The Number Nobody Printed

A complete row always adds to the same amount. That is free information, it is on every board ever made, and it is what hard means here.

Spending a number that was always there

A row on a board of side six holds 1, 2, 3, 4, 5 and 6 exactly once, so it adds to twenty-one. Not usually, not on this board — always, on every six ever printed. A row of a nine adds to forty-five and a row of a four adds to ten. Nobody wrote those numbers on the puzzle and nobody had to.

The way to spend it is to account for the row a cage at a time. Find the cages lying wholly inside the row, note what each of them must contribute, and subtract the lot from the row’s total. What remains is what the leftover cells must add up to between them — and when there is exactly one leftover cell, that subtraction has just written a digit.

When there is more than one leftover cell the same arithmetic still narrows each of them. Take the most the others could possibly be, subtract from the shortfall, and you have a floor for this one; take the least they could be and you have a ceiling. That is rarely a decision on its own and it is frequently the elimination that lets a cage below finish the job.

Before — three cells of a 9+ cage lie along the top row
	1	2	3	4
1	cage adding to 91234	1234	1234	cage adding to 111234
2	cage adding to 71234	1234	1234	1234
3	1234	cage adding to 131234	1234	1234
4	1234	1234	1234	1234

After — the fourth cell of the row loses a candidate
	1	2	3	4
1	cage adding to 91234	1234	1234	cage adding to 11234
2	cage adding to 71234	1234	1234	1234
3	1234	cage adding to 131234	1234	1234
4	1234	1234	1234	1234

The 9+ cage has three cells in the top row and one directly below them. Whatever that stray cell holds is between one and four, so the three cells in the row contribute between five and eight. The row itself must total ten, which leaves the fourth cell somewhere between two and five — and since the board only goes up to four, the 1 is gone from it.

what the reasoning read what it decides what it wrote

The cage that sticks out

A cage lying wholly inside a row is a gift and there are never many of them. The technique becomes worth the trouble when you allow cages with a single cell outside the row, because those are everywhere — and the arithmetic barely changes. The cage’s contribution to the row is its own total minus whatever that one stray cell holds, and even without knowing the stray cell you know it is somewhere between one and the board’s largest digit.

That gives a range rather than a number, and a range is usually enough. A 13+ cage with three cells in the bottom row and one in the row above contributes at least nine to the bottom row, because the stray cell cannot exceed four on a board of side four. Subtract nine from the row’s total of ten and the last cell of that row can be at most one — which on a board whose digits start at one is a decision.

The same argument runs in the other direction for a cage sticking INTO the row from outside, and it runs on columns exactly as it does on rows. Working around a board asking the question of each line in turn is slow, which is why it belongs on this rung, and it is exhaustive, which is why it is worth doing properly when the free sweeps have dried up.

Before — a 13+ cage with three cells in the bottom row
	1	2	3	4
1	cage adding to 91234	1234	1234	cage adding to 11234
2	cage adding to 71234	1234	1234	1234
3	1234	cage adding to 131234	1234	1234
4	1234	1234	1234	1234

After — the last cell of that row is decided outright
	1	2	3	4
1	cage adding to 9234	1234	1234	cage adding to 11234
2	cage adding to 7234	1234	1234	1234
3	234	cage adding to 131234	1234	1234
4	1	234	234	234

The 13+ cage puts three cells in the bottom row and one in the row above. That stray cell is at most four, so the three cells down here contribute at least nine of the row’s total of ten — leaving at most one for the cell outside the cage. On a board whose digits start at one, at most one means exactly one, and the cell is written without any cage being consulted.

What hard means on a cage square

A hard board here is one where the solver, always taking the cheapest move available, was at some point forced to add a line up. Nothing was printed, no cage had a short list, no cell and no digit had run out of room. The only thing left to spend was the fixed total of a complete line.

It is also why side and difficulty stay separate choices on this site. A 9×9 with nothing but two-cell cages never needs this argument once, and a 5×5 with two awkward columns may need it three times. Most players find the second takes longer despite being a third of the size.

Where this sits on the ladder

Rung 1Free cells
A one-cell cage is its own answer, and a row with one gap left has no choice about it.
Rung 2Cage combinations
List every way a small cage can be filled and keep whatever all of them agree on.
Rung 3Singles
A cell with one digit left, or a digit with one cell left in its row or column.
Rung 4Unit totals
Every row adds up to the same known number, so the cages inside it price the rest.
Rung 5Cage interaction
What a group of cells can hold between them, whether they share a cage or only a row.
Rung 6Assumption
Write a digit, follow it until something breaks, and rub it out again.

Common questions

What does a row of a calcudoku add up to?

The sum of 1 to n, so ten on a 4×4, twenty-one on a 6×6 and forty-five on a 9×9. It is the same on every row and every column of every board of that size.

What if no cage lies wholly inside the row?

Use the ones with a single cell outside it. Their contribution to the row is their total minus that one cell, which is a range rather than a number and is very often narrow enough to decide something.

Does the same trick work with multiplication?

In principle — a complete row always has the same product too — but the numbers get large quickly and the arithmetic stops being something anybody wants to do by hand. Sums are where the value is.