Ka Mok - 1 year ago 38

Ruby Question

The following method generates 81

`Cell`

`row`

`column`

`blocks`

`Cell`

The

`1.times do`

`if`

`until`

`cell_counter`

`def initialize_default_cells`

cell_counter, row, column = 0,0,0

block = 1

until cell_counter == 81

1.times do

break if cell_counter == 0

if cell_counter % 1 == 0

column += 1

end

if cell_counter % 3 == 0

block += 1

end

if cell_counter % 9 == 0

column -= 9

row += 1

block -= 3

end

if cell_counter % 27 == 0

block += 3

end

end

@cells << Cell.new(ROW_ID[row], COLUMN_ID[column], block)

cell_counter += 1

end

end

Answer Source

I concluded that it was easiest to calculate `row`

, `column`

and `block`

from scratch for each `i = 0,..,80`

.

```
def initialize_default_cells
(0..80).each do |i|
@cells << Cell.new(ROW_ID[i/9], COLUMN_ID[i%9], 1 + (i%9)/3 + 3*(i/27))
end
end
```

The key for `COLUMN_ID`

(`i%9`

) is reduced from `i-9*(i/9)`

and the last argument (`1 + (i%9)/3 + 3*(i/27)`

) is reduced from `1 + i/3 - 3*(i/9) + 3*(i/27)`

Consider three examples.

`i=0`

```
@cells << Cell.new(ROW_ID[0/9], COLUMN_ID[0%9], 1 + (0%9)/3 + 3*(0/27))
#=> << Cell.new(ROW_ID[0], COLUMN_ID[0], 1)
```

`i=6`

```
@cells << Cell.new(ROW_ID[6/9], COLUMN_ID[6%9], 1 + (6%9)/3 + 3*(6/27))
#=> << Cell.new(ROW_ID[0], COLUMN_ID[6], 3)
```

`i=29`

```
@cells << Cell.new(ROW_ID[29/9], COLUMN_ID[29%9], 1 + (29%9)/3 + 3*(29/27))
# << Cell.new(ROW_ID[3], COLUMN_ID[2], 4)
```

When `i=6`

, `6/3 #=> 2`

is the number of positive numbers that are divisible by `3`

, `6/9 #=> 0`

is the number of positive numbers that are divisible by `9`

and `6/27 #=> 0`

is the number of positive numbers that are divisible by `27`

. The arguments of `Cell::new`

are then computed with these values.