1. Determine whether or not the following binary relations are equivalence relations. Be sure to justify your answers.

Transcription

1 Mth 0 Exm - Prti Prolm Solutions. Dtrmin whthr or not th ollowing inry rltions r quivln rltions. B sur to justiy your nswrs. () {(0,0),(0,),(0,),(,),(,),(,),(,),(,0),(,),(,),(,0),(,),(.)} on th st A = {0,,,,} First noti tht this rltion is rlxiv, sin (0, 0),(, ),(, ),(, ),(, ) R. Nxt, on n lso s tht this rltion is symmtri, sin th rvrs o vry non-rlxiv lmnt is prsnt. Finlly, on n vriy tht this rltion is lso trnsitiv y hking ll ovrlpping pirs. Thus this rltion is n quivln rltion. () {(,),(,),(,),(,),(,),(,),(,),(,)} A = {,,,} First noti tht this rltion is rlxiv, sin (, ),(, ),(, ),(, ) R. Nxt, on n lso s tht this rltion is symmtri, sin th rvrs o vry non-rlxiv lmnt is prsnt. Howvr, this rltion is not trnsitiv, sin (,) n (,) r in R ut (,) / R. Thus this rltion is not n quivln rltion. () {(x,y) y is iologil prnt o x} on th st o ll popl. This rltion is not rlxiv, sin prson is not prnt o thmslvs. It is lso not symmtri, sin i is prnt o, thn is not prnt o Finlly, trnsitivity lso ils, sin grnprnts r not prnts. Thus this rltion is not n quivln rltion. () {(x,y) N N lm(x,y) = 0} Noti tht this rltion is givn y {(,0),(,5),(,0),(5,),(5,0),(0,),(0,),(0,5),(0,0)} This rltion is symmtri, ut it is not rlxiv sin (,) / R, n it is not trnsitiv sin (5,0),(0,) R ut (5,) / R Thus this rltion is not n quivln rltion. () {(x,y) R R y = x + } This rltion is not rlxiv sin (, ) / R. This rltion is lso not symmtri sin (,5) R, ut (5,) / R. This rltion is not trnsitiv sin (,),(,5) R, ut (,5) / R Thus this rltion is not n quivln rltion. () or h o ()-() tht r quivln rltions, list th quivln lsss or th rltion. Noti tht () is th only quivln rltion. It quivln lsss r [0] R = {0,,} n [] R = {,}.. Din rltion R on R y {((x,y ),(x,y )) (x + y ) = (x + y )} () Show tht R is n quivln rltion. First noti tht this rltion is rlxiv, sin or vry orr pir (x,y ), (x + y ) = (x + y ). Nxt, on n lso s tht this rltion is symmtri, sin i (x + y ) = (x + y ) thn (x + y ) = (x + y ) Finlly, on n vriy tht this rltion is lso trnsitiv y osrving tht i (x +y ) = (x +y ) n (x +y ) = (x + y ), thn (x + y ) = (x + y ). Thus this rltion is n quivln rltion. () Dsri th quivln lsss o R. Suppos (x + y ) = r or som r R. thn [(x,y )] R = {(x i,y i ) (x i + y i ) = r }. Tht is, quivln lsss r sts o points tht li on th sm irl ntr out th origin.

