1. CS 341: Algorithms Lecture 5: divide and conquer
Slides by Trevor Brown (some material from Doug Stinson’s slides) (

2. Finishing up last time: solving recurrence relations

3. Worked Examples Recall: simplified master theorem
b=2;
y=1;
x=1
Recall: simplified master theorem
Θ 𝑛 𝑥 log 𝑛
=Θ 𝑛 log 𝑛
a=3;
b=2;
y=1;
x=log2 3
Θ 𝑛 𝑥
=Θ 𝑛 log 2 3
a=4;
b=2;
y=1;
x=log2 4
Θ 𝑛 𝑥
=Θ 𝑛 2
Questions: a=? b=? y=? x=?
which Θ function?
a=2;
b=2;
y=3/2;
x=1
Θ 𝑛 𝑦
=Θ 𝑛 3/2

4. General master theorem
Example recurrence:
Arbitrary function of 𝒏
(not just 𝑐 𝑛 𝑦 )
Must reason about relationship between 𝑓(𝑛) and 𝑛 𝑥

5. Revisiting the recursion tree method
Some recurrences with complex 𝑓(𝑛) functions (such as 𝒇(𝒏) = 𝒍𝒐𝒈⁡𝒏) can still be solved “by hand”
Example: Let 𝑛= 2 𝑗 ; 𝑇 1 =1 ; 𝑻 𝒏 =𝟐𝑻 𝒏 𝟐 +𝒏 𝐥𝐨𝐠 𝒏
There may have been a problem with this slide and the next one… covering next time instead…

6. More Divide and conquer memes algorithms

7. Problem: Non-dominated points
So, I am a non-dominated point
I dominate everything to my southwest
No other point dominates me +𝑦
So are we!
So are we!
So are we! +𝑥

8. What’s an easy (brute force) algorithm for this?
More formally +𝑦
Given two points ( 𝑥 1 , 𝑦 1 ) and ( 𝑥 2 , 𝑦 2 ), we say ( 𝑥 1 , 𝑦 1 ) dominates ( 𝑥 2 , 𝑦 2 ) if 𝒙 𝟏 > 𝒙 𝟐 and 𝒚 𝟏 > 𝒚 𝟐
Input: a set S of n points,
Output: all non-dominated points in S, i.e., all points in S that are not dominated by any point in S +𝑥
What’s an easy (brute force) algorithm for this?

9. Let’s come up with a better algorithm
Brute force algorithm
For each point p
dominated[p] = false
For each other point q
If q.x > p.x and q.y > p.y then dominated[p] = true
If dominated[p] = false then add p to the output
Running time?
𝑶(𝒏𝟐)
Let’s come up with a better algorithm

11. Problem decomposition

12. Combining to get Non-dominated points
Let 𝑄 1 , 𝑄 2 , …, 𝑄 𝑘 be the non-dominated points in 𝑺 𝟏
Let 𝑅 1 , 𝑅 2 , …, 𝑅 𝑚 be the non-dominated points in 𝑺 𝟐
Just need to find rightmost 𝑸 𝒊 that is not dominated (that has 𝑦-coordinate ≥ 𝑅 1 .𝑦)
𝑆 1
𝑆 2

13. Running time complexity?

14. Multiprecision multiplication
Input: two 𝒌-bit positive integers X and Y
With binary representations: X=[X[k-1], …, X[0]] Y=[Y[k-1], …, Y[0]]
Output: The 2k-bit positive integer Z=XY
With binary representation: Z=[Z[2k-1], …, Z[0]]

15. Grade school multiplication
How to multiply 𝑘-bit numbers:
Multiply all 𝑘2 pairs of digits
This creates 𝑘 partial products, each with between k and 2𝑘 bits
Sum the partial products
Total cost:
bit-wise multiplication: 𝑂(𝑘2) bit operations
𝒌 sums of 𝑂(𝑘)-bit numbers: 𝑂(𝑘)∗𝑂(𝑘)=𝑂(𝑘2) bit operations

18. Complexity 4 recursive calls on 𝑘 2 bit numbers
combining takes Θ 𝑘 steps
𝑇 𝑘 =4𝑇 𝑘 2 +Θ(𝑘)
Master theorem says 𝑇 𝑘 = 𝑘 log = 𝑘 2
Same complexity as naïve grade school multiplication!
But with a more complicated recursive algorithm…

19. An algorithm that’s actually better

22. We did not get past here

23. Matrix multiplication
Input: A and B
Output: their product C=AB
Word-RAM model (64-bit ints)
Naïve algorithm for 𝑛×𝑛 matrices:
For each output cell 𝑪𝒊𝒋
𝐶 𝑖𝑗 =𝐷𝑜𝑡𝑃𝑟𝑜𝑑(𝑟𝑜 𝑤 𝑖 (𝐴), 𝑐𝑜 𝑙 𝑗 𝐵 𝑇 )
= 𝑘=1 𝑛 𝐴 𝑖𝑘 𝐵 𝑘𝑗
Running time?

24. Attempting a better solution
What if we first partition the matrix into sub-matrices
Then divide and conquer on the sub-matrices
Example of partitioning: 4x4 matrix into four 2x2 matrices

26. Deriving a recurrence
MatrixMultiply performs 𝑻(𝒏) work, consisting of:
Eight recursive calls to perform multiplications on 𝑛 2 × 𝑛 2 matrices
Four additions of 𝑛 2 × 𝑛 2 matrices
Work: 4 𝑛 2 ⋅ 𝑛 2 ∈Θ( 𝑛 2 )
𝑇 𝑛 =8𝑇 𝑛 2 +Θ( 𝑛 2 )
Recurse to multiply
Adding is “combining”

27. Solve using the (Simp.) Master theorem
a=? b=? y=? x=?
Recurrence: 𝑇 𝑛 =8𝑇 𝑛 2 +Θ( 𝑛 2 )
No better than the naïve algorithm!
a=8
b=2
y=2
x= log 2 8=3
x>y so 𝑻 𝒏 ∈𝚯 𝒏 𝒙 =𝚯( 𝒏 𝟑 )

28. Strassen matrix multiplication
For example, by definition 𝒕=𝒄𝒆+𝒅𝒈
And according to Strassen, 𝒕= 𝑷 𝟑 + 𝑷 𝟒
Plugging in 𝑃 3 , 𝑃 4 , 𝒕= 𝒄+𝒅 𝒆+𝒅(𝒈−𝒆)
This simplifies to 𝒕=𝒄𝒆+𝒅𝒆+𝒅𝒈−𝒅𝒆
=𝒄𝒆+𝒅𝒈

29. How much better is Θ( 𝑛 2.81 ) than Θ( 𝑛 3 )?
Let n=10,000
𝑛 2.81 ≈174 billion
𝑛 3 =1 trillion (~6x more)
How much better is Θ( 𝑛 ) than Θ( 𝑛 3 )?
Let n=10,000
𝑛 ≈3.2 billion
𝑛 3 =1 trillion (~312x)