BUG+n

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This page discusses ways of exploiting BUG+n patterns, but it assumes that you know the difference between a BUG and a BUG+n, that you know what a Hero is, and that you know how to determine which candies are the Heroes in a BUG+n. Read the page on Deadly Patterns before starting to work your way through the material below.

In all the Grids below, the Heroes are shown in green and written last in a Cell's list of candies — this is done to make the Heroes more visible for purposes of discussion. (Since the Cell background colour is often blue, the Heroes won't look distinctly green, but at least they'll have a lighter colour.)

To keep track of what you're doing when you exploit a BUG+n with n greater than 1, you'll often find it helpful to make use of a set of coloured pencils. Circle all the Heroes in colour — that will allow you to not lose track of the candies that are the focus of all your efforts. If you're constructing a long Chain and you want to label the Chain Cells (for purposes of verification), label them with coloured letters — those will stand out, but they won't obscure the primary markup (the candies themselves).

BUG+1

A BUG+1 has one or more Heroes in (only) one Cell of the pattern.

In the Grid below, there is one Hero, shown in green in the blue Cell; the deadly candies are left in black.

BUG+1.

The blue Cell is the only non-Pair Cell. It contains {478}.

Look at the blue Cell's Block: the 4s and 7s are each Twins; there is a Triplet of 8s, which is one too many 8s. So it looks like the 8 in the blue Cell is the Hero that breaks the BUG.

This is confirmed by the fact that there are also one too many 8s in the blue Cell's Row and in its Column. (The other candies are all Twins, as would be expected in a BUG.)

In the blue Cell, you can kill the deadly 4 and 7 (leaving only the 8), which will prevent this BUG+1 from degenerating into a BUG.

That newly crowned 8 gives rise to a cascade of moves which cracks the Sudoku.

Usually a BUG+1 is fairly big, like that one. The following is the smallest BUG+1 you will find: it has only 12 Cells and only 4 deadly candies:

Smallest BUG+1.

The blue Cell is the only non-Pair Cell. It contains {123}.

Look at the blue Cell: in that Cell's Block, in its Row, and in its Column, the 3 is a Triplet (all the other candies are Twins). So the 3 in the blue Cell is the Hero that breaks the BUG.

In the blue Cell, you can kill the deadly 1 and 2 (leaving only the 3), which will prevent this BUG+1 from degenerating into a BUG.

That newly crowned 3 gives rise to a cascade of moves which cracks the Sudoku.

A BUG+1 is obviously very easy to exploit, and it always cracks the Sudoku.

The rest of this page will deal with BUG+n patterns where n is greater than 1. As you will see, these require a lot more thinking.

BUG+2

The pattern below contains two Cells that are not Pairs. That doesn't guarantee that it's a BUG+2, so that's the first issue we have to address:

BUG+2: verification.

Consider the {364} blue Cell in the Upper Left Block: in that Cell's Block, in its Row, and in its Column, the 4 is a Triplet (all the other candies are Twins). So it looks like the 4 is a Hero.

But we have to check the other {364} blue Cell, the one in the Middle Right Block: again, the 4 is a Triplet in all three of its Houses.

This means that each of the two blue Cells contains a candie that qualifies as a Hero. That's what allows us to declare this pattern to be a BUG+2.

You recall that a Sudoku cannot contain a BUG, because a BUG is a Deadly Pattern which would cause a puzzle to have no solution at all, whereas a Sudoku is guaranteed to have exactly one Solution.

Now, in order for the BUG+2 above not to degenerate into a BUG, at least one of the two Heroes must be true. We can make use of that fact to derive a very useful result.

We're going to make two different arguments that lead to the desired result. (You don't need both arguments — either one suffices.)

Same BUG+2 as above: exploitation.

There are two Heroes, and they both happen to be 4. At least one of these two Heroes must be true.

Consensus: If either 4 Hero is true, the 4 in the purple Cell gets killed. So we have agreement, and we can in fact kill the 4 in the purple Cell.

