Theoretical Computer Science Cheat Sheet - Theory of Algorithms | CSC 721, Study notes of Algorithms and Programming

Download Theoretical Computer Science Cheat Sheet - Theory of Algorithms | CSC 721 and more Study notes Algorithms and Programming in PDF only on Docsity! Theoretical Computer Science Cheat Sheet De nitions Series f(n) = O(g(n)) i 9 positive c; n0 such that 0  f(n)  cg(n) 8n  n0. nX i=1 i = n(n+ 1) 2 ; nX i=1 i2 = n(n+ 1)(2n+ 1) 6 ; nX i=1 i3 = n2(n+ 1)2 4 : In general: nX i=1 im = 1 m+ 1  (n+ 1)m+1 1 nX i=1 (i+ 1)m+1 im+1 (m+ 1)im
 n1X i=1 im = 1 m+ 1 mX k=0  m+ 1 k  Bkn m+1k: Geometric series: nX i=0 ci = cn+1 1 c 1 ; c 6= 1; 1X i=0 ci = 1 1 c ; 1X i=1 ci = c 1 c ; c < 1; nX i=0 ici = ncn+2 (n+ 1)cn+1 + c (c 1)2 ; c 6= 1; 1X i=0 ici = c (1 c)2 ; c < 1: Harmonic series: Hn = nX i=1 1 i ; nX i=1 iHi = n(n+ 1) 2 Hn n(n 1) 4 : nX i=1 Hi = (n+ 1)Hn n; nX i=1  i m  Hi =  n+ 1 m+ 1  Hn+1 1 m+ 1  : f(n) = (g(n)) i 9 positive c; n0 such that f(n)  cg(n)  0 8n  n0. f(n) = (g(n)) i f(n) = O(g(n)) and f(n) = (g(n)). f(n) = o(g(n)) i limn!1 f(n)=g(n) = 0. lim n!1 an = a i 8 2 R, 9n0 such thatjan aj < , 8n  n0. supS least b 2 R such that b  s, 8s 2 S. inf S greatest b 2 R such that b  s, 8s 2 S. lim inf n!1 an lim n!1 inffai j i  n; i 2 Ng. lim sup n!1 an lim n!1 supfai j i  n; i 2 Ng. n k
Combinations: Size k sub- sets of a size n set. n k  Stirling numbers (1st kind): Arrangements of an n ele- ment set into k cycles. 1.  n k  = n! (n k)!k! ; 2. nX k=0  n k  = 2n; 3.  n k  =  n n k  ; 4.  n k  = n k  n 1 k 1  ; 5.  n k  =  n 1 k  +  n 1 k 1  ; 6.  n m  m k  =  n k  n k m k  ; 7. X kn  r + k k  =  r + n+ 1 n  ; 8. nX k=0  k m  =  n+ 1 m+ 1  ; 9. nX k=0  r k  s n k  =  r + s n  ; 10.  n k  = (1)k  k n 1 k  ; 11.  n 1  =  n n  = 1; 12.  n 2  = 2n1 1; 13.  n k  = k  n 1 k  +  n 1 k 1  ;  n k Stirling numbers (2nd kind): Partitions of an n element set into k non-empty sets. n k 1st order Eulerian numbers: Permutations 12 : : : n on f1; 2; : : : ; ng with k ascents. n k 2nd order Eulerian numbers. Cn Catlan Numbers: Binary trees with n+ 1 vertices. 14.  n 1  = (n 1)!; 15.  n 2  = (n 1)!Hn1; 16.  n n  = 1; 17.  n k    n k  ; 18.  n k  = (n 1)  n 1 k  +  n 1 k 1  ; 19.  n n 1  =  n n 1  =  n 2  ; 20. nX k=0  n k  = n!; 21. Cn = 1 n+ 1  2n n  ; 22.  n 0  =  n n 1  = 1; 23.  n k  =  n n 1 k  ; 24.  n k  = (k + 1)  n 1 k  + (n k)  n 1 k 1  ; 25.  0 k  = n 1 if k = 0, 0 otherwise 26.  n 1  = 2n n 1; 27.  n 2  = 3n (n+ 1)2n +  n+ 1 2  ; 28. xn = nX k=0  n k  x+ k n  ; 29.  n m  = mX k=0  n+ 1 k  (m+ 1 k)n(1)k; 30. m!  n m  = nX k=0  n k  k nm  ; 31.  n m  = nX k=0  n k  n k m  (1)nkmk!; 32.  n 0  = 1; 33.  n n  = 0 for n 6= 0; 34.  n k  = (k + 1)  n 1 k  + (2n 1 k)  n 1 k 1  ; 35. nX k=0  n k  = (2n)n 2n ; 36.  x x n  = nX k=0  n k  x+ n 1 k 2n  ; 37.  n+ 1 m+ 1  = X k  n k  k m  = nX k=0  k m  (m+ 1)nk; Theoretical Computer Science Cheat Sheet Identities Cont. Trees 38.  n+ 1 m+ 1  = X k  n k  k m  = nX k=0  k m  nnk = n! nX k=0 1 k!  k m  ; 39.  x x n  = nX k=0  n k  x+ k 2n  ; 40.  n m  = X k  n k  k + 1 m+ 1  (1)nk; 41.  n m  = X k  n+ 1 k + 1  k m  (1)mk; 42.  m+ n+ 1 m  = mX k=0 k  n+ k k  ; 43.  