Here, e(a, B) is defined as the number of edges between A and B in the n dimensional boolean hypercube.

Transcription

1 Lecture 2 Topics i Complexity Theory ad Pseudoradomess (Sprig 2013) Rutgers Uiversity Swastik Kopparty Scribes: Amey Bhagale, Mrial Kumar 1 Overview I this lecture, we will complete the proof of formula lower bouds via Krapcheko s method ad the look at formula lower bouds via Subbotovskaya s method ad Adreev s method. We will also see oe lower boud o the size of formula over full biary basis give by Nechiporuk. We will fially colcude with a itroductio to the otio of radomized circuits. 2 Krapcheko s method Krapcheko s method is based upo the idea of formal complexity measure, which is defied as follows. Defiitio 1. A fuctio F C from the set of all boolea fuctios to the set of real umbers is said to be formal complexity measure if it satisfies the followig properties. 1. F C(x i ) = 1 for all variables x i 2. F C(f) = F C( f) for every boolea fuctio f 3. Sub-additivity: F C(f g) F C(f) + F C(g) for all boolea fuctios f ad g We also proved the followig theorem i the last lecture which relates the formula complexity of a fuctio to its formal complexity, for ay formal complexity measure, over the De Morga s basis. Theorem 2. For ay boolea fuctio f ad ay formal complexity measure F C, the followig is true over the De Morga s basis F C(f) L(f) (1) We will ow defie Krapcheko s formal complexity measure ad the use it to prove lower bouds. Defiitio 3. (Krapcheko s measure) For a boolea fuctio f, we defie e(a, B) 2 F C(f) = max A f 1 (0),B f 1 (1) A B Here, e(a, B) is defied as the umber of edges betwee A ad B i the dimesioal boolea hypercube. Before we go o ad show that F C as defied above is ideed a formal complexity measure ad satisfies all the properties i defiitio 1, we will see some applicatios. (2) 1

2 2.1 Formula Lower Boud for Parity over De Morga s basis We will ow show ay formula computig the Parity of bits must be of size Ω( 2 ). We first prove the followig lemma. Lemma 4. For the formal complexity defied i defiitio 3, F C(P ARIT Y ) 2. Proof. To prove the statemet, it is sufficiet to show a set A P ARIT Y 1 (0) ad a set B P ARIT Y 1 (1), such that e(a,b)2 A B 2. Let us take A = P ARIT Y 1 (0) ad B = P ARIT Y 1 (1). Now, for ay x, such that P ARIT Y (X) = 0, all its eighbours i the boolea hypercube will have parity 1 ad vice versa. Therefore, e(a, B) = 2 ( 1), where A = B = 2 1. Subsitutig back i the defiitio of F C, we obtai F C(P ARIT Y ) 2. Now, from the lemma above ad theorem 2, we have the followig theorem. Theorem 5. (Krapcheko) Ay formula which computes a P ARIT Y of bits over the De Morga s basis, must have size Ω( 2 ). 2.2 Formula lower boud for Majority over De Morga s basis We will ow show a formula lower vboud for the majority of bits, via provig that the formal complexity of the majority fuctio is high. More precisely, we will prove the followig lemma. Lemma 6. For the formal complexity defied i defiitio 3, F C(MAJ) Ω( 2 ). Proof. The proof idea is agai very similar to the proof for the parity fuctio. We first check if the sets defied i lemma 4 work. For the majority fuctio, the oly strigs x MAJ 1 (1), which have a eighbour i MAJ 1 (0) are the oes which have 2 1 oes. Moreover, the umber of such eighbours is precisely For the simplicity of presetatio, let us assume that is eve. The asymptotics remai the same eve for a odd. Now, if we take our set A = MAJ 1 (0) ad set B = MAJ 1 1, we get e(a, B) = ( 2 + 1)( ). Hece, we get 2 1 F C(MAJ) (( 2 + 1)( ) 2 ) ( 1) (3) which is just liear i. So, we do ot get aythig o trivial. It ca be observed that all the edges across A ad B are cocetrated betwee the strigs which have just less tha half 1 s ad the strigs which have oe more 1 tha them. Keepig this i cosideratio, let us redefie the sets A ad B as follows. Let A = ( [] ) ( ad let B = [] ) 2 1. Clearly the umber of edges betwee A 2 ad B remais the same while their sizes becomes smaller. So, we get which is Ω( 2 ). F C(MAJ) (( 2 + 1)( ) 2 ) 2 ( 1 )( ) (4)

