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# Tutorial (Expert)

## Expert-Level Tips

So you’re breezing through the easier puzzles, huh? Ready to step up to the big leagues? Before you do, there are some more advanced solving techniques that will be very helpful to have in your arsenal. Here, we present several of them as we walk through solving this 6 x 6 grid, step-by-step.
This tutorial, by David Levy, KENKEN’s technology expert, appears in KENKEN: Books 1 - 4: The New Brain-Training Puzzle Phenomenon, which HarperCollins in Britain published. Step 1

We first fill in the given numbers:  2 in J and 3 in V.

Step 2

Next we look to see if there are any squares for which only one number is possible. Here we can be sure that FF is 5. Can you see why?

The reasoning goes like this. The six numbers in the bottom row add up to 21 (1+2+3+4+5+6). But GG, HH and JJ total 8, so KK+LL+MM must total 13 (i.e. 21-8). Now we know that FF+KK+LL+MM = 18, and by subtraction we can see that FF must be 5 (18-13).

Step 3

Now we look for any cages in which only one combination of numbers will work.

N+U+AA = 15, and the only way to use three different numbers in this puzzle to total 15 is 4+5+6. No other combination of three different numbers from 1 to 6 will do. All three squares N, U and AA therefore have the three “candidates” 4, 5 and 6. But wait a minute! FF is 5, so AA can not be 5, it can only be 4 or 6. The candidates for the three squares in this cage are therefore:

N                     4, 5, 6

U                     4, 5, 6

AA                   4, 6

There is one more cage that can have only one possible combination of numbers: the L S cage. The only two numbers that are possible in this puzzle, that multiply together to give 20, are 4 and 5. So both of these numbers are candidates in squares L and S.

At this stage we can take stock of what we have learned so far. Step 4

What can we deduce immediately, based on the information in this diagram? There are a few things.

The three numbers 4, 5 and 6 are all in the N U AA cage, so the remaining numbers in that column, those in A, G and GG, are 1, 2 and 3.These three numbers are therefore candidates for all three squares: A, G and GG.

Step 5

Now let us look a little more deeply into cage GG HH JJ. There are only two possible combinations of three different numbers that add to 8: 1, 2 and 5; or 1, 3 and 4. If GG HH JJ is 1, 2, 5, then KK, LL, MM is 3, 4, 6; and if GG HH JJ is 1, 3, 4, then KK LL MM is 2, 5, 6. We should make a note of these two possibilities.

Step 6

Similarly, if we look closely at cage F M T, we can see that there are only two possible groups of three numbers that can be used in this cage. Why is that? After all, we can make a total of 9 by adding:

1 2 6

1 3 5

2 3 4

But the 1, 3, 5 combination is impossible here, because FF is 5, so only the two combinations 1, 2, 6 and 2, 3, 4 are possible. It is often very helpful to know that only two combinations of numbers are possible for a particular cage, because it can be relatively quick to prove that one of them does not work and therefore the other one must be correct.

Note that, because this cage must be 1, 2, 6 or 2, 3, 4, we know that 2 must be in this cage.

Step 7

What else can we deduce from the information in the previous diagram? Do you remember how we used the total of 21 for the bottom row to deduce (in step 2) that FF is 5? We can use the total of 21 again, in the rightmost column. This is how. We are told that F+M+T = 9 and we know that FF is 5, so Z+MM must total 7 (the difference between 21 and 14 (from 9+5). But if Z+MM = 7 what are the possibilities?

1+6 = 7

2+5 = 7

3+4 = 7

But FF is 5 so the 2, 5 combination does not work for Z and MM. We can therefore be sure that Z and MM use either 1, 6 or 3, 4.

Step 8

Can we narrow down these possibilities even further? Yes we can! Look at the cage Y Z DD EE, which has four numbers adding up to 9. We have shown in step 7 that each of Z and MM is 1, 3, 4 or 6. But if Z is 6 then what can the other squares in this cage be? They must all be 1s, to make a total of 9 (6+1+1+1) but that would mean two 1s in the same row (Y and Z) and two 1s in the same column (Y and EE), both of which are impossible. Therefore Z cannot be 6, and if Z is not 6 then MM cannot be 1 because Z+MM=7.

Also we know that Z cannot be 3 because V=3, so Z must be 1 or 4, and to make a total of 7 with Z we can now see that MM must be 6 or 3 (since 6+1=7 and 3+4=7).

It is time to add all this new information into our diagram.

