Alexander G. Chefranov, 26.11.2014, CMPE Dept., EMU, Famagusta, TRNC
Some material from
was used.
CMPE371 Fall 2014
Problems to be discussed
3.1, 3.2, 3.3, 3.4
3.2.
lg k n  cn
a. m  lg n
m k  c 2m
2m / k
m  c 2 ;1  c
m
m m
(
 1)
m
c(1 
 k k
)
2m / k
k
2
c
m
m
m / k
m m
(
 1)
m
 m
 m
k
k
c(1 

)
(
 1)
(
 1)
c
c

c

k
2
 
k k

k k
1
m
m k
2
k
2
m(
2(1 

c
)
k  1) k

k
n
n/2
2
2
n/2
d. 2   (2 ) ; m  2 ; m   (m) : m  cm; m  c
lg c
lg n
e. n  c
n
e
n
n
f. lg( n!)   (lg n ) ; n!  2n ( ) e n ,

1
1
 n 
12n  1
12n
n
lg( n!)  lg( 2n ( ) n e n )  0.5  lg   lg n  n lg n  n lg e   n lg e   (n lg n)
e
a.
b.
a.1.
22
n1
 (22 ) ; 22
n
n1
 222  (22 )2  (22 )
n
n
n
a.2. 22  ((n  1)!) ;Prove by mathematical induction: n  1;22  4  (1  1)! 2 ;
n
22
n1
1
 222  (22 )2  ((n  1)!)2  (n  1)!(n  1)! (n  1)!(n  1)n  (n  1)!(n  2)  (n  2)!, n  2
n
n
(n  1)n  n  2 ; n 2  2 ; n  2
a.3
4.1, f and g; 4.3, d,e,h,i,j recursion and master method, solve both of them by substitution method also
f.
f.1.Use recursion tree
N0.5
0.5
(N/4)
0.5
0.5
0.5
(N/4)
T(1)
(N/4)
(N/4)
(N/4)
0.5
T(1)
0.5
(N/4)
T(1)
T(1)
On the i-th level, there are 2i nodes. The bottom level has number log 4 n . Each bottom level node has
log n
log 2
0.5
complexity T(1)=1, hence the bottom level complexity is 2 4  n 4  n . Sum of complexities of
the tree nodes is
T (n) 
log4 n 1

i 0
log4 n 1
n
2i ( i )0.5 n0.5  n0.5 ( 1  1)  n0.5 (log 4 n  1)
4
i 0
(1)
f.2. Use substitution method to prove (1)
1, n  1
T (n)   0.5
 n (log 4 n  1), n  1
0.5
Let T (n / 4)  (n / 4) (log 4 (n / 4)  1) . Then
T (n)  2(n / 4)0.5 (log 4 (n / 4)  1)  n0.5  n0.5 (log 4 n  1  1)  n0.5  n0.5 (log 4 n  1) , qed
f.3. Use master method: a=2, b=4, f(n)=n0.5=  (n
T (n)   (n0.5 lg n)
0.5
Prove that (1) is O(n lg n) :
n0.5 (log 4 n  1)  cn0.5 lg n
Hence
logb a
)   (n0.5 ) . So case 2 takes place, and
log 4 n
1
lg n
1
1
1



