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.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
6 Comments »

6 Responses to “Base Conversion”

  1. James Curtis-Smith said
    May 2, 2017 at 9:44 AM 
    %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
    May 2, 2017 at 10:26 AM

    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
    May 2, 2017 at 12:19 PM

    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
    May 2, 2017 at 2:15 PM

    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
    May 2, 2017 at 3:19 PM

    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
    May 3, 2017 at 9:40 PM

    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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