1 Summary: Binary and Logic

Transcription

1 1 Summary: Biary ad Logic Biary Usiged Represetatio : each 1-bit is a power of two, the right-most is for 2 0 : = = = Usiged Rage o bits is [0...(2 1)]. For =4 we have the rage [0=0000, 1=0001,..., 15=1111] Biary Two s Complemet Represetatio : same as usiged, except the power of two for the first bit (the sig bit) is with egative sig. Whe the siged bit is 0 its positive idetical with usiged trasformatio; whe its 1 that power of two is egative: = = = = = = = it is essetial to kow how may bits to use, to kow which bit is the sig bit for egatives: 7 bits two s complemet : -27 = bits two s complemet: -27 = Two s Complemet Rage o bits is [ (2 1 1)]. For = 4 we have the rage [-8=1000, -7=1001, -6=1010,... -1=1111, 0=0000, 1=0001,..., 7=0111] 3-Step rule to covert from base 10 to two s complemet egatives - write i biary 27 = flip the bits add oe additio works like i base 10 with carry bit 1+1 = write 0, carry 1 ; = write 1 carry 1 etc subtractio base2 : use additio of the egative value base 4 = 2 2 use every two bits i base 2 to make a bit i base 4: = = 0 (val = 2) (val = 1) (val = 1) 4 = = base 8 = 2 3 use every three bits i base 2 to make a bit i base 8: = = 0 (val = 4) (val = 5) 8 = 45 8 = base 16 = 2 4 use every four bits i base 2 to make a bit i base 16: = = (val = 2) (val = 5) 16 = = = = (val = 1) (val = 11) 16 = 1B 16 = = ay base, like base=3 : break the umber ito powers of 3, with coefficiets {0,1,2} = = =

2 LOGIC TABLE A B A B (AND) A B (OR) A B (XOR) (A B) (NAND) DNF (OR betwee clauses for out=1); CNF (OR betwee egated clauses for out=0). EXAMPLE: A B C output DN F CN F A B C A B C A B C A B C A B C A B C A B C A B C DNF = ( A B C) ( A B C) (A B C) (A B C) CNF = (A B C) (A B C) ( A B C) ( A B C) BOOLEAN ALGEBRA LAWS (1=true=T; 0=false=F) 2

3 LOGIC GATES 3

4 2 Summary : Mod, Number Theory, RSA divisio a to b 2 : r = a mod b a = bq+r; with quotiet q ad remaider r Z b = {0, 1, 2,..., b 1} recap mod operatios fast expoetiatio by writig expoet a sum of powers of 2. Example 3 2 = 9; 3 4 = 9 2 = 81 = 1 mod 10; 3 8 = 1 2 = 1 mod 10 ; 3 16 = 1 2 = 1 mod mod 10 = mod 10 = = 3 mod 10 = p e1 1 p e1 1 p e1 1...p et t uique decompositio ito primes gcd(a, b) = all commo (itersectio) primes (each with mi expoet) lcm(a, b) = uio of primes (each with max expoet) ab = all primes together (with sum of expoets) gcm(a, b) lcm(a, b) = ab a b meas a divides b same as a is factor of b same as b is multiple of a same as b = ak for some iteger k a, b have the same remaider mod if ad oly if divides their differece : a mod = b mod a b if prime p ab; the p a p b a, b are coprimes (or relatively prime) if they have o commo prime factors; the gcd(a, b) = 1 if ab ad a, coprimes gcd(a, ) = 1; the b if a ad m a ad gcd(, m) = 1; the m a after dividig a, b by their d = gcd(a, b), oe gets coprime umbers: gcd( a d, b d ) = 1 a has multiplicative iverse b = a 1 mod meas ab mod = 1. gcd(a, ) = 1 That is possible if ad oly if a iverse mod (if exists) ca be foud as a v 1 for iteger v with property a v = 1 mod (v = order of a). Tryig powers to obtai the order is iefficiet, ot practical for large. gcd-bezout-coefficiets (x, y) for (a, b) always exist to give the gcd(a, b) = ax + by. if a, b coprime, 1 = gcd(a, b) = ax + by. The x, y are the two iverses x = a 1 mod b ad y = b 1 mod a Euclid-Exteded fids x, y coefficiets by trasformig the problem(a, b) ito problem(b, r) recursively, ad the recursively-back computig the coefficiets. It is efficiet, eve for large a, b. Liear cipher y = ax + b mod works if ad oly if (a, ) are coprime, that is a 1 mod. decoder is x = (y b)a 1 mod The 4

