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### For Wednesday

Things to Desire in a Program

- Correctness
- Robustness
- Maintainability
- Efficiency

Measuring Efficiency

- Empirical - use clock to get actual running times for different inputs
- Problems:
- Different machines have different running times
- Some running times are so long that it is impractical to check them

Measuring Efficiency

- Analytical - use pencil and paper to determine the abstract amount of work that a program does for different inputs
- Advantages:
- Machine independent
- Can predict running times of programs that are impractical to run

Complexity Analysis

- Pick an instruction that will run most often in the code
- Determine the number of times this instruction will be executed as a function of the size of the input data
- Focus on abstract units of work, not actual running time

Example: Search for the Minimum

defourMin(lyst):

minSoFar = lyst[0]

for item inlyst:

minSoFar = min(minSoFar, item)

returnminSoFar

We focus on the assignment (=) inside the loop and ignore

the other instructions.

for a list of length 1, one assignment

for a list of length 2, 2 assignments

.

for a list of length n, n assignments

Big-O Notation

- Big-O notation expresses the amount of work a program does as a function of the size of the input
- O(N) stands for order of magnitude N, or order of N for short
- Search for the minimum is O(N), where N is the size of the input (a list, a number, etc.)

Common Orders of Magnitude

Constant O(k)

Logarithmic O(log2n)

Linear O(n)

Quadratic O(n2)

Exponential O(kn)

Graphs of O(n) and O(n2)

Common Orders of Magnitude

n O(log2n) O(n) O(n2) O(2n)

2 1 2 4 4

4 2 4 16 64

8 3 8 64 256

16 4 16 256 65536

32 5 32 1024 4294967296

64 6 64 4096 19 digits

128 7 128 16384 yikes!

256 8 256 65536

512 9 512 262144

1024 10 1024 1048576

Approximations

- Suppose an algorithm requires exactly 3N + 3 steps
- As N gets very large, the difference between N and N + K becomes negligible (where K is a constant)
- As N gets very large, the difference between N and N / K or N * K also becomes negligible
- Use the highest degree term in a polynomial and drop the others (N2 – N)/2 N2

Example Approximations

n O(n) O(n) + 2 O(n2) O(n2) + n

2 2 4 4 6

4 4 6 16 20

8 8 10 64 72

16 16 18 256 272

32 32 34 1024 1056

64 64 66 4096 5050

128 128 130 16384 16512

256 256 258 65536 65792

512 512 514 262144 262656

1024 1024 1026 1048576 1049600

Example: Sequential Search

defourIn(target, lyst):

foritem inlyst:

ifitem == target:

returnTrue# Found target

returnFalse# Target not there

Which instruction do we pick?

How fast is its rate of growth as a function of n?

Is there a worst case and a best case? An average case?

Improving Search

- Assume data are in ascending order
- Goto midpoint and look there
- Otherwise, repeat the search to left or to right of midpoint

target

89

34

41

56

63

72

89

95

0 1 2 3 4 5 6

midpoint

3

left

right

Improving Search

- Assume data are in ascending order
- Goto midpoint and look there
- Otherwise, repeat the search to left or to right of midpoint

target

89

34

41

56

63

72

89

95

0 1 2 3 4 5 6

midpoint

5

left

right

Example: Binary Search

defourIn(target, sortedLyst):

left = 0

right = len(sortedLyst) - 1

while left <= right:

midpoint = (left + right) // 2

if target == sortedLyst[midpoint]:

returnTrue

elif target < sortedLyst[midpoint]:

right = midpoint - 1

else:

left = midpoint + 1

returnFalse

Analysis

while left <= right:

midpoint = (left + right) // 2

if target == sortedLyst[midpoint]:

return True

elif target < sortedLyst[midpoint]:

right = midpoint - 1

else:

left = midpoint + 1

How many times will == be executed in the worst case?

Selection Sort

- For each position i in the list
- Select the smallest element from i to n - 1
- Swap it with the ith one

Trace

i

Step 1: find the smallest element

89

56

63

72

41

34

95

0 1 2 3 4 5 6

smallest

i

Step 2: swap with first element

34

56

63

72

41

89

95

0 1 2 3 4 5 6

i

Step 3: advance i and goto step 1

34

56

63

72

41

89

95

0 1 2 3 4 5 6

Design of Selection Sort

for each i from 0 to n - 1

minIndex = minInRange(lyst, i, n)

ifminIndex != i

swap(lyst, i, minIndex)

- minInRange returns the index of the smallest element
- swap exchanges the elements at the specified positions

Implementation

defselectionSort(lyst):

n = len(lyst)

foriinrange(n):

minIndex = minInRange(lyst, i, n)

ifminIndex != i:

swap(lyst, i, minIndex)

Implementation

defselectionSort(lyst):

n = len(lyst)

foriinrange(n):

minIndex = minInRange(lyst, i, n)

ifminIndex != i:

swap(lyst, i, minIndex)

defminInRange(lyst, i, n):

minValue = lyst[i]

minIndex = i

for j inrange(i, n):

iflyst[j] < minValue:

minValue = lyst[j]

minIndex = j

returnminIndex

Implementation

defselectionSort(lyst):

n = len(lyst)

foriinrange(n):

minIndex = minInRange(lyst, i, n)

ifminIndex != i:

swap(lyst, i, minIndex)

defminInRange(lyst, i, n):

minValue = lyst[i]

minIndex = i

for j inrange(i, n):

iflyst[j] < minValue:

minValue = lyst[j]

minIndex = j

returnminIndex

defswap(lyst, i, j):

lyst[i], lyst[j] = lyst[j], lyst[i]

Analysis of Selection Sort

- The main loop runs approximately n times
- Thus, the function minInRange runs n times
- Within the function minInRange, a loop runs n - i times

Analysis of Selection Sort

Overall, the number of comparisons performed

in function minInRange is

n - 1 + n - 2 + n - 3 + . . + 1 = (n2 – n) / 2

n2

Finding Faster Algorithms

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