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# What is the next step of this Kenken?

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# What is the next step of this Kenken?

January 30, 2010 6:07 PM Subscribe

What is the next step of this Kenken?

Players are supposed to be able to figure out the next step of every Kenken by logic. I can't figure this one out.

After being stumped on it for a couple of days, I started pencilling in scenarios very lightly and finally stumbled onto the solution after the fourth or fifth attempt.

Then I went back to it as a blank puzzle. Still can't figure it out. Obviously it's a 1,3 or 3,1 in the rightmost column at the bottom, with a 2,4 or 4,2 above it, but I can't figure out what order they are in--or anything else for that matter.

I think there is some sort of principle or tactic involved in solving these that I haven't yet learned. This is the 68th puzzle out of 500, so imagine the author thought it was easy.

Players are supposed to be able to figure out the next step of every Kenken by logic. I can't figure this one out.

After being stumped on it for a couple of days, I started pencilling in scenarios very lightly and finally stumbled onto the solution after the fourth or fifth attempt.

Then I went back to it as a blank puzzle. Still can't figure it out. Obviously it's a 1,3 or 3,1 in the rightmost column at the bottom, with a 2,4 or 4,2 above it, but I can't figure out what order they are in--or anything else for that matter.

I think there is some sort of principle or tactic involved in solving these that I haven't yet learned. This is the 68th puzzle out of 500, so imagine the author thought it was easy.

I was wondering why gleuschk said there must be a 2 in the 9+ region.

I think if you follow the 'and so on.', it doesn't work.

The 9 can be made by 4-3-2 or by 4-4-1.

I used to pencil in values on the grid.

posted by MtDewd at 6:45 PM on January 30, 2010

I think if you follow the 'and so on.', it doesn't work.

The 9 can be made by 4-3-2 or by 4-4-1.

I used to pencil in values on the grid.

posted by MtDewd at 6:45 PM on January 30, 2010

I wonder if the 1- is a mistake. With

4132

2314

1243

3421

it solves everything but 1- that and if that is supposed to be 2-1, 3-2 or 4-3, then it conflicts with 6x and 9+, because they both need 2's and 3's in at least one of the first two columns.

posted by stavrogin at 6:46 PM on January 30, 2010

4132

2314

1243

3421

it solves everything but 1- that and if that is supposed to be 2-1, 3-2 or 4-3, then it conflicts with 6x and 9+, because they both need 2's and 3's in at least one of the first two columns.

posted by stavrogin at 6:46 PM on January 30, 2010

My understanding of the rules of KenKen was that the regions can't contain duplicates. So 4,4,1 would be out. But checking on kenken.com (surely the world's experts) I see that "A number can be repeated within a cage [region] as long as it is not in the same row or column." So you're right, unless this particular incarnation of kenken uses a different rule. Sorry about that.

posted by gleuschk at 6:49 PM on January 30, 2010

posted by gleuschk at 6:49 PM on January 30, 2010

Hmm. I don't understand why there can't be a 4 in the top row of the 9+ region and a 4,1 in the bottom two squares. Perhaps gleuschk is seeing something I'm not.

posted by Mr. Justice at 6:50 PM on January 30, 2010

posted by Mr. Justice at 6:50 PM on January 30, 2010

There can, Mr. Justice.. I was able to solve it with 4 4 1 in the 9+ region. The rules just prohibit duplicates within rows and columns, not regions.

posted by alligatorman at 6:51 PM on January 30, 2010

posted by alligatorman at 6:51 PM on January 30, 2010

You already figured out the 2, so the other two in that region must be 1 & 3.

So the left two on the bottom row must either be 1 & 4 or 3 & 4.

So the top square of the 9+ must either be a 2 or a 4.

The other 6x must be 1, 2 & 3 so I'd pencil in those values.

My first try was

3214

1342

2431

4123

It looks like that works too.

You;re not supposed to have multiple answers.

posted by MtDewd at 6:54 PM on January 30, 2010

So the left two on the bottom row must either be 1 & 4 or 3 & 4.

So the top square of the 9+ must either be a 2 or a 4.

The other 6x must be 1, 2 & 3 so I'd pencil in those values.

My first try was

3214

1342

2431

4123

It looks like that works too.

You;re not supposed to have multiple answers.

posted by MtDewd at 6:54 PM on January 30, 2010

stavrogin's doesn't work -- there are two 1s in the bottom row. Instead:

3214

1342

2431

4123

posted by gleuschk at 6:55 PM on January 30, 2010

3214

1342

2431

4123

posted by gleuschk at 6:55 PM on January 30, 2010

I didn't even notice that I had multiple numbers in the same row. oops.

posted by stavrogin at 6:56 PM on January 30, 2010

posted by stavrogin at 6:56 PM on January 30, 2010

But he wasn't asking us to solve it.

Now we ruined it.

Sorry.

posted by MtDewd at 6:58 PM on January 30, 2010

Now we ruined it.

