Frameless Repeat

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Frameless Repeat

Postby HATMAN » Mon Sep 12, 2011 9:10 pm

Frameless Maximum Repeat 2

It is frameless so each number refers to the cage sum (of unknown length) in one of the two diagonal directions.

Cages are allowed to cross each other.

Within a cage singletons are not allowed all must repeat so a five cage will be forced to be {XY}X{XY}.

It is maximum repeat: so you are required to find a solution with the maximum number of repeats. Where cages cross the cell at the crossing counts as two repeats. My Solution has
63 repeats. You may be able to beat this - udosuk once did.

This one is reasonably hard but I found it flowed.

If anyone does solve it please post that you have.


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Re: Frameless Repeat

Postby dyitto » Fri Sep 16, 2011 10:20 pm

Could you demonstrate the constraints using a complete example grid?
evert on the crashed forum
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Re: Frameless Repeat

Postby simon_blow_snow » Sat Sep 17, 2011 4:08 pm

dyitto wrote:Could you demonstrate the constraints using a complete example grid?


The solution grid itself is a plain/vanilla Sudoku grid. There is no constraint like Non-Consecutive or Anti-Whatever.

The (invisible) cages are all diagonal, straight line cage with an unknown length. Clues at the corners has 1 possible direction only. Clues at the edges has 2 possible directions (e.g. a clue at the top edge can only lead to 2 different directions: towards bottom-left or towards bottom-right).

The most important rule is the "no singleton" rule. All values in any cage must appear at least twice within that cage. In poker terms, for a five-cell cage only "full house" is allowed but not "four of a kind" or any other combinations.

For example, the 7 clue at the top middle indicate that it must be a five-cell cage (not possible for 3-cell or 4-cell without singleton), so there are only two possibilities: R1C5-R2C4-R3C3-R4C2-R5C1 or R1C5-R2C6-R3C7-R4C8-R5C9.

I have solved the puzzle, and have posted the solution in the RCBroughton forum. If you want to peek at the solution you can go there. I will also write a complete walkthrough later. Once you get hold of all the rules it isn't really that difficult.
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Re: Frameless Repeat

Postby simon_blow_snow » Tue Sep 20, 2011 6:16 pm

Here is the complete walkthrough: (I also posted this on the R C Broughton forum)

Complete walkthrough (12 steps): Show
Code: Select all
21             07
  +--+--+--+--+--+--+--+--+--+
  |        |        |        |
30|        |        |        |
  |        |        |        |25
  +--+--+--+--+--+--+--+--+--+
  |        |        |        |46
11|        |        |        |12
  |        |        |        |
  +--+--+--+--+--+--+--+--+--+
  |        |        |        |38
  |        |        |        |
  |        |        |        |
  +--+--+--+--+--+--+--+--+--+
      30       07    27

Step 1:
The two 7(?) cages must be of length 5 (length 3 must have singleton, length 4 must be even-total)
The two 7(5) cages must be in the form [{12}1{12}]
Their starting ends must be R19C5, their stopping ends must be R5C19
(stopping ends must be different, otherwise two 1s on the same column)
--> R19C5={12} (C5), R5C19={12} (R5)
--> The two 7(5) cages together force D\37=[11] or D/37=[11]

Step 2:
The 11(?) cage also must be of length 5, from R5C1 to R1C5 or R9C5, with the middle cell in R3C3 or R7C3
The 11(5) cage must be in the form [{ab}a{ab}] --> 11=3a+2b --> a must be odd and a<5
Since one of R37C3 is [1] and belongs to a 7(5) cage, a cannot be 1 --> a must be 3
11(5)=[{13}3{13}]=[13331] (Step 1: R5C1 and R19C5 must be from {12})

Step 3:
Summarising steps 1 & 2, we have the following 2 possible cases:
R1C5 -7(5)- R5C1 -11(5)- R9C5 -7(5)- R5C9: [2112133312112]
R9C5 -7(5)- R5C1 -11(5)- R1C5 -7(5)- R5C9: [2112133312112]
--> R5C19=[12], R46C2={23} (C2,N4), R37C3={13} (C3), R28C4={13} (C4)
--> [2] of C6 locked in R28C6, [1] of C7 locked in R37C7, [1] of C8,N6 locked in R46C8

Step 4:
The 21(?) cage must be of length 7/8/9 (length 3/4/5 must have singleton, length 6 must be even-total)
--> 21(?) must include D\456 and all of these 3 cell values must repeat at least once
--> D\46 must contain at least one of {12}
(otherwise min D\456={345} and when these cell values repeat the total will exceed 21)
--> R4C2+R6C8<>[21] --> R6C2+R4C8=[21]
--> (step 3) R1C5..R5C1=[13331], R1C5..R5C9=[12112], R5C1..R9C5=[12112]
Hidden singles: D\2469=[1211]

