Theoretical Computer Science Cheat Sheet
                            Definitions                                           ...
Theoretical Computer Science Cheat Sheet
                                                              Identities Cont.   ...
Theoretical Computer Science Cheat Sheet
                                                                                 ...
Theoretical Computer Science Cheat Sheet
                       Trigonometry                                              ...
Theoretical Computer Science Cheat Sheet
              Number Theory                                                      ...
List of Software V1 00
List of Software V1 00
List of Software V1 00
List of Software V1 00
List of Software V1 00
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List of Software V1 00

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List of Software V1 00

  1. 1. Theoretical Computer Science Cheat Sheet Definitions Series iff ∃ positive c, n0 such that f (n) = O(g(n)) n n n n2 (n + 1)2 n(n + 1) n(n + 1)(2n + 1) 0 ≤ f (n) ≤ cg(n) ∀n ≥ n0 . i2 = i3 = i= , , . 2 6 4 i=1 i=1 i=1 iff ∃ positive c, n0 such that f (n) = Ω(g(n)) In general: f (n) ≥ cg(n) ≥ 0 ∀n ≥ n0 . n n 1 (n + 1)m+1 − 1 − (i + 1)m+1 − im+1 − (m + 1)im im = f (n) = Θ(g(n)) iff f (n) = O(g(n)) and m+1 i=1 i=1 f (n) = Ω(g(n)). n−1 m 1 m+1 im = Bk nm+1−k . f (n) = o(g(n)) iff limn→∞ f (n)/g(n) = 0. m+1 k i=1 k=0 iff ∀ > 0, ∃n0 such that lim an = a Geometric series: n→∞ |an − a| < , ∀n ≥ n0 . ∞ ∞ n cn+1 − 1 1 c |c| < 1, ci = ci = ci = , c = 1, , , least b ∈ R such that b ≥ s, sup S c−1 1−c 1−c ∀s ∈ S. i=0 i=0 i=1 ∞ n ncn+2 − (n + 1)cn+1 + c c greatest b ∈ R such that b ≤ |c| < 1. ici = ici = inf S , c = 1, , (c − 1)2 (1 − c)2 s, ∀s ∈ S. i=0 i=0 Harmonic series: lim inf{ai | i ≥ n, i ∈ N}. lim inf an n n n(n − 1) 1 n(n + 1) n→∞ n→∞ Hn − Hn = , iHi = . lim sup{ai | i ≥ n, i ∈ N}. i 2 4 lim sup an i=1 i=1 n→∞ n→∞ n n i n+1 1 n Hi = (n + 1)Hn − n, Hn+1 − Combinations: Size k sub- Hi = . k m m+1 m+1 sets of a size n set. i=1 i=1 n n Stirling numbers (1st kind): n n! n n n 1. 2. 3. k = 2n , = , = , Arrangements of an n ele- (n − k)!k! n−k k k k k=0 ment set into k cycles. n n−1 n−1 n−1 n n 4. 5. = , = + , n k k−1 k−1 Stirling numbers (2nd kind): k k k k Partitions of an n element n n−k n m n r+k r+n+1 6. 7. = , = , set into k non-empty sets. m−k m k k k n k=0 n 1st order Eulerian numbers: n n k n+1 r s r+s k 8. 9. = , = , Permutations π1 π2 . . . πn on n−k m m+1 k n {1, 2, . . . , n} with k ascents. k=0 k=0 k−n−1 n n n 10. 11. k n = (−1) , = = 1, 2nd order Eulerian numbers. k k 1 n k Cn Catalan Numbers: Binary n−1 n−1 n n 12. = 2n−1 − 1, 13. =k + , trees with n + 1 vertices. k−1 2 k k n n n n n 14. = (n − 1)!, 15. = (n − 1)!Hn−1 , 16. 