2 . Givn tht A = {0,,,,} () Fin th smllst quivln rltion on A ontining th orr pirs {(,),(,),(,),(,0)} Th smllst quivln rltion ontining th givn points is: {(0,0),(0,),(0,),(,),(,),(,),(,),(,0),(,),(,),(,0),(,),(,)} () Drw th grph o th quivln rltion or oun in prt (). R 0 () List th quivln lsss o th rltion or oun in (). Th quivlnv lsss r: [0] R = {0,,} n [] R = {,}.. For h o th ollowing olltions o susts o A = {,,,,5}, trmin whthr o not th olltion is prtition. I it is, list th orr pirs in th quivln rltion trmin y th prtition. () {{,}, {,}, {5}} This olltion o susts is prtition. Noti tht th sts r isjoint n tht thir union is ll o A. Th orr pirs in th quivln rltion gnrt y this prtition r: {(,),(,),(,),(,),(,),(,),(,),(,5),(5,5)} () {{,,}, {}, {5}} This olltion o susts is prtition. Noti tht th sts r isjoint n tht thir union is ll o A. Th orr pirs in th quivln rltion gnrt y this prtition r: {(,),(,),(,),(,),(,),(,),(,),(,),(,)(,),(5,5)} () {{,,,}, {5}} This olltion o susts is prtition. Noti tht th sts r isjoint n tht thir union is ll o A. Th orr pirs in th quivln rltion gnrt y this prtition r: {(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(,),(5,5)} () {{,}, {}, {5}} This olltion o susts is not prtition. Th sts r isjoint ut thir union is missing. () {{,}, {,,}, {5}} This olltion o susts is not prtition. Noti tht th sts r not isjoint sin two o th sts ontin. 5. Dtrmin whthr or not th ollowing inry rltions r prtil orrs. B sur to justiy your nswrs. () {(0,0),(0,),(0,),(0,),(0,),(,),(,),(,),(,),(,),(,),(,),(,),(,)} on th st A = {0,,,,} Noti tht this rltion is rlxiv, sin it ontins ll rlxiv pirs. This rltion is lso ntisymmtri, sin thr r no rvrss o non-rlxiv lmnts prsnt in th rltion. Finlly, this rltion is trnsitiv [hk ll ovrlpping pirs]. Thus this is prtil orr.

3 () {(,),(,),(,),(,),(,),(,),(,)} on th st A = {,,,} Noti tht this rltion is rlxiv, sin it ontins ll rlxiv pirs. This rltion is lso ntisymmtri, sin thr r no rvrss o non-rlxiv lmnts prsnt in th rltion. Finlly, this rltion is trnsitiv [in t, th only ovrlpping pirs h involv rlxiv pir]. Thus this is prtil orr. () {(x,y) y is iologil prnt o x} This rltion is not rlxiv sin prson is not thir own prnt, so it is not prtil orr. () {(x,y) R R y = x + } This rltion is not rlxiv sin (, ) / R, so it is not prtil orr. () or h o ()-() tht r posts, rw th Hss igrm or th post (or thos tht r in on ininit sts, only rw init suprt o th igrm). 0 () () 6. Init whih lmnt is grtr or h givn pir using th stnr lxiogrphi orring. () (, 7) n (, ) (,7) (,) sin <. () (,7,,9) n (,,7,9) (,,7,9) (,7,,9) sin = n < 7. () (,,,) n (i,,,) (,,,) (i,,,) sin < i. () (,,n,,,n,) n (,,n,,n,,s) (,,n,,n,,s) (,,n,,,n,) sin =, =, n = n, n <. 7. Drw th Hss Digrm or th post (P({,,}), ) Ο {} {} {} {,} {,} {,} {,,}

4 8. Drw th Hss Digrm or th post (P({0,,,}), ) {0,,,} {0,,} {0,,} {0,,} {,,} {0,} {0,} {0,} {,} {,} {,} {0} {} {} {} Ο 9. Givn th post ({,,,5,6,7,0,0,0,60,70}, ) () Drw th Hss Digrm or this post () Fin th lst lmnt or xplin why thr is no lst lmnt. is th lst lmnt () Fin ll uppr ouns o {,5} 0, 0, 0, 60, () Fin th mximl lmnts. 60, 70 () Fin th miniml lmnts. () Fin th grtst lmnt or xplin why thr is no grtst lmnt. Thr is no grtst lmnt, sin thr r two mximl lmnts. (g) Fin th lst uppr oun o {,5} (i it xists). 0 is th lst uppr ouun. (h) Fin ll lowr ouns o {6,0}, (i) Fin th grtst lowr oun o {6,0} (i it xists). is th grtst lowr oun. (j) is this post ltti? Justiy your nswr. No. Sin th pir {60,70} hs no uppr oun, this post nnot ltti. 0. Suppos th ollowing grph rprsnts th rsult o roun roin vollyll tournmnt. Giv th won lost ror or h tm who prtiipt. Who won th tournmnt? A D C B Rll tht th in-gr o vrtx in irt grph rprsnting roun roin tournmnt givs th numr o losss or tht tm. Similrly, th out-gr o vrtx in irt grph rprsnting roun roin tournmnt givs th numr o wins or tht tm. Using this, Tm A hs ror o ( ), Tm B hs ror o (0 ), Tm C hs ror o ( 0), n Tm D hs ror o ( ).

