## Multiples Killer

For fans of Killer Sudoku, Samurai Sudoku and other variants

### Multiples Killer

In this puzzle, all cage sums are multiples of a certain positive integer M. It is actually very easy to deduce the value of M.

Special properties:

Touchless: aka Anti-King (AK), each cell must have a value different to its diagonal neighbours. E.g. r4c5 must be different to all 4 of r35c46.

Windoku: each window (r234c234, r234c678, r678c234, r678c678) must have 9 different values. Note there are also 5 hidden groups.

X: each diagonal must have 9 different values.

Additionally, there are (at least) 2 extra special properties, but they are not required to give it a unique solution (and you don't really need them if you know the secret key to unlock this puzzle):

Triple click to see the 2 extra special properties I wrote:Disjoint Group (DG): each of the following 9 groups: r147c147, r147c258, r147c369, r258c147, r258c258, r258c369, r369c147, r369c258, r369c369 must have 9 different values.

Anti-Elephant (AE): Each cell must have a value different to any cell 2 squares away diagonally. E.g. r4c5 must be different to all 4 of r26c37.

Okay, here is a very important piece of information: this puzzle has a unique solution. Using this information wisely, you can solve this puzzle very easily. Without using this information, it will (probably) take a lot of T&E/forcing chains to work out the solution.

This is another educational puzzle that grasping a key concept will help immensely in the solving process. Let's see if any of you smart people can work that out.
udosuk

Posts: 2698
Joined: 17 July 2005

4M can't be >(9+8+7)
& M can't be <(1+2+3)

====>

tarek

tarek

Posts: 3759
Joined: 05 January 2006

### Re: Multiples Killer

Obviously, I wrote:It is actually very easy to deduce the value of M.

Now could you go ahead and tackle the actual puzzle?
udosuk

Posts: 2698
Joined: 17 July 2005

Here is the big hint on how to get started, which I have posted on another forum:

Triple click to read what I wrote:First, suppose the solution grid looks like this:

r1c1 r1c2 r1c3 | r1c4 r1c5 r1c6 | r1c7 r1c8 r1c9
r2c1 r2c2 r2c3 | r2c4 r2c5 r2c6 | r2c7 r2c8 r2c9
r3c1 r3c2 r3c3 | r3c4 r3c5 r3c6 | r3c7 r3c8 r3c9
---------------+----------------+---------------
r4c1 r4c2 r4c3 | r4c4 r4c5 r4c6 | r4c7 r4c8 r4c9
r5c1 r5c2 r5c3 | r5c4 r5c5 r5c6 | r5c7 r5c8 r5c9
r6c1 r6c2 r6c3 | r6c4 r6c5 r6c6 | r6c7 r6c8 r6c9
---------------+----------------+---------------
r7c1 r7c2 r7c3 | r7c4 r7c5 r7c6 | r7c7 r7c8 r7c9
r8c1 r8c2 r8c3 | r8c4 r8c5 r8c6 | r8c7 r8c8 r8c9
r9c1 r9c2 r9c3 | r9c4 r9c5 r9c6 | r9c7 r9c8 r9c9

Now, subtract each cell values from 10, and then rotate the whole grid 180 degrees:

10-r9c9 10-r9c8 10-r9c7 | 10-r9c6 10-r9c5 10-r9c4 | 10-r9c3 10-r9c2 10-r9c1
10-r8c9 10-r8c8 10-r8c7 | 10-r8c6 10-r8c5 10-r8c4 | 10-r8c3 10-r8c2 10-r8c1
10-r7c9 10-r7c8 10-r7c7 | 10-r7c6 10-r7c5 10-r7c4 | 10-r7c3 10-r7c2 10-r7c1
------------------------+-------------------------+------------------------
10-r6c9 10-r6c8 10-r6c7 | 10-r6c6 10-r6c5 10-r6c4 | 10-r6c3 10-r6c2 10-r6c1
10-r5c9 10-r5c8 10-r5c7 | 10-r5c6 10-r5c5 10-r5c4 | 10-r5c3 10-r5c2 10-r5c1
10-r4c9 10-r4c8 10-r4c7 | 10-r4c6 10-r4c5 10-r4c4 | 10-r4c3 10-r4c2 10-r4c1
------------------------+-------------------------+------------------------
10-r3c9 10-r3c8 10-r3c7 | 10-r3c6 10-r3c5 10-r3c4 | 10-r3c3 10-r3c2 10-r3c1
10-r2c9 10-r2c8 10-r2c7 | 10-r2c6 10-r2c5 10-r2c4 | 10-r2c3 10-r2c2 10-r2c1
10-r1c9 10-r1c8 10-r1c7 | 10-r1c6 10-r1c5 10-r1c4 | 10-r1c3 10-r1c2 10-r1c1

Now try to put the cage layouts of the original puzzle onto this grid. What do you notice about the cage sums?

