April Fools Suduro Logic Explanation

Here's the discussion of the logic behind the April-Fools puzzle. First, it is interesting to note that the two puzzles that supplied the sums clues for the April Fools puzzle are in a sort of logical sense almost the same puzzle - if you switch the answer key changing every number "n" into "10-n", then create sums clues for the new puzzle, then the new puzzle can be solved just as easily (or with just as much difficulty) as the old puzzle, using virtually identical logic. For example, if a pair of gray squares add up to 4 in the original puzzle, we know that the two numbers in the gray squares must be 1 and 3, in some order, and in the new puzzle, the sums clue for those gray squares will be 16, and the numbers must be 9 and 7, in some order, so every sums clue in the new puzzle will be precisely equally restrictive as the sums clue in the old puzzle, logically speaking. The puzzles are, in a sort of logical sense, mirror images of one another.

Now, on to the actual April Fools solution. If a sum of 3 gray squares is s, adding the specific 3 numbers n1, n2, and n3 in the original puzzle, then the sum S for the new puzzle must be (10-n1)+(10-n2)+(10-n3) which equals 30-(n1+n2+n3) which equals 30-s, since (n1+n2+n3)=s. In general, in the left black column, you can recover the orignal sums clue by counting the gray squares, setting that count to c, and calculating (c*10-s), where s is the clue for the second puzzle.

The 3-by-3 white sums clues in the upper left are a little trickier: In each case, you can find the count c of gray squares in the puzzle's 3-by-3 grid that the clue refers to. (For example, in the upper-left, the sums clue is 16, and c=3, for the 3 gray squares in the upper-left subgrid.) If you look at the sums clue s, for that subgrid, you know that the original puzzle's sums clue must be *either* s or (c*10-s), since the white sums clues are completely scrambled. However, you also know that the original sums clues for all three vertical 3-by-3 subgrids in a column must equal the sum of the three columns that fill those same three subgrids. For example, the left-most three subgrids (which have the scrambled clues 16,6.6) *should* have clues that add up to (18+19+17), the first three sums clues in the top black strip. If you consider all of the ways that the current scrambled clues 16, 6, 6 might each be replaced by potential clues (c*10-s), there will turn out to be only one set of clues that add up to (18+19+17), and so forth for the other 3 vertical columns of subgrid clues. (There are some Suduros where an April-Fools version like this would not yield just one possible set of 3-by-3 subgrid clues consistent with the 3 sets of columns' clues, but most work fine, including this puzzle that I'd originally scheduled for April 1, anyway.)

Have fun!