Probabilistic Methods in Combinatorics: Homework Assignment Number 1 Noga Alon
Solutions will be collected in class on Wednesday, March 25, 2009. n−1
1. Suppose n > 4 and let H be an n-uniform hypergraph with at most 4 3n edges. Prove that there is a coloring of the vertices of H by 4 colors so that in every edge all 4 colors are represented. 2. Prove that there is an absolute constant c > 0 with the following property. Let A be an n by n matrix with pairwise distinct entries. Then there is a permutation of the rows of A so that no √ column in the permuted matrix contains an increasing sub-sequence of length at least c n. 3. (i) Prove that every set A of n nonzero integers contains two disjoint subsets B1 , B2 ⊂ A, so that |B1 | + |B2 | > 2n/3 and each set Bi is sum-free (that is, there are no b1 , b2 , b3 ∈ Bi so that b1 + b2 = b3 .) (ii) Prove that the same conclusion holds for any set A of n nonzero reals. 4. Let {(Ai , Bi ), 1 ≤ i ≤ h} be a family of pairs of subsets of the set of integers such that |Ai |+|Bi | = k for all i, Ai ∩ Bi = ∅ and (Ai ∩ Bj ) ∪ (Aj ∩ Bi ) 6= ∅ for all i 6= j. Prove that h ≤ 2k . 5. Let X be a collection of pairwise orthogonal unit vectors in Rn and suppose the projection of each of these vectors on the first k coordinates is of Euclidean norm at least ǫ. Show that p |X| ≤ k/ǫ2 , and this is tight for all n = 2r , and ǫ = k/2r < 1, with r ≥ 1 an integer.
1 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 uxna 25
1 dl`y
e
m`
xe
= 0-e
mirav 4-a dreav `l
e
xe
m`
e
xicbp
= 1
zyw lkl .z"ae cig` ote`a erav z` lixbp znev lkl -y xexa .mirav drax`a dreav
E [xe ] = 4
n
n
3
2
6
4
n 1
+4
4
4
<4
n 3
4
:miiw .mipey mirav drax`a zereav `ly zezywd xtqn
E [x] =
X
e
E [xe ] <
X
e
3
n 4
n 1
4
n
1
n 3
3
n 4
n 1
x=
P
e xe
idi
1
x < 1 dxear driav zniiw okle
.mirav 4-a dreav `ly zyw da oi`y driav zniiw xnelk ,
2 dl`y
.c e xgap pA-n zlawznd dvixhnd z` A -a onqp .zixwn dxenz idz , i n lkl c n jxe`a dler dxcq-zz yi A ly i dcenra m` x
onqp . -l m`zda zexeyd selgy i"r
xi
zexazqdd .zxg`
k
= 0-e
E [xi ] k
.
k
,
:=
jxe`a dler dxcq-zz zniiwy zexazqdd okle dl`k zexcq
f pn
n k! k
1
en k k k k e
=
E [x]
lw .
(
ez ) ik xexa f`e f (z ) = z e 2 2
)
(
)
=
k
yi . !
i
1
`id dler didz
= 1
k
2
1
lceba dxcq-zzy :zniiwn
2 cpn e 2cpn p en e2 n k 2e2 n = = = e 2 2 k cn c
x=
Pn
Stirling itl oekp oexg`d oeieey-i`d) dtivx divwpet xicbp .x = 0 dxear zniiw f`e E [x] < 1 ik d`xp
jxe`a dler dxcq-zz oda yiy zecenrd xtqn z` 2
n k
(
i=1 xi -a
onqp (
:ik ze`xl 2 f 0 (z ) = 2ze 2e z (1
,
n
lkl xnelk ,
z
lkl
f (z ) < 1-y
f e
o`kn . (
2 f 00 (z ) = 2e 2e z (2z 2 e4
4
ok enk ,meniqkn zcewp efe
z
ze2 ) 2
)
<1
e2 z + 1) =
e
2
-a lawzn oeviwdy o`kn .yexck ,
E [x] < 1
` 3 dl`y
p > 2 maxa2A jaj miiwnd (mixf 6-e 5 ik Dirichlet itl dl`k seqpi` yi) ipey`x p = 6k + 5 idi
:ik ze`xl lw .
K (1) = f2k + 2; : : : ; 4k + 3g
K (2) = fk + 1; : : : ; 2k + 1; 4k + 4; : : : ; 5k + 4g
(2) (i) A(1) x \ Ax = ;-y xexa .Ax = fy 2 A j xy (mod p) 2 K (i) g xicbp 0 < x < p lkl .p elecen zeiyteg x-yk p - y -e li`ed .zxg` vy(i) = 0-e y 2 A(xi) m` vy(i) = 1 xicbp y 2 A lkl .ziyteg A(xi) ,ok enk .K (1) \ K (2) = ; (i) jK (i) j = 2k+2 > 1 :okl .mixtqn mze` lr weica xaer bx (mod p) ,f1; : : : ; p 1g lr xaer :miiw .Pr(vy ) = p 1 6k+4 3
ixdy
h
i
(2) E jA(1) x j + jAx j
X =
y 2A
h
E vy(1)
i
X +
y 2A
h
i
E vy(2) > 2 n = n 3 3
jB1 j jB2 j > 23 n mr B2
.
