introduction to algorithms 6.046j/18.401j/sma5503 lecture 1 prof. charles e. leiserson
TRANSCRIPT
![Page 1: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/1.jpg)
Introduction to Algorithms
6.046J/18.401J/SMA5503
Lecture 1Prof. Charles E. Leiserson
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Welcome to Introduction toAlgorithms, Fall 2001
Handouts
1. Course Information
2. Calendar
3. Registration (MIT students only)
4. References
5. Objectives and Outcomes
6. Diagnostic Survey
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Course information
1. Staff 8. Website2. Distance learning 9. Extra help3. Prerequisites 10.Registration (MIT
only)4. Lectures 11.Problem sets5. Recitations 12.Describing algorithms6. Handouts 13.Grading policy7. Textbook (CLRS) 14.Collaboration policy
▲Course information handout
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Analysis of algorithms
The theoretical study of computer-program
performance and resource usage.
What’s more important than performance?
• modularity • user-friendliness
• correctness • programmer time
• maintainability • simplicity
• functionality • extensibility
• robustness • reliability
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Why study algorithms andperformance?
• Algorithms help us to understand scalability.
• Performance often draws the line between what
is feasible and what is impossible.
• Algorithmic mathematics provides a language
for talking about program behavior.
• The lessons of program performance generalize
to other computing resources.
• Speed is fun!
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The problem of sorting
Input: sequence <a1, a2, …, an> of numbers.
Output: permutation <a'1, a'2, …, a'n> such
that a'1 ≤ a'2 ≤ … ≤ a'n
Example:
Input: 8 2 4 9 3 6
Output: 2 3 4 6 8 9
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Insertion sort
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Example of insertion sort
8 2 4 9 3 6
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Example of insertion sort
8 2 4 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
2 4 8 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
2 4 8 9 3 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
2 4 8 9 3 6
2 3 4 8 9 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
2 4 8 9 3 6
2 3 4 8 9 6
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Example of insertion sort
8 2 4 9 3 6
2 8 4 9 3 6
2 4 8 9 3 6
2 4 8 9 3 6
2 3 4 8 9 6
2 3 4 6 8 9 done
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Running time
• The running time depends on the input: an
already sorted sequence is easier to sort.
• Parameterize the running time by the size of
the input, since short sequences are easier to
sort than long ones.
• Generally, we seek upper bounds on the
running time, because everybody likes a
guarantee.
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Kinds of analyses
Worst-case: (usually)
• T(n) = maximum time of algorithm
on any input of size n.
Average-case: (sometimes)
• T(n) = expected time of algorithm
over all inputs of size n.
• Need assumption of statisticaldistribution of inputs.Best-case: (bogus)
• Cheat with a slow algorithm thatworks fast on some input.
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Machine-independent time
What is insertion sort’s worst-case time?
• It depends on the speed of our computer:
• relative speed (on the same machine),
• absolute speed (on different machines).
BIG IDEA:
• Ignore machine-dependent constants.
• Look at growth of T(n) as n → ∞ .
“Asymptotic Analysis”
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Θ-notation
Math:
Θ(g(n)) = { f (n) : there exist positive constants c1, c2, and n0 such that 0 ≤ c1 g(n)
≤ f (n) ≤ c2 g(n) for all n ≥ n0 }
Engineering:
• Drop low-order terms; ignore leading constants.
• Example: 3n3 + 90n2 – 5n + 6046 = Θ(n3)
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Asymptotic performance
When n gets large enough, a Θ(n2) algorithmalways beats a Θ(n3) algorithm.
• We shouldn’t ignore asymptotically slower algorithms, however.• Real-world design situations often call for a careful balancing of engineering objectives.• Asymptotic analysis is a useful tool to help to structure our thinking.
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Insertion sort analysis
Worst case: Input reverse sorted.
[arithmetic series]
Average case: All permutations equally likely.
Is insertion sort a fast sorting algorithm?
• Moderately so, for small n.
• Not at all, for large n.
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Merge sort
MERGE-SORT A[1 . . n]
1. If n = 1, done.
2. Recursively sort A[ 1 . . n/2 ] and A[ n/2 +1 . . n ] .
3. “Merge” the 2 sorted lists.