3 From the lemma above ad theorem 2, we get the followig theorem. Theorem 7. Ay formula which computes a MAJ of bits over the De Morga s basis, must have size Ω( 2 ). 2.3 FC is a formal complexity measure We will ow verify that the fuctio F C, as defied i defiitio 3 ideed satisfies all the properties metioed i defiitio 1. To this ed, we will verify the properties sequetially. 1. F C(x i ) = 1, for all variables x i Proof. For the fuctio f(x) = x i, f 1 (0) = {x : x i = 0} ad f 1 (1) = {x : x i = 1}. We ca observe that the umber of eighbours that ay strig i f 1 (0) ca have i f 1 (1) is at most 1, ad vice-versa. So, for ay sets A f 1 (0) ad B f 1 e(a,b) (1), A 1 1. Multiplyig together, we have F C(x i ) 1. Now,cosider A = {0 } ad ad e(a,b) B B = {0 i 10 i 1 }. Clearly, e(a, B) = A = B = 1. So, we have F C(x i ) 1. Cosequetly, we get F C(x i ) = F C(f) = F C( f) for all boolea fuctios f. Proof. The proof follows from the fact that the defiitio of F C is symmteric with respect to 0 ad Sub-additivity: F C(f g) F C(f) + F C(g) for all boolea fuctios f ad g Proof. Let h = f g. For the proof, it would suffice if for every pair of sets A h 1 (0) ad B h 1 (1), we could show the existece of sets A f f 1 (0), A g g 1 (0), B f f 1 (1) ad B g g 1 (1) which satisfy e(a, B) 2 A B e(a f, B f ) 2 A f B f + e(a g, B g ) 2 A g B g (5) We also observe the followig relatioships. h 1 (0) = f 1 (0) g 1 (0) h 1 (1) = f 1 (1) g 1 (1) Now takig these ito cosideratio, for ay A h 1 (0), take A f = A g = A h 1 (0) = f 1 (1) g 1 (1) ad take B f = B f 1 (1) ad take B g = B \ B f. So, B f, B g is a partitio of B. Therefore, we also have e(a, B) = e(a, B f ) + e(a, B g ). Now, ivokig the iequality i claim 8, which is proved below with a = e(a, B f ), b = e(a, B g ), c = B g ad d = B f, we get ( e(a, B f ) 2 B f + e(a, B g) 2 ) (e(a, B f ) + e(a, B g )) 2 B g ( B f + B g ) (6) 3

4 Dividig by A both sides, we get ( e(a, B f ) 2 A B f + e(a, B g) 2 ) (e(a, B f ) + e(a, B g )) 2 A B g A ( B f + B g ) (7) which is the same as Hece, we obtai ( e(a, B f ) 2 A B f + e(a, B g) 2 (e(a, B)2 ) A B g A B F C(H) F C(f) + F C(g) (9) (8) Claim 8. For ay four positive real umbers a, b, c, d, (a + b) 2 (c + d) a2 d + b2 c (10) Proof. Sice a, b, c, d are positive real umbers, usig the AM GM iequality o c.a2 d ad d.b2 c, we get c.a 2 d + d.b2 2. ( c.a2 d.b 2 ) = 2ab (11) c d c Addig a 2 + b 2 o both sides of the iequality, we obtai (c + d).a 2 + d (c + d).b2 c Dividig both sides by the positive quatity (c + d), we get a 2 + b 2 + 2ab (12) (a + b) 2 (c + d) a2 d + b2 c (13) Exercise: Show that usig Krapcheko s method oe caot get a lower boud better that Ω( 2 ). 3 Subbotovskaya s method We will ow look at aother formula lower boud proved via a differet method. Before movig oto the method, let us defie a few ideas which will be cruicially used i the rest of the lecture. Defiitio 9. Restrictios of a fuctio: Give a boolea fuctio f, a restrictio of f is a fuctio which ca be obtaied from f by settig a subset of variables to some fixed values. Observe that, give the formula for a fuctio f, we ca obtai the formula for ay restrictio of f by settig a subset of variables to some fixed value i the formula. This procedure reduces the size of the formula i followig ways. 4

