(5 votes, average: 5.00 out of 5)

## Project Euler 112: Investigating the density of "bouncy" numbers.

#### Problem Description

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.

Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

#### Analysis

It was a surprise how long this program took to run. The is_bouncy() routine seemed to slow things down. We begin at the 90% point specified in the problem description and wait for the ratio to hit 99%.

#### Project Euler 112 Solution

Runs < 2.8 seconds in Python 2.7.

Use this link to get the Project Euler 112 Solution Python 2.7 source.

#### Afterthoughts

Project Euler 112 Solution last updated

## Discussion

### One Response to “Project Euler 112 Solution”

1. A better way would be generate all the increasing(symmetrically decreasing) numbers. So the rest are bouncy numbers. No need to check all the numbers one by one

Posted by Yang Yang | March 29, 2011, 12:25 AM