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A A nalisis de lgoritmos Informacion tomada del libro de Cormen.
OUTLINE ,[object Object],[object Object],[object Object]
ASYMPTOTIC NOTATION
Asymptotically  N o Negative Functions f(n)  is asymptotically no negative if there exist  n 0    N  such that for every  n    n 0 , 0     f(n) n 0 f
Theta      ( g(n) )  = {   f  : N   R *  | (    c 1 , c 2     R   + )   (    n 0   N) (    n    n 0 )  ( 0   c 1  g(n)    f(n)    c 2  g(n) )   } c 1      lim  f(n)      c 2 n     g(n )   n 0 g c 1 g f c 2 g
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Example Lets show that We have to find  c 1 , c 2  and   n 0   such that
For all  n     n 0 , Dividing by  n 2  yields We have that  3/n  is a decreasing sequence  3, 3/2, 1, 3/4, 3/5, 3/6, 3/7… and then  1/2 - 3/n  is increasing sequence  -5/2, -1, -1/2, -1/4, -1/10, 0, 1/14 that is upper bounded by  1/2 .
The right hand inequality can be made to hold for  n     1  by choosing  c 2     1/2 . Likewise the left hand inequality can be made to hold for  n     7  by choosing  c 1     1/14 .
Thus by choosing  c 1  =  1/14 ,  c 2  =  1/2  and  n 0  =7 then we can verify that
Big  O   O (g(n))   = { f : N   R *  | (    c  R  + ) (  n 0  N) (  n   n 0 ) (  0     f(n)     cg(n)  ) } lim  f(n)     c n     g(n)   n 0 g cg f
[object Object],[object Object],[object Object],[object Object],[object Object]
Example Lets show that We have to find  c   and   n 0   such that
For all  n     n 0 , Dividing by  n   yields The inequality can be made to hold for  n     1  by choosing  c      a+b .  Thus by choosing  c   =  a+b  and  n 0  =1 then we can verify that
n 0 =1 a+b b a
Big Omega      ( g(n) )  = {   f : N   R *  | (    c  R  + ) (  n 0  N) (    n   n 0 ) (  0    cg(n)     f(n)   )   } c      lim  f(n)  n     g(n)   n 0 g f cg
[object Object],[object Object],[object Object],[object Object],[object Object]
Little  o o   (g(n))  =  {  f  : N   R *  | (  c   R  + ) (  n 0  N) (    n   n 0 ) (  0     f(n) < cg (n)  )  } lim  f(n)  = 0 n     g(n)  “ o(g(n))  functions that grow slower than  g  ”   g f
[object Object],[object Object],[object Object]
n 2 n Examples Lets show that We only have to show
Lets show that We have 3n n
Little Omega      ( g(n) )  =  {   f  : N   R *  | (  c  R  + ) (  n 0  N) (    n   n 0 ) (  0     c g(n) < f(n ) )  } lim  f(n)  =     n     g(n)   “    (g(n))  functions that grow faster than  g  ” f g
[object Object],[object Object],[object Object]
Analogy with the comparison of two real numbers Asymptotic   Real Notation numbers f(n)    O(g(n))    f     g f(n)     (g(n))    f     g f(n)     (g(n))    f =  g f(n)    o(g(n))    f <  g f(n)     (g(n))    f >  g Trichotomy does not hold
Not all functions are asymptotically comparable Example f(n)=n g(n)=n (1+sin n) (1+sin(n))    [0,2]
[object Object],[object Object],[object Object],[object Object],   (f(n))=O(f(n))      (f(n))  Properties
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[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],o (f(n))        (f(n)) =  
Examples ,[object Object],[object Object],[object Object],[object Object],[object Object]
Examples ,[object Object],[object Object],[object Object],[object Object],[object Object]
Examples ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Asymptotic notation two variables O(g(m, n))= {  f: N   N     R *  | (    c   R  + )(    m 0  , n 0    N)   (    n    n 0 ) (    m    m 0 )   (f(m, n)     c g(m, n))  }
COMMON FUNCTIONS
Monotonicity ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Floors and Ceilings ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],-1 0 1  x  -2 2  x  x
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Modular arithmetic For every integer  a  and any possible positive integer  n, a mod n  is the   remainder   of  ( or  residue )  of the quotient  a/n a mod n = a -   a  / n    n.
