Solutions for HW11. Exercise 34. (a) Use the recurrence relation t(g) = t(g e) + t(g/e) to count the number of spanning trees of v 1

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1 Solutions for HW Exris. () Us th rurrn rltion t(g) = t(g ) + t(g/) to ount th numr of spnning trs of v v v u u u Rmmr to kp multipl gs!! First rrw G so tht non of th gs ross: v u v Rursing on = (v, u ): u v u t(g) = t(g ) + t(g/) = 6 +. () Wht is th Prüfr o for th following ll tr? Chk your nswr y rvrsing th pross n uiling th tr from th o. Answr:,7,,,. 7 () Drw th tr whos Prüfr o is,,,, 6. Chk your nswr y lulting th Prüfr o tht gos with your tr () Drw ll K (ll with,, ), n list ll th spnning trs, n th orrsponing Prüfr o. Vrify tht thr is ijtion twn th ll trs on vrtis n th lngth- Prüfr os. () How mny spnning trs os K 7 hv? (f) How mny ll trs r thr on vrtis?

2 Exris. () For th following prmuttions w, rw T (w) n T (w). Vrify tht T (w) n T (w) r inrsing. (i) w = 687 T (687) = T (687) = (ii) w = T () = T () = 0 (iii) w = 0 T () = T () =

3 () Lt T = If T = T (w), wht is w? Vrify tht th oul riss, oul snts, vllys, n pks of w orrspon to th orrt hvior of sussors. () Lt If T = T (w), thn w = T = If T = T (w), wht is w? Vrify th following for T (w). If T = T (w), thn w = 876. (i) Th sussors of 0 r just th lft-to-right minim of w. (ii) Th lvs r w(i) i D(w) n}}. w(i) i D(w)} =, 8, 7, 6} () Brifly justify h of () () in Proposition.. of EC. () Brifly justify h of () () in Proposition.. of EC. Exris 6. Lt A, B, C, n D th grphs A = B = g f f

4 C = f () Clult th hromti numrs for A, B, C, n D. For h, giv n xmpl of vrtx oloring of th orrsponing grph using xtly χ olors. Colorings init y olors,,,... : D = A = B = C = D = Sin A, B, n B ll hv liqus of siz, thir hromti numrs r t lst, n D hs gs, so its hromti numr is t lst. Thus, y th olorings provi ov, w hv χ(a) = χ(b) = χ(c) =, χ(d) =. () Wht r th liqu n inpnn numrs of A, B, C, n D? How o ω n V /α ompr to χ for h grph? Th liqu numrs r givn y ω(a) = ω(b) = ω(c) =, ω(d) =, whih xtly mth th hromti numr. Th inpnn numrs r givn y α(a) =,, } =, α(b) = f,, } =, α(c) =, f} =, α(d) =,, } =, so tht V (A) /α(a) = 6/ =, V (B) /α(b) = 7/ =., V (C) /α(c) = 6/ =, V (D) /α(d) = / =. 6, whos ilings giv shrp ouns for th hromti numrs of B, C, n C, ut not A.

5 () Wht r th hromti numrs of (i) K m,n, (ii) C n, (iii) W n, n (iv) Q n? n vn, (i) χ(k m,n ) =, (ii) χ(c n ) = n o, n vn, (iii) χ(w n ) = n o,, n (iv) χ(q n) =. () Wht r th liqu n inpnn numrs of (i) K m,n, (ii) C n, (iii) W n, (iv) Q n? How o ω n V /α ompr to χ for h grph? (You my n to rk into ss.) ω(k m,n ) = n =, ω(c n ) = n > n = ω(w n ) = n > ω(q n ) =. α(k m,n ) = mx(m, n) α(c n ) = n/ α(w n ) = n/ α(q n ) = n. () Wht r nssry n suffiint onitions for grph to hv hromti numr (i), n (ii). Th hromti numr of G is if n only if G hs no gs. Th hromti numr of G is if n only if G is iprtit n ontins n g. (f) For A, B, C, n D, pik n orring of th vrtis n olor th grph using th gry lgorithm. Compr th olors you us to th hromti numr of th grph. Cn you fin n orring tht yils oloring with too mny olors? (g) Explin why th liqu numr of th omplmnt of iprtit is no smllr thn th numr of vrtis in h prt. (Rll tht th prts of iprtit grph r th two olltions of pirwis non-jnt vrtis.) Sin h prt of iprtit grph is n inpnnt st, tht st forms liqu in th omplmnt.

6 (h) Noti tht th grph x x x x 7 G = x x x 6 x 8 is iprtit, so shoul hv hromti numr. Now olor this grph using th gry lgorithm. How mny olors i you n? Answr: Exris 7. () Comput th numr of wys to olor th grph with h of,,..., V olors for G = n H = χ(, 0) χ(, ) χ(, ) χ(, ) χ(, ) G H () Clult th hromti polynomil for G ov. Answr: χ(g, t) = t(t ) (t ), χ(g, t) = t(t )(t ). () Th hromti polynomil for th yl C n is χ(c n, k) = (k ) n + ( ) n (k ). (i) Drw ll th wys of oloring th -yl with olors. Thn omput χ(c, ) n ompr your nswrs. Orr th vrtis in lokwis orr. Choos th olor for th first vrtx ( hois), thn th son ( hois), n finlly th thir vrtx ( hoi), giving 6 olorings. This mths χ(c, ) = ( ) + ( ) ( ) = = 6. (ii) How mny wys r thr to olor th -yl with olors? χ(c, ) = ( ) + ( ) ( ) =. (iii) How mny wys r thr to olor th 6-yl with olors? χ(c, ) = ( ) 6 + ( ) 6 ( ) = + =.

7 (iv) Us χ(c n, k) to vrify tht vn yls r iprtit n o yls r not. For n vn or o, χ(c n, ) = ( ) n + ( ) n ( ) = 0. For n vn, χ(c n, ) = ( ) n + ( ) n ( ) = + = 0, so χ(c n ) =. For n o, χ(c n, ) = ( ) n + ( ) n ( ) = = 0, so χ(c n ) >.