### Open Source Your Knowledge, Become a Contributor

Technology knowledge has to be shared and made accessible for free. Join the movement.

## Euclidean division

Now we are able to read an interger. The next step would be ability to *print* an integer.

However, we first need to have an algorithm for euclidean division in order to implemented the "print integer" program.

Euclidean division is an operation on A (dividend) and B (divisor) that returns Q (quotient) and R (remainder), with

- A=B * Q + R
- 0 <= R < B

In BrainFuck, this can be implemented like this

- Decrease dividend
- Increase remainder
- Decrease divisor
- If divisor = 0 then
- note that now, remainder = initial divisor
- so: rebuild initial divisor from remainder
- reset remainder
- increase quotient

and repeat as long as dividend is not null. When it reaches 0, then we have the remainder, the quotient,....

## Let's start

- Memory: A, 0, 0, 0, B, 0
- Cursor: on A
- Input: any

## Process

*Invariant : memory = dividend, remainder, 0, else flag, divisor, quotient with B = divisor + remainder and A = quotient * B + dividend + remainder*- while dividend is not null
- decrease dividend
- increase remainder
- set else flag
- decrease divisor
*invariant is still valid here*- If divisor is not null reset else flag
- Go left
- divisor not null : between remainder and else flag (0)
- divisor null : on else flag (1)

- if not null (divisor null, on else flag)
- rebuild divisor (move remainder to divisor)
- increment quotient
- reset else bit
- move left

*again invariant is still valid here*- In all cases we are between remainder and else flag

- Loop
*Memory = 0, remainder, 0, 0, divisor, quotient; with A = quotient * B + remainder*

## Code

```
[ while dividend is not null
- decrease dividend
>+ increase remainder
>>+ set else flag
>- decrease divisor
[<-] if divisor is not null reset else flag
<[ and if divisor is null
<<[->>>+<<<] rebuild divisor
>>>>+ increase quotient
<<-<] move between remainder and else flag (reset it btw)
<<] loop
```

## Minified version

```
[->+>>+>-[<-]<[<<[->>>+<<<]>>>>+<<-<]<<]
```

## Final state

- Memory: 0, R, 0, 0, B', Q with A = B * Q + R and B = B'+R
- Cursor: on first cell
- Input: unchanged
- Output: unchanged

#Test program

This code divides 131 by 2 and print quotient (65, ASCII code A) as many times as the remainder

```
>++++++++++[-<+++++++++++++>]<+ (131)
>>>>++<<<< (2)
[->+>>+>-[<-]<[<<[->>>+<<<]>>>>+<<-<]<<] division
>[>>>>.<<<<-] and print quotient as many times as the remainder
```

Result : A

You can replace first two lines by these examples

- +++++[->+++++<]->[-<++++++>]>>>+++<<<< : 149 = 49 x 3 + 2, it prints "1" 2 times
- ++++++++[->++++<]+++>[-<+++++>]>>>++++<<<< : 163 = 40 x 4 + 3, it prints "(" 3 times
- etc...

Suggested playgrounds

Open Source Your Knowledge: become a Contributor and help others learn. Create New Content