Because I've specifically stated that the original puzzle has a unique solution, what does it tells you about the relationship between r[i]c[j] & r[10-i]c[10-j]? And what does it tells you about the cell value of r5c5?

Now I'm sure this can help you fill in the 2M & 3M cages easily.

Multiples Killers #2 & #3 to be followed shortly.
udosuk

Posts: 2698
Joined: 17 July 2005

Okay, here is Multiple Killer #2. Very challenging with the "secret key". Almost impossible without!

Special properties:

Disjoint Group (DG): each of the following 9 groups: r147c147, r147c258, r147c369, r258c147, r258c258, r258c369, r369c147, r369c258, r369c369 must have 9 different values.

Windoku: each window (r234c234, r234c678, r678c234, r678c678) must have 9 different values. Note there are also 5 hidden groups.

X: each diagonal must have 9 different values.
udosuk

Posts: 2698
Joined: 17 July 2005

I'll show my walkthroughs for the first 2 puzzles, before posting the 3rd, which is the series finale...

(In the walkthroughs all cage sums are shown explicitly, as they're pretty obvious.)

Triple click below to see the walkthrough for Multiples Killer 1 which I wrote:(Notation: SYM=symmetrical property, AK=touchless)

6/3 @ r1c5={123} (NT @ c5,n2)
6/3 @ r2c2={123} (NT @ d\,w1)
24/3 @ r6c6={789} (NT @ d\,w4)
24/3 @ r7c5={789} (NT @ c5,n8)

SYM: r5c5=5 (NE @ d\/)
=> r46c5={46} (NP @ n5)
=> \19={46} (NP @ r159c159)

Now {46} @ r5 locked @ 12/3+18/3
SYM: each of 12/3 & 18/3 @ r5 can't have both of {46}
=> One of these 2 cages has 4, the other has 6
=> 12/3 @ r5c1 can't be {156|246|345}
=> 12/3 @ r5c1={147} (NT @ r5,n4)
SYM: 18/3 @ r5c7={369} (NT @ r5,n6)
=> r5c46={28} (NP @ n5)

Similar line of reasoning:
12/3 @ r6c4={147} (NT @ d/,w3)
18/3 @ r2c8={369} (NT @ d/,w2)
/19={28} (NP @ r159c159)

9 @ c6,n5 locked @ r46c6
6 @ r5,n6,r159c678 locked @ r5c78
4 @ d/,n7,w3 locked @ /78
=> 4 @ c1,n1 locked @ r123c1
=> 4 @ c4,n2,w1 locked @ r23c4
=> r123c6={578} (NT @ c6,n2)
AK: r2c7 can't have {578}

r456c6=[329]
=> r456c4=[187]
AK: r37c5=[28] (2 NE @ r234c159, 8 NE @ r678c159)
AK: r5c37=[46], r4c7 can't be 2
=> HS @ r4: r4c8=2
=> r6c789={145} (NT @ r6,n6)
AK: r7c8 can't have {145}

r46c5=[46]
6/3 @ r2c2=[231], 18/3 @ r2c8=[693]
12/3 @ r6c4=[714], 24/3 @ r6c6=[978]
=> r7c8=3

AK: r123c4=[946], r6c123=[382], r789c6=[461]
AK: r7c4+r8c3=[59]
AK: r4c3=5

All naked singles from here.

Eeeasy, peasy? Just a brief entrée to get you warmed up on the symmetry tricks...

Then along comes the main course:

Triple click below to see the super tricky walkthrough for Multiples Killer 2 which I wrote:(Notation: SYM=symmetrical property)

6/3 @ r1c5={123} (NT @ c5,n2)
24/3 @ r7c5={789} (NT @ c5,n8)
SYM: r5c5=5 (NE @ d\/)
=> r46c5={46} (NP @ c5,n5)

Now {46} @ r5 locked @ 18/3+12/3
SYM: each of 18/3 & 12/3 @ r5 can't have both of {46}
=> One of these 2 cages has 4, the other has 6
=> 18/3 @ r5c1 can't be {459|468|567}
=> 18/3 @ r5c1={369} (NT @ r5,n4)
SYM: 18/3 @ r5c7={147} (NT @ r5,n6)
=> r5c46={28} (NP @ n5)

9/3 @ /2 from {12346789}={126|234}=[{26}1|{24}3]
=> /23={24|26} (2 @ d/,n3,w2 locked)
21/3 @ /6 from {12346789}={489|678}=[7{68}|9{48}]
=> /78={48|68} (8 @ d/,n7,w3 locked)
/2378={2468} (NQ @ d/)

