(Latest Revision: 03/14/04)

Example Recurrence Problem And Solution

Solve the following recurrence exactly for n a power of 2.
T(1)=1; T(2)=2; T(4)=3

T(n) = 5T(n/2) - 7T(n/4) + 3T(n/8) + nlog(n), for n>4
SOLUTION

Step One - Change Variable
(Note: All logarithms in this discussion are base-2) 

n=2k; k=log(n); (n/2)=2k-1; (n/4)=2k-2; (n/8)=2k-3; 

nlog(n) = 2k*k = k*2k; 

T(2s) = ts

tk = 5tk-1 - 7tk-2 + 3tk-3 + k*2k

tk - 5tk-1 + 7tk-2 - 3tk-3 = k*2k
Step Two - Find Characteristic Polynomial

On the right hand side above we have k*2k which is of the form p(k)*bk where b=2 and p(k)=k is a polynomial of degree d=1. Accordingly the characteristic polynomial of the recurrence is 
  (x3 - 5x2 + 7x - 3)(x-2)d+1 
= (x3 - 5x2 + 7x - 3)(x-2)2 
= (x-1)2(x-3)(x-2)2
Step Three - Apply Formula And Write Solution
tk = c0*1k + c1*k*1k + c2*3k + c3*2k + c4*k*2k
   = c0 + c1*k + c2*3k + c3*2k + c4*k*2k
Step Four - Use Intial Values To Get Equations In The Unknown Constants
1 = t0 = c0 + c2 + c3
2 = t1 = c0 + c1 + c2*3 + c3*2 + c4*2
3 = t2 = c0 + c1*2 + c2*9 + c3*4 + c4*8
Step Five - If There Are Not Yet As Many Equations As Unknown Constants, Make More Equations Using the Recurrence Relation

The recurrence is: tk = 5tk-1 - 7tk-2 + 3tk-3 + k*2k. We get
t3 = 5(t2)-7(t1)+3(t0)+24 = 15-14+3+24=28, and
t3 = c0 + c1*3 + c2*27 + c3*8 + c4*24
and also
t4 = 5(t3)-7(t2)+3(t1)+64 
   = 5(28)-7(3)+3(2)+64 = 140-21+6+64 = 189, and 
t4 = c0 + c1*4 + c2*81 + c3*16 + c4*64
which leads to:
 28 = c0 + c1*3 + c2*27 + c3*8  + c4*24, and
189 = c0 + c1*4 + c2*81 + c3*16 + c4*64
The complete set of 5 equations in 5 unknowns is:
1   = c0        + c2    + c3
2   = c0 + c1   + c2*3  + c3*2  + c4*2
3   = c0 + c1*2 + c2*9  + c3*4  + c4*8
28  = c0 + c1*3 + c2*27 + c3*8  + c4*24
189 = c0 + c1*4 + c2*81 + c3*16 + c4*64
Step Six - Solve The Set Of Equations For The Unknown Constants and Check The Values By Substituting Back Into The Equations

One method of solution is to start this way: subtract the appropriate multiple of the first equation from the others in order to eliminate c0 from them:
1   = c0        + c2    + c3
1   =    + c1   + c2*2  + c3    + c4*2
2   =    + c1*2 + c2*8  + c3*3  + c4*8
27  =    + c1*3 + c2*26 + c3*7  + c4*24
188 =    + c1*4 + c2*80 + c3*15 + c4*64
Next subtract an appropriate multiple of the second equation from all the equations below it to eliminate c1 from all those equations:
1   = c0        + c2    + c3
1   =    + c1   + c2*2  + c3    + c4*2
0   =    +      + c2*4  + c3    + c4*4
24  =    +      + c2*20 + c3*4  + c4*18
184 =    +      + c2*72 + c3*11 + c4*56
Now continue the pattern - you get:
1   = c0        + c2    + c3
1   =    + c1   + c2*2  + c3       + c4*2
0   =    +      + c2*4  + c3       + c4*4
24  =    +      +       + c3*(-1)  + c4*(-2)
184 =    +      +       + c3*(-7)  + c4*(-16)
and then:
1   = c0        + c2    + c3
1   =    + c1   + c2*2  + c3       + c4*2
0   =    +      + c2*4  + c3       + c4*4
24  =    +      +       + c3*(-1)  + c4*(-2)
16 =     +      +       +          + c4*(-2)
Now the bottom equation says: c4 = -8. If you then plug in the value c4 = -8 into the fourth equation, it is easy to get c3 = -8. Then plugging the values of c3 and c4 into the third equation we easily find the value of c2. Similarly we find the values of c1 and c0. We get:
 