5 A leaf gets set to either 0 or 1 ad so the umber of leaves which are idexed by variables reduces. Settig a variable to some value might directly or covert a existig o trivial gate i the circuit ito a trivial gate. Here, a gate is said to be trivial if it computes a costat fuctio, it is o trivial otherwise. If a gate trivializes due to settig oe of its iput variables, the umber of leaves reduces by 2, for every leaf which is set to a costat. We also observe the followig properties of a optimal formula F computig a fuctio. Without loss of geerality, we ca assume that all the leaves i the formula are labelled by iput variables. Otherwise, the formula ca be reduced i size further, which would cotradict optimality of F. For example, f 1 ca be replaced by 1 ad f 0 ca be replaced by f, ad similarly for gates. Let there be a gate i F such that its output h = x i g. The, without loss of geerality, we ca assume that the sub-formula computig g does ot deped upo x i. The observatio follows from the fact that if x 1 is 0, the h = 0 ad so h depeds o g oly if x i = 1. So, the subformula for g ca be replaced by a formula where every occurece of x i i the subformula for g is replaced by 1. This etire procedure leads to o icrease i size of f. A gate of the form x i g ca be hadled similarly. For the rest of the lecture, we would be workig with a formula satisfyig these two properties. We will call such a formula to be ice. Defiitio 10. Nice formula: A formula computig a fuctio is said to be ice if it satisfies the above two properties. We ca observe that a arbitary formula for a fuctio ca be coverted ito a a ice formula without a icrease i size. Ituitio: The basic idea of Subbotovskaya s proof is the followig. If there is a formula computig a fuctio f, we ca compute a formula for ay restrictio of f from it. Now, if we somehow kew that this process of restrictio leads to a drastic decrease i size, yet the restrictio obtaied is a o trivial fuctio i the remaiig variables ad hece has a o-trivial size, we will get a lower boud. For ay ice formula havig s leaves has a iput variable labellig at least s leaves, if it computes a fuctio of variables. So, settig this variable to either 0 or 1, reduces the size of the formula by a factor of at least (1 1 ). If we cotiue restrictig the formula, till it depeds o just oe variable, we get Π 2 i= (1 1 i )s 1, which just gives us a trivial boud s. So, we eed to prove a better shrikage factor if we hope to obtai a otrivial lower boud. The followig lemma helps us achieve this goal. Lemma 11. Radom Restrictios: For ay boolea fuctio f, let F be the optimal sized formula computig f over the De Morga basis. Cosider the followig radom process. We select a subset of variables of size k uiformly at radom ad idepedetly set each of these variables to 0 or 1, each with probability 0.5. Let us deote this proess by ρ ad the resultig formula obtaied from this process by F ρ. The, E[ F ρ ] ( k ) 1.5 F (14) 5

6 To prove the lemma, we will aalyse the effect of radomly restrictig oe variable i the formula, chose uiformly at radom. We will first prove the followig lemma. Lemma 12. For ay boolea fuctio f, let F be the optimal sized ice formula computig f over the De Morga basis. Let us defie the followig radom process ρ. Let us pick oe variable uiformly at radom ad set it to 0 or 1 with probability 0.5 each. The, the expected size of the formula computig the fuctio f ρ is at most F (1 1.5 ). Proof. For each i [], let s i be the umber of leaves labelled with x i. Now, whe we apply the radom restrictio, the formula F is affected i the followig two ways. Ay variable which is set to a costat leads to a decrease i the umber of leaves. If oe of the iputs of a gate gets set to 1, the it ca be replaced by a costat 1 ad the umber of leaves decreases by at least 2. Similarly, if a iput to a gate gets set to 0, the it ca be replaced by 0 ad the umber of leaves falls by at least 2. Therefore, for every leaf i the origial formula, this secod decrease of 1 happes with a probability 1 2. So, expected decrease i the umber of leaves per leaf of the origial formula is = 3 2. Therefore, by the liearity of expectatios, the expected decrease i the umber of leaves i the formula is s 3 2. Observe that sice we bega with a ice formula, so there is o overcoutig beig doe here. At the ed of the above process, the expected umber of leaves survivig i the formula is s(1 1.5 ), ad we have the lemma. We ow use the above argumet sequetially to prove lemma 11. Proof. of lemma 11: Now, we will apply the lemma agai ad agai k times to obtai stroger lemma 11. After each step, we will simplify the resultig formula ito a ice formula so that the premises of lemma 12 are met. Hece, at the ed of k steps, the size of the resultig formula is at most Π k i= (1 i ). For every i, the i th term i the above product is positive ad is upper bouded by (1 1 i )1.5. Therefore, the product is at most ( k ) Lower bouds for Parity Let us ow apply lemma 11 to obtai lower bouds for P ARIT Y. We ivoke the lemma for k = 1. Sice the parity fuctio depeds upo all its iput, eve whe we restrict a formula computig parity to just oe live variable, we still have a o trivial fuctio. Now, lemma 11 tells us that there exists a restrictio ρ such that F ρ F ( 1 )1.5 ad F ρ is o trivial. So, the size of F ρ is at least 1. So, we get F ( 1 ) This implies F

