Puzzle

Normally, when something is amiss in a Toronto Star puzzle, I just contact the author and let it go at that. However, in the last two days there have been at least three mistakes, assuming that I am seeing the paper correctly.

The Saturday chess problem, #1928, looked eerily familiar. Yep, it is. I remember the king trapped by his own bishop and the hostile knight at h1. It took me awhile to remember the solution (won’t spoil it here) but not nearly as long as the first time this puzzle was run. It’s fairly hard, imho. I am sure this is a duplicate of a puzzle a year or more ago.

The Saturday Challenger is a number puzzle. You are given four numbers in a four-by-four grid. You are also given the column totals, the row totals, and the diagonal totals. You are only allowed to place single digits 1 through 9 in the missing spaces, and the result should add up to the given totals. You are allowed to re-use digits (this isn’t like Kakuro).

In the Saturday February 18 version there is a column (2) with a given number of 6 and a total of 11. This means the largest possible number in this column is 3. (assume 1’s in the other missing spaces and the total is 11, eh?). There is also a row (2) with a given number of 3 and a total of 25. This means the smallest possible number in this row is 4. (assume 9’s in the other missing spaces and the total is 25, eh?)

Therefore where row 2 crosses column 2 you need a number at least as large as four and at least as small as 3. This is not possible.

Now for the Sunday February 19 Stickeler. (Sorry, Terry). I think what happened here is the opening sentence got corrupted. It reads, in my copy,

if X > 1, is the following true or false?

X |X|^3 < |x|^X

The claim is that the above relation is always true. However if X > 1 I can choose X = 4 and the left hand side becomes equal to 4^4 as does the right hand side. So they are equal, not unequal, in this specific case. Clearly for larger X the right hand side grows faster than the left.

If however the initial constraint were X < 1 then the claim might be correct.

What I find weird about this is the coincidence. Errors do occur in the chess column, but quite rarely. Errors do occur in Stickelers, but also quite rarely. I’ve never before seen an error in the Challenger (Linus Maurer) and I always do this puzzle.

Rather than laugh at the Star and its puzzle makers, to whom I owe a debt for some fun and intellectual stimulation, I ask a dumb question:

Was this set of puzzles rushed, due to staffing or other limitations, surrounding the holiday weekend? It doesn’t really matter, so that does qualify as a sort-of dumb question.

4 thoughts on “Puzzle

  1. The Challenger puzzle in the Toronto Star has become a formula puzzle. I can usually complete it in about 30 sec. because there is always a row that includes two 9’s. Once you find that row the rest is just filling in the blanks. Had you noticed that too? I wanted to write to Linus Maurer about it, but couldn’t find a way to contact him. Maybe an old fashioned letter addressed to him at the Toronto Star?

    • I enjoy doing the challenger puzzle, and can generally beat the stated times, but not at the thirty second level. What I do is annotate each row, diagonal, and column with GE (greater than or equal to) and LE (less than or equal to) by imagining nines or ones in all but one missing space. If a crossing row / column / diagonal have the same limit but opposite qualifier, then that limit goes in the intersection and ones and nines go in the other spaces. Sometimes this requires a re-evaluation as additional constraints have been put on rows, columns, et cetera.
      As for contactin Linus Maurer, I thought I had his eMail but it appears that I do not.
      I googled Linus Maurer and found King Features Syndicate. They have a ‘contact us’ page and I asked them for contact information to Mr. Maurer on our behalf. I’ll post here if I find anything useful

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