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PostPosted: Tue Apr 29, 2008 3:15 pm 
Grand Master
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Seeing that Richard created this section upon my request :salute:, I feel I owe him and the community here a puzzle at the very least. So here I will post one of my new creations here. It's also my favourite of the five ones in my latest series. And hopefully the great players here will not be baffled too much by the rules (Fers Non-Consecutiveness should not be too hard to figure out for players who have tried normal Non-Consecutive puzzles). :study:


Note the "Equality Killer" concept comes from Ms Miyuki Misawa's website in 2005. Also players who have visited know they publish an "Inequality Killer" every week.

This puzzle is not too hard once the players figure out the cage sum. :ugeek:

Just for completeness I posted 4 other puzzles in this series elsewhere, available from these links:



And I also posted this puzzle in the Sudocue forum, as a tribute to Ruud:




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PostPosted: Thu May 01, 2008 8:20 pm 

Joined: Mon Apr 21, 2008 8:12 pm
Posts: 90
Location: London, UK
Ok Matt

Not a complete walkthrough yet, but enough to give the required cage sum. Don't know how to do the triple click trick so it's in Tiny type.

Proposition 1 All 2-cell cages are {78} consecutive digits if 2-cell cage sum is 15.
Assume cage sum is 15. 5-cell cages are {12345}. TopRightPane=>r2c8&r4c678<>1,2,3,4,5 => r1c7=1,2,3,4 or 5=> r1c8=6,7,8 or 9.
r1c8+r2c7=15, only combo is r1c8,r2c7={69} ({87} blocked by Ferz)
r89c7={78} ({69} combo blocked by r2c7)
Top Right Pane {78} locked in r4c68
Cage sum 15 => r45c6={78}
r5c5 sees the whole cage r5c234&r6c45={12345} => r5c5=6|7|8|9.
On Row 5 Positions of 7,8 r5c9 blocked by pair r12c9, r5c8 blocked by pair r5c6+r9c7 (Right Stiles) and r5c5 by r45c6 => r5c16={78} => r56c1={78}
Proposition 1 is true

Proposition 2 All 2-cell cages cannot be consecutive digits A,B
Set r12c9=[AB] => r56c1=[BA] => r45c6=[BA] =>r89c7=[AB]
Consider Top Stiles r2c9&r3c5=[BA]
Examining r3c5&r4c6=[AB] consecutive digits blocked by Ferz.
Proposition 2 is true

1)Cage sum cannot be 17 blocked by Proposition 2.
2)Cage sum cannot be 15 blocked by Proposition 1 and 2.
3)Cage sum equals 16.

I have 81 brain cells left, I think.

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PostPosted: Mon May 05, 2008 4:04 am 
Grand Master
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Finally, I've written my own complete walkthrough for this, aided with a few diagrams. :ugeek:

Note I've used one proposition instead of two by Glyn, and then used some coloring tricks to determine the cage sum out of 2. Would have been nicer to do it without any proposition but don't think it's viable. :scratch:

Here is the complete walkthrough in hidden text:

Hidden Text:
Stage 1:

If all 2-cell cages have the same pair of numbers, then these numbers can't be consecutive.

Suppose all 2-cell cages are {AB}. Then all 5-cell cages can't have {AB}.
(Otherwise sum for the 5-cell cage >= 1+2+3+4+A|B = 10+A|B > A+B.)
Now r12c9=r45c6=r56c1=r89c7={AB}
NP r45c6={AB} @ c6,n5 => r45c5+r89c6 can't be from {AB}
NP r1c9+r5c1={AB} @ r159c159 => r9c5 can't be from {AB}
HP @ c5: r38c5={AB} => NP r8c57={AB} @ r8 => r8c4 can't be from {AB}
=> Either r7c4+r8c5 or r8c5+r9c4={AB} @ n8
=> FNC: A & B can't be consecutive


The common cage sum can't be 17, as 2-cell cages can't be all {89}.

Therefore the common cage sum must be 15 or 16:

5-cell cage = {12345|12346}
2-cell cage = {69|78|79}

Stage 2:

Let the 5-cell cages be {PQRST}, and the other 4 candidates be {WXYZ}.
Then all 2-cell cages must be from {WXYZ}.

Now colour all known cells with {PQRST} in orange, and those with {WXYZ} in purple.

r12c5+r123c6, r2c8+r3c789+r4c9, r5c234+r6c45, r567c7+r7c56, r789c1+r8c23 are all orange
r12c9, r45c6, r56c1, r89c1 are all purple
=> n2,n7,r234c678: r123c4+r3c5, r79c23, r2c7+r4c78 are all purple
=> c7: r1c7 is orange => n3: r1c8 is purple

Now n8 must have 4 purple cells, with at most one @ r789c4 (c4) & at most one @ r9c456 (r9).