+
1
1
=
2
(1) A(2) x -e B1 = Ax
exear
x miiw okl
a 3 dl`y
fa ; : : : ; an g = A Rnf0g dveaw lkl ik d`xp .mcewd sirqdn diral zeliwy d`xp fPa01 ; : : : ; a0n g = zniiw n lceba Pn 1 n 0 0 . i=1 ai i ly oniql deey i=1 ai i ly ( 1; 0; 1) oniqd 1 ; : : : ; n 2 f 1; 0; 1g lkly jk n lceba A Znf0g P
n if` 0 `ed oniqd m` . 3n -n zg` lkl i=1 ai i ly oniql m`zda oeieey(-i`) xicbp 1 ; : : : ; n xegal zeiexyt`d Pn Pn -i` xicbp . q -y jk q miiw .iaeig `ed ik k"da gipp ,0-n dpey oniqd m` . x = 0 xicbp i=1 ai i Pni=1 i i x q :oeieey mincwnd lky oeeikn .iynn oexzt zlra mipzyn n-a miix`pil mipeieey(-i`) zkxrn eplaiw . i=1 i i
n
2Q
lawp ,mi`zn irah xtqna ilpeivxd oexzta mixtqnd zltkd i"r .zkxrnl ilpeivx oexzt mb miiw ,miilpeivx
0
0 0 0 0 0 0 0 0 0 0 enk .ziyteg Bi -e B1 \ B2 = ;-y jk B1 ; B2 A zeniiwy mcewd sirqa epi`xd .f (ai ) = ai 2 A ,ai 2 A lkl xicbp 0 0 0 0 0 0 0 ai +aj = ak -y xexa .zeyexcd zeveawd od B1 ; B2 f`e .i = 1; 2 xear Bi = ff (a ) j a 2 Bi g xicbp .jB1 j+jB20 j > n 3 ,ok Pn Pn 0 . a = 0 = a -l f`e xg` p lkl = 0 -e = 1 , = = 1 xicbdl xyt` ixdy a + a = a m"m` p k i j i j k p=1 p p p=1 p p 2n sqepa .r"gg f ik B \ B = ; ,ok enk .i = 1; 2 xear ziyteg B okl 0 0 .jB1 j + jB2 j = jB1 j + jB2 j > 1 2 i .yexck ,a1 ; : : : ; an mixtqn
3
4 dl`y Sh Ai ixai` lky rxe`nd zeidl Ei xicbp .z"ae ixwn ote`a legke mec`a Y ixai` z` ravp .Y = i=1 (Ai Bi ) idz .x Ai Bj ik k"da gipp .`ly dlilya gipp .mixf Ej -e Ei zerxe`nd .legka mireav Bi ixai` lke mec`a mireav
2
\
[
:milawne Pr(Ei ) = 2
h2
k
=
k
ik xexa .Ej llba legka reav
h X
k
2
=
h X
h [
Pr(Ei ) = Pr
ipy cvne
!
Ei
i=1
i=1
i=1
x
Ei
llba mec`a reav
x
cg` cvn if`
1 =) h 2k 5 dl`y
ozipy xexa .X ly mixehweed od dly zecenrdy ly dnxepd hxta ilnxepezxe` qiqa zeedn mb
n
AB
m xcqn AX = (aij ) dvixhn xicbp X lkl .m = jX j onqp ly zexeyd f`e
B
ilnxepezxe` qiqal
X
dveawd z` milydl
.1 lr dler `l AX ly zexeyd ly dlnxepd ,zecenrdn wlg zwign i"r AB -n zlawzn AX -e li`ed .1 `id dxey lk Pk 2 2 zlgezd zxcbdn . 2 miiw 1 j m dcenr lkl .cig`e ixwn ote`a mly 1 s k xgap i=1 aij k :milawn .1 j m lkl
E [a2sj ]
1
2 3 m m X X E4 a2sj 5 = E a2sj j =1 j =1
2
m k
.yexck ,m
Ar
:onwlck iaihwecpi` ote`a zxcben
A1 =
1 1
xy`k
1
1
Br = 2
r=2
Ar
Ar 1 Ar = Ar 1
ly zecenrd zveaw zeidl
Ar 1 Ar 1
Xr
k2
z` xicbp
:o`kn
r
lkl
T r T al miyp .Ar Ar = 2 I2r xnelk ,Br Br =
r
I2r ik ze`xdl witqn miilnxepezxe` Xr -a mixehweedy ze`xdl zpn lr 2 1 1 T r 2 0 = [ 0 2 ] :r = 1 xear .A2 1-l gipp .yexck , 1 1 r = 2 I2r ik r lr divwecpi`a d`xp ,okl .r lkl Ar = Ar ik
:miwelaa letkp .r -l gikepe
Ar 1 Ar 1
Ar 1 Ar 1
.yexck ,
2
=
2Ar 1 0
0 2Ar 1
=
r 2 I2r 1
2r I2r 1
r
= 2 I2r
Pk 2r 2 2 r r :k lkly o`kn bij = 2 :i; j lkl .Br = (bij )i;j =1 onqp .yexck i=1 bij = k2
2
Probabilistic Methods in Combinatorics: Homework Assignment Number 2 Noga Alon
Solutions will be collected in class on Wednesday, April 22, 2009. 1. (i) Prove that there is a positive constant c so that for all n > 1 and every n-uniform hypergraph H with at most cn1/4 2n edges there is an ordering of the vertices of H such that there are no two edges A and B that intersect in a unique element, and all members of A − B precede all those of B − A, while the unique element in A ∩ B appears after all those of A − B and before all those of B − A. (ii) Apply (i) to conclude that for c, n and H as above, H is two-colorable. 2. (Every monotone property has a threshold). Let F be a family of graphs on n labeled vertices, and suppose F is monotone, that is, if F ∈ F and G contains all edges of F , then G ∈ F. Suppose ǫ > 0, 0 < p < 1 and suppose that the probability that the random graph G(n, p) ⌉}, the probability that G(n, q) ∈ F is at belongs to F is ǫ. Show that for q = min{1, p⌈ ln(1/ǫ) ǫ least 1 − ǫ. 3. Let v1 = (x1 , y1 ), . . . , vn = (xn , yn ) be n two dimensional vectors, where each xi and each yi is 2n/2 √ . Show that there are two disjoint nonempty sets a positive integer that does not exceed 100 n I, J ⊂ {1, 2, . . . , n} such that X X vi = vj . i∈I
j∈J
4. Call a family F of subsets of [n] = {1, 2, . . . , n} distinguishing if for every two distinct subsets A and B of [n] there is an F ∈ F so that |A ∩ F | = 6 |B ∩ F |. (i) Show that there exists such an F of size |F| ≤ (2 + o(1)) logn n . 3
(ii) Show that any such F is of size at least (2 − o(1)) logn n . 2
5. Show that there is a positive constant c such that the following holds. For any n vectors P a1 , a2 , . . . , an ∈ R2 satisfying ni=1 ||ai ||2 = 1 and ||ai || ≤ 1/10, where || · || denotes the usual Euclidean norm, if (ǫ1 , . . . , ǫn ) is a {−1, 1}-random vector obtained by choosing each ǫi randomly and independently with uniform distribution to be either −1 or 1, then P rob(||
n X
ǫi ai || ≤ 1/3) ≥ c.
