Process of Elimination in Puzzles: Rule Out Until One Is Left

The process of elimination in a puzzle means listing every option that could be true, then crossing out each one a rule or clue makes impossible, until a single option is left standing. When the crossing out stalls before that point, you add one careful guess-and-check step: try one option, follow it, and if it leads to a conflict, cross it out too.

This guide covers how to set out your options, the handful of moves that do almost all the work, a full worked grid, an example where a guess is needed, and a practice puzzle with its answer.

What does the process of elimination mean in a puzzle?

Most puzzles hide the answer among a fixed set of possibilities. A grid cell can hold one of four letters. A secret number lies in a known range. A set of people can be matched to a set of items in a limited number of ways.

Instead of hunting for the answer directly, you work on what is false. Each clue rules some possibilities out. You never need to see why the answer is right, only that nothing else survives.

Why does crossing options out lead to the answer?

It works because a fair puzzle has exactly one answer. If every wrong option is removed, whatever remains must be that answer, even if you never tested it directly.

The same idea warns you when something has gone wrong: a cell with no option left means a mark was placed in error, or the puzzle has no answer.

There is one more condition: you must only cross out what a rule truly forbids. A hunch is not a reason. One option removed by mistake can make the true answer vanish, and then the puzzle seems impossible.

How do you use the process of elimination, step by step?

1. List every option. For a grid, write all possible symbols small in the corner of each empty cell. For a number puzzle, write the range of possible values.
2. Apply the plain rules. Remove anything the basic rules forbid, such as a letter that already sits in the same row or column.
3. Apply each clue. Cross out what every clue rules out. Work through the clues in order and tick each one off once used.
4. Look for a single option. A cell with one option left is solved. So is a letter that fits in only one cell of a row or column.
5. Spread every new answer. Each time you write an answer, remove it from the rest of its row and column, then return to step four.
6. If nothing moves, test one option. Pick a cell with only two options, try one and follow the chain. If it ends in a conflict, the other option is correct.
7. Check the finished answer. Read every rule and clue once more against your result.

Steps four and five form a loop, and most of the solve is simply going round it.

Which elimination moves do the most work?

The first two moves are two views of the same idea. One looks at a cell and asks which symbols can still go there. The other looks at a symbol and asks where in a row or column it can still go. Good solvers keep moving from one view to the other.

How does elimination solve a full worked example?

Here is an original grid. Fill it with the letters A, B, C and D so that each letter appears once in every row and once in every column. A cell marked not C may hold anything except C.

Try it before reading on. This grid has exactly one filling, and every given letter and barred mark is needed to make it so.

Step one: place the last B. Rows one, three and four already have B, in columns two, one and four. Row two needs a B, and column three is the only column still without one. Write B in row two, column three.

Step two: finish row four. Row four has A and B, so its open cells take C and D. Column three is not C, so it holds D, and column two holds C.

Step three: use column three. It now holds B and D, so rows one and three take A and C. Row one, column three is not C, so it holds A, and row three, column three holds C.

Step four: finish row three. Row three has B and C, so A and D remain for columns two and four. Column four is not A, so it holds D, and column two holds A.

Step five: column two. It now holds B, A and C, so the only letter left for row two is D.

Step six: row two. It has D and B, leaving A and C. Column one already has B and A below, so row two, column one cannot be A and must be C. Row two, column four is then A.

Step seven: row one. Column one now has C, B and A, so row one, column one is D. The last cell, row one, column four, is C.

Every step was a single option left or a single place left. No guess was needed. That is the mark of a puzzle where elimination alone is enough.

When is elimination alone enough?

Elimination alone is enough when, at every stage, at least one cell has one option left or one symbol has one place left. Each answer you write removes more options, which creates the next single, and the chain carries you to the end.

A setter who wants a smooth solve can build a puzzle this way on purpose, by checking that the simple moves never run dry.

Elimination alone runs out when every open cell still holds two or more options and every symbol still has two or more places. You are then stuck, even though the puzzle may still have just one answer. That is the moment for a guess-and-check step.

When do you need a guess-and-check step?

Here is a second original grid with the same rules: A, B, C and D once in every row and column, with barred letters in some cells.

