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 [ ]. 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 [ ]. 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 = = 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 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 gcda, b = all commo itersectio primes each with mi expoet lcma, b = uio of primes each with max expoet ab = all primes together with sum of expoets gcma, b lcma, 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 gcda, b = 1 if ab ad a, coprimes gcda, = 1; the b if a ad m a ad gcd, m = 1; the m a after dividig a, b by their d = gcda, b, oe gets coprime umbers: gcd a d, b d = 1 a has multiplicative iverse b = a 1 mod meas ab mod = 1. gcda, = 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 gcda, b = ax + by. if a, b coprime, 1 = gcda, 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 problema, b ito problemb, 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 ba 1 mod The 3

4 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 = EuclidExteda,. 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 expoetiatiov 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, ϕp = 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 testsa gives NO, the certaily ot prime if all testsa gives YES, is likely prime rare exceptios: Carmichael umbers = pq two primes the ϕ = p 1q 1; so if a coprime with the a ϕ = a p 1q 1 = 1 mod or a p 1q 1k+1 = a mod for ay k RSA. if = pq two large primes; e ad d = e 1 are each other iverse mod p 1q 1 meas ed = 1 mod p 1q 1. The a ed = a p 1q 1k+1 = a mod. is kow but the prime factors p, q are ot ad hard to fid. RSA public key for ecryptio is e. ENCRYPTa = a e mod RSA secret key for decryptio is d. DECRYPTa 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 4

5 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 PA = the set of all subsets of A, icludig A ad ; PA = 2 A. Example: A = {x, y, z}. The PA = { ; {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 = k + 1 =! k! Combiatios C, k = k ways to choose a set of k items out of. ORDER DOESNT MATTER. k = 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 5

6 x = 1, y = 1 : 2 = = k=0 k = x = 1, y = 1 : 0 = = k=0 k 1 k = Pascal Formula +1 k = k + k 1 Pascal Triagle applies his formula at every row: 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 fx fy - surjectivity all B are possible fuctio values: v B, x A, fx = v 6