1. CSCE 411 Design and Analysis of Algorithms
Set 10: Lower Bounds
Dr. J. Michael Moore
Fall 2014
Based primarily on slides by Prof. Jennifer Welch
CSCE 411, Fall 2014: Set 10

2. What is a Lower Bound?
Provides information about the best possible efficiency of ANY algorithm for a problem
Tells us whether we can improve an algorithm or not
When lower bound is (asymptotically) the same as the best upper bound (provided by an algorithm), the bound is TIGHT
If there is a gap, there might be room for improvement
CSCE 411, Fall 2014: Set 10

3. Lower Bound Proof Technique #1: Trivial
Use size of input or output
Ex: Any algorithm to generate all permutations of n elements requires time Ω(n!) because there are n! outputs
Ex: Computing the product of two n-by-n matrices requires time Ω(n2) because the output has n2 entries
Ex: Any TSP solution for n cities requires Ω(n2) time because there are Θ(n2) inputs to be taken into account.
CSCE 411, Fall 2014: Set 10

4. Lower Bound Proof Technique #2: Information-Theoretic
Consider the amount of information that the solution must provide
Show that each step of the algorithm can only produce so much information
Uses mechanism of decision trees
Example next…
CSCE 411, Fall 2014: Set 10

5. Comparison-Based Sorting
We’ve seen mergesort, insertion sort, quicksort, heapsort,…
All these algorithms are comparison-based
the behavior depends on relative values of keys, not exact values
behavior on [1,3,2,4] is same as on [9,25,23,99]
Fastest of these algorithms is O(nlog n).
We will show that's the best you can get with comparison-based sorting.
CSCE 411, Fall 2014: Set 10

6. Decision Tree Consider any comparison based sorting algorithm
Represent its behavior on all inputs of a fixed size with a decision tree
Each non-leaf tree node corresponds to the execution of a comparison
Each non-leaf tree node has two children, one for the case when the comparison is true and one for false
Each leaf tree-node represents correct sorted order for that path
CSCE 411, Fall 2014: Set 10

7. Decision Tree Diagram first comparison: check if ai ≤ aj YES NO
second comparison
if ai ≤ aj : check if
ak ≤ al
second comparison
if ai > aj : check if
am ≤ ap
YES NO
YES NO
third comparison
if ai ≤ aj and ak ≤ al : check if ax ≤ ay NO
YES
CSCE 411, Fall 2014: Set 10

8. Insertion Sort comparison for j := 2 to n to key := a[j] i := j-1
while i > 0 and a[i] > key do
a[i+1] := a[i]
i := i -1
endwhile
a[i+1] := key
endfor
comparison
CSCE 411, Fall 2014: Set 10

9. Insertion Sort for n = 3 a1 ≤ a2 ? YES NO a2 ≤ a3 ? a1 ≤ a3 ? NO YES
CSCE 411, Fall 2014: Set 10

10. Insertion Sort for n = 3 a1 ≤ a2 ? YES NO a2 ≤ a3 ? a1 ≤ a3 ? YES NO
CSCE 411, Fall 2014: Set 10

11. How Many Leaves?
Must be at least one leaf for each permutation of the input
otherwise there would be a situation that was not correctly sorted
Number of permutations of n keys is n!.
Idea: since there must be a lot of leaves, but each decision tree node only has two children, tree cannot be too shallow
depth of tree is a lower bound on running time
CSCE 411, Fall 2014: Set 10

12. Key Lemma Height of a binary tree with n! leaves is (n log n).
Proof: Maximum number of leaves in a binary tree with height h is 2h.
h = 3, 23 leaves
h = 2, 22 leaves
h = 1,
21 leaves
CSCE 411, Fall 2014: Set 10

13. Proof of Lemma Let h be height of decision tree.
Number of leaves in decision tree, which is n!, is at most 2h.
2h ≥ n!
h ≥ log(n!)
= log(n(n-1)(n-2)…(2)(1))
≥ (n/2)log(n/2) by algebra
= (n log n)
CSCE 411, Fall 2014: Set 10

14. Finishing Up Any binary tree with n! leaves has height (n log n).
Decision tree for any comparison-based sorting algorithm on n keys has height (n log n).
Any comparison-based sorting algorithm has at least one execution with (n log n) comparisons
Any comparison-based sorting algorithm has (n log n) worst-case running time.
CSCE 411, Fall 2014: Set 10

15. Lower Bound Proof Technique #3: Problem Reduction (Transformation)
We have already seen how to use problem transformations in a “positive” way
use a pre-existing algorithm for one problem Q to get an algorithm for another problem P
So an upper bound for Q helps us get an upper bound for P
We can also use this idea in a “negative” way
use an already-known lower bound for P to get a lower bound for Q (note change in direction!)
CSCE 411, Fall 2014: Set 10

16. Using a Transformation to Get an Algorithm for P from an Algorithm for Q
Recall transforming problem P to problem Q:
transform P-input to Q-input
run (known) Q-algorithm on Q-input
transform Q-output to P-output
If we have an algorithm for Q already, and we can do steps 1 and 3, then we get an algorithm for P
Ex: transform max-flow to LP
CSCE 411, Fall 2014: Set 10

17. Using a Transformation to Get an Algorithm for P from an Algorithm for Q
P-algorithm
Q-input
Q-output
P-output
P-input
transform
Q-algorithm
transform
Touttransform
Tintransform TQ
TP = TQ + (Tintransform + Touttransform)
= TQ + Ttransform
CSCE 411, Fall 2014: Set 10

18. Using a Transformation to get a Lower Bound for Q from a Lower Bound for P
transform P-input to Q-input
run any Q-algorithm on Q-input
transform Q-output to P-output
If we have a lower bound for P and if steps 1 and 2 can be done “quickly enough”, we get a lower bound for Q
CSCE 411, Fall 2014: Set 10

19. Using a Transformation to Get a Lower Bound for Q from a Lower Bound for P
P-algorithm
Q-input
Q-output
P-output
P-input
transform
Q-algorithm
transform
Touttransform
Tintransform TQ
TP = TQ + Ttransform
If TP = Ω(X) and Ttransform = o(X), then TQ = Ω(X)
CSCE 411, Fall 2014: Set 10

20. Use Ω(n log n) Sorting Lower Bound to Get Ω(n log n) Convex Hull Lower Bound
Idea: If we could solve convex hull faster than Θ(n log n), then we could solve sorting faster than Θ(n log n)
Need to figure out how to use a convex hull “subroutine” to sort numbers
Key observation: given n numbers x1,…,xn, the points (x1,x12),…,(xn,xn2) are on a parabola and thus are corners of a convex region
CSCE 411, Fall 2014: Set 10

21. Transform Sorting to Convex Hull
Given input x1,…,xn, compute (x1,x12),…(xn,xn2)
Run a convex hull algorithm on (x1,x12),…(xn,xn2), resulting in a sequence of points Y
Return sequence of first coordinates in Y, starting at point with smallest first coordinate and wrapping around if necessary
Note: the convex hull algorithm must be comparison-based and must produce its output points in counter-clockwise (or clockwise) order
O(n)
Ω(n log n)
CSCE 411, Fall 2014: Set 10