The simple moves start well. Rows one, two and four have B in columns one, four and two, so row three's B goes in column three. Row four then needs A and D in columns one and three, and column three is not D, so it takes A, which puts D in column one.

Then everything stops. Row one, column two can be A or C. Row one, column four can be A or D. Every other open cell also has two or three options, and no letter has a single place left in any row or column.

The test. Suppose row one, column two is A. Row one then needs C and D in columns three and four. Column four cannot hold C, because row four already has C there, so it takes D and column three takes C.

Column three now holds C, B and A, so row two, column three must be D. Row two then has D and B, and its column two cell is not C, so it must be A. But column two already has A in row one. Two A marks in one column is a conflict.

So row one, column two is not A. Cross it out, and the only option left is C. From here, the simple moves run to the end without any further test.

Notice that the test was not a blind guess. It was a short demonstration that one option fails, which is just elimination done one level deeper. Always pick a cell with only two options, so a single conflict settles it.

How does the sum and product puzzle use elimination?

The sum and product puzzle is a puzzle about knowledge that runs on pure elimination. Two whole numbers are chosen, both greater than 1, the second larger than the first, with a sum of at most 100 ^[1]. One person is told only the sum, another only the product ^[1].

Then three remarks are made. The sum holder says the product holder cannot know the numbers. The product holder replies that now the numbers are known. The sum holder then says that now the numbers are known too ^[1].

Each remark is a filter. Before anyone speaks, every allowed pair is a possible answer, and after each remark, only the pairs that agree with it stay on the list ^[1]. A paper published on 18 July 2023 wrote these filters as rules for an answer set solver, and running the solver left exactly one pair: 4 and 13 ^[1].

The puzzle looks like a riddle about what people know, yet its solution is the same loop as a grid: write down the options, apply a rule, cross out what fails, repeat.

Here is a smaller riddle that rewards the same habit. List the possible numbers, then let each line of the riddle cross some out.

Can elimination show that a puzzle has no answer?

Yes. If you keep crossing out and some cell ends with no option at all, and every step was sound, then the puzzle has no answer. Elimination can show that a puzzle is impossible as well as find its answer.

Grids with barred entries have been studied closely ^[2]. A grid of barred marks is called avoidable when the grid can be filled, each letter once per row and column, without any cell holding its barred letter ^[2]. The question of which grids can be avoided was first asked in 1989 ^[2].

A paper published on 12 May 2006 states, citing earlier work, that every part-filled grid of this kind, with each cell barring at most one letter and no letter barred twice in a row or column, can be avoided once the grid is at least 4 by 4 ^[2]. The same paper notes that this is not true for 3 by 3 grids, where some unavoidable examples exist ^[2].

Its main result goes further: for any whole number m there is a constant such that, whenever k is larger than that constant, every grid with 2mk rows and 2mk columns, in which each cell bars at most m letters and no letter is barred more than m times in any single row or column, can be filled while dodging all the barred letters ^[2].

Here is a three by three grid of the stubborn kind. Fill it with A, B and C once in every row and column, but no cell may hold the letter written in it.

Try it, and elimination will close every route. A three by three grid of this kind can be filled in twelve ways, and each of them puts some letter in a cell that bars it.

What mistakes break the process of elimination?

• Crossing out on a hunch. Only remove an option a rule truly forbids. One wrong cross can erase the real answer.
• Forgetting to spread an answer. Every letter you write should be removed from its whole row and column straight away.
• Looking only at cells. Ask where each symbol can go, not just what each cell can hold. The only place left move is easy to miss.
• Guessing in a crowded cell. Test a cell with exactly two options, so one conflict decides it.
• Skipping the final check. Read every rule once more against your filled grid.

Can you solve this practice puzzle?

Fill the grid with A, B, C and D, once in every row and once in every column. A cell marked not A may hold anything except A, and so on.

Elimination alone is enough for this one. Start with column one, then think about where row two can put its A.

Answer:

Column one has B, A and C, so row two, column one is D. Row two cannot put A in columns two or three, so A goes in column four. From there, each row and column falls to the only option left or the only place left.

Now try a live riddle about three people and three fruits. Cross out what each clue forbids and see which option is left.