cs 23022 discrete mathematical structures

CS 23022Discrete Mathematical Structures

Mehdi GhayoumiMSB rm 132mghayoum@kent.edu Ofc hr: Thur, 9:30-11:30a

Announcements

There is no Homework!!!

Next session we have a quiz for Induction

Mathematical Induction - why does it work?

Definition:

A set S is “well-ordered” if every non-empty subset of S has a least element.

Examples:

the set of natural numbers (N) is well-ordered.

Is the set of integers (Z) well ordered?

No. { x Z : x < 0 }

has no least element.

Mathematical Induction - why does it work?

Is the set of non-negative reals (R) well ordered?

No. { x R : x > 1 }

has no least element.

Strong Mathematical InductionAn example.

Given n blue points and n orange points in a plane with no 3 collinear, prove there is a way to match

them, blue to orange, so that none of the segments between the pairs intersect.

Strong Mathematical InductionBase case (n=1):

Assume any matching problem of size less than (k+1) can be solved.

Show that we can match (k+1) pairs.

Strong Mathematical InductionShow that we can match (k+1) pairs.

Suppose there is a line partitioning the group into a smaller one of j blues and j oranges, and another smaller one of (k+1)-j blues and (k+1)-j oranges.

OK!! (by IH)

OK!! (by IH)

Strong Mathematical InductionHow do we know such a line always exists?

Consider the convex hull of the points:

If there is an alternating pair of colors on the hull, we’re done!

OK!! (by IH)

OK!! (by IH)

Strong Mathematical InductionIf there is no alternating pair, all points on hull are

the same color.

Notice that any sweep of the hull hits an orange point first and also last. We sweep on some slope not given by a pair of points.

OK!! (by IH)

OK!! (by IH)

Keep score of # of each color seen. Orange gets the early lead, and then comes from behind to tie at the end.

There must be a tie along the way

Strings and Inductive DefinitionsLet be a finite set called an alphabet.

The set of strings on , denoted * is defined as:

*, where denotes the null or empty string.

If x , and w *, then wx *, where wx is the

concatenation of string w with symbol x.

Strings and Inductive Definitions

Countably infinite

Example: Let = {a, b, c}. Then

* = {, a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc, aaa, aab,…}

How big is *?

Is there a largest string in *? No.

Strings and Inductive DefinitionsInductive definition of the length of strings

(the length of string w is |w|.):

|| = 0

If x , and w *, then |wx| = |w| + 1

I point this out because the length of strings is something

we might like to use for an inductive argument.

Strings and Inductive DefinitionsInductive definition of the reversal of a string

(the reversal of string w is written wR.):R =

If x , and w *, then (wx)R = ? For example (abc)R = cbax(w)R

For example (abc)R = c(ab)R

= cb(a)R

= cba()R

= cba= cba

Strings and Inductive DefinitionsA Theorem:

x,y * (xy)R = yRxR

Proof (by induction on |y|):Base Case (|y| = 0): If |y| = 0, y = , then (xy)R = (x)R = xR = xR =

yRxR.IH: If |y| n, then x *, (xy)R = yRxR.

Prove: If |y| = n+1, then x *, (xy)R = yRxR.

Strings and Inductive DefinitionsIH: If |y| n, then x *, (xy)R = yRxR.

Prove: If |y| = n+1, then x *, (xy)R = yRxR.If |y| = n+1, then a , u *, so

that y = ua, and |u| = n.Then, (xy)R = (x(ua))R by substitution

= ((xu)a)R by assoc. of concatenation = a(xu)R by inductive defn of reversal = auRxR by IH

= (ua)RxR by inductive defn of reversal = yRxR by substitution

AlgorithmsAn iterative algorithm is one that repeats the same sequence of

steps a number of times.

for loops

while loops

repeat loops

goto??

The running time of an iterative algorithm depends on the number of times the loop is invoked.

AlgorithmsHow many times does “twiddle-thumbs” happen?

1. for i = 1 to n2. for j = 1 to m

3. twiddle-thumbs

The “time complexity” of an algorithm is a measure of its running time.

But different machines run at different speeds!

So we give running times in terms of big-oh, since different machines affect run times by constant factors.

Complexity is O(mn)

Inductive DefinitionsWe completely understand the function f(n) = n!,

right?

As a reminder, here’s the definition:n! = 1 · 2 · 3 · … · (n-1) · n, n 1

Inductive (Recursive) Definition

But equivalently, we could define it like this:

n!n (n 1)!, if n 11, if n 1

Recursive Case

Base Case

Inductive DefinitionsAnother VERY common example:

Fibonacci Numbers

Recursive Case

Base Cases

f (n) 0 if n 01 if n 1f (n 1) f (n 2) if n 1

Is there a non-recursive definition for the Fibonacci

Numbers?

Inductive DefinitionsFibonacci Numbers

double fib(int n) { double prev = -1; double result = 1; double sum; int i; for(i = 0;i <= n;++ i) { sum = result + prev; prev = result; result = sum; } return result; }

AlgorithmsAlgorithm MAX

Input: x1, x2, …, xn, an array of numbersOutput: y, the maximum of x1, x2, …, xn

1. for j = 1 to n-12. if xj > xj+1 then3. temp = xj+1 4. xj+1 = xj

5. xj = tempComplexity is O(n)

vars x1 x2 x3 x4

input 3 2 4 1j = 1 3 2 4 1j = 2 2 3 4 1j = 3 2 3 4 1final 2 3 1 4

AlgorithmsAlgorithm BUBBLE

Input: x1, x2, …, xn, an array of numbersOutput: ??

1. for j = n downto 22. MAX(x1,x2,…,xj)

3. output ??

A reasonable output for this function would be:A. x1, the minimum element of the arrayB. xn, the max element of the arrayC. The entire array… it has now been sortedD. j the loop counter

Algorithms

Algorithms

Algorithmsint main() { int array[BUBBLE], i, j,temp = 0; for (i = 0; i < BUBBLE; i ++) { cout<<“…”;for (j = 0; j < BUBBLE; j++); { temp = array[j+1]; array[j+1] = array[j]; array[j] = temp; } } return 0; }

AlgorithmsThe running time of this algorithm is:

A. O(n log n)

B. O(n)

C. O(n2)

D. None of the above.

Running timesIt’s fun to make comparisons about the running times of

algorithms of various complexities. Inp size

compxity

10 20 30 40 50 60

n .00001s .00002s .00003s .00004s .00005

s .00006s

n2 .0001s .0004s .0009s .0016s .0025s .0036s

n5 .1s 3.2s 24.3s 1.7m 5.2m 13m

3n .059s 58m 6.5y 3855c 2x108c 1.3x1013c

But computers are getting

faster! Maybe we can do

better.