DISCRETE STRUCTURES
SOLUTION 4
1. Call a partition P of An special, if no subset in P contains two consecutive
integers. Denote by S '(n, r ) the number of special partitions of An
containing exactly r non-empty subsets. First I prove by induction on n, that
S '(n, r )  S (n  1, r  1) for all n  1 and for all r in the range 1  r  n . For n =
1 we have S '(1,1)  1  S (0, 0) and the induction basis is proved. So take
n  2 and 1  r  n , and assume that the claim holds for S '( n  1, r ') for all
legitimate indices r ' . Consider the special partition of An1 into r-1 non
empty subsets. Adding {n} as singleton gives a special partition of An .
Then consider a special partition of An1 into r non-empty subsets. Adding
n to any one the r-1 subsets not containing n-1 yields a special partition of
An . All special partitions of An into r non-empty subsets can be generated
in this way. Subsequently:
S '(n, r )  S '(n  1, r  1)  (r  1) S '(n  1, r )
 S (n  2, r  2)  (r  1) S (n  2, r  1) [by the Induction hypothesis]
 S (n  1, r  1) [by the recurrence for Stirling numbers]
And the inductive step is established. For n  1 it then follows that
n
n
Bn'  S '(n, 0)   S '(n, r )  S '(n, r ) [Since S '(n, 0)  0 ]
r 1
r 1
n
n 1
r 1
r 1 0
  S (n  1, r  1)   S (n  1, r  1) Bn 1
2. The characteristic equation is
 ( X )  X 3  5 X 2  3X  9  X 3  3X 2  2 X 2  6 X  3X  9
  X  3  X 2  2 X  3   X  3 X  1 X  3   X  1 X  3  0
2
Therefore, a general solution of the given recurrence is:
Tn  c(1) n  c '3n  c "3n
Plugging in the initial values gives:
c  c '  1,
c  3c ' 3c "  2 ,
c  9c '18c "  28
The solution of this system is c 
Tn 
25
9
7
, c '   , c "  . Then we have
16
16
4
25
9
7
(1) n   n   3n .
16
16 
4
3. We prove all the three properties as follows:
[Reflexive] For any a  A we have min(a) = min(a).
[Symmetric] For any a, b  A , if min(a) = min(b), then min(b) = min(a)
[Transitive] For any a, b, c  A , if min(a) = min(b), min(b) = min(c), then
min(a) = min(c).
4. The proof is as follows:
n
 
Given L(1) = 3 and L(n)  L    1
2
n
n
So L    L    1
2
4
 n 
   
n
i.e. L(n)  L    2
4
n
Similarly, L(n)  L    3 and so on.
8
 n 
In general, L(n)  L  a   a for any positive integer a
2 
a
Let 2  n , or a  log 2 n
i.e. L(n)   L    1  1
4
n
L(n)  L    log 2 n
n
i.e. L(n)  L 1  log 2 n
i.e. L(n)  log 2 n  3 .
1 5
. So a particular
2
solution is of the form   1n . Substitution gives       d , i.e.   d .
5. The characteristic equation x 2  x  1 has roots x 
Therefore, a general solution is of the form
n
n
 1 5 
 1 5 
Tn  
 u  
 v  d
 2 
 2 
where u, v are constants to be determined from the initial conditions
 1 5 
 1 5 
T1  a  
 u  
 v  d ,
 2 
 2 
 3 5 
 3 5 
T2  b  
 u  
 v  d
 2 
 2 
Solving this system gives,

 


 

1 
3  5 a  1  5 b  2d 

2 5
1 
v
3  5 a  1  5 b  2d 

2 5
u
Notice that the exact values of the constants u, v, w are not very important
here. It suffices to know the form of the solution. In particular, since a, b, d
are positive, it follows that u > 0. Also
1 5
 1 . Consequently,
2
  1  5 n 
Tn    
 .
  2  


S
6. Let n ,3 denote the total number of bit strings of length n that have three
consecutive 0s. Now we have two possibilities, if 1 occurs on the first place
or 0 occurs in the first place. If 1 occurs, then the problem reduces to
finding the number of bit strings of length n-1 that have three consecutive
0s i.e. S n 1,3 and if 0 occurs in the first place then we look for the 2nd digit.
Similarly if 1 occurs in the 2nd digit as well, then we have to solve the
relation S n  2,3 , and if again 0 occurs we see the 3rd digit, if 1 occurs here
then we solve S n 3,3 otherwise the remaining n-3 places can be either filled
with 1 or 0, giving a total number of ways as 2n3 Hence the recurrence
relation becomes
Sn,3  Sn1,3  Sn2,3  Sn3,3  2n3
S 2,3  0 , S1,3  0 , S 0,3  0
From the above recurrence relation, we get
S3,3  1
S 4,3  3
S5,3  8
S 6,3  20
S7,3  47
7. If n is odd, the f(n+2) = 2f(n+1) + 1 and f(n+1) = 2f(n); if n is even, f(n+2) =
2f(n+1) and f(n+1) = 2f(n) + 1. So the result holds in either case.
This is not a “pure” recurrence relation because of added 1. But if we set
g(n) = f(n) + ½, the g(n+1) + 2g(n) = f(n+1) + 2f(n) + 3/2 = g(n+2). The
characteristic equation is x 2  x  2 , with solutions 2 and -1, so
g (n)  a2n  b(1)n . The initial values g(1) = 3/2 and g(2) = 5/2 yield a = 2/3
2
3
1
6
1
2
and b = -1/6. So f (n)  2n  (1) n  .