Math 316 Solutions To Sample Final Exam Problems

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1 Solutions to Smpl Finl Exm Prolms Mth 16 1 Mth 16 Solutions To Smpl Finl Exm Prolms 1. Fin th hromti polynomils o th thr grphs low. Clrly show your stps. G 1 G G () p = p p = k(k 1) k(k 1)(k ) () Atr simpliying, w s tht p G1 (k) = k(k 1) ( (k 1) (k ) ). p(g ) = p p = p p p + p = k(k 1) k(k 1) k(k 1)(k )(k 1) + k(k 1)(k ) = k(k 1) k(k 1) k(k 1) (k ) + k(k 1)(k ) () Not tht rmoving on o th intrior gs o G prous grph tht is isomorphi to G rom th prvious prt o th prolm. Thror, w hv p(g ) = p p = p(g ) p + p = p(g ) k k(k 1)(k ) + k(k 1)(k ), whih, whn omin with our nswr rom prt () n simplii it, yils p G (k) = k(k 1) k(k 1) k(k 1) (k ) + k(k 1)(k ) k (k 1)(k ).

2 Solutions to Smpl Finl Exm Prolms Mth 16. Givn to th right is grph (not rwn to sl) rprsnting ntwork o irt ros onnting 10 towns. Th g lls rprsnt istns twn towns, in mils. () Whih ros shoul pv so tht th istn rom th town ll u to ny othr town long pv ros is s smll s possil? Also init th minimum istn tht h town is wy rom u. u () Whih ros shoul pv so tht it is possil to riv twn ny two towns on only pv ros, n so tht th highwy prtmnt pvs th minimum totl istn? Wht is th minimum totl istn tht must pv? Figur 1. Distn tr Figur. Minimum wight spnning tr u 6 (6) () (5) 7 (1) (1) (1) 5 1 (1) 10 () 1 u () () Hr, w r ing sk to in minimum istn tr rom th vrtx u. Atr pplying Dijkstr s lgorithm, w n up pving th ros init y th sh gs in Figur 1 ov, n th istns rom u to h town r init y th numrs in prnthss. () In this s, w r ing sk to in minimum wight spnning tr. Applying Prim s lgorithm, w n up pving th ros init y th sh gs in Figur ov. By ing th g wights in th spnning tr, w s tht th totl istn tht ns to pv is = mils. Not tht th g wights ov r list in th orr in whih th gs wr slt y Prim s lgorithm.. For h o th itms low, w will rstrit ourslvs to grphs, not multigrphs or gnrl grphs. In othr wors, loops n/or multipl gs r not llow. () Consir th ollowing proprtis tht grph G my or my not hv: (i) G hs Hmilton yl, (ii) G is iprtit, (iii) G hs rig. Drw Vnn igrm to illustrt th rltionship twn ths thr proprtis. Thn, in n xmpl o grph in h o th istint zons rt y your Vnn igrm. (For xmpl, in grph whih is iprtit n hs rig, grph tht is not iprtit ut hs rig, t., until you v xhust ll possil omintions o proprtis.) () Rpt prt () with th ollowing proprtis: (i) G is plnr, (ii) G is iprtit, (iii) G is isomorphi to K n or som n.