17. ≥ = 1, , 1 2 n k k n n−1 n−1 n n n n n 1 2n 18. = (n − 1) 19. 20. 21. Cn = + , = = , = n!, , k−1 n−1 n−1 k k 2 k n+1 n k=0 n−1 n−1 n n n n n 22. 23. 24. + (n − k) = = 1, = , = (k + 1) , n−1 n−1−k k−1 0 k k k 0 n n n+1 1 if k = 0, 25. 26. = 2n − n − 1, 27. = 3n − (n + 1)2n + = , k 1 2 2 0 otherwise n m n n x+k n n+1 n n k 28. 29. (m + 1 − k)n (−1)k , 30. m! xn = , = = , n−m k n m k m k k=0 k=0 k=0 n n−k n n n n 31. 32. 33. (−1)n−k−m k!, = = 1, = 0 for n = 0, m k m 0 n k=0 n n−1 n−1 (2n)n n n 34. + (2n − 1 − k) 35. = (k + 1) , = , k−1 2n k k k k=0 n n x+n−1−k x n n+1 n k k 36. 37. (m + 1)n−k , = , = = x−n k 2n m+1 k m m k=0 k k=0
  2. 2. Theoretical Computer Science Cheat Sheet Identities Cont. Trees n n n Every tree with n n+1 n k k n−k 1k x n x+k 38. 39. = = n = n! , = , vertices has n − 1 x−n m+1 k m m k! m k 2n k k=0 k=0 k=0 edges. n n k+1 n n+1 k 40. 41. (−1)n−k , (−1)m−k , = = Kraft inequal- m k m+1 m k+1 m k k ity: If the depths m m m+n+1 n+k m+n+1 n+k of the leaves of 42. 43. = k , = k(n + k) , m k m k a binary tree are k=0 k=0 d1 , . . . , dn : n n+1 k n n+1 k 44. 45. (n − m)! for n ≥ m, (−1)m−k , (−1)m−k , = = n m k+1 m m k+1 m 2−di ≤ 1, k k m−n m−n n m+n m+k n m+n m+k i=1 46. 47. = , = , n−m n−m m+k n+k k m+k n+k k and equality holds k k n−k n−k only if every in- n +m k n n +m k n 48. 49. = , = . ternal node has 2 +m m k +m m k k k sons. Recurrences Master method: Generating functions: 1 T (n) − 3T (n/2) = n a ≥ 1, b > 1 T (n) = aT (n/b) + f (n), 1. Multiply both sides of the equa- tion by xi . 3 T (n/2) − 3T (n/4) = n/2 If ∃ > 0 such that f (n) = O(nlogb a− ) 2. Sum both sides over all i for . . . then . . . . . . which the equation is valid. T (n) = Θ(nlogb a ). 3log2 n−1 T (2) − 3T (1) = 2 3. Choose a generating function ∞ If f (n) = Θ(nlogb a ) then G(x). Usually G(x) = i=0 xi gi . T (n) = Θ(nlogb a log2 n). Let m = log2 n. Summing the left side 3. Rewrite the equation in terms of we get T (n) − 3m T (1) = T (n) − 3m = the generating function G(x). If ∃ > 0 such that f (n) = Ω(nlogb a+ ), T (n) − nk where k = log2 3 ≈ 1.58496. 4. Solve for G(x). and ∃c < 1 such that af (n/b) ≤ cf (n) Summing the right side we get 5. The coefficient of xi in G(x) is gi . for large n, then m−1 m−1 ni Example: 3i 3 =n . T (n) = Θ(f (n)). 