By Stephen Meskin

This column could stop here with just the title. But I get paid by the word, so I’ll ramble on a bit before we get to Josh’s solution to the problem he posed in the last issue.

The common crossword puzzle (CWP) in the U.S. has a 15 × 15 grid. If there are no squares colored in, the common CWP would have 30 words—15 across and 15 down. Most common CWPs have more than twice that number. Colored-in squares are called “blocks.” If a block is added into a grid without touching any other block or a side, it breaks the horizontal row into two words and simultaneously breaks the vertical column into two words. Thus it adds two words. Adding a block that touches another block, or touches the side, might add only one word, or none—or it might even delete a word. Indeed, if the grid is totally filled with blocks, there would be no words at all.

Answers to CWP clues can be more than one literary word, but here we don’t care about literary words—we are interested in geometric words, for which there is a one-to-one correspondence with the clues. To avoid this ambiguity, I will henceforth use “answer” or “clue,” and you should interpret my use of “word” in the prior paragraph in that geometric sense. For common CWPs, there are no one-letter or two-letter answers. It clearly makes sense to exclude one-letter answers; how many ways are there to make up a clue for “a”? Excluding one-letter answers implies that every square must satisfy both an Across clue and a Down clue. Excluding two-letter answers is a little shakier—but considering that there are only 676 letter pairs, many of which are undefinable, it makes sense.

The limitation on answer length for linguistic reasons has significant geometric impact. It is easy to see that a common 15×15 CWP must have fewer than 120 answers. Indeed, there can be no more than four 3-letter answers in each of its 15 rows, and in each of its 15 columns. So, there must be fewer than 120 answers in total. “Fewer” because if there were 4 answers in each row, there could not be 4 answers in each column.

The problem in the title should be restated as:

**A.** **What is the maximum possible number of clues in a common 15 **×

**15 CWP?**

But to answer the question, you need to know the full definition of a common CWP. The CWP gods seem to have set four rules that define a common CWP in the U.S.

- 15×15 grid (other sizes appear but are less common)
- No answers of length 1 or 2
- Connectivity
- Only 180-degree rotational symmetry

Items 1 & 2 have been discussed already. *Connectivity*
means that there are no self-contained sub-CWPs, i.e., the white space is
connected. Item 4 means that the grid looks
the same if it is rotated 180° but **not** 90° or
270° **nor** if it is flipped about a horizontal axis, vertical axis, 45°
line or 135° line. (Variations can be observed in the wild; I have seen a grid
that looked like a smiley face, i.e., it had symmetry only about a vertical
line.) The last two rules seem to be primarily for aesthetic, that is
geometric, reasons. The puzzle on page 34 satisfies all four rules.

Some of you will quickly run off to program your computers to solve this problem. And there is nothing wrong with that; in fact, writing a computer program to solve a problem can be a very interesting proposition in and of itself. If you use a brute-force technique, though, the computer could take days to solve it.

It turns out, actually, that the best way to use your computer to solve this problem is to do an internet search.

Once you know the answer to Problem A, the more interesting problem is to prove your answer is correct. But I think that would be too hard for you to do and definitely too hard for me to review. However, if we use something smaller than 15, it would be doable.

**B. Same as Problem A, but change 15 to 7—and prove your answer correct.**

Other variations are also interesting.

**C. Same as Problem A (n=15) but change the symmetry rule (allow a different single symmetry, or multiple symmetries, or none at all)—no proof required.**

**D. Same as Problem A (n=15) but don’t require connectivity—no proof required.**

I only know the answers to some of these questions. But please don’thave any cross words for me; I do know that they all have answers. I’m counting on you to tell me what the answers are.

**Bonus Puzzle: **Why are CWP and this question called *puzzles*
and A to D *problems?* This question has no definitive right or wrong
answer.

**Solutions may be emailed to cont.puzzles@gmail.com.**

**In order to make the solver list, your solutions must be
received by June 1, 2021.**

**Answer to Prior Puzzle: Road Race Shenanigans**

How much farther did Batty run compared to those on the starting line? Let x equal the distance the first group moves. Let V be the velocity of the first group. Let v be Batty’s velocity. And let t be the time it takes for Batty to reach the front runners originally. We can write in equations the distance the first group moves: x=Vt (equation 1) and the distance Batty moves: 150+x=vt (equation 2). Dividing equation 2 by equation 1 yields (150+x)/x = v/V (equation 3).

Next, let’s do the same exercise for the runners at the back of the pack, and let T= the time it takes the back of the pack to reach the starting line. We know that 150=VT (equation 4), and since Batty needs to double back, 150+2x=vT (equation 5). Dividing equation 4 by equation 3 gives you (150+2x)/150=(v/V) (equation 6). Now we just set equations 3 and 6 equal to each other, and then solve the resulting quadratic equation. As we know x has to be a positive number, we see that x=106.06. Finally, the extra distance Batty ran is simply 2x+150=362.13 meters.

**Solvers**

*Anthony Amodeo, Robert Bartholomew, Bob Byrne, Lois
Cappellano, Samantha Casanova, Bob Conger, Andrew Dean, Sam Ellis, David
Engelmayer, Alex Esche, Bill Feldman, Rui Guo, Pat Johnston Clive Keating,
Timothy Mosler, Paul Navratil, Suzy Poole, David Promislow, Anthony Salis, Noam
Segal, Tomasz Serbinowski, John Snyder, Al Spooner, Daniel Wade, and Gary Wang.*