Base Conversion

May 2, 2017

I continue cleaning out my list of saved homework questions:

Given a number represented as a string in base 2, convert the number to a string in base 4. For instance, the number 110110002 = 31204.

Your task is to write a program that converts numbers from base 2 to base 4; for extra credit, write a program that converts from any base to another. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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6 Responses to “Base Conversion”

  1. %X=qw(00 0 01 2 10 1 11 3);
    sub base2to4 {
      return join q(),
             reverse
             map { $X{$_}//$_ }
             ( (reverse $_[0]) =~ m{(..?)}g );
    }
    
    print base2to4('11011000'),"\n";'
    
  2. Rutger said

    In Python.

    def base10toBase(num, base):
        result = ''
        current = num
        while current != 0:
            remainder = current%base
            result = str(remainder)+result
            current = current//base
        return result
    
    def baseBaseto10(num, base):
    	result = 0
    	exp = 1
    	for d in reversed(num):
    		result += int(d) * exp 
    		exp *= base
    	return result
    
    
    for i in [1, 11, 101, 1101, 1111, 1000]:
    	print(i, 'base2 = \n', base10toBase(baseBaseto10(str(i), 2),4), 'base4\n', baseBaseto10(str(i), 2), 'base10\n',)
    
    #outputs:
    #   1 base2 = 
    #   1 base4
    #   1 base10
    #
    #   11 base2 = 
    #    3 base4
    #    3 base10
    #
    # 101 base2 = 
    #   11 base4
    #     5 base10
    #
    #1101 base2 = 
    #    31 base4
    #    13 base10
    #
    #1111 base2 = 
    #   33 base4
    #   15 base10
    #
    #1000 base2 = 
    #    20 base4
    #      8 base10
    
  3. Paul said

    Up to base 64, so we cover also the Babylonians.

    import string
    from functools import reduce
    
    # up to base 62 (10 + 26 + 26)
    digits = string.digits + string.ascii_lowercase + string.ascii_uppercase
    N = len(digits)
    D = dict(enumerate(digits))
    I = {v:k for k, v in D.items()}
    
    def check_base(base):
        if not 2 <= base <= N:
            raise ValueError("base must be in between {} and {}".format(2, N))
    
    def tobase10(astring, base=2):
        check_base(base)
        return reduce(lambda acc, i: acc*base + I[i], astring, 0)
    
    def frombase10(anumber, base=2):
        check_base(base)
        x, result = anumber, []
        while x:
            x, m = divmod(x, base)
            result.append(D[m])
        return "".join(reversed(result))
    
    def convert_base(s, base_from, base_to):
        return frombase10(tobase10(s, base_from), base_to)
    
    print(int("11011000", 2))  # 216
    print(convert_base("11011000", 2, 60))  # 3A
    print(convert_base("3A", 60, 2))  # 11011000
    print(convert_base("3A", 60, 10))  #  216
    
  4. Jussi Piitulainen said

    A web search indicates to me that the English word for base 4 is “quaternary”.

    ;; cannot bother with strings - put the digits as numbers in a vector
    ;; in their usual order, the most significant digit first - assume
    ;; base greater than one, and no empty vectors so that zero is a bit
    ;; special - also, error checking is inadequate, as usual
    
    (define (vector->natural vec base)
      (do ((nat 0 (+ (* base nat) (vector-ref vec pos)))
           (pos 0 (+ pos 1)))
          ((= pos (vector-length vec)) nat)))
    
    ;; first I had remainder and quotient the wrong way around
    
    (define (natural->vector nat base)
      (do ((lis '() (cons (remainder nat base) lis))
           (nat nat (quotient nat base)))
          ((zero? nat)
           (if (null? lis)
               (vector 0)
               (list->vector lis)))))
    
    (write (natural->vector (vector->natural (vector 1 1 0 1 1 0 0 0) 2) 4)) (newline)
    (write (natural->vector (vector->natural (vector 3 1 2 0) 4) 2)) (newline)
    (write (natural->vector (vector->natural (vector 0) 10) 10)) (newline)
    
    (write (vector->natural (natural->vector 31415926 3) 3)) (newline)
    
  5. john said

    C11:


    #include <stdio.h>
    #include <stdlib.h>

    int main(int argc, char **argv) {
      int ch;
      
     initial: ch = getchar();
      if (ch == '0') goto i0;
      if (ch == '1') goto i1;
      else goto end;
      
     i0: ch = getchar();
      if (ch == '0') goto i00;
      if (ch == '1') goto i01;
      else goto end;

     i1: ch = getchar();
      if (ch == '0') goto i10;
      if (ch == '1') goto i11;
      else goto end;

     i00: putchar('0');
      goto initial;

     i01: putchar('1');
      goto initial;

     i10: putchar('2');
      goto initial;

     i11: putchar('3');
      goto initial;
      
     end: puts("");
      exit(0);
    }

  6. Steve said

    MUMPS solution:

    BASECHG (STR,BASE1,BASE2) ; Convert number to different base
    N ACC,I,LEN,REM,REMS,STR2
    S LEN=$L(STR),ACC=0
    F I=1:1:LEN-1 S ACC=ACC+$E(STR,I),ACC=ACC*BASE1
    S ACC=ACC+$E(STR,LEN)
    I ACC<BASE2 S STR2=ACC
    E D
    . S I=0 F Q:ACC<BASE2 S REM=ACC#BASE2,ACC=ACC\BASE2,I=I+1,REMS(I)=REM
    . S STR2=ACC,I="" F S I=$O(REMS(I),-1) Q:I="" S STR2=STR2_REMS(I)
    W !,STR," (",BASE1,") = ",STR2," (",BASE2,")"
    Q

    D ^BASECHG("11011000",2,4)

    11011000 (2) = 3120 (4)

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