i=1
6. The Haj´ os number of a graph G is the maximum number k such that there are k vertices in G with a path between each pair so that all the k2 paths are internally pairwise vertex disjoint (and no vertex is an internal vertex of a path and an endpoint of another). Is there a graph whose chromatic number exceeds twice its Haj´os number ?
2 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 lixt`a 22
zeillk zexrd
jXi j 6 i
.Pr [
.
;
(
= 1 2)]
>
3 :miiw 2 2
1
t t t 2t s 6 bt=2c miiw s lkl ok enk . bt=2c 6 pt+1
,
2 zepeye 0 zlgez ilra X1 ; X2 n"n bef xear :1 Berge htyn m
t > 3 xeare 22pm 6 2
m 22m m 6 p2m
2
miiw :ifkxn inepia mcwn zkxrd
` 1 dl`y
A \ B = fxg-y jk B -e A zezyw bef lkl .miznvd ly ixwn xeciq idz :ik xexa ."B A-a miznvd lkl mcewy x-l ( itl) mincew A B -a miznvd lk" :rxe`nl XA;B
xehwicpi` dpzyn zeidl
XA;B = 1] =
Pr [
n (2n
[(
z` xicbp
2
1)!]
1)!
=
2
n
1
k-n
nn [(
1
2
1)!]
(2
2)!
=
2
n
1
nn 2
1
2
1
6 n
1
2(
1
1)
p n 2
1 2
n
2 2
n
3 2
6 pn 2
P
H -a zezyw k := cn1=4 2n -n xzei `l yi m` .X = XA;B xicbp k 23p 2n < c2 23 = 1 :zlgezd zeix`piln if` 3=2 xeciq xnelk ,X < 1 exear ,xeciq miiw okle c = 2 xear ,E [X ] 6 2 2 n ,
A; B
dl`k (mixecq) zebef 2
2
xzei `l okle
.yexck
a 1 dl`y
xeciqd itl zyw idyefi`a oey`xd znevd `ed v m` legka ravi v znev .l"pk xeciq idz A ly oexg`d xai`d hxta .legka dreav A ik xexa .zipeb-cg A gipp .mec`a .dxizqa ,XA;B = 1 f` la` .B ly
eravi miznvd x`y .
oey`xd xai`d `edy oeeikn legka reav
2 dl`y
:=
dyrpy ieqipd lr xefgp . miieqipd
l
ln(1
=)
lky zexazqdd .minrt
m xear
(F -l
q
=
p ik gipp okl
.1 zexaqzda
jiiy sxbd m`d dwicae
p
G(n; q) = Kn 2 F
zexazqda zezyw dxiga
q = 1 m` l"x) G(n; p)-a f`
p divwpetd ik al miyp .(1 6 exp ( ) 6 exp ln(1=) = lr dler dpi` elyki 0 zexazqdd xnelk ,1 (1 p) 6 p = q ik milawn .f (0) = 0 ,ok enk .[0; 1] rhwa dler dpi` f okle f 6 0 zniiwn dphw dpi`y zexazqda F -l jiiy epx`izy ieqipdn lawznd sxbd .q lr dler dpi` miieqipdn cg`a xgaiz zywy .1 -n dphw dpi`y zexazqda F -l jiiy ,dphw `l zyw zxigal zexazqdd eay ,G(n; q) okle ,1 -n
f (p) = 1
(1
p)
)
3 dl`y xexa .
Berge
V
=
itl
X = P v ok enk ;Y = P y -e X = P x onqp .z"ae cig` ote`a ; : : : ; 2 f0; 1g xgap 1 n i i i i i i i i i Y P 2 P P 1 1 2n 1 2n :dnec ote`a .V ar [X ] = .V ar [Y ] 6 i xi 6 41002 :oke E [Y ] = 2 i yi ,E [X ] = 2p i xi -y 41002 4 :( = 100 2 xear) h Pr
jX E X j; jY E Y j 6 [
]
[
]
2
(
n
=
1) 2
i
>
1
4
3 10
4
Berge, P.O (1937), A note on a form of Tchebyche's theorem for two variables. Biometrika 29:405-4061
1
2 1
1 1 4103 4 -y o`kn lawzny jxr miiw okle ,mikxr 2 -n zegt milawn minekq 2 4 0 > 1-n xzei i"r P P 0 0 0 0 0 xexa .J = J K -e I = I K ,K = I \ J xicbp . 2 0 v = 2 0 v -y jk I 6= J [n] zeniiw xnelk .minekq 3 4 10
n
n
i
j
0 did ynn miiaeig mixehwe ly mekq zxg`] I; J 6= ; ,I \ J = ; ik X X X X X X v = v v = v v = v i
I
j
J
:oke [
2
i
i
i
20
I
i
2
I
i
i
K
j
20
j
J
j
2
j
K
j
2
j
J
(zeillk zexrd) 4 dl`y
A (A)-a onqp .A ly A zecenrdn zakxeny dvixhnd z` A
log2 liaya lg onqp
.