 
c
lg n
lg n lg 4 lg n lg n 2 lg n
Hence, 1  c, if n  n0  4
(2)
0.5
Now, prove that (1) is (n lg n)
n0.5 (log 4 n  1)  n0.5 log 4 n  0.5n0.5 lg n  cn0.5 lg n
Hence, 0.5  c, n  n0  1
(2), (3) prove that (1) is  (n
(3)
0.5
lg n) .
2
g) T (n)  T (n  2)  n
g.1. Use recursion tree:
T (n)  T (n  2)  n 2  T (n  4)  (n  2)2  n 2  T (n  6)  (n  4)2  (n  2)2  n 2 
k
T (1)  32  ..  (n  4) 2  (n  2) 2  n 2  1  32  ..  (n  4)2  (n  2) 2  n 2   (1  2i ) 2 , n  2k  1
i 0
g.2. Substitution method.
T (n)   (n3 )
Show first that
k
T (n)  O(n3 ) : T (n)   (1  2i ) 2  cn3
i 0
k
k
i 0
i 0
T (n)   (1  2i)2   n2 
n 2 n3
1
n   cn3 , c 
2
2
2
k
T (n)  (n3 ) : T (n)   (1  2i ) 2  cn3 ; n=1: T (1)  1  c
i 0
k 1
1
2
1
T (n  2)   (1  2i ) 2  cn3  1  2(k  1) 2  cn3  (n  1) 2  cn3  n 2  2n  1  c(n3  n 2  n  ) 
c
c
c
i 0
1
c(n3  6n 2  12n  8)  c(n  2)3 , c 
24
D.
T (n)  3T (n / 3  2)  n / 2
d.1. Recursion tree
N/2
n/6-1
n/18-4/3
T(1)
0
1
2
3
4
n/18-4/3
n/6-1
n/18-4/3
T(1)
n
n/3-2
n/9-8/3
n/27-26/9
n/81-80/27
n/6-1
n/18-4/3
T(1)
n/3-2
n/9-8/3
n/27-26/9
n/81-80/27
n/243-242/81
n/2
n/6-1
n/18-4/3
n/54-13/9
n/162-40/27
T(1)
n/3i-(3i-1)/
3i-1
i
n/3i+1-(3i+1-1)/ 3i
n/(2*3i)-(3i1)/ (2*3i-1)
m 1
3i  1
1
m 1
m 1 3
T (n)  n / 2  3(n / 6  1)  9(n / 36  4 / 3)  ..  3 (n /( 2 3 ) 
)  ..  3 (n / 3  m  2 )  3m
i 1
23
3
m
n  3 n 3 1
m  log 3
; ( m  m 1  1, n  3(3m  1)  3m ,4  3m  n  3; m  log 3 (n  3)  log 3 4)
4
3
3
i
i
3i  1
3m 1  1
m 1
m 1
)

..

3
(
n
/(
2

3
)

)  3m 
2  3i 1
2  3m  2
m 1
n
3i  1
n  3 m 1 n 3i  1 n  3 (3n  1)
n  3 1 m 1 i
i
3
(

)


(

)


m

 3 


2  3i 2  3i 1
4
23
4
6
4
6 i 0
i 0
i 0 2
T (n)  n / 2  3(n / 6  1)  9(n / 36  4 / 3)  ..  3i (n /( 2 3i ) 
(3n  1)
n  3 1 1  3m (3n  1)
n  3 1 n 1
m


(log 3 (n  3)  log 3 4) 

  (n lg n)
6
4
6 1 3
6
4
12 4
d.2. Substitution method
T (n)  3T (n / 3  2)  n / 2  cn lg n  dn, c  1
T (3n)  3T (n  2)  3n / 2  3c(n  2) lg( n  2)  d 3n  3n / 2  c(3n) lg n  d 3n  c(3n) / 2  c(3n)(lg n  0.5) 
c(3n)(lg n  lg 3)  d 3n  c(3n) lg( 3n)  d 3n
T (n)  3T (n / 3  2)  n / 2  cn lg n  dn
T (3n)  3T (n  2)  3n / 2  3c(n  2) lg( n  2)  d (n  2)  3n / 2  3c(n  2) lg( n / 2)  3d (n  2)  3n / 2 
3c(n  2) lg n  3c(n  2)  3d (n  2)  3n / 2  3cn lg n  6c lg n  3cn  6c  3dn  6d  3n / 2 
3cn lg 3n  3cn lg 3  6c lg n  3cn  6c  3dn  6d  3n / 2  (c3n lg 3n  3dn)  (3cn lg 3
 6c lg n  3cn  6c  6d  3n / 2)  c3n lg 3n  3dn
if
 3cn lg 3  6c lg n  3cn  6c  6d  3n / 2  0
2c  n / 2  cn lg 3  2c lg n  cn  2d
1
2c lg n
1
1
n(  c lg 3  c 
)  n(  c lg 3  2c)  2(d  c), n  n0  4, d  c, c 
2
n
2
2(2  lg 3)
log a
e. T (n)  2T (n / 2)  n / lg n, a  2, b  2, f (n)  n / lg n, log b a  1, n b  n
f (n)  O(nlogb a ) but not f (n)  O(nlogb a  ),   0 as it is required by Master method (MM), case 1.
Hence, MM is not applicable.
e.1 Recursion tree. T (n) 
logb a 1

