Integer roots: Difference between revisions

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=={{header|Phix}}==

<lang Phix>include ~~bigatom~~.e

<lang Phix>include mpfr.e

function integer_root(integer n, object b)

function integer_root(integer n, object A)

-- yields the nth root of A, adapted from https://en.wikipedia.org/wiki/Nth_root_algorithm

bigatom r = BA_ONE

b = ~~ba_new~~(b)

mpz res = mpz_init(1),

x = mpz_init(),

delta = mpz_init()

A = mpz_init(A)

while true do

~~bigatom~~ x ~~= b~~

mpz_set(x,A)

~~for i=1 to~~ n-1 do

mpz_pow_ui(delta,res,n-1)

x ~~= ba_floor(ba_div(~~x,r))

mpz_fdiv_q(x, x, delta)

~~end for~~

mpz_sub(delta,x,res)

~~bigatom~~ ~~delta :~~= ~~ba_floor~~(~~ba_div(ba_sub(x~~,r),n))

{} = mpz_fdiv_q_ui(delta, delta, n)

if delta=~~BA_ZERO~~ then exit end if

if mpz_cmp_si(delta,0)=0 then exit end if

~~r = ba_add~~(r,delta)

mpz_add(res,res,delta)

end while

return ~~ba_sprint~~(r)

return mpz_get_str(res)

end function

printf(1,"3rd root of 8 = %s\n", {integer_root(3,8)})

printf(1,"3rd root of 9 = %s\n", {integer_root(3,9)})

string s = integer_root(2,"~~2e200~~") ~~-- best I could manage...~~

string s = integer_root(2,"2"&repeat('0',4000))

printf(1,"First digits of the square root of 2: %s\n", {shorten(s)})

integer l = length(s)

s = integer_root(3,"2"&repeat('0',6000))

s[20..-20] = " ... "

printf(1,"First %d digits of the ~~square~~ root of 2: %s\n", {l,s})</lang>

printf(1,"First digits of the cube root of 2: %s\n", {shorten(s)})</lang>

<pre>

3rd root of 8 = 2

3rd root of 9 = 2

First ~~101~~ digits of the square root of 2: 1414213562373095048 ... ~~8503875343276415727~~

First digits of the square root of 2: 1414213562373095048...7107578486024636008 (2001 digits)

First digits of the cube root of 2: 1259921049894873164...2546828353183047061 (2001 digits)

</pre>