Now we'll start all over with a different argument.

Troublemaker: what if the 4 in the purple Cell were true? It would kill both the 4 Heroes. But you can't kill all the Heroes, because that would make the BUG+2 become a BUG (which would mean that the Sudoku has no solution, which is never the case). So we eliminate the Troublemaker — we kill the 4 in the purple Cell.

Killing the 4 in the purple Cell (and crowning the 5) unleashes a series of moves which cracks the Sudoku. So the BUG+2 Tactic turns out to be exceedingly effective.

We used two different arguments:

Consensus: if the assumed truth of every Hero (one by one) implies that one specific candie in one specific Cell is killed, then that candie must be killed (since at least one of the Heroes must be true).
Troublemaker: if some candie's being true would mean that all the Heroes were killed, then that candie must be false (since at least one of the Heroes must be true).

You can see that the Consensus approach and the Troublemaker approach are logically equivalent, and of course you only need to use one of them. In simple cases, both approaches are obvious. But in more complex situations, one of the two approaches may be easier than the other.

In practice, Consensus is sometimes harder, since you might not see what candie all the Heroes would kill, and you have to keep adjusting your sights to find it.

For the Troublemaker approach, of course you don't know which candie will be a good Troublemaker, but you know what your objective is — to find a Troublemaker that kills all the Heroes — so you can try likely Troublemakers and quickly discard the ones that get you nowhere.

I tend to favour the Troublemaker approach, but we'll do one more BUG+2 example using both approaches:

BUG+2: verification.

Consider the {356} blue Cell: in that Cell's Block, in its Row, and in its Column, the 6 is a Triplet (all the other candies are Twins). So it looks like the 6 is a Hero.

But we have to check the other {674} Cell: this time, it's the 4 that's a Triplet in all three of its Houses (the other candies being Twins).

So each of the two blue Cells contains a candie that qualifies as a Hero, which confirms that this pattern is a genuine BUG+2.

In order for the BUG+2 above not to degenerate into a BUG, at least one of the two Heroes must be true. And again, we can make use of that fact to derive a useful result:

Same BUG+2 as above: exploitation.

There are two Heroes, a 6 and a 4, and at least one of these Heroes must be true.

Consensus: If the 6 Hero is true, the 6 in the purple Cell gets killed; but if the 4 Hero is true, then look at the yellow Cell — the 6 in the purple Cell will still be killed. So we have agreement, and we can in fact kill the purple Cell's 6.

The other argument:

Troublemaker: what if the 6 in the purple Cell were true? It would kill the 6 Hero in the blue Cell on the left. And in fact, because of the yellow Cell, it would also kill the 4 Hero in the blue Cell on the right. But you can't kill all the Heroes, since that would cause the BUG+2 to degenerate into a BUG. So we eliminate the Troublemaker — we kill the 6 in the purple Cell.

Killing the 6 in the purple Cell (and crowning the 3) cracks that Sudoku.

One more BUG+2:

BUG+2: verification.

Consider the {3856} blue Cell: in that Cell's Block, in its Row, and in its Column, the 5 is a Triplet and the 6 is a Triplet (all the other candies are Twins). So it looks like the 5 and 6 are Heroes.

But we have to check the {265} Cell: this time, it's the 5 that's a Triplet in all three of its Houses (the other candies being Twins).

So each of the two blue Cells contains candies that qualify as Heroes, which confirms that this pattern is a genuine BUG+2.

Notice that the 2 in BUG+2 refers to the number of non-Pair Cells present. It does not refer to the number of Heroes (there are 3 Heroes here).

As usual, we can exploit this BUG+2:

Same BUG+2 as above: exploitation.

There are a total of three Heroes in the two blue Cells; at least one of these Heroes must be true.

