Design and Analysis of Algorithms (Autumn 2017)

Transcription

1 Din an Analyi o Alorim (Auumn 2017) Exri 3 Soluion 1. Sor pa Ain om poiiv an naiv o o ar o rap own low, o a Bllman-For in a or pa. Simula ir alorim a ru prolm o a layr DAG ( li), or on a an riv rom rurrn. Sow alo ow o raak an opimal pa. Soluion. For air rawin, ollowin omi ar (aum i ain i nou o o a on an u i anno or pa). L u n ain ollowin o or or ar: Now layr vrion o rap wi V = 4 layr look lik i: Fir w iniializ ian in ir layr a 0 or, an or or no.

2 Tn w ompu ian or or no wi Bllman-For rurrn (v, k) = min v N (v){(v, k 1) + (v, v)}. Lookin a valu a no 1, 2, 3 an 4 w orv a or pa rom o u 3 an a o 2. Train ak pa a prou i valu w oain:

3 Wi ak o oriinal rap i a. 2. Pa in rap II L G = (V, E) a ir ayli rap an l w : E N a union ainin wi o o G. Rall rom lur a olnk o a pa P in G i in a minimum valu o wi o i. Givn wo vri an o G, ow a amon all pa rom o, w an in on o maximum olnk in im O( E ). Soluion. L u all (v) maximum olnk rom our o no v. T prour o ompu valu i own in Alorim 1. Runnin im analyi (w aum a V E ). T im i O( E ) au opoloial or ak O( E ), an in Alorim 1, in wo or loop, a o G i inp on. Alorim 1 Maximum Bolnk in a DAG 1: prour MaxBolnk(G = (V, E))) 2: opoloial or(g) 3: or (u V \ {}) o 4: (u) 5: () + 6: or (v opoloial orr(g)) o 7: or (u N (v)) o 8: (v) max((v), min((u), w(u, v))) 9: rurn () 3. Flow an inraliy Prov inraliy orm o low: I apaiy union ak only inral valu, n maximum low prou y For-Fulkron mo a propry a i an inr. Morovr, or all (u, v), valu o (u, v) i an inr. Soluion. Ti an provn y inuion: i a a ivn i- p o alorim w aum a low ain o a vrx i inr, i i ay o a riual nwork will av only inral apaii. Tror, any pa oon in riual nwork will av inral olnk, an aumnin pa will inra low y an inral valu. Tror, rulin low in i + 1- p will alo inral. In iniial a, a o 0- iraion, laim ol au nwork a low qual o zro or all. 4. Bipari main uin For-Fulkron Conir ipari rap low.

4 a Simula ruion onir a Wnay lur (alo in our ook S. 27.3) rom maximum ipari main o maximum low, AND n imula For-Fulkron alorim on i inpu. Explain ow o xra maximum main rom maximum low oluion. Soluion. T rulin low nwork i a ollow: a Ar runin For-Fulkron alorim a poil rul i a ollow: 1 a

5 Tror, oon main i {a, }, {, }, {, }. 5. Max low - Jo ainmn wi onrain loa L A a o ay o (ink o ompur o), A a o ar o, an l B a o main. For a o a A A, w know on wi main i an xu. Morovr, a main B an xu: a mo a ivn numr () o ay o; a mo a ivn numr () o ar o; a mo a ivn numr () o oal o (ay or ar). W n o in an ainmn o (a u o) o o main a oy xuion limi on main, an maximiz numr o o ain or xuion. Giv a ruion rom i prolm o a maximum low prolm an aru a i i orr. Soluion. W will uil a nwork N wi ollowin no: Sour, ink, on no a i pr a o (ay or ar), an r no M H i, M E i an M T i or a main. W will a ollowin wi apaii: (M H i, M T i ) = ( i ), (M E i, M T i ) = ( i ), (M T i, ) = ( i ). For a ay o a i a an xu in main w ra an (a i, M E) wi (a i, M E) = 1, an or a ar o a i a an xu in main w ra an (a i, M H) wi (a i, M E ) = 1. Finally, or a o a i w ra an (, a i ) wi (, a i ) = 1. W mu prov a o ainmn prolm an ainmn o k o i an only i r i an inral low in N o valu k. : Suppo w av a ainmn o k o: or a ay/ar o a i, rpivly, ain o main, w a low 1 on (, a i ), (a i, M E/H ), (M E/H, M T ), (M T, ). By onruion, in w ar wi a vali ainmn o o, i i low aii apaii o N an a valu k. : L a low o valu k in N. By onruion, in a only inr valu, r ar xaly k o yp (a i, M E/H ) wi low valu 1. For a o, w ain o a i o main M (k ainmn in oal). By way apaii in N ar in, i i alo a vali o ainmn.