Notes on the Combinatorial Nullstellensatz

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1 Notes o the Combiatorial Nullstellesatz Costructive ad Nocostructive Methods i Combiatorics ad TCS U. Chicago, Sprig 2018 Istructor: Adrew Drucker Scribe: Roberto Ferádez For the followig theorems ad examples we let F be a field ad P[x 1,..., x ] be a formal polyomial over F. Note that i our defiitio of formal polyomials we cosider a polyomial as beig uiquely defied by its coefficiets. 1 Schwartz-Zippel Lemma The Schwartz-Zippel lemma [4, 6] provides us with sufficiet coditios uder which we ca boud the probability that a specific evaluatio of a polyomial vaishes, ad thus allows us to make statemets about whether a polyomial is idetically zero. Theorem 1 (Schwartz-Zippel Lemma). Let P F[x 1,..., x ] be a o-zero polyomial with deg(p) = d 0. We the cosider a fiite subset S F. If we radomly ad idepedetly sample r 1,..., r from S the P[P(r 1,..., r ) = 0] d S. Proof. We prove this by iductio o. The base case is immediate sice = 1 gives us a uivariate polyomial, for which we kow deg(p) = d implies at most d roots. Now we assume it holds for all polyomials with degree 1. Give that P / 0 we ca, without loss of geerality, assume that x 1 occurs i some ozero moomial. We ca the cosider each moomial ad factor out this x 1 compoet so that we ca write P as a polyomial i x 1 : P(x 1,..., x ) = d x i 1 Q i(x 2,..., x ). By our assumptio that x 1 occurs i a ozero moomial we kow that, for some i, we have Q i / 0, let k deote such moomial with the highest degree. We kow that deg(x1 kq k(x 2,..., x )) d ad thus deg(q k ) d k. Now let A deote the evet that Q k (y 2,..., y ) = 0, where y 2,..., y is radomly sampled from S 1. It follows by our iductive hypothesis that P[A] d k S. If we the have such y y 2,..., y with Q k (y) / 0 the it follows that P(x 1, y 2,..., y ) is uivariate i x 1, ozero, ad of degree k. If we the coditio this polyomial o y we ca use the base case ad get that P[P(x 1, y 2,..., y ) = 0 (y 2,..., y )] k S. Ultimately, we ca write: P[P 0] (1 P(A)) (1 k S ) = (1 d k ) (1 k S S ) 1 d k k S S = 1 d S. It the clearly follows that P[P 0] = 1 (1 d S ) = d S, just as desired. 1

2 The above theorem thus tells us that, if applicable, we ca fid a x S such that P(X) 0 with high probability after O ( S d ) trials. We ow cosider our first applicatio of the Schwartz-Zippel Lemma, Polyomial Idetity Testig (PIT): Example. Let our iput be a arithmetic circuit represetig a formal polyomial over variables x 1,..., x i a field F. We are the iterested o whether P 0. This ca be used to test the equivalece of two polyomials P 1, P 2 over this same field sice we ca simply write Q = P 2 P 2 ad test whether Q 0. Give our above result we ca provide a algorithm which shows that PIT co-rp, at least i cases where our polyomial has bouded degree. If we have P(x 1,..., x ) over our field F such that deg(p) we ca use the followig algorithm. 1. Radomly ad uiformly sample y y 1,..., y from F. 2. If P(y) 0 the coclude P is ot idetically 0 ad coclude executio. 3. If P(y) = 0 retur that P is idetically zero with a probability error of at most d < 1, due to Schwartz-Zippel. We ca thus repeat this procedure util this error has reached a appropriate boud. This boud o our error gives us, eve i the worst case uder our assumptios of d = 1, that our boud decreases by ( 1 ) (1 1 ) = e 1. We ca ow look at aother applicatio of the Schwartz-Zippel Lemma: checkig for the existece of perfect matches i bipartite graphs. [5] Example. Let G = (U, V, E) be a bipartite graph with V = U = ad E U V. We thus wat to check if there exists a set of edges M E such that every vertex is cotaied i exactly oe edge of M. We call such a subset M a perfect matchig of G. Defiitio 2. Let G = (U, V, E) be a bipartite graph. We the defie the tutte matrix, M G, of our graph as follows: x i,j (i, j) E M i,j = 0 (i, j) / E where x i,j is a fresh symbolic variable at each iteratio (i.e. each x i,j value is uique.) Lemma 3. The determiat of M G, which is a polyomial i (x i,j ) with degree at most, is ozero as a formal polyomial if ad oly if G has a perfect matchig. Proof. We ca use the Leibiz Formula for the determiat: det(m G ) = (sg(σ) M i,σ(i) ) σ S ad we the have a 1-1 correspodece betwee σ S ad the possible perfect matchigs i G which ca be writte as {(u 1, v σ(1) ),..., (u, v σ() }. By the defiitio of the tutte matrix we ca see 2