Troublemaker: there are a lot of 5 and 6 candies on the Grid that could conceivably be Troublemakers, but most of them don't work. If you look long enough, you'll find the 6 in the purple Cell: if that 6 were true then, with the help of the yellow Cell, all three Heroes would get killed. Since that can't happen, the 6 in the purple Cell must be false, and we can kill it.

As usual, that cracks the Sudoku. (The reason for that, by the way, is the overwhelming predominance of Pair Cells on the Grid: crack one Pair, and a cascade of moves leading to the Solution is no surprise.)

If you're not used to doing BUG+2s, then that example looks easier than it really is. Finding a successful Troublemaker is not trivial; but the more you do it, the better you'll get at it.

BUG+3

The pattern below contains three Cells that are not Pairs. But of course we have to check whether this pattern amounts to a BUG with legitimate Heroes in the blue Cells — if so, it will be a BUG+3.

It's harder to do that check here, since some of the blue Cells see each other.

BUG+3: verification.

If we just check each blue Cell's Row, we will avoid the problem of trying to check a blue Cell that has another blue Cell in the House we're looking at.

In the {153} Cell's Row, the 3 is the only Triplet, so it looks like the 3 is a Hero.

In the {123} Cell's Row, again the 3 is the only Triplet, so it looks like the 3 is a Hero in that Cell too.

In the {457} Cell's Row, the 7 is the only Triplet, so it looks like the 7 is a Hero in that Cell.

Now, ignoring all three presumed Heroes, we have to check that every Row, Column, and Block containing a blue Cell contains only Twin candies (like a BUG would). This check succeeds, so we have a BUG+3.

Doing that check is a lot of work, but you have to do it — if it fails, then you don't have a BUG+n, you have no Heroes, and you can't apply the BUG+n Tactic. But after you've done it a dozen times, you'll find it much easier to carry out.

We can exploit this BUG+3, but it's getting harder: now we have to find a Troublemaker that would wipe out all three Heroes:

Same BUG+3 as above: exploitation.

There are three Heroes; at least one of these Heroes must be true.

There are a lot of 3 and 7 candies on the Grid that could conceivably be Troublemakers, but many of them don't work, so you have to do a lot of looking.

But the 3 in the purple Cell directly kills both 3 Heroes, and with the help of the yellow Cells, it also kills the 7 Hero. Since we can't have all three Heroes wiped out, the 3 in the purple Cell must be false, and we can kill it.

There are other Troublemakers that work: the 7 in R5C6, or the 3 in R1C9. Any one of them will do.

Killing that 3 in the purple Cell (and crowning the 5) cracks the Sudoku.

One more BUG+3. The initial check for this one is even harder:

BUG+3: verification.

The blue Cells all see each other, so we have to be smart about checking for Heroes.

In the {563} Cell's Column, only the 3 is a Triplet, so it looks like the 3 is a Hero.

In the {283} Cell's Row, again only the 3 is a Triplet, so it looks like that 3 is a Hero.

Now, ignoring the two presumed "3" Heroes, look at the {367} Cell's Block: only the 7 is a Triplet, so it looks like that 7 is a Hero.

Finally, ignoring all three presumed Heroes, we have to check that every Row, Column, and Block containing a blue Cell contains only Twin candies (like a BUG would). This check succeeds, so we have a BUG+3.

I suppose that checking process looks pretty awful, but after awhile you get used to it, and it goes fast.

Finding a Troublemaker that would kill all three Heroes is not so easy for this one:

Same BUG+3 as above: exploitation via Troublemaker.

There are three Heroes; at least one of these Heroes must be true.

Troublemaker: after a lot of looking, you will find the 7 in the purple Cell. It directly kills the 7 Hero. And if you follow the Chain ABCDE with Cell A = 7, you will get Cell E = 3, which kills the two 3 Heroes. Since we can't lose all three Heroes, the 7 in the purple Cell must be false, and we can kill it.

That cracks the Sudoku.

As Chains go, that one was neither short nor long. You really can learn to find Chains like that if you keep working at it.

