Problem #112 concerns numbers which are neither increasing nor decreasing. The question reads:
Project Euler Problem 112: Bouncy numbers
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%.
My solution for this question can be summarized as a brute force approach that relies on the percentage being an integer. Here’s my solution:
Solution #1: Brute Force Approach
We simply check every interval of 100 consecutive positive integers until the cumulative total of bouncy numbers is 99% of the cumulative total of the number of integers in all of the intervals. Here is an implementation of this approach in Python 2.7:
'''
Author: ArrGeeStee
Brute Force Approach to Project Euler Problem #112
'''
import time
def isBouncy(n):
a = str(n)
curr = int(a[0])
increasing = True
decreasing = True
for c in range(1,len(a)):
v = int(a[c])
if(v>curr):
decreasing = False
if(v<curr):
increasing = False
if not decreasing and not increasing:
return True
curr = v
return False
def projectEulerProblemOneHundredTwelve(n):
complement = 100-n
notBouncy = 100
total = 100
while(complement*total!=notBouncy*100):
total+=100
for x in range(100):
if not isBouncy(total-x):
notBouncy+=1
return total
start = time.time()
print projectEulerProblemOneHundredTwelve(99)
print ("--- %s seconds ---" % (time.time()-start))
'''
Prints
1587000
--- 2.88207697868 seconds ---
for input of n = 99.
'''
And with that, we’re done. This brute force approach wasn’t very efficient as it took nearly 3 seconds to run. I may come back to this problem at some point to search for a more efficient solution.
Thanks for reading! See you tomorrow.
The Inner Machinations of Walker 