Probabilistic Methods in Combinatorics

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1 Probabilistic Methods i Combiatorics Po-She Loh Jue 2009 Warm-up 2 Olympiad problems that ca probably be solved. (Leigrad Math Olympiad 987, Grade 0 elimiatio roud Let A,..., A s be subsets of {,..., M}, ad suppose that oe of the A i are subsets of each other. For each idex i, let a i = A i. Prove that s i= ( M a i. Remark. This is also kow as the Lubell-Yamamoto-Meshalki iequality. Solutio: Geerate a radom permutatio σ = (x,..., x M of [M]. Let each E i be the evet that the set A i appears as the iitial segmet of σ, i.e., that {x,..., x ai } = A i. Sice oe of the A i cotai each other, the E i are mutually exclusive. But each P [E i ] = ( M, a i so the LHS is P [ Ei ], ad probabilities are always. 2. (Very slight simplificatio of a problem i the MathLiks.ro 2008 cotest Let A,..., A ad B,..., B be distict fiite subsets of N such that: for every i, A i B i =, ad for every i j, (A i B j (A j B i. Prove that for every real umber 0 p, p Ai ( p Bi. Remark. This result is due to Tuza (985. i Solutio: Sample a radom subset of the uiverse by takig each elemet with probability p. Let E i be the evet that everythig i A i got picked while everythig i B i was missed. These are disjoit evets, so the LHS is the probability that some E i occurred, hece.

2 3. (IMO Shortlist 2006/C3 Let S be a fiite set of poits i the plae such that o three of them are o a lie. For each covex polygo P whose vertices are i S, let a(p be the umber of vertices of P, ad let b(p be the umber of poits of S which are outside P. Prove that for every real umber x: x a(p ( x b(p =, P where the sum is take over all covex polygos with vertices i S. Importat: a lie segmet, a poit, ad the empty set are cosidered to be covex polygos of 2,, ad 0 vertices, respectively. Solutio: Radomly color the poits i black ad white, with a poit receivig black with probability x. For each covex polygo P, let E P be the evet that all vertices o the perimeter of P are black, ad all vertices i P s exterior are white. These evets are mutually exclusive, so the LHS is the probability that some E p holds. But there is always some E P that always holds: cosider the covex hull of all of the black poits. 3 Liearity of expectatio Defiitio. Let X be a radom variable which takes values i some fiite set S. The the expected value of X, deoted E [X], is: E [X] = x S x P [X = x] Use the followig exercises to get used to the cocept of expected value.. What is the expected umber of heads i 0 tosses of a fair coi? Prove this formally from the defiitio. Solutio: 0 ( 0 0 i = 2 0 i 0 ( ( 00 = = = 0/2. i 20 i i j i=0 i= 2. What is the expected umber of heads i toss of a coi which lads heads with probability /0? 3. Ca you calculate the expected umber of heads i 0 tosses of a coi which lads heads with probability /0? Oe of the most useful facts about the expected value is called liearity of expectatio. Propositio. For ay two ot ecessarily idepedet radom variables X ad Y, we have E [X + Y ] = E [X] + E [Y ]. Proof. This is essetially a cosequece of the commutativity of additio. Followig the defiitio: E [X + Y ] = x,y (x + y P [X = x, Y = y] j=0 = x P [X = x, Y = y] + y P [X = x, Y = y] x,y x,y = ( x P [X = x, Y = y] + ( y P [X = x, Y = y] x y y x = x (xp [X = x] + y (yp [Y = y] = E [X] + E [Y ]. 2