3 that if for some i we have (u i, v σ(i) ) (i.e. the possible perfect matchig is ot i G) the that term of our summatio will be 0 sice M i,σ(i) = 0. We ca the restrict our summatio rage to solely cosider σ P, where P is the set of perfect matchigs i G. We thus write det(m G ) = (sg(σ) M i,σ(i) ) = (sg(σ) x i,σ(i) ) σ P σ P ad usig this expressio we ca show the lemma. If the determiat is 0 the it is clear that P = ad thus G has o perfect matchig. If P, however, we kow that for σ P we have sg(σ) M i,σ(i) = sg(σ) x i,σ(i) 0 ad by our defiitio of x i,σ(i) always beig a fresh variable we kow that there does ot exist aother term i our sum that has the same set of variables ad thus this value is ot cacelled out ad we must have a o-zero determiat. Give the above result ad the PIT algorithm we ca defie a algorithm, i RP, for determiig whether G has a perfect matchig. Furthermore, we ca also desig a algorithm to retur a perfect matchig. Oe major advatage of this tutte matrix approach compared to the traditioal max flow approach is that of easy parallelizatio. Refer to [3]. 2 Combiatorial Nullstellesatz We ow move o to the related Combiatorial Nullstellesatz, [1, 2] which gives us coditios uder which we are assured a poit withi a defied sub-cube for which our polyomial does t vaish. Theorem 4 (Combiatorial Nullstellesatz). Let P(x 1,..., x ) be a polyomial i F[x 1,..., x ] with deg(p) = k i where each k i N ad suppose that the coefficiet x k 1 1 xk 2 2 xk i P is o-zero. The for ay collectio of subsets S 1,..., S F such that S i > k i for all i we have that there exists x S 1 S such that P(x) 0. Proof. We show the above result through iductio o d deg(p). The base case of d = 1 is trivial sice a polyomial of degree d will have at most d roots. Now let d > 1 ad for cotradictio s sake assume that P satisfies all of our theorem s assumptios ad yet P 0 o S 1 S 2 S. We ca assume without loss of geerality that α 1 > 0. Now fix ay s S 1. We ca the use the divisio algorithm to give us the followig polyomial: P(x 1,..., x ) = (x 1 s)q(x 1,..., x ) + R(x 2,..., x ). Now let x {s} S 2 S. We the have that P(x) = 0 implies that R(x) = 0. Sice R does ot cotai x 1 we ca also say that R vaishes o (S 1 {s}) S 2 S. Now if we fix x (S 1 {s}) S 2 S it follows that x 1 s is o-zero but sice we assumed the polyomial vaishes it must be true that Q(x) = 0 ad we thus have that Q vaishes o (S 1 {s} S 2 S. Sice we also have deg(q) = deg(p) 1 this vaishig implies a smaller couterexample which cotradicts our iductive hypothesis. We ow look at some possible applicatios of the above result, startig with the Cauchy- Daveport Theorem. 3