3 Solutions to Smpl Finl Exm Prolms Mth 16 () Th Vnn igrm is shown to th right. Not tht th irls or proprtis (i) n (iii) o not intrst us it is not possil or grph to hv oth rig n Hmilton yl. Ths irls ivi th pln into 6 rgions, whih r numr in th igrm. W will now in xmpls o grphs tht it into h o th six zons. 1. C hs Hmilton yl ut is not iprtit n hs no rigs.. C hs Hmilton yl n is iprtit ut hs no rigs.. K, hs no Hmilton yl, is iprtit, ut hs no rigs.. K hs no Hmilton yl, is iprtit, n hs rig. 5. Th grph G 1 shown to th right hs no Hmilton yl, is not iprtit, ut hs rig. 6. Th grph G shown to th r right hs no Hmilton yl, is not iprtit, n hs no rig. (iii) (ii) (i) G 1 G () Th Vnn igrm is shown to th right. Not tht thr r svn, not ight zons us it is not possil or grph to isomorphi to K n, iprtit, n not plnr ll t th sm tim. This is us, or K n to not plnr, w must hv n 5, mning tht K n will hv o yls (lik K ) s sugrphs, prvnting it rom ing iprtit. W will now in xmpls o grphs tht it into h o th svn zons. (i) 6 (iii) (ii) 1. C 5 is plnr, not iprtit, n not omplt.. C is plnr, iprtit, ut not omplt.. K, is not plnr, is iprtit, ut is not isomorphi to K n or ny n.. K is plnr, not iprtit, ut is omplt. 5. K is plnr, iprtit, n omplt. 6. K 5 is not plnr, not iprtit, ut is omplt. 7. Th grph G shown to th right, whih is simply K 6 with on g rmov, is not plnr, not iprtit, n not omplt. Not tht it is not plnr y Kurtowski s Thorm (it hs K 5 s sugrph). G. Prov tht, i tr hs vrtx o gr p, thn it hs t lst p pnnt vrtis. Lt T ny tr o orr n tht hs vrtx o gr p. Lt k th numr o pnnt vrtis in T. W will show tht k p. Lt 1,,..., n rprsnt th vrtx grs, n suppos tht 1 through k giv th grs o th pnnt vrtis, n tht n orrspons to th vrtx hving gr p. Thn 1 = = = k = 1, n n = p, n i or ll i stisying k +1 i < n. W thror hv (n 1) = n = ( k ) + ( k+1 + k+ + + n 1 ) + n k + (n 1 k) + p = k + (n 1) k + p Atr sutrting (n 1) rom oth sis o th ov inqulity n simpliying, w otin 0 k +p, whih is quivlnt to k p. This omplts th proo.

4 Solutions to Smpl Finl Exm Prolms Mth Dtrmin whih o th ollowing grphs r plnr. I thy r plnr, rw plnr rprsnttion. I thy r not plnr, show tht thy r not plnr using Kurtowski s Thorm or som othr mtho. () () () This grph is not plnr. To s this, not tht th grph hs = 1 gs n n = 6 vrtis, so n 6 = 1, whih mns tht > n 6. () g g () h i () This grph is plnr, s illustrt y th plnr rprsnttion to th right. Not tht this plnr rprsnttion n otin rom th originl grph y simply rrrnging th positions o th vrtis n llowing th gs to strth or shrink oringly. () This grph is not plnr. To s this, w highlight th gs o K 5 suivision sugrph in th originl grph (s th originl grph givn ov with th sttmnt o th prolm); th ol, soli gs r tul gs in th sir sugrph, whil th sh gs r th suivi gs. Not tht this sugrph is isomorphi to th K 5 suivision shown to th right, whih mns tht th givn grph is not plnr y Kurtowski s Thorm. g () This grph is not plnr. To s this, w highlight th gs o K, suivision sugrph in th originl grph (s th originl grph givn ov with th sttmnt o th prolm); th ol, soli gs r tul gs in th sir sugrph, whil th sh gs r th suivi gs. Not tht this sugrph is isomorphi to th K, suivision shown tothright,whihmnsthtthgivngrphisnotplnrykurtowski s Thorm. g h i 6. In th gm o Yhtz, iv stnr six-si i r roll simultnously. For simpliity, lt us ssum th i r irnt olors so tht w n tll thm prt. Dtrmin th numr o irnt wys in whih h o th ollowing outoms n our i th iv i r roll on. () ull hous (thr i shring on numr, two i shring son numr) () our o kin (our i shring th sm numr) () thr o kin (ut not our o kin, iv o kin, or ull hous) () lrg stright (iv onsutiv numrs) () smll stright (our onsutiv numrs, ut not iv onsutiv) () Yhtz (iv o kin) () W rk th tsk into svrl prts:

