amortized and master method.pdf
TRANSCRIPT
-
7/24/2019 Amortized and Master method.pdf
1/63
Amortized Analysis and Master Method
-
7/24/2019 Amortized and Master method.pdf
2/63
Deliverables
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Amortized Analysis
Aggregate Analysis
Accounting Method
Potential Analysis
Solving Recurrence Relations
Master Method
-
7/24/2019 Amortized and Master method.pdf
3/63
Amortized Analysis
Analyze a sequence of operations on a data structure.
Goal is to show that although some individual operations mbe expensive, on average the cost per operation is small.
Such analysis is not immediately obvious and needs deeknowledge of problem parameters and complexity bounds.
No probability is involved. It is average cost in the worst c
Aggregate analysis, Accounting method, and Potential meth
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
4/63
Aggregate Analysis
It requires knowledge of which series of operations possible. It is common case with data structures, whhave state that persists between operations.
Idea is that a worst case operation can alter the state iway that worst case cannot occur for a long time, thamortizing its cost. E.g. doubling an array will be operation for a heavy cost.
Aggregate analysis determines the upper bound T(n) on total cost of a sequence of n operations, then calcula
average cost to be T(n)/n
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
5/63
Aggregate Analysis-Example
k-bit binary counter A[0 . . k 1] of bits, where A[0] is thleast significant bit and A[k 1] is the most significant biCounts upward from 0.
Value of counter is
Initially, counter value is 0, so A[0 . . k 1] = 0.
Copyright @ gdeepak.com
1
[ ].2k
i
i o
A i
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
6/63
Binary counter algorithm
Increment(A, k)i0
while (i < k and A[i ] = 1)
do A[i ]0
ii + 1if i < k
then A[i ]1
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
7/63
Binary Counter example k = 3
Copyright @ gdeepak.com
0
1
2
34
5
6
7
0 0 0
0 0 1
0 1 0
0 1 11 0 0
1 0 1
1 1 0
1 1 1
Counter Value
0
1
3
47
8
10
11
cost
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
8/63
Binary counter- analysis
Cost of Increment = # of bits flippedEach call could flip k bits, so n Increments takes O(nk) time.
Observation is that not every bit flips every time.
bit flips how often times in n Increments
0 every time n
1 1/2 the time n/22 1/4 the time n/4
...
i 1/2i the time n/2i
i k never 0
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
9/63
Binary counter-Aggregate analysis
Therefore, total # of flips =< n =
= 2n .
Average cost per operation = O(1)
Copyright @ gdeepak.com
1
2ii o
1
2
n
i
i o
n
11
2
n
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
10/63
Stack Operations example
Stack operations Push(S, x): O(1) O(n) for any sequence of n operatio
Pop(S): O(1) O(n) for any sequence of n operations
Multipop(S, k)
whileS is not empty and k > 0
do Pop(S)
kk 1
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
11/63
Stack Complexity for Multipop
Running time of Multipop is linear in #of Poperations
# of iterations of while loop is min(s, k), where s =of objects on stack therefore, total cost = min(s, k)
Worst-case cost of Multipop is O(n)
For Sequence of n Push, Pop, Multipop operatio
worst-case cost of sequence is O(n2
).Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
12/63
Aggregate Analysis for stacks
Each object can be popped only once per time thais pushed.
Have n Pushes n Pops, including those Multipop.
Therefore, total cost = O(n) and average overoperations O(1) per operation
No probability is involved and shows worst-ca
O(n) cost for sequenceCopyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
13/63
Accounting method
A set of elementary operations are chosen to be used in
algorithm and their cost is set to 1.
Actually, cost differs because a prepayment may necessary to complete an operation with some payment over to be used later from savings.
Assign different charges to different operations. Some charged more than actual cost and some are charged less
Amortized cost = amount we charge
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
14/63
Accounting Method
Copyright @ gdeepak.com
When amortized cost > actual cost, store the differencespecific objects in the data structure as credit.
Use credit later to pay for operations whose actual cosamortized cost.
Differs from aggregate analysis because differoperations can have different costs while in aggreganalysis, all operations have same cost.
Need credit to never go negative
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
15/63
Stack
Intuition: When pushing an object, pay `2. ` 1 pays for the PUSH.
` 1 is prepayment for it being popped by either POP or Multipop.
