## Posts Tagged ‘Feynman’

### Programming Praxis – Feynman’s Puzzle

June 12, 2009

Today’s Programming Praxis problem is about a long division puzzle by Richard Feynman. The provided solution is 14 lines of code. Since both his and my solution are little more than a simple list comprehension, there is not going to be much room for improvement, but let’s see what we can do.

We need a way to refer to specific digits of numbers. The provided solution does so by converting the number to a list of digits. This function just returns the nth digit of a number, starting with the least significant one. For instance, digit 1 1234 equals 4.

```digit :: Int -> Int -> Int
nth `digit` n = n `mod` 10 ^ nth `div` 10 ^ (nth - 1)```

As mentioned, this problem is most naturally solved with a list comprehension, so that is what we will use as well. The actual conditions are of course nearly identical to the scheme version, with the exception that the condition that a * n1 has four digits is unnecessary, and therefore removed in this version.

```feinman :: [(Int, Int)]
feinman = [ (n1, n2)
| b <- [1..9], a <- [0..9], c <- [0..9],
d <- [1..9], e <- [0..9], f <- [0..9],
a /= b, a /= c, a /= d, a /= e, a /= f, e < d,
let n1 = 100 * b + 10 * a + c,
let n2 = 1000 * d + 100 * e + 10 * a + f,
n1 * n2 > 999999, n1 * n2 < 10000000,
digit 3 (n1 * n2) == a,
digit 1 (d * n1) == a, digit 2 (d * n1) == a,
digit 1 (e * n1) == a, digit 3 (a * n1) == a]```

To test, we just print the list.

```main :: IO ()
main = print feinman```

As expected, we get the single result of (484, 7289). Runtime is about 8.5 seconds in ghci and 60 ms compiled (You’ve got to love compilers). The end result is 11 lines of code, so a slight improvement over the scheme version.