Billboard Challenge, Part 2

June 26, 2012

In the previous exercise, we looked at a programming challenge that involved the digits of e. Solving that exercise and clicking on the indicated web site took the solver to this page:

Congratulations. You've made it to level 2. Go to www.Linux.org and enter Bobsyouruncle as the login and the answer to this equation as the password.

f(1)= 7182818284 f(2)= 8182845904 f(3)= 8747135266 f(4)= 7427466391 f(5)= __________

Don’t try to login; the account no longer exists.

Your task is to write a program to find the next number in the sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Posted by programmingpraxis

Filed in Exercises

4 Comments »

4 Responses to “Billboard Challenge, Part 2”

Programming Praxis – Billboard Challenge, Part 2 « Bonsai Code said
June 26, 2012 at 7:29 PM
[…] today’s Programming Praxis challenge, our goal is to solve the second part of the billboard test, which is […]
Remco Niemeijer said
June 26, 2012 at 7:30 PM
My Haskell solution (see http://bonsaicode.wordpress.com/2012/06/26/programming-praxis-billboard-challenge-part-2/ for a version with comments):
```
main :: IO ()
main = print $ [x | x <- map (take 10) $ tails stream_e, sum x == 49] !! 4
```
Mike said
June 27, 2012 at 2:14 AM
The difficulty with these kinds of problems is that the given terms often do not identify a unique sequence.

For example, I determined that the values for f(1), f(2), f(3), and f(4) start at position 2, 6, 24, and 100, respectively, in the digits of e. After a few minutes at OEIS.org, I discovered that the differences between successive indices–2, 4, 18, 76–correspond to the first few even Lucas numbers. So I whipped up the following Python code to calculate f(5):
```
from itertools import ifilter, islice
from myutils import accumulate, even

def Lucas():
    '''Lucas numbers are given by: L(0) = 2, L(1) = 1, and L(n) = L(n-1) + L(n-2).  Similar to Fibonacci numbers
    '''
    a, b = 2, 1

    while True:
        yield a
        a, b = b, a+b

for i, n in enumerate(islice(accumulate(ifilter(even, Lucas())), 5),1):
    print "f({}) = {}".format(i, ''.join(map(str, islice(e_digits(), n-1, n+9))))
```
Which results in the following output:

f(1) = 7182818284 f(2) = 8182845904 f(3) = 8747135266 f(4) = 7427466391 f(5) = 4637721112

If you’ve looked at the Praxis solution, this is not the answer they were looking for.
programmingpraxis said
June 27, 2012 at 5:30 PM
Mike: Yes, that’s a problem.

My solution also differs from the suggested solution at A095926, because I require all numbers to be ten digits long, as in the first challenge problem, but the OEIS solution permits leading zero.

Programming Praxis