2 2i gi+1 = 2gi + 1, g0 = 0. i=0 i=0 Substitution (example): Consider the Let c = 3 . Then we have Multiply and sum: following recurrence 2 gi+1 xi = 2gi xi + xi . i m−1 Ti+1 = 22 · Ti2 , T1 = 2. c −1m i n c =n i≥0 i≥0 i≥0 c−1 Note that Ti is always a power of two. i=0 We choose G(x) = i≥0 xi gi . Rewrite = 2n(clog2 n − 1) Let ti = log2 Ti . Then we have in terms of G(x): ti+1 = 2i + 2ti , t1 = 1. = 2n(c(k−1) logc n − 1) G(x) − g0 xi . = 2G(x) + Let ui = ti /2i . Dividing both sides of x = 2nk − 2n, i≥0 the previous equation by 2i+1 we get and so T (n) = 3n − 2n. Full history re- 2i k Simplify: ti+1 ti = i+1 + i . G(x) 1 currences can often be changed to limited i+1 2 2 2 = 2G(x) + . 1−x x history ones (example): Consider Substituting we find i−1 ui+1 = 1 + ui , u1 = 1 , Solve for G(x): 2 2 Ti = 1 + Tj , T0 = 1. x G(x) = . (1 − x)(1 − 2x) which is simply ui = i/2. So we find j=0 i−1 that Ti has the closed form Ti = 2i2 . Note that Expand this using partial fractions: i Summing factors (example): Consider 2 1 − Ti+1 = 1 + Tj . G(x) = x the following recurrence 1 − 2x 1 − x   j=0 T (n) = 3T (n/2) + n, T (1) = 1. Subtracting we find = x 2 xi  2i xi − Rewrite so that all terms involving T i i−1 are on the left side Ti+1 − Ti = 1 + Tj − 1 − i≥0 i≥0 Tj T (n) − 3T (n/2) = n. − 1)x i+1 i+1 j=0 j=0 = (2 . = Ti . Now expand the recurrence, and choose i≥0 a factor which makes the left side “tele- So gi = 2i − 1. And so Ti+1 = 2Ti = 2i+1 . scope”
  3. 3. Theoretical Computer Science Cheat Sheet √ √ 1+ 5 1− 5 ˆ π ≈ 3.14159, e ≈ 2.71828, γ ≈ 0.57721, ≈ 1.61803, ≈ −.61803 φ= φ= 2 2 2i i pi General Probability Continuous distributions: If 1 2 2 Bernoulli Numbers (Bi = 0, odd i = 1): b 1 1 1 −2, − 30 , 2 4 3 B0 = 1, B1 = B2 = 6, B4 = Pr[a < X < b] = p(x) dx, 1 1 5 B6 = 42 , B8 = − 30 , B10 = 66 . 3 8 5 a then p is the probability density function of Change of base, quadratic formula: 4 16 7 √ X. If −b ± b2 − 4ac loga x 5 32 11 Pr[X < a] = P (a), logb x = , . loga b 2a 6 64 13 then P is the distribution function of X. If Euler’s number e: 7 128 17 P and p both exist then 1 1 1 1 + 24 + 120 + · · · e=1+ + a 8 256 19 2 6 P (a) = p(x) dx. xn = ex . 9 512 23 −∞ lim 1 + n n→∞ Expectation: If X is discrete 10 1,024 29 1n 1 n+1 1+ n <e< 1+ n . E[g(X)] = g(x) Pr[X = x]. 11 2,048 31 e 11e 1 1n x =e− −O 1+ + . 12 4,096 37 2 n3 n 2n 24n If X continuous then ∞ ∞ 13 8,192 41 Harmonic numbers: E[g(X)] = g(x)p(x) dx = g(x) dP (x). 14 16,384 43 1, 3 , 11 , 25 , 137 , 49 , 363 , 761 , 7129 , . . . −∞ −∞ 2 6 12 60 20 140 280 2520 15 32,768 47 Variance, standard deviation: VAR[X] = E[X 2 ] − E[X]2 , 16 65,536 53 ln n < Hn < ln n + 1, 17 131,072 59 1 σ = VAR[X]. Hn = ln n + γ + O . n 18 262,144 61 For events A and B: Pr[A ∨ B] = Pr[A] + Pr[B] − Pr[A ∧ B] Factorial, Stirling’s approximation: 19 524,288 67 Pr[A ∧ B] = Pr[A] · Pr[B], ... 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 20 1,048,576 71 iff A and B are independent. 