U nicod e cha ra cte r. S o the p a ge would re quire a bout 1 0, 0 0 0 bytes or 1 0 sectors of siz e 1 02 4 bytes.

The nove l wou ld re quir e a bout 1. 4 M B using A S CII a nd a bout 2.

8 M B if two byte U nicod e cha ra cte rs were used. The la tency time of a d i sk spi nni ng a t 3 60 0 re voluti ons pe r min ute is o nly 0. 00833 second s. 4 millisec ond s.

A bout 7 yea rs! Wha t d oes it sa y?

He x nota tio n 2 1. D o e s 1 0 0. 8 0 1 0 00 0 10 0 1 1 01 1 11 0 1 1 00 1 01 0 1 1 10 0 11 0 0 1 00 0 00 0 0 1 10 0 01 0 0 1 10 0 00 0 0 1 10 0 00 / 5 = 2 0 0 1 00 0 00 0 0 1 01 1 11 0 0 1 00 0 00 0 1 1 01 0 10 0 0 1 00 0 00 0 0 1 11 1 01 0 0 1 00 0 00 0 0 1 10 0 10 0? 0 0 1 10 0 00 0 0 1 11 1 11 b. T h e t o t a 0 1 0 10 1 00 0 1 1 01 0 00 0 1 1 00 1 01 0 0 1 00 0 00 0 1 1 10 1 00 0 1 1 01 1 11 0 1 1 10 1 00 0 1 1 00 0 01 l c o s t i 0 1 1 01 1 00 0 0 1 00 0 00 0 1 1 00 0 11 0 1 1 01 1 11 0 1 1 10 0 11 0 1 1 10 1 00 0 0 1 00 0 00 0 1 1 01 0 01 s $ 7.

9 0 1 1 10 0 11 0 0 1 00 0 00 0 0 1 00 1 00 0 0 1 10 1 11 0 0 1 01 1 10 0 0 1 10 0 10 0 0 1 10 1 01 0 0 1 01 1 10 2 2. 4 2 6 F 6 5 7 3 2 0 3 1 30 30 2 0 2 F 2 0 3 5 20 3 D 20 3 2 3 0 3 F b. 5 4 6 8 6 5 20 74 6 F 7 4 6 1 6C 2 0 63 6 F 74 2 0 6 9 7 3 2 0 24 3 7 2 E 3 2 3 5 2 E 2 3. 10 0 0, 1 00 1, 1 0 10, 10 1 1, 11 0 0, 1 1 01, 11 1 0, 1 11 1, 1 0 00 0, 1 0 0 01, 10 0 10 2 4. 0 01 1 00 1 0 0 0 11 0 01 1 b. 1 0 11 1 2 5. They a re the power s of t wo.

1 1 0 10 0 1 0 00 10 0 00 1 0 0 00 0 2 6. 1 11 1 1 2 8. 0 00 1 10 1 b. 1 11 0 01 1 c.

1 1 11 1 11 d. 0 0 00 0 00 e. 0 0 10 0 0 0 3 2.

1 00 0 0 ( incorr ec t) d. 10 0 11 ( incorr ec t) g. 10 0 00 ( incorr ec t) j.

1 1 11 1 3 3. 5 00 101 + 1 bec om es + 0 000 1 00 110 whic h r epr es en ts 6 b. 5 001 01 00 101 - 1 bec om es - 0000 1 whic h c o nv erts t o + 1 111 1 00 100 whic h r epr es en ts 4 c.

12 011 00 011 00 - 5 bec om es - 0010 1 whic h c o nv erts to + 1 101 1 001 11 whic h r epr es en ts 7 d. 8 010 00 010 00 - 7 bec om es - 0011 1 whic h c o nv erts to + 1 100 1 000 01 whic h re pres ents 1 e. 12 0 110 0 + 5 b ec om es + 0010 1 10 001 whic h repr es ents - 15 ( ov erf low) f. 5 001 01 00 101. 1 0 - 11 bec om es - 0101 1 whic h c o nv erts to + 1 010 1 11 010 whic h re pres ents - 6 3 4. 2 1 / 4.

1 1 3 5. 11 0.1 0 1 3 6. 1 11 1 11 1 1 b. 01 0 0 10 0 0 c. 1 11 0 11 1 1 d. 0 01 0 11 1 0 e.

00 0 11 1 11 ( trunca tio n) 3 8. 00 1 11 1 00, 01 0 00 1 10, a nd 0 1 0 10 0 11 3 9. The be st a ppr oxima tio n of the squa re root of 2 is 1 3 /8 re pr ese nted a s 0 10 1 101 1. The squa re of this va lue w hen re pr ese nte d in f loa t ing - p oi nt f or ma t is 0 1 0 1 1 11 1, which i s the re pr ese nta tio n of 1 7 /8.