A -a
onqp
A [n] dveawe k n xcqn A dvixhn xear A ly zecenrd mekq z` (A) heyt e`)
:zelewy ze`ad zeprhd ik orhp .
A
jF \ Aj =6 jF \ B j zniiwnd F 2 F zniiw ,A 6= B [n] lkly jk k lceba F 2[ ] zniiw .(A) = 6 (B ) ,A =6 B [n] lkly jk 0=1 ly k n xcqn A dvixhn zniiw (*) (A) = ik ze`xl lw .A = 6 B [n] dpiidz .jtidle ,j 2 F m"m` A = 1 xicbp F = fF1 ; : : : ; F g xear ,ok`e .ipyd geqipd xear dl`yd z` ,ok m` ,xeztp .jF \ Aj = jF \ B j ,i lkl m"m` (A) = (B ) okle .jF \ Aj n
.
i
i
ij
i
k
i
i
` 4 dl`y
k) k n xcqn (1=2 zexazqdae z"a ote`a 0 zeidl xgap da xai` lk xnelk) zixwn 0=1 zvixhn A idz X P = 1 m`y xexa ."(A) = (B )" :rxe`nd ly X xehwicpi` dpzyn xicbp A; B [n] bef lkl .(jynda 0 0 .E [X ] := \ =; E [X ] lr dler dpi` miiwzn `l (*)-y zexazqddy o`kn .X 0 0 = 1-y jk zexf A ; B mb zeni rawii
-iw f`
A;B
A
A;B
A ;B
A;B
B
2
E [X
A;B
fp;qg X
min
] = 42
p
q
q `
p `
`=0
:ik xexa
3k 5
q = jB j-e p = jAj xy`k
:okle .
E [X ] =
2
n! 42 p ! q !( n p q)! 16 + 6 X
p
q
fp;qg X
min p
q
`=0
n
q `
p `
3k
X
=
5
n! 2 p!q!(n p q)!
p
q
:lawpe
n!
k
1
k+1
p+q p
k
t = p + q onqp
t n 2 2 st = s t ( n t )! s !( t s )! =0 =0 =1 =1 :f`e . = (n) = 2 + 1= lg lg n = 2 + o(1) xy`k k = lg3
X n
E [X ] =
X t
X n
t
s
t
X t
tk
s
t
n
2
4n 2k
A(n) 6
n
lg n 2
n lg2 n n
:ixdy
n
2 lg2 n 2
lgnn
6 n 2e lg2 n
n2
lg n
2
E [X ] 6
P lgn2 n
A(n) =
lgnn
t
s
t=3
6
t
2
4n 2k
= o(1)
2 (t + 1) ( 1) 2 = o(1) 2 + n lg2 n lg(2 lg ) = o(1)
b 2c t=
= o(1) ixdy
xgap
n
t
k
=
t
lgnn +lg n
=2
:o`kn
lgnn +2 lg(2n)
=2
n
e
n
t :miiw t > 3 lkl ik reci 6 p2+1 t
4n2 + X n 2 (t + 1) 2 + 6 4n2 + X n 2 6 k+1 2 t 2 t 2 (t + 1) (t + 1) k 2 1 =3 =3 n
tk
t
t
tk
k
k
t
2 6 42n +
n 2n lg X n
k
= 2
n
t=3
k 1 (2 lg lg n
2
lg2 n 2 + t n t
lg n)+n lg 3
t
k 1
2
t=
+ o(1) 6 2 .yexck
n X
k
n
lg2 n
lg lg n+
lg2 n n 2 6 n t t
21 lg n
( 2 1)
n lg
3
k21
=2
3 + o(1) = n
n lg 3 lg lg n +o( lg nlg n )
= o(1)
A dvixhn zniiw ziaeig zexazqda (lecb witqn n xear) okle 2
a 4 dl`y .z"ae cig` ote`a
i
i
1 2
okl . -e
i
Pr xen`dn ."si
2 f0; 1g xgap
1 ; : : : ; n
:= E [v ] = 12 s
si
= ([n])
vi
n xcqn zizexixy dvixhn A idz = (A)-e A = fj j = 1g xicbp
k
onqpe ,yexck
ik xexa .v
j
Cherno itl ,o`kn . 2 := V ar[v ] = 41 s -e
:
h vi
.s
mixhnxt mr zinepia bltzn
p
si
2 >
> pn log n mbe v
si
i
log s 6 2 exp
2s log s i
i
2 >
si
i
i
6 2 exp ( 2 log s ) = 2s i
si
ps log s " (*) :rxe`nl xehwicpi` dpzyn zeidl X i
i
i
i
i
i
i
2
1 6 i 6 k lkl
xicbp
:lirl
2 (n log n) 1 = 2k(n log n) 1 6 log2 n = o(1) `l yi zeveawd xzi xear .(*) i`pzd z` miiwnd aikx likn (A) oxear A [n] zeveaw o(2 )-n xzei `l yi okl p p :o`kn ,2 6 (2 n log n) + o(2 ) :dl`yd ii`pz z` zniiwn A-e li`ed .aikx lkl mipey mikxr 2 n log n-n xzei n(1 o(1)) k> > (2 o(1)) lg n(1 n+ o(1)) = (2 o(1)) lgnn lg +lg lg 1+ 2 [ ] := E
EX
hX
Xi
i
=
X
X
[ ]6
E Xi
n
k
n
n
n
n
5 dl`y X
:mihpnen xtqn aygp .
