general announcements

General AnnouncementsGeneral Announcements

Project Due Friday, 1/30Labs start Wednesday & Thursday

– Java review– Weiss 1.19, 1.20– You may show up & hand in

Workshops start Sunday– Why do we do workshops

WORKSHOPSWORKSHOPS

“Analysts say tech employers today seek workers who are well educated in math and science but they also want them to have intangible skills, such as the ability to work well in teams.”

-”New Tech Products Mean New Tech Jobs”-Brian Deagon, IBD, 1/20/04

PROOF PROOF IN IN

COMPUTER SCIENCECOMPUTER SCIENCE

CSC 172 SPRING 2004

LECTURE 2

ExampleExample

Write a method to compute an

public static double power(int a , int n)

You have 5 minutes

Possible solutionPossible solution

public static double power(int a, int n) {

double r = 1; double b = a; int i = n ;

while (i>0){

if (i%2 == 0) { b = b * b; i = i / 2;}

else { r = r * b; i--; }

return r;

}

Does it work?Does it work?

SURE! TRUST ME!

Well, look at a100 if you don’t believe me!– Note, less loops!

Can you “prove” that it works?

b i r

a 100 1

a2 50

a4 25

24 a4

a8 12

a16 6

a32 3

2 a36

a64 1

0 a100

Loop invariantsLoop invariants

In order to verify loops we often establish an assertion (boolean expression) that is true each time we reach a specific point in the loop.

We call this assertion, a loop invariant

AssertionsAssertions

When ever the program reaches the top of the while loop, the assertion is true

INIT

BODY

TEST

INVARIANT

What is the loop invariant?What is the loop invariant?

At the top of the while loop, it is true that

r*bi = an

It is?– Well, at the top of the first loop

r==1 b==a i==n

So, if it’s true at the startSo, if it’s true at the start

Even case

rnew= rold

bnew == (bold)2

inew==(iold)/2Therefore,

– rnew * (bnew)inew == rold * ((bold)2)iold/2

– == rold * ((bold)2)iold

– == an

So, if it’s true at the start IISo, if it’s true at the start II

Odd case

rnew= rold*bold

bnew == bold

inew==iold-1Therefore,

– rnew * (bnew)inew == rold *bold* (bold)iold-1

– == rold * (bold)iold

– == an

So, So,

If it’s true at the startAnd every time in the loop, it remains trueThen, it is true at the end

r*bi = an

And, i == 0 ( the loop ended)What do we know?

Correctness ProofsCorrectness Proofs

Proof are more valuable than testing– Tests demonstrate limited correctness– Proofs demonstrate correctness for all inputs

For some time, people hoped that all formal logic would replace programming

The naïve idea that “programming is a form of math” proved to be an oversimplification

Correctness ProofsCorrectness Proofs

Unfortunately, in practice, these methods never worked very well.– Instead of buggy programs, – people wrote buggy logic

Nonetheless, the approach is useful for program analysis

The take away message?The take away message?

In the end, engineering and (process) management are at least as important as mathematics and logic for the successful completion of large software projects

ExampleExample

One dimensional pattern recognitionInput: a vector x of n floating point

numbersOutput: the maximum sum found in any

contiguous subvector of the input.

X[2..6] or 187How would you solve this?

31 -41 59 26 -53 58 97 -93 -23 84

Obvious solutionObvious solution

Check all pairs

int sum; int maxsofar = 0;

for (int i = 0; i<x.length;i++)

for (int j = i; j<x.length;j++){

sum = 0;

for (int k = i;k<=j;k++) sum += x[k];

maxsofar = max(sum,maxsofar);

}

An improved solutionAn improved solutionint maxSum = 0 ;

for (int i = 0; i<a.length;i++) {

int thisSum = 0;

for (int j = i; j<a.length;j++){

thisSum += a[j];

if (thisSum > maxSum)

maxSum = thisSum;

}

}

return maxSum;

How many loops ?How many loops ?int maxSum = 0 ;

for (int i = 0; i<a.length;i++) {

int thisSum = 0;

for (int j = i; j<a.length;j++){

thisSum += a[j];

if (thisSum > maxSum)

maxSum = thisSum;

}

}

return maxSum;

Total number of comparisonsTotal number of comparisons

N + (N-1) + (N-2) + . . . + 1

Reversing

1 + 2 + 3 + . . . + N =

n

i

i1

Prove

In order to calculate workIn order to calculate work

2

)1(

1

nni

n

i

Simple InductionSimple Induction Three Pieces

1. A statement S(n) to be proved The statement must be about an integer parameter n

2. A basis for the proof The statement S(b) for some specific integer b Often b==0 or b==1

3. An inductive step for the proof Show that “If S(n) is true, then S(n+1) must also be true” Prove the statement “S(n) implies S(n+1)” for any n. For this part, you get to “suppose” S(n) is true

– “For the sake of argument”– Aka the inductive hypothesis

Prove: Prove:

1. Statement:

S(n) : For any n>=1

2

)1(

1

nni

n

i

2

)1(

1

nni

n

i

Prove: Prove:

2. Basis

Select n == 1

2

)1(

1

nni

n

i

2

)11(11

1

i

i

Prove: Prove:

2. Alternate Basis

Select n == 2

2

)1(

1

nni

n

i

2

)12(22

1

i

i

212

1

i

i

Prove: Prove:

2. Alternate Basis

Select n == 3

2

)1(

1

nni

n

i

2

)13(33

1

i

i

3213

1

i

i

Prove: Prove:

3. Inductive Step

Assume:

To show:

2

)1(

1

nni

n

i

2

)1(

1

nni

n

i

2

)2)(1(1

1

nni

n

i

Inductive StepInductive Step

We know, by definition:

Rewrite it:

)1(...321)1(

1

nnin

i

)1(1

)1(

1

niin

i

n

i

Inductive StepInductive Step

“By the Induction hypothesis”

(we can make the following substitution)

)1(1

)1(

1

niin

i

n

i

)1(2

)1()1(

1

nnn

in

i

Inductive StepInductive StepThe rest is just algebra

)1(2

)1()1(

1

nnn

in

i

2

)1(2

2

)1()1(

1

nnni

n

i

2

)1(2

2

)1()1(

1

nnni

n

i

2

)1(2)1()1(

1

nnni

n

i

2

222)1(

1

nnni

n

i

2

232)1(

1

nni

n

i

2

)2)(1()1(

1

nni

n

i

Which, of course, is what we set out to prove!

So, what did we doSo, what did we do

We showed that it worked for 1And that if it worked for n, it must work for n+1

So, is it true for n==7?Why?

Is it true for n==984375984237598437594373457?

Template for Simple InductionTemplate for Simple Induction1. State what S(n) is.

2. Explain what n represents. “any positive integer” or “length of the string”

3. Tell what the value of n is for the basis case n==b

4. Prove S(b)

5. State that you are assuming n>=b and S(n)

6. Prove S(n+1) using the assumptions (say: “B.T.I.H.”)

7. State that due to (4) and (6) you conclude S(n) for all n>=b

Interesting Aside: Visual ProofInteresting Aside: Visual Proof

Proof is “convincing prose” Not all proof is “mathematical”

Visual ProofVisual Proof

1 2 3 . . n0

123

n

nin

i

...3211


1 .. n/2. . n0

123

n


1 .. n/20

123

n


1 .. n/20

123

nn+1 2

)1(

1

nni

n

i


1 ..(n+1)/2. .n0

123

n

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