Step 5:
The 12(?) cage must be of length 4/5 (length 3 must have singleton)
If the length is 4, with [2] already in R5C9, 12(4)=[2424]
If the length is 5, with [2] already in R5C9+R9C5, 12(5)=[23232]
--> D\7=[2], R6C8+R8C6=[33/44] (<>{56789})

Step 6:
Now the 21(?) cage already has the cell values {123} within D\123456, each must repeat at least once
--> 6 cells of 21(?) must be {112233} totalling 12, 9 short of 21
--> At least one value from {456789} must appear in 21(?) and repeat at least once
(otherwise max total = 1+1+1+2+2+2+3+3+3 = 18 < 21)
--> To make up the remaining total of 9, there is only one possible way: {144}
--> 21(?) must be of length 9 --> 21(9)=D\={111223344}
--> D\1=[4], D\58={34} (<>{56789})

Step 7:
Consider the bottom 30(?) cage: it must start from R8C1 and extend towards R2C9 with a min length of 4
Now [2] of N1 is locked in R1C3+R3C1
--> R3C8+R8C3, collectively seeing R1C3+R3C1, cannot be both [2]
--> 30(?) R9C2 cannot contain [2] as a repeating cell value
--> R8C3<>[2]
Hidden singles: R1C3=R8C1=[2] --> R3C8=[2]

Step 8:
Consider the 46(?) cage: it must start from R4C9 and extend towards R9C4 with a length of 6
--> The only possiblility is 46(6)=[{689}{689}] (<>{34578})
--> R4C9+R5C8+R6C7={689} (N6), R7C6+R8C5=R9C4={689} (N8)

Step 9:
Consider the 38(?) cage: it must start from R7C9 and extend towards R1C3 with a min length of 5
It cannot be of length 7 (no second [2] to match R1C3=[2])
It cannot be of length 5 (max R6C8=4 and it must repeat, so max total would be 4+4+9+9+9=35<38)
--> 38(?) must be of length 6, from R7C9 to R2C4 --> R2C4=[3] must repeat within the cage
--> R6C8<>[4] (otherwise max total would be 3+3+4+4+9+9=32<38)
--> (step 5) R6C8+R8C6=[33] --> (step 6) D\58=[34]
--> with R2C4+R6C8=[33], remaining total=38-3-3=32
--> with max R5C7=7, the only way to make this up is {7799}
--> R5C7=[7], R3C5+R4C6+R7C9 must be from {79} (<>{34568})
Hidden singles: R1C9=R7C1=R9C7=[3]

Step 10:
Consider the 27(?) cage: it must start from R9C7 and extend towards R3C1 with R4C2+R8C6+R9C7=[333]
It cannot be of length 4/5/6 (to avoid singletons max total would be 3+3+3+5+5+5=24<27)
--> 27(?) must be of length 7, from R9C7 to R3C1 --> R3C1+R5C3+R6C4+R7C5=27-3-3-3=18
--> To make up this total of 18 without singletons, there is only one possible way: {4455}
--> R3C1=[5], R5C3+R6C4+R7C5 must be from {45} (<>{6789})

Step 11:
Consider the 25(?) cage: it must start from R3C9 and extend towards R9C3 with R4C8+R5C7+R6C6+R8C4=[1711]
It cannot be of length 3/5/6 (singletons would appear)
It cannot be of length 4 (max length would be 1+1+7+7=16<25)
--> 25(?) must be of length 7, from R3C9 to R9C3 --> 25(7) = {1114477} (only possible way without singleton)
--> R7C5=[4], R3C9+R9C3={47} (<>{5689})
--> [4] of N2 locked in R3C46 (R3) --> R3C9+R9C3=[74]
--> (step 10) R5C3+R6C4=[54] --> R4C7+R6C9=[45]
--> (step 9) R3C5+R4C6+R7C9=[979] --> R7C4+R9C6=[75]
Hidden singles: R2C9=R3C6=R5C2=[4], R1C4=R4C5=[5], R1C2=R2C5=R9C8=[7], R5C6=R9C4=[9]

Step 12:
Consider the top 30(?) cage: it must start from R2C1 and extend towards R9C8 with a min length of 4
It cannot be of length 5/6/7/8 (min total would be 6+8+6+6+8=34>30)
--> 30(?) R2C1 must be of length 4, from R2C1 to R5C4 --> 30(4) R2C1 = [9696]
(only possible way without singleton)
--> 30(?) R9C2 must be of length 4, from R9C2 to R6C5 --> 30(4) R9C2 = [8778]

Naked singles to finish.
Code: Select all
472516893
918372654
563894127
839257416
145639782
726481935
351748269
297163548
684925371

21(9) = [413231241]
7(5) = [12112]
30(4) = [9696]
25(7) = [7171414]
46(6) = [689869]
11(5) = [13331]
12(5) = [23232]
38(6) = [397739]
30(4) = [8778]
7(5) = [12112]
27(7) = [5354433]

Total = 63 repeats
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