This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 6.042/18.062J Mathematics for Computer Science March 15, 2005 Srini Devadas and Eric Lehman Lecture Notes Sums, Approximations, and Asymptotics II 1 Block Stacking How far can a stack of identical blocks overhang the end of a table without toppling over? Can a block be suspended entirely beyond the tables edge? Table Physics imposes some constraints on the arrangement of the blocks. In particular, the stack falls off the desk if its center of mass lies beyond the desks edge. Moreover, the center of mass of the top k blocks must lie above the ( k + 1)-st block; otherwise, the top k would fall over. In order to find the best configuration of blocks satisfying these constraints, well need a fact about centers of mass. Fact 1. If two objects have masses m 1 and m 2 and centers-of-mass at positions z 1 and z 2 , then the center of mass of the two objects together is at position: z 1 m 1 + z 2 m 2 m 1 + m 2 Define the offset of a particular configuration of blocks to be the horizonal distance from its center of mass to its rightmost edge. offset ? center of mass s- ? Table 2 Sums, Approximations, and Asymptotics II The offset determines precisely how far that configuration can extend beyond the desk since at best the center of mass lies right above the desks edge. So hereafter we can focus on maximizing the offset of a stack of n blocks rather than the overhang . As a result, we need only be concerned with whether the stack is internally stable and not with whether the whole configuration is too far to the right and falls off the table. We can establish the greatest possible offset of a stack of n blocks with an inductive argument. This is an instance where induction not only serves as a proof technique, but also turns out to be a nice tool for reasoning about the problem. Theorem 1. The greatest possible offset of a stack of n 1 blocks is: 1 1 1 1 X n = + + + . . . + 2 4 6 2 n Proof. We use induction on n , the number of blocks. Let P ( n ) be the proposition that the greatest possible offset of a stack of n 1 blocks is 1 / 2 + 1 / 4 + . . . + 1 / (2 n ) . Base case: For a single block, the center of mass is distance X 1 = 1 / 2 from its rightmost edge. So the offset is 1 / 2 and P (1) is true. Inductive step: For n 2 , we assume that P ( n 1) is true in order to prove P ( n ) . A stack of n blocks consists of the bottom block together with a stack of n 1 blocks on top. In order to acheive the greatest possible offset with n blocks, the top n 1 blocks should themselves have the greatest possible offset, which is X n 1 ; otherwise, we could do better by replacing the top n 1 blocks with a different configuration that has greater offset....
View Full Document
This note was uploaded on 02/08/2011 for the course EECS 6.042 taught by Professor Dr.ericlehman during the Spring '11 term at MIT.
- Spring '11
- Computer Science