m+ n+ 1 m  = mX k=0 k(n+ k)  n+ k k  ; 44.  n m  = X k  n+ 1 k + 1  k m  (1)mk; 45. (nm)!  n m  = X k  n+ 1 k + 1  k m  (1)mk; for n  m, 46.  n nm  = X k  m n m+ k  m+ n n+ k  m+ k k  ; 47.  n nm  = X k  m n m+ k  m+ n n+ k  m+ k k  ; 48.  n `+m  `+m `  = X k  k `  n k m  n k  ; 49.  n `+m  `+m `  = X k  k `  n k m  n k  : Every tree with n vertices has n 1 edges. Kraft inequal- ity: If the depths of the leaves of a binary tree are d1; : : : ; dn: nX i=1 2di  1; and equality holds only if every in- ternal node has 2 sons. Recurrences Master method: T (n) = aT (n=b) + f(n); a  1; b > 1 If 9 > 0 such that f(n) = O(nlogb a) then T (n) = (nlogb a): If f(n) = (nlogb a) then T (n) = (nlogb a log2 n): If 9 > 0 such that f(n) = (nlogb a+), and 9c < 1 such that af(n=b)  cf(n) for large n, then T (n) = (f(n)): Substitution (example): Consider the following recurrence Ti+1 = 2 2i
T 2i ; T1 = 2: Note that Ti is always a power of two. Let ti = log2 Ti. Then we have ti+1 = 2 i + 2ti; t1 = 1: Let ui = ti=2 i. Dividing both sides of the previous equation by 2i+1 we get ti+1 2i+1 = 2i 2i+1 + ti 2i : Substituting we nd ui+1 = 1 2 + ui; u1 = 12; which is simply ui = i=2. So we nd that Ti has the closed form Ti = 2 i2i1 . Summing factors (example): Consider the following recurrence Ti = 3Tn=2 + n; T1 = n: Rewrite so that all terms involving T are on the left side Ti 3Tn=2 = n: Now expand the recurrence, and choose a factor which makes the left side \tele- scope" 1 T (n) 3T (n=2) = n
3 T (n=2) 3T (n=4) = n=2
... ... ... 3log2 n1 T (2) 3T (1) = 2
3log2 n T (1) 0 = 1
Summing the left side we get T (n). Sum- ming the right side we get log2 nX i=0 n 2i 3i: Let c = 32 and m = log2 n. Then we have n mX i=0 ci = n  cm+1 1 c 1  = 2n(c
clog2 n 1) = 2n(c
ck logc n 1) = 2nk+1 2n  2n1:58496 2n; where k = (log2 3 2 ) 1. Full history recur- rences can often be changed to limited his- tory ones (example): Consider the follow- ing recurrence Ti = 1 + i1X j=0 Tj ; T0 = 1: Note that Ti+1 = 1 + iX j=0 Tj : Subtracting we nd Ti+1 Ti = 1 + iX j=0 Tj 1 i1X j=0 Tj = Ti: And so Ti+1 = 2Ti = 2 i+1. Generating functions: 1. Multiply both sides of the equa- tion by xi. 2. Sum both sides over all i for which the equation is valid. 3. Choose a generating function G(x). Usually G(x) = P1 i=0 x i. 3. Rewrite the equation in terms of the generating function G(x). 4. Solve for G(x). 5. The coecient of xi in G(x) is gi. Example: gi+1 = 2gi + 1; g0 = 0: Multiply and sum:X i0 gi+1x i = X i0 2gix i + X i0 xi: We choose G(x) = P i0 x i. Rewrite in terms of G(x): G(x) g0 x = 2G(x) + X i0 xi: Simplify: G(x) x = 2G(x) + 1 1 x : Solve for G(x): G(x) = x (1 x)(1 2x) : Expand this using partial fractions: G(x) = x  2 1 2x 1 1 x  = x 0 @2X i0 2ixi X i0 xi 1 A = X i0 (2i+1 1)xi+1: So gi = 2 i 1. Theoretical Computer Science Cheat Sheet Number Theory Graph Theory The Chinese remainder theorem: There ex- ists a number C such that: C  r1 mod m1 ... ... ... C  rn mod mn if mi and mj are relatively prime for i 6= j. Euler's function: (x) is the number of positive integers less than x relatively prime to x. If Qn i=1 p ei i is the prime fac- torization of x then (x) = nY i=1 pei1i (pi 1): Euler's theorem: If a and b are relatively prime then 1  a(b) mod b: Fermat's theorem: 1  ap1 mod p: The Euclidean algorithm: if a > b are in- tegers then gcd(a; b) = gcd(a mod b; b): If Qn i=1 p ei i is the prime factorization of x then S(x) = X djx d = nY i=1 pei+1i 1 pi 1 : Perfect Numbers: x is an even perfect num- ber i x = 2n1(2n1) and 2n1 is prime. Wilson's theorem: n is a prime i (n 1)!  