5
\$\begingroup\$

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.

\$\endgroup\$
8
  • 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, 2012 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, 2012 at 23:32
  • \$\begingroup\$ @Gareth: Yeah, it's probably 12345678, I probably copied it from WA wrong. \$\endgroup\$
    – beary605
    Aug 4, 2012 at 23:53
  • \$\begingroup\$ @Matt: No, as long as you don't use any command that explicitly sets the precision. \$\endgroup\$
    – beary605
    Aug 4, 2012 at 23:55
  • 3
    \$\begingroup\$ What should 2 10000 output? 100? \$\endgroup\$
    – Inkbug
    Aug 5, 2012 at 6:54

6 Answers 6

7
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\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, 2012 at 18:54
  • \$\begingroup\$ @Matt: fixed... \$\endgroup\$ Aug 6, 2012 at 22:55
5
\$\begingroup\$

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!

\$\endgroup\$
3
\$\begingroup\$

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!

\$\endgroup\$
1
\$\begingroup\$

C# 41 chars

Math.Sqrt(n).ToString().Substring(0,m+1);
\$\endgroup\$
0
\$\begingroup\$

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);
        }
    }       
}
\$\endgroup\$
2
  • \$\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, 2012 at 21:02
  • \$\begingroup\$ tx @beary605 for pointing out issues around #4, updated code with new oneide link \$\endgroup\$
    – Saumil
    Aug 6, 2012 at 21:57
0
\$\begingroup\$

Pascal, 142 bytes

This full program requires a processor supporting features of “Extended Pascal” (ISO standard 10206). The program will crash if m is negative.

program p(input,output);var n,m:integer;p:string(maxInt);begin read(m,n);writeStr(p,sqrt(abs(n)):1:m);write(p:m+ord(index(p,'.')in[1..m]))end.

Ungolfed:

program findFirstNDigitsOfSquareRootOfANumber(input, output);
    var
        number, digit: integer;
        print: string(maxInt);
    begin
        readLn(digit, number);
        
        { `Sqrt` only works on non-negative integers and always returns
          the principal root as a `real` value. The `writeStr` routine is
          defined by Extended Pascal. The `:1` specifies the minimum width
          of the given argument. The write routines in combination with a
          `real` argument accept a second format specifier (here `:digit`)
          that disables scientific notation and determines the number of
          post-decimal (i.e. fractional) digits. }
        writeStr(print, sqrt(abs(number)):1:digit);
        
        { The format specifier `:digit…` in combination with a `string`
          argument specifies the _exact_ width. We will need to take
          account of the radix mark (`'.'`) _if_ it is printed, though. }
        writeLn(print:digit + ord(index(print, '.') in [1..digit]));
    end.

It is not possible to request maxInt digits since one char value will be occupied by the radix mark (.). You will probably need to replace maxInt by a sufficiently small natural number, though.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.