updates hw#1 has been delayed until next monday. there were two errors in the assignment. 2-1. merge...
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Updates
• HW#1 has been delayed until next MONDAY. There were two errors in the assignment.
• 2-1. Merge sort runs in Θ(n log n). Insertion sort runs in Θ(n2). Consider a modification to Merge sort where n/k sublists of length k are sorted using Insertion sort and then merged by Merge sort.
Updates
• A) Show each n/k sublist can be sorted by Insertion Sort in Θ(n/k) worst case time.
• B) Show that the sublists can be merged in Θ(n lg(n/k)) worst case time.
• C) Skip.
• D) How should k be chosen in practice? (hint: philosophical)
Lecture 3: Solving Recurrences
Recurrences
• Recurrence: An equation that describes a function in terms of its value on smaller inputs.
• Example (MergeSort):T(n) = Θ(n) if n = 1,
2T(n/2) + Θ(n) if n>1
Recurrences
• We will study three techniques:1. The substitution method
2. The recursion-tree method
3. The master method
• Sometimes more than one works well, sometimes none work well.
The Substitution Method
The most general method:1. Guess the form of the solution.
2. Verify using induction.
3. Solve for the constants.
Substitution Method: Example 1
• Example: T(n) = 4T(n/2) + 100n
• Guess: T(n) = O(n3)
Substitution Method: Example 1• Example: T(n) = 4T(n/2) + 100n• Proof:
1. Assume that T(1) = Θ(1).2. Guess T(n) = O(n3). We want to prove T(n) < cn3 for some c>0.3. Verify by substituting into the recurrence.
We assume this holds for T(n/2) < c (n/2)3, thusT(n) < 4c(n/2)3 + 100n = cn3/2 + 100n= (c/2)n3 + 100n= cn3 - (c/2)n3 + 100n= cn3 - ((c/2)n3 - 100n) “residual”< cn3
when (c/2)n3 - 100n > 0, that is solving for constants, when c > 200 and n > 1.
Always verify the base case(s) as well. Here Θ(1) < cn3 for sufficiently large c.
A Tighter Bound?
• Example: T(n) = 4T(n/2) + 100n
• Guess: T(n) = O(n2)
A Tighter Bound?
• Example: T(n) = 4T(n/2) + 100n• Proof:
1. Assume that T(1) = Θ(1).2. Guess T(n) = O(n2). We want to prove T(n) < cn2 for some c>0.3. Verify by substituting into the recurrence.
We assume this holds for T(n/2) < c (n/2)2, thusT(n) < 4c(n/2)2 + 100n = cn2 + 100n= cn2 – (-100n) residual?< cn2
when -100n > 0. No valid assignment of c and n > 0 can be made! We have made a bad guess.We MUST prove the EXACT form of the inductive hypothesis (our guess).
Subtleties..
• Example: T(n) = 4T(n/2) + 100n
• Guess T(n) = O(n2 -n).
Subtleties..
• Example: T(n) = 4T(n/2) + 100n• Proof:
1.Assume that T(1) = Θ(1).
2.Guess T(n) = O(n2 -n). We want to prove T(n) < c1n2 -c2 for some c1>0 and c2>0.
3.Verify by substituting into the recurrence.We assume this holds for T(n/2) < c1(n/2)2-c2n/2, thusT(n) < 4c1(n/2)2 - 4c2n/2 + 100n = c1n2 - 2c2n + 100n= c1n2 - c2n - (c2n - 100n) residual< c1n2 - c2n when c2n-100n > 0, that is, c2>100.
Recursion Tree Method
• The Recursion Tree Method doesn’t necc. build a proof, but a good guess. You might use this method with the substitution method.
• The idea is to draw the tree and do the “accounting” the way we did with MergeSort last time.
• This can be unreliable, but it provides good intuition.
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
Example of recursion tree
T(n)
Solve T(n) = T(n/4) + T(n/2) + n2:
Example of recursion tree
T(n/4) T(n/2)
n2
Solve T(n) = T(n/4) + T(n/2) + n2:
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
n2
(n/4)2 (n/2)2
T(n/16) T(n/8) T(n/8) T(n/4)
Example of recursion tree
(n/16)2 (n/8)2 (n/8)2 (n/4)2
(n/4)2 (n/2)2
(1)
…
Solve T(n) = T(n/4) + T(n/2) + n2:
n2
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
(n/16)2 (n/8)2 (n/8)2 (n/4)2
(n/4)2 (n/2)2
(1)
…
2nn2
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
(n/16)2 (n/8)2 (n/8)2 (n/4)2
(n/4)2 (n/2)2
(1)
…
2165 n
2nn2
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
(n/16)2 (n/8)2 (n/8)2 (n/4)2
(n/4)2
(1)
…
2165 n
2n
225625 n
n2
(n/2)2
…
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
(n/16)2 (n/8)2 (n/8)2 (n/4)2
(n/4)2
(1)
…
2165 n
2n
225625 n
13
1652
165
1652 n
…
Total == (n2)
n2
(n/2)2
geometric series
The Master Method
• The Master Theorem: Let a > 1 and b > 1 be constants. Let f(n) be a function and T(n) be defined on the non-negative integers as,T(n) = a T(n/b) + f(n).
• T(n) can be asymptotically bounded as follows:1. If f(n) = O(nlogba-ε), then T(n) = Θ(nlogba)
2. If f(n) = Θ(nlogba), then T(n) = Θ(nlogbalg n)
3. If f(n) = Ω(nlogba+ ε) and if af(n/b)<cf(n) for c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).