5 Solutions to Smpl Finl Exm Prolms Mth 16 5 Choos ommon numr or th thr i 6 wys (W n hoos thr 1 s, thr s, thr s,..., or thr 6 s, giving 6 hois.) Choos ommon numr or th two i 5 wys (W n hoos ny numr xpt th on hosn t th prvious stp.) Choos olors or th thr i C(5, ) wys Choos olors or th two i C(, ) wys (Thr r only olors lt to hoos rom tr th thr olors r hosn in th prvious stp.) 5 Thror, our nswr is 6 5. () W orgniz th tsk s ollows: Choos ommon numr or th our i 6 wys Choos olors or th our i C(5, ) wys Choos numr or th ith i 5 wys Choos olor or th ith i only 1 wy 5 Thror, our nswr is 6 5. () W orgniz th tsk s ollows: Choos ommon numr or th thr i 6 wys Choos two irnt numrs or th rmining two i C(5, ) wys Choos olors or th thr i hving ommon numr C(5, ) wys Choos olors or th two rmining i 1 wys (Atr th thr olors r hosn or th i in th prvious stp, thr r olor hois or th ourth i, n only 1 hoi or th ith i.) Thror, our nswr is 6 ( 5 () W rk into ss s ollows: ) ( 5 ). Cs 1. (Th lrg stright involvs th numrs 1,,,,5.) Sin th numrs r trmin, th only thing rmining to o is to ssign olors. Thr r 5 olor hois or th i with on on it, rmining olor hois or th i with two on it, t., own to only on olor hoi or th i with iv on it. Thror, thr r 5! suh lrg strights. Cs. (Th lrg stright involvs th numrs,,,5,6.) By th sm logi s ov, thr r 5! o this typ o lrg stright. Comining Css 1 n, w s tht our nswr is 5!. () W rk into ss s ollows: Cs 1. (Th smll stright involvs th numrs 1,,,.) W orgniz th tsk s ollows: Choos olors or th our squntil i P(5, ) wys Choos numr or th ith i 5 wys (Th ith i nnot iv; othrwis, w woul hv lrg stright. Thror, thr r only 5 numrs to hoos rom.) Choos olor or th ith i only 1 wy Thror, thr r 5 P(5,) suh smll strights. Cs. (Th smll stright involvs th numrs,,,5.) In this s, th tsk n rokn into th sm thr prts s in Cs 1 ov. Th only irn is in th son stp, whn w hoos numr or th ith i. This tim, it nnot on n it nnot six, sin oth o ths outoms woul giv us lrg stright. Thror, thr r only hois or th numr on th ith i, whih yils n nswr o P(5,) suh smll strights. Cs. (Th smll stright involvs th numrs,,5,6.) Using nlogous rsoning s in Cs 1, thr r 5 P(5,) suh smll strights. Comining th ov ss, our nswr is 5P(5,) + P(5,) + 5P(5,) = 1 P(5,).

6 Solutions to Smpl Finl Exm Prolms Mth 16 6 () Th only wys to gt iv o kin r iv ons, iv twos, iv thrs, iv ours, iv ivs, or iv sixs, so our nswr is just Lt P st o ny n intgrs, whr n. () Prov tht i non o th intgrs in P r multipls o n, thn thr r t lst two intgrs whos irn is multipl o n. (Hint: Think out rminrs whn iviing y n.) () Now, suppos tht w llow intgrs in P tht r multipls o n. Cn w still onlu tht thr r two intgrs whos irn is multipl o n? Justiy your nswr. () Lt P = {k 1,k,...,k n } ny st o n intgrs suh tht no k i is multipl o n. For h i, ivi th intgr k i y n to otin quotint q i n rminr r i, so tht k i = nq i + r i. Osrv tht, sin non o th k i s r multipls o n, non o th rminrs r i n zro, mning tht 1 r i n 1 or ll i. Thror, w hv n rminrs (r 1,r,...,r n ), ut only n 1 irnt vlus tht thy n hv. Thus, y th Pigonhol Prinipl, w must hv r i = r j or som i j, so k i k j = n(q i q j ) + (r i r j ) = n(q i q j ), sin r i = r j n w s tht th irn twn k i n k j is multipl o n. () No, w nnot rw th sm onlusion. For xmpl, i w hoos P = {1,,,...,n}, thn th irn twn ny two o th intgrs in P is twn 1 n n 1, so non o th irns is multipl o n.. For n, th whl grph, W n, is grph tht onsists o th yl C n 1 on th outsi, with th nth vrtx jnt to ll vrtis on th yl. Th grphs W n W 5 r shown to th right. In this prolm, you will in th hromti polynomil o W n. Assum tht thr r k olors vill. W W 5 () I w strt y oloring th ntr vrtx, how mny olor hois r thr? () Cn th sm olor tht ws us or th ntr vrtx us or ny o th rmining n 1 vrtis? () Using th multiplition prinipl n th rsult o Prolm 1.1 rom your txtook, writ own th hromti polynomil or W n. () Thr r k olor hois. () No, it nnot us, y inition o W n, th ntr vrtx is jnt to ll othr vrtis in th grph. () By Prolm 1.1, w hv p Cn (k) = (k 1) n + ( 1) n (k 1). Atr oloring th ntr vrtx o W n, thr r only k 1 olors lt with whih to olor th rmining n 1 vrtis o W n, mning tht th outr C n 1 sugrph n olor in p Cn 1 (k 1) wys. Thror, w hv p Wn (k) = (# wys to olor ntr vrtx) (# wys to olor outr C n 1 sugrph) = k p Cn 1 (k 1) = k ( (k ) n 1 + ( 1) n 1 (k ) ).

