This preview shows page 1. Sign up to view the full content.
Unformatted text preview: k IS: Suppose we have a pile of k + 1 stones. Then if we break it into piles of sizes k 1 , k 2 we incur a cost of k 1 k 2 . By the IH, it costs k 1 ( k 11) 2 and k 2 ( k 21) 2 to fully break apart each of these. Thus, the total cost is: k 1 k 2 + k 1 ( k 11) + k 2 ( k 21) 2 = 2 k 1 k 2 + k 2 1k 1 + k 2 2k 2 2 = k 2 1 + 2 k 1 k 2 + k 2 2( k 1 + k 2 ) 2 = ( k 1 + k 2 ) 2 + ( k 1 + k 2 ) 2 = ( k + 1) 2( k + 1) 2 = ( k + 1)( k + 11) 2 as was to be shown. Thus, by strong induction, the result holds for all n ....
View
Full
Document
This note was uploaded on 02/06/2012 for the course MATHEMATIC 55 taught by Professor Robbayer during the Summer '09 term at University of California, Berkeley.
 Summer '09
 RobBayer
 Math

Click to edit the document details