\ Problem: \ The 5-digit number, 16807=75, is also a fifth power. Similarly, the \ 9-digit number, 134217728=89, is a ninth power.
\ How many n-digit positive integers exist which are also an nth power? \ They don't count 0^1 (0 is not deemed to be positive)
\ Solution: \ 10^n produces a n+1-digit integer, so we need only look at bases 0-9.
: n-digits { n -- n2 } \ number of n-th powers with n digits \ for n=1, it does not count 0 and 1 (I correct for the 1 below) \ this works by just seeing what the lowest x is that fits in \ 10^(n-1), and then rounding x up to an integer n 1- 0 d>f n 0 d>f f/ falog f>d drop 9 swap - ;
: solve ( -- ) 1 1 begin ( n sum ) over n-digits dup 0> while + swap 1+ swap repeat drop nip ;
> \ Problem: > \ The 5-digit number, 16807=75, is also a fifth power. Similarly, the > \ 9-digit number, 134217728=89, is a ninth power.
> \ How many n-digit positive integers exist which are also an nth power? > \ They don't count 0^1 (0 is not deemed to be positive)
> \ Solution: > \ 10^n produces a n+1-digit integer, so we need only look at bases 0-9.
> : n-digits { n -- n2 } > \ number of n-th powers with n digits > \ for n=1, it does not count 0 and 1 (I correct for the 1 below) > \ this works by just seeing what the lowest x is that fits in > \ 10^(n-1), and then rounding x up to an integer > n 1- 0 d>f n 0 d>f f/ falog f>d drop 9 swap - ;
Clever, but the stack makes it hard to follow the code.
> : solve ( -- ) > 1 1 begin ( n sum ) > over n-digits dup 0> while > + swap 1+ swap repeat > drop nip ;
Again, the stack makes it very difficult to follow. When one sees the +, he can't easily tell what two items are being added together. This code is hard to understand and to modify.
Ruby:
def n_digits n 9 - ( 10 ** ((n-1.0) / n) ).to_i end
sum = 1 (1 .. 9999).each{|i| n = n_digits(i) break if n < 1 sum += n } p sum
William James <w_a_x_...@yahoo.com> writes: >On May 5, 11:49 am, an...@mips.complang.tuwien.ac.at (Anton Ertl) >> : n-digits { n -- n2 } >> \ number of n-th powers with n digits >> \ for n=1, it does not count 0 and 1 (I correct for the 1 below) >> \ this works by just seeing what the lowest x is that fits in >> \ 10^(n-1), and then rounding x up to an integer >> n 1- 0 d>f n 0 d>f f/ falog f>d drop 9 swap - ;
>Clever, but the stack makes it hard to follow the code.
I find that the conversion between integer and FP, with a double-cell intermediate step makes it a little bit convoluted. One can fix that with factoring (and using FLOOR, which I have hitherto overlooked):
: u>f 0 d>f ; : f>u f>d drop ; : nnf/ ( n1 n2 -- f ) swap u>f u>f f/ ; : n-digits1 { n -- r } 9e n 1- n nnf/ falog floor f- ;
>> : solve ( -- ) >> 1 1 begin ( n sum ) >> over n-digits dup 0> while >> + swap 1+ swap repeat >> drop nip ;
>Again, the stack makes it very difficult to follow. >When one sees the +, he can't easily tell what two >items are being added together. This code is hard >to understand and to modify.
With N-DIGITS1, it becomes a little easier:
: solve ( -- r ) 1 1e begin ( n rsum ) dup n-digits1 fdup f0> while ( n rsum r ) f+ 1+ repeat fdrop drop ;
Or you can use locals to reduce the stack load:
: solve ( -- r ) 1 1e begin { n f: rsum } n n-digits1 { f: r } r f0> while n 1+ r rsum f+ repeat rsum ;
But given the tone of your posting, I guess you think that the stack makes that difficult to follow, too. Then you have two options:
1) Get some more practice in reading Forth or other stack-based languages. You may eventually get the hang of it.
2) Conclude that stack-based languages are not for you and stick with whatever other language you like.
>> \ Problem: >> \ The 5-digit number, 16807=75, is also a fifth power. Similarly, the >> \ 9-digit number, 134217728=89, is a ninth power.
>> \ How many n-digit positive integers exist which are also an nth power? >> \ They don't count 0^1 (0 is not deemed to be positive)
>> \ Solution: >> \ 10^n produces a n+1-digit integer, so we need only look at bases 0-9.
>> : n-digits { n -- n2 } >> \ number of n-th powers with n digits >> \ for n=1, it does not count 0 and 1 (I correct for the 1 below) >> \ this works by just seeing what the lowest x is that fits in >> \ 10^(n-1), and then rounding x up to an integer >> n 1- 0 d>f n 0 d>f f/ falog f>d drop 9 swap - ;
>Clever, but the stack makes it hard to follow the code.
>> : solve ( -- ) >> 1 1 begin ( n sum ) >> over n-digits dup 0> while >> + swap 1+ swap repeat >> drop nip ;
>Again, the stack makes it very difficult to follow. >When one sees the +, he can't easily tell what two >items are being added together. This code is hard >to understand and to modify.
>sum = 1 >(1 .. 9999).each{|i| > n = n_digits(i) > break if n < 1 > sum += n } >p sum
Please don't publish spoilers without the words spoiler in the subject line.
Groetjes Albert
-- -- Albert van der Horst, UTRECHT,THE NETHERLANDS Economic growth -- like all pyramid schemes -- ultimately falters. albert@spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst
