Download Divide and conquer

Recap: Oh, Omega, Theta
• Oh (like ≤)
O(n) is asymptotic upper bound 0 ≤ f(n) ≤ cg(n)
• Omega (like ≥)
Ω(n) is asymptotic lower bound 0 ≤ cg(n) ≤ f(n)
• Theta (like =)
Θ(n) is asymptotic tight bound 0 ≤ c1g(n) ≤ f(n) ≤ c2g(n)
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More on Oh, Omega, Theta
Theorem: f(n) = Θ(n) if and only if (iff)
f(n) = O(n) and f(n) = Ω(n)
Question: Is f(n) = n2 + 5 Ω(n3) ?
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Answer: NO (why?)!
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Question: Is f(n) = n2 + 5 Ω(n3) ?
Answer: NO!
Proof by contradiction
Definition: Ω (g(n)) = f(n): There exist positive constants
c, such that 0 ≤ cg(n) ≤ f(n) for all n ≥ n0
Therefore:
If f(n) = Ω(n3), then there exists n0; c such that for all n ≥ n0
n2 + 5 ≥ cn3
Remember from before: n2 + 5n2 ≥ n2 + 5
n2 + 5n2 ≥ cn3
6 ≥ cn Can’t be true for all n ≥ n0
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Some properties of Oh, Omega, Theta
Transitivity: f(n) = Θ(g(n)) and g(n) = Θ(h(n))
then f(n) = Θ(h(n))
(same for O and Ω?)
Reflexivity: f(n) = Θ(f(n))
(same for O and Ω?)
Symmetry:
f(n) = Θ(g(n)) iff g(n) = Θ(f(n))
(same for O and Ω?)
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Final thoughts: Oh, Omega, Theta
• Useful comparison formula of functions in book (section3.2)
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What kind of recurrences arise in algorithms and how do we
solve more generally (than what we saw for merge sort)?
• More recurrence examples
• Run time not always intuitive, so need tools
Divide and Conquer approach
Max Subarray Problem: Can buy stock once, sell stock once.
Want to maximize profit; allowed to look into the future
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Max Subarray Problem: Can buy stock once, sell stock once.
Want to maximize profit; allowed to look into the future
Sell
Buy
Buy low, sell high …
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Brute force: Try every possible pair of buy and
sell dates
æn æ
n!
n(n-1)(n- 2)! n(n-1)
=
=
= Q(n2 )
æ æ=
(n- 2)!2!
2
æ 2 æ (n- 2)!2!
Hmm, can we do better?
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Let’s reframe the problem as greatest sum of any contiguous array
Efficiency? Still brute force
Divide and conquer anybody?
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Where could max subarray be?
low
high
middle
low
i
j
i
j
i
j
high
middle
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Where could max subarray be?
low
high
middle
1. DIVIDE chop subarray in two equal subarrays
A[low,…,middle] and A[middle+1,..., high]
2. CONQUER find max of subarrays
A[low,…,middle] and A[middle+1,..., high]
3. COMBINE find the best solution of:
a. the two solutions found in conquer step
b. solution of subarray crossing the midpoint
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Keep recursing until low=high (one element left)
1. DIVIDE chop subarray in two equal subarrays
A[low,…,middle] and A[middle+1,..., high]
2. CONQUER find max of subarrays
A[low,…,middle] and A[middle+1,..., high]
3. COMBINE find the best solution of:
a. the two solutions found in conquer step
b. solution of subarray crossing the midpoint
low
i
j
high
middle
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i
low
j
high
middle
•
•
•
•
Start from the middle
Go to the left until hit max sum.
Go to the right until hit max sum.
Return total left and right sum.
COMPLEXITY?
Θ(n)
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float recmax(int l, int u)
if (l > u) /* zero elements */
return 0;
if (l == u) /* one element */
return max(0, A[l]);
m = (l+u) / 2;
/* find max crossing to left */
lmax = sum = 0;
for (i = m; i ≥ l; i--) {
sum += A[i];
if (sum > lmax)
lmax = sum;
}
/* find max crossing to right
*/
rmax = sum = 0;
for (i = m+1; i ≤ u; i++) {
sum += A[i];
if (sum > rmax)
rmax = sum;
}
return max(max(recmax(l, m),
recmax(m+1, u)),
lmax + rmax);
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COSTS:
1. DIVIDE T = Θ(1)
2. CONQUER T = 2T(n/2)
3. COMBINE T = Θ(n)+Θ(1)
comparison
Subarray
crossing
TOTAL :
T(n) = 2T(n/2) + Θ(n) = Θ(n log n)
Merge sort anyone?
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CLASSICAL EXAMPLE: MATRIX MULTIPLICATION
N
cij = å aik bkj
k=1
Run time: O(n3)
in the naïve
implementation
1. n = A.rows
2. Let C be a new n by n matrix
3. for i=1 to n
4. for j=1 to n
5.
cij = 0
6.
for k=1 to n
7.
cij = cij+ aik bkj
8. return C
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Can we do better?
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