# Can you solve this grid?

I'm not completely sure there is enough information here to solve these puzzles, if not I'll add hints until someone solves it.

Rules:

• The numbers $$1$$ to $$12$$ appear once each.
• The numbers on the perimeter are the row/column totals
• The greater/lesser signs indicate just that.

The grid:

Explanation:

The second column numbers add to 11, in increasing order. The only possibilities that satisfy this are 1-2-8, 1-3-7, 2-3-6, 1-4-6, and 2-4-5. But the largest of these (placed in the bottom position) cannot be 7 or 8, as that would force the bottom row (the other three numbers of which are all greater) to be larger than 33. So only 2-3-6, 1-4-6, or 2-4-5 are possibilities.

The 1 can only go in two places, based on the greater-than / less-than signs: second column top, or fourth column middle. If 4th column middle, the other numbers in the column must be 10 and 12. 10 would have to be the top one, as the number in that position is less than the number to its immediate left. That number to the left would then have to be 11, but the remaining numbers in that top row, no matter what they are, would exceed 23, so this is not a possibility. Therefore the 1 can only be placed in the second column top, and the second column must therefore contain 1-4-6, in that order top to bottom.

Knowing that the second column bottom number is 6, we can determine the possibilities for the other three numbers in the bottom row. They must add to 27 (33-6), and all three are greater than 6, based on the gt/lt signs. The only possibilities are 7-8-12, 7-9-11, or 8-9-10.

The 12 can only go in three places in the grid, based on the gt/lt signs: left column middle, lower right, or third column top. We can rule out third column top, as in that case the other two numbers in the column would have to add to ten, with the smaller of these (middle position) being greater than 4. This is impossible.

Now consider the left column middle position for the 12. This is more complicated. Based on the column total, the other numbers must add to ten, with the bottom one larger than 6. The only possibilities are 2-8 or 3-7. These options would only allow 9-10 or 9-11, respectively, in the rightmost two slots of the bottom row, in that order. Either way, a 9 would be in the third-column bottom slot. Then, the other numbers in its column would be 11-2, 10-3, 9-4, 8-5, or 7-6. Based on the other numbers already in use, only 8-5 would be a possibility. This is only possible with 3-7 in the top and bottom of the left column, but would require an 11 in EACH of the top and bottom slots of the right column, and so is impossible. So 12 must be in lower right.

With 6 and 12 in position, the other two numbers in the bottom row must be 7-8, in either order.

Now let's look at the right column. The 12 is placed, so the other two numbers must add to 11. They can only be 1-10, 2-9, 3-8, 4-7, or 5-6. The 1, 4, and 6 are already placed elsewhere, as are the 7 and 8, so that only leaves 2-9 as a possibility; based on the gt/lt sign, 9 must be in the top row of this 4th column, and 2 below it. Since 2 is used, the upper left must be 3, the lower left 7, and the bottom row 3rd position an 8. That leaves 5 and 10, which are easily placed.

## Step by step

First, look at the second cell in the bottom row, at the intersection of $$11$$ and $$33$$.

This is the largest number among three summing to $$11$$, so it must be one of $$5,6,7,8$$.
It is the smallest number among four summing to $$33$$, so it can't be $$7$$ or $$8$$.
If it is $$5$$, then above it must be $$4$$ and then $$2$$. In that case, consider the third cell in the middle row. It's bigger than $$4$$, so it must be at least $$6$$, but it's also the smallest number among three summing to $$22$$. Contradiction.
So this number is $$6$$, and above it is either $$4$$ then $$1$$ or $$3$$ then $$2$$, while the rest of the bottom row is either $$12,8,7$$ or $$11,9,7$$ or $$10,9,8$$.

Now consider the middle row and the third column.

The second cell in the middle row is either $$3$$ or $$4$$. The third cell in the middle row is bigger than that, but also the smallest number among three summing to $$22$$, and it's not $$6$$. So it must be either $$4$$ or $$5$$. If it's $$4$$, then the second column is $$2,3,6$$ (in order) and the middle row is $$?,3,4,1$$ summing to $$22$$, contradiction.
So the third cell in the middle row is $$5$$, and the other two must be either $$8,9$$ or $$10,7$$ in some order.
In the middle row, we have $$5$$ and two smaller numbers ($$4,3$$ or $$4,2$$ or $$3,1$$) and something else summing to $$22$$. So the first cell must be $$10$$ or $$11$$, and the $$3,1$$ option is impossible.
Assume the second cell in the middle row is $$3$$. Then the second column is $$2,3,6$$ (in order) and the middle row is $$10,3,5,4$$ (in order) and the third column is either $$8,5,9$$ or $$9,5,8$$. So the bottom row is either $$7,6,8,12$$ or $$7,6,9,11$$. But then the first column is $$5,10,7$$, contradiction.
So the second column is $$1,4,6$$ (in order), and we have:

If the last cell in the middle row is

$$3$$, then the middle row is $$10,4,5,3$$ (in order) so the other two cells in the first column must sum to $$12$$. But no possibilities are left for that, since $$5,4,3,10,1$$ are all used up. Contradiction.
So the last cell in the middle row is $$2$$, which means the middle row is $$11,4,5,2$$ in order, and again there's only one possibility for the first column:

Then the last four cells are $$7,9,10,12$$, with $$7,12$$ in one row and $$9,10$$ in the other. So the final solution must be: