Question

This is probably a quite exotic question.

My Problem is as follows:

The TI 83+ graphing calculator allows you to program on it using either Assembly and a link cable to a computer or its built-in TI-BASIC programming language.

According to what I’ve found, it supports only 16-Bit Integers and some emulated floats.

I want to work with a bit larger numbers however (around 64 bit), so for that I use an array with the single digits:

{1, 2, 3, 4, 5}

would be the Decimal 12345.

In binary, that’s 110000 00111001, or as a binary digit array:

{1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1}

which would be how the calculator displays it.

How would i go about converting this array of decimal digits (which is too large for the calculator to display it as a native type) into an array of decimal digits?

Efficiency is not an issue. This is NOT homework.

This would leave me free to implement Addition for such arrays and such.

thanks!

Answer 1

Thought about it and I think I would do it with the following ‘algorithm’

• check the last digit (5 in the example case)

• if it is odd, store (from the reverse order) a 1 in the binary array

• now divide the number by 2 through the following method:

• begin with the first digit and clear the ‘carry’ variable.
• divide it by 2 and add the ‘carry’ variable. If the remainder is 1 (check this before you do the divide with an and&1) then put 5 in the carry
• repeat untill all digits have been done

repeat both steps again untill the whole number is reduced to 0’s.

the number in your binary array is the binary representation

your example:
1,2,3,4,5

• the 5 is odd so we store 1 in the binary array: 1
• we divide the array by 2 using the algorithm:
• 0,2,3,4,5 => 0,1+5,3,4,5 => 0,6,1,4,5 => 0,6,1,2+5,5 => 0,6,1,7,2

and repeat:

0,6,1,7,2 last digit is even so we store a 0: 0,1 (notice we fill the binary string from right to left)

etc

you end up with a binary

EDIT:
Just to clarify above: All I’m doing is the age old algorithm:

 int value=12345;
 while(value>0)
 {
      binaryArray.push(value&1);
      value>>=1;     //divide by 2
 }

except in your example we don’t have an int but an array which represents a (10 base) int ;^)