7 4 Adreev s method I this sectio, we will use Subbotovskaya s method of radom restrictios to obtai a stroger lower boud for formula s i De Morga s basis. Let us first defie the fuctio for which we obtai the lower boud. Defiitio 13. Cosider the fuctio f : {0, 1} 2 {0, 1} def as follows. First, partitio the iput 2 bits as follows: Let the first bits be idicated by φ. These bits will be treated as the truth table of a boolea fuctio o log() bits. The ext bits should be viewed as sittig i a matrix X with log() rows of size log log The, f(φ, X) = φ( j=1 X 1j, j=1 X 2j,..., log j=1 X (log )j). The followig observatio follows from the defiitio. log each. Observatio 14. The fuctio f defied as above is explicitely computable for ay iput x {0, 1} 2. Key Idea for the lower boud : The proof builds up o Subbotovskaya s idea. Cosider a strig φ which is the truth table of a hard fuctio o log bits. Now, if we radomly restrict some variables i the matrix X, such that each row of X has at least oe live radom variable, we ca obtai a resultig formula which ca be used to compute the fuctio φ over all iputs. Sice φ was choose to be a hard fuctio o log bits, the resultig formula should have size at least log log(), ad we will ow use the iequalities obtaied to get a lower boud. We will ow formalize this proof sketch to prove the followig theorem. Theorem 15. Ay formula computig the fuctio f as defied i defiitio 13 i the De Morga basis is of size Ω( 2.5 ɛ ) for every epsilo > 0. Proof. Cosider the optimal sized formula F for f. Without loss of geerality, we ca assume that it is a ice formula. Now, let us restrict the first log() bits to the truth table of a hard fuctio φ. 2 By a hard fuctio, we mea ay fuctio with formula size Ω( log log log() ). We kow from Shao s theorem that such fuctios exist. Let us call the resultig fuctio f φ ad the resultig formula F φ. We ca preprocess F φ ow to obtai a ice formula for f φ. For the sake of simplicity, let us agai call it F φ. Now, set k = 10 log() log log() ad apply lemma 11 to F φ. Claim 16. All the rows of the matrix X cotai at least oe live variable with probability at least 0.9. log ) ) k Proof. For a fixed row i of X, the probability that row i has o live variable is equal to ((1. ( k) Let us call this p. Now, expadig the biomial coefficiets i the umerator ad the deomiator, we get p = Π k 1 (1 log ) i i=0 i. Now, usig the iequality (1 log ) i i (1 log ) (1 log )i i, we get that p (1 1 log )k. For, k = 10 log() log log(), we obtai p 1 (log ) 10. Now, by the uio boud, 7

8 there exists a row i X which cotais o live variables with probability at most log 1 (log ) for large eough. Now, from lemma 11, the expected size of the formula after radomly restrictig to k variables is at most F ( k )1.5. Sice formula size is a o egative radom variable, Markov s iequality tells us that the formula size is larger tha 20 F ( k )1.5 with a probability at most Hece, with probability at most 0.15 either of these bad evets happe. So, with positive probability, the formula shriks a lot ad all of the rows i the matrix X have at least oe live variable. The, formula obtaied at the ed of the process ca be used to compute the fuctio φ, which we kow is hard. So, we have log log() 20 F ( k )1.5. This implies F Ω( 2.5 ɛ ), for every positive ɛ. Exercise: For the fuctio f defied i defiitio 13, show that f ca be computed by a formula of size O( 3 ). 5 Lower Bouds over Full Biary Basis All the above methods give a lower boud o the formula size over De Morga s basis. A atural questio would be to ask for a lower boud o the formula size of some explicit fuctio over full biary basis. Recall that i the formula over full biary basis we are allowed to use a gate which computes arbitrary biary boolea operatio. 5.1 Nechiporuk s Method Let us start by defiig oe otio. Defiitio 17. Let f : {0, 1} {0, 1} be a fuctio o variables ad Y be a subset of variables. N f (Y ) is the umber of differet fuctios g : {0, 1} Y {0, 1} we ca get whe we fix all iput bits ot i Y. Ituitively, larger the quatity N f (Y ) for a fuctio f, harder is the fuctio f ad hece formula computig f should require large size. Nechiporuk s Method uses the above otio of hard fuctio ad gives a lower boud. We are ow ready to defie a fuctio for which Nechiporuk obtaied a lower boud. Defiitio 18. f : {0, 1} {0, 1} be a boolea fuctio o variables. We ca group variables to form a block i.e. f(x 1, x 2,, x ) = f(x 1,, x c, x c+1,, x 2c,, x (b 1)c+1,, x bc ) So there are b blocks (Y 1, Y 2,, Y b ) each havig c variables. The fuctio f is give as: { 1 if all Y f(y 1, Y 2,, Y b ) = i s are distict 0 otherwise 8

9 Theorem ( 19. Size ) of formula over full biary basis computig fuctio f defied above is at least Ω 2 log 2 log log. Proof. It is easy to get a boud o N f (Y i ) where f is defied as above. Cosider fixig variables i the blocks {Y 1, Y 2,, Y b }\{Y i }. If at least two blocks have the same assigmet the the fuctio defied i Y i is uique which is a zero fuctio. O the other had, if all blocks have differet assigmets the it defies uique fuctio i Y i for every such assigmet (upto permutatio of these assigmets betwee blocks). Sice there are 2 c differet biary strigs of legth c ad ad if we choose ay b 1 differet strigs as a assigmet for variables i blocks {Y 1, Y 2,, Y b } \ {Y i }, it gives followg simple relatio: ( ) 2 c N f (Y i ) = + 1 (15) b 1 Cosider a formula F for computig f. Let l(y i ) be umber of leaves labelled by variables i Y i i F. Whe we fix the variables i {Y 1, Y 2,, Y b } \ {Y i }, we ca assume without loss of geerality that the formula iduced by fixig these varaibles has leaves labelled by variables i Y i oly ad o leaf is labelled by 0 or 1. If we get a upper boud o umber of fuctios computable o variables i Y i by fixig variables i other bolcks i F, the usig equatio 15 it gives us a lower boud o umber of leaves labelled by Y i i F. Sice umber of iteral odes roughly equals umber of leaves i a formula, if we restrict F to variables i Y i, the total umber of iteral odes is roughly l(y i ). Sice the formula F is over full biary basis, we have 16 differet possibilities of fuctio that a particular iteral ode computes. i, Number of differet fuctios computable o variables i Y i } 16 l(y i) (16) From equatio 15 ad 16, ( ) 2 c i, l(y i ) log b 1 Sice the formula size is atleast the umber of leaves labelled by its variables, We kow that bc =, by settig b = F log b l(y i ) i=1 ( ) 2 c blog b 1 ( ) 2 c b(b 1)log b ad c = log, we get ( ) 2 F = Ω log 2 log log 9

10 ( ) By settig the values of b ad c optimally i the above proof you ca get a lower boud of Ω 2 log. Nechiporuk boud is the best kow lower boud o the size of formulae over full biary basis. Exercise : Show that by this method oe caot get a lower boud better tha Ω ( 2 log ). 6 Radomized Circuits Use of radomizatio is foud to be useful i may applicatios. We ca ask the same questio about circuit complexity. Let us first formally defie what radomized circuit is: Defiitio 20. Radomized circuits: Let C(x 1,, x, r 1,, r m ) be a circuit. computes fuctio f : {0, 1} {0, 1} with error probability ɛ if We say C the C is a radomized circuit for computig f. x, P r r {0,1} m [C(x, r) f(x)] ɛ We ca ask several questios about radomized circuits 1. Ca we have a fuctio whose radomized circuit complexity is expoetial i the umber of variables? 2. Are radomized circuits more powerful tha determiistic circuits? 3. Give a radomized circuit for a fuctio, ca we deradomized it? 6.1 Hard fuctios for radomized circuits Followig theorem gives a affirmative aswer for questio 1. Theorem 21. There exists a fuctio f o variables such that for all radomized circuits C of size << 2, C caot compute f. Proof. Shao s coutig argumet for determiistic circuits works i this case also sice ay fixed radomized circuits caot compute two differet fuctios. 6.2 Deradomizatio From defiitio 20, we ca make the followig observatio: Observatio 22. If C is a radomized circuit computig f with some error probability ɛ < 1/2 the rs.t.p r x {0,1} [C(x, r) = f(x)] (1 ɛ) 10

