Uses:
� Use recursion tree to determine a good asymptotic bound on the recurrence T(n) = …� Sum the costs within each level of the tree to obtain a set of per-level costs.
� Sum all the per-level costs to determine the total cost of all levels of recursion.
� Best if used to generate a good guess for the closed form bounds of the recurrence T(n).
� Guess is verified by using Substitution Method or Master Method.
Note: the bound sought will be one of the following:
� “asymptotic upper bound” means you’re looking for Big-O � “tight asymptotic bound” means you’re looking for Θ � “asymptotic lower bound” means you’re looking for Ω
Steps:
1. Draw the tree based on the recurrence
2. From the tree determine: a. # of levels in the tree
b. cost per level
c. # of nodes in the last level
d. cost of the last level (which is based on the number found in 2c)
3. Write down the summation using ∑ notation – this summation sums up the cost of all the levels in the recursion tree
4. Recognize the sum or look for a closed form solution for the summation created in 3). Use Appendix A.
5. Apply that closed form solution to your summation coming up with your “guess” in terms of Big-O, or Θ, or Ω (depending on which type of asymptotic bound is being sought).
6. Then use Substitution Method or Master Method to prove that the bound is correct.
Example - Show Factorial is O(n)
For the recursive n! algorithm:
n! is defined by the recurrence:
The run time cost, T(n), is defined by the recurrence:
where 1 is the running time for each execution of the factorial function.
Backward Substitution method for exact closed-end equation - T(n) = n
Chapter 4 - Recursion TreeDocument last modified: 02/09/2012 18:42:34
Search C455
int F(int n) {
if (n == 1) return 1;
return F (n-1) * n;}
F(1) = 1 for n=1
F(n) = F(n-1)*n for n>1
T(1) = 1 for n=1
T(n) = T(n-1) + 1 for all n>1
T(n) = T(n-1) + 1 = [ T(n-2)+1 ] + 1 Substitute T(n-2)+1 for T(n-1)
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After i substitutions:
T(n) = T(n-i) + T(n-i+1) + ... + 1 + 1
Initial condition: T(1) = T(n-i)
n-i=1 or i = n-1
T(1) = T(n-i) + i
= T(1) + n-1
= 1 + 0 = 1
Closed-end formula:
T(n) = T(1) + n-1 = 1 + n-1 = n
Prove by induction:
T(n) = T(n-1)+1 = n
Base: T(1) = 1 by recurrence definition T(2) = T(2-1) + 1 = T(1) + 1 = 2
IH: T(k) = k
Show: T(k+1) = k+1
Guess method using Recursion Tree - T(n) = n
Backward substitution gave exact solution of: T(n) = n
Recursion tree provides guess: T(n) = n = O(n)
Guess is sum of costs at each level.
T( n ) = 1 + 1 + ... + 1 = n
Show: T(n) = O(n)
By definition of Big-O:
0 ≤ T(n) ≤ cn must hold
For O(n) must prove: 0 ≤ T(n) ≤ cn for c > 0, for all n ≥ n0
Inductive Hypothesis:
= T(n-2) + 2 = [ T(n-3)+1 ] + 2 Substitute T(n-3)+1 for T(n-2)
= T(n-3) + 3 = [ T(n-4)+1 ] + 3 Substitute T(n-4)+1 for T(n-3)
= T(n-4) + 4
T(k+1) = T((k+1)-1) + 1 Recurrence definition
= T(k) + 1
= k + 1 IH: T(k) = k
T(1) = 1 for n=1
T(n) = T(n-1) + 1
for all n>1
Recursion Tree
Height 3
Levels 4
Level n Cost
0 4 → 1
1 3 → 1
2 2 → 1
3 1 → 1
Total 4
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T(k-1) ≤ c(k-1)
Show:
T(k) = T(k-1) + 1 ≤ ck
Base or boundary conditions: Now find c and n0 to satisfy O definition