3 Now use that fact to solve the followig problems.. Calculate the expected umber of heads i 0 tosses of a coi which lads heads with probability /0. Solutio: This is the sum of 0 radom variables, each of which has expectatio /0. 4 Olympiad problems oe ca expect to solve The key lemma is the followig apparetly trivial result. Lemma. Let X be a radom variable. The there is some poit i the probability space where X E [X], ad also some poit i the probability space where X E [X]. Try these problems.. (Ira Team Selectio Test 2008/6 Suppose 799 teams participate i a touramet i which every pair of teams plays agaist each other exactly oce. Prove that there exist two disjoit groups A ad B of 7 teams each such that every team from A defeated every team from B. Solutio: Sample A as a radom 7-set. Let X be the umber of guys that are totally domiated by A. Lettig d v deote the i-degree of v, we have E [X] = ( d ( v v 7 / But v d v = ( 799 2, which meas that the average i-degree is exactly 399. By covexity, E [X] 799 (399 ( 7 / (/ , which is eough sice X is a iteger. Pick 7 teams B from the domiated group. 2. (Russia 996/4 I the Duma there are 600 delegates, who have formed 6000 committees of 80 persos each. Prove that oe ca fid two committees havig at least four commo members. Solutio: Simply select a radom 2-set of committees. Calculate expected umber of people i both committees, ad use covexity of x 2 i (avg2. More ( carefully: Sample a pair of committees uiformly at radom (i.e., radomly pick oe of the possible pairs. Let X be the umber of people who are i both chose committees. Note that X = X + + X 600, where each X i is the {0, }-radom variable tellig whether the i-th perso was i both chose committees. By liearity of expectatio, E [X] = E [X ] + + E [X 600 ]. The magic is that each E [X i ] is easy to calculate! Let i be the umber of committees that the i-th perso belogs to. The, each E [X i ] = P [i-th perso is i both picked committees] = ( i ( 2 / The oly piece of iformatio we kow about the { i } is that their sum i i = , so this suggests that we use covexity to boud E [X] i terms of the average of { i }, which we deote by = ( /600 = 800: ( ( ( ( E [X] 600 / = 600 / = = (Oe could see that sice ad , the last fractio is roughly /400. But by the Lemma, we kow that some outcome of the probabilistic samplig produces a X Sice X is always a iteger, that outcome must i fact have X 4. I particular, we coclude that some pair of committees has 4 commo members. 3. (MOP Test 2008/7/2 Suppose that a, b, c are positive real umbers such that for every iteger, Prove that at least oe of a, b, c is a iteger. a + b = c. 3

4 Remark. You may use the followig well-kow umber-theoretic fact: if x is irratioal, the the fractioal parts of the multiples of x are equidistributed over the iterval [0, ]. I particular, if we choose uiformly amog {,..., N}, the E [{x}] /2 as N. Solutio: Suppose oe of a, b, c are itegers. Divide both sides by ad take the limit. This gives a + b = c, so we also kow that the sum of fractioal parts: {a} + {b} = {c}. ( If x is irratioal, the {x} is equidistributed over the iterval [0, ]. I particular, if we choose uiformly amog {,..., N}, the E [{x}] /2 as N. O the other had, if x is ratioal with reduced form p/q, the {x} has expectatio tedig to q 2q = 2 2q. So, it is i the iterval [/4, /2. Coclusio: for ay oiteger x, E [{x}] t, with t [/4, /2]. Takig the expectatio of equatio (??, ad takig limits, we immediately see that the oly way to have the equality is if E [{a}] ad E [{b}] both ted to /4, ad E [{c}] /2. But the oly way to get expecatio /4 is whe a, b are ratioal, ad the oly way to get the full expectatio /2 is whe c is irratioal. Yet our first deductio was that a + b = c, so we caot have two ratioals summig to oe irratioal. Cotradictio. 4. (MOP Test 2007/7/ I a array, each of the umbers, 2,..., appears exactly times. Show that there is a row or a colum i the array with at least distict umbers. Solutio: Choose a radom row or colum (2 choices. Let X be the umber of distict etries i it. Now X = I i, where each I i is the idicator variable of i appearig (possibly more tha oce i our radom row or colum. Clearly, each E [I i ] = P [I i ]. To lower-boud this, observe that the worst-case is if all appearaces of i are i some submatrix, which gives P [I i ] 2 /(2 = /. Hece by liearity, E [X]. 5. (IMO Shortlist 999/C4 Let A be ay set of residues mod 2. Show that there is a set B of residues mod 2 such that at least half of the residues mod 2 ca be writte as a + b with a A ad b B. Solutio: Make idepedet uiformly radom choices from the 2 residues, ad collect them ito a set B. Note that sice we use idepedece, this fial set may have size <. But if we still have A + B occupyig at least half of the residues, the this is okay (we could arbitrarily augmet B to have the full size. Let X be the umber of residues achievable as a + b. For each potetial residue i, there are exactly ways to choose some b for which A + b i, sice A =. Therefore, the probability that a give residue i appears i A + B is precisely ( 2. The E [X] is exactly 2 times that, because there are 2 total residues. Hece it suffices to show that ( 2 /2. But this follows from the boud e /, usig e (Austria-Polish Math Competitio 997/8 Let be a atural umber ad M a set with elemets. Fid the biggest iteger k such that there exists a k-elemet family of 3-elemet subsets of M, o two of which are disjoit. Solutio: We eed to be careful about case whe is small: for 5, we ca simply take everythig. Otherwise, it is Erdős-Ko-Rado. 5 Real problems oe ca expect to prove These are iterestig results from research mathematics (as opposed to Olympiad mathematics that have very elegat probabilistic proofs. 4