GG HH  JJ = 1, 2, 5 or 1, 3, 4             F M T = 1,  2, 6 or 2, 3, 4. Step 9

We now know something useful about the Y Z DD EE cage, so let us try to take advantage of it. We know that Z is 1 or 4, so let us consider the implications of each of those possibilities.

If Z is 1 then Y+DD+EE = 8. What could Y be?

We should start considering possible values of Y from the largest to the smallest because the largest numbers impose the most restrictions on the other numbers in the cage. So what about Y = 6 ? In that case Y+Z = 7, in which case the only way to make a total of 9 would be for DD and EE both to be 1, which is not allowed. So Y cannot be 6.

Y cannot be 5 or 4 because both of these are in the L S cage. Y cannot be 3 because V=3.

If Y is 2 then Y+Z = 3, in which case DD+EE = 6, which can only be made with 1+5 or 2+4, but FF is 5 so the idea of 1+5 will not work for DD+EE, so it must be 2+4. If one of DD and EE is 4 it cannot be EE because 4 is in the L S cage, so it would have to be DD which is 4 and EE which is 2, but we started this little part of the investigation with Y=2, so it is not possible for EE also to be 2, and of course Y cannot be 1 because we are considering what happens if Z=1.

We have therefore proved that no values of Y work when Z is 1, so Z cannot be 1 and it must therefore be 4. This is a big step forward!

Step 10

Since Z=4 we know that MM=3 (because Z+MM=7).

And we also know that F M T must be 1, 2, 6 because the other possibility (2, 3, 4) uses numbers which we now know to be Z and MM. Also, M cannot be 2 because of the 2 in J, so M must be 1 or 6.

Furthermore, since MM=3, the cage GG HH JJ cannot be 1, 3, 4, so it must be 1, 2, 5. Therefore GG must be 1 or 2 (since 5 is already in this column in the N U AA cage); HH is 1, 2 or 5; and JJ is 1 or 5 (since 2 is in this column at J).

And with 1, 2, 3 and 5 already accounted for in the bottom row, we know that KK and LL must use 4 and 6. But LL cannot be 4 because 4 is used in the L S cage, so KK must be 4 and LL is 6.

With LL known to be 6, and both 4 and 5 in the L S cage, it is clear that the other three squares in this column: E, Y and EE, must use 1, 2 and 3. But Y cannot be 3 because V=3, so Y is 1 or 2; E is 1, 2 or 3; and EE is also 1, 2 or 3.

Finally, in this step, since Z is 4, U must be 5 or 6.

So by discovering that Z is 4 we have also revealed quite a few additional pieces of information. Step 11

Let us now revisit our helpful friend, the Y Z DD EE cage. Since Z is 4, Y+DD+EE = 5 so as to create a total of 9 for that cage. The only ways we can do this are 1+1+3 or 1+2+2, which gives us two possibilities:

Y=2     DD=2  EE=1

or                     Y=1     DD=1  EE=3

So either           DD=2  and EE=1

or                     DD=1  and EE=3

Let us examine each of these possibilities in turn and see where they lead us.

If DD=2 and EE=1, then the BB CC cage must use two of the numbers 3, 4, 6, and one is twice the other (from the 2÷ total), so BB CC must use 3 and 6, and therefore AA must be 4. BB cannot be 3 because V=3, so in this case BB is 6 and CC is 3.

If DD=1 and EE=3, the BB CC cage muse use two of the numbers 2, 4, 6, and again one is twice the other (because of the 2÷ total), so BB CC must use 2 and 4, and AA must be 6. CC cannot be 2 because J=2, so in this case CC is 4 and BB is 2.

We now know that either: BB is 2 and CC is 4 (and therefore AA is 6), or BB is 6 and CC is 3 (and therefore AA is 4).

So BB is 2 or 6, and CC is 3 or 4.

And DD is 1 or 2, while EE is 1 or 3. Step 12

It does not look easy to make any more progress at the moment in the lower half of the grid, so let us turn our attention to the upper half, and in particular to the A B G H square where we already have a complete list of candidates for A and G.

We know that A is 1, 2 or 3, and G is 1, 2 or 3. But we know more than this – one of them must be 3 because GG can only be 1 or 2. So A+G is either 3+1 or 3+2, in other words A+G is 4 or 5.

Since the total of the A B G H cage is 13, B+H must therefore be 8 or 9.

