Complt Solutions or MATH 012 Quiz 2, Otor 25, 2011, WTT Not. T nswrs ivn r r mor omplt tn is xpt on n tul xm. It is intn tt t mor omprnsiv solutions prsnt r will vlul to stunts in stuyin or t inl xm. In w pls, t worin o prolm is n slitly to rlt t moi lyout. A tl proviin point vlus or t prolms is ivn t t vry n. 1. Dv lims tr is trinl-r rp wit 8 vrtis n 412 s. Yoln sys is wron. Explin wo is rit. T intnt o tis qustion is to trmin wtr you know Turán s torm, wi stts tt t mximum numr o s in trinl-r rp on n vrtis is n 2 /4. Tis vlu is t numr o s in omplt iprtit rp in wi t two prt sizs r s ln s possil. In tis s, sin n = 8, t two prts woul ot v 19 vrtis, so t mximum numr o s, provi tr wr no trinls, is 19 2 = 61. Sin t rp Dv rrs to s 412 s, it must v trinl. So Yoln is rit. 2. Zori lims tt tr is plnr rp wit 1012 vrtis n 672 s. Crlos sys s is wron. Explin wo is rit. T intnt r is to trmin wtr you rmmr tt plnr rp on n vrtis s t most n 6 s wn n. W sow tis in lss y strtin wit n ritrry plnr rp n sowin tt i ny is not trinl, tn w n lwys n n prsrv plnrity. Wn ll s (inluin t xtrnl ) r trinls, tn Eulr s ormul will sow tt t numr o s is xtly n 6. In t s t n, Crlos is rit. T rp s 1012 vrtis n 672 s. Sin 672 > (1012) 6, it n t plnr.. Vriy Eulr s ormul or tis plnr rp. W ount V, t numr o vrtis; E, t numr o s; n F, t numr o s, n s tt V = 9, E = 10 n F =. So, V E + F = 9 10 + = 2. 4. How mny symmtri inry rltions r tr on {1, 2,..., n}? O ts ow mny r rlxiv? A inry rltion R on st X is symmtri wn (x, y) R i n only i (y, x) R, or vry x, y X. Tr r n pirs o t orm (x, x), wr x {1, 2,..., n}. For o ts pirs, w i wtr to put (x, x) in t rltion or not. Similrly tr r ( n 2) 2-lmnt susts o t orm {x, y} wit x, y {1, 2,..., n} n x y. For su pir, w must itr put ot (x, y) n (y, x) in t rltion or put nitr in t rltion. So t totl numr o symmtri rltions is 2 n 2 (n 2). For t son prt, inry rltion R is rlxiv i (x, x) R or vry x X. Now, w los t issu o oi. All pirs (x, x) wit x {1, 2,..., n} must lon to R, n w v oi only or t istint 2-lmnt susts. So t totl numr o rltions wi r ot symmtri n rlxiv is 2 2) (n.
5.. Lt X = {,,,, } n lt P = {(, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, )}. Drw irm or t post (X, P ). 6. Sow tt t ollowin rp is omprility rp y trnsitivly orintin t s. In t sp to t rit, rw t irm o t ssoit post. T iur on t lt is t oriinl rp ivn on t tst. In t ntr is trnsitiv orinttion. I strt wit t wi I orint s (, ), i.., irt rom to. Tis totr wit orms V, sin is not n in t rp. Tis ors t irt (, ). T (, ) ors (, ), wi tn ors (, ). In turn, (, ) ors (, ). Now (, ) ors (, ) n (, ) ors (, ). It is sy to s tt t orinttion is ssntilly uniqu, i.., t only option is to rvrs t irtions on ll t s. T iur on t rit is irm or t rsultin post. O ours, i t rrows r rvrs in t mil iur, tn t irm on t rit is turn upsi own. 7. Consir t post sown on t lt. j 2 j 2 k k 2 2 1. Fin ll points omprl to. i 1 1 1 A point u is omprl to wn itr u < or u >. In tis post, is miniml point, so no otr point is lss tn. Howvr,, n r ll ir tn. Answr: {,, }.. Fin ll points wi ovr. A point u ovrs wn u > n tr is no point v twn tm. Hr n ovr. Not tt > ut os not ovr sin is twn tm. Answr: {, }. i
. Fin mximl in o siz 2. A in C is mximl wn tr is no point u wi is not in C or wi C {x} is lso in. In tis post {, }, i, j} n {, i} r 2-lmnt mximl ins.. Fin mximl ntiin o siz. In tis post, {, i, k} is -lmnt mximl ntiin.. Fin t st o ll miniml lmnts. An lmnt u is miniml wn tr is no v wit v < u in t post. Hr t st o miniml lmnts is {,,, i}.. Usin t loritm tut in lss (rursivly rmovin t st o miniml lmnts), in t it o t post n prtition o P into ntiins. Also in mximum in. You my init t prtition y writin irtly on t im. T ntiin prtition is rwn to t rit o t oriinl post, i.., or i = 1, 2,, t ntiin A i onsists o t points lll wit ol i. Fin t it o tis post n in mximum in. To nswr tis qustion, you tk n lmnt rom t ist ntiin, n tn prorm k-trkin. Hr you mit strt wit wi oms rom A. Wy is not in A 2? Bus is ovr somtin in A 2. Hr w s is in A 2 n >. Wy is not in A 1? Bus it is ovr somtin in A 1. Now w s tt is in A 1 n >. So t it is n {,, } is mximum in. Not tt tr r svrl mximum ins in tis post. 8. In t sp to t rit, rw t irms o tr irnt posts ll vin t ollowin rp s tir ovr rps () it 2, wit post; () it, wit post; n () it 4, wit 2 post. T nswr or prt () is uniqu up to ulity, i.., t post sown n turn upsi own. Also, tr r mny wys to rw t irm or t sm post, so orrt nswrs n ppr to somwt irnt ut ty r rlly rwins o t sm post. On t otr n, tr r multipl orrt nswrs or ot prt () n or prt (). Tis rlts t t tt it is iiult to trmin wtr rp is ovr rp. Also, it is iiult to ount t numr o istint posts wit t sm ovr rp.