1 mod n: Mobius inversion: (i) = 8>< >: 1 if i = 1. 0 if i is not square-free. (1)r if i is the product of r distinct primes. If G(a) = X dja F (d); then F (a) = X dja (d)G a d  : Prime numbers: pn = n lnn+ n ln lnn n+ n ln lnn lnn +O  n lnn  ; (n) = n lnn + n (lnn)2 + 2!n (lnn)3 +O  n (lnn)4  : De nitions: Loop An edge connecting a ver- tex to itself. Directed Each edge has a direction. Simple Graph with no loops or multi-edges. Walk A sequence v0e1v1 : : : e`v`. Trail A walk with distinct edges. Path A trail with distinct vertices. Connected A graph where there exists a path between any two vertices. Component A maximal connected subgraph. Tree A connected acyclic graph. Free tree A tree with no root. DAG Directed acyclic graph. Eulerian Graph with a trail visiting each edge exactly once. Hamiltonian Graph with a path visiting each vertex exactly once. Cut A set of edges whose re- moval increases the num- ber of components. Cut-set A minimal cut. Cut edge A size 1 cut. k-Connected A graph connected with the removal of any k 1 vertices. k-Tough 8S  V; S 6= ; we have k
c(G S)  jSj. k-Regular A graph where all vertices have degree k. k-Factor A k-regular spanning subgraph. Matching A set of edges, no two of which are adjacent. Clique A set of vertices, all of which are adjacent. Ind. set A set of vertices, none of which are adjacent. Vertex cover A set of vertices which cover all edges. Planar graph A graph which can be em- beded in the plane. Plane graph An embedding of a planar graph.X v2V deg(v) = 2m: If G is planar then nm+ f = 2, so f  2n 4; m  3n 6: Any planar graph has a vertex with de- gree  5. Notation: E(G) Edge set V (G) Vertex set c(G) Number of components G[S] Induced subgraph deg(v) Degree of v
(G) Maximum degree (G) Minimum degree (G) Chromatic number E(G) Edge chromatic number Gc Complement graph Kn Complete graph Kn1;n2 Complete bipartite graph r(k; `) Ramsey number Geometry Projective coordinates: triples (x; y; z), not all x, y and z zero. (x; y; z) = (cx; cy; cz) 8c 6= 0: Cartesian Projective (x; y) (x; y; 1) y = mx+ b (m;1; b) x = c (1; 0;c) Distance formula, Lp and L1 metric:p (x1 x0)2 + (x1 x0)2;jx1 x0jp + jx1 x0jp1=p; lim p!1 jx1 x0jp + jx1 x0jp1=p: Area of triangle (x0; y0), (x1; y1) and (x2; y2): 1 2 abs x1 x0 y1 y0x2 x0 y2 y0 : Angle formed by three points:B cos  = (x1; y1)
(x2; y2) `1`2 : Line through two points (x0; y0) and (x1; y1): x y 1 x0 y0 1 x1 y1 1 = 0: Area of circle, volume of sphere: A = r2; V = 43r 3: If I have seen farther than others, it is because I have stood on the shoulders of giants. { Issac Newton Theoretical Computer Science Cheat Sheet  Calculus Wallis' identity:  = 2
2
2
4
4
6
6


