Runtime with Functions CSC 172 SPRING 2002 LECTURE 9.

Runtime with Functions

CSC 172

SPRING 2002

LECTURE 9

Programs with Method Calls

Establish a size measure n (for each function) Let Tf(n) be the running time of function f

Include the running time of any function calls when evaluating a simple or a compound statement i.e. int y = fibonacci(x);

Do not put big-Oh around the functions runtimeWe must keep track of how many calls

Non-recursive Functions

As long as there is no recursion we can use a structure tree

Recursion

Define running time Tf(n) recursively In terms of itself with argument smaller than n

Solve the recurrence relationRepeated expansionGuess and check – based on common forms

Tf(n) subsumes O(1)

In general, big-Oh and T(n) cannot be combinedWe need to make the combination explicit in the proof

Example: printing a list

public void printList(Node l) {

if (l == null) return;

else {

System.out.println(l.getData());

printList(l.getNext());

}

}

A linked List

6 1 3 2 5 9

A linked List in Printlist

6 1 3 2 5 9

PrintList l


6 1 3 2 5 9

PrintList l

Output:6


6 1 3 2 5 9

PrintList l

Output:61


6 1 3 2 5 9

PrintList l

Output:613


6 1 3 2 5 9

PrintList l

Output:6132


6 1 3 2 5 9

PrintList l

Output:61325


6 1 3 2 5 9

PrintList l

Output:613259

Example: printing a linked list

public void printList(Node l) {

if (l == null) return;

else {

System.out.println(l.getData());

printList(l.getNext());

}

}

//O(1)

//O(1)

//T(n-1)

// n == l.length// TprintList(n) == ?

// TprintList(n) == O(1) + O(1) + TprintList(n-1)

Recurrence Relation for printList

Basis

n == 0

TprintList(0) = O(1)

Induction

TprintList(n) = O(1) + TprintList(n-1)

Replace O(1)s by concrete constants

TprintList(0) = a

TprintList(n) = b + TprintList(n-1)

Repeated expansion:

TprintList(n) = b + (b + TprintList(n-2) )

TprintList(n) = b + (b + (b + TprintList(n-3) )

TprintList(n) = b + (b + (b + (b + TprintList(n-4) )

Repeated ExpansionTprintList(n) = b + TprintList(n-1)

TprintList(n) = b + (b + TprintList(n-2) )

TprintList(n) = b + (b + (b + TprintList(n-3) )

TprintList(n) = b + (b + (b + (b + TprintList(n-4) )

= 2b + TprintList(n-2)



= (n-1)b + (b + TprintList(0))

. . .

Repeated Expansion

TprintList(n) = (n-1)b + (b + TprintList(0))

TprintList(n) = nb + a

Now, we have eliminated the recursive callSo, we can replace the unknown constants by Big-Oh

TprintList(n) = nb + a

TprintList(n) = nO(b) + O(a)

TprintList(n) = nO(1) + O(1)

TprintList(n) = O(n)

What is the RR for split?public static Node split(Node head){

Node secondNode;if (head == null) return null;else if (head.getNext() == null) return null;else {

secondNode = head.getNext();head.setNext(secondNode.getNext());

secondNode.setNext(split(secondNode.getNext());return secondNode;

}}

//O(1)//O(1)//O(1)

//O(1)//O(1)

//O(1)

What is the RR for split?public static Node split(Node head){

Node secondNode;if (head == null) return null;else if (head.getNext() == null) return null;else {

secondNode = head.getNext();head.setNext(secondNode.getNext());

secondNode.setNext(split(secondNode.getNext());return secondNode;

}}

//O(1)//O(1)//O(1)

//O(1)//O(1)

//O(1)

//O(1) + T(n-1)

What is the RR for Merge?public static Node merge(Node list1, Node list2){

if (list1 == null) return list2;else if (list2 == null) return list1;else if (list1.getData.compareTo(list2.getData()) < 1) {

list1.setNext(merge(list1.getNext(),list2);return list1;

} else {list2.setNext(merge(list1,list2.getNext());return list2;

}}

What is the RR for MergeSort?

public static Node mergeSort(Node list){

Node secondList;

if (list == null) return null;

else if (list.getNext() == null) return list;

else {

secondList = split(list);

return merge(mergeSort(list),mergeSort(secondList));

}

}

//O(1)//O(1)//O(1)

//O(n)



Node secondList;



else {



}

}

//O(1)//O(1)//O(1)

//O(n)

What is the length of list? secondList?

length(list) == length(secondList) = n/2



Node secondList;



else {



}

}

//O(1)//O(1)//O(1)

//O(n)

What is the length of list? secondList?

length(list) == length(secondList) = n/2

So, what is the cost of the call to merge?

//O(n)



Node secondList;



else {



}

}

//O(1)//O(1)//O(1)

//O(n)

If the cost of mergeSort is TmergeSort(n):

What is the cost of mergeSort is TmergeSort(list)?

TmergeSort(secondlist)?//O(n)



Node secondList;



else {


return merge(mergeSort(list),mergeSort(secondList) );

}

}

//O(1)//O(1)//O(1)

//O(n)

//O(n)

Tmergesort(n) = O(1)+O(n)+2Tmergesort(n/2)

Recurrence Relation for MergeSort

Basis

n == 0

TmergeSort(0) = O(1)

Induction

TmergeSort(n) = O(n) + 2TmergeSort(n/2)

Replace O(1)s by concrete constantsTmergeSort(1) = a

TmergeSort(n) = bn + 2TmergeSort(n/2)

TmergeSort(n)/n = b + TmergeSort(n/2)/(n/2)

TmergeSort(n/2)/(n/2) = b + TmergeSort(n/4)/(n/4)

TmergeSort(n/4)/(n/4) = b + TmergeSort(n/8)/(n/8)

TmergeSort(n/8)/(n/8) = b + TmergeSort(n/16)/(n/16)…

TmergeSort(2)/(2) = b + TmergeSort(1)/(1) How many calls?

How many calls?

log2n calls

TmergeSort(n)/n = blog2n + TmergeSort(1)/(1)

TmergeSort(n) = nblog2n + na

TmergeSort(n) = O(nlog2n)

Guess and Check Approach

Use domain knowledge to “guess” a bound

Try to prove the bound inductively

Tms(n) <= c nlog2n + dn

assume n a power of 2

discover c and d

Basis

If n=1, Tms(n) = a

If a = Tms(1) <= c(1)log21+d(1),

Then d >= a since log21=0

InductionAssume: Tms(n/2) <= (cn/2)log2(n/2) + dn/2

Tms(n) = bn + 2Tms(n/2) <= bn + cn(log2n –1) + dn

We want to show

Tms(n) <= cnlog2n+dn

By showing

bn + cnlog2n – cn + dn <= cnlog2n + dn

i.e. bn<=cn

So, the proof holds if (c >= b) && (d >= a)

Let d == a and c == b

An Exponential Recurrence

How many strings of length n over symbols a,b,c have no identical consecutive symbols?

Basis: T(1) = 3 , they are “a”, “b”, and “c”Induction: Each string ends in a symbol, all strings

of length n-1 are represented, so each string can give rise to two more of length n by adding one of two more symbols on the endT(n) = 2T(n-1)

Expand

T(n) = 2T(n-1)

T(n) = 4T(n-2)

T(n) = 8T(n-3)

T(n) = 16T(n-4)

T(n) = 32T(n-5)

…

T(n) = 2n-1 T(1) = 3 * 2n-1

Runtime with Functions CSC 172 SPRING 2002 LECTURE 9.

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