Each object has ` 1, which is credit, credit can never go negative
Total amortized cost, = O(n), is an upper bound on total actual co
Copyright @ gdeepak.com
operation Actual cost Amortized Cost
PUSH 1 2
Pop 1 0
Multipop min(k, s) 0
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
16/63
Accounting Method
Let ci= actual cost of ith
operation,= amortized cost of ithoperation.
Then require
for all sequences of n operations.Total credit stored = - 0
Copyright @ gdeepak.com
ic
1
n
i
i
c
1
n
i
i
c
1
n
i
i
c
1
n
i
i
c
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
17/63
Binary Counter-Example
Charge `2 to set a bit to 1. `1 pays for setting a bit to 1. `1 is prepayment for flipping it back to 0. Have ` 1 of credit for every 1 in the counter. Therefore, credit 0. Amortized cost of increment: Cost of resetting bits to
paid by credit. At most 1 bit is set to 1. Therefore, amortized cost
For n operations, amortized cost = O(n).
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
18/63
Potential Method
Like the accounting method, but think of creditpotential stored with entire data structure. Accounting method stores credit with specific objects. Potential method stores potential in the data structure
a whole. Can release potential to pay for future operations. Most flexible of the amortized analysis methods. Let Di= data structure after ith operation ,
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
19/63
Potential Method
D0= initial data structure ci= actual cost of ith operation ,
= amortized cost of ith operation .
Potential function : DiR
(Di) is the potential associated with data structure D
= ci+ (Di) (Di1)
Increase in potential due to ith operation
Copyright @ gdeepak.com
ic
ic
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
20/63
Potential Method
Total amortized cost
= + DiDi-1
telescoping: every term except D0and Dn is added and subtrac
= + DnD0
If we require that (Di) (D0) for all i , then amortized co
an upper bound on actual cost (D0) = 0, (Di) 0
Copyright @ gdeepak.com
1
n
i
ic
1
n
i
i
c
1
n
i
i
c
1
n
i
i
c
1
n
i
i
c
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
21/63
Stack Example
= # of objects in stack
D0= empty stack (D0) = 0.
Since # of objects in stack is always 0, (Di) 0 = (D0) for al
Therefore, amortized cost of a sequence of n operations = O(n).
Copyright @ gdeepak.com
Operation Actual cost s=#of objects initially
AmortizedCost
Push 1 (s + 1) s =1 1+ 1 = 2
Pop 1 (s 1) s = 1 1 1 = 0
Multipop k=min(k, s) (sk) s = k k k = 0
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
22/63
Binary Counter example k = 3
Copyright @ gdeepak.com
0
1
2
34
5
6
7
0 0 0
0 0 1
0 1 0
0 1 11 0 0
1 0 1
1 1 0
1 1 1
Counter Value0
1
3
4
7
8
10
11
cost
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
23/63
Potential Method
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
24/63
Potential Method
Copyright @ gdeepak.com
ic
6/5/2012 7:15 PM
S l i R E i f
-
7/24/2019 Amortized and Master method.pdf
25/63
Solving Recurrence Equations forComplexity 3 methods
Guess the solution and prove
Using Recursion Tree Method
Master Method
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
26/63
Guess the solution and prove
1. Guess the form of the solution2. Verify by induction and show that the solution works f
a set of constants
T(n)=2T(n/2) + n Which means that now the problem h
been divided into two sub problems and the size of the suproblem is n/2
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
27/63
Guess and Substitute by Example
We guess that it works for T(n) = O(n lg n)
To prove that T(n) cn lgn for an appropriate c.
