I have convinced myself that the quantity d in nh4
is boolean, enabling further simplifications. Thus:
nh5=: 3 : 0
hi=. 1.693-~ 3%:*/6,y,'ln2 ln3 ln5'=. ^.2 3 5
k=. i.1+<.hi%ln5 NB. exponents for 5
m=. 1+<.ln3%~hi-k*ln5
j=. ;i.&.>m NB. exponents for 3
k=. m#k
s=. <.p=. ln2%~hi-(j*ln3)+k*ln5
i=. >.p-0.01%ln2 NB. exponents for 2
z=. (i=s)#i,.j,.k
c=. y -~ (#s)++/s
assert. (0<:c)*0<#z
*/2 3 5x^c{z\:z+/ .*^.2 3 5
)
----- Original Message -----
From: Roger Hui <[email protected]>
Date: Thursday, January 7, 2010 12:19
Subject: Re: [Jprogramming] Rosetta code: hamming numbers
To: Programming forum <[email protected]>
> nh4 obtained by manipulation of the J code for nthHam
> without much understanding of the underlying mathematics.
>
> nh4=: 3 : 0
> lo=. 0.01 -~ hi=. 1.693-~ 3%:*/6,y,'ln2 ln3 ln5'=. ^. 2 3 5
> k=. i.1+<.hi%ln5
> NB. exponents for 5
> m=. 1+<.ln3%~hi-k*ln5
> j=.
> ;i.&.>m NB. exponents for 3
> k=. m#k
> r=. >.ln2%~lo-q=. (j*ln3)+k*ln5
> s=. <.ln2%~hi-q
> d=. 0>.1+s-r
> i=. (d#r)+;i.&.>d-.0 NB. exponents for 2
> z=. i,.d#j,.k
> c=. y -~ (#s)++/s
> assert. 0 <: c
> assert. c < #z
> */2 3 5x^c{z\:z+/ .*^.2 3 5
> )
>
>
>
> ----- Original Message -----
> From: Roger Hui <[email protected]>
> Date: Thursday, January 7, 2010 11:38
> Subject: Re: [Jprogramming] Rosetta code: hamming numbers
> To: Programming forum <[email protected]>
>
> > The }.z (and }.t) in your code are necessary because
> > z and t are initialized incorrectly, engendering extra
> > leading rows of 0. If instead you initialize
> > t=. i.0 5 and z=. i.0 4 then drops
> > can be removed.
> >
> > Anyway I have a non-looping and clearer version
> > that provides a factor of 10 improvement for y=1e6.
> > I will post that version in a little while.
> >
> >
> >
> > ----- Original Message -----
> > From: Aai <[email protected]>
> > Date: Thursday, January 7, 2010 11:21
> > Subject: Re: [Jprogramming] Rosetta code: hamming numbers
> > To: Programming forum <[email protected]>
> >
> > > Correction to make the last assert work correctly:
> > >
> > > nthHam=: 3 : 0
> > > lns=. 'ln2 ln3 ln5'=. ^. tdv=.2 3 5
> > > lo=. 0.01 -~ hi=. 1.693-~(*/6,y,lns)^%3
> > > t=.,:0$0
> > > for_k. i. 1+ <. hi % ln5 do.
> > > for_j. i. 1+ <. ln3 %~ hi - p=. k * ln5 do.
> > > t=. t, j,k,(>.ln2%~lo-q),(<.ln2%~hi-
> > > q), q=. p + j * ln3
> > > end.
> > > end.
> > > c=. +/ 1 + 3{"1 }. t
> > > z=.,:0$0
> > > for_r. (#~(2&{ <: 3&{)"1) }. t do.
> > > for_i. ([+i.@>:@-~)/2 3{r do. z=.z, i,(2{.r),
> > > ({:r)+i*ln2 end.
> > > end.
> > > assert. 0 <: c =. c - y
> > > assert. c < # }. z
> > > __ q: inv tdv ,: x: 3 {. c { (\: {:"1) }. z
> > > )
> > >
> > > It shouldn't be:
> > >
> > > nthHam 31
> > > 1
> > >
> > > but:
> > >
> > > nthHam 31
> > > |assertion failure: nthHam
> > > | c<#}.z
> > >
> > >
> > > Anyway, if you dare using it, use nthHam for N>:50000
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