11 It meas if a radomized circuit C computes a fuctio f the there exists oe fixed strig r that works for most of the iputs. Whe we talk about deradomizig a give circuit, we oly allow blow up i size of circuit which is polyomial i size of give circuit. So the problem of deradomizig a circuit is equivalet to askig followig questio 1. Give a radomized circuit C computig a fuctio f, ca we fid a cricuit C of size O( C k ) for some costat k ad a strig r such that C (x, r ) = f(x) for all x? Surprisigly, the aswer to the above questio is also yes, give by followig theorem Theorem 23. For every radomized circuit of size s computig a fuctio f, there exists a determiistic circuit of size at most polyomial i s computig the same fuctio. Proof Idea : We will prove above theorem by showig the existece of a circuit C that outputs correct aswer for every iput, give a radomized circuit computig f. We will use the method of amplifyig the success probability of a circuit by repetitio. It is eough to show existece of a circuit computig f with error probability less tha 2, sice i this case by observatio 22, there exists a strig r that works for more tha (1 2 ) fractio of iputs. It meas it works for every iput! So by fixig such strig r i circuit C we get a determiistic circuit computig f. Before provig the theorem we prove the followig lemma, the proof of the theorem follows directly for the lemma Lemma 24. Suppose there exists a circuit C(x 1, x 2, x, r 1,, r m ) computig a fuctio f : {0, 1} {0, 1} with error probability 0.1 (0.1 ca be replaced with ay positive costat less tha 0.5), the circuit C (x, r ) such that C computes f with error probability δ, where r = (r 1,, r m ), r i = m m = O ( log ( 1 δ )) C = poly ( log( 1 δ )) + C.O ( log( 1 δ )) Proof. We costruct circuit C as follows. C cosists of k = O(log(1/δ)) copies of circuit C, each of these circuits C 1, C 2,, C k has their ow seperate radom bits as a iput. The k output bits is fed to MAJ ad output of MAJ is the output of circuit C. Clearly, circuit C satisfies above three properties. We will ow show that the cricuit C has desired error probability. Let Z i be a idicator radom variable which is 1 if C i (x, s i ) f(x) ad 0 otherwise. P r[z i = 1] = 0.1 Let The the expected value of Z is: Z = k i=1 Z i E[Z] = 0.1k 11

12 Figure 1: Costructio of determiistic circuit from radomized circuit Circuit C outputs wrog aswer iff more tha half of the circuits C 1, C 2,, C k output wrog aswer. So Z > 0.5k is the evet that circuit C outputs wrog aswer. Sice the evets Z 1, Z 2,, Z k are idepedet evets, we ca apply Cheroff bouds. The probability that C outputs wrog aswer is bouded by : P r[ C outputs wrog aswer] = P r[z > 0.5k] P r [ Z 0.1k > 4 0.1k] e O(k) = e O(log( 1 δ )) < δ (17) From the above lemma we ca ow prove theorem 23 Proof of theorem 23 : Settig δ = 2, we get ( ( )) 1 m = O log δ C poly() + O( C ) = O() So the size of determiistic circuit is at most poly i size of origal radomized circuit ad the error probability of circuit C is less tha 2. Thus, the theorem follows. 6.3 Pseudoradom Set Motivatio : Now that we kow that every fuctio beig computed by a radomized circuit ca also be computed by a determiistic circuit of about the same size, a atural questio would be to fid this circuit give the radomized circuit. From the proof discussed above, oe way of 12

13 solvig this problem would be to fid the good strig r, which ca be substituted for the radom strig for every iput, while keepig the output the same. The aim i the ext sectio is to show the existece of a small set of strigs such that a strig choose uiformly at radom from this set behaves like a truly radom strig for the give circuit. Let us formalize the otio below. Defiitio 25. The set {r 1, r 2,, r t }, where r i {0, 1} m is said to be pseudoradom set for a radomized circuits C(x, r), x {0, 1}, r {0, 1} m computig some fuctio f : {0, 1} {0, 1} with error probability 0.1, if x, P r i [t] [C(x, r i ) f(x)] 0.2 Followig claim shows the existece of small pseudoradom set for family of circuits Claim 26. c,, m = c, r 1, r 2,, r t {0, 1} m, t 2c, such that for all radomized circuits C(x, r), x {0, 1}, r {0, 1} m computig some fuctio f : {0, 1} {0, 1} with error probability 0.1, ad C c x, P r i [t] [C(x, r i ) f(x)]