Frictional Games Forum (read-only)

We have a seven link chain - numbered 1 to 7.
We divide it into three groups, A, B and C.

Suppose the innkeeper will accept change.

On Day 1, Peter gives him Link 1 (Group A).
On Day 2, Peter gives him Links 2 and 3 (Group B).
But this would mean Peter's paying in advance and violating the rule of one link per day.
So he takes Link 1 back off the innkeeper.

Now Peter has five links and the innkeeper has two.

On Day 3, Peter gives him Link 1 (Group A) again.
On Day 4, Peter gives him Links 4 - 7 (Group C).
Once again, this would violate paying in advance and one link per day.
So he takes Groups A and B back (Links 1 - 3).

Now Peter has three links and the innkeeper has four.

On Day 5, Peter gives him Link 1 (Group A) again.
On Day 6, Peter gives him Links 2 - 3 (Group B).
Once again, this would violate paying in advance and one link per day.
So he takes Link 1 back off the innkeeper.

Now Peter has one link and the innkeeper has six.

So on Day 7, Peter just has to give the first link.

He minimises the damage done by making three cuts.
He gives one each day, and compensates by taking the required change back.
He does not pay in advance, nor late.
The innkeeper is happy. Peter is happy.
Peter can go home without paying money for a chain that probably costs $x*7.

2 right (land on 3)
3 down (land on 3)
2 right + 1 down (land on 4)
3 right + 1 down (land on 4)
4 right (land in red square)
Total: 16

2 minutes later, found;
2 right (land on 3)
3 right (land on 4)
1 left + 2 down + 1 right (land on 3)
3 down (land in red square)
Total: 12

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