Enigma 1722

Spoiler for New Scientist Enigma 1722: “Double Latin” (Follow the link to see the puzzle.)

Label the rows 1 through 5 (bottom to top) and the columns a through e (left to right) so that, for instance, the red side of card e4 is 3.

Now examine red card sides to see what numbers they can have on them. At e1 and e5, for instance, we can have 1 or 2. Let’s put 1 at e5 and 2 at e1.

Now d2 can only be 1 and d4 can only be 5. From there d5 can only be 4,  d1 must be 3, c3 must be 3. And so on; every card can have only one value and the complete red side is:

5 3 2 4 1
2 4 1 5 3
1 5 3 2 4
3 2 4 1 5
4 1 5 3 2

On the other hand if we put 1 at e1 and 2 at e5, we can again get a unique set of values for the red side:

4 1 3 5 2
5 2 4 1 3
1 3 5 2 4
2 4 1 3 5
3 5 2 4 1

Now consider the green side. In the first case, b1 has 2 on the green side and 1 on the red. None of the rest of row 1, column b, a2, or c2 can be 2 on the green side, but also neither can d2, a3, or c4 because they also have a 1 on the red side. This means e2 must be 2, after which c3 must be 2, and then there is nowhere in column d to put a 2. So that case is impossible.

In the other case, not only can none of the rest of row 1, column b, a2, or c2 can be 2 on the green side, but also neither can e2, c3, or a4 because they also have a 5 on the red side. So d2 must be 2, and then a3 must be 2. b4 must be 1, and e3 must be 1. Now e4 can’t be 2, because there’s a 3 on the red side, and that combination has already been used at d2. So c4 and e5 are 2. Likewise a1 and c2 are 1.

That accounts for all the 1s and 2s. Now there’s a problem, because there’s nothing left to constrain the rest. So for example we can put 3 at a5, b3, d4, c1, and e2; 4 at a4, c5, b2, d3, and e1; and 5 at a2, b5, c3, d1, and e4. That satisfies the constraints. But we also could interchange all the 3s with all the 4s for another valid solution, or with all the 5s for another — in fact there are six solutions corresponding to all the permutations of {345}.

So is this a problem? The enigma asks for “the numbers hidden under L A T I N”. I interpreted that as meaning the numbers on the hidden side — the green side. There is no unique answer if so.

But if this means the letters L A T I N are hiding the red-side numbers, and these are the numbers to be given, then there is a solution: 4 5 1 2 3. I suppose this must be what was meant, but if so I don’t like the wording of the question much.

Then again, I could be pedantic and say we do know the green side numbers; they’re 1 2 3 4 and 5. It didn’t say we had to specify them in any particular order!



Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s