1. Master Theorem Yin Tat Lee
CSE 421
Master Theorem
Yin Tat Lee

2. Almost 1D Problem
Partition each side of 𝐿 into 𝛿 2 × 𝛿 2 squares Claim: No two points lie in the same 𝛿 2 × 𝛿 2 box. Say we want to check if there is a point < 𝛿 close to point 31. Only need to check points with indices roughly 31. (coz we sorted). How many? Not too many coz every box only have 1 point.  29 30 31 28 26 25 49 27
>𝛿

3. Closest Pair (2 dimension)
if(𝒏≤𝟐) return | 𝒑 𝟏 − 𝒑 𝟐 |
Compute separation line 𝑳 such that half the points are on one side and half on the other side.
𝜹 𝟏 = Closest-Pair(left half)
𝜹 𝟐 = Closest-Pair(right half)
𝜹 = min( 𝜹 𝟏 , 𝜹 𝟐 )
Delete all points further than  from separation line L
Sort remaining points p[1]…p[m] by y-coordinate.
for 𝒊=𝟏,𝟐,⋯,𝒎
for k = 𝟏,𝟐,⋯,𝟏𝟏
if 𝒊+𝒌≤𝒎
 = min(, distance(p[i], p[i+k]));
return . }

4. Closest Pair Analysis
Let 𝐷(𝑛) be the number of pairwise distance calculations in the Closest-Pair Algorithm
𝐷 𝑛 ≤ if 𝑛=1 2𝐷 𝑛 𝑛 o.w. ⇒𝐷 𝑛 =O(𝑛log 𝑛)
BUT, that’s only the number of distance calculations
What if we counted running time?
𝑇 𝑛 ≤ if 𝑛=1 2𝑇 𝑛 2 +𝑂(𝑛 log 𝑛) o.w. ⇒𝐷 𝑛 =O(𝑛 log 2 𝑛)

5. Closest Pair (2 dimension) Improved
if(𝒏≤𝟐) return | 𝒑 𝟏 − 𝒑 𝟐 |
Compute separation line 𝑳 such that half the points are on one side and half on the other side.
( 𝜹 𝟏 , 𝒑 𝟏 ) = Closest-Pair(left half)
( 𝜹 𝟐 , 𝒑 𝟐 ) = Closest-Pair(right half)
𝜹 = min( 𝜹 𝟏 , 𝜹 𝟐 )
𝒑 𝒔𝒐𝒓𝒕𝒆𝒅 = merge( 𝒑 𝟏 , 𝒑 𝟐 ) (merge sort it by y-coordinate)
Let 𝒒 be points (ordered as 𝒑 𝒔𝒐𝒓𝒕𝒆𝒅 ) that is 𝜹 from line L.
for 𝒊=𝟏,𝟐,⋯,𝒎
for k = 𝟏,𝟐,⋯,𝟏𝟏
if 𝒊+𝒌≤𝒎
 = min(, distance(q[i], q[i+k]));
return  and 𝒑 𝒔𝒐𝒓𝒕𝒆𝒅 . }
𝑇 𝑛 ≤ if 𝑛=1 2𝑇 𝑛 2 +𝑂 𝑛 o.w. ⇒𝐷 𝑛 =𝑂(𝑛 log 𝑛)

6. Master Theorem Suppose 𝑇 𝑛 =𝑎 𝑇 𝑛 𝑏 +𝑐 𝑛 𝑘 for all 𝑛>𝑏. Then,
If 𝑎< 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘
If 𝑎= 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘 log 𝑛
If 𝑎> 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 log 𝑏 𝑎
Works even if it is 𝑛 𝑏 instead of 𝑛 𝑏 .
We also need 𝑎≥1,𝑏>1 , 𝑘≥0 and 𝑇 𝑛 =𝑂(1) for 𝑛≤𝑏.

7. Master Theorem Suppose 𝑇 𝑛 =𝑎 𝑇 𝑛 𝑏 +𝑐 𝑛 𝑘 for all 𝑛>𝑏. Then,
If 𝑎< 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘
If 𝑎= 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘 log 𝑛
If 𝑎> 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 log 𝑏 𝑎
Example: For mergesort algorithm we have
𝑇 𝑛 =2𝑇 𝑛 2 +𝑂 𝑛 .
So, 𝑘=1,𝑎= 𝑏 𝑘 and 𝑇 𝑛 =Θ(𝑛 log 𝑛)