15/3 @ \2 & \8 from {12346789}={168|249|267|348}
=> 15/3 @ \2=[{68}1|{48}3|{26}7|{24}9]
=> 15/3 @ \8=[1{68}|3{48}|7{26}|9{24}]
=> \23={24|26|48|68}, \78={24|26|48|68}
=> \2378={2468} (NQ @ d\)

r19c19={1379} (NQ @ r159c159)
=> r19c5=[28], r5c19=[64]
=> 2 NE @ s2, 8 NE @ s8, 6 NE @ s4, 4 NE @ s6
=> r23c5={13} (NP @ r234c159), r78c5={79} (NP @ r678c159)
=> r5c23={39} (NP @ r159c234), r5c78={17} (NP @ r159c678)

1 @ n3 locked @ r1c9+r2c7+r3c8 (pointing @ r4c6)
=> 9/3 @ /2=[{24}3] (NT @ d/,w2; 4 @ n3 locked)
=> 21/3 @ /6=[7{68}] (NT @ d/,w3; 6 @ n7 locked)
1 @ n3,w2 locked @ r2c7+r3c8
=> /19=[91]
3 @ n1,w1 locked @ r2c3+r3c2
=> \19=[73]

HS @ s3: r4c3=7
HS @ s7: r6c7=3
HS @ n3: r1c8=3
HS @ n7: r9c2=7

1 @ r1,n1 locked @ r1c23
9 @ r9,n9 locked @ r9c78
2 @ n1,w1 locked @ r23c23
8 @ n9,w4 locked @ r78c78

Critical Step #1:
1 @ r4,w1 locked @ r4c24
15/3 @ \2=[{24}9|{68}1]
But \234+r4c2 can't be [{68}18] (all 4 cells @ w1)
=> r4c24 can't be [81]
=> r4c2 can't be 8

Tricky Symmetry Step #1:
2 @ r4 locked @ r4c19
SYM: r4c19 can't be {28}
=> r4c19 can't have 8

HS @ n4: r6c1=8
SYM: r4c9=2
=> r234c1={459} (NT @ c1,r234c159)
=> r46c5=[64] (6 NE @ r234c159)

6 @ r2 locked @ r2c236 (pointing @ r3c4)
4 @ r8 locked @ r8c478 (pointing @ r7c6)
6 @ n1,w1 locked @ r23c23
4 @ n9,w4 locked @ r78c78

Tricky Symmetry Step #2:
6 @ c4 locked @ r19c4
SYM: r19c4 can't be {46}
=> r19c4 can't have 4
4 @ c6 locked @ r19c6
SYM: r19c6 can't be {46}
=> r19c6 can't have 6

6 @ r1,s1 locked @ r1c47
4 @ r9,s9 locked @ r9c36
=> 15/3 @ \2=[429|681|861] (\2 can't be 2)
=> 15/3 @ \6=[186|924|942] (\8 can't be 8)

Tricky Symmetry Step #3:
r28c2 from {468}
SYM: r28c2 can't be {46}
=> 8 @ c2 locked @ r28c2

Critical Step #2:
r1c2+r23c1 from {1459} must have 1|4 (with 1 @ r1c2 only)
=> r24c2 can't be [41]
=> \234+r4c2 can't be [4291]
=> r4c2 can't be 1

HS @ r4: r4c4=1
15/3 @ \2: \23={68} (NP @ d\,n1,w1)
HS @ s9: r6c3=1
HS @ n4: r6c2=2
HS @ r3: r3c7=2
=> r7c7=4, SYM: r7c3=8
HS @ n7: r9c3=4

All naked singles from here.

The last dish, Multiples Killer 3, will come later, and by no means will it be a cheesecake or any other dessert!
udosuk

Posts: 2698
Joined: 17 July 2005

Here is Multiples Killer 3, the last one in the series:

Special properties:

Touchless-2: each cell must have a value different to all cells one or two steps away diagonally. E.g. r4c5 must be different to all 8 of r26c37+r35c46. (In JSudoku you can implement this property by using AK (Anti-King) & AE (Anti-Elephant).)

Windoku: each window (r234c234, r234c678, r678c234, r678c678) must have 9 different values. Note there are also 5 hidden groups.

X: each diagonal must have 9 different values.

There are 2 more difficult versions of this:

1. You only use AE but not AK. But it probably takes quite a few forcing-chains.

2. You only use AK but not AE. I still haven't been able to solve that version.

Note there are some extra symmetry properties to explore about, if you work hard enough!
udosuk

Posts: 2698
Joined: 17 July 2005

Return to Sudoku variants