Data Str & Algorithm HW Solutions 12

Data Str & Algorithm HW Solutions 12 - n holes,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
12 Chap. 2 Mathematical Preliminaries = 8 3 ( 5 3 ) n 2 < ( 5 3 ) 2 ( 5 3 ) n 2 = 5 3 n . Thus, the theorem is proved by mathematical induction. 2.24 Proof : (a) Base case. For n =1 , 1 3 = 1 2 (1+1) 2 4 =1 . Thus, the formula is correct for the base case. (b) Induction Hypothesis. n 1 X i =0 i 3 = ( n 1) 2 n 2 4 . (c) Induction Step. n X i =0 i 3 = ( n 1) 2 n 2 4 + n 3 = n 4 2 n 3 + n 2 4 + n 3 = n 4 +2 n 3 + n 2 4 = n 2 ( n 2 +2 n +2) 4 = n 2 ( n +1) 2 4 Thus, the theorem is proved by mathematical induction. 2.25 (a) Proof : By contradiction. Assume that the theorem is false. Then, each pigeonhole contains at most 1 pigeon. Since there are
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n holes, there is room for only n pigeons. This contradicts the fact that a total of n + 1 pigeons are within the n holes. Thus, the theorem must be correct. (b) Proof : i. Base case. For one pigeon hole and two pigeons, there must be two pigeons in the hole. ii. Induction Hypothesis. For n pigeons in n 1 holes, some hole must contain at least two pigeons....
View Full Document

Ask a homework question - tutors are online