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Frame CanCan 4
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=833		 Page 1 of 1
Author:  HATMAN [ Mon Oct 25, 2010 10:55 am ]
Post subject:  Frame CanCan 4
Frame CanCan 4

I've not posted one of these for a while. However I have done quite a few as I find it a simple way of making puzzle book ones harder.

The "frame" numbers are the horizontal or vertical sums of the cage totals.

This one is human solvable in style, so with the right approach it's not too hard.

Image
Author:  simon_blow_snow [ Sun Oct 31, 2010 8:08 am ]
Post subject:  Re: Frame CanCan 4
Thanks HATMAN, interesting puzzle.

I think in a sense it is "paper solvable" (no pencil marks needed), even though a little bit of short term memory is necessary. ;-)

short walkthrough:
Min cage indices in R3C2,R3C4 sum to 1+1+2+3+4+1=12
--> max cage index in R3C3=14-12=2
--> min cage index in R5C3=10-2=8
--> max cage index in R5C2=13-8-1=4
--> R56C2 must include {1} of C2

Max cage indices in R3C2,R3C4 sum to 14-1=13
Min cage indices in R3C2,R3C4 sum to 2+1+2+3+4+1=13
--> R2C2=2, R4C4=1, R3C1245={1234} --> R3C36={56}
--> cage index in R3C3=14-13=1
--> cage index in R5C3=10-1=9
Code:
......
.2....
......
...1..
......
......

Max cage index in R5C2=13-9-1=3
Min cage index in R5C2=1x3x1=3
--> 3-(3) cage in R56C2+R6C3 = [131] --> R3C2=4
Code:
......
.2....
.4....
...1..
.1....
.31...

C15 cage indices sum to 16+20=36
--> cage indices in R1C1,R1C5,R5C5 sum to 36-32=4
--> R12C1,R56C5 cages are both 1-(2), R1C56 cage is 2x(2)
--> cage index in R4C1=16-1=15, cage index in R4C5=20-2-1=17
--> R456C1+R4C2 cage is 15+(4), R4C5+R456C6 cage is 17+(4)

Min R4C2=5 --> Max R456C1=15-5=10={234/235}
--> 1-(2) cage in R12C1 <> {12/23/34/45}, = {56}
--> R3C1=1, R456C1={234} --> R4C2=15-2-3-4=6 --> R1C2=5
--> 1-(2) cage in R12C1 = [65]
Code:
65....
52....
14....
.6.1..
.1....
.31...

Cage index in R3C2=2+1+4=7 --> cage index in R1C2=22-7-3=12
--> R12C3=12-5=7 <> {16/25}, = {34}
--> 1-(2) cage in R34C3 <> {12/23/34/45}, = [65]
--> R3C6=5, R5C3=2
--> R56C4=9-2=7 must include {5} of C4, = [52]
Code:
65....
52....
146..5
.651..
.125..
.312..

(Finish)
R456C1=[234], R123C4=[463], R3C5=2
--> R1C356=[312] --> R2C356=[431]
--> R456C5=[465], R456C6=[346]
Code:
653412
524631
146325
265143
312564
431256
Author:  HATMAN [ Sun Oct 31, 2010 1:16 pm ]
Post subject:  Re: Frame CanCan 4
Simon

I had about six attempts at creating this puzzle, I was trying to make it insolvable if you did not use the primary solution. However I failed, which on thinking about it is probably automatic.

Your's was a neat way of solving it, however the primary solution starts:

Sum of cage totals = 93, sum of all cells = 126 shortfall 33
There are 3 two cell minuses which as 56's deliver a shortfall of 30 - 3 left to find
The double times only delivers a shortfall if one cell is a 1 (which gives a shorfall of 1) and Triple times only delivers a shortfall if it is 1X1 (shortfall of 2)
All others are sum cages which gives the total of 126 - 10*3 - 1 - 2 = 93
You then follow on with a shortened version of your solution.

You can see why I called it human solvable in style.
Author:  simon_blow_snow [ Sun Oct 31, 2010 2:05 pm ]
Post subject:  Re: Frame CanCan 4
Thanks HATMAN, your intended approach was a neater way to go about it.

However I think you miscalculated the total of the sum cages. It should be 8 less than your number.

Here is how I would express your logic:

logic:
Total sum of grid=126
Total sum of cage indices=93
Discrepancy=33
Max discrepancy between a -(2) cage index and the sum of its cell values=5+6-1=10
Max discrepancy between a x(2) cage index and the sum of its cell values=a+1-a*1=1
Max discrepancy between a x(3) cage index and the sum of its cell values=b+1+1-b*1*1=2
Therefore the five non-sum cage indices must all achieve their max limits.

With the three 1-(2) cages on C135, the two +(4) cage indices on R4 are 15 & 17, and the +(3) cage index on R5 is 9.
--> The x(3) cage index=13-9-1=3, the x(2) cage index=20-17-1=2.

Therefore the total of sum cage indices=93-1-1-1-3-2=85.
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