Assuming that the guess works we will have
T(n) = 2T(n/2)+ n
= n lg(n/2) +n + n
= n lg n n lg 2 + n +n
= n lg n + n cnlg n
this is true for a specific value of c
Copyright @ gdeepak.com
2( lg )2 2 2
n n nn
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
28/63
Guess by reasoning and experience
Looks similar to binary search, try O(logn)
Assume T(k) = O(log2k) for all k < n
Show that T(n) = O(log2n)
We need to prove T(n/2) c log2(n/2)
Copyright @ gdeepak.com
dnTnT )2/()(
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
29/63
Guess by reasoning and experience
Copyright @ gdeepak.com
dnTnT )2/()(dnc )2/(log2
dcnc 2loglog 22
nc 2log
if c d
Residualdcnc 2log
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
30/63
Recursion Tree
Guessing may be difficult for recurrences which does notresemble any common recurrence
Need to check the cost of tree at each level of recursion asum costs of all the levels. For this information of depth the tree, no of leaves of the tree is required
In the end answer can be verified by substitution using th
above as a guess.Copyright @ gdeepak.com
2)4/(3)( nnTnT
cnnTnTnT )3/2(2)3/()(
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
31/63
Recursion Tree (Level 1)
Copyright @ gdeepak.com
2)4/(3)( nnTnT cn2
T(n/4 ) T(n/4 )T(n/4 )
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
32/63
Recursion Tree (Level 2)
Copyright @ gdeepak.com
2)4/(3)( nnTnT
cn2
2
4
nc
2
4
nc
2
4
nc
16
nT
16
nT
16
nT
16
nT
16
nT
16
nT
16
nT
16
nT
16
nT
3/
6/5/2012 7:15 PM
2)4/(3)( TT
-
7/24/2019 Amortized and Master method.pdf
33/63
Copyright @ gdeepak.com
2)4/(3)( nnTnT
cn2
2
4
nc
2
4
nc
2
4
nc
2
16
nc
3/
2
16
nc
2
16
nc
2
16
nc
2
16
nc
2
16
nc
2
16
nc
2
16
nc
2
16
nc
(3/1
What is the cost at each level?2
16
3
cn
d
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
34/63
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
35/63
Important to grasp
Data is divided into 4 parts at each level
Copyright @ gdeepak.com
14
d
n
0
4
log
d
n
04loglog dn
nd log4log
nd4
log
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
36/63
Important to grasp
No of leaves in the tree with three children of everynode at each level
Copyright @ gdeepak.com
nd 4log33
6/5/2012 7:15 PM
12 d
-
7/24/2019 Amortized and Master method.pdf
37/63
Copyright @ gdeepak.com
)3(16
3...
16
3
16
3)( 4
log2
1
2
2
22 n
d
cncncncnnT
)3(16
34
4log
1log
0
2 nn
i
i
cn
xx
k
k
1
10
let x = 3/16
)3(16
34log
0
2 n
i
i
cn
)3()16/3(1
14log2 n
cn
)3(13
164log2 ncn )(
13
16)( 3log2 4ncnnT
)()( 2nOnT 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
38/63
Proving after Guess through recursion tre
Now we have reasonable guess with the help of recursiotree , let us prove it by guessing
Assume T(k) = O(k2) for all k < n Show that T(n) = O(n
Given that T(n/4) = O((n/4)2), then T(n/4) c(n/4)2
Copyright @ gdeepak.com
2
)4/(3)( nnTnT 22)4/(3 nnc 22 16/3 ncn
2cn13
16c
6/5/2012 7:15 PM
i h
-
7/24/2019 Amortized and Master method.pdf
39/63
Recurrence equations-Master Metho
Where a 1 and b>1 are constants and f(n) is asymptotically positive function
We are dividing a problem of size n into a sub problems, eof size n/b, where a and b are positive constants. a problems are solved recursively each in time T(n/b) wheand b are positive constants. Cost of dividing the problem combining the results of the sub problems is described byfunction f(n) which can be written as D(n)+C(n).
Copyright @ gdeepak.com
dnnfbnaTdncnT
if)()/(if)(
6/5/2012 7:15 PM
h f h d
-
7/24/2019 Amortized and Master method.pdf
40/63
Three cases of Master Method
Compare f (n) with nlogba:
1. f (n) = O(nlogba ) for some constant > 0.
f (n) grows polynomially slower than nlogba (by an nfacto
Solution:T(n) = (nlogba) .
2. f (n) = (nlogbalgkn) for some constant k 0.f (n) and nlogbagrow at similar rates.