c0 = -1; c1 = 5; c2 = 10; c3 = c4 = -8
We check by verifying that
1   = -1       + 10    + (-8)
2   = -1 + 5   + 10*3  + (-8)*2  + (-8)*2
3   = -1 + 5*2 + 10*9  + (-8)*4  + (-8)*8
28  = -1 + 5*3 + 10*27 + (-8)*8  + (-8)*24
189 = -1 + 5*4 + 10*81 + (-8)*16 + (-8)*64
Step Seven - Plug The Values Of The Constants Into The Formal Expression For tk

We get 
tk = c0 + c1*k + c2*3k + c3*2k + c4*k*2k
   = -1 + 5k + 10*3k - 8*2k - 8*k*2k
Step Eight - Rewrite The Solution By Doing The Reverse Of The Change Of Variable
T(n) = tk; n=2k; k=log(n)

tk   = -1 + 5k       + 10*3k        -8*2k  -8*2k*k
T(n) = -1 + 5log(n) + 10*3log(n)  -8n   - 8nlog(n)
Also, since 
log(3log(n)) = log(n)log(3) = log(nlog(3)),
we see that 
3log(n) = nlog(3)
so we can write the solution as:
T(n) = -1 + 5log(n) + 10*nlog(3) -8n - 8nlog(n)
(Since log(3) ~ 1.585, we see that T(n) is Q(nlog(3)), i.e. ‘about’ Q(n1.585).)

Step Nine - Final Check: Verify That The Original Relation And Initial Values Are Consistent With The Answer 
T(1) = -1 + 5log(1) + 10*3log(1) - 8*1 - 8*1*log(1)
     = -1 + 0       + 10        - 8   - 0
     =  1  (checks)

T(2) = -1 + 5log(2) + 10*3log(2) - 8*2 - 8*2*log(2)
     = -1 + 5       + 30        - 16  - 16
     =  2  (checks)

T(4) = -1 + 5log(4) + 10*3log(4) - 8*4 - 8*4*log(4)
     = -1 + 10      + 90        - 32  - 64
     =  3  (checks)

T(n)   = -1 + 5log(n) + 10*3log(n)  -8n   - 8nlog(n)

T(n/2) = -1 + 5log(n/2) + 10*3log(n/2)  -8n/2   - 8n/2log(n/2)
       = -1 + 5log(n) - 5 + 10*3log(n)-1  -4n   - 4nlog(n) +4n
       = -6 + 5log(n) + (10/3)*3log(n)  - 4nlog(n)

T(n/4) = -1 + 5log(n/4) + 10*3log(n/4)  -8n/4   - 8n/4log(n/4)
       = -1 + 5log(n)  -10 + 10*3log(n)-2  -2n  - 2nlog(n) +4n
       = -11 + 5log(n) + (10/9)*3log(n)  +2n  - 2nlog(n) 

T(n/8) = -1 + 5log(n/8) + 10*3log(n/8)  -8n/8   - 8n/8log(n/8)
       = -1 + 5log(n) - 15 + 10*3log(n)-3  -n   - nlog(n) + 3n
       = -16 + 5log(n) + (10/27)*3log(n)  +2n   - nlog(n)

 5T(n/2) = -30 + 25log(n) + (50/3)*3log(n)        - 20nlog(n)
-7T(n/4) =  77 - 35log(n) - (70/9)*3log(n)  -14n  + 14nlog(n) 
 3T(n/8) = -48 + 15log(n) + (10/9)*3log(n)  + 6n  -  3nlog(n)

5T(n/2) - 7T(n/4) + 3T(n/8) + nlog(n) 
= -1 + 5log(n) + 10*3log(n) - 8n - 8nlog(n) 
= T(n) (checks)