Key subroutine: MERGE
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Merging two sorted
20 1220 12
13 1113 11
7 97 9
2 12 1
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Merging two sorted
20 1220 12
13 1113 11
7 97 9
2 12 1
11
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Merging two sorted
20 12 20 1220 12 20 12
13 11 13 1113 11 13 11
7 9 7 97 9 7 9
2 1 2 2 1 2
11
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Merging two sorted
20 12 20 1220 12 20 12
13 11 13 1113 11 13 11
7 9 7 97 9 7 9
2 1 2 2 1 2
1 1 2 2
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Merging two sorted
20 12 20 12 20 1220 12 20 12 20 12
13 11 13 11 13 1113 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9
2 1 2 2 1 2
1 21 2
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Merging two sorted arrays
20 12 20 12 20 1220 12 20 12 20 12
13 11 13 11 13 1113 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9
2 1 2 2 1 2
1 1 2 2 7 7
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Merging two sorted arrays
20 12 20 12 20 12 20 1220 12 20 12 20 12 20 12
13 11 13 11 13 11 13 1113 11 13 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9 9 9
2 1 2 2 1 2
1 1 2 2 7 7
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○
Merging two sorted arrays
20 12 20 12 20 12 20 1220 12 20 12 20 12 20 12
13 11 13 11 13 11 13 1113 11 13 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9 9 9
2 1 2 2 1 2
1 1 2 2 7 7 99
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Merging two sorted arrays
20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12
13 11 13 11 13 11 13 11 13 1113 11 13 11 13 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9 9 9
2 1 2 2 1 2
1 1 2 2 7 9 7 9
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Merging two sorted arrays
20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12 20 12
13 11 13 11 13 11 13 11 13 1113 11 13 11 13 11 13 11 13 11
7 9 7 9 7 97 9 7 9 7 9 9 9
2 1 2 2 1 2
1 1 2 2 7 9 7 9 11 11
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Merging two sorted arrays
20 12 20 12 20 12 20 12 20 12 20 1220 12 20 12 20 12 20 12 20 12 20 12
13 11 13 11 13 11 13 11 13 11 13 13 11 13 11 13 11 13 11 13 11 13
7 9 7 9 7 9 97 9 7 9 7 9 9
2 1 2 2 1 2
1 1 22 7 9 7 9 11 11
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Merging two sorted arrays
20 12 20 12 20 12 20 12 20 12 20 1220 12 20 12 20 12 20 12 20 12 20 12
13 11 13 11 13 11 13 11 13 11 13 13 11 13 11 13 11 13 11 13 11 13
7 9 7 9 7 9 97 9 7 9 7 9 9
2 1 2 2 1 2
1 1 22 7 9 7 9 11 11 12 12
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Merging two sorted arrays
20 12 20 12 20 12 20 12 20 12 20 1220 12 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 13 13 11 13 11 13 11 13 11 13 11 13 7 9 7 9 7 9 97 9 7 9 7 9 9 2 1 2 2 1 2
1 1 22 7 9 7 9 11 11 12 12
Time = Θ(n) to merge a total of n elements (linear time).
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Analyzing merge sort
T(n) MERGE-SORT A[1 . . n]
Θ(1) 1. If n = 1, done.
2T(n/2) 2. Recursively sort A[ 1 . . n/2 ]Abuse and A[ n/2 +1 . . n ] .
Θ(n) 3. “Merge” the 2 sorted lists
Sloppiness: Should be T( n/2 ) + T( n/2 ) ,
but it turns out not to matter asymptotically.
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Recurrence for merge sort
T(n) = Θ(1) if n = 1;
2T(n/2) + Θ(n) if n > 1.
• We shall usually omit stating the base
case when T(n) = Θ(1) for sufficiently
small n, but only when it has no effect on
the asymptotic solution to the recurrence.
• CLRS and Lecture 2 provide several ways
to find a good upper bound on T(n).
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
T(n)
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
T(n/2) T(n/2)
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/2 cn/2
T(n/4) T(n/4) T(n/4) T(n/4)
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/2 cn/2
T(n/4) T(n/4) T(n/4) T(n/4)
Θ(1)
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/2 cn/2
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4)
Θ(1)
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Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
cn/2 cn/2
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4)
Θ(1)
![Page 48: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/48.jpg)
Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
cn/2 cn/2 cn
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4)
Θ(1)
![Page 49: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/49.jpg)
Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
cn/2 cn/2 cn
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4) cn
Θ(1)
![Page 50: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/50.jpg)
Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
cn/2 cn/2 cn
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4) cn
Θ(1) Θ(n)#leaves = n
![Page 51: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/51.jpg)
Recursion tree
Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
cn/2 cn/2 cn
h = lg n
T(n/4) T(n/4) T(n/4) T(n/4) cn
Θ(n)
Θ(1) Θ(n lg n)#leaves = n
![Page 52: Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 1 Prof. Charles E. Leiserson](https://reader033.vdocuments.net/reader033/viewer/2022061605/56649eb55503460f94bbd8b9/html5/thumbnails/52.jpg)
Conclusions
• Θ(n lg n) grows more slowly than Θ(n2).
• Therefore, merge sort asymptotically
beats insertion sort in the worst case.
• In practice, merge sort beats insertion
sort for n > 30 or so.
• Go test it out for yourself!