5 . (Erdős, 965 A set S is called sum-free if there is o triple of (ot ecessarily distict elemets x, y, z S satisfyig x + y = z. Prove that every set A of ozero itegers cotais a subset S A of size S > A /3 which is sum-free. Solutio: Choose a prime p of the form 3k + 2 such that p is greater tha twice the maximum absolute value of ay elemet i A. Let C = {k +,..., 2k + }, which is sum-free modulo p. The pick a uiformly radom x {,..., p} ad let B be the set obtaied by multiplyig each elemet of A by x, modulo p. For each elemet, probability of mappig ito C is C /(p > /3. So expected umber of elemets mappig ito C is > A /3, ad we ca take S to be those that do. 2. (Turá s Theorem, umerical boud Let G be a graph with vertices ad average degree d. Prove that oe ca fid a set S of pairwise o-adjacet vertices of size at least Remark. This is called a idepedet set, ad the result is tight because a disjoit uio of copies of K d+ has o idepedet set larger tha Solutio: Start with a radom orderig of the vertices. If ay vertex precedes all of its eighbors accordig to the orderig, the take it for S. Clearly this is a idepedet set. Let I v be the evet that vertex v is chose. E [I v ] = d, so by liearity of expectatio, E [ S ] = v+ v d v+. By covexity, this is at least 3. (Compoet of research problem of Tao Jiag s masters studet Let G be a graph whose vertices have bee properly colored. Nothig is assumed about the umber of colors used, except for the local fact that for every vertex v, the umber of differet colors which appear o v s eighbors is at most M. Prove that there is a set S of pairwise o-adjacet vertices, of size at least Solutio: (Proof due to Bey Sudakov. Start with a radom orderig of the color classes. If ay vertex s color class precedes all of its eighbors color classes accordig to the orderig, the take the vertex for S. Agai this is clearly a idepedet set. A proper colorig has the property that ay v is differetly colored from all of its eighbors, so at most M + total colors appear o v ad its eighbors. Thus the probability that v s color is earliest i the permutatio is at least M+. Fiish by liearity of expectatio. 4. (Crossig Lemma No matter how you draw a graph with V vertices ad E edges i the plae, there will be E3 64V pairs of crossig edges, as log as E 4V. 2 Solutio: Sice plaar graphs have E 3V 6, we automatically fid that the crossig umber is always E (3V 6 > E 3V. Now take a drawig with, say, t crossigs, ad sample vertices radomly with probability p. V goes dow to pv, E goes dow to p 2 E, ad cr goes dow to p 4 t. But this ew drawig eeds to satisfy the above, so Substitutig p = 4V/E, we get the desired result. p 4 t > p 2 E 3pV. 6 Real problems oe does ot expect to prove, but are probably true May of these come from the excellet book titled The Probabilistic Method, by Noga Alo ad Joel Specer.. (Sperer s Lemma Prove that the maximum atichai i 2 [] has size ( /2. That is, show that if F is a collectio of subsets of {,..., } such that o two distict sets A, B F satisfy A B, the F ( /2. Solutio: Note that this is a special case of oe of the first problems i this hadout. 5