1
3
3
5
5
7


Brouncker's continued fraction expansion:  4 = 1 + 12 2 + 3 2 2+ 5 2 2+ 7 2 2+


Gregrory's series:  4 = 1 13 + 15 17 + 19


Newton's series:  6 = 1 2 + 1 2
3
23 + 1
3 2
4
5
25 +


Sharp's series:  6 = 1p 3  1 1 31
3 + 1 32
5 1 33
7 +


 Euler's series: 2 6 = 1 12 + 1 22 + 1 32 + 1 42 + 1 52 +


2 8 = 1 12 + 1 32 + 1 52 + 1 72 + 1 92 +


2 12 = 1 12 122 + 132 142 + 152


Derivatives: 1. d(cu) dx = c du dx ; 2. d(u+ v) dx = du dx + dv dx ; 3. d(uv) dx = u dv dx + v du dx ; 4. d(un) dx = nun1 du dx ; 5. d(u=v) dx = v du dx
u dvdx
v2 ; 6. d(ecu) dx = cecu du dx ; 7. d(cu) dx = (ln c)cu du dx ; 8. d(ln u) dx = 1 u du dx ; 9. d(sinu) dx = cosu du dx ; 10. d(cosu) dx = sinudu dx ; 11. d(tanu) dx = sec2 u du dx ; 12. d(cotu) dx = csc2 u du dx ; 13. d(secu) dx = tanu secu du dx ; 14. d(cscu) dx = cotu cscudu dx ; 15. d(arcsinu) dx = 1p 1 u2 du dx ; 16. d(arccosu) dx = 1p 1 u2 du dx ; 17. d(arctanu) dx = 1 1 u2 du dx ; 18. d(arccotu) dx = 1 1 u2 du dx ; 19. d(arcsecu) dx = 1 u p 1 u2 du dx ; 20. d(arccscu) dx = 1 u p 1 u2 du dx ; 21. d(sinhu) dx = coshu du dx ; 22. d(coshu) dx = sinhu du dx ; 23. d(tanhu) dx = sech2 u du dx ; 24. d(cothu) dx = csch2 udu dx ; 25. d(sechu) dx = sechu tanhudu dx ; 26. d(cschu) dx = cschu cothudu dx ; 27. d(arcsinhu) dx = 1p 1 + u2 du dx ; 28. d(arccoshu) dx = 1p u2 1 du dx ; 29. d(arctanhu) dx = 1 1 u2 du dx ; 30. d(arccothu) dx = 1 u2 1 du dx ; 31. d(arcsechu) dx = 1 u p 1 u2 du dx ; 32. d(arccschu) dx = 1 jujp1 + u2 du dx : Integrals: 1. Z cu dx = c Z u dx; 2. Z (u+ v) dx = Z u dx+ Z v dx; 3. Z xn dx = 1 n+ 1 xn+1; n 6= 1; 4. Z 1 x dx = lnx; 5. Z ex dx = ex; 6. Z dx 1 + x2 = arctanx; 7. Z u dv dx dx = uv Z v du dx dx; 8. Z sinx dx = cosx; 9. Z cosx dx = sinx; 10. Z tanx dx = ln j cosxj; 11. Z cotx dx = ln j cosxj; 12. Z secx dx = ln j secx+ tanxj; 13. Z cscx dx = ln j cscx+ cotxj; 14. Z arcsin xadx = arcsin x a + p a2 x2; a > 0; Partial Fractions Let N(x) and D(x) be polynomial func- tions of x. We can break down N(x)=D(x) using partial fraction expan- sion. First, if the degree of N is greater than or equal to the degree of D, divide N by D, obtaining N(x) D(x) = Q(x) + N 0(x) D(x) ; where the degree of N 0 is less than that of D. Second, factor D(x). Use the follow- ing rules: For a non-repeated factor: N(x) (x a)D(x) = A x a + N 0(x) D(x) ; where A =  N(x) D(x)  x=a : For a repeated factor: N(x) (x a)mD(x) = m1X k=0 Ak (x a)mk+ N 0(x) D(x) ; where