21 2,097,152 73 √ n n 1 Pr[A ∧ B] n! = 2πn 1+Θ . 22 4,194,304 79 e n Pr[A|B] = Pr[B] 23 8,388,608 83 Ackermann’s function and inverse:  For random variables X and Y : 24 16,777,216 89  2j i=1 E[X · Y ] = E[X] · E[Y ], a(i, j) = a(i − 1, 2) j=1 25 33,554,432 97  a(i − 1, a(i, j − 1)) i, j ≥ 2 if X and Y are independent. 26 67,108,864 101 E[X + Y ] = E[X] + E[Y ], α(i) = min{j | a(j, j) ≥ i}. 27 134,217,728 103 E[cX] = c E[X]. Binomial distribution: 28 268,435,456 107 Bayes’ theorem: n k n−k 29 536,870,912 109 q = 1 − p, Pr[X = k] = pq , Pr[B|Ai ] Pr[Ai ] k Pr[Ai |B] = n . 30 1,073,741,824 113 j=1 Pr[Aj ] Pr[B|Aj ] n n k n−k 31 2,147,483,648 127 E[X] = k pq = np. Inclusion-exclusion: k 32 4,294,967,296 131 n n k=1 Pr Xi = Pr[Xi ] + Poisson distribution: Pascal’s Triangle e−λ λk i=1 i=1 1 Pr[X = k] = , E[X] = λ. n k k! (−1)k+1 Pr Xij . 11 Normal (Gaussian) distribution: ii <···<ik j=1 k=2 121 1 2 2 e−(x−µ) /2σ , E[X] = µ. p(x) = √ Moment inequalities: 1331 2πσ 1 Pr |X| ≥ λ E[X] ≤ The “coupon collector”: We are given a , 14641 λ random coupon each day, and there are n 1 5 10 10 5 1 1 Pr X − E[X] ≥ λ · σ ≤ different types of coupons. The distribu- . λ2 1 6 15 20 15 6 1 tion of coupons is uniform. The expected Geometric distribution: 1 7 21 35 35 21 7 1 number of days to pass before we to col- q = 1 − p, Pr[X = k] = pq k−1 , lect all n types is 1 8 28 56 70 56 28 8 1 ∞ 1 nHn . 1 9 36 84 126 126 84 36 9 1 kpq k−1 = E[X] = . p 1 10 45 120 210 252 210 120 45 10 1 k=1
  4. 4. Theoretical Computer Science Cheat Sheet Trigonometry Matrices More Trig. C Multiplication: n (0,1) C = A · B, ci,j = ai,k bk,j . a b h b (cos θ, sin θ) k=1 C θ Determinants: det A = 0 iff A is non-singular. A (-1,0) (1,0) A c B det A · B = det A · det B, Law of cosines: c a c2 = a2 +b2 −2ab cos C. n (0,-1) det A = sign(π)ai,π(i) . B Area: π i=1 Pythagorean theorem: 2 × 2 and 3 × 3 determinant: C 2 = A2 + B 2 . A = 1 hc, 2 ab = ad − bc, = 1 ab sin C, Definitions: cd 2 sin a = A/C, cos a = B/C, c2 sin A sin B ab c = . b c ac ab csc a = C/A, sec a = C/B, −h 2 sin C f =g de +i e f df de sin a A cos a B Heron’s formula: gh i tan a = =, cot a = =. cos a B sin a A √ aei + bf g + cdh A = s · sa · sb · sc , = − ceg − f ha − ibd. Area, radius of inscribed circle: s = 1 (a + b + c), AB 1 Permanents: 2 AB, . 2 A+B+C n sa = s − a, ai,π(i) . perm A = Identities: sb = s − b, π i=1 1 1 sc = s − c. sin x = , cos x = , Hyperbolic Functions csc x sec x 1 More identities: Definitions: sin2 x + cos2 x = 1, tan x = , ex − e−x −x x 1 − cos x cot x e +e sin x = sinh x = , cosh x = , , 1 + tan2 x = sec2 x, 1 + cot2 x = csc2 x, 2 2 2 2 −x e −e x 1 tanh x = x , csch x = , 1 + cos x e + e−x −x , sin x = sin(π − x), π cos x = sinh x sin x = cos , 2 2 2 1 1 sech x = , coth x = . 