# Morose Stronginthearm

Inventor of Hnaflbaflsniflwhifltafl, the precursor to Thud, at the bequest of Hugen the then-Low King of the Dwarfs. Stronginthearm as a surname is hugely common amongst dwarfs. Morose, on the other hand, is a typically TP-esque spin on the Disney convention. Instead of "Happy" or "Dozy" we have a name which is a synonym for "miserable".

As a reward for inventing the game, Hugen asked Morose what he wanted. The inventor is on record as saying: "If it please you, your majesty, I ask for nothing more than that you should place one plk [a small gold piece then in general circulation] on the first square, two on the second, four on the third and so on until the board is filled." (This is a reference to the Roundworld legend of the Chinese sage who requested one piece of rice on the first, two on the second &c.)

After being threatened with an axe by the King, Morose hastily amended his request to "as much gold as he could carry", whereupon Hugen agreed and merely had one of his arms broken. "For," he said, "all should know that while Hnaflbaflsniflwhifltafl teaches preparedness, strategy, boldness and quick thinking, it is also important to know when not be too drhg'hgin clever by half."

## Annotation

To work out what Morose had asked for, if you number the squares of a chessboard from 0 to 63, the amount of gold on square n would then be 2^n. The amount of gold on the final square alone would therefore be 2^63 coins, which would require a very big chessboard.

The number of coins on #62 would be half that many, the number on #61 would be half again, and so on. If this series is continued to square #0, the sequence would add up to

N = 2^63 * sum ((1/2)^k, k=0, 63)

I won't prove it here but if |s| < 1, then

sum (s^k, k=0, ∞) = 1 / (1 - s)

It can also be shown that

sum (s^k, k=0, n)

= sum (s^k, k=0, ∞) - sum (s^k, k=n+1, ∞)

= sum (s^k, k=0, ∞) - s^(n+1) * sum (s^k, k=0, ∞)

= (1 - s^(n+1)) * sum (s^k, k=0, ∞)

= (1 - s^(n+1)) / (1 - s)

For our particular problem, s = 1/2 and n = 63. Therefore,

N = 2^63 * (1 - (1/2)^64) / (1 - 1/2)

which simplifies to

N = 2^64 - 1 coins, which is **18,446,744,073,709,551,615** coins.

"Not enough gold in the universe" doesn't cover it by any stretch. As of 2006, if you added all the gold *ever* mined on Roundworld together, the total weight would be 145,000 tonnes. Now, let's think about a gold coin. Even the purest on Roundworld are about 977 parts gold per 1000 parts - and how pure would a Discworldian coin be? Ankh-Morporkian coins we know have "all the gold content of seawater", but we can realistically expect a dwarfish gold coin to be pretty much what it says it is. Now a *small* gold piece would weigh perhaps 1/4 ounce, which equals 7.0875g.

And that means **14,175,000,000,000,000 tonnes of gold!!**

Which is 9,775,862,069 - nearly **ten billion** - times more gold than has **ever** been mined on Roundworld. He's lucky he only had an arm broken.

In stories of the Roundworld grains of rice or wheat are usually used instead, the most common being the so-called Indian inventor of chess, Sessa, who was a lord in charge of farms (a Vellalor) in India. This story may have been linked to Sessa wanting outrageous prices from the peasants who worked on his land.