=> r8c56 are purple => r9c4 is orange, both r78c4 & r9c56 are one orange, one purple
=> r159c234: r1c23 are one orange, one purple => r1: r1c1 is orange
=> r678c159: r678c9 are one orange, two purple => c9: r59c9 are both orange
=> r159c159: r59c5 are both purple => c5: r4c5 is orange => r5: r5c8 is purple
=> r9: r9c68 are both orange => c6: r6c6 is purple => n5: r4c4 is orange


Stage 3:

Let C be the common cage sum.

Observe that if W+X=C, then all of W+Y, W+Z, X+Y, X+Z must not be C.

Let r12c9={WX} (without loss of generality), so W+X=C.

Now colour all known cells with {WX} in pink, and those with {YZ} in blue.

Suppose a 2-cell cage contains a pink cell, i.e. one cell is W|X.
Then the other cell = C-W|C-X = X|W, i.e. must be pink.
Thus if a 2-cell cage contains a pink cell, the other cell must also be pink.

r12c9 are both pink => n3: r2c8+r3c7 are both blue
=> c7: at least one of r89c7 must be pink => r89c7 are both pink
=> c7: r4c7 is blue => r234c678: r4c68 are both pink => r5c6 is pink
=> c6: r68c6 are both blue => n5: r5c5 is blue => r159c678: r5c8 is blue
=> r5: r5c1 is pink => r6c1 is pink => r159c159: r9c5 is blue => c5: r38c5 are both pink

Now n9 must have 2 blue cells, but they can't be @ r78c8 (c8).

=> r78c9 are both blue => c9: r6c9 is orange => n6: r6c8 is pink
=> c8: r78c8 are both orange => r8: r8c4 is orange => n8: r7c4 is pink


Observe that @ n8, there are 2 diagonally adjacent pink cells & 2 such blue cells.

=> {WX} & {YZ} both can't be {78} => C can't be 15
=> C=16 => {PQRST}={12346}, {WX}={79}, {YZ}={58}

Stage 4:


Apply FNC on all {79} cells: r59c5=r86c6=r42c7=r15c8=r78c9=[58]
Apply FNC on all {5} cells: r3c68, r46c4, r7c57, r8c8 all from {123}
r6c1 from {79} => FNC: r7c2 can't be 8 => r7: r7c3=8, r7c2 from {79}
r5c1 from {79} => FNC: r4c2 can't be 8 => r4: r4c1=8, r4c23 from {12346}


r7c7+r8c8 from {123} => FNC: r7c7+r8c8={13} (NP @ n9,r678c678)
=> FNC @ r8c8: r9c9 from {46} => 2 @ c8,n9 locked @ r79c8 => r38c8={13} (NP @ c8)
=> FNC @ r3c8: r4c9 can't be 2 => 2 @ r3,n3,5-cell cage locked @ r3c79
=> r3c68={13} (NP @ r3,r234c678) => r2c8+r3c79={246} (NT @ n3,5-cell cage)
=> FNC @ r1c7: r2c6 can't be 2 => r2c68={46} (NP @ r2,r234c678) => r3c7=2
=> c7: r56c7={46} (NP @ n6,5-cell cage) => r7c6=2, r7c57={13} (NP @ r7) => n9: r9c8=2
=> FNC @ r3c6: r24c5 can't be 2 => r2c5+r3c6={13} (NP @ n2) => n2: r1c5=2
=> c5: r46c5={46} (NP @ n5) => c4: r89c4={46} (NP @ n8) => r159c159: r5c9 can't be 2
=> n6: r6c9=2 => r234c159: r2c1=2 => c1: r3c1=5 => c4: r2c4=5 => FNC: r13c3 can't be {46}
=> r3: r3c2 from {46} => r3: r3c4=8 => r1c4, r3c3 from {79}, FNC: r2c3 can't be {79}
=> r2: r2c2 from {79} => r2: r2c3 from {13} => r1: r1c2=8 => r1: r1c3 from {13}
=> c2: r9c2=5 => r9c3 from {79} => c3: r6c3=5 => r6: r6c24={13} (NP @ r678c234)
=> FNC @ r9c2: r8c3=2 => c3: r45c3={46} (NP @ n4) => n9: r89c1={13} (NP @ c1)


4 @ r2 locked @ r2c68 => FNC: r1c7 can't be 3

The rest are singles under FNC.



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