Solution: T(n) = (nlogbalgk+1n)
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Th C f M M h d
-
7/24/2019 Amortized and Master method.pdf
41/63
Three Cases of Master Method
Compare f (n) with nlogba:
3. f (n) = (nlogba+ ) for some constant > 0.
f (n) grows polynomially faster than nlogba(by an nfacto
and f (n) satisfies the regularity condition that
a f (n/b) c f (n) for some constant c < 1.Solution: T(n) = ( f (n))
Copyright @ gdeepak.com 6/5/2012 7:15 PM
M Th
-
7/24/2019 Amortized and Master method.pdf
42/63
Master Theorem
There is a gap between cases 1 and 2 when f (n) is smaller tbut not polynomially smaller. Similarly, there is a gap betwcases 2 and 3 whenf (n) is larger but not polynomially largethe functionf (n) falls into these gaps, or if regularity conditin case 3 fails to hold, master method cannot be used.
Copyright @ gdeepak.com
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
dnnfbnaT
dncnT
if)()/(
if)(
6/5/2012 7:15 PM
E W
-
7/24/2019 Amortized and Master method.pdf
43/63
Easy Way
Copyright @ gdeepak.com
( ) ( ) kn
T n aT nb
blog ak
k k
k k
if a > b then T(n) = O(n ) (Bottom Heavy)
if a b then T(n) = O(n log(n)) (Steady State)
if a < b then T(n) = O(n ) (Top Heavy)
6/5/2012 7:15 PM
T( ) T( / )
-
7/24/2019 Amortized and Master method.pdf
44/63
T(n)=4T(n/2)+n
a=4 = n2b=2
f(n)=n
We are in case 1 so T(n) is (n2)
Copyright @ gdeepak.com
abn
log
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
2
2
2
is f( ) ( )?
is f( ) ( log )?
is f( ) ( )?
k
n O n
n n n
n n
6/5/2012 7:15 PM
T( ) T( / ) l
-
7/24/2019 Amortized and Master method.pdf
45/63
T(n)=2T(n/2)+nlogn
a=2 = nb=2
f(n)=nlogn
We are in case 2 and k=1 so T(n) is (n log2 n)
Copyright @ gdeepak.com
abn
log
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
1
1
1
is f( ) ( )?is f( ) ( log
is f( ) ( )?
k
n O n
n n
n n
6/5/2012 7:15 PM
T( ) 4T( /2)+ 3
-
7/24/2019 Amortized and Master method.pdf
46/63
T(n)=4T(n/2)+n3
a=4 = 2
b=2
f(n)=nlogn
We are in case 3 says T(n) is (n3).For n=1/2 4*(1/2)3/2 (1/2)3 (1/8) for = 1/
Copyright @ gdeepak.com
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
ab
n
log 2
2
2
is f( ) ( )?
is f( ) ( log
is f( ) ( )?
k
n O n
n n
n n
6/5/2012 7:15 PM
T(n) 8T(n/2)+n2
-
7/24/2019 Amortized and Master method.pdf
47/63
T(n)=8T(n/2)+n2
Solution: logba=3, so case 1 says T(n) is (n3).
Copyright @ gdeepak.com
.1somefor)()/(provided)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnafnfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
dnnfbnaT
dnc
nT if)()/(
if
)(
6/5/2012 7:15 PM
T(n) 9T(n/3)+n3
-
7/24/2019 Amortized and Master method.pdf
48/63
T(n)=9T(n/3)+n3
Solution: logba=2, so case 3 says T(n) is (n3)
Copyright @ gdeepak.com
.1somefor)()/(provided)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnafnfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
dnnfbnaT
dnc
nT if)()/(
if
)(
6/5/2012 7:15 PM
Master Theorem
-
7/24/2019 Amortized and Master method.pdf
49/63
Master Theorem
T(n)=T(n/2)+1 (binary search)
Solution: logba=0, so case 2 says T(n) is (log n).
Copyright @ gdeepak.com
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
6/5/2012 7:15 PM
T(n) = 4T(n/2) + n2/lgn
-
7/24/2019 Amortized and Master method.pdf
50/63
T(n) = 4T(n/2) + n2/lgn
a = 4, b = 2 nlogba= n2;f (n) = n2/lgn.