Sorry.

posted by MtDewd at 6:58 PM on January 30, 2010

I think a good answer does more than provide a 16-digit matrix. Instead, it should show how you deduce that certain numbers definitely go into certain boxes.

posted by Mr. Justice at 6:58 PM on January 30, 2010

posted by Mr. Justice at 6:58 PM on January 30, 2010

There doesn't have to be a 2 in that square (e.g. 4, 1, 4 works). This one is pretty tough, but here it is:

- The bottom row of 9+ is either (1,4) or (3,4). It can't be (1,3). That means the other square of that region is either a 4 or a 2.

- The 6x region above that is (1,2,3).

- The 1- region above that must have either a 2 or 3.

This is where it gets complicated, but the short of it is that neither 3,4 or 4,3 will work in the bottom row for the 9+ region (if you work it out). You end up running out of 2's or 3's, depending on which order you put the 3 & 4 in. Basically, the 2's and 3's for the first 2 columns are already accounted for to the extent that they BOTH can't be in the 9+ region.

posted by Khalad at 7:09 PM on January 30, 2010

- The bottom row of 9+ is either (1,4) or (3,4). It can't be (1,3). That means the other square of that region is either a 4 or a 2.

- The 6x region above that is (1,2,3).

- The 1- region above that must have either a 2 or 3.

This is where it gets complicated, but the short of it is that neither 3,4 or 4,3 will work in the bottom row for the 9+ region (if you work it out). You end up running out of 2's or 3's, depending on which order you put the 3 & 4 in. Basically, the 2's and 3's for the first 2 columns are already accounted for to the extent that they BOTH can't be in the 9+ region.

posted by Khalad at 7:09 PM on January 30, 2010

Starting with all possible numbers for all possible boxes:

1234 1234 431 42

321 321 431 42

321 24 431 13

431 431 2 13

A little checking will show you that neither of the two top left boxes can be 4 (if one is four, then the other is 3, and so on:

--12

--34

3241

--23

Leaving the bottom left two to be 1 and 4, but 1+4+2!=9)

Eliminate 4 as a choice for the top left two, then one of them MUST be 2. That means the top right must be 4, and so on. I didn't have any trouble from there.

I got (spoiler):

3214

1342

2431

4123

posted by anaelith at 7:09 PM on January 30, 2010

1234 1234 431 42

321 321 431 42

321 24 431 13

431 431 2 13

A little checking will show you that neither of the two top left boxes can be 4 (if one is four, then the other is 3, and so on:

--12

--34

3241

--23

Leaving the bottom left two to be 1 and 4, but 1+4+2!=9)

Eliminate 4 as a choice for the top left two, then one of them MUST be 2. That means the top right must be 4, and so on. I didn't have any trouble from there.

I got (spoiler):

3214

1342

2431

4123

posted by anaelith at 7:09 PM on January 30, 2010

I feel that the exact line between "pencilling scenarios" and "logic" is pretty fuzzy. I know the OP wanted to avoid trial and error, but ultimately the logic in these puzzles is just trial and error on a small scale.

posted by mrgoldenbrown at 7:29 PM on January 30, 2010

posted by mrgoldenbrown at 7:29 PM on January 30, 2010

My personal line is where you can't say that you eliminated some option at each step and only filled something in if it was the only option. Even if you had to do penciling over the whole puzzle in order to eliminate something. Basically, by the end of the puzzle I want to be sure that it is, in fact, the only answer, based only on the work that I did to solve it.

posted by anaelith at 7:36 PM on January 30, 2010

posted by anaelith at 7:36 PM on January 30, 2010

Figured this out lying in bed.

Here's my idea:

In a 4x4, 9+ can only be (4,3,2) or (4,4,1) and 6x in 3 can only be (1,2,3)

The left two columns contain the following numbers:(4,4,3,3,2,2,1,1)

If you put the (1,2,3) above (4,3,2) the top is left with (1,4)

If you put the (1,2,3) above (4,4,1) the top is left with (2,3)

Since it's 1-, only (2,3) works, and the top row is (2,3)14

That gives you

--14

---2

-4--

412-

which I think is easy enough to finish.

But this is pretty tough for a 4x4.

posted by MtDewd at 3:13 AM on January 31, 2010

Here's my idea:

In a 4x4, 9+ can only be (4,3,2) or (4,4,1) and 6x in 3 can only be (1,2,3)

The left two columns contain the following numbers:(4,4,3,3,2,2,1,1)

If you put the (1,2,3) above (4,3,2) the top is left with (1,4)

If you put the (1,2,3) above (4,4,1) the top is left with (2,3)

Since it's 1-, only (2,3) works, and the top row is (2,3)14

That gives you

--14

---2

-4--

412-

which I think is easy enough to finish.

But this is pretty tough for a 4x4.

posted by MtDewd at 3:13 AM on January 31, 2010

I don't think I'm repeating anyone else's strategy here, but this is how I got it, and to refer to the boxes I'll call them by letters, arranged left-right, top-bottom like so:

ABCD

EFGH

IJKL

MNOP (so the starting 2 is in box O)

Others have said that the 9+ (JMN) could be J4-M4-N1 or many combinations of 4-3-2. M can only be 4 or 3. Then Look at eliminating possibilities in AB (the 1-): If A4-B3, then M3. So in some combination, E & I are left to be 1 & 2, so F3...which contradicts B3.

