HHC 2015 RPN programming Contest is now open

[Egan Ford]

(10-02-2015 01:01 PM)David Hayden Wrote:  

(10-01-2015 04:30 PM)Peter Murphy Wrote:  Would anyone be willing to discuss the scheme or schemes they used to solve this?

Egan's first solution works by taking each unique die value X and summing up 10^X. For example, if the roll is 4 3 5 3 1, it will add up 10⁴ + 10³ + 10⁵ + 10¹ = 111,010. It then looks for 1111 in the sequence. In this example, 1111 doesn't appear so there is no small straight.

Others took this approach as well including Werner's 42-byter. However Werner looked for >= 1111 so that nothing was required to remove dups. I was working on a similar solution, but decided to pursue different approaches, including your flag approach, but I never finished it.