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Colouring with four colours?

 
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Gabriele
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Joined: 07 Jan 2007
Posts: 1
Location: Deutschland/Augsburg

PostPosted: Sat Jul 14, 2007 11:50 am    Post subject: Colouring with four colours? Reply with quote

Hello everybody!

I know colouring with one chain (colour) and two chains (colours). Andrew Stuart writes in his solving guide that colouring with three chains (colours) is theoretically possible, but he has no example. My be I have found one accidentally. Please can anyone proof my “discovery” ?

Is it really true that I discovered an example for colouring with four colours?
Does colouring with more than two colours really exist?


This is Ruuds One Trick Pony no. 222 from 27/06/2007
..7..829...4.21...8............5......6....39.9.....4...8.....34..7..5...6..13.8.
617538294954621378823974651741359826586142739392867145178295463439786512265413987

Elementary steps took me to this point.
Code:

.-----------------------.------------------------.----------------------.
|  136    135     7     |  3456    346     8     |  2      9      146   |
| *369    35      4     | *3569    2       1     |  3678   567    678   |
|  8      1235   *1239  |  34569  ~34679   45679 |  1346   156    146   |
:-----------------------+------------------------+----------------------:
|  137    48      13    | +134689  5      +4679  |  1678   1267   12678 |
|  5      48      6     |  1248    47      247   |  178    3      9     |
|  1237   9       123   |  1368    367     67    |  1678   4      5     |
:-----------------------+------------------------+----------------------:
|  129    7       8     |  24569  ~469     24569 | $49     126    3     |
|  4      123    *1239  |  7       8      *269   |  5      126    126   |
|  29     6       5     |  249     1       3     | $479    8      47    |
'-----------------------'------------------------'----------------------'

You can mark four chains:
chain 1 (*): [r2c4]=9=[r2c1]=9=[r3c3]=9=[r8c3]=9=[r8c6]
chain 2 (+): [r4c4]=[r4c6]
chain 3 (~): [r3c5]=9=[r7c5]
Chain 4 ($): [r7c7]=9=[r9c7]

The underlined cells mean for example “true” and the not underlined cells “false”.
One of the cells with ~-sign and +-sign must be true and the other false. One cell of both chains faces r2c4 (underlined) and the other r8c6 (underlined), which is a contradiction.
So the underlined cells of chain 1 (*) can’t contain the nine, the others can be solved with nine.
We found: r3c3<>9, r2c4<>9, r8c6<>9 and r2c1=9, r8c3=9
This results according colouring Type 2.

Now you can also make reductions according colouring Type 1.
One of the cells r7c5 and r7c7 must be false, the other true. That’s why all other nines in row 7 can get eliminated.

Ruuds solver makes a reduction according to the skyscraper, which is hidden in the cells r8c3+r3c35+r7c5 and makes a headstand.
It follows: r7c1<>9, r8c6<>9

If you only look at the skyscraper, you don’t use the full information, which sticks in all the nines. But it’s enough to solve the puzzle and that’s the target.

Gabriele
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Ruud
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Joined: 30 Dec 2005
Posts: 601

PostPosted: Sat Jul 14, 2007 1:48 pm    Post subject: Reply with quote

Hi Gabriele,

Only chains 1 and 2 are required.

I marked them a-A and b-B in this diagram:
Code:
.-----------------------.------------------------.----------------------.
|  136    135     7     |  3456    346     8     |  2      9      146   |
| a369    35      4     | A3569    2       1     |  3678   567    678   |
|  8      1235   A1239  |  34569   34679   45679 |  1346   156    146   |
:-----------------------+------------------------+----------------------:
|  137    48      13    | b134689  5      B4679  |  1678   1267   12678 |
|  5      48      6     |  1248    47      247   |  178    3      9     |
|  1237   9       123   |  1368    367     67    |  1678   4      5     |
:-----------------------+------------------------+----------------------:
|  129    7       8     |  24569   469     24569 |  49     126    3     |
|  4      123    a1239  |  7       8      A269   |  5      126    126   |
|  29     6       5     |  249     1       3     |  479    8      47    |
'-----------------------'------------------------'----------------------'

because color A can see both colors b & B, it can be eliminated.
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