congruency  or equivalence mod  n If ( a mod n) =  ( b mod n)  we write  a     b  ( mod n)  and we say that  a  is  equivalent  to  b  modulo  n  or that a  is  congruent  to  b  modulo  n. In other words  a     b  ( mod n)  if  a  and  b  have the same remainder when divided by  n . Also  a     b  ( mod n)  if and only if  n  is a divisor of  b-a.
0 1 2 3 -4 -3 -2 -1 -8 -7 -6 -5 -12 -11 -10 -9         12 13 14 15 8 9 10 11 4 5 6 7     Example     (mod 4)    (mod n)   defines a equivalence  relation in   Z  and produces a  partitioned set called  Z n =   Z /n  = {0,1,2,…, n -1} in which can be defined  arithmetic operations a+b (mod n)  a*b (mod n)   [0] [1] [2] [3] Z /4 = {[0],[1],[2],[3]} = {0,1,2,3} 4+1 ( mod  4) = 1 5*2 ( mod  4) = 2
Polynomials Given a no negative integer  d,  a  polynomial  in n of degree d  is a function  p(n)  of the form: Where  a 1  , a 2  , …,   a d  are the  coefficients   and  a d     0.
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Exponentials ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
Logarithms ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
A function  f(n)   is  polylogaritmically bounded  if  f(n)= O ( lg k  n)  for some constant  k. We have the following relation between polynomials  and polylogarithms: then   lg k  n = o ( n k  )
Factorials Definition   (no recursive)     (recursive) Weak upper bound  n!    n n
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[object Object],[object Object]
Functional iteration Given a function  f  ( n ) the i-th functional iteration of  f  is  defined as : with  I  the identity function. For a particular  n , we have,
Examples: f(n) = 2n  then f(n) = n 2   then
f(n) = n n   then
Iterated logarithm ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
In general k

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Clase3 Notacion

  • 1. A A nalisis de lgoritmos Informacion tomada del libro de Cormen.
  • 2.
  • 4. Asymptotically N o Negative Functions f(n) is asymptotically no negative if there exist n 0  N such that for every n  n 0 , 0  f(n) n 0 f
  • 5. Theta   ( g(n) ) = { f : N  R * | (  c 1 , c 2  R + ) (  n 0  N) (  n  n 0 ) ( 0  c 1 g(n)  f(n)  c 2 g(n) ) } c 1  lim f(n)  c 2 n  g(n ) n 0 g c 1 g f c 2 g
  • 6.
  • 7. Example Lets show that We have to find c 1 , c 2 and n 0 such that
  • 8. For all n  n 0 , Dividing by n 2 yields We have that 3/n is a decreasing sequence 3, 3/2, 1, 3/4, 3/5, 3/6, 3/7… and then 1/2 - 3/n is increasing sequence -5/2, -1, -1/2, -1/4, -1/10, 0, 1/14 that is upper bounded by 1/2 .
  • 9. The right hand inequality can be made to hold for n  1 by choosing c 2  1/2 . Likewise the left hand inequality can be made to hold for n  7 by choosing c 1  1/14 .
  • 10. Thus by choosing c 1 = 1/14 , c 2 = 1/2 and n 0 =7 then we can verify that
  • 11. Big O O (g(n)) = { f : N  R * | (  c  R + ) (  n 0  N) (  n  n 0 ) ( 0  f(n)  cg(n) ) } lim f(n)  c n  g(n) n 0 g cg f
  • 12.