And just in case you prefer the Consensus approach, here's the same BUG+3 done that way:

Same BUG+3 as above: exploitation via Consensus.

There are three Heroes; at least one of these Heroes must be true.

Consensus: if the 7 Hero is true, then the 7 in the purple Cell gets killed; but if either of the 3 Heroes is true, then look at the Chain AABCDE: with either Cell A = 3, we get Cell E = 7, which also kills the 7 in the purple Cell. So we have agreement, and we can in fact kill the 7 in the purple Cell.

Hard work, but that cracks the Sudoku.

BUG+4 and Higher

Well, in principle you could apply this Tactic to a big BUG+n, with n greater than 3. But there are disadvantages for large n:

The Hero verification process is less likely to result in a confirmation that you really have a BUG+n.
Finding a Troublemaker that will succeed in killing all the Heroes in all n non-Pair Cells can get to be impossible.

This is not the end of the world. Consult the page on Chains for a Tactic that will let you reach a Solution.

Not a BUG+2

Well, so what happens when the Hero verification process fails, and you don't have a BUG+n? Here's a Grid like that:

Not a BUG+2: verification.

Look at the {147} Cell: its Row says that the 7 is a Triplet, but its Block says that the 1 is a Triplet. Failure. No Hero. No underlying BUG pattern. No BUG+2 here.

We don't have to look at the {123} Cell, but it has the same problem: its Row says that the 1 is a Triplet, but its Block says that the 7 is a Triplet, and nominally extra 7 doesn't even lie in the gray Cell!

This is not a BUG+2.

Recommendation? Forget the two gray Cells — they're of no particular interest. Pick a Pair Cell; choose one of its candies and try to construct a Wrong Chain that will show that candie is wrong because it would lead to a Conflict.

The idea is not too hard, so we'll give an example. But you may want to read the page on Chains for a detailed discussion.

This is what it looks like:

Same not-a-BUG+2 as above: Solution via Wrong Chain.

We are ignoring the gray Cells.

There's no way to know which Pair Cell is a good one to try, or which of the two candies in the Cell is actually the wrong one. There are many Pair Cells on the Grid which will work, but you have to do a lot of trial and error to find one.

The purple Cell has a fairly simple Wrong Chain for its 3 candie.

Follow the Chain ABCDE with Cell A = 3, and you will get Cell E = 3. That's impossible, because then you would have two 3s in Row 5.

So the 3 in the purple Cell must be false, and we can kill it (and crown the 7).

That result generates a cascade of moves that cracks the Sudoku.

You shouldn't get the idea that a Wrong Chain necessarily cracks a Sudoku: all it does is eliminate one candie. But if almost all the empty Cells in the Grid are Pair Cells, then cracking one Pair Cell often leads to a Solution.

In general when you have a Grid whose empty Cells are mostly Pair Cells but you don't have a BUG+1 or a tractable BUG+2 or BUG+3, then a Wrong Chain (or a Gotcha Chain) is the way to go. See Chains.

There is another possibility that frequently arises specifically for a not-a-BUG+2. I hesitate to mention it because the logic is so tricky, but this is it:

Same not-a-BUG+2 as above: tricky exploitation.

Same Grid as above.

The 1 in the yellow Cell has a peculiar set of properties: if we take that 1 to be true then the following things happen —

The 1 is crowned and the 7 disappears (in the yellow Cell).
The 1s in the gray Cells disappear, and the gray Cells become Pair Cells.
No other kills occur.
The resulting pattern is a BUG (all candies are Twins).

Since a Sudoku cannot contain a BUG, the 1 in the yellow Cell must be false, and we can kill it. That will crack this Sudoku.

The conditions listed above are very restrictive. I've only seen this happen for a not-a-BUG+2, but I have seen it arise several times. You shouldn't use this gimmick unless the logic is crystal clear to you. In general a Wrong Chain (as shown previously) is less tricky.

This page was last updated on 2010 November 28.

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