6 The idea is give as a hit i Alo-Specer. Take a radom permutatio σ ad let the radom variable X = #{i : σ(,..., σ(i F}. Cosider E [X]. By defiitio of F, X is bouded by, ad the evets {σ(,..., σ(k} F are disjoit for distict K. Let N k be the umber of subsets of size k i F. E [X] = P [{σ(,..., σ(k} F] = k= k= N k ( k k= N ( k. /2 Sice E [X], we coclude that N k is bouded by ( /2, ad we are doe. 2. (Corollary: Littlewood-Offord Lemma Let x,..., x be ozero real umbers, ot ecessarily distict. Suppose that c,..., c are idepedet radom variables, each of which is ± with equal probability. Prove that P [c x + + c x = 0] O(. Solutio: First, we may assume all x i > 0 by flippig the sig of ay egative guy. (Sice the c i are ± with equal probability, it wo t make a differece. Cosider ay combiatio of (c i which makes c i x i = 0. Idetify this with a subset of [], based o which idices have c i = +. Let the F be the collectio of all such subsets of []. This is a atichai because if we had some S strictly cotaied i a larger T, the the sum c i x i correspodig to T would be strictly larger sice all x i > 0. By Sperer s Lemma, we coclude that the umber of ways to have (c i makig zero-sum is at most, ad so the desired probability boud is this divided by 2, which is of order ( /2 3. (Erdős-Ko-Rado Theorem Let 2k be positive itegers, ad let C be a collectio of pairwiseitersectig k-elemet subsets of {,..., }, i.e., every A, B C has A B. Prove that C ( k. Remark. This correspods to the costructio which takes all subsets that cotai the elemet, say. Solutio: Pick a radom k-set A from 2 [] by first selectig a radom permutatio σ S, ad the pickig a radom idex i []. The defie A = {σ(i,..., σ(i + k }, with idices after wrappig aroud, of course. It suffices to show that P [A C] k/. Let us show that coditioed o ay fixed σ, P [A C σ] k/, which will fiish our problem. But this is equivalet to the statemet that C ca oly cotai k itervals (wrappig after of the form {i,..., i + k }, which is easy to show. 4. (ChipLiar game, Alo-Specer Theorem 4.2. Paul ad Carole play a game o a board with positios labeled {0,,..., k}. Iitially, stoes are at positio k. Paul ad Carole play for r rouds, where each roud has the followig structure: Paul ames a subset S of the stoes o the board, ad the, Carole either moves all stoes i S oe positio to the left, or moves all stoes i S c oe positio to the left. Ay stoe that is moved leftwards from 0 is discarded. If the umber of stoes o the board becomes or 0, Carole loses. Prove that if k ( r i=0 i 2 r >, the Carole has a wiig strategy. Solutio: Perfect iformatio game, so oe of them has a wiig strategy. Suppose it is Paul, ad let him play it. Carole will respod by playig radomly. Note that if Paul ideed has wiig strategy, the it ca eve always beat a radom strategy. Calculate expected umber of stoes left o the board after r rouds. By liearity, this is times the probability that a chip survives for r rouds. But with radom play, each chip moves left Bi(r, /2 may times. So, expected umber of survivig chips is: P [Bi(r, /2 k], which we are give to be greater tha. Hece some play sequece will leave Carole with 2 stoes, cotradictig existece of Paul s perfect strategy. 5. (Bollobás, 965 Let A,..., A ad B,..., B be distict subsets of N such that:. 6

7 every A i = r ad every B i = s, for every i, A i B i =, ad for every i j, A i B j. Prove that ( r+s r. Solutio: (Proof from Alo-Specer. Let the uiverse X be the uio of all A i ad B j. Take radom permutatio of X. Defie evet X i to be that all elemets of A i precede all elemets of B i i permutatio. Easy to check that all of the evets X i are pairwise disjoit, ad each P [X i ] = ( r+s. r But sum of probabilities is, so doe. 6. (Lovász, 970 Let A,..., A ad B,..., B be distict subsets of N such that: every A i = r ad every B i = s, for every i, A i B i =, ad for every i < j, A i B j. Prove that ( r+s r. Remark. This is much more difficult, ad the proof uses liear algebra. 7. (Alo-Specer, Exercise.7 Let A,..., A ad B,..., B be distict subsets of N such that: every A i = r ad every B i = s, for every i, A i B i =, ad for every i j, (A i B j (A j B i. Prove that (r+sr+s r r s s. Solutio: Plug ito Tuza s result above with p = r r+s or Alteratively, we ca do this directly as follows. Let X := i= A i B i be the base set. Defie p := r/(r + s, ad cosider a coi which has oe side sayig A ad oe side sayig B, with the A-side appearig with probability p. For each elemet i X, idepedetly flip the coi. This defies a mappig f : X {A, B}. Defie the family of evets {E i } by havig E i occur whe all elemets x A i have f(x = A, ad all elemets y B i have f(y = B. By defiitio of the family of sets, it is impossible for E i ad E j to occur simultaeously if i j, because i particular, there would exist some elemet i either A i B j or A j B i, ad it could either be A or B. So, just as i the previous problem, cosider the probability of ay E i occurrig. Trivially, P (E i = p r ( p s for ay i. Sice the evets are disjoit, the total probability is p r ( p s. Yet all probabilities are bouded by, so just as i the previous problem, p r ( p s, which turs out to be the desired boud. (I fact, this choice of p is optimal. 7 Really harder problems s r+s.. Prove that the Riema Hypothesis has probability > 0 of beig true. Refereces [] N. Alo ad J. Specer, The Probabilistic Method, 2d ed., Wiley, New York,