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# Computer Science 112 - PowerPoint PPT Presentation

Computer Science 112. Fundamentals of Programming II Searching, Sorting, and Complexity Analysis. Things to Desire in a Program. Correctness Robustness Maintainability Efficiency. Measuring Efficiency. Empirical - use clock to get actual running times for different inputs Problems:

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### Computer Science 112

Fundamentals of Programming II

Searching, Sorting, and Complexity Analysis

• Correctness

• Robustness

• Maintainability

• 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

• Analytical - use pencil and paper to determine the abstract amount of work that a program does for different inputs

• Machine independent

• Can predict running times of programs that are impractical to run

• 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

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 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.)

Constant O(k)

Logarithmic O(log2n)

Linear O(n)

Exponential O(kn)

Graphs of O(n) and O(n2)

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

• 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

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

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?

• 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

• 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

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

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?

89

56

63

72

41

34

95

0 1 2 3 4 5 6

sort

34

41

56

63

72

89

95

0 1 2 3 4 5 6

• For each position i in the list

• Select the smallest element from i to n - 1

• Swap it with the ith one

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

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

defselectionSort(lyst):

n = len(lyst)

foriinrange(n):

minIndex = minInRange(lyst, i, n)

ifminIndex != i:

swap(lyst, i, minIndex)

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

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]

• The main loop runs approximately n times

• Thus, the function minInRange runs n times

• Within the function minInRange, a loop runs n - i times

Overall, the number of comparisons performed

in function minInRange is

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

 n2

### For Wednesday

Finding Faster Algorithms