9. By insption (not y loritm), in our points in t ollowin post tt orm 2 + 2. T our points in {,,, } orm two 2-lmnt ins wit ot points in on in inomprl wit ot points in t otr. 10. Fin y insption t wit w o t post sown low on t lt n in prtition o t post into w ins. Also in mximum ntiin. You my init t prtition y writin irtly on t irm. i 1 2 i 2 1 1 4 T in prtition is init y t iur on t rit. Fin t wit w o tis post n in mximum ntiin. By insption, t wit is 4 n {,,, } is 4-lmnt ntiin. Tr r otr 4-lmnt ntiins in tis post. On su is {,,, }. Not tt xpt or wn P is n intrvl orr, w o not v n loritm or inin t wit w o post P n prtition o P into w ins. From Dilwort s torm, w know tt su prtition xists. W just on t know ow to in it. Tis issu will rss rit t t n o our ours. 11. Sown low is t irm o n intrvl orr. Us t loritm tut in lss to in n intrvl rprsnttion. Tn us t First Fit olorin loritm to in t wit w n prtition o t post into w ins. Also, in mximum ntiin. W list low t own sts n t up sts. Tr r 5 istint own sts n 5 istint up sts (ts numrs will lwys t sm. Cn you xplin wy?), n w ll t own sts rom littl to i n t up sts rom i to littl. Tis lllin is sown in ol typ.
D() = {,,, } 5 D() = {, } D() = {,, } 4 D() = 1 D() = 1 D() = {} 2 D() = {, } U() = 5 U() = 5 U() = 5 U() = {,,,, } 1 U() = {,,, } 2 U() = {} 4 U() = {, } W sow low t intrvl rprsnttion wi tis lllin prous. Not tt tis rprsnttion is uniqu n it uss t wst numr o npoints. 1 2 [ ] [ ] 2 [ ] [ ] [ ] 1 1 [ ] [ ] 1 2 4 5 Fin t wit w o tis post n in mximum ntiin. In ontrst to t sitution in Prolm 10, w v ln loritm or inin t wit w, mximum ntiin n prtition o t post into w ins. Tis is omplis y pplyin t irst it olorin lorit in t orr o lt n points. Tis is illustrt in t iur y t olorin wit t ol olors pl nr t lt n points. Hr I v rokn tis y proin rom top to ottom in t pitur. Furtrmor, w s tt w = n {,, } is mximum ntiin. To s ow tis is on, lt k t lrst olor ssin. Wn vrtx x is ssin olor k, tn or i = 1, 2,..., k 1, t vrtx x must jnt to vrtx x i olor i wi s lry n olor. Tis mns tt t intrvl or x i ontins t lt npoint o t intrvl or x, so {x, x 1, x 2,..., x k 1 } is n ntiin o siz k. 12. Lt 2 1 t post onsistin o ll susts o {1, 2,,..., 1}, orr y inlusion.. Wt is t it o tis post? W not tt ll mximl ins r mximum n ll v 14 points. T rson is tt mximl in strts wit (0, 0, 0,..., 0 s its lst lmnt (1 zros in tis strin). Tn s w pro up t in, t stp zro is swpp out or on. Tis is rpt until w t to t vry top lmnt (1, 1, 1,..., 1).. Wt is t wit o tis post? T intnt o tis qustion is to s i you rmmr Sprnr s torm wi stts tt t wit o t ltti o ll susts o {1, 2,,..., n} is t mil inomil oiint C(n, n/2 ). So in tis s, t wit is C(1, 6) wi o ours n lso writtn s ( 1 6 ).
. How mny mximl ins os t post v? W v lry ommnt on t strutur o mximl ins. Usin tis strutur, w s tt tr r n! mximl ins. Tis ollows rom t t tt t stp 1, w n oos ny o t n zros in (0, 0, 0,..., 0) to tol to on. For t son stp, w oos on o t rminin n 1 zros to lip. Tn n 2, t.. How mny mximl ins in tis post pss trou t st {2, 5, 8, 10}? To nswr tis qustion, w pply t rsonin o t proin nswr in ot irtions, so t nswr is 4!9!. 1. Extr Crit. Sow tt t post sown low s imnsion 2 y inin two linr xtnsions wos intrstion is t prtil orr. i A rprsnttion o P usin qurtrplns is sown low. i From tis rprsnttion, w s tt w my tk: L 1 = [,,,,,,,, i] n L 2 = [,,,,, i,,, ].
Point Totls 1. Svn points. 2. Svn points.. Svn points. 4. Six points. 5. Six points. 6. Eit points. 7. Sixtn points. 16 = 2 + 2 + 2 + 2 + 2 + 6. 8. Six points. 6 = 2 + 2 + 2. 9. Six points. 10. Eit points. 11. Fitn points. 12. Eit points. 8 = 2 + 2 + 2 + 2. T totl is 100 n t Extr Crit prolm oul rn up to +10 points.