4
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Given two integers m and n, return the first m digits of sqrt(n), with the decimal point. They will be given with a space in between.

You only have to produce m digits: so if m=5, n=500, then the output will be 22.360, not 22.36067.

Do not use anything that will increase the precision of any operation.

Test Cases:
20 99 -> 9.9498743710661995473
15 12345678 -> 3513.64170057221
16 256 -> 16.00000000000000
2 10000 -> 10

Shortest code wins.

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  • 1
    \$\begingroup\$ Looks like your second test case gives the square root of 12345678 not 1234567 (which is 1111.11070555548 according to my J program). \$\endgroup\$ – Gareth Aug 4 '12 at 23:26
  • \$\begingroup\$ When you say 'do not use anything that will increase the precision of floating points.' does that disqualify arbitrary precision languages (such as bc) \$\endgroup\$ – Matt Aug 4 '12 at 23:32
  • \$\begingroup\$ @Gareth: Yeah, it's probably 12345678, I probably copied it from WA wrong. \$\endgroup\$ – beary605 Aug 4 '12 at 23:53
  • \$\begingroup\$ @Matt: No, as long as you don't use any command that explicitly sets the precision. \$\endgroup\$ – beary605 Aug 4 '12 at 23:55
  • 3
    \$\begingroup\$ What should 2 10000 output? 100? \$\endgroup\$ – Inkbug Aug 5 '12 at 6:54
7
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Python, 143 chars

m,n=map(int,raw_input().split())
d=10**m
n*=d*d
a=0
b=n
while a<b-1:c=(a+b)/2;a,b=[[c,b],[a,c]][c*c>n]
print('%d.%0*d'%(a/d,m,a%d))[:m+(a/d<d)]

Computes the answer by multiplying n by 10^2m, doing an integer square root (using binary search), then "dividing" the result by 10^m.

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  • \$\begingroup\$ This doesn't work for large perfect squares that request small numbers of digits. ex: m=2 n=1000000 will print 100 instead of 10 \$\endgroup\$ – Matt Aug 5 '12 at 18:54
  • \$\begingroup\$ @Matt: fixed... \$\endgroup\$ – Keith Randall Aug 6 '12 at 22:55
5
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Mathematica 66 272 240 chars

New approach

This uses the same, reasonably efficient (59 chars), method for obtaining the smallest useful convergent of Sqrt[n]. It takes a slightly different approach for dividing the numerator by the denominator, accurate to m places.

t = ToString; q = QuotientRemainder;
w = FixedPoint[(# + n/# )/2 &, 1, SameTest -> (Abs[#1 - #2] < 10^(-m) &)];
r = q[Numerator@w, k = Denominator@w];
h[{c_, d_, e_}] := {Append[c, q[d, e][[1]]], 10 q[d, k][[2]], k};
t@r[[1]] <> "." <> t@FromDigits@Nest[h, {{}, 10 r[[2]], k}, m][[1]]

Example: Find the Square root of 5 accurate to 18 places

n=5; m=18;
<run the above code>

(* out *)
"2.236067977499789696"

By the way, the convergent, w, for the above case is given below.

This is still long-winded but it works.


Old approach

The following 59 chars suffice to produce a fraction that will, in decimal form, solve the problem, assuming m, n are entered programmatically:

FixedPoint[(# + n/# )/2 &, 1, SameTest -> (Abs[#1 - #2] < 10^(-m) &)]

When m=18, n=5, here's the fraction:

(* out *)
562882766124611619513723647/251728825683549488150424261

The trick is to convert this fraction into a decimal. The easy way is to use N;

N[%, m+1]
(* out *)
2.236067977499789696

However, N violates the rules by specifying the precision to work with.


Back to the drawing board:

q = FixedPoint[(# + n/# )/2 &, 1, SameTest -> (Abs[#1 - #2] < 10^(-m) &)];
f[{a_, n_, d_}] := 
   With[{q = QuotientRemainder[n, d]}, {Append[a, q[[1]]], q[[2]], d/10}]
   StringInsert[IntegerString@FromDigits@#[[1]],  ".", -1/Log[Denominator@#[[3]], 10]] 
   &[NestWhile[f, {{}, Numerator@q, Denominator@q}, Length@#[[1]] < m &]]

Unfortunately, it takes another 205 characters (by my reckoning) to generate a decimal expression from the fraction. Surely there must be a more direct way to divide one integer by another to m decimal places!

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3
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Haskell 126

main=interact$(\[x,y]->(\s->if '.'`elem`s then(x+1)`take`s else x`take`s)$(show.sqrt.fromIntegral)y++cycle"0").map read.words

Darn sqrt not taking Ints, and fromIntegral being so long!

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1
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C# 41 chars

Math.Sqrt(n).ToString().Substring(0,m+1);
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0
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Version-1

C# 364 Characters (Short Version)

using System;namespace X{class T{static void Main(string[] a){int m=int.Parse(Console.ReadLine()),n=int.Parse(Console.ReadLine());Console.WriteLine("{2}",f(n,m));}static string f(long s,int l){decimal x=s;for(int i=0;i<20;x=(((x*x)+s)/(2*x)),i++);var b=x.ToString(string.Format("F{0}",l));return(x.ToString().Contains("."))?b.Substring(0,l+1):b.Substring(0,l);}}}

code can be ran from OneIDE - http://ideone.com/9tZsD.

C# Normal Version

using System;

namespace X
{
    class T
    {
        static void Main(string[] a)
        {
            int m = int.Parse( Console.ReadLine()), n = int.Parse( Console.ReadLine());
            Console.WriteLine("m:{0}, n:{1} -> {2}", m, n, f(n,m));
        }
        static string f(long s,int l)
        {
            decimal x = s;
            for(int i = 0; i < 20; x = (((x * x) + s) / (2 * x)), i++) ;
            var b=x.ToString(string.Format("F{0}", l));
            return(x.ToString().Contains("."))?b.Substring(0,l+1):b.Substring(0,l);
        }
    }       
}
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  • \$\begingroup\$ It doesn't work for all the test cases: test case #4 (2, 10000) should output 10, not 100.00. As well, #1 and #2's outputs are cut off at 13 digit precision. \$\endgroup\$ – beary605 Aug 6 '12 at 21:02
  • \$\begingroup\$ tx @beary605 for pointing out issues around #4, updated code with new oneide link \$\endgroup\$ – Saumil Aug 6 '12 at 21:57

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