5 Euler totiet ϕ() is the size of the set C ={remaiders coprime with }; i other words ϕ() = umber of coprimes smaller tha. Euler s theorem : a ϕ() = 1 mod for ay a C. So we have four ways to fid a 1, the iverse of a mod : 1) Brute force. Try differet values b < util oe works (ab = 1 mod ) 2) Best i practice. x, y = EuclidExted(a, ). The x = a 1 is the iverse mod. 3) Fid order v for a, so a v = 1 mod the a v 1 mod is the iverse of a. Cat do fast expoetiatio(v ukow); still usually faster tha method 1) 4) Best if ϕ() kow. ϕ() acts as a order for a (a ϕ() = 1), so the iverse is a 1 = a ϕ() 1. Power modulo is efficiet with fast expoetiatio. For primes p, ϕ(p) = p 1 so that theorem becomes Fermats theorem a p 1 = 1 mod p whe (a, p) coprimes Primality Test for. Try for several a < to see if a 1 mod = 1. if ay of the tests(a) gives NO, the certaily ot prime if all tests(a) gives YES, is likely prime (rare exceptios: Carmichael umbers) = pq (two primes) the ϕ() = (p 1)(q 1); so if a coprime with the a ϕ() = a (p 1)(q 1) = 1 mod or a (p 1)(q 1)k+1 = a mod for ay k RSA. if = pq (two large primes); e ad d = e 1 are each other iverse mod (p 1)(q 1) meas ed = 1 mod (p 1)(q 1). The a ed = a (p 1)(q 1)k+1 = a mod. is kow but the prime factors p, q are ot ad hard to fid. RSA public key for ecryptio is e. ENCRYPT(a) = a e mod RSA secret key for decryptio is d. DECRYPT(a e ) = (a e ) d mod = a RSA sigature: verify that oe has the correct secret key, by receivig (a, b = a d ) ad decryptig b with public key b e = (a d ) e mod = a Chiese Remider : if p, q are coprime, ay pair of remaiders (a Z p, b Z q ) correspods uiquely to a remaider x Z pq such that x mod p = a ad x mod q = b 5

6 3 Summary: Sets, Coutig, Permutatios & Combiatios Set builder otatio A = {positive itegers smaller tha 100 divisible with 7} = {x x Z; 0 < x < 100; 7 x} cout A via idexig A = {x x Z; 0 < x < 100; 7 x} = {7 i i Z; 1 i 14} A = 14 uio, itersectio, set differece, set symmetric differece A B = {x x A OR x B} A B = {x x A AND x B} A B = {x x A AND x / B} A B = (A B) (B A) = (A B) (A B) Cartesia Product A B = {(x, y) x A; y B} Product Rule: if ay elemet from A ca be combied with ay elemet from B, the the umber of combiatios is A B = A B power set of A is P(A) = the set of all subsets of A, icludig A ad ; P(A) = 2 A. Example: A = {x, y, z}. The P(A) = { ; {x}; {y}; {z}; {x, y}; {x, z}; {y, z}; {x, y, z}} iclusio-exclusio priciple : A B C = A + B + C A B A C B C + A B C Sum (Partitio) Rule : if = A B = A C = B C the A B C = A + B + C Pigeohole Priciple: if items are put ito k boxes, the at least oe box cotais at least k items Permutatios P (, k) ways to choose a sequece of k items out of. ORDER MATTERS. P (, k) = ( 1) ( 1)... ( k + 1) =! ( k)! Combiatios C(, k) = ( ( k) ways to choose a set of k items out of. ORDER DOESNT MATTER. ) k = ( 1) ( 1)... ( k + 1)/( k) =! k!( k)! P (, k) ca be thought of as choosig a set of k, the permute the elemets chose i all k! ways. Thus P (, k) = ( ) k k! =!! k! = k!( k)! ( k)! ( ) ( k = k) sice choosig k items to take is equivalet to choosig k to remai Biomial Theorem (x + y) = ( ) 0 x 0 y + ( ) 1 x 1 y ( ) x y 0 = k=0 ( k ) x k y k 6

