I think I’ve found a straightforward solution. - Emmanuel Van Hese - Medium

I think I've found a straightforward solution. - Emmanuel Van Hese - Medium

I think I've found a straightforward solution. It's linear, but very simple to implement. There are certain symmetries that allow us to group the dial keys into categories, that I will label A, B, C, D, and E:

Category A: numbers 1, 3, 7, and 9

Category B: numbers 4 and 6

Category C: numbers 2 and 8

Category D: the number 0

Category E: the number 5

I will associate which each category a corresponding value, which I will initialize as a=b=c=d=e=1. Then, with no hops, we can dial 4a + 2b + 2c + d + e = 10 numbers. Now let's hop 1 time:

Every number in category A is connected to 1 number in category B and 1 number in category C. Every number in category B is connected to 2 numbers in category A and 1 number in category D. Every number in category C is connected to 2 numbers in category A. The number in category D is connected to 2 numbers in category B. The number in category E isn't connected to anything. We can represent this as follows:

a -> b + c

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