. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, x;-1 >= X¡ <= Xj+1- The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them. 'Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the
cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, xj-1 = X; <= Xj+1.
The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only
responsible for finding one of them.
Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by
turning over only O(log n) cards.
Transcribed Image Text:3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, xj-1 = X; <= Xj+1. The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them. Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.
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