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"If he succeeds in finding a pattern in prime numbers, he will hold the key to every computer in the world."
Definition: If every vertex in a graph is connected to every other (different) vertex by a single arc, we call it a complete graph.
Each number, considered as a complete graph, encompasses all the numbers that go before it:
Name of graph: | ||||
Number of arcs: | \(T_0=\frac{0(1)}{2}\) | \(T_1=\frac{1(2)}{2}\) | \(T_2=\frac{2(3)}{2}\) | \(T_3=\frac{3(4)}{2}\) |
and we drew a few more (\(K_7\) and \(K_8\)).
But how might "primitive" people have counted? (Ernie's not primitive enough; neither is he a person.) You won't be surprised to learn that we will be using a tree to help us.
You make a note of whether there is a leftover sheep or not -- maybe you make a mark, like a "1" or a "0". This is all you have to do to communicate the number to the King!
But you must also pay attention to the direction in which you write the marks.
Let's see how we might use a tree to represent the solution to the "22" counting problem: we'll use a ternary tree (three children branch off of a parent, rather than just two -- as we've seen in the binary trees for prime factorization).
I say "Whoever's doing this..." -- I mean, "Whoever's doing this primitive counting...." This is presumably someone who doesn't know how to count -- at least not the way we do -- but they can tell if there's one left over (and write "1") or not (and write "0").
Notice that we eventually have a single sheep in a pen, and that's when we're done. We have to write a "1" for the final sheep, to indicate that there's "one leftover".
Then the answer will be written as 1, 0, 1, 1, 0. That is, from the bottom up, left to right. This is important! We have to have a consistent scheme for writing (or otherwise recording our results -- perhaps on a knotted string).
So how do we write
(Notice that, while there can be either a 0 or 1 at the right end, there is always the "1" on the left -- meaning the last sheep standing!
Can you rebuild the tree using these "tally sticks"? If so, you can get a job in the King's counting house.
But rather than count out fifty pennies, I'd make a stack of 10, then four more stacks of exactly the same height (one-to-one correspondence). Five stacks of 10 makes 50. This is a similar idea....
If you have a lot of pennies, you could divide them in half, count half, and then multiply by two. But if you have one left over, you'd have to add that one.
If the half is still too many to count, do it again (on half), and so on -- until you get to the point where you can count "the half" (like if it gets down to 1 coin, say!:).
Watch Dave Brubeck's leg, counting out the rhythm....