7 Solutions to Smpl Finl Exm Prolms Mth In h o th ollowing itms, in th numr o 0-igit squns tht only ontin 0 s, 1 s, s, n/or s n hv th spii proprtis. () Th squn ontins t lst 1 zro. () Th squn ontins 5 zros, ons, twos, n thrs. () Th squn os not ontin onsutiv zros. (Hint: First, in n solv n pproprit rurrn rltion. To mk th lultions sir, you my us th t tht 0 = 1.) () First, w not tht, without ny rstritions, thr r hois or h igit, giving 0 totl squns. On th othr hn, thr r 0 totl squns with no zros, sin thr r only thr hois or h o th 0 igits. Thror, w hv # squns with t lst l zro = 0 (# squns hving no zros) = 0 0. () Think o rking this pross into squntil tsks: irst, w hoos 5 o 0 vill positions or th zros, whih n on in C(0,5) wys. Son, w hoos o th 15 rmining positions or th ons, whih n on in C(15,) wys. Continuing in this shion, thr r C(11,) wys to pl th twos, n C(,) = 1 wy to pl th thrs. On wy to omplt this pross is shown low: This givs totl o C(0,5) C(15,) C(11,) C(,) totl squns. Anothr wy to look t this sm prolm is to osrv tht thr r 0! wys to rrng th 0 igits in row, ut 5!!!! o th rrngmnts will look th sm us o rpt igits. Thror, th totl numr o prmuttions n writtn s ithr ( 0 5 )( 15 )( 11 )( ) or 0! 5!!!!. () Lt n rprsnt th numr o th sir typs o igit strings o lngth n tht n m. First, w trmin our initil onitions. W wr sk to ssum tht 0 = 1, n not tht 1 = us thr r our squns o lngth on: 0, 1,, or. To trmin rurrn ormul, w us th igrm to th right s n i. I string o lngth n strts with ny igit othr thn 0, thn th rmining n 1 sps n ill in n 1 wys. On th othr hn, i string o lngth n strts with 0, thn th irst two igits must look lik 01,0, or 0 in orr to voi rpt zros. Eh o ths thr onigurtions n omplt in n wys, yiling th ollowing rursion ormul: n = n 1 + n or n. Lngth n string String Count 1 n 1 n 1 n n 0 n 0 n Th hrtristi qution o this rurrn is x x = 0, whih, tr using th qurti ormul, yils th ollowing roots: x = ± (1)( ) = ± 1 Thror, th gnrl solution hs th orm ( + ) n ( 1 ) n 1 n = +.