Verify: TR(n) ≤ T
C(n) Recurrence ≤ Closed-end
TR(n) = T
R(n-1) + 1 Recurrence
TC(n) = cn Closed-end bounds solution for O(n)
O(n): 0 ≤ T(n) ≤ cn for c ≥ 1, for all n ≥ n0 = 1 or 2
Example -Recursion Tree - Finding O
T(k) = T(k-1) + 1 Recurrence
≤ c(k-1) + 1 Substitute IH
= ck - c + 1
≤ ck
Solve ck - c + 1 ≤ ck
-c + 1 ≤ 0
-c ≤ -1
c ≥ 1
Try n0 = 1
Succeeds for n0 = 1
TR(n) ≤ T
C(n)
TR( 1 ) ≤ T
C( 1 )
TR( 1-1 ) + 1 ≤ T
C( 1 ) Substitute
recurrence
1 ≤ c(1) T( 0 ) doesnot exist
1 ≤ c
c ≥ 1
Try n0 = 2
Succeeds for n0 = 2
TR(n) ≤ T
C(n)
TR( 2 ) ≤ T
C( 2 )
TR( 2-1 ) + 1 ≤ T
C( 2 ) Substitute
recurrrence
T( 1 ) + 1 ≤ c(2) T( 1 ) = 1
1 + 1 ≤ 2c
c ≥ 1
T(1) =
1 n=1
T(n) = 2T(n-1) + 1 n>1
Recursion Tree
Height 3
Levels 4
Level n Cost
0 4 20=1
1 3 21=2
2 2 22=4
3 1 23=8
Total 24-1 = 15
n=4
a) # levels = n
b) # nodes/level = 2i, i=0 to n-1
c) # bottom nodes = 2n-1
d) Cost/level =2i, i=0 to n-1
e) Bottom cost = 2n-1
f)
Cost
Level Nodes
0 20
1 21
2 22
n-1 2n-1
Total
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Recursion tree provides guess: T(n) = O(2n)
Show: T(n) = O(2n)
By definition of Big-O:
T(n) = 2T(n-1) + 1 ≤ c2n must hold
Show:
T(n) = 2T(n-1) + 1 ≤ c2n
Inductive Hypothesis:
T(n-1) ≤ c(2n-1)
Substitution Subtlety
Show:
T(n) = 2T(n-1) + 1 ≤ c2n - b
Inductive Hypothesis:
T(n-1) ≤ c(2n-1) - b
Base or boundary conditions: Now find c and n0 to satisfy O definition
Verify: TR(n) ≤ T
C(n) Recurrence ≤ Closed-end
TR(n) = 2T
R(n-1) + 1 Recurrence
TC(n) = c2n - b Closed-end bounds solution for O(2n)
by A5, page 1147
T(n) = 2T(n-1) + 1 Recurrence
≤ 2c(2n-1) + 1 Substitute IH
= c2n + 1
≤ c2n Fails
T(n) = 2T(n-1) + 1 Recurrence
≤ 2[c(2n-1) - b] + 1 Substitute IH
= c2n -2b + 1
≤ c2n - b
Solve c2n -2b + 1 ≤ c2n - b
1 ≤ b
Try n0 = 2
TR(n) ≤ T
C(n)
TR( 2 ) ≤ T
C( 2 )
2TR( 2-1 ) + 1 ≤ T
C( 2 ) Substitute
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O(n): T(n) ≤ c2n - b ≤ c2n for c ≥ 1, for all n ≥ n0 = 2
Question 4.12.1
Draw the recursion tree.
Determine the number of nodes and cost at each level.
Determine the total cost for all levels (i.e. in summation form).
Example - Merge Sort - Finding O
Recall that a balanced binary tree of height k has k + 1 levels. We've proven that.
A problem of size n=16 divided as
described by the following recurrence equation would then
be represented by the tree at right.
Use the implied coefficient c for arithmetic purposes:
Steps using recursion tree.
1. Draw the tree based on the recurrence. Recursion tree shows successive expansions of the recurrence.
Succeeds for n0 = 2
2T( 1 ) + 1 ≤ c22 - b T( 1 ) = 1
2 + 1 ≤ 4c - b Let b = 1
4 ≤ 4c
c ≥ 1
| 2 if n = 1T(n) = | | 3T(n-1) + 2 if n > 1
| O(1) if n = 1T(n) = | | 2T(n/2) + O(n) if n > 1
| c if n = 1T(n) = | | 2T(n/2) + cn if n > 1
Root cost = cn
2T(n/2) + cn
For each of size n/2 subproblems, cost:
cn/2 + 2T(n/4)
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Each level cost sums to cn, for example:
cn = cn/2 + cn/2
Continue expanding tree until problem sizes are 1, each with a cost of c
cn = c + c + ... + c
Height = lg n
Levels = lg n + 1
Cost/level = cn
Bottom node number n = 2lg n for height lg n dividing problem by 2 each time.
