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CMPT 128: Introduction to Computing Science for Engineering StudentsPowerPoint Presentation

CMPT 128: Introduction to Computing Science for Engineering Students

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CMPT 128: Introduction to Computing Science for Engineering Students

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CMPT 128: Introduction to Computing Science for Engineering Students

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CMPT 128: Introduction to Computing Science for Engineering Students

Running Time

Big O Notation

- We have seen that there are many algorithms to solve any problems.
- How to we choose the most efficient?
- What is efficient?
- One measure is how fast our algorithm can determine the solution
- This is not the only measure, nor is it always the best measure
- How do we measure ‘how fast’

- What contributes to how fast a program runs?
- The speed the CPU can process operations
- The efficiency of your code (the number of operations needed to complete your calculations)
- This depends on the algorithm used
- This depends of the particular implementation of the algorithm
- This may depend on the size of the data set being analyzed

- How many other things your computer is doing at the same time

- Two approaches
- Analyze your algorithm/code
- determine an upper limit on the number of operations needed
- Know the speed of your CPU
- Calculate an upper limit on running time for an input data set of a particular size

- Implement your algorithm and make measurements of how long it takes to run for data sets of varying sizes
- Create a common baseline, run tests on same machine with same background load
- Disadvantage: you already have spend the time coding and testing if the algorithm is not practical this may have been wasted

- Consider the operation used in your code
- +, -, *, /, %, <, <=, >, >=, ==, =, !=, &, !, &&, || …
- Make a simplifying assumption that each of these operations take the same length of time to execute
- Now we just need to count the operations in your program to get an estimate of ‘how fast’ it will run
- This estimate is independent of the machine on which the code runs.
- Once we know the time take by an ‘operation’ on our machine we immediately know how long our code will take

- Simple linear or branching code
if( neighborcount > 3 || neighborcount < 2 )

{ nextGenBoard[indexrow][indexcol] = '.'; }

else if( neighborcount == 3 )

{ nextGenBoard[indexrow][indexcol] = organism;}

else

{ nextGenBoard[indexrow][indexcol] = lifeBoard[indexrow][indexcol]; }

- The first if executes three operations, >, ||, and <
- If the first if is true then the block of code above executes with 6 operation (3 in the if and [ ] , [ ], and =)
- If the first if is false then the elseif adds one more operation, == (giving a total of 3+1=4 operations)

if( neighborcount > 3 || neighborcount < 2 )

{ nextGenBoard[indexrow][indexcol] = '.'; }

else if( neighborcount == 3 )

{ nextGenBoard[indexrow][indexcol] = organism;}

else

{ nextGenBoard[indexrow][indexcol] = lifeBoard[indexrow][indexcol]; }

- If the first if and the else if are both false the last line of code executes with 5 operations ([ ], [ ], =, [ ], [ ]). In this case the block of code above executes with a total of 4+5= 9 operations
- If the elseif is true then the block of code above executes with a total of 4+3 = 7 operations (3 operations [ ], [ ], =)
- Take the worst case number of operations is 9

- While loop
count = 0;

while (count < n )

{

localSum = dataArray[count] + 2 * localSum;

count++;

}

- n determine the number of times through the while loop.
- Each time through the while loop, the body of the loop executes 5 operations( =, [ ], +, *, ++)
- Each time through the while loop the test is executed which adds another operation (Total operations per time through loop is 5+1=6)

- While loop
count = 0;

while (count < n )

{

localSum = dataArray[count] + 2 * localSum;

count++;

}

- Total operations each time through loop is 6
- The initialization of count takes one operation before the loop begins executing
- The loop is executed n times
- The number of operations is 6*n + 1

- While loop
count = 0;

while (count < n )

{

localSum = dataArray[count] + 2 * localSum;

count++;

}

- The number of operations is 6*n + 1
- The test in the while is executed one additional time at the end of the loop
- The number of operations is 6*n + 2

- While loop
count = 0;

while (count < n )

{

localSum = dataArray[count] + 2 * localSum;

count++;

}

- Total operations each time through loop is 6
- The initialization of count takes one operation before the loop begins executing
- The loop is executed n times
- The number of operations is 6*n + 1

- Nested Loops
for( k=0; k<n; k++)

{

for( j=0; j<n; j++)

{

matOut[k] = matIn1[k] * matIn2[j] + matin1[j] * matin2[k];

}

}

- The number of rows and columns determine the number of operations.
- A count of the number of operations will be a function of n
- The inner loop contains 9 operations so the number of operations inside the inner loop will be 9*n
- In order the operations are [ ], =, [ ], *, [ ], +, [ ], * [ ]

- Nested loops
for( k=0; k<n; k++)

{

for( j=0; j<n; j++)

{

matOut[k] = matIn1[k] * matIn2[j] + matin1[j] * matin2[k];

}

}

- The number of rows and columns determine the number of operations.
- Each time the inner for statement is executed 2 more operations occur (<, ++), so each time through the inner loop takes 11*n operations
- Before starting through the inner loop the initialization j=0 takes place, one more operation
- Therefore, each time through the body of the outer loop 11*n+1 operations

- Nested loops
for( k=0; k<n; k++)

{

for( j=0; j<n; j++)

{

matOut[k] = matIn1[k] * matIn2[j] + matin1[j] * matin2[k];

}

}

- The number of rows and columns determine the number of operations.
- Each time the outer for statement is executed 2 more operations occur (<, ++), so each time through the outer loop takes 3 +11*n operations or a total of (3 + 11*n) * n operations
- Before starting through the inner loop the initialization j=0 takes place, one more operation
- Total number of operation 1+(3 + 11*n) * n = 1+3n+11n2

- Nested loops
for( k=0; k<n; k++)

{

for( j=0; j<n; j++)

{

matOut[k] = matIn1[k] * matIn2[j] + matin1[j] * matin2[k];

}

}

- The number of rows and columns determine the number of operations.
- Each time the outer for statement is executed 3 more operations occur (<, ++, and one more test in inner loop), so each time through the outer loop takes 4 +11*n operations or a total of (4 + 11*n) * n operations
- Before starting through the inner loop the initialization j=0 takes place, one more operation, and one additional test occurs
- Total number of operation 2+(4 + 11*n) * n = 2+4n+11n2

- Estimate the order of the number of calculations needed
- Order is the largest power of n in the estimated upper limit of the number of operations

- For most n (amount of data) it is generally true that an order n algorithm is significantly faster than an order n+1 algorithm
- An algorithm with order n operations is said to run in linear time
- An algorithm with order n2 operations is said to run in quadratic time

- Looking for a ‘good’ upper limit
- Just consider the Order.
- The order is the largest power of n

- First example: 9 operations
- O(9) = 0 Order 0 (not a function of n)

- Second example: 6*n +1 operations
- O(6*n +1) = n Order 1 (largest power of n is 1)

- Third example: 1+3n+11n2
- O(1+3n+11n2) = n2 Order 2 (largest power of n is 2)

- How good are our estimates
- The estimates we have made are worst case estimates.
- In some cases algorithms will finish much faster if input data has particular properties
- Be careful the measurement is only as good as the assumptions

- We can directly measure ‘how fast’ for particular types of data sets of particular sizes
- You are doing this is your lab
- This is still a way to approximate performance in a general case on a wider variety of sizes
- Measure for some sizes
- Fit results with a curve, then predict using the curve what the performance would be expected to be