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What well studyGeneral Techniques
Divide and Conquer, Dynamic Programming, Hashing,
Greedy Algorithms, Reduction to other problems, ...Specific Problems
Sorting, sorting, shortest paths, max flow, sorting, ...
Various Paradigms
Probabilistic algorithms
Alternate models of computationNP Completeness
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Sounds like my undergrad course ...
Going over same material twice is good!
Well probably go deeper mathematical formalisms
modified assumptions
assorted topics every computer scientist shouldknow.
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Logistics Textbook: Introduction to Algorithms, 2ndedition
Cormen, Leiserson, Rivest & Stein
First edition probably OK (new chapter K ~ old chapter (K+1).)
Written work Individual homeworks
Group homeworks
In-class Midterm Final (oral? take-home?)
Grades
A: Demonstrate mastery of material. B (or B+ or B-): Typical grade. Understand most of material,
solve most routine problems and some hard ones.
C: Really dont get it.
D,F: Gave up.
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Logistics Classes will include lecture (with overheads),
discussion (at board), and exercises (at your seat).
Id prefer you pay attention, think, ask questions, andparticipate in discussions rather than taking notes.
Website: //www.cs.ucsd.edu/classes/fa02/cse202-b
.pdf and .ps of lectures. (Sometimes available before class.)
Homeworks, announcements, etc. too You are responsible for checking website
Be sure to register for the class email list
Ill use it to send out urgent messages, like HW corrections
Directions on class webpage
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Formal Analysis of Algorithms
Algorithm: a step-by-step method of producingoutput of instances given input
The steps are instructions for some model of computation. Standard model is the Random Access Machine (RAM)
Memory
ALU
One needs to be careful about what operations are allowed- E.g. operations on very long numbers arent one step.
- For this course anything reasonable is OK.
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Mathematical notation
P.T. Barnum asserted, You can fool all of the peoplesome of the time.
Suppose on Monday, I fooled all the men, and on Tuesday,I fooled all the women (but not vice versa).
Is this an example of Barnums assertion?
Mathematics is (or should be) preciseLet P = set of all the people, T = {Monday,Tuesday},
Let F(p,t) mean I fooled person p at time t
Is "p e P $t e T F(p,t) true?
Is $t e T "p e P F(p,t) true?
" means for all
$ means there exists
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Classifying (complexity) functionsWe focus on functions from to , though definitions are general.
Given function g, O(g) (pronounced Oh of g or
Big Oh of g) is the set of functions (from to )for which g is an asymptotic upper bound. Thismeans:
O(g) = {f : $n0e $c>0 "n>n0 0 f(n) c g(n) }
Example: 3n log n + 20 e O(n2)
Note: Since O(n2) is a set,
using e is technically correct.
But people often use = .
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Classifying (complexity) functions
W(g) (pronounced Omega or Big Omega) is theset of functions for which g is an asymptotic
lower bound. Formally:W(g) = {f : $n0e $c>0 "n>n0 0 c g(n) f(n) }
Q(g) (Theta) is the set of functions for which gis an asymptotic tight bound. This means:
Q(g) = {f : $n0e $c1,c2>0 "n>n0 0 c1g(n) f(n) c2g(n) }
Theorem: Q(g) = W(g) O(g).
Note: we wont use little-o or little-omega notation.
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Review
Consider the decision problem, Is x prime?
How should we define size?
Consider the algorithm:if(i
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What is a proof?
Informal Proof (Hand waving) Anything that convinces the reader
Formal Proof (First few classes) A sequence of statements, each being:
a hypothesis of the theorem
a definition, axiom, or known theorem a statement that follows from previous statements via a
rule of inference
Proof (Rest of course) A subset or summary of a formal proof that convinces a
literate but sleepy reader that the writer could write aformal proof if forced to.
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Notes about proofs
Use complete sentences. Sentences have a subject (you may be understood), a
verb, a period at the end.
If sentence is too long, introduce notation or definitions.
Each statement should indicate whether its anassumption, a definition, a known fact, ...
Dont just write, x e A. Instead write:
Let x e A. if youre introducing x and want it to be in A.
Suppose x e A. if x has already been introduced, and
youre seeing what would happen if it were in A. Thus, x e A. if it follows from earlier statements.
Make sure each variable is properly introduced Like declaring variables in a program.