E X2
[ ] =
2 3 X E4 xi xj i j 5
=
i;j
E X
4
X
=
X
[
xi xj E i j
]=
X
i;j
[
xi1 xi2 xi3 xi4 E i1 i2 i3 i4
]=
[
X
] =
x4i
i
:z` aygp .E
[X 2 + Y 2 ] = 1 hxtae ,Y [
xi1 xi2 yi3 yi4 E i1 i2 i3 i4
]=
i1 ;i2 ;i3 ;i4
X
=
Pn
=1 a = i
i
Pn
=1
i
i
i
xi yi
onqp
x2i
i
X
i1 ;i2 ;i3 ;i4
E X 2Y 2
Y
+ 42
X
x2i x2j
i
ly mihpnend z` milawn dnec ote`a
x2i yi2
+4
X
xi yi xj yj
i
i;j
:ik al miyp
E2 X 2
+Y2 =
X
x2i
+
X
yi2
2
=
X
x4i
+2
X
x2i x2j
+2
X
x2i yj2
yi2 yj2
+
X
yi4
i
i;j
i
+2
X
:okle
V ar X 2
+Y2
h
i 2 2
i
i
j
i
i
Cauchy
Bunyakovsky
= E X2 + Y E2 X 2 + Y 2 = X X x2 x2 + 2x y x y + y 2 y 2 = 4 = 4 ha ; a i2 6
6 4
X
ka k2 k a k2 6 i
j
j
j
i
i
j
j
i
2 2 X ka k2 6 2 2 10 10 i
i
i
.)-: xzei e` zegt dfd libxzd lr cibdl il yiy dn lk dfe
6 dl`y sxb-zz
H
= G[S ]-e dl`yak micewcw ly klceba dveaw S idz .G ly Haj o s xtqn z` k-e miznv v lr sxb G idi 2 1 zegtl H -a okle ,e 6 1 2 ,T uran htyn itl if` K +1 * G m` .zezyw e lra S lr yxtpd 2 k(k 1) 1 k k k 1 m 2 =2 m 1 2 k
m
m
3
S
:o`kn . -a `ly miznvd cg`a zegtl ynzyny lelqn i"r mixaegn dxqg zyw ly zeevwd ipy .zexqg zezyw
v>
k 2
k
m
+1
>
k2 2m
Km+1 likn `ly miznv 2m=2plr sxb miiw m lkl xnelk - r(m + 1; m + 1) > 2m=2 ,m > 2 lkl ,Erdos htyn itl > 2m=2 2m ! 1 :milawn .(G) > 2m=2 :ok enk .k2 < 2m2m=2 :l"pd q"r .K ,m = 8 xear hxta , m+1 likn `le k m+1 (m+1)2m=4 . > 3k
4
Probabilistic Methods in Combinatorics: Homework Assignment Number 3 Noga Alon
Solutions will be collected in class on Wednesday, May 6, 2009. 1. Let G = (V, E) be a graph with maximum degree d and let V = V1 ∪ V2 . . . ∪ Vs be a partition of V into s pairwise disjoint sets. Suppose, further, that |Vi | ≥ 2ed for all i. Prove that there is an independent set of G containing precisely one vertex from each Vi . 2. Let G = (V, E) be a simple graph and suppose each v ∈ V is associated with a set S(v) of colors of size at least 10d, where d ≥ 1. Suppose, in addition, that for each v ∈ V and c ∈ S(v) there are at most d neighbors u of v such that c lies in S(u). Prove that there is a proper coloring of G assigning to each vertex v a color from its class S(v). 3. A simple path of an even length P = v1 v2 · · · v2k in a graph G = (V, E) with a vertex coloring f : V 7→ [r] = {1, 2, . . . , r} is periodic if f (vj ) = f (vk+j ) for all j, 1 ≤ j ≤ k. Prove that there is a finite r so that every graph G with maximum degree 3 admits a vertex coloring with r colors in which no simple path (of any even length) is periodic. 4. Prove that there is a positive constant c so that every d-regular graph, where d ≥ 2, contains a spanning subgraph in which every connected component is a star with at least c logd d leaves. 5. Show that the probability that in the random graph G(2k, 1/2) the maximum degree is at most k − 1 is at least 1/4k . 6. Let G be a graph and let P denote the probability that a random subgraph of G obtained by picking each edge of G with probability 1/2, independently, is connected (and spanning). Let Q denote the probability that in a random 2-coloring of G, where each edge is chosen, randomly and independently, to be either red or blue, the red graph and the blue graph are both connected (and spanning). Is Q ≤ P 2 ? (Prove, or supply a counter-example).