Ak = 1 k!  dk dxk  N(x) D(x)  x=a : The reasonable man adapts himself to the world; the unreasonable persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable. { George Bernard Shaw Theoretical Computer Science Cheat Sheet Calculus Cont. 15. Z arccos xadx = arccos x a p a2 x2; a > 0; 16. Z arctan xadx = x arctan x a a2 ln(a2 + x2); a > 0; 17. Z sin2(ax)dx = 12a ax sin(ax) cos(ax)
; 18. Z cos2(ax)dx = 12aax+ sin(ax) cos(ax)
; 19. Z sec2 x dx = tanx; 20. Z csc2 x dx = cotx; 21. Z sinn x dx = sin n1 x cosx n + n 1 n Z sinn2 x dx; 22. Z cosn x dx = cosn1 x sinx n + n 1 n Z cosn2 x dx; 23. Z tann x dx = tann1 x n 1 Z tann2 x dx; n 6= 1; 24. Z cotn x dx = cot n1 x n 1 Z cotn2 x dx; n 6= 1; 25. Z secn x dx = tanx secn1 x n 1 + n 2 n 1 Z secn2 x dx; n 6= 1; 26. Z cscn x dx = cotx csc n1 x n 1 + n 2 n 1 Z cscn2 x dx; n 6= 1; 27. Z sinhx dx = coshx; 28. Z coshx dx = sinhx; 29. Z tanhx dx = ln j coshxj; 30. Z cothx dx = ln j sinhxj; 31. Z sechx dx = arctan sinhx; 32. Z cschx dx = ln tanh x2 ; 33. Z sinh2 x dx = 14 sinh(2x) 12x; 34. Z cosh2 x dx = 14 sinh(2x) + 1 2x; 35. Z sech2 x dx = tanhx; 36. Z arcsinh xadx = x arcsinh x a p x2 + a2; a > 0; 37. Z arctanh xadx = x arctanh x a + a 2 ln ja2 x2j; 38. Z arccosh xadx = 8< : x arccosh x a p x2 + a2; if arccosh xa > 0 and a > 0, x arccosh x a + p x2 + a2; if arccosh xa < 0 and a > 0, 39. Z dxp a2 + x2 = ln  x+ p a2 + x2  ; a > 0; 40. Z dx a2 + x2 = 1a arctan x a ; a > 0; 41. Z p a2 x2 dx = x2 p a2 x2 + a22 arcsin xa ; a > 0; 42. Z (a2 x2)3=2dx = x8 (5a2 2x2) p a2 x2 + 3a48 arcsin xa ; a > 0; 43. Z dxp a2 x2 = arcsin x a ; a > 0; 44. Z dx a2 x2 = 1 2a ln a+ xa x ; 45. Z dx (a2 x2)3=2 = x a2 p a2 x2 ; 46. Z p a2  x2 dx = x2 p a2  x2  a22 ln x+pa2  x2 ; 47. Z dxp x2 a2 = ln x+px2 a2 ; a > 0; 48. Z dx ax2 + bx = 1 a ln xa+ bx ; 49. Z x p a+ bx dx = 2(3bx 2a)(a+ bx)3=2 15b2 ; 50. Z p a+ bx x dx = 2 p a+ bx+ a Z 1 x p a+ bx dx; 51. Z xp a+ bx dx = 1p 2 ln p a+ bxpap a+ bx+ p a ; a > 0; 52. Z p a2 x2 x dx = p a2 x2 a ln a+ p a2 x2 x ; 53. Z x p a2 x2 dx = 13 (a2 x2)3=2; 54. Z x2 p a2 x2 dx = x8 (2x2 a2) p a2 x2 + a48 arcsin xa ; a > 0; 55. Z dxp a2 x2 = 1 a ln a+ p a2 x2 x ; 56. Z x dxp a2 x2 = p a2 x2; 57. Z x2 dxp a2 x2 = x 2 p a2 x2 + a22 arcsin xa; a > 0; 58. Z p a2 + x2 x dx = p a2 + x2 a ln a+ p a2 + x2 x ; 59. Z p x2 a2 x dx = p x2 a2 a arccos ajxj ; a > 0; 60. Z x p x2  a2 dx = 13 (x2  a2)3=2; 61. Z dx x p x2 + a2 = 1a ln xa+pa2 + x2 ;