2i
i 0
lg n 1
n
2i lg(
n
)
2i
 2lg n  n 
i 0
lg n
1
1
 n  n(  1)   (n lg lg n)
lg n  i
i 1 i
e.2. Substitution method. Prove that T (n)   (n lg lg n) . Let
T (n)  cn lg lg n
T (2n)  2T (n)  2n / lg 2n  2cn lg lg n  2n /(lg n  1)  2cn(lg lg n  1 /(lg n  1)), c  1
lg n 1
1
1
1
2cn(lg lg n  1 /(lg n  1))  2cn( 
)  2cn   c(2n) lg(lg n  1)  c(2n) lg lg( 2n)
lg n  1
i 1 i
i 1 i
lg n
T (n)  cn lg lg n
T (2n)  2T (n)  2n / lg 2n  2cn lg lg n  2n /(lg n  1)  2cn(lg lg n  1 /(lg n  1)), c  1
lg n 1
1
1
1
2cn(lg lg n  1 /(lg n  1))  2cn( 
)  2cn   c(2n) lg(lg n  1)  c(2n) lg lg( 2n)
lg n  1
i 1 i
i 1 i
lg n
h. T (n)  T (n  1)  lg n
h.1. MM not applicable
lg n
2
h.2. Recursion tree method. T (n)  lg n  lg( n  1)  ..  1   i  (lg n)
i 1
2
h.3. Substitution method. T (n)   (lg n)
T (n)  c lg 2 n
T (n  1)  T (n)  lg( n  1)  c lg 2 n  lg( n  1)  c(lg 2 n  lg( n  1)), c  1
lg n
c(lg n  lg( n  1))  c( i  lg( n  1))  c
2
i 1
lg(n 1)
 i  c lg
i 1
T (n)  c lg 2 n
T (n  1)  T (n)  lg( n  1)  c lg 2 n  lg( n  1)
2
(n  1)
c lg 2 n  lg( n  1)  c(lg 2 n  lg( n  1)), c  1
lg n
c(lg n  lg( n  1))  c( i  lg( n  1))  c lg 2 (n  1)
2
i 1
i. T (n)  T (n  2)  1 / lg n
n/2
1
1
1 n2
1
1

 ..   

  (lg lg n)
lg n lg( n  2)
1 i 0 lg( n  2i) i 1 lg 2i
i.1. Recursion tree. T (n) 
i.2. Substitution method. T (n)   (lg lg n)
T (n)  c lg lg n
1
1
1
 c lg lg( n  2) 
 c(lg lg( n  2) 
), c  1
lg n
lg n
lg n
T (n)  T (n  2) 
c(lg lg( n  2) 
n / 2 1
n/2
1
1
1
1
)  c( 

) c
 c lg lg n
lg n
lg n
i 1 lg 2i
i 1 lg 2i
T (n)  c lg lg n
1
1
1
 c lg lg( n  2) 
 c(lg lg( n  2) 
), c  1
lg n
lg n
lg n
T (n)  T (n  2) 
c(lg lg( n  2) 
n / 2 1
n/2
1
1
1
1
)  c( 

) c
 c lg lg n
lg n
lg n
i 1 lg 2i
i 1 lg 2i
j. T (n)  nT ( n )  n
j.1. Recursion method. Let n  2
m1
2m
m1
T ( 22 )  22 T ( 22 )  22
m
m
There will be a tree of the height m with (m+1) levels. In the i-th level, i=1,m, number of
nodes is ni 
i
2
2 m j
(it is equal to 1, if i=0), a node on the i-th level has the weight of
j 1
mi
wi  22 ,i=0,m-1; let T(2)=2.
m 1
i
m
m
i
T (22 )   ni wi   22
m
i 0
m
2
i 0
2 m i
1 2 i
 2 m i
1 2
m j
22
i  0 j 1
m
 22
m i
( 2 i 1)  2 mi
i 0
m
 2 m  j  2 m i
m
 2 j 1
i 0
m
 22
i 0
j.2. Substitution method. T (22 )   (22 m)
m
m i
m
 2 m i  2 m i
 2 j  2 m i
m
  2 j  m i
i 0
m
 22  22
i 0
m
m
 2
2 m i
i 1
 2 j  2 m i
j 0
i 0
m
m
1  2
i 0
2m
(m  1)

T (22 )  c22 m
m
m
m 1
m
T ( 22 )  22 T ( 22 )  22
m 1
c 22 m  22
m 1
m
m 1
 c 22 22 m  22
m
m
m 1
 c( 22 ) 2 m  22
m 1
 c 2 2 2 m  2 2
m
m 1

m 1
 c( 22 ) 2 m  22
m 1
 c 2 2 2 m  2 2
m 1

m
m 1
 c 22 (m  1), c  1
T (22 )  c22 m
m
m
m 1
m
T ( 22 )  22 T ( 22 )  22
m 1
c 22 m  22
m 1
m
m 1
m 1
 c 22 22 m  22
m
 c 22 (m  1), c  1
m
m
m