Enigma 1368 from New Scientist
Background
The New Scientist magazine runs a regular series of little combinatorial puzzles, most of which are clearly designed with computer solutions in mind. Sometimes you just look at them and think “That looks like a one-liner”and don’t actually bother to code them up, and sometimes you get sufficiently interested to flip open your laptop for some gentle fireside programming. This being the Ken Iverson tribute edition of Vector, I thought I could make a small contribution by logging my attack on one of these puzzles in APL. Why do I think this is appropriate? Basically because without Ken’s work, I would never have used the computer to have fun. Writing C# is something you do to earn a living; fooling around in an APL session is something you can do of an evening, with Bach on the radio and a nice log fire at your feet. So much of today’s software leaves you feeling frustrated, angry and stressed; APL and its many descendants are doing their little bit to redress the balance, and Ken deserves all our thanks for that.
The Puzzle
EEN, VIER and NEGEN are Dutch for 1,4,9. Replacing letters by numbers we get three perfect squares (no number starts with zero). What is the value of the square root of ( EEN × VIER × NEGEN )?
The Attack
This one is interesting, in being largely just a logic puzzle rather than a huge combinatoric. It is a nice example of where a ‘desk calculator’ APL can be really helpful in expressing the tests you need to reduce the lists of numbers to a manageable size. No programming was required at all here, just typing.
The first stage is obvious – how many perfect squares are there with the pattern EEN. I immediately thought of 225, but let’s list them to be sure:
(10↓⍳31)*2
121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961
Damn, forgot about 441. However there are only two of these, so we have only two choices for E and N. Let’s go with E=2, N=5 and see how we fare. Here are the candidates for NEGEN, created by copy/paste in the session:
2⍕ 52025 52125 52225 52325 52425 52525 52625 52725 52825 52925 * 0.5
228.09 228.31 228.53 228.75 228.97 229.18 229.40 229.62 229.84 230.05
Nope, none of those look very integer. Let’s try E=4, N=1 and play the same game:
2⍕ 14041 14141 14241 14341 14441 14541 14641 14741 14841 14941 * 0.5
118.49 118.92 119.34 119.75 120.17 120.59 121.00 121.41 121.82 122.23
121*2
14641
Got it! Now we just need a 4-digit perfect square with 4 as the third digit. There are only 9000 or so candidates, but at this point a bit of computer assistance is quite welcome!
qq← (4=( 10 10 10 10⊤ (⍳9999))[3;])/⍳9999
⍴qq
1000
qq/⍨←qq>1000
⍴qq
900
qq/⍨ {⍵=⌊⍵} qq*0.5
1444 1849 3249 3844 5041 6241 7744 8649
Visual inspection (and the backspace key) can do the rest. Anything with repeated digits is out of it, for starters. Then we can remove anything with 1 or 6 as these are already allocated other letters. So the answer is ...
( 3249 × 441 × 14641 )*0.5 144837
No, I didn’t win the £15 book token, but at least I could paste my entry directly from the APL session to save a bit of typing in Outlook. It really was rather a simple one, and they must have had plenty of correct solutions.