5 Thror, Tm C won th tournmnt.. Stt wht typ o grph mol you woul us or h o th ollowing. Mk it lr wht th vrtis in your mol rprsnt, wht th gs rprsnt, whthr th gs r irt or unirt, n whthr not not loops n multigs r llow. () A grph mol or th strts in Moorh, Minnsot. W will tk vrtis to intrstions n gs to ros (or lns o ros) twn intrstions. Sin Moorh hs som on-wy strts, w n irt grph. Loops o not mk sns in this ontxt. W oul hoos to us multigs o w wish to ount or th strts tht hv mor thn on ln hing in prtiulr irtion. () A grph tht mols th molulr strutur o rtin protin molul. W will tk vrtis to toms n gs to hmil ons twn toms. Chmil ons r gnrlly thought o s ing twn pir o toms rthr thn rom on tom to nothr, so on woul normlly us n unirt grph. W gnrlly o not think o ons rom n tom to itsl, so thr woul no loops. Sin thr multipl ons in som moluls, w woul mk us o multi-gs. () A grph moling ll th phon lls m on rtin rrir uring th lst hour. I w tk our vrtis to phon numrs, n gs to lls originting rom on numr n onnting with nothr numr, thn w lrly n irt gs. W oul l with onrn lls y rwing n g rom th prson originting th onrn ll to h o th othr iniviul lins involv in th ll. Loops o not mk sns in this ontxt, sin prson nnot ll himsl (t lst not t th sm numr). W woul llow multi-gs, sin pron oul ll th sm prson mor thn on uring singl hour. () A grph moling omputr ntwork. Hr, loops o not mk sns. Whthr on uss multigs n/or multi-gs pns on how prisly you think out how onntions r m twn omputrs (srvr/slv rltionships n/or multipl ports oul tkn into ount).. Givn th ollowing unirt grph: g i h j () Fin E E = () Fin V V = 0 () Fin g(), g(), n g(g) g() = 5, g() =, n g(g) = () List ll isolt vrtis o j is n isolt vrtx. () List ll pnnt vrtis o is pnnt vrtx.

6 () Is onnt? Justiy your nswr. How mny onnt omponnts os hv? is not onnxt, sin it hs n isolt vrtx. hs two onnt omponnts sin vry pir o vrtis not involving j is onnt. (g) List ll rig gs in. hs two rig gs: {,} n {,g} (h) List ll ut vrtis in.,, n g r ut vrtis. (i) Fin th st o vrtis jnt to vrtx. {,,,,}. (j) Fin th gr squn or. 5,,,,,,,,,0 (k) Vriy tht th Hnshking Thorm hols or. Aoring to th Hnshking Thorm, v V g(v) = E. Hr, g(v) = = = () = E. Thror, th Hnshking Thorm is vrii or this v V grph. (l) Form th jny mtrix or with th vrtis orr lphtilly (m) Form n inin mtrix or with th vrtis orr lphtilly n th gs n orr o your hoosing Givn th ollowing irt grph: g i h () Fin E E = () Fin V V = 9