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Question 4.12.2 - Draw the recursion tree.
Can ignore ⌊ ⌋ by simplifying assumption that n is a power of 2.
Steps using recursion tree continued.
Recall: alogan = n
so 2log2n =
2lg n =n
Bottom cost = T(1) * n = cn
Note that T(1) is problem size n=1, there are n subproblems at bottom.
| c if n = 1T(n) = | | 4T(⌊n/2⌋) + cn if n > 1
2. From the tree determine: a. # of levels in the tree of height lg n:
lg n + 1
b. cost per level:
cn
c. # of nodes in the last level, problem size n=1:
n
d. cost of the last level (which is based on the number found in 2c times base cost):
T(1) * n = cn
3. Write down the summation using ∑ notation. Summation of cost for lg n levels in the recursion tree + bottom level.
4. Find closed form solution for the summation created in 3) often using Appendix A.
Total: cost of each level * the number of levels + cost of bottom level.
5. Use closed form solution as “guess” in terms of Big-O, or Θ, or Ω (depending on which type of asymptotic
bound is being sought).
| c if n = 1T(n) = | | 2T(n/2) + cn if n > 1
T(n) = cn lg n + cn
= cn lg n + O(n)
= O(cn lg n) + O(n)
= O(n lg n)
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6. Then use Substitution Method or Master Method to prove that the bound is correct.
By definition of Big-O, 0 ≤ T(n) ≤ d * n lg n
Substitution Method for O(n lg n): 0 ≤ T(n) ≤ dn lg n
For O(n lg n) must prove: 0 ≤ T(n) ≤ dn lg n for d > 0, for all n ≥ n0
Assume n is a power of 2 to avoid floor and ceil complications to our guess.
Inductive Hypothesis:
Assume: T(k/2) ≤ d k/2 lg k/2
Show: T(k) = 2T(k/2) + ck ≤ d * k lg k
Find d that satisfies last two lines.
Satisfied by d ≥ c
Basis:
T(1) = 2T(1/2) + c*1 = c ≤ d1 lg 1 = d1 lg 20 = 0 at odds with d > 0, d ≥ c
Since need n ≥ n0 for n a power of 2, choose n
0 = 2
Use as basis:
T(2) ≤ d2 lg 21 = 2d
By the recurrence, where c is the constant divide and combine time:
| c if n = 1T(n) = | | 2T(n/2) + cn if n > 1
T(k) = 2T(k/2) + ck Recurrence
≤ 2( dk/2 lg k/2 ) + ck Substitute IH
= dk lg k/2 + ck
= dk lg k - dk lg 2 + ck
= dk lg k - dk + ck
≤ dk lg k
dk lg k - dk + ck ≤ dk lg k
- dk + ck ≤ 0
ck ≤ dk
c ≤ d
| c if n = 1T(n) = | | 2T(n/2) + cn if n > 1
T(2) = 2T(2/2) + 2c
= T(1) + T(1) + 2c
= c + c + 2c
= 4c
Need T(2) = 4c ≤ d2 lg 2 = 2d
4c ≤ 2d
so let d = 2c
Satisfies d = 2c ≥ c
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Example - Merge Sort - Find O
The following is a Merge_Sort that divides each problem into 3 subproblems
yielding the following recurrence:
where bn is the cost to divide and the combine cost of Merge.
