Lecture 4. RAM Model, Space and Time Complexity 1

Lecture 4.

RAM Model, Space and Time Complexity

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Recap

Time and space tradeoff factors should be focus during designing of algorithms,

Performance of algorithm should be measure rather than performance of its implemented program.

Switching between CPU and IO circuit should be in mind before writing the instructions of an algorithm.

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The RAM Model

Random Access Machine (not R.A. Memory)An idealized notion of how the computer

works– Each "simple" operation (+, -, =, if) takes exactly

1 step. – Each memory access takes exactly 1 step– Loops and method calls are not simple

operations, but depend upon the size of the data and the contents of the method.

Measure the run time of an algorithm by counting the number of steps.

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Random Access Machine

A Random Access Machine (RAM) consists of:– a fixed program– an unbounded memory– a read-only input tape– a write-only output tape

Each memory register can hold an arbitrary integer (*).

Each tape cell can hold a single symbol from a finite alphabet s.

. . .

Program

Memory

...

. . .

output tape

input tape

Instruction set:x y, x y {+, -, *, div, mod} zgoto labelif y {<, , =, ,> , } z goto labelx input, output yhalt

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Space Complexity

The amount of memory required by an algorithm to run to completion– The term memory leaks refer to the

amount of memory required is larger than the memory available on a given system

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Space Complexity (Cont !!!)

Fixed part: The size required to store certain data/variables, that is independent of the size of the problem:

– Such as int a (2 bytes, float b (4 bytes) etc

Variable part: Space needed by variables, whose size is dependent on the size of the problem:– Dynamic array a[]

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Space Complexity (cont’d)

S(P) = c + S(instance characteristics)– c = constant

Example:void float sum (float* a, int n) {

float s = 0; for(int i = 0; i<n; i++) { s+ = a[i]; } return s;}Constant Space: one for n, one for a [passed by reference!], one for s, one for

I , constant space=c=47

Running Time of Algorithms

Running time– depends on input size n

size of an array

polynomial degree

# of elements in a matrix

# of bits in the binary representation of the input

vertices and edges in a graph

– number of primitive operations performed Primitive operation

– unit of operation that can be identified in the pseudo-code

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Steps To determine Time Complexity

Step-1. Determine how you will measure input size. Ex: – N items in a list– N x M table (with N rows and M columns)– Two numbers of length N

Step-2. Choose the type of operation (or perhaps two operations)– Comparisons– Swaps– Copies– Additions

Note: Normally we don't count operations in input/output.

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Steps To determine Time Complexity (Cont !!!)

Step-3. Decide whether you wish to count operations in the– Best case? - the fewest possible operations

– Worst case? - the most possible operations

– Average case? This is harder as it is not always clear what is meant by an "average case". Normally calculating this case requires some higher mathematics such as probability theory.

Step-4. For the algorithm and the chosen case (best, worst, average), express the count as a function of the input size of the problem.

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Steps To determine Time Complexity (Cont !!!)

5. Given the formula that you have determined, decide the complexity class of the algorithm.

Q. What is the complexity class of an algorithm?

A. a complexity class is a set of problems of related resource-based complexity, such as P, NP classesQ. Is there really much difference between

3n

5n + 20

and 6n -3

A. Yes but when n is large?

Primitive Operations in an algorithm

Assign a value to a variable (i.e. a=5) Call a method (i.e. method()) Arithmetic operation (i.e. a*b, a-b*c) Comparing two numbers ( i.e. a<=b,

a>b &&a>c) Indexing into an array (i.e. a[0]=5) Following an object reference (i.e. Test

obj) Returning from a method (i.e. return I )

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Assignment Statement

The running time of a an assignment is considered as constant

Examples:– A=5; – A[5]=7– C=a+b;

Note:- Running time of assignments can be considered as 1 or C (Refer to constant values)

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for Loops

The running time of a for loop is at most the running time of the statements inside the for loop (including tests) times the number of iterations.

ExampleFor(i=0;i<=n-1;i++)A[i]=0; Note: (number of basic steps in loop body) * (number of iterations)

Let1.running time of basic operations is constant C 2.and loop is iterated n times thenTotal Running time will be

n*c

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Nested for Loops

The total running time of a statement inside a group of nested for loops is the running time of the statement multiplied by the product of the sizes of all the for loops.

Example for(i=0;i<=n-1;i++) for(j=0;j<=m-1;j++) K++;

Let1.running time of basic operations is constant C 2.Running time of outer loop (i.e. i) is n3.And running time of Inner loop (i.e. j) is m4.Total Running time will be

m*n*c

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Consecutives for Loops

The total running time of consecutive loops is the sum of running of all consecutives loops

Example for(i=0;i<=n-1;i++) K++; for(j=0;j<=m-1;j++) K++;

Let1.running time of basic operations is constant C 2.Running time of loop (i.e. i) is n3.And running time of loop (i.e. j) is m4.Total Running time will be

n*c + m * c

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Conditional Statements

The running time of an if-else statement is never more than the running time of the test plus the larger of the running times of S1 and S2.

Exampleif(Condition)

S1; else

S2;Note: Number of basic steps on branch that is executed

Let1.Condition is true and path have one basic operation (i.e. S1) then Ruining time will be constant C12.Similarly, if condition is false and path have one basic operation (i.e. S2) then Ruining time will be constant C2

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Primitive Operations: Example -1

Count the number of primitive operations in this program :Int method(){

i = 0; a = 0; for (i=1;i<=n;i++)

{ printf(“%d”,i);a=a+i; }

return I}

Primitive Operations

Yes/No

Total

Assign a value to a variable

Call a method

Arithmetic operation

Comparing two numbers

Indexing into an array

Following an object reference

Returning from a method

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Primitive Operations: Example -1 (Cont !!!)


Yes/No

Total


Yes 5

Call a method No 0


Yes 2


Yes 1


No 0


No 0


Yes 1

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Primitive Operations - Example 2

Count the number of primitive operations in this program :

Int method(){ i = 0;

a = 0; for (i=1;i<=n;i++)

{ printf(“%d”, i); for (a=1;a<=n;a++)

printf(“%d”,a); }

return I}


Yes/No

Total


Call a method






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Primitive Operations - Example 2 (Cont !!!)


Yes/No Total


yes 6

Call a method No 0

Arithmetic operation yes 2


yes 2

Indexing into an array No 0


No 0


yes 1

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Summary

RAM model is used to measure the run time of an algorithm by counting the number of steps.Space complexity of an algorithm refer to the amount of memory required to run.Time complexity of an algorithm refer to its running time, which depends on input size.Primitive operations refer to the unit of operation that can be identified in the pseudo-code

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In Next Lecturer

In next lecture, we will talk about the types of algorithm complexity and how to find running time of different pseudo codes.

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Lecture 4. RAM Model, Space and Time Complexity 1