Strategy

A worked Master solve: cracking a cross-math puzzle step by step

The best way to learn the no-guessing method is to watch it work. Here is a compact cross-math puzzle solved end to end, with every step forced by the one before it.

The board

Here is the puzzle. Four empty cells — call them A and B on the top row, C and D on the bottom — and a tray holding exactly the numbers 2, 3, 6, and 7. Four equations must all be true at once: the two rows, and the two columns.

Look closely and you will notice something that marks this as a Master board: no equation has a single blank to fill. Every one has two unknowns. You cannot simply drop in a forced number to start — you first have to reason about which numbers are even possible. That extra layer is what deduction depth means.

Tray:  2   3   6   7

   [ A ]  +  [ B ]   =  9     (row 1)
   [ C ]  ×  [ D ]   =  18    (row 2)

   column 1:   A + C  =  8
   column 2:   B × D  =  21

Step 1 — factor the product

When no cell is directly forced, start with the most restrictive equation. A product is usually it, because multiplication has the fewest ways to land on a target. Look at row 2: C times D must equal 18, using two different tiles from {2, 3, 6, 7}.

Run the tray: 2 times anything available cannot reach 18, and 7 divides nothing here cleanly to 18 either. Only 3 and 6 multiply to 18. So C and D are 3 and 6 in some order — and that immediately forces A and B to be the leftovers, 2 and 7.

A, B  ∈  { 2, 7 }      (the leftovers)
C, D  ∈  { 3, 6 }      (only factors of 18 in the tray)

Step 2 — disambiguate with a column

Now use a crossing equation to pin the order. Column 1 says A plus C equals 8, with A from {2, 7} and C from {3, 6}. Test the four combinations: 2 plus 6 is 8, which works; 2 plus 3 is 5, 7 plus 3 is 10, and 7 plus 6 is 13, which do not.

Only one survives. A is 2 and C is 6 — both forced, with no guessing.

   [ 2 ]  +  [ B ]   =  9
   [ 6 ]  ×  [ D ]   =  18

Step 3 — the leftovers fall

The rest is forced by elimination. A was 2, so B must be 7, the other member of {2, 7}. C was 6, so D must be 3, the other member of {3, 6}. Every cell now has a value, and not one of them was a guess.

   [ 2 ]  +  [ 7 ]   =  9
   [ 6 ]  ×  [ 3 ]   =  18

Step 4 — confirm every equation

Finish by checking all four equations, including the one we never leaned on. Row 1: 2 plus 7 is 9. Row 2: 6 times 3 is 18. Column 1: 2 plus 6 is 8. Column 2: 7 times 3 is 21. All four are true together.

That last column is the reward of the no-guessing method: we never used B times D equals 21 to deduce anything, yet it falls out true. Because Lattice proved the board unique before release, this is not one answer among several — it is the answer.

   [ 2 ]  +  [ 7 ]   =  9     ✓
   [ 6 ]  ×  [ 3 ]   =  18    ✓
   column 1:   2 + 6  =  8    ✓
   column 2:   7 × 3  =  21   ✓

What made it Master

Notice that no first cell was ever directly forced — every equation began with two unknowns, so you had to reason about candidate sets before a single tile could land. Factor the product down to one pair, intersect it with a column sum, then let elimination finish. Depth, not size.

The same three moves scale to far larger Master boards: find the most restrictive equation, usually a product or a division; reduce it to a small set of possibilities; then use a crossing equation to disambiguate. Repeat, and even a daunting grid closes one forced deduction at a time.

Play today's Lattice