4 Theorem 5 (Cauchy-Daveport). Let p be a prime ad A, B be two o-empty subsets of F p ad defie A + B = {a + b a A, b B}. The, A + B mi{p, A + B 1}. Proof. The A + B > p case is straightforward sice for ay x F p there are p distict expressios of the form a + b that equal x. Sice the sum of elemets i our sets is greater tha p we must the have that these expressios overlap ad thus A + B = F p. Now we cosider the A + B p case. Assume for cotradictio s sake that A + B A + B 2 ad let C F p such that A + B C ad C = A + B 2. Now defie P(x, y) = c C (x + y c) be a polyomial over F p. (Note that P(a, b) = 0 for all a A, b B a + b C.) Now let S 1 A, S 2 B ad sice the coefficiet of x A 1 y B 1 is give by ( A + B 2 ) ad sice p is a prime by assumptio we kow that this coefficiet is ozero i F p ad thus by the CNSS we have that there exists (x, y) A B such that P(x, y) 0. Sice A 1 our polyomial was costructed such that it vaishes for all poits i C we reach a cotradictio sice A + B / C. Now we look at the Chevalley-Warig Theorem, which cocers algebraic varieties i F p. Defiitio 6. Let P 1,..., P m be a collectio of polyomials i F p. We the defie the vaishig set: V({P i } i ) {x F p P i (x) = 0 for all i}. Give the above defiitio, however, we ote that differet collectios ca have the same vaishig set, e.g. for c = {c 1,..., c } we could cosider A = {x c 1, x c 2,..., x c }, B = (x c i ) ad it is clear that V(A) = V(B) = c. We ca thus associate a cost measure to the collectios of polyomials, amely the sum of the degrees of each polyomial i the collectio, which defie a sigleto vaishig set. Further, we ca boud the miimum of this cost measure required to specify such a vaishig set. Theorem 7 (Chevalley-Warig Theorem). Let P 1,..., P m F[x 1,..., x ] ad assume that V({P i } i ) = {c} where c F p. The, m deg(p i ). Proof. Assume for sake of cotradictio that we have the desired sigleto vaishig set but d i <. Let P m [1 P p 1 ] δ (x j a) a c j where we choose δ such that P(c) = 0. (Note that the LHS evaluates to 1 o c so δ 0.) Let S i = F for all i ad defie α (p 1, p 2,..., p ) ad cosider x α. We claim that F 0 o S 1 S ad to show this we will prove the existece of some leadig term with degree p 1 i x 1,... x. I the x = c case it is clear that P 0 by defiitio. Now we cosider the x c ad assume WLOG that x k c k. I this case we ote that x c implies that for some i we have P i (x) 0 which implies that the LHS evaluates to 0 sice 1 P i (x) p 1 = 0 ad we must also have that the RHS evaluates to 0 sice by our assumptio x k c k ad thus a ck (x k a) will have a zero term. Sice the degree of the LHS is (p 1) d i < (p 1) but it has a o-zero coefficiet of the term x p 1 i it thus gives us a cotradictio through CNSS, which assures us a o-zero value of F. 4 j=1

5 Our fial applicatio is that of coverig the boolea cube, C {0, 1}, with hyperplaes. We defie a hyperplae, H, as follows: H { x a, x = b} for a give a, b Theorem 8. Let H 1,..., H m cover all of C except for oe poit. The we have that m. Proof. Assume without loss of geerality that the oe poit ot covered by our hyperplaes is 0 ad let our collectio of hyperplaes, {H i }, be defied such that H i = { a i, x = b i }. For cotradictio s sake assume that m < ad cosider P(x) ( 1) +m+1 m b j ( j=1 (x i 1)) m ( a i, x b i ). From the left term we clearly have that deg(p) = ad the coefficiet of it is ( 1) m++1 m j=1 b j 0. By CNSS we the have that there exists x {0, 1} such that P( x) 0. This poit x, however, is ot 0 sice the LHS of our polyomial is desiged to cacel the RHS uder 0, ad thus P( 0) = 0. We kow that our LHS vaishes for x 0 ad thus P( x) 0 implies that the RHS does t vaish. We thus have a i, x b i 0 for all i ad thus x is ot i ay H i, givig us the desired cotradictio. 5

6 Refereces [1] Noga Alo. Combiatorial ullstellesatz. Comb. Probab. Comput., 8(1-2):7 29, Jauary [2] M. Michalek. A short proof of Combiatorial Nullstellesatz. ArXiv e-prits, April [3] Keta Mulmuley, Umesh V. Vazirai, ad Vijay V. Vazirai. Matchig is as easy as matrix iversio. I Proceedigs of the Nieteeth Aual ACM Symposium o Theory of Computig, STOC 87, pages , New York, NY, USA, ACM. [4] J. T. Schwartz. Fast probabilistic algorithms for verificatio of polyomial idetities. J. ACM, 27(4): , October [5] W. T. Tutte. The factorizatio of liear graphs. Joural of the Lodo Mathematical Society, s1-22(2): [6] Richard Zippel. Probabilistic algorithms for sparse polyomials. I Edward W. Ng, editor, Symbolic ad Algebraic Computatio, pages , Berli, Heidelberg, Spriger Berli Heidelberg. 6