Question

Define the graph Gn to have the 2n nodes 20, 21,...,an-1, bo, b1, ..., bn-1 and the following edges. Each node ai, for i = 0,1,...,n - 1, is connected to the nodes b; and bk, where j = 2i mod n and k = (2i + 1) mod n (a) Prove that for every n, G, has a perfect matching. (b) How many different perfect matchings does G100 have?

Answer 1

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theorem - A) proove Hall's marriage theorem state that Halls marriage I a perfect matching it for W cx or wly, WI SING (W)). NG (w] =) nel bowywed Le) w Now if let I wc sao, a ... Such that (W! > ING (will as each of the verises of w has 2 edge posible an vertices lt Nis) That have a degree of 3 [aldeast] Say Cinsi Osie? Ikit.). Ouist. without loss of generality, Soinicsck - then odd. withoutless of geniarility solution 1-2 s. modn 1-2 i mod in 2 we can choose this because 72 vertices among ij,k, such that they come from Same equation as I only 2 egn to come do modna 23 mod n. 2 Ci-j) mod no. 2j-1). En j - #0; <,,Olicin =) jich = 2 (j=1) 2an

but in is odd Lлода се weget contradiction case 2 - let u even Subcase even 1 then lazimod n=2j mod n=2.k mod n. 2 imod na 2 as ait mod in 8. add. for ti j modn. :-) 2(3-7). En by omilior argument? 2 j mod nark mod n. Sby Similor 2 CK-j) en argument 2 (57) +2 (K-j) - ntn. '} = 2 (1-1)=2.11 Krian. contradiction ما د ملي مانية ken, ">,0 Subcar 2 N la odda then l-2i+1) mod n (2jt i) moda (2 K-timedia by Similarly taking ist two and two equalis We will get - (-) = x 2 ( x ) = h . ad arrived +0 the same conclution kin which s. a contradiction Hence of such that fwl > Norw?l. Hence. By : Hall's marriage theoren. will perfect matching