  • 13. Example Lets show that We have to find c and n 0 such that
  • 14. For all n  n 0 , Dividing by n yields The inequality can be made to hold for n  1 by choosing c  a+b . Thus by choosing c = a+b and n 0 =1 then we can verify that
  • 15. n 0 =1 a+b b a
  • 16. Big Omega   ( g(n) ) = { f : N  R * | (  c  R + ) (  n 0  N) (  n  n 0 ) ( 0  cg(n)  f(n) ) } c  lim f(n) n  g(n) n 0 g f cg
  • 17.
  • 18. Little o o (g(n)) = { f : N  R * | (  c  R + ) (  n 0  N) (  n  n 0 ) ( 0  f(n) < cg (n) ) } lim f(n) = 0 n  g(n) “ o(g(n)) functions that grow slower than g ” g f
  • 19.
  • 20. n 2 n Examples Lets show that We only have to show
  • 21. Lets show that We have 3n n
  • 22. Little Omega   ( g(n) ) = { f : N  R * | (  c  R + ) (  n 0  N) (  n  n 0 ) ( 0  c g(n) < f(n ) ) } lim f(n) =  n  g(n) “  (g(n)) functions that grow faster than g ” f g
  • 23.
  • 24. Analogy with the comparison of two real numbers Asymptotic Real Notation numbers f(n)  O(g(n)) f  g f(n)   (g(n)) f  g f(n)   (g(n)) f = g f(n)  o(g(n)) f < g f(n)   (g(n)) f > g Trichotomy does not hold
  • 25. Not all functions are asymptotically comparable Example f(n)=n g(n)=n (1+sin n) (1+sin(n))  [0,2]
  • 26.
  • 27.
  • 28.
  • 29.
  • 30.
  • 31.
  • 32.
  • 33. Asymptotic notation two variables O(g(m, n))= { f: N  N  R * | (  c  R + )(  m 0 , n 0  N) (  n  n 0 ) (  m  m 0 ) (f(m, n)  c g(m, n)) }
  • 35.
  • 36.
  • 37.
  • 38. Modular arithmetic For every integer a and any possible positive integer n, a mod n is the remainder of ( or residue ) of the quotient a/n a mod n = a -  a / n  n.
  • 39. congruency or equivalence mod n If ( a mod n) = ( b mod n) we write a  b ( mod n) and we say that a is equivalent to b modulo n or that a is congruent to b modulo n. In other words a  b ( mod n) if a and b have the same remainder when divided by n . Also a  b ( mod n) if and only if n is a divisor of b-a.
  • 40. 0 1 2 3 -4 -3 -2 -1 -8 -7 -6 -5 -12 -11 -10 -9         12 13 14 15 8 9 10 11 4 5 6 7     Example  (mod 4)  (mod n) defines a equivalence relation in Z and produces a partitioned set called Z n = Z /n = {0,1,2,…, n -1} in which can be defined arithmetic operations a+b (mod n) a*b (mod n) [0] [1] [2] [3] Z /4 = {[0],[1],[2],[3]} = {0,1,2,3} 4+1 ( mod 4) = 1 5*2 ( mod 4) = 2
  • 41. Polynomials Given a no negative integer d, a polynomial in n of degree d is a function p(n) of the form: Where a 1 , a 2 , …, a d are the coefficients and a d  0.
  • 42.
  • 43.
  • 44.
  • 45.
  • 46.
  • 47.
  • 48.
  • 49. A function f(n) is polylogaritmically bounded if f(n)= O ( lg k n) for some constant k. We have the following relation between polynomials and polylogarithms: then lg k n = o ( n k )
  • 50. Factorials Definition (no recursive) (recursive) Weak upper bound n!  n n
  • 51.
  • 52.
  • 53. Functional iteration Given a function f ( n ) the i-th functional iteration of f is defined as : with I the identity function. For a particular n , we have,
  • 54. Examples: f(n) = 2n then f(n) = n 2 then
  • 55. f(n) = n n then
  • 56.