Have I got an interesting game for you!

Imagine that you have an infinite row of squares, numbered by non-negative integers. Square #0 is the first square. To its right is square #1, then square #2, and so on forever. Each square can hold as many (or as few) pennies as you like. There are only two rules.

First, the *fission rule* states that you may replace a penny by a pair of pennies, placing one in each of the immediately adjacent squares. For example:

The *fusion rule*, on the other hand, states that if there are two pennies separated by exactly one intervening square, you may replace them by a single penny in that middle square. For example:

In a sense, there’s only *one* rule, but you’re allowed to apply it either forwards or backwards. Note also that there’s no requirement for any of the squares involved to be empty or anything like that.

Here’s the challenge: starting with just a single penny in square #7,

can you end up with just a single penny in square #1,

making only moves allowed by the rules above? What’s the fewest number of moves that you need?

Once you’ve solved that one, try starting with a penny in square #7 and ending up with a penny in square #2.

Tune in next time for the answer… and an amazing mathematical analysis of the Nuclear Pennies Game! (Note: I didn’t come up with the Nuclear Pennies Game; I’ll give proper attribution next time, but for now I don’t want to give away the answer by linking to it.)

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About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

Is it allowed to have a negative number of pennies in a square?

Good question, I should have been more clear; negative numbers of pennies are not allowed. You should be able to simulate it (except for the “infinite strip” part) with actual pennies and some squares drawn on a piece of paper.

i say this problem is impossible! if it has an infinite number of spaces there cant be a beginning nor an end to the strip!

I did the first task in 18 moves. I’d say the second one is impossible–I can’t do it. Looks like 5 spaces away is too close to reduce to one penny. I’m not going to attempt the analysis, though.

OT: Happy National Mole Day.

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Dave C: Good work, as far as I know 18 moves is the best possible. And you’re absolutely right, the second one is impossible… stay tuned for proof. =)

jon: Maybe my explanation wasn’t clear, but there

isa beginning to the strip; the leftmost square is the first. However, there isn’t an end. In any event, that doesn’t prevent the problem from being solvable.