# Basic Algorithms

================ Start Lecture #3 ================

Remark: I changed my mind about homework. Too many to have each one really graded. We now have homeworks and problem sets as explained here.

#### What is "fast" or "efficient"?

If the asymptotic time complexity is bad, say n5, or horrendous, say 2n, then for large n, the algorithm will definitely be slow. Indeed for exponential algorithms even modest n's (say n=50) are hopeless.

Algorithms that are o(n) (i.e., faster than linear, a.k.a. sub-linear), e.g. logarithmic algorithms, are very fast and quite rare. Note that such algorithms do not even inspect most of the input data once. Binary search has this property. When you look a name in the phone book you do not even glance at a majority of the names present.

Linear algorithms (i.e., Θ(n)) are also fast. Indeed, if the time complexity is O(nlog(n)), we are normally quite happy.

Low degree polynomial (e.g., Θ(n2), Θ(n3), Θ(n4)) are interesting. They are certainly not fast but speeding up a computer system by a factor of 1000 (feasible today with parallelism) means that a Θ(n3) algorithm can solve a problem 10 times larger. Many science/engineering problems are in this range.

### 1.2.3: The Importance of Asymptotics

It really is true that if algorithm A is o(algorithm B) then for large problems A will take much less time than B.

Definition: If (the number of operations in) algorithm A is o(algorithm B), we call A asymptotically faster than B.

Example:: The following sequence of functions are ordered by growth rate, i.e., each function is little-oh of the subsequent function.
log(log(n)), log(n), (log(n))2, n1/3, n1/2, n, nlog(n), n2/(log(n)), n2, n3, 2n.

#### What about those constants that we have swept under the rug?

Modest multiplicative constants (as well as immodest additive constants) don't cause too much trouble. But there are algorithms (e.g. the AKS logarithmic sorting algorithm) in which the multiplicative constants are astronomical and hence, despite its wonderful asymptotic complexity, the algorithm is not used in practice.

#### A Great Table

See table 1.10 on page 20.

Homework: R-1.7

## 1.3: A Quick Mathematical Review

This is hard to type in using html. The book is fine and I will write the formulas on the board.

### 1.3.1: Summations

Definition: The sigma notation: ∑f(i) with i going from a to b.

Theorem: Assume 0<a≠1. Then ∑ai i from 0 to n = (1-an+1)/(1-a).

Proof: Cute trick. Multiply by a and subtract.

Theorem: ∑i from 1 to n = n(n+1)/2.

Proof: Pair the 1 with the n, the 2 with the (n-1), etc. This gives a bunch of (n+1)s. For n even it is clearly n/2 of them. For odd it is the same (look at it).

### 1.3.2: Logarithms and Exponents

Recall that logba = c means that bc=a. b is called the base and c is called the exponent.

What is meant by log(n) when we don't specify the base?

• Some people use base 10 by default.
• Mathematicians use base e.
• We will use base 2 (common in computer science)

I assume you know what ab is. (Actually this is not so obvious. Whatever 2 raised to the square root of 3 means it is not writing 2 down the square root of 3 times and multiplying.) So you also know that ax+y=axay.

Theorem: Let a, b, and c be positive real numbers. To ease writing, I will use base 2 often. This is not needed. Any base would do.

1. log(ac) = log(a)+log(c)
2. log(a/c) = log(a) - log(c)
3. log(ac) = c log(a)
4. logc(a) = (log(a))/log(c): consider a = clogca and take log of both sides.
5. clog(a) = a log(c): take log of both sides.
6. (ba)c = bac
7. babc = ba+c
8. ba/bc = ba-c

#### Examples

• log(2nlog(n)) = 1 + log(n) + log(log(n)) is Θ(log(n))
• log(log(sqrt(n))) = log(.5log(n)) = log(.5)+log(log(n)) = -1 + log(log(n)) = Θ(log(log(n))
• log(2n) = n = 2log(n)

Homework: C-1.12

#### Floor and Ceiling

⌊x⌋ is the greatest integer not greater than x. ⌈x⌉ is the least integer not less than x.

⌊5⌋ = ⌈5⌉ = 5

⌊5.2⌋ = 5 and ⌈5.2⌉ = 6

⌊-5.2⌋ = -6 and ⌈-5.2⌉ = -5

### 1.3.3: Simple Justification Techniques

#### By example

To prove the claim that there is an n greater than 1000, we merely have to note that 1001 is greater than 1001.

#### By counterexample

To refute the claim that all n are greater than 1000, we merely have to note that 999 is not greater than 1000.

#### By contrapositive

"P implies Q" is the same as "not Q implies not P". So to show that no prime is a square we note that "prime implies not square" is the same is "not (not square) implies not prime", i.e. "square implies not prime", which is obvious.

Assume what you want to prove is false and derive a contradiction.

Theorem: There are an infinite number of primes.

Proof: Assume not. Let the primes be p1 up to pk and consider the number A=p1p2…pk+1. A has remainder 1 when divided by any pi so cannot have any pi as a factor. Factor A into primes. None can be pi (A may or may not be prime). But we assumed that all the primes were pi. Contradiction. Hence our assumption that we could list all the primes was false.

#### By (complete) induction

The goal is to show the truth of some statement for all integers n≥1. It is enough to show two things.

1. The statement is true for n=1
2. IF the statement is true for all k<n, then it is true for n.

Theorem: A complete binary tree of height h has 2h-1 nodes.

Proof: We write NN(h) to mean the number of nodes in a complete binary tree of height h. A complete binary tree of height 1 is just a root so NN(1)=1 and 21-1 = 1. Now we assume NN(k)=2k-1 nodes for all k<h and consider a complete binary tree of height h. It is just a complete binary tree of height h-1 with new leaf nodes added. How many new leaves?
Ans. 2h-1 (this could be proved by induction as a lemma, but is fairly clear without induction).

Hence NN(h)=NN(h-1)+2h-1 = (2h-1-1)+2h-1 = 2(2h-1)-1=2h-1.

Homework: R-1.9

#### Loop Invariants

Very similar to induction. Assume we have a loop with controlling variable i. For example a "for i←0 to n-1" loop. We then associate with the loop a statement S(j) depending on j such that

1. S(0) is true (just) before the loop begins
2. IF S(j-1) holds before iteration j begins, then S(j) will hold when iteration j ends.
By induction we see that S(n) will be true when the nth iteration ends, i.e., when the loop ends.

I favor having array and loop indexes starting at zero. However, here it causes us some grief. We must remember that iteration j occurs when i=j-1.

Example:: Recall the countPositives algorithm

```Algorithm countPositives
Input: Non-negative integer n and an integer array A of size n.
Output: The number of positive elements in A

pos ← 0
for i ← 0 to n-1 do
if A[i] > 0 then
pos ← pos + 1
return pos
```

Let S(j) be "pos equals the number of positive values in the first j elements of A".

Just before the loop starts S(0) is true vacuously. Indeed that is the purpose of the first statement in the algorithm.

Assume S(j-1) is true before iteration j, then iteration j (i.e., i=j-1) checks A[j-1] which is the jth element and updates pos accordingly. Hence S(j) is true after iteration j finishes.

Hence we conclude that S(n) is true when iteration n concludes, i.e. when the loop terminates. Thus pos is the correct value to return.