1 − cos x cos x = − cos(π − x), −x , π cosh x tanh x tan x = cot 2 tan x = , 2 1 + cos x Identities: cot x = − cot(π − x), csc x = cot x − cot x, 1 − cos x 2 cosh2 x − sinh2 x = 1, tanh2 x + sech2 x = 1, = , sin x sin(x ± y) = sin x cos y ± cos x sin y, sin x coth2 x − csch2 x = 1, sinh(−x) = − sinh x, = , 1 + cos x cos(x ± y) = cos x cos y sin x sin y, tanh(−x) = − tanh x, cosh(−x) = cosh x, 1 + cos x tan x ± tan y cot x = , tan(x ± y) = 2 , 1 − cos x sinh(x + y) = sinh x cosh y + cosh x sinh y, 1 tan x tan y 1 + cos x cot x cot y 1 = , cot(x ± y) = cosh(x + y) = cosh x cosh y + sinh x sinh y, , sin x cot x ± cot y sin x 2 tan x = , sinh 2x = 2 sinh x cosh x, 1 − cos x sin 2x = 2 sin x cos x, sin 2x = , 1 + tan2 x eix − e−ix cosh 2x = cosh2 x + sinh2 x, cos 2x = cos2 x − sin2 x, cos 2x = 2 cos2 x − 1, sin x = , 2i cosh x − sinh x = e−x , 1 − tan2 x cosh x + sinh x = ex , eix + e−ix cos 2x = 1 − 2 sin2 x, cos 2x = , 1 + tan2 x cos x = , n ∈ Z, (cosh x + sinh x)n = cosh nx + sinh nx, 2 cot2 x − 1 eix − e−ix 2 tan x tan x = −i ix 2 sinh2 x 2 cosh2 x tan 2x = 2, cot 2x = , , = cosh x − 1, = cosh x + 1. 1 − tan x e + e−ix 2 cot x 2 2 e2ix − 1 sin(x + y) sin(x − y) = sin2 x − sin2 y, = −i 2ix , θ sin θ cos θ tan θ . . . in mathematics e +1 cos(x + y) cos(x − y) = cos2 x − sin2 y. you don’t under- sinh ix 0 0 1 0 sin x = , √ √ stand things, you 1 3 3 i π Euler’s equation: 6 2 2 3 just get used to √ √ = −1. cos x = cosh ix, eix = cos x + i sin x, iπ e 2 2 π them. 1 4 2 2 √ tanh ix √ – J. von Neumann v2.02 c 1994 by Steve Seiden 3 1 tan x = . π 3 i 3 2 2 sseiden@acm.org ∞ π 1 0 2 http://www.csc.lsu.edu/~seiden
  5. 5. Theoretical Computer Science Cheat Sheet Number Theory Graph Theory Notation: Definitions: The Chinese remainder theorem: There ex- ists a number C such that: E(G) Edge set Loop An edge connecting a ver- V (G) Vertex set tex to itself. C ≡ r1 mod m1 c(G) Number of components Directed Each edge has a direction. .. . G[S] Induced subgraph Simple Graph with no loops or .. . .. . deg(v) Degree of v multi-edges. C ≡ rn mod mn ∆(G) Maximum degree Walk A sequence v0 e1 v1 . . . e v . δ(G) Minimum degree Trail A walk with distinct edges. if mi and mj are relatively prime for i = j. χ(G) Chromatic number Path A trail with distinct Euler’s function: φ(x) is the number of χE (G) Edge chromatic number vertices. positive integers less than x relatively Gc Complement graph n Connected A graph where there exists prime to x. If i=1 pei is the prime fac- i Kn Complete graph a path between any two torization of x then Kn1 ,n2 