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Some proof schemas Literate readers even sleepy ones know that:
To show A B, can write Let x e A. (blah ...). Thus, x e B.
To show A = B (for sets A and B), show A B and B A.
To show P implies Q, write Assume P. (blah ...). Thus Q.
To show $x P, write Let x=(whatever). (blah ...) Thus P.
To show "x P, write Given any x, (blah ...). Thus P.
To show an algorithm has complexity O(f), you willprobably construct a positive constant c and an integer n0and then show that if n>n0 and I is an instance of size n,then the algorithm requires time at most c f(n) on I.
These schemas can be mostly implicit. Thus, after writingLet x e A. (blah blah). Thus, x e B. you often dont needto write This shows that A B.
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Example formal proof
Thm: Suppose feO(g) and geO(h). Then feO(h).
Proof:
Since fe O(g), $n0e $c0>0 "n>n0 0 f(n) c0 g(n).Similarly, $n1e $c1>0 "n>n1 0 g(n) c1 h(n).
Let n2 = max(n0, n1) and c2 = c0 c1. Note that c2 is positive
since both c0 and c1 are.Suppose n>n2. Then n>n0 and so 0 f(n) c0 g(n).
But we also have n>n1 and so g(n) c1 h(n).Thus, 0 f(n) c0 c1 h(n) = c2 h(n).
Q.E.D.
Just the definition of O(g), but introduces c0 and n0.
Setup for proving $n2e $c2>0 ...
Setup for proving "n>n2 ...
Means, Weve proved what we intended to.
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Your turn ...
Thm: If feO(g) then geW(f).
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Mathematical Induction
Suppose for " ie, Pi is a statement.
Suppose also that we can prove P0, and we canprove " ie, Pi implies Pi+1
Then the Principle of Mathematical Induction
allows us to conclude "
ie
Pi.
P3P0
P1P2
P4...
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Mathematical Induction
Alternatively, suppose we can prove P0, and wecan prove " ie, (P0 & P1 & ... & Pi) implies Pi+1
Again, the Principle of Mathematical Inductionallows us to conclude " ie Pi.
P3P0
P1
P2
P4
...P5
P6
P7
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Induction Example
Definitions from CLRS pg. 1088:
A binary tree is a structure on a finite number of
nodes that either contains no nodes or is composedof three disjoint subsets: a root node, a binarytree called the left subtree and a binary treecalled the right subtree.
The height of a non-empty tree is the maximumdepth of its nodes. The depth of a node is thelength of the path from the root to the node.
Thm: If T is a non-empty binary tree of height h,then T has fewer than 2h+1 nodes.
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Induction Example
Let Ph be the statement, If T is a binary tree ofheight h, then T has at most 2h+11 nodes.
We will prove " he Ph by induction.
Base Case (h=0): T is non-empty, so it has a root
node r. Let s be any node of T. Since the heightof T is 0, the depth of s must be 0, so s = r.Thus, T has only one node (which is 20+1-1 =1).
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Induction Example
Let Ph be the statement, If T is a binary tree ofheight h, then T has at most 2h+11 nodes.
We will prove " he Ph by induction.
Base Case (h=0): T is non-empty, so it has a root
node r. Let s be any node of T. Since the heightof T is 0, the depth of s must be 0, so s = r.Thus, T has only one node (which is 20+1-1).
Obviously, the only binary tree of height 0 isthe tree of one node, so P0 is true.
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Induction Example
Induction step: Assume that Ph is true.
Let T be a tree of height h+1. Then the left subtree L is a
binary tree.If L is empty, it has 0 nodes.
Otherwise, each node in L is has depth one less than its
depth in T. Thus, L is a non-empty binary tree of depthat most h. By assumption
OOOPS!Ive only assumed Ph. But L may have smaller height.
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Induction Example
Induction step: Assume that Pi is true " ih.
Let T be a tree of height h+1. Then the left subtree L is a
binary tree.If L is empty, it has 0 nodes.
Otherwise, each node in L is has depth one less than its
depth in T. Thus, L is a non-empty binary tree of depthat most h. By assumption, L has at most 2h+11 nodes.
Similarly, the right subtree R has at most 2h+11 nodes.
Thus, T has at most 1 + 2h+11 + 2h+11 = 2 2h+11= 2(h+1)+11 nodes.
This shows that Ph+1 is true, and completes our proof.
for the root for L for R