7 () Fin g (), g (), g + (), n g + (). g () =, g () =, g + () =, n g + () =. () Is strongly onnt? Justiy your nswr. How mny strongly onnt omponnts os hv? is not strongly onnt us thr r no pths into vrtx. hs two strongly onnt omponnts: {} n {,,,,, g, h, i} () wkly onnt? Justiy your nswr. Ys. Th unrlying grph is onnt, so is wkly onnt. () Fin th st o vrtis jnt to vrtx. Th vrtis {,,} r jnt to vrtx. (g) Fin th st o vrtis jnt rom vrtx. Th vrtis {,} r jnt rom vrtx. (h) Form th jny mtrix or with th vrtis orr lphtilly Us th Hnshking Thorm to prov tht n unirt grph hs n vn numr o vrtis o o gr. Proo: Aoring to th Hnshk Thorm, or ny unirt grph, v V o th gr squn o grph is vn. g(v) = E. Tht is, th sum o th lmnts Lt V th st o vrtis o with vn gr, n lt V th st o vrtis o o gr. Lt k = v V g(v). Sin th sum two vn numrs is vn, k is vn. Lt m = v V g(v) Thn, E k = m is lso vn, sin th irn o two vn numrs is vn. I w suppos tht V is o, thn m = v V g(v) is o, sin th sum o n o numr o o intgrs is o. This is ontrition, sin m hs lry n shown to vn. Hn V must vn. Tht is, must hv n vn numr o vrtis with o gr. 5. () Drw h o th ollowing grphs:

8 i. K ii. C 8 iii. W 5 K C 8 W 5 iv. K 6 v. Q vi. C 7 K 6 Q C 7 vii. K, viii. K, ix. K, K, K, K, () Whih o th grphs ov r iprtit? Justiy your nswr. ii, v, vii, viii, n ix r iprtit. A -oloring hs n inlu in th grphs shown ov. All th othrs ontin t lst on o yl, so tht nnot iprtit. 6. Drw ll sugrphs o th ollowing grph. [***** I ll ths ltr *****] 7. From th list o sugrphs you oun ov, rw on rprsnttiv o h isomorphism lss. [***** I ll ths ltr *****] 8. Fin th union o th ollowing grphs:

9 U 9. 5 oys (A, Bn, Chuk, Donl, Elmr) n 6 girls (Frnin, Grthn, Hthr, Irn, Jnnir, n Kti) r trying to in ts to th junior prom. Frnin is willing to go with Bn, Chuk, Donl, or Elmr. Grthn is willing to go with A or Donl. Hthr is willing to go with Bn or Chuk. Irn is willing to go with Chuk or Donl. Jnnir is willing to go with Bn, Chuk, or Donl. Kti is willing to go with Donl or Bn. () Drw grph moling this sitution. F G H I J K A B C D E () Fin mthing or whih vry oy hs t to th prom. F G H I J K A B C D E This mthing orrspons to A going with Grthn, Bn going with Hthr, Chuk going with Jnnir, Donl going with Kti, n Elmr going with Frnin. 0. Drw simpl grph with h o th ollowing gr squns or stt why on is not possil. (),,, (),,, Not possil. Sin thr r only vrtis totl in this vrtx squn, thr r only thr othr gs tht ny vrtx n jnt to, sin w r insisting on simpl grph. Thror, w nnot hv vrtx with gr. In t, ny vrtx in simpl grph with n vrtis hs gr t most n. (),,,,

10 () 6,,,,,, () 6,,,, Not possil. As ov, sin thr r only 5 vrtis in this gr squn, thr nnot hv vrtx with gr 6. In t, ny vrtx in simpl grph with n vrtis hs gr t most n.. For h pir o grphs, prov tht th grphs r isomorphi, or prov tht thy nnot isomorphi. 5 () Ths grphs r isomorphi. To s this, onsir th ollowing ijtion twn th vrtx sts: Vrtx in Vrtx in 5 It is sy to vriy tht this ijtion prsrvs jny, sin it prsrvs 5-yl prsnt in oth grphs. () Ths grphs r isomorphi. To s this, onsir th ollowing ijtion twn th vrtx sts: Vrtx in Vrtx in In t, sin vry pir o istint vrtis r jnt in ths grphs, ny ijtion twn th vrtis is n isomorphism [so thr r n! =! = irnt ijtions twn ths grphs]. 6 () 5 Ths grphs r not isomorphi. To s this, noti tht hs gr, whil no vrtx in hs gr.