3 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 i`na 6
1 dl`y
10 [ [ V 0 lr yxtpd sxb xear gikepe k lcebn V 0 V xgap zxg` ,jV j = k ea sxb xear gikepe k = d2ede onqp .(v1 ; : : : ; v ) 2 V1 V miznv zxcq cig`e ixwn ote`a xgap .[ixewnd sxba mb z"a ,ycgd sxba z"a dveaw]
V
s
ea ,sxb
i
i
i
H = (A E ) idi ." 2 ( [ ])" :rxe`nd zeidl xicbp 2 zyw xear . = f j 1 6 6 g onqp 0 g 2 E -e A = f g H ik xexa . dveaw dze`a cg` dvw zegtl 0 -e -l m"m` f 2 2 .zraep dprhd zilweld dnldn . 2 6 1 :o`kne .Pr[ ] 6 ik ze`xl lw .(H) 2 ( ) 6 2 s
s
;
f
E G U
miiwe zeielz sxb `ed
Af
f
E
Vi
U
f
ek
kd
Af
f
vi
i
s
Af
Af ; Af
k
<
k
G
kd
2 dl`y
10 lcebn 0( ) ( ) xgap zxg` ,j ( )j = 10 xear gikep :rxe`nd zeidl xicbp 2 ( ) \ ( ) rave ( ) = 2 zyw xear . ( ) 2 ( ) rav z"ae cig` ixwn 0 0 j 2 0 0 2 ( )g onqp . -e ly mrava wx ielz ( ) ik xexa ." ( ) = ( ) = " f`e , ( ) = f ( ) :miiw . ( ) [ ( )-a wx ielz
ote`a v znev lkl xgap .el` zeveaw xear gikepe Af;c
A u;v ;c
c
N w
S u
w
Af ;c
d
S v
f ;c
S
u; v
S w
v
f
v
S v
S v
E
c v
u
d
S v
A u;v ;c
c u
N u
e
Pr [
Af;c
c v
c
N v
] (j ( ) [ ( )j + 1) 6 (10 ) 2 (2 10 + 1) 3 1 N u
N v
e
d
d
d
<
e
<
.zilweld dnldn zraep dprhde
3 dl`y
= 18 16 miravdn cg`a z"ae ixwn ote`a znev lk ravp .zeielz sxb H ik xexa . \ 6= ; m"m` f g 2 E -e A = f g ea ,sxb H = (A E ) idi ."[dxgapy driavd itl] S ( ) mb onqp ;H-a ly oky y jk 2 jxe`n milelqnd lk z` ( )-a onqp 2 lra lelqn . ( ) = . ( ) 6 4 9 =: miiw j j = 2 lkl okle .miznv 2 ilra milelqn 4 32 -n xzei `l mr znev wleg miznv :oeieey-i`l al miyp .1 > exp ( 2 ) :miiw . = xicbp 2 jxe`a lelqn xear f`e = 18 xicbp lkl P Y Y Y Y Y (1 ) = 2 = (1 )j ( )j > (1 ) = (1 ) = 1
ixefgn P " :rxe`nd zeidl AP xicbp P lelqn lkl . ; : : : ; r P
i
N P
Nj P
ij
j
Q
j
Nj P
Q
2 ( )
xQ
j
N P
= 18
i
e
Q
8
xP
N
Q
18
aj
ai
P
ij
9 > j
h
18 8 e
P
j
i
ai
Nj P
ij
aj
ai
i
aj
ai e
ij
j
j
P
j
i
Nj P
j
ij
ai
xQ
2 j( ) P
j
j
ai
xP
;
AQ
i
ai
xP
AP
AP
P
ij
e
AP ; AQ
2
j
i
i
> 18
16
i
e
=
r
i
= Pr[ ] AP
.zraep dprhd (illkd gqepd) zilweld dnldn okle
4 dl`y ote`a da xag zeidl xgap znev lky jk miznv ly zixwn dveaw (ixlebx d-d sxbd `ed G xy`k) V e` Xv
= 0" :rxe`nd
Av
0 -e V -a v
ly mipkyd xtqn Xv eidi v
2
V
znev lkl .p
( ) \ ( ) = ; xy`k z"a
mr v xear miiwzn `l oexg`d oeieeyde li`ed .N u
N v
= (2 + 2ln )
d =d
Av -e Au
(
Pr [ ] 6 (1 ) + Pr( 3 ) 6 (1 ) + ( 2 27)2+2 ln 6 + .0 6 6 + 6 ln =: -y jk 0 dveaw zniiw zilweld dnld itl okle p
< Xv
d
X >
d
pd
k
p
V
1
d
e =
d
e
pd
zexazqdae z"a
3 1) 2 xzeid lkl 1 2 Pr[ ] 2 1 :miiw
zerxe`nd ik xexa ."Xv >
.u miznv d d Av
0 V [G] idz
e
pd
< d
<
e
=ed
Av d
<
pd
v
m"m`
f(
g2
v; i); u
=
E -e X
f(
v; i)
j 2 v
xzeid lkl `id Y -a znev lk ly dbxcde d milrd zveaw z` xicbp ,v
2
V
0
0
V ; 0 < i < d=k
g,
Y
=
V
lkl .X z` deexny H -a f : X
!
Y
ik xexa .S
S
.onwck H = (X
[
Y; E)
v"ec sxb xicbp
zegtd lkl `id X -a znev lk ly dbxcd ik al miyp .G-a u ly oky
k
j vj =
0
V
d k
d
1=
1
6 + 6 ln d
beeif miiw ,H all q"r okl .