7 x = 1, y = 1 : 2 = (1 + 1) = ( ( k=0 k) = ( 0) + ( 1) ) x = 1, y = 1 : 0 = ( 1 + 1) = ( ) k=0 k ( 1) k = ( ( 0) ( 1) ) ( 1) Pascal Formula ( ) ( +1 k = ( k) + ) k 1 Pascal Triagle applies his formula at every row: ( ( ( ( ( ( 0) 1) 2) 3) 4) 5) Balls i Bis: The umber of ways to throw idistiguishable balls ito m distiguishable bis B 1, B 2,..B m is couted by choosig m 1 bi-separator locatios out of + m 1 spots for balls ad separators. For example for = 10, m = 6 the throw with bi couts correspods to the followig choices of m 1 = 5 separator locatios amog m + 1 = 15 balls+separators: Balls i Bis cout is ( ) +m 1 m 1 circular permutatio : whe sittig people at a roud table the umber of ways to sit them (ot icludig rotatios) is by keepig oe chair-fixed ad permute the rest; thus ( 1)! ways. coutig with bijective fuctios: if a bijectio (pairig) ca be established betwee set A ad set B the they have the same size, A = B. A bijectio fuctio f has two properties - ijectivity (differet argumets produce differet values): x, y A; x y f(x) f(y) - surjectivity (all B are possible fuctio values): v B, x A, f(x) = v 7

8 4 Summary: Probability if Ω is a set of outcome/evets ad A Ω the uiform probability P r[x A] = A / Ω. There are 2 optios here for each outcome, either x A or x / A radom variables X, Y, X each partitio the space Ω by a certai criteria (for example X = object color, Y= object price, Z = object shape). They are called radom because ay object pulled at radom ca have ay of the values of X as color, ay value of Y as price, ad ay Z as shape. For short otatio we say P r(x, Y ) = P r(x = X, y = Y ) is the probability that a radom object has color x ad price y Bayes Theorem P r(y X) P (X) = P r(x, Y ) = P r(x Y ) P (Y ) which meas probability to have a particular (X = red, Y = 100) object is the probability to have (X = red) times probability to have (Y = 100, give that X = red) or vice versa. The equality is the same as sayig P r(y X) = margializatio of variable Y over variable X: P r(y ) = x P r(x = X) P r(x, Y ) = x P r(x = X) P r(y X). If X is biary with oly two possible values (X; X) (for example pass vs fail ) the we have P r(y ) = P r(x) P r(y X) + P r( X) P r(y X) with margializatio of Y over biary radom variable X we ca write Bayes as P r(x Y ) = P r(y X) P (X) P (Y ) = P r(y X) P (X) P r(x) P r(y X)+P r( X) P r(y X) P r(x Y ) P (Y ) P (X) Idepedet variables X, Y meas P (Y X) = P (Y ) which is to say Y does ot deped o X (price does ot deped o color). From Bayes this meas also P (X Y ) = P (X) ad P (X, Y ) = P (X) P (Y ) expected value (or mea) for a umeric-value radom variable is the average of values weighted by probabilities: E[X] = x x P r(x = X) = x x P r(x) expectatio ALWAYS distributes over sum, eve if variables are ot idepedet. But they have to have umeric values E[X + Y + Z] = E[X] + E[Y ] + E[Z] variace = avg distace-to-mea 2 weighted by probabilities var[x] = x (x mea)2 P r(x = X) = E[(X E[X]) 2 ] = E[X 2 + E 2 [X] 2XE[X]] = E[X 2 ] + 2E 2 [X] 2E 2 [X] = E[X 2 ] E 2 [X] variace distributes over sum ONLY IF X, Y are idepedet X, Y idepedet var(x + Y ) = var(x) + var(y ) etropy = radomess of X (the more radom, the higher the etropy) H[X] = x P r(x) log( 1 P r(x) ) Markov Chais 8