Start over with A3-B4: then M4, and B4 precludes the 4+4+1=9, so J & N are some combination of 3 & 2. That leaves F1, so E & I must be 2 & 3 to complete 1*2*3=6...which contradicts A3!

Therefore: neither A nor B can be 4. Since the 6X cannot have a 4, both 4's must be in the 9+, so we can, with complete certainty, fill in M4-J4-N1. After that, the rest should work out with (relative) ease.

Not an easy 4x4! BTW, NYTimes.com and Kenken.com have several free puzzles every day:

http://www.nytimes.com/ref/crosswords/kenken.html

http://www.kenken.com/playnow.html

-----

SPOILERS FOR THE REST: If anyone is stuck with the rest, I might as well finish:

P3 since it's the only thing left on the bottom, and L1 to complete that 6X. The third row needs 2 & 3, but K can't be 2 based on O2, so K3, and then I2. To complete the EFI 6X we need 1 & 3; that makes G 2 or 4, but O2 means it must be G4, with C1 following to complete the third column. H2 & D4 follow. A & B must be 3 & 2, with I2 forcing A3 then B2. E1 then F3 follow and we're done!

posted by bah213 at 8:37 AM on January 31, 2010

ABCD

EFGH

IJKL

MNOP (so the starting 2 is in box O)

Others have said that the 9+ (JMN) could be J4-M4-N1 or many combinations of 4-3-2. M can only be 4 or 3. Then Look at eliminating possibilities in AB (the 1-): If A4-B3, then M3. So in some combination, E & I are left to be 1 & 2, so F3...which contradicts B3.

Start over with A3-B4: then M4, and B4 precludes the 4+4+1=9, so J & N are some combination of 3 & 2. That leaves F1, so E & I must be 2 & 3 to complete 1*2*3=6...which contradicts A3!

Therefore: neither A nor B can be 4. Since the 6X cannot have a 4, both 4's must be in the 9+, so we can, with complete certainty, fill in M4-J4-N1. After that, the rest should work out with (relative) ease.

Not an easy 4x4! BTW, NYTimes.com and Kenken.com have several free puzzles every day:

http://www.nytimes.com/ref/crosswords/kenken.html

http://www.kenken.com/playnow.html

-----

SPOILERS FOR THE REST: If anyone is stuck with the rest, I might as well finish:

P3 since it's the only thing left on the bottom, and L1 to complete that 6X. The third row needs 2 & 3, but K can't be 2 based on O2, so K3, and then I2. To complete the EFI 6X we need 1 & 3; that makes G 2 or 4, but O2 means it must be G4, with C1 following to complete the third column. H2 & D4 follow. A & B must be 3 & 2, with I2 forcing A3 then B2. E1 then F3 follow and we're done!

posted by bah213 at 8:37 AM on January 31, 2010

Thanks, everyone. After reading these over and looking at the puzzle, I see how each step can be deduced logically.

Penciling in pairs of options might be the way for me to go, as I seem to have trouble working it all out in my mind if there are more than three pairs of options.

posted by johnofjack at 3:16 AM on February 1, 2010

Penciling in pairs of options might be the way for me to go, as I seem to have trouble working it all out in my mind if there are more than three pairs of options.

posted by johnofjack at 3:16 AM on February 1, 2010

This is simpler than most of you are making this seem. The 6x cage toward the northwest can only be 1-2-3 in some order. Since the sum of two rows in a 4x4 must be twenty, the sum of the numbers in the 1- cage must be 5. So 1- must be 2-3 in any order. This leaves 1-4 in any order for the remain boxes in the top row. In either case the second box in the 2/ cage must be 2. This leaves only one box left for the 2 in the 6x cage and fixes the order of the 2-3 in the 1- cage and allows you to complete the northwest 6x cage.

posted by DanSachs at 8:41 AM on February 1, 2010 [2 favorites]

posted by DanSachs at 8:41 AM on February 1, 2010 [2 favorites]

When I said "Since the sum of two rows ..." make that "Since the sum of two columns ...".

posted by DanSachs at 8:42 AM on February 1, 2010

posted by DanSachs at 8:42 AM on February 1, 2010

DanSachs method is even easier than mine.

It's helpful to remember that all KenKen rows & columns add up to a constant.

4x4=10, 5x5=15, 6x6=21...

Sometimes that can be used to directly get a number.

Sometimes, like here, you use 2 rows or columns together.

(I don't think I've ever done 3 together, but you could do it)

posted by MtDewd at 2:13 PM on February 1, 2010

It's helpful to remember that all KenKen rows & columns add up to a constant.

4x4=10, 5x5=15, 6x6=21...

Sometimes that can be used to directly get a number.

Sometimes, like here, you use 2 rows or columns together.

(I don't think I've ever done 3 together, but you could do it)

posted by MtDewd at 2:13 PM on February 1, 2010

This thread is closed to new comments.

posted by gleuschk at 6:32 PM on January 30, 2010