8. Proving Master Theorem𝑛
Problem size
𝑛/𝑏
𝑛/ 𝑏 2 𝑏 1
𝑇 𝑛 =𝑎𝑇(𝑛/𝑎)+𝑐 𝑛 𝑘
# probs 𝑎2 𝑎 1 𝑎𝑑
cost
𝑐 𝑛 𝑘
𝑎⋅𝑐 𝑛/𝑏 𝑘
𝑎 2 ⋅𝑐 𝑛/ 𝑏 2 𝑘
𝑎 𝑘 ⋅𝑐 𝑛/ 𝑏 𝑘 𝑘
𝑑= log 𝑏 𝑛 𝑎
𝑇 𝑛 = 𝑖=0 𝑑= log 𝑏 𝑛 𝑎 𝑖 𝑐 𝑛 𝑏 𝑖 𝑘

9. Master Theorem Suppose 𝑇 𝑛 =𝑎 𝑇 𝑛 𝑏 +𝑐 𝑛 𝑘 for all 𝑛>𝑏. Then,
If 𝑎< 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘
If 𝑎= 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 𝑘 log 𝑛
If 𝑎> 𝑏 𝑘 then 𝑇 𝑛 =Θ 𝑛 log 𝑏 𝑎
# of problems increases slower
than the decreases of cost.
First term dominates.
# of problems increases faster
than the decreases of cost
Last term dominates.

10. A Useful Identity
Theorem: 1+𝑥+ 𝑥 2 +…+ 𝑥 𝑑 = 𝑥 𝑑+1 −1 𝑥−1 Proof: Let 𝑆=1+𝑥+ 𝑥 2 +…+ 𝑥 𝑑 Then, 𝑥𝑆=𝑥+ 𝑥 2 +…+ 𝑥 𝑑+1 So, 𝑥𝑆−𝑆= 𝑥 𝑑+1 −1 i.e., 𝑆 𝑥−1 = 𝑥 𝑑+1 −1 Therefore, 𝑆= 𝑥 𝑑+1 −1 𝑥−1 Corollary: 1+𝑥+ 𝑥 2 +…+ 𝑥 𝑑 = 𝑂 𝑥 1 if 𝑥<1 𝑑+1 if 𝑥=1 𝑂 𝑥 𝑥 𝑑+1 if 𝑥>1
𝑂 𝑥 means the hidden
constant depends on 𝑥

11. Solve: 𝑇 𝑛 =𝑎𝑇 𝑛 𝑏 +𝑐 𝑛 𝑘
Corollary: 1+𝑥+ 𝑥 2 +…+ 𝑥 𝑑 = Θ 𝑥 1 if 𝑥<1 Θ 𝑑 if 𝑥=1 Θ 𝑥 𝑥 𝑑+1 if 𝑥>1 Going back, we have 𝑇 𝑛 = 𝑖=0 𝑑= log 𝑏 𝑛 𝑎 𝑖 𝑐 𝑛 𝑏 𝑖 𝑘 =𝑐 𝑛 𝑘 𝑖=0 𝑑= log 𝑏 𝑛 𝑎 𝑏 𝑘 𝑖 Hence, we have 𝑇 𝑛 =Θ 𝑛 𝑘 1 if 𝑎< 𝑏 𝑘 log 𝑏 𝑛 if 𝑎= 𝑏 𝑘 𝑎 𝑏 𝑘 log 𝑏 𝑛 if 𝑎> 𝑏 𝑘
constant depends on 𝑎,𝑏,𝑐
When we can hide constant?

12. Solve: 𝑇 𝑛 =𝑎𝑇 𝑛 𝑏 +𝑐 𝑛 𝑘
𝑇 𝑛 =Θ 𝑛 𝑘 1 if 𝑎< 𝑏 𝑘 log 𝑏 𝑛 if 𝑎= 𝑏 𝑘 𝑎 𝑏 𝑘 log 𝑏 𝑛 if 𝑎> 𝑏 𝑘 For 𝑎< 𝑏 𝑘 , we simply have 𝑇 𝑛 =Θ 𝑛 𝑘 . For 𝑎= 𝑏 𝑘 , we have 𝑇 𝑛 =Θ 𝑛 𝑘 log 𝑏 𝑛 =Θ( 𝑛 𝑘 log 𝑛 ). For 𝑎> 𝑏 𝑘 , we have 𝑇 𝑛 =Θ 𝑛 𝑘 𝑎 𝑏 𝑘 log 𝑏 𝑛 =Θ( 𝑛 log 𝑏 𝑎 ).
𝑎 log 𝑏 𝑛
= (𝑏 log 𝑏 𝑎 ) log 𝑏 𝑛
= (𝑏 log 𝑏 𝑛 ) log 𝑏 𝑎
= 𝑛 log 𝑏 𝑎
𝑏 𝑘 log 𝑏 𝑛
= 𝑏 log 𝑏 𝑛 𝑘
= 𝑛 𝑘

13. Example