d k
1
k
=
d
k
v = ff (v; i) j 0 < i < d=kg :ezail v -y akeka
> 7 ln d
d
.izexixy ote`a miniiw miakekl xagp Y -a mixzepd miznvd z` .H all q"r mikzgp mpi` miakekd
5 dl`y ik xexa .d
i6k
1 dpekzd zeidl
Mi ] = 21
Pr[
.Pr
hT
2k i=1
i
k Pr [M ] > 2 Mi > Q2i=1 i
2k = 4
Mi xicbp ,1 6 6 2 i
2k
kX1 2k i=0
k
1 i
=
k
znev lkl .1; : : : ; 2k :miznvd z` onqp
1 2
:K leitman itl .(zecxei) zeipehepen
Mi zepekzdy al miyp 6 dl`y
(V [G]; N mb ik xexa
E)
m"m` E
2 C -e (yxete) xiyw ( [ ] ) m"m` 2 A-y jk A C 2N dpiidz . = [ ] idz .ok jAj = 2jN j -y oezp .zcxei zipehepen C -e dler zipehepen A ik xexa .(yxete) xiyw jN j . 6 2 okle 22jN j = 2jN j jA \ Cj 6 jAjjCj = 22jN j 2 : itl .jCj = 2 V G ;E
jN j Q-e .jA \ Cj = 2
E
;
N
E G
P
Q
P
Q
P
2
K leitman
P
Probabilistic Methods in Combinatorics: Homework Assignment Number 4 Noga Alon
Solutions will be collected in class on Wednesday, June 3, 2009. 1. Let G1 , G2 , . . . , Gm be m graphs on the same set of vertices [n] = {1, 2, . . . , n}, and suppose that the chromatic number of each graph Gi is exactly k. Show that there is a partition of the set of vertices [n] into two disjoint sets A1 , A2 so that for every i, 1 ≤ i ≤ m, and for every j, 1 ≤ j ≤ 2, √ p the chromatic number of the induced subgraph of Gi on Aj is at least k/2 − 2 ln(2m) k. Hint: use an appropriate martingale to show that this holds with positive probability for a random partition. 2. Prove that there exists a positive constant δ > 0 and an integer n0 = n0 (δ) so that for all n > n0 and every collection S1 , S2 , . . . , Sm , where m ≤ 2δn , of subsets of [2n] = {1, 2, . . . , 2n}, satisfying |Si | = n for all i, there is a function f : [2n] 7→ [n] = {1, 2, . . . , n} so that for every i, 1 ≤ i ≤ m, 0.63n ≤ |f (Si )| ≤ 0.64n. Hint: 0.63 < 1 −
1 e
< 0.64.
3. Show that for any ǫ > 0 there is a C = C(ǫ) such that every set S of at least ǫ3n vectors in Z3n contains three vectors so that the Hamming distance between any pair of them is at least √ n − C n. Hint: use an appropriate martingale to show that more than 2/3 of the vectors are within √ distance C n/2 of S. 4. Using Janson’s Inequality, find a threshold function for the property: G(n, p) contains at least n/10 pairwise vertex disjoint copies of K5 . 5. (i) Is it true that for every ǫ > 0 there is a finite constant C = C(ǫ) such that every set X of n points in the plane contains a subset Y of size at most C with the following property: any convex set K which is the intersection of 10 half-planes and contains at least ǫn points of X, contains at least one point of Y ? Prove, or give a counterexample. (ii) Is it true that for every ǫ > 0 there is a finite constant C = C(ǫ) such that every set X of n points in the plane contains a subset Y of size at most C with the following property: any convex set K which contains at least ǫn points of X, contains at least one point of Y ? Prove, or give a counterexample.
4 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 i`na 22
1 dl`y S
`id
mb ik xexa .(dcig` zebltzdae z"a ote`a da xag zeidl xgap xai` lky) zixwn dveaw-zz miiw
E [(Gi [S ])] = E [(Gi [S ])]
(Gi [S ]) + (Gi [S ]) > (Gi ) = k-e
oke
S
[n] idz
li`ed .zixwn dveaw-zz
= E [(Gi [S ])] > k=2 xear
fv j ci (v) 6 j g onqp .fgg zeivwpet zveawe Gi [S ] mitxbd oia zedfl ozip .Gi zriav ci : [n] ! [k] idz :Azuma q"r okle Lipschitz i`pz z` zniiwn L ik ze`xl lw .L = -e 0 6 j 6 k h h p p pi pi ln(2m) 2 ln(2m) k 6 Pr (Gi [S ]) < 2 ln(2m) k < e = 1= 2 m Pr (Gi [S ]) < k=2
Bj
=
6 i 6 m xear Gi [S ]; Gi [S ] mitxbdn cg` zegtly zexazqdd o`kn
okle ,1-n (ynn) dphw yexcd z` miiwi `l 1
.A2 =
S -e A1
=
S
xicbdl xzep .yexck
S
zniiw
2 dl`y