9 5 Summary: Order of Growth For Positive Fuctios i order of growth we care about the geeral asymptotic (large ) behaviour of f(). f() = 3 2 2log() 5 grows like 2 For example f = O(g) BIG-O meas f less or equal with g asymptotically. Formally f() C g() for 0 ad a costat C > 0. We ca pick C ad 0 to aythig we wat but oce picked they cat chage Example: f() = log(); g() = 2 f = O(g) Example: f() = ; g() = f = O(g) f = Ω(g) BIG-OMEGA meas f bigger or equal with g asymptotically. Formally f() D g() for 0 ad a costat D > 0. We ca pick D ad 0 to aythig we wat but oce picked they cat chage Example: f() = 2 log(); g() = 2 f = Ω(g) Example: f() = 2 ; g() = f = Ω(g) f = Θ(g) BIG-THETA meas f grows equal with g asymptotically. Formally D g() f() C g() for 0 ad a costats C > D > 0. We ca pick C > D > 0 ad 0 to aythig we wat. The iequality ca be stated i reverse D f() g() C f() for differet costats C, D it is logically equivalet. Example: f() = ; g() = f = Θ(g) Example: f() = ; g() = f = Θ(g) Example: f() = log a (); g() = log b ()(a > b > 1) f = Θ(g) f = o(g) small-o meas f strictly less tha g asymptotically. Formally f() < C g() for ANY costat C > 0 ad a properly chose C. Such iequality must be prove for ay costat C; equivaletly it meas lim f()/g() = 0 Example: f() = log(); g() = 2 f = o(g) Example: f() = ; g() = f = o(g) Example: f() = 2 ; g() = f = o(g) Example: f() = ; g() =! f = o(g) f = ω(g) small-omega meas f strictly bigger tha g asymptotically. Formally f() > D g() for ANY costat D > 0 ad a properly chose D. Such iequality must be prove for ay costat D; equivaletly it meas lim f()/g() = Example: f() = log(); g() = f = ω(g) Example: f() = ; g() = f = ω(g) i geeral if < asy stads for asymptotic behaviour we have log c () < asy < asy log c () < asy 2 < asy 3 < asy... < asy < asy 2 < asy 3 < asy... < asy! < asy 9

10 6 Summary: Sequeces, Series, Iductio liear sum of idices k=1 k = (+1) 2 arithmetic progressio sequece x k = ak + b = (a + b, 2a + b, 3a + b, 4a + b,...) Characteristic : differece betwee values (delta) is costat a Partial sum is k=1 x k = k=1 (ak + b) = a k=1 k + b = a (+1) + b 2 quadratic sum of idices k=1 k2 = (+1)(2+1) 6 quadratic progressio sequece x k = ak 2 + bk + c = (a + b + c, 4a + 2b + c, 9a + 3b + c, 16a + 4b + c,...) Characteristic : differece betwee values (delta) is arithmetic progressio 2ak + (a + b) Partial sum is k=1 x k = k=1 (ak2 + bk + c) = a k=1 k2 + b k=1 k + c = a (+1)(2+1) + b (+1) + c 6 2 geometric sum of idices with base r 1: k=0 rk = r+1 1 r 1 - if base r > 1: r+1 1 r 1 = Θ(r ) - if base r < 1: k=0 rk = lim > r +1 1 r 1 = 1 - if base r = 1: k=0 rk = = Θ(1) 1 r k=0 1 = + 1 = Θ() geometric progressio sequece x k = ar k = (a, ar, ar 2, ar 3,...) Characteristic : ratio betwee values is costat r Partial sum for r 1 is k=1 x k = k=1 ark = a k=1 rk = a r+1 1 r 1 harmoic series = k=1 k l() = Θ(log()) telescopig series k=1 (a k a k+1 ) = (a 1 a 2 ) + (a 2 a 3 ) + (a 4 a 5 ) (a a +1 ) = a 1 a +1-1 k=1 = k(k+1) k=1 ( 1 1 ) = 1 1 k k log( k k=1 ) = k+1 k=1 log( k ) = k+1 k=1 (log(k) log(k + 1)) = log(1) log( + 1) = log( + 1) Iductio formula proofs S = k=1 (...) = f() For example S = (+1) k=1 k = f() = 2 -Base case = 1, 2 : verify maually that S 1 = f(1), S 2 = f(2), etc -Iductio Step (weak). Statemet : S = f() S +1 = f( + 1) -Iductio Step proof. Idea: relate S +1 to S S +1 = S + delta(s +1 S ) = f() + delta(s +1 S ) =...(algebra)... = f( + 1) -Iductio Step (strog). Statemet : {S 1 = f(1); S 2 = f(2);...; S = f()} S +1 = f( + 1) -Iductio Step proof. Idea: relate S +1 to several previous S k (k ) S +1 = S +delta(s +1 S, S 1, etc) = f()+delta(s +1 S S 1...) =...(algebra)... = f(+1) 10

