Question

I am trying to understand how to solve recurrence relations. I understand it to the point where we have to simplify.

T(N) = T(N-1) + N-1              Initial condition: T(1)=O(1)=1
T(N) = T(N-1) + N-1
T(N-1) = T(N-2) + N-2
T(N-2) = T(N-3) + N-3
……
T(2) = T(1) + 1


**Summing up right and left sides**

T(N) + T(N-1) + T(N-2) + T(N-3) + …. T(3) + T(2) =

= T(N-1) + T(N-2) + T(N-3) + …. T(3) + T(2) + T(1) +

(N-1) + (N-2) + (N-3) + …. +3 + 2 + 1


** Canceling like terms and simplifying **

T(N) = T(1) + N*(N-1)/2 1 + N*(N - 1)/2

T(N) = 1 + N*(N - 1)/2

I really don’t understand the last part. I understand canceling like terms but don’t understand how the simplification below works:

T(N) = T(1) + (N-1) + (N-2) + (N-3) + …. +3 + 2 + 1
T(N) = T(1) + N*(N-1)/2 1 + N*(N - 1)/2

How is the second line derived from the first? Doesn’t make any sense to me.

Would be a great help if someone can help me understand this. Thanks =)

Answer 1

In your second-to-last-line:

 S = (N-1) + (N-2) + (N-3) + ... + 3 + 2 + 1

You can say:

2S = S + S
   = (N-1) + (N-2) + (N-3) + ... +   3   +   2   +   1
       1   +   2   +   3   + ... + (N-3) + (N-2) + (N-1)
   =   N   +   N   +   N   + ... +   N   +   N   +   N
       |__________________ N-1 times ________________|

You’re counting from N - 1 to 1, so there are N - 1 terms in the sequence. But the whole sequence is just N so you can say:

2S = N * (N - 1)
 S = (N * (N - 1)) / 2

So in your last chunk:

T(N) = T(1) + (N-1) + (N-2) + (N-3) + ... + 3 + 2 + 1
     = T(1) + (N * (N - 1)) / 2