8 Solutions to Smpl Finl Exm Prolms Mth 16 Using th initil onitions 0 = 1 n 1 =, w otin th ollowing: ( + ) = 1 = 1 = + = 1 (+ 1) + ( 1) = Solving th irst qution or yils = 1, n sustituting into th son qution, w hv Thror, (+ 1) + ( 1)(1 ) = = = ( ) = 1 = 1 = 1 = = = Thror, th numr o 0-igit squns with no onsutiv zros is givn y 0 = ( ) ( 1 5 ) Suppos tht w hv plntiul supplis o ppls, lry stiks, wlnuts, n grps. Assum tht th ojts in h tgory r intil. () Us ounting thniqus to in th numr o irnt sltions o 0 o th vill oo itms in whih no mor thn ppls r hosn. () Fin gnrting untion or th numr o irnt sltions o n oo itms tht ontin no mor thn ppls. () Us lgr n your nswr to prt () to onirm your nswr to prt (). () W rk th pross into ss oring to how mny ppls r hosn. Cs 1. (0 ppls) In this s, w r trying to hoos 0 oo itms rom irnt tgoris: lry stiks, wlnuts, or grps. W thror hv 0 strs n rs, giving C(, 0) sltions. Cs. (1 ppl) W rk th tsk into two prts: Choos 1 ppl only 1 wy Choos th othr 1 itms C(1, 1) wys (In this s, thr r 1 rmining oo itms (strs) to pik, n sin th ppls r lry pik, thr r oo tgoris, giving rs.) Thror, thr r C(1, 1) suh sltions. Cs. ( ppls) W rk th tsk into two prts: Choos th ppls only 1 wy Choos th othr 1 itms C(0, 1) wys (Rsoning s in Cs, thr r 1 strs n rs.) Thror, thr r C(0, 1) suh sltions. 1 0 Aing, w otin n nswr o

9 Solutions to Smpl Finl Exm Prolms Mth 16 () Our gnrting untion is givn y ppls lry wlnuts grps (x) = (1+x+x ) (1+x+x +x + ) (1+x+x +x + ) (1+x+x +x + ) = (1+x+x ) (1+x+x +x + ) = 1+x+x (1 x). () Our inl nswr will th oiint o th x 0 trm in our ormul or (x) rom prt (). Clulting, w hv 1 (1 x) ( +k 1 = (1+x+x ) k (x) = 1+x+x (1 x) = (1+x+x ) = (1+x+x ) k=0 k + x k, k n w not tht thr r thr wys to gt n x 0 trm in th ov xprssion: y tking 1 rom th (1+x+x ) portion o th xprssion tims th trm with k = 0 in th ininit summtion, y tking x rom th (1+x+x ) portion o th xprssion tims th trm with k = 1 in th ininit summtion, n y tking x rom th (1+x+x ) portion o th xprssion tims th trm with k = 1 in th ininit summtion. In othr wors, th thr trms in th xpnsion o (x) tht involv x 0 n orgniz s ollows: 1 0 x 0 + x 1 1 x 1 + x 0 x 1 = 1 k=0 [ + 0 ) x k ] 0 x 0 1 W thror s tht th oiint on th x 0 trm grs with our nswr to prt (). 11. Lt m,n, n k positiv intgrs with k m n k n. Giv omintoril proo o th ollowing intity: m+n m n m n m n m n = k k 0 k 1 1 k 0 k Hint: Think o hoosing ommitts rom group o m mn n n womn. W lim tht oth sis o th intity ount th numr o wys to hoos ommitt o k popl rom group o m mn n n womn vill to srv on th ommitt. W monstrt low: Lt si: By inition, C(m+n, k) ounts th numr o wys to hoos k popl rom group o m+n totl popl, whih is xtly wht w lim ov. Right si: This si ounts th sir quntity y ss oring to how mny womn r on th ommitt. To s this, onsir ny vlu o i, whr 0 i k. W rk th tsk o hoosing k-prson ommitt with i womn into two prts: Choos th womn C(n, i) wys (Hr, w r hoosing i womn rom th n totl womn vill.) Choos th mn C(m,k i) wys (To omplt our ommitt, w n to hoos th rmining k i ommitt mmrs rom th m vill mn.)

10 Solutions to Smpl Finl Exm Prolms Mth Thror, thr r ( m k i)( n i) suh ommitts with xtly i womn. Sin k-prson ommitt n hv nywhr rom i = 0 to i = k womn srving on it, w gt th totl numr o ommitts y lulting k i=0 ( m k i )( n i Sin this sum is qul to th right-hn si o our intity, this omplts our proo. ).

Cayley s Formula. Graphs - II The number of labeled trees on n nodes is n n-2. Planar Graphs. Is K 5 planar? Outline. K 5 can be embedded on the torus

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