Complete bipartite graph n vertices. pi i −1 (pi − 1). e r(k, ) Ramsey number φ(x) = Component A maximal connected i=1 subgraph. Geometry Euler’s theorem: If a and b are relatively Tree A connected acyclic graph. Projective coordinates: triples prime then Free tree A tree with no root. (x, y, z), not all x, y and z zero. 1 ≡ aφ(b) mod b. DAG Directed acyclic graph. (x, y, z) = (cx, cy, cz) ∀c = 0. Eulerian Graph with a trail visiting Fermat’s theorem: Cartesian Projective each edge exactly once. 1 ≡ ap−1 mod p. Hamiltonian Graph with a cycle visiting (x, y) (x, y, 1) The Euclidean algorithm: if a > b are in- y = mx + b (m, −1, b) each vertex exactly once. tegers then (1, 0, −c) Cut A set of edges whose re- x=c gcd(a, b) = gcd(a mod b, b). moval increases the num- Distance formula, Lp and L∞ ber of components. metric: n If i=1 pei is the prime factorization of x i Cut-set A minimal cut. (x1 − x0 )2 + (y1 − y0 )2 , then pi i +1 − 1 n Cut edge A size 1 cut. e 1/p |x1 − x0 |p + |y1 − y0 |p , S(x) = d= . k-Connected A graph connected with pi − 1 the removal of any k − 1 p 1/p i=1 d|x lim |x1 − x0 |p + |y1 − y0 | . vertices. p→∞ Perfect Numbers: x is an even perfect num- ∀S ⊆ V, S = ∅ we have Area of triangle (x0 , y0 ), (x1 , y1 ) k-Tough ber iff x = 2n−1 (2n −1) and 2n −1 is prime. k · c(G − S) ≤ |S|. and (x2 , y2 ): Wilson’s theorem: n is a prime iff x1 − x0 y1 − y0 (n − 1)! ≡ −1 mod n. k-Regular A graph where all vertices 1 2 abs x − x . 0 y2 − y0 have degree k. 2 M¨bius  o inversion: k-Factor A k-regular spanning Angle formed by three points: 1 if i = 1.  subgraph. 0 if i is not square-free. µ(i) = Matching A set of edges, no two of  (−1)r if i is the product of (x2 , y2 )  which are adjacent. r distinct primes. 2 Clique A set of vertices, all of If θ which are adjacent. G(a) = F (d), (x1 , y1 ) (0, 0) 1 Ind. set A set of vertices, none of (x1 , y1 ) · (x2 , y2 ) d|a which are adjacent. cos θ = . then Vertex cover A set of vertices which 12 a F (a) = µ(d)G . Line through two points (x0 , y0 ) cover all edges. d and (x1 , y1 ): d|a Planar graph A graph which can be em- beded in the plane. xy1 Prime numbers: ln ln n Plane graph An embedding of a planar x0 y0 1 = 0. pn = n ln n + n ln ln n − n + n graph. ln n x1 y1 1 n Area of circle, volume of sphere: +O , deg(v) = 2m. ln n A = πr2 , V = 4 πr3 . 3 v∈V n n 2!n If G is planar then n − m + f = 2, so π(n) = + + ln n (ln n)2 (ln n)3 If I have seen farther than others, f ≤ 2n − 4, m ≤ 3n − 6. it is because I have stood on the n +O . Any planar graph has a vertex with de- shoulders of giants. (ln n)4 gree ≤ 5. – Issac Newton

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