11 Iductio logic proofs S = predicate(). For example S = Ay biary three with vertices has depth at least log() -Base case = 1, 2,etc : verify maually the predicates S(1), S(2), etc. Might eed more tha 2 base cases -Iductio Step (weak). Statemet : S TRUE S +1 TRUE -Iductio Step proof. Idea: relate S +1 to S Start with a cofiguratio appropriate for S +1 (for example a tree with + 1 vertices). The reduce coveietly the setup to the previous size (for example igore a vertice). Apply S for this reduced problem, the usig the result coveietly deduce S +1 -Iductio Step (strog). Statemet : {S, S 1,,,, S base } TRUE S +1 TRUE -Iductio Step proof. Idea: relate S +1 to several previous S k, k Start with a cofiguratio appropriate for S +1 (for example a tree with + 1 vertices). The reduce coveietly the setup to several prior sizes k (for example remove a edge to get two smaller trees). Apply S k for these reduced problems, the usig the result coveietly deduce S +1 11

12 7 Summary : Algorithms ad Recurreces Liear Uordered Search : search the list/array util value foud or array fiished -best case : foud o first elemet Θ(1) -average case : foud about i the middle Θ(/2) = Θ() -worst case : ot foud after checkig all values = Θ() Liear Ordered Search : search the sorted list/array util value foud or array fiished -best case : foud o first elemet Θ(1) -average case : foud about i the middle, or value checked is already too high so give up Θ(/2) = Θ() -worst case : ot foud after checkig all values = Θ() Biary Search : search the sorted array by checkig the middle value ad recurse o appropriate half - T () = 1 + T (/2): 1 for the compariso ad T (/2) for the recursio o oe of the halves -best case : foud o first elemet Θ(1) -worst case : ot foud after halvig all the way to oe elemet = Θ(log()) -average case : foud after about half of the recursive calls Θ(log()) Bubble Sort: sort a array by fixig bubbles A[i], A[i + 1] that are i icorrect order, as log as they exist. -best case : o bubbles eed fixig, so a pass through is ecessary to verify, thus Θ() -worst case : all ( 2) = ( 1)/2 bubbles eeds fixig (array is fully backwards) = Θ( 2 ) -average case : about half of bubbles eeds fixig Θ( 2 ) Selectio Sort: fid array miimum, output it, remove it, the repeat for remaiig elemets, etc -fidig miimum i k elemets is liear Θ(k) -for each k =, 1, 2,..., 1 we eed liear effort, so total that is about + ( 1) = Θ( 2 ) -T () = + T ( 1); for fidig miimum, ad T ( 1) for recursig to a problem of size 1 Isert Sort: keep previously elemets sorted ad isert the ew elemet i the correct spot by swappig it from the right, util all elemets are sorted -OUT P UT k = [a 1, a 2,...a k ] sorted -OUT P UT k+ext = [a 1, a 2,...a k, v] compare v to the left ad swap if ecessary, util v is i the correct spot, i.e. whe the elemet o the left is ot bigger tha v -repeat for all elemets i iput order -best case: already sorted, o swaps; Θ() -worst case: backwards sorted, all swaps; Θ( 2 ) -average case: about half of all swaps; Θ( 2 ) Merge Sort: split i half, recurse o both sides to sort them, the merge the sorted halves ito a sorted output -T () = + 2T (/2); for mergig sorted halves, ad twice T (/2) for recursig to both half-sides -best/worst/average case: solvig the recurrece gives T () = Θ( log()) - optimal sortig time based o comparisos is Θ( log()), so Merge Sort is optimal i that sese. 12