Recursion Tree
Diagram recursion tree to generate a guess at a closed form bounds for: T(n) = 3T(n/3) + bn
Height: log3n
Levels: log3n + 1
Cost per level: bn
Bottom (size n=1 problem):
Cost per node: b
Nodes: 3log3n = n
7. O(n lg n): 0 ≤ T(n) ≤ dn lg n for d > 0, for all n ≥ n0
satisfied by d ≥ 2c > 0, for all n ≥ n0 = 2
Merge_Sort (A, p, r)
1 if p < r then
2 q ← floor ((p + r) / 3)
3 Merge_Sort (A, p, q)
4 Merge_Sort (A, q + 1, 2*q)
5 Merge_Sort (A, 2*q + 1, r)
6 Merge (A, p, q, 2*q, r)
| b if n = 1T(n) = | | 3T(n/3) + bn if n > 1
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3 - the number of subproblems created each recurrence (level)
log3n - the number of recurrences (levels preceding bottom)
O(n log3n): 0 ≤ T(n) ≤ cn log
3n
Recurrence:
Guess: T(n) = bn log3n + bn = O(n log
3n)
Basis:
T(1) = 3T(1/3) + b*1 = b ≤ cn log3n = c1 log
3 1 = 0 at odds with T(1) = b
and b > 0
Need to establish inductive basis for n ≥ n0 for n
0 of our choosing:
Choose n0=3
Use as inductive basis (not the recurrence basis which is T(1)=b):
T(3) ≤ c3 log33 = 3c
Prove basis holds for n0
By the recurrence: T(n) = 3T(n/3) + bn
IH: Assume holds for n/3, that is:
T(n/3) ≤ c(n/3) log3(n/3) is true
Show: T(n) ≤ cn log3n
| b if n = 1T(n) = | | 3T(n/3) + bn if n > 1
T(3) = 3T(3/3) + 3b
= T(1) + T(1) + T(1) + 3b
= b + b + b + 3b
= 6b
Need 6b = T(3) ≤ c3 log33 = 3c
6b ≤ 3c
2b ≤ c
so let c = 2b
T(n) = 3T(n/3) + bn Recurrence
≤ 3[c(n/3) log3(n/3)] + bn IH substitution
= 3[c(n/3) log3n - c(n/3) log
33] + bn Recall that log a/b = log a - log b
= 3[c(n/3) log3n - c(n/3)] + bn log
33=1
= 3c(n/3) log3n - 3c(n/3) + bn
= cn log3n - cn + bn
≤ cn log3n
cn log3n - cn + bn ≤ cn log
3n
-cn + bn ≤ 0
-c + b ≤ 0
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Therefore:
T(n) = O(n log3n) for c = 2b and n
0=3
Question 4.13 - For:
2. From the tree determine: a. # of levels in the tree of height lg n:
b. cost per level:
c. # of nodes in the last level, problem size n=1:
d. cost of the last level (which is based on the number found in 2c times base cost):
3. Write down the summation using ∑ notation. Summation of cost for lg n levels in the recursion tree + bottom level.
4. Find closed form solution for the summation created in 3) often using Appendix A.
5. Use closed form solution as “guess” in terms of Big-O, or Θ, or Ω (depending on which type of
asymptotic bound is being sought).
6. Then use Substitution Method or Master Method to prove that the bound is correct.
See Question 4.10 and 4.11 for Substitution Method.
Example - Unbalanced Recursion Tree - Find Θ
T(n) = T(n/3) + T(2n/3) + Θ(n)
O upper bound: rewrite as T(n) ≤ T(n/3) + T(2n/3) + cn
Ω lower bound: rewrite as T(n) ≥ T(n/3) + T(2n/3) + cn
By summing across each level, the recursion tree shows the cost at each level of recursion (minus the costs
of recursive calls, which appear in subtrees):
b ≤ c, that is c ≥ b
Note that c can be chosen while b is determined by algorithm.
Recall to satisfy the induction basis we have already chosen
c = 2b ≥ b
| c if n = 1T(n) = | | 4T(⌊n/2⌋) + cn if n > 1
T(n) = O( ? )
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� There are log3n full levels,
after log3/2
n levels, the problem size is down to 1;
log3n ≤ log
3/2n
� Height log3/2n
� Recall that dividing n = 2i size problem into 1/2 size problems has a tree height:
lg n = log2/1
n = log2/1
(2/1)i=log22i = i
� T(2n/3) means that the n size problem is reduced to 2/3 the size but double the size of T(n/3).
� T(2n/3) path is the longest from root to leaf.
� A leaf is a problem of size 1.
