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As an intro, I know how the numbers are represented, how to do it if I can calculate powers of the base, and then move between base $m$ to base $10$ to base $n$. I feel that this is overly "clunky" though, and would like to do it in such a way that the following conditions are met:

  1. No need to calculate the powers of the base explicitly
  2. No need for intermediate storage (i.e. no conversion to base ten required if base ten is not one of the bases)

I am pretty sure that the only operations that I strictly need to use are modulo, division and concatenation, but I can't seem to figure it out.

Any pointers?

3 Answers 3

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Let $x$ be a number. Then if $b$ is any base, $x \% b$ ($x$ mod $b$) is the last digit of $x$'s base-$b$ representation. Now integer-divide $x$ by $b$ to amputate the last digit.

Repeat and this procedure yields the digits of $x$ from least significant to most. It begins "little end first."

EDIT: Here is an example to make things clear.

Let $x = 45$ and $b = 3$.

x   x mod 3 45    0 15    0                (integer divide x by 3)   5    2  1    1 

We see that $45 = 1200_3$. Read up the last column to get the base-3 expansion you seek. Let us check.

$1\cdot 3^3 + 2\cdot 3^2 + 0 + 0 = 27 + 18 = 45.$

I hope this helps you.

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    Math Gems has a nice exposition below. BTW, +1 for you Math Gems.2012-02-20
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You can perform base conversion directly by representing radix notation in horner (nested) form. Let's work a simply example. We convert $\:1213_{\:6}\:$ from radix $6$ to radix $8$

$ 1{\color{red}2}{\color{blue}1}{\color{orange}3}_{\:6}\ =\ ((1\cdot 6+{\color{red}2})\:6+{\color{blue}1})\:6 + {\color{orange}3}$

Now perform the computation inside-out in radix $8$:

$ 1\cdot 6+ {\color{red}2} = 10)\: 6 = 60) + {\color{blue}1}) = 61)\: 6 = 446) + {\color{orange}3} = 451$

Hence $\:1213_{\:\!6} = 451_{8}$

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To convert from one base to another is pretty simple and will work for any base:

value = 1024  base 2: log 1024 / log 2 = 10 ; 2 ^ 10 = 1024   base = 10 ^ ( log 1024 / 10 ) = 2  base 10: log 1024 / log 10 = 3.0103 ; 10 ^ 3.0103 = 1024   base = 10 ^ ( log 1024 / 3.0103 ) = 10  base 6: log 1024 / log 6 = 3.8685 ; 6 ^ 3.8685 = 1024   base = 10 ^ ( log 1024 / 3.8685 ) = 6  base x: log VALUE / log x = y ; x ^ y = VALUE  x = 10 ^ ( log VALUE / y ) 

To do this in C++ : http://www.cplusplus.com/reference/cmath/log10/

#include       /* printf */ #include        /* log10 */ int main () {   double result;   result = log10 (1024) / log10 (2);   printf ("log10 (1024) / log10 (2) = %f\n",  result );   printf ("2 ^ %f = %f\n",  result, 2.0 ^ result );   return 0; } 
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    I just used the change of base formula: http://www.mathwords.com/c/change_of_base_formula.htm log2 1024 = log10 1024/log10 2 = 10; 2^10 =1024 (bin: 1 with 8 0's) log2 2048 = log10 2048/log10 2 = 11; 2^11 = 2^10 + 2^10 = 2(1024)=2048 (bin: 1 with 9 0's) log2 1448.155 = log10 1448.155/log10 2= ~10.5 = (2^10)(2^0.5) =1448.155 (bin: 10110101000) = 2^3+2^5+2^7+2^8+2^10=1448 Using this formula is how you represent a number in an arbitrary base. For example: 2^A = 1024; A=log10 1024/ log10 2=10 B^C = 1024; B=10^((log10 C)/C)=2 (It's just basic algebra)2018-04-03