The va lue one - eight h, w hich wou ld be re pr ese nted a s 0 0 1 01 0 00. S ince the va lue one - ten th ca nn ot be re pr ese nted a cc ura tely, such re cord i ng s wou ld suff er f rom trunca tio n er rors. The va lue is either el ev en or nega tive f ive.

A va lue re pr ese nted i n tw o' s co mple ment no ta tion ca n be c ha nged t o ex c ess n ota ti on b y cha ngi ng t he hi gh - ord er bit, a nd vice ve rsa. The va lue is two; the pa tter ns a re ex ce ss, f loa tin g - p oint, a nd t wo' s co mpleme n t, re spec tively. B would re quire to o ma ny sig nifica nt d i gits.

Manual

C wou ld re quire too la rge of a n ex pone nt. D would re quire to o ma n y si gnifica n t d igi ts.

When u sin g bi na ry not a tion, t he la rge st va lue tha t could be re pr ese nted w o uld cha n ge f ro m 1 5 to 63. When using two 's comple men t nota ti on the la rge st va lue tha t could be rep re sented woul d cha nge f ro m 7 to 31. 11 2 32 2 13 4 34 3 5 48.

Yyxy xx yyxy xyx xx xy x 49. S ta rting with t he f irst e n tries, they wou ld be x, y, spa ce, x xy, yy x, a nd x x yy. Not a cha nce. M PEG r equire s tra n sfe r ra tes of 4 0 M bps.

D o e s 1 0 0 0 0 1 00 0 01 0 0 0 1 10 1 11 1 0 0 1 10 0 10 1 1 0 1 11 0 01 1 1 0 0 10 0 00 0 1 0 0 11 0 001 0 0 0 11 0 00 0 0 0 0 11 0 00 0 / 5 = 2 1 0 0 10 0 00 0 1 0 0 10 1 11 1 1 0 0 10 0 00 0 0 0 0 11 0 10 1 1 0 0 10 0 00 0 1 0 0 11 1 10 1 1 0 01 0 00 0 0 1 00 11 0 0 10. 1 2 0? 0 0 0 11 0 00 0 0 0 01 1 11 1 1 b. T h e t o t a 1 0 1 01 0 10 0 1 0 1 10 1 00 0 0 0 1 10 0 10 1 1 0 0 10 0 00 0 0 0 1 11 0 10 0 0 0 1 10 1 11 1 0 0 1 11 0 10 0 1 0 1 10 0 00 1 l c o s t i 1 0 1 10 1 10 0 1 0 0 10 0 00 0 0 0 1 10 0 01 1 0 0 1 10 1 11 1 1 0 1 11 0 01 1 0 0 1 11 0 10 0 1 0 0 10 0 00 0 0 0 1 10 1 00 1 s $ 7. 1 0 1 11 0 01 1 1 0 0 10 0 00 0 1 0 0 10 0 10 0 1 0 0 11 0 11 1 0 0 0 10 1 11 0 1 0 0 11 0 01 0 0 0 0 11 0 10 1 0 0 0 10 1 11 0 5 2. The und er lined stri ngs d ef initely c onta i n er rors.

1 1 0 01 11 0 11 10 1 10 00 0 00 1 1 1 11 10 0 01 10 1 01 00 1 00 0 1 1 10 53. The code would ha ve a Ha mmi ng d is ta nce of 3.

Thus, by usin g it, one co uld d ete ct up to 2 er rors pe r cha ra cte r a nd corr ec t up to 1 er ror pe r cha ra cte r. CA B B A GE e. A nswer s will va ry a s th e ex cha nge ra tes c ha nge d a ily. A nswer s will va ry d ep ending on t he cur re ncy sele cte d. This is a n inter a ctive ex er cise, re sults will d ep end on t he br ow ser. A cha nge to the d olla r a mou nt w ould nee d t o be ma d e in ea c h of the ex pr ess ion s.

NoB = 123234234 noKB = noB / 10 24 noMB = noKB / 1024 noGB = noMB / 1024 noTB = noTB / 1024 print(str(noB) + ' B ' ) print(str(noKB) + ' KB ' ) print(str(noMB) + ' MB ' ) print(str(noGB) + ' GB ' ) print(str(noTB) + ' TB ' ) print(str(noKB) + ' KB ' ) print(str(noKB) + ' KB ' ) 60. NoTB = 1.123 noGB = noTB.

1024 noMB = noGB. 1024 noKB = noMB. 1024 noB = noKB. 1024 print(str(noB) + ' B ' ) print(str(noKB) + ' KB ' ) print(str(noMB) + ' MB ' ) print(str(noGB) + ' GB ' ) pr int(str(noTB) + ' TB ' ) print(str(noKB) + ' KB ' ) print(str(noKB) + ' KB ' ) m ins = 50 s ecs = 23 t ot s ecs = m ins. 60 + s ecs samplesper s ec = 44100 bytesper s ample = 2.

1 4 t o tbytes = totsecs. samplespersec. bytespersample print(totbytes) 61.

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Brookshear Computer Science 11th Edition Solutions Manual

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