## Stuck on this one

Post the puzzle or solving technique that's causing you trouble and someone will help
Sorry for the mistake in my first post and do allow me to try again!

Further to the work done on the Eppstein's hard puzzle, I constructed a combined bivalue & bilocation graph for this puzzle. From this graph, (correct me if I am wrong as I am still not 100% confident with the technique) a chain was identified. This chain consists of a combination of bivalue and bilocation edges.

Using Scott's notation:
[A:78:4*]<-4>[99:4*]<4>[29:42*]<2>[39:2*]<-2>[36:2*]<2>[15:23]<3>[11:34*]<4>[71:4*]<-4>[B,78:4*]
This chain enforced A<>4 or B<>4, Since A=B, therefore A=r9c9<>4 => r7c8=8.

Using short hand notation:
[A,78:4]-[99:4]=[29:42]=[39:2]-[36:2]=[15:23]=[11:34]=[71:4]-[B,78:4]
=> A=B<>4 => r9c9<>4 => r7c8=8.

Using general forcing chain notation:
r9c9=4 => r7c8<>4
r9c9=1 => r2c9=4 => r3c9=2 => r3c6<>2 => r1c5=2 => r1c1=3 => r7c1=4 => r7c8<>4
Therefore r7c8<>4 => r7c8=8.

I don't know whether this chain has been identified by other solvers. Although the construction of the graph is quite tedious, the chain is nevertheless eventually identifiable within human ability.
Jeff

Posts: 708
Joined: 01 August 2005

Jeff wrote:
I don't know whether this chain has been identified by other solvers. Although the construction of the graph is quite tedious, the chain is nevertheless eventually identifiable within human ability.

For hard puzzles, it's certainly a point to find the shortest or easiest or fastest solution path, according to some measure. The measures 'easiest to find', 'easiest to describe' and 'easiest to recognise' are all important and do not always coincide.
Ocean

Posts: 442
Joined: 29 August 2005

### Re: Stuck on this one

Verity: {7,2}=2, {7,2}=4, {7,2}=6, {7,2}=8 all imply {9,2}<>6
AMcK, I don't get this one. Can you explain in more details?
Jeff

Posts: 708
Joined: 01 August 2005

### stuck on this one

Hi Glenn
In your original post you say you have progressed to this point.What were the original starting number in the grid before any entries were made.Maybe starting back at the beginning may throw a different light on the problem.Could you give us the original grid?
Many thanks .
thefitter

Posts: 2
Joined: 03 September 2005

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