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Project Euler Solutions

Project Euler 112 Solution

Project Euler 112 Solution

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%.


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.
def is_bouncy(n):
  inc, dec, s = False, False, str(n)
  for i in range(len(s)-1):
    if s[i+1] > s[i]: inc = True
    elif s[i+1] < s[i]: dec = True
    if inc and dec: return True
  return False

n, p = 21780, 0.90
b = n * p
while p != 0.99:
  n += 1
  if is_bouncy(n): b += 1
  p = b / n
print "Project Euler 112 Solution =", n
download arrowUse this link to get the Project Euler 112 Solution Python 2.7 source.


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Project Euler 112 Solution last updated


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

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