Hi,
I don't know if this is the right place to post this, but here I go.
I translated a BC-program found on the home page of Keith Matthews,
http://www.numbertheory.org/keith.html.
It's about the expension of a (not necessarily reduced) fraction m/n in
base b.
Looking at the result of my rewriting gives me the idea it isn't
Haskelly enough.
Anyways, here's my interpretation:
-- period m/n base = (period length, preperiod digits, period digits)
period :: Integer -> Integer -> Integer -> (Int, ([Integer], [Integer]))
period m n base | m' >= n' = error $ show m ++ " >= " ++ show n
| m' < 1 = error $ "m (" ++ show m ++ ") < 1"
| n' < 1 = error $ "n (" ++ show n ++ ") < 1"
| base <= 1 = error $ "base (" ++ show base ++ ") < 2"
| otherwise = decimals [] [m']
where g = gcd m n
m' = m `div` g
n' = n `div` g
decimals qs rs | i' /= -1 = (length rs - i', splitAt i' qsi)
| otherwise = decimals qsi (rs++[ri])
where (qi,ri) = divMod (base * last rs) n
i = findIndex (==ri) rs
i' = if i == Nothing then -1 else fromJust i
qsi = qs++[qi]
Which comes from the BC program:
(Keith Matthews)
define period(m,n,b){
auto a[],g,i,p,j,c[]
g=gcd(m,n)
m=m/g;n=n/g
if(m>=n){"m>=n:"; return(0)}
if(m<1){"m<1:"; return(0)}
if(n<1){"n<1:"; return(0)}
if(b<=1){"b<=1:"; return(0)}
i=0
c[0]=m
flag=0
while(flag==0){
p=b*c[i]
i=i+1
c[i]=p%n
a[i]=p/n
/*print "a[",i,"]=",a[i],"\n" /* the i-th digit */
if(i==1000){
print "limit of 1000 digits produced\n"
return(0)
}
for(j=0;j