� The longest path is then: n → (2/3)n → (2/3)2n → ... → 1
� (2/3)kn=1 when k=log3/2
n the height is log3/2
n
� Solve for k in (2/3)kn=1
� n = (3/2)k
� log3/2
n=log3/2
3/2k=k
� Cost ≤ cn log3/2n
� Expect cost to be at most: # levels * cost at each level = cn log3/2n when total cost evenly
distributed across all levels. � Because some internal leaves are absent (more absent nearer bottom), some levels contributes ≤
cn.
� A complete binary tree of height log2n has 2log
2n = nlog
22 = n leaves (at bottom).
� The complete binary tree leaves are: 2log3/2
n = nlog3/2
2
� Upper bound guess: dn log3/2
n = O(n lg n) for some positive constant d.
� Note: log3/2
n/log2 n= (log
2 n*log
3/2 2)/ log
2 n = log
3/2 2 = 0.585
� n log3/2
n / n lg n = c or about 0.585 suggesting dn log3/2 n = O(n lg n)
O upper bound: rewrite as T(n) ≤ T(n/3) + T(2n/3) + cn
Guess: T(n) = O(n lg n)
IH: T(n/3) ≤ d(n/3) lg(n/3)
T(2n/3) ≤ d(2n/3) lg(2n/3)
Substitution: Recall: log a/b = log a - log b
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Therefore, T(n) = O(n lg n).
Ω lower bound: rewrite as T(n) ≥ T(n/3) + T(2n/3) + cn
Guess: T(n) ≥ dn lg n.
Substitution: Same as for the upper bound, but replacing ≤ by ≥.
Need:
0 < d ≤ c / (lg 3 − 2/3)
Therefore, T(n) = Ω(n lg n)
Θ: T(n) = O(n lg n) and T(n) = Ω(n lg n), conclude that T(n) = Θ(n lg n)
Finding the closed form for a sequence
Suppose we have an algorithm to compute all permutations of n things. The order is listed in the tree by
position:
as a1, b2, c3, d4
as a3, b1, c2, d3
as a4, b3, c2, d1
The algorithm generates the following recursion tree:
Question: What is the complexity of the algorithm?
Obviously the bottom cost is n! since we are generating all permutations so Ω(n!).
The other n-1 levels are:
n+n(n-1)+n(n-1)(n-2)+...+n(n-1)(n-2)*...*3*2
The actual number of calls for n=4 is: 64
4 + 4(4-1) + 4(4-1)(4-2) + 4(4-1)(4-2)(4-3) =
4 + 4(3) + 4(3)(2) + 4! =
4 + 12 + 24 + 24 = 64
The actual number of calls for n=5 is: 325
5 + 5(5-1) + 5(5-1)(5-2) + 5(5-1)(5-2)(5-3) + 5! =
T(n) ≤ T(n/3) + T(2n/3) + cn
≤ d(n/3) lg(n/3) + d(2n/3) lg(2n/3) + cn IH
= (d(n/3) lg n − d(n/3) lg 3) + (d(2n/3) lg n − d(2n/3) lg(3/2)) + cn
= dn lg n − d((n/3) lg 3 + (2n/3) lg(3/2)) + cn
= dn lg n − d((n/3) lg 3 + (2n/3) lg 3 − (2n/3) lg 2) + cn
= dn lg n − dn(lg 3 − 2/3) + cn
≤ dn lg n if −dn(lg 3 − 2/3) + cn ≤ 0 ,
d ≥ c / (lg 3 − 2/3)
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5 + 5(4) + 5(4)(3) + 5(4)(3)(2) + 5! =
5 + 20 + 60 + 120 + 120 = 325
Determining a tight upper-bound requires finding a closed form representation of the summation of
the recursion tree calls defined by:
There is an on-line encyclopedia of integer sequences, that can be searched by providing it with a few terms of a sequence. For example, we can search 1, 4, 15, 64, 325 (terms 1 through 5 of the
sequence).
http://www.research.att.com/~njas/sequences/A007526
According to the site, the sequence is the number of permutations of non-empty subsets of
{1,2,3,...,n}.
If we define the sequence a(n) to be the expression given, we find that it obeys the recursion:
a(n) = n + n*a(n-1).
It also has the closed formula a(n) = [e*n! – 1], where the symbol [x] is the nearest integer to x, and e is the usual constant 2.718...
The upper-bound is then: O(e*n!) = O(n!)
See http://www.research.att.com/~njas/sequences/index.html for the site's main page.
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