Notes Midterm Review Tyler Moore CSE 3353, SMU, Dallas, TX , 2013 Portions of these slides have been adapted from the slides written by Prof. Steven Skiena at SUNY Stony Brook, author of Algorithm Design Manual. For more information see http://www.cs.sunysb.edu/~skiena/ Administrivia Notes Extra office hours next week Monday 12:30pm-1:30pm Monday 5-5:30pm You may pick up graded HW2 and answer key then Midterm next Tuesday March 5 You may use one side of 1/2 sheet of letter paper for handwritten notes No calculators Review today 2 / 34 Defining bounding functions Notes f ( n ) = O ( g ( n )) means c · g ( n ) is an upper bound on f ( n ). Thus there exists some constant c such that f ( n ) is always ≤ c · g ( n ) for n ≥ n o for some constant n 0 . f ( n ) = Ω( g ( n )) means c · g ( n ) is a lower bound on f ( n ). Thus there exists some constant c such that f ( n ) is always ≥ c · g ( n ) for n ≥ n o . for some constant n 0 . f ( n ) = Θ( g ( n )) means c 1 · g ( n ) is an upper bound on f ( n ) and c 2 · g ( n ) is a lower bound on f ( n ). Thus there exists some constant c 1 and c 2 such that f ( n ) ≤ c 1 · g ( n ) and f ( n ) ≥ c 2 · g ( n ) for n ≥ n o for some constant n 0 . 3 / 34 Dominance and little oh Notes g ( n ) We say that f ( n ) dominates g ( n ) if lim n →∞ f ( n ) = 0. Otherwise f ( n ) does not dominate g ( n ). We say that f ( n ) = o ( g ( n )) ⇐ ⇒ g ( n ) dominates f ( n ) So n 2 = o ( n 3 ) since n 3 dominates n 2 . 4 / 34
Dominance examples Notes f ( n ) = n 2 , g ( n ) = n . Does f ( n ) dominate g ( n )? Is n 2 o ( n )? Is n o ( n 2 )? Is n o ( n )? 5 / 34 Dominance relations Notes You should come to accept the dominance ranking of the basic functions: n ! ≫ 2 n ≫ n 3 ≫ n 2 ≫ n log n ≫ n ≫ √ n ≫ log n ≫ 1 6 / 34 Discussion on Analysis Notes If the question asks you to explain why f ( n ) = Ω( g ( n )), etc., then provide a value for c and n 0 where the relationship holds, perhaps doing a bit of algebra to make the point clear. Best-case/worst-case/average-case does not correspond to Oh/Omega/Theta. Why do we see each of the following cost functions? (log n , n , n 2 , 2 n , n !) 7 / 34 Useful rules on exponents and logarithms Notes log a ( xy ) = log a ( x ) + log a ( y ) log a ( b ) = log c b log c a log a ( n b ) = b · log a ( n ) x a x b = x a + b x a x b = x a − b x a · y a = ( xy ) a x a y a = ( x y ) a 8 / 34
Python Notes If you have to write code on the exam, in most circumstances pseudo-code or Python is fine There are a few Python-specific points worth noting Python’s handling of mutable/immutable objects List comprehensions You do NOT need to worry about user-defined classes in Python (e.g., the code I showed on Binary Search Trees) 9 / 34 Variables in Python Notes Better thought of as names or identifiers attached to an object. A nice explanation: http://python.net/~goodger/projects/pycon/2007/ idiomatic/handout.html#other-languages-have-variables 10 / 34 Key distinction: mutable vs. immutable objects Notes Immutable: objects whose value cannot change Tuples (makes sense) 1 Booleans (surprise?) 2 Numbers (surprise?) 3 Strings (surprise?) 4 Mutable: objects whose value can change Dictionaries 1 Lists 2 User-defined objects (unless defined as immutable) 3 This distinction matters because it explains seemingly contradictory behavior 11 / 34 Variable assignment in action Notes >>> #variables are really names ... c = 4 >>> d = c >>> c+=1 >>> c 5 >>> d #d does not change because numbers are immutable 4 >>> #lists are mutable ... a = [1,4,2] >>> b = a #so this assigns the name b to the object attached to name a >>> a.append(3) >>> a [1, 4, 2, 3] >>> b #b still points to the same object, its contents have just changed. [1, 4, 2, 3] 12 / 34
