Question

pen minimum ranbot of dons so tat a gesn con sthart isit all the rns at least ence once circuat

Answer 1

uton does hot clealy mention nhehe the Thus, Je o3S ume thot the does no indicote o belween every Pais, of odjacent is a roon o Moo rni, we model the Pioblem as follows in groph theory >oom i5 a Vey tex. Thre exists an E veau, eween two vertice there is a door between beloi ern. The diaztom wil be modelled a3 follows 0 Foch om should be visited ot least once. ? poss thoa doo-s ex octl once'. Here pass thouah all the door s Eulev ian Civcuit !we have to then' exa utl once , t rding Pvoblern. Ofhoevise 's we hae some doors exactly once than roblem Hamiltonion cycle finsrdi Y O

obout it. Hence we consrd cl eoo not ver boh Hhe cases case-i 。- Eulerian. Circuit Problem h has, vev texc I Since ot γ10phane of even not au tho Vevtices ее Thus roph is not Ewlohian Hence Eulevian esrcuit does not, cxist、 Problem is not solvable ,_ Hani | to n an Cycle Problem Here we tavel eachnoms exactty once so that is becomes the min rinium COS e Some doos exactly once The Hamiltonion cle is Hence m inimum number dloórs othing but , the number dogs opened ỉs nothi ín the above cycle whith s 9 Thus ot et1 dloors have to be e openad