Master method does not apply
Copyright @ gdeepak.com
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.
log
1loglog
loglog
nfbnaf
nfnTnnf
nnnTnnnfnnTnOnf
a
kaka
aa
b
bb
bb
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
51/63
Copyright @ gdeepak.com
)()/()( nfbnaTnT
)(nf
a
)/( bnf
)/( bnf )/( bnf
)/( 2bnf
a
)/( 2bnf )/( 2bnf )/(2
bnf
a
)/( 2bnf )/( 2bnf )/( 2bnf
a
)/( 2bnf )/(2bnf
f
(naf
(2 nfa
6/5/2012 7:15 PM
Depth of the tree
-
7/24/2019 Amortized and Master method.pdf
52/63
Depth of the tree
Data is divided by b at each level of the tree
Copyright @ gdeepak.com
1db
n
0log
db
n
log log 0dn b
nbd loglog
nd blog
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
53/63
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Proof-Master Method
-
7/24/2019 Amortized and Master method.pdf
54/63
Proof-Master MethodCase 1: The Weight decreases
Geometrically from the root tothe leaves. The root hold aconstant fraction of the totalweight. First term is dominating.Case 2: Weight is approximatelysame on all levels. It ispolynomial or slowly decreasingfunction.Case 3: The Weight increasesGeometrically from the root tothe leaves. The leaves hold aconstant fraction of the totalweight. Last term is dominating.
Copyright @ gdeepak.com
1)(log
0
log
1)(log
0
log
2233
22
2
)/()1(
)/()1(
...
/()/()/(
)()/()/(
))/())/(()()/()(
n
i
iia
n
i
iin
b
b
b
b
bnfaTn
bnfaTa
bnafbnfabnTa
nfbnafbnTa
bnbnfbnaTanfbnaTnT
6/5/2012 7:15 PM
( ) 4 ( / 2)T n T n n
-
7/24/2019 Amortized and Master method.pdf
55/63
Copyright @ gdeepak.com
cn3
cn
2
cn
4 c
n
4 c
n
4 c
n
4
cn
2
cn
4 c
n
4 c
n
4 c
n
4
cn
2
cn
4 c
n
4 c
n
4 c
n
4
cn
2
cn
4 c
n
4 c
n
4
cn
cn
2 c
n
2 c
n
2 c
n
2
cn Total cn
Total 4c
Total 16c
6/5/2012 7:15 PM
3( ) 4 ( / 2)T n T n n
-
7/24/2019 Amortized and Master method.pdf
56/63
Copyright @ gdeepak.com
cn3
cn
2
3
cn
8
3
cn
8
3
cn
8
3
cn
8
3
cn
2
3
cn
8
3
cn
8
3
cn
8
3
cn
8
3
cn
2
3
cn
8
3
cn
8
3
cn
8
3
cn
8
3
cn
2
3
cn
8
3
cn
8
3
cn
8
cn3
cn
2
3
cn
2
3
cn
2
3
cn
2
3
cn3 Total cn3
4c
3
=1
16c
3
= 1
6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
57/63
Copyright @ gdeepak.com 6/5/2012 7:15 PM
-
7/24/2019 Amortized and Master method.pdf
58/63
Copyright @ gdeepak.com
cn
cn
cn
2 c
n
2
cn
cn
2
cn
4 c
n
4
cn
2
cn
4 c
n
4
Total cn
Total cn
Total cn
T(n)=2T(n/2) + n
6/5/2012 7:15 PM
T(n)= T(n/3)+T(2n/3)+n
-
7/24/2019 Amortized and Master method.pdf
59/63
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Questions, Comments and Suggestion
-
7/24/2019 Amortized and Master method.pdf
60/63
Questions, Comments and Suggestion
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Question 1
-
7/24/2019 Amortized and Master method.pdf
61/63
Question 1
For recurrence T(n) = T(n-1) + n what will be the big oh
complexity.
A) O(n)
B) O(nlogn)
C) O(n2)
D) O(2n)
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Question 2
-
7/24/2019 Amortized and Master method.pdf
62/63
Question 2
The recurrence relation that arises in relation to
complexity of binary search is
(A) T(n) = T(n/2) + k, k is a constant
(B) T(n) = 2T(n/2) + k, k is a constant
(C) T(n) = T(n/2) + log n
(D) T(n) = T(n/2) + n
Copyright @ gdeepak.com 6/5/2012 7:15 PM
Question 3
-
7/24/2019 Amortized and Master method.pdf
63/63
Q 3
Which of the following sorting algorithms has the low
worst-case complexity?
(A) Merge sort
(B) Bubble sort
(C) Quick sort
(D) Selection sort
Copyright @ gdeepak.com 6/5/2012 7:15 PM