Correct solutions for Problem 114 were received from Brent Erickson, Kristin Hellem, Rich Bernstein, Shreeti Shrestha, Peter Petrany, Marcus Hazel, James Sichaleune, Kenny Albright, Amanda Parsons, Nicole Fruitman, Jason Doll, Emily Hebeler, Eric Peuschel, Ray Kremer.  Further correct solutions came from Brian Laughlin, Steve Prowse, Ahron Teitelman, Bryan Fluhere, Burkart Venzke, Jens Voss, Sudipta Das, Philippe Fondanaiche, Steve Smith, Mike Lynch, Vinco Marinkovic, Adnan Al Ayoub, Don Cherf, Rick Hoselton.

Correct solutions for Problem 115 were received from Marcus Hazel, Shreeti Shrestha, Nicole Fruitman, Rich Bernstein, Emily Hebeler, Ray Kremer, Eric Peuschel, Peter Petrany.  Further correct solutions came from Jens Voss, Philippe Fondanaiche, Steve Prowse, Ivan Lisac, Mike Lynch, Tom Lynch, Burkart Venzke, Vinco Marinkovic, Don Chef.

You can reach the word Z starting with A using all four rules, but you cannot reach it using only the first three rules.

Jason Doll's solution to problem 114 was among the shortest:
A Þ AAA Þ AAAAAAAAAAAAA Þ AAAAZZZAAA Þ AAAAAAAAAA Þ AZAAZAZ Þ Z.

A special mention goes to Jens Voss who showed that, in fact, every word is in the dictionary and can be made using only rules 1, 2 and 4.

For the second part, note that the initial word has an odd number of A's, namely one A, and that rules 1, 2, and 3 transform a word with an odd number of A's into another word with an odd number of A's.  Since the word Z has an even number of A's, namely zero A's, it can never be reached starting from A.

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Page last updated 15 May 2001.

ã 2001 Alberto L Delgado