Im/mutablility and function calls Notes >>> #let’s try this in a function ... def increment(n): #n is a name assigned to the function argument when called ... #because numbers are immutable, the following ... #reassigns n to the number represented by n+1 ... n+=1 ... return n ... >>> a = 3 >>> increment(a) 4 >>> #a does not change ... a 3 13 / 34 Im/mutablility and function calls Notes >>> def sortfun(s): ... s.sort() ... return s ... >>> def sortfun2(s): ... l = list(s) ... l.sort() ... return l ... >>> a = [1,4,2] >>> sortfun(a) [1, 2, 4] >>> a [1, 2, 4] >>> b = [3,9,1] >>> sortfun2(b) [1, 3, 9] >>> b [3, 9, 1] 14 / 34 Im/mutablility and function calls Notes def selection_sort(s): """ Input: list s to be sorted Output: sorted list """ for i in range(len(s)): #don’t name min since reserved word minidx=i for j in range(i+1,len(s)): if s[j]<s[minidx]: minidx=j s[i],s[minidx]=s[minidx],s[i] return s >>> b [3, 9, 1] >>> selection_sort(b) [1, 3, 9] >>> b [1, 3, 9] 15 / 34 List comprehensions Notes Recall set-builder notation from Discrete Math: S = { 3 x | x ∈ N , x > 5 } We can approximate that in Python > range (1 ,11) > > [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10] > { 3 ∗ x f o r x i n range (1 ,11) i f x > 5 } > > s e t ( [ 2 4 , 18 , 27 , 21 , 3 0 ] ) > [3 ∗ x f o r x i n range (1 ,11) i f x > 5] > > [18 , 21 , 24 , 27 , 30] Comprehensions arise in very common coding scenarios 16 / 34
List comprehensions Notes Here’s a common coding task: iterate over some list, perform some action on each element of that list, and store the results in a new list. Here’s an example: > c h e e s e s = [ ’ s w i s s ’ , ’ g r u y e r e ’ , ’ cheddar ’ , ’ s t i l t o n ’ , > > ’ r o q u e f o r t ’ , ’ b r i e ’ ] > c h e e s e l e n =[] > > > f o r c i n c h e e s e s : > > . . . c h e e s e l e n . append ( l e n ( c )) . . . > c h e e s e l e n > > [ 5 , 7 , 7 , 7 , 9 , 4] We can do this on a single line: c h e e s e l e n =[ l e n ( c ) c c h e e s e s ] f o r i n > c h e e s e l e n > > [ 5 , 7 , 7 , 7 , 9 , 4] 17 / 34 List comprehensions Notes But wait, there’s more! Suppose you only want to add items to the list if they meet a certain condition, say if the item begins with the letter s. Well here’s the long way: > s c h e e s e l e n =[] > > > f o r c i n c h e e s e s : > > . . . c[0]== ’ s ’ : i f . . . s c h e e s e l e n . append ( l e n ( c ) ) . . . > s c h e e s e l e n > > [ 5 , 7] You can add a condition at the end of the list comprehension: c h e e s e l e n =[ l e n ( c ) f o r c i n c h e e s e s i f c[0]==” s ” ] > s c h e e s e l e n > > [ 5 , 7] 18 / 34 More list comprehension examples Notes 1 > > s p o r t s > 2 [ ’ f o o t b a l l ’ , ’ t e n n i s ’ , ’ i c e hockey ’ , ’ l a c r o s s e ’ , ’ f i e l d hockey ’ , ’ b a s k e t b a l l ’ , ’ b a s e b a l l ’ , ’ swimming ’ ] 3 > > [ s s s p o r t s l e n ( s ) > 8] f o r i n i f > 4 [ ’ i c e hockey ’ , ’ f i e l d hockey ’ , ’ b a s k e t b a l l ’ ] 5 > > [ s f o r s i n s p o r t s i f ’ b a l l ’ i n s ] > 6 [ ’ f o o t b a l l ’ , ’ b a s k e t b a l l ’ , ’ b a s e b a l l ’ ] 7 8 > > s p o r t s l o c > 9 [ ( ’ f o o t b a l l ’ , ’ out ’ ) , ( ’ t e n n i s ’ , ’ both ’ ) , ( ’ i c e hockey ’ , ’ i n ’ ) , ( ’ l a c r o s s e ’ , ’ out ’ ) , ( ’ f i e l d hockey ’ , ’ out ’ ) , ( ’ b a s k e t b a l l ’ , ’ i n ’ ) , ( ’ b a s e b a l l ’ , ’ out ’ ) , ( ’ swimming ’ , ’ i n ’ ) ] 10 > > [ s [ 0 ] f o r s i n s p o r t s l o c ] > 11 [ ’ f o o t b a l l ’ , ’ t e n n i s ’ , ’ i c e hockey ’ , ’ l a c r o s s e ’ , ’ f i e l d hockey ’ , ’ b a s k e t b a l l ’ , ’ b a s e b a l l ’ , ’ swimming ’ ] 12 > > o u t d o o r s p o r t s =[ s [ 0 ] f o r s i n s p o r t s l o c i f s [1]== ’ out ’ ] > 13 > > o u t d o o r s p o r t s > 14 [ ’ f o o t b a l l ’ , ’ l a c r o s s e ’ , ’ f i e l d hockey ’ , ’ b a s e b a l l ’ ] 19 / 34 List comprehension exercise Notes Write a function that squares each element of a list so long as the elements are positive using a list comprehension. Complete this code: s q u a r e L i s t ( l ) : def return # f i l l i n l i s t comprehension 20 / 34
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