Programming Enigma Puzzles
From New Scientist #2670, 23rd August 2008
My old mini-computer screen displayed the entire alphabet in order (followed by other miscellaneous symbols) in a rectangle with the first two rows:
Below the screen were five buttons:
At the start of typing each new word the “A” was highlighted. You then moved to any other letter, one box at a time, using the arrows and pushing the OK button to type the highlighted letter. So typing “THIRTY” could be done in 30 button pushes.
My new computer is of a similar style, but with more letters per row. Now typing LOW can be done in a dozen or fewer button pushes and there are three numbers which, if spelled out, can be done in a dozen or fewer pushes each.
What are those three numerals?
[enigma1508]
 | Frits on Enigma 764: Increasing de… |
| Frits on Enigma 764: Increasing de… |
 | Jim Randell on Enigma 764: Increasing de… |
 | GeoffR on BrainTwister #22: Even ro… |
| Jim Randell on BrainTwister #22: Even ro… |
| GeoffR on BrainTwister #22: Even ro… |
| Frits on BrainTwister #22: Even ro… |
| GeoffR on BrainTwister #22: Even ro… |
Enigmatic Code 
The following Python program runs in 40ms.
from enigma import irange, printf LETTERS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" # count the number of key presses required to type <word> # where the letters are arranged in rows of <n> letters def press(word, n): # cursor starts at "A", return value is 0 (cx, cy, r) = (0, 0, 0) for i in word: p = LETTERS.index(i) (py, px) = divmod(p, n) # add horizontal moves, vertical moves, OK r += abs(cx - px) + abs(cy - py) + 1 (cx, cy) = (px, py) return r # verify the initial condition assert press("THIRTY", 4) == 30 # we're interested in numbers that can be typed in 12 or fewer presses # so, anything with more than 6 different letters is out NUMBERS = ( "ZERO", "ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX", "SEVEN", "EIGHT", "NINE", "TEN", "ELEVEN", "TWELVE", "TWENTY", "THIRTY", "FORTY", "FIFTY", "SIXTY", "NINETY" ) # "LOW" can be done in 12 or fewer presses for n in irange(5, 26): p = press("LOW", n) if p > 12: continue printf("LOW, {n} => {p}") # find how many of the numbers can be done in 12 or fewer presses ns = list(i for i in NUMBERS if press(i, n) < 13) if len(ns) != 3: continue printf("{n} => {ns}")Solution: The numerals are: ONE, TWO, TEN.