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## Pascal's triangle

Pascal's triangle is a triangular array of binomial coefficients: cell k of row n indicates how many combinations exist of n things taken k at a time. It looks like this

- 1
- 1 1
- 1 2 1
- 1 3 3 1
- 1 4 6 4 1
- ...

One can prove that a given cell is the sum of 2 cells from previous row: the one just above and the one on top left.

Given that, let's see how we can generate the Nth row of Pascal's triangle

## Let's start

- Memory: empty
- Cursor: first cell
- Input: a decimal number N (greater than 0)

## Process

- Read a decimal number
- Init
*line 0*: just 1 - While counter is not null
- Decrease counter
- Generate next line
- Go to row last cell (e.g. mem : A B C D 0 0 0
- Move it to the next 2 cells (e.g. mem : A B C 0 D D)
- Go to previous cell and loop
- mem : A B 0 C C+D D
- then A 0 B B+C C+D D
- and finally 0 A A+B B+C C+D D

- Move all values back on the left

- Print last line, comma separated. Line is symetric : can be print from right to left as well
- Print '1' and ignore first value (it's always 1)
- While there is an item
- Print comma
- Move / Print item
- Go left

- Loop

## Code

```
>,[>++++++[-<-------->]>+++++++++[- read an integer
<<<[->+>+<<]>>[-<<+>>]>]<<[-<+>],] ** part 2 **
>+ Initialize first row
<<[ while counter is not null
- decrease counter
>>[>]< go to last cell
[[->+>+<<]<] while there is a cell: move to the next 2 ones on the left and repeat
>>[[-<+>]>] Move whole array to the left
<<[<]< back to counter
] loop
>+++++++[-<+++++++>]<.----- print 49 (='1') and change into comma (44)
>>[>]<-< go to last cell (1) then reset; then go to previous cell
[ for each cell in the row
[<]<.>>[>]< print comma (first memory cell) then go back to the current cell
[->+<]> move to the right (at least one 0 needed to print a decimal value)
[>>>>++++++++++<<<<[->+>>+>-[<-]<[- Print decimal value
>>+<<<<[->>>+<<<]>]<<]>+[-<+>]>>>[- ** part 2 **
]>[-<<<<+>>>>]<<<<]<[>++++++[<+++++ ** part 3 **
+++>-]<-.[-]<] ** part 4 **
<]
```

## Minified version

```
>,[>++++++[-<-------->]>+++++++++[-<<<[->
+>+<<]>>[-<<+>>]>]<<[-<+>],]>+<<[->>[>]<[
[->+>+<<]<]>>[[-<+>]>]<<[<]<]>+++++++[-<+
++++++>]<.----->>[>]<-<[[<]<.>>[>]<[->+<]
>[>>>>++++++++++<<<<[->+>>+>-[<-]<[->>+<<
<<[->>>+<<<]>]<<]>+[-<+>]>>>[-]>[-<<<<+>>
>>]<<<<]<[>++++++[<++++++++>-]<-.[-]<]<]
```

## Final state

- Memory: 44, 0, 0, 0
- Cursor: on second cell
- Input: read (empty)
- Output: Nth row from Pascal's triangle (modulo 256)

*Note*: because of the nature of the algorithm, if a cell equals 0 on a row it will break the loop. And modulo 256, a cell can actually be null.

However, the first cell that will be a multiple of 256 in standard Pascal's triangle appears on row 256, and the counter itself, from user input, cannot be more than 255.

In other words: this algorithm is valid as long as we rely on a counter. Having an infinite loop to generate more than 255 rows will fail. We can of course introduce 2-cell-long arrays but it's not really important here.