For B+H to total 8, there might appear to be two possibilities: 6+2 and 5+3. But 3 is already used in this column, so only 6+2 will work if B+H = 8, which would mean that B = 2 and H = 6. (Note that H cannot be 2 because J is 2.) However, we already know from step 11 that BB is 2 or 6, so it is not possible for B and H to use 2 and 6.

Therefore B+H cannot total 8, which means that B+H must total 9, which in turn means that A and G must use 1 and 3 (so that A+B+G+H = 13). Therefore GG must be 2. Also, B and H must use either 4 and 5, or 6 and 3 (to make a total of 9), but the 6+3 combination is impossible because V is 3, and so B and H must use 4 and 5.

Since GG is 2, HH can only be 1 or 5. But 5 is already used in B H, so HH = 1 and therefore JJ = 5. Furthermore, since HH is 1, P can only be 2 or 6, and therefore Q can only be 1 or 3 in order to produce the total 6 when multiplying P x Q.

So we have discovered in this step that:

A =      1 or 3

G =      1 or 3

GG =   2

HH =   1

JJ =      5

P=        2 or 6

Q =      1 or 3

B and H use 4 and 5.

Now we add all this information from step 12 into a new diagram Step 13

Now let us see if we can discover anything useful about the C D E K cage. We know that exactly one number in this cage must be 5 in order to create a total, by multiplication, that ends in 80. E cannot be 5 because 5 is used in the L S cage. C cannot be 5 because 5 is already in this column. K cannot be 5 because 4 and 5 are used in H and L, so no other square in this row could possibly be 4 or 5. Having eliminated C, E and K as possible locations for 5, it is now clear that D must be 5, which means that B is 4, which in turn means that H is 5, which in turn means that L is 4, which means that S is 5, which means that N is 4 or 6. And since 4 and 6 are used in N and AA, we also know that U must be 5.

Having discovered all these new numbers, it’s time for another diagram! Step 14

We should now start to list the candidate for all those unsolved squares that do not yet have any candidates: C, K, R, W and X.

C cannot be 4 or 5 because both numbers are already in the same row, and C cannot be 2 because J=2. So C is 1, 3 or 6.

K cannot be 2, 4 or 5 because of clashes in the same row, so K is 1, 3 or 6.

R cannot be 4 or 5 because both numbers are already in the same column, so R is 1, 2, 3 or 6.

W cannot be 3, 4 or 5 (all three are used in the same row), nor can W be 2 (which is used in the same column). So W is 1 or 6.

X cannot be 3, 4 or 5, all of which are already used in the same row, so X is 1, 2 or 6.

Once again we have quite a lot of new information, so let us take  a look at the situation after the dust has settled. Step 15

Now look at the column with C, Q, W and CC. The three numbers 1, 3 and 6 are the only three used in C, Q and W, so those three numbers must all be used in those three squares. This means that CC must be 4, and therefore AA is 6. This illustrates a very common and important strategy in KenKen – if two squares in the same row or column use two particular numbers, and if neither of those numbers appears anywhere else in the same row or column, we can be certain that those two squares must be taken up by those two numbers. Similarly for three numbers in three squares in the same row or column, four numbers in four squares, and so on.

Since AA is 6, N must be 4. And since CC is 4,  BB must be 2. And since BB is 2, P must be 6, so Q is 1 (to make a total of 6 from 1 x 6), therefore T is 2 (since 1 is already used in Q, and 6 is in P,) and so F can only be 1 or 6. Furthermore, since Q is 1, W must be 6, which means that C must be 3 (since both 1 and 6 are now in the same column as C).

Since C is 3, E can only be 1 or 2.

As usual after a flurry of discoveries we should have a new diagram. Step 16

It is now easy to see that we can immediately eliminate some more candidates, with decisive effect

DD cannot be 2,  because 2 is already used in the same row. So DD must be 1.

EE must be 3, which is the only unused number in this row.

A must be 1 because there is already a 3 in C, and since A is 1, F is 6 and E is 2.

Also, since F is 6, M must be 1.

And since E is 2, Y must be 1, and therefore EE can only be 3.

And since A is 1, G must be 3, so K can only be 1 or 6. But we have just discovered that in the same row M is 1, so K must therefore be 6.

R must be 3, which was hitherto the only unused number in this row.

X can only be 2 because we already have 1 and 6 in this row.

Finally, DD must be 1 because it is the only unused number in this row (and in this column). Puzzle solved!

Copyright (c) 2009 David N. L. Levy

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