11 g i () Ths grphs r isomorphi. To s this, onsir th ollowing ijtion twn th vrtx sts: Vrtx in Vrtx in i g h It is sy to vriy tht this ijtion prsrvs jny, i you n visuliz th trnormtion y imgining strightning out th right ngl in orm y n thn rotting th grph 80 ir.. For h o th ollowing grphs trmin: (i) Whthr or not th grph hs n Eulr Ciruit. (ii) Whthr or not th grph hs n Eulr Pth. (iii) Whthr or not th grph hs Hmilton Ciruit. (iv) Whthr or not th grph hs Hmilton Pth. B sur to justiy your nswrs. h () () Grph () hs vrtis o o gr, so it os not hv n Eulr Ciruit, ut it os hv n Eulr Pth. Grph () os not hv Hmilton Ciruit. To s this, noti tht oth gr vrtis r ut vrtis, ut on ths gs r rmov, th only othr g onnt to th ntr vrtx is rig g, so on nnot orm Hmilton iruit. Grph () os hv Hmilton Pth (strt in th lowr lt ornr vrtx, trvl lokwis though th othr 6 vrtis on th primtr o th grph, n thn go up to th lst vrtx). Grph () hs vrtis with vn grs, so it os hv n Eulr Ciruit. Grph () os not hv Hmilton Ciruit. To s this is it sutl, ut visiting ll o th vrtis in on o th min igonl isonnts th grph so tht iruit nnot omplt. Grph () os hv Hmilton Pth, s xhiit y th ollowing:

12 () () Grph () hs vrtis o o gr, so it os not hv n Eulr Ciruit, ut it os hv n Eulr Pth. Grph () os not hv Hmilton Ciruit. To s this, noti th rig g in th grph. Grph () os hv Hmilton Pth (ltrnt trvling lt n thn right through h lvl o th grph). Grph () hs vrtis o o gr, so it os not hv n Eulr Ciruit, ut it os hv n Eulr Pth. Grph () os not hv Hmilton Ciruit. To s this, noti th rig g in th grph. Grph () os not hv Hmilton Pth. To s this, noti tht th vrtx o gr 5 is ut vrtx whos rmovl sprts th grph in to omponnts, only two o whih n visit y n pth tht os not rpt this vrtx.. Drw grph tht stisis th hypothss o Dir s Thorm. Explin how you know tht h hypothsis is stisi. Noti tht this grph hs 6 vrtis (so mor thn ). To stisy Dir s Thorm, h vrtx ns to hv gr t lst 6 =, n w n s tht h vrtx hs gr. [Not: I w think o ginning with just ths six vrtis n ing gs, you shoul l to irst onvin yoursl tht you nnot mk ll vrtis hv gr without th grph ing onnt. Nxt, onvin yoursl tht thr must Hmilton Ciruit.]. Drw grph tht stisis th hypothss o Or s Thorm. Explin how you know tht h hypothsis is stisi. (your xmpl my not omplt grph.) Noti tht th sm grph lso stiis Or s Thorm. Th grph hs 6 vrtis (so mor thn ). To stisy Dir s Thorm, h nonjnt pir o vrtis ns to hv grs tht up to t lst n = 6, n w n s tht sin h vrtx hs gr, th sum o th grs o ny pir o nonjnt vrtis is 6.

13 5. For h o th ollowing wight grphs, us Dijkstr s Algorithm to in shortst pth rom to z. () 8 5 z (,,,,) () (,,,) z 8 5 9(,) 0(,,) (,,,) 7(,) () Th igur ov shows th rsult o rrying out Dijkstr s Algorithm on th givn wight grph. From this, w s tht th shortst pth hs wight: n ollows th vrtx squn,,,,,z. () g 6 h z () (,,,) 5 5 g 5 (,) 7(,,) 6 h 5() (,,,) (,) 5(,,,,,g) 6(,,,,) z (,,,) Th igur ov shows th rsult o rrying out Dijkstr s Algorithm on th givn wight grph. From this, w s tht th shortst pth hs wight: 5 n ollows th vrtx squn,,,,,g,z. Noti tht,,,,h,z is lso shortst pth, ut it is not th pth oun y th lgorithm. 6. Us rut or lgorithm to in n optiml solution or th trvling slsmn prolm or th givn grph strting t hom vrtx : () [***** I ll ths ltr *****]

14 () 5 [***** I ll ths ltr *****]