Tirgul%202

1
Tirgul 2

• Subjects of this tirgul
• Short discussion about standard output/error.
• Short discussion about comparators.
• Main subject asymptotic analysis.

2
Default I/O Streams

• Class System has 3 default streams, available to
  every Java program
• Input from the keyboard goes into the standard
  input. This is the data member System.in of
  type java.io.InputStream
• Output (usually to the terminal window) written
  through 2 streams
• Standard output System.out of type
  java.io.PrintStream
• Standard error System.err of type
  java.io.PrintStream

3
Default I/O Streams

• The standard output and error are directed by the
  Operating System. By default - both to the
  terminal.
• The convention - standard error for error
  messages, standard output for regular output.
• In UNIX, the user can redirect to a file
• standard output by adding gt my_out.txt. For
  example java MyClass param1 gt out.txt
• both to the same file, by adding gt my_out.txt
• You cant redirect only the standard error, but
  redirecting to different files is possible (by
  outsmarting) (java MyClass gt out.txt) gt err.txt

4
Comparators

• You already know interface Comparable to compare
  objects.
• Problem some classes may be compared by
  different criteria (participants may be compared
  by ID, or by name)
• Solution create a Comparator for this specific
  class.
• In the constructor specify the comparison
  criteria.
• The method compare() receives two objects of this
  class and compares them (similar to CompareTo()).
• Many Java API data structures can compare classes
  either by Comparable or by a Comparator.

5
Asymptotic Analysis

• Motivation Suppose you want to evaluate two
  programs according to their run-time for inputs
  of size n. The first has run-time ofand the
  second has run-time ofAnswer For small
  inputs, it doesnt matter, both programs will
  finish before you notice. What about (really)
  large inputs?

6
Big - O

• Definition
  
  if there exists constants cgt0 and n0 such
  that for all ngtn0,
• In other words, g(n) bounds f(n) from above (for
  large ns) up to a constant.
• Examples

7
Big - Omega

• Definition
  
  if there exists constants cgt0 and n0 such
  that for all ngtn0,
• In other words, g(n) bounds f(n) from below (for
  large ns) up to a constant.
• Examples

8
Big - Theta

• Definition if
  and
• In other words, g(n) is a tight estimate of f(n)
  (in asymptotic terms).
• Examples

9
Example 1(question 2-4-e. in Cormen)
Claim If (for ngtn0)
then Proof Take . Thus for
ngtn0, Note if there is no such for f(n)
then the claim is not true take for example f(n)
1/n. For any constant c, take ngtc and so
f(n) gt (c/n)f(n) c (f(n))2
10
Example 2(question 2-4-d. in Cormen)

• Does imply
• Answer No. For example, Then,
  so but, for any constant
  cso

11
Summations(from Cormen, ex. 3.2-2., page 52)

• note how we got rid of the integer rounding
• The first term is n so the sum is also
• Its an example how the largest item dominates
  the growth of the term in an exponential
  decrease/increase.

12
Summations (example 2)(Cormen, ex. 3.1-a., page
52)

• (r is a constant)
  note that
• But it seems that we can do better by tightening
  our analysis a bit more...

13
Example 2 (Cont.)

• For an even n
• For an odd n
• Thus so our
  upper bound was tight!

14
Recurrences

• What is the running time of the Towers of Hanoi
  ??
• Reminder
• We can express the running-time as a
  recurrence h(n) 2h(n-1) 1 h(1) 1
• How do we solve this ?

H(s,t,m,k) if (k gt 1) H(s,m,t,k-1)
moveDisk(s,t) H(m,t,s,k-1) else
moveDisk(s,t)
15
Step 1 guessing the solution

• h(n) 2h(n-1) 1
• 22h(n-2)1 1 4h(n-2) 3
  42h(n-3)1 3 8h(n-3) 7
• When repeating k times we get h(n)2k h(n-k)
  (2k - 1)
• Now take kn-1. Well geth(n) 2n-1
  h(n-(n-1)) 2n-1 - 1 2n-1 2n-1 -1 2n - 1

16
Step 2 proving by induction

• If we guessed right, it will be easy to prove by
  induction that h(n)2n - 1
• For n1 h(1) 2-11 (and indeed h(1)1)
• Suppose h(n-1) 2n-1 - 1. Then, h(n) 2h(n-1)
  1 2(2n-1 - 1) 1 2n -2 1 2n
  -1
• So we conclude that h(n) O(2n)

17
Another Example

• And we get T(n) k T(n/k)(k-1)For kn we get
  T(n) n T(1)n-12n-1Now proving by induction is
  very simple.
• Another way guess right away T(n) lt c n - b
  (for some b and c we dont know yet), and try to
  prove by inductionFor n1 T(1)c-b, which is
  true when c-b1The induction stepT(n) lt 2
  (c(n/2) - b) 1 c n - 2b 1 lt c n - b(the
  last step is true if bgt1). Conclusion T(n)
  O(n)

T(n) 2T(n/2) 1 2 (2T(n/4) 1) 1
4T(n/4) 3 4 (2T(n/8) 1) 3 8T(n/8) 7
T(n) 2 T(n/2) 1 T(1) 1
18
Common errors

• A precise guess is important for the induction to
  workh(n) 2n - 1, and not just h(n)O(2n) or
  h(n)2n.
• For example, the method from the previous
  slideIf we guess just T(n)lt cn the induction
  step wont work. Lets try by induction
  T(n/2) lt c(n/2) T(n) 2T(n/2) 1 c n 1 gt
  c n (!!!)
• Notice that it is not correct to sayc n 1
  O(n) so we proved T(n)O(n).The induction step
  must be precise. If guessingT(n) lt cn then this
  is what you have to show.

19
Recursion Trees
The recursion tree for the towers of Hanoi

• For each level we write the time added due to
  this level. In Hanoi, each recursive call adds
  one operation (plus the recursion). Thus the
  total is