13 Quick Sort: pick a radom a elemet v ad put it i its correct sorted positio p (v = A[p]); rearrage the array so the elemet left to the pivot p are smaller tha A[p], ad the oes right to the pivot are larger tha A[p] - the rearragemet takes Θ() - recurse o both sides, which ow ca be sorted idepedetly - T () = + T (p 1) + T ( p); for rearragemet, T (p 1) for recurrece o left side; T ( p) for recurrece o right side -best/avg case: pivot roughly i the middle; Θ( log()) -worst case: pivot always at extremes; Θ( 2 ) Sovig recurreces of the form T () = at (/b) + c for the order of growth: apply the recurrece iteratively few times to fid a patter for k iteratios, the figure out how the patter looks for the last/biggest k T () = at ( ) + b c = a[at ( ) + ( b 2 b )c ] + c = a 2 T ( ) + a( b 2 b )c + c = a 2 [at ( ) + ( ) c ] + a( b 3 b 2 b )c + c = a 3 T ( ) + a 2 ( ) c + a( b 3 b 2 b )c + c = a 3 [at ( ) + ( ) c ] + a 2 ( ) c + a( b 4 b 3 b 2 b )c + c = a 4 T ( ) + a 3 ( ) c + a 2 ( ) c + a( b 4 b 3 b 2 b )c + c (geeral patter after k iteratios) = a k T ( ) + k 1 a i ( ) c b k b i i= (last k = log b ()) log = a log b () b () 1 T (1) + a i ( ) c b i i=0 log = log b (a) T (1) + c b () 1 ( a ) i b c i=0 -if a = b c (c = log b (a)) that becomes Θ( c + c log b ()) = Θ( c log()) -if r = a/b c 1, applyig geometric series formula for base r, that becomes Θ( log b (a) + c rlog b () 1 r 1 ) -if r = a/b c < 1 a < b c c > log b (a)), it becomes Θ( log b (a) + c 1 1 r ) = Θ(log b (a) + c ) = Θ( c ) - if r = a/b c > 1 a > b c c < log b (a)) it becomes Θ( log b (a) + c r log b () ) = Θ( log b (a) + c log b (r) ) = = Θ( log b (a) + c log b (a) log b (bc) ) = Θ( log b (a) + c+log b (a) c ) = = Θ( log b (a) ) Fiboacci base F 0 = 0; F 1 = 1 ad recurrece F +1 = F + F 1, satisfies F = Θ(φ ) where φ is the positive root of quadratic equatio φ 2 = φ

14 8 Summary: Graphs Graph G = (V, E) has set of vertices V ad set of edges E; each edge e = (u, v) is a pair of vertices. Graphs are udirected if edges are udirected (u, v) = (v, u) (for all edges) or directed if ((u, v) (v, u)). Graphs ca be represeted by adjacecy matrix where A uv = 1 idicates edge (u, v) exists, or by liked lists per vertex. degree(u) = umber of edges icidet vertex u. Sum off all degrees is twice the umber of edges, because each edge (u, v) cotributes +1 to both degrees of u ad v path(u, v) = a sequece of edges startig from u edig i v. If graph is directed, edges have to be cosidered i their directio (see picture left for path ) cycle is a path that begis ad eds at the same vertex u (see picture right for cycle 2-5-4). udirected cycle that passes through u ad v ca be thought tas two differet paths betwee u ad v A coected compoet : a subgraph where there is a path betwee ay two odes. Graphs ca have several coected compoets, which of course are ot coected betwee them (left picture for udirected compoets; right for directed compoets) tree= coected graph with o cycles. A tree with V vertices must have exactly V 1 edges: less would discoect it; more would form cycles. Spaig tree = a tree that icludes all graph vertices. two below draw with red edges) There ca be multiple spaig trees (see 14

15 BFS=Breadth-First-Search traversal : start at a root-vertex (wave 0), look for immediate eighbors (wave 1); look at their ew eighbors (wave 2) etc. Edges used to discover eighbors (blue i the picture below) are called the BFS-edges ad form the BFS-tree BFS: the shortest path from u to v (by umber of edges) ca be foud ruig BFS from root u ad check the wave umber that fids v. If v ot foud, it meas that there is o path from u to v, or v is ureachable from u. DFS=Depth-First-Search traversal: go as deep as possible o the curret brach, before comig back to follow other braches. A immediate eighbor to the root is discovered oly oly after the etire subtree of the curret eighbor is explored (see picture). Edges used for DFS advace are marked i red ad form the DFS tree ; the picture below cotais two DFS trees because the first oe startig i A could ot explore the whole graph. Each vertes is marled with a discovery time ad a fiishig time. 15