4 6 0:64-y jk xgap 0 6 j 6 2n xear Bj = [j ] onqp .Si -a opeazp .zeivwpet n jezn ixwn ote`a f : [2n] ! [n] divwpet xgap 1 n okle Lipschitz i`pz z` zniiwn Li ik ze`xl lw .i = E [jf (Si )j] = n 1 1 ik xexa .Li (f ) = jf (Si )j-e n .jynda xgai
n0 .e
2
= 2 -y jk
xicbpe 0:63
61
1=e
2n
h Pr oekp ipyd oeieey-i`d)
0:63
jLi (f )
61
1=e
i j 6
jLi (f )
i j >
6
p
p
2 4n ln(4e n )
4 = lim
i n
p
p
4
.n0 = maxi
2 ln(4m)
2n
2 4n ln(e4 n )
i
:Azuma q"r
< 1=m
6 4 n :1 6 i 6 m lkly jk f zniiwy o`kn zrk .(milecb witqn mi-n xear
6 lim Lin(f ) 6 lim ni + 4 = 1
1=e + 4
6 0:64
fn0i g xicbp .0:63 6 Li (f ) 6 0:64 ,n > n0i lkly jk n0i miiw xnelk 3 dl`y
S -a
xehweel
g -n
:mipeieey-i`d
Hamming wgxn `ed L(g ) ,Zn3 -a xehwe zbviin g miniiw Azuma q"r okle Lipschitz i`pz z` zniiwn L ik
:
ixrfn
Pr
p
L(g ) > + n < e
2 =2
Pr
L(g ) <
B
! A ,B = [n] ,A = [3] xicbp 6 i 6 n xear Bi = [i]-e
ze`xl lw .0
p
n
2 =2
:ipnid oeieey-i`dn zxg`
h
= Pr [L(g ) = 0] 6 Pr L(g ) < C
=2
p
2 ln +
p
2 ln 3
p
p
p
p
2 ln
i
6
p
pn ik orhp
2 ln
n <
pni < 1=3 :il`nyd oeieey-i`dn f`e .dxizqa pnC=2 mixehweedn 2=3 ik lawpe 0 .S -a mixehwe ly ef dveaw onqp .S -l miaexw h
onqp .Pr
L(g ) >
2 ln +
1
2 ln 3
wegx
n-y
mixehweed lr xearp exqei okl .dl`k
z0
2 S -e y-l y0
.n
C
pnC=2-y dyly zniiw ik orhp .S -l daexw
2n yi Zn3 -a mixehweed 3n -n cg` lkl n n .dl`k zeyly 3 2 yiy o`kn .mdn cg` lkn
(weica) cg` xehwe yi dfk bef lkle mdn miwegx-n-y mixehwe
3 2n -n xzei `l yi u lkly xexa .u z` zelikny zeyly oze` lk z` xiqp ,u xehwe lkle Zn3 S 0 -a 2 S ,x-l xzeia aexw x0 2 S idi .(x; y; z ) zg` dyly zegtl x`yize ,3n 1 3 2n -n (ynn) zegt :(y -e x oia Hamming wgxn `ed (x; y )) miiw .z -l p n = (x; y ) 6 (x; x0 ) + (x0 ; y 0 ) + (y 0 ; y ) 6 C n + (x0 ; y 0 )
pn-n ohw `l z 0 -e y0 oiae z 0 -e x0 oia wgxnd dnec ote`a .n
C
pn-n ohw `l y0 -e x0 oia wgxnd ik eplaiw .miyexcd mixehweed md
x0 ; y 0 ; z 0
4 dl`y .dievxd dpekzd xear sq ziivwpet `id
5
:okl `id mi-K -d xtqn zlgez
( )=n
t n
2=5 ik orhp
(5) `id G[S ] ' K5 zniiwn 5 lceba S dveawy zexazqdd f` p n
p 2
n
5
o(n) n=10
2=5 m`
(5) 6 n5p10 = o(n)
p 2
= o(1) lr dler dpi` mi-K5 n=10 zegtl yiy zexazqdd ,M arkov q"r o`kn 2=5 ln n ik (dler zipehepen dpekzd ixdy) k"da gipp f` p n 2=5 m` .n=2 lceba miznv zveawa opeazp .p 6 n f`e "G[S ] ' K5 " :rxe`nd zeidl BS z` xicbp ef dveawa miznv dying za S dveaw lkl X n=2 10 n=2 (5) 2 = Pr[BS ] = 5 p = 5 p S V;jS j=5 4 X n=2 X 5 (n=2 i) p2(52) (2) Pr [BS ^ BS0 ] = 5 = i S S 0 i=2 V :lr dler epi` K5 z` zelikn opi`y n=2 lceba zeveawd xtqn okl .Pr B S < e +=2 :J anson q"r .yexck ,
i
n n=
2
e +=2 < :okle
lim (n
+ =2) = lim
en n=
2
n6 p14
n
n=2
=O
e +=2 < en +=2
n n6 ln n 5:6
= o (n) ik al miyp .zniiw dti`yd ik d`xp Cn5 p10 + C 0 n6 p14 6 lim C 00 n Cn5 p10 14
p>n
:mialwne
lim (n dveaw
V1
+ =2) 6 lim
C 00 n
!0
Cn5
1 + C 00 n C
4
2=5 1+C 00 1=10 hxtae p n 2=5 la`
C
= lim((C 00 1
) )= 1
C 00 n
V (G) iaihwecpi` ote`a zeveaw ly dxcq xicbp .K5 z` dlikn n=2 lceba dveaw lk hrnk ,xnelk G[S2 ] ' K5 -y jk S2 V2 -e V2 V (G) V1 .G[S1 ] ' K5 -y jk S1 V1 -e n=2 lceba idylk .yexck ,k > n=10 ,jSi j = 5-e li`ed .jV (G) S1 [ [ Sk j > n=2
cer lk d`ld jke
` 5 dl`y mixeyin-ivgd lk ly
V C -d
cnin ik dcaera aygzdae ,Spencer -e
Alon
ly mxtq jezn 14.4.3 dpwqn itl .oekp
10 ly jezig opidy (zexenwd) zeveawd lk ly V C -d cnin ik lawp ,3 `ed xeyina zecewp lrn ( ) divwpet zniiw ,14.4.5 htynn okl .iteq hxtae ,2 3 10 log(3 10) lr dler epi` xeyina zecewp
lrn mixeyin-ivg
X
lky jk
C
jY j 6 C mr Y
.
zyx- dlikn
a 5 dl`y ipt lr zecewp
n
> 2C ly dveaw zeidl X z` xgap
.edylk
C
.X -n zecewp
jX
Y
j>n
C
= C ()-e = 1=2 idi .oky dlilya gipp .oekp `l
idi .jY j 6 C -y calae idylk Y X idz .lbrn > n=2 = n likne Y -n dcewp s` likn `l K ,ok enk .xenw K ik xexa .X Y -n
zecewpd md eicewcwy (eil` zkiiy dneqgd d`tdy jk) rlevn
2
K