FIXED

Hi.

I know this is an odd request, but can I, using only operators, take a number counting up as shown:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20..80,81 (the variable i)

and output a sequence as shown?

1,1,1,2,2,2,3,3,3,1,1,1,2,2,2,3,3,3,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,4,4,4,5,5,5,6,6,6,4,4,4,5,5,5,6,6,6,7,7,7,8,8,8,9,9,9,7,7,7,8,8,8,9,9,9,7,7,7,8,8,8,9,9,9

Thanks,
~Molive

Actually your request doesn't belong to help with scripts, it is pretty much belonging to help with maths. Try it in Advanced Topics.

asivi wrote:

Actually your request doesn't belong to help with scripts, it is pretty much belonging to help with maths. Try it in Advanced Topics.

ok.

asivi wrote:

Actually your request doesn't belong to help with scripts, it is pretty much belonging to help with maths. Try it in Advanced Topics.

ok.
https://scratch.mit.edu/discuss/topic/215806/

The expression “(floor ((i-1)/27) )*3 + 1” gives you 27 ones followed by 27 fours, then 27 sevens, etc.

Now you need to add zero, one or two, going up by one every three items, but back down to zero after 9 items.

So add something like “(floor ((i-1)/3)) mod 3” to above.


EDIT: oops, started typing this before replies above…

Here, done it in scratchblocks:

((( [floor v] of (((i)-(1))/(27)) ) * (3)) + ( (([floor v] of (((i)-(1))/(3)) ) mod (3)) + (1) ) )

Hope I got all the brackets in the right places!

TheLogFather wrote:

The expression “(floor ((i-1)/27) )*3 + 1” gives you 27 ones followed by 27 fours, then 27 sevens, etc.

Now you need to add zero, one or two, going up by one every three items, but back down to zero after 9 items.

So add something like “(floor ((i-1)/3)) mod 3” to above.


EDIT: oops, started typing this before replies above…

Here, done it in scratchblocks:

((( [floor v] of (((i)-(1))/(27)) ) * (3)) + ( (([floor v] of (((i)-(1))/(3)) ) mod (3)) + (1) ) )

Hope I got all the brackets in the right places!

Looks right, Thanks

TheLogFather wrote:

The expression “(floor ((i-1)/27) )*3 + 1” gives you 27 ones followed by 27 fours, then 27 sevens, etc.

Now you need to add zero, one or two, going up by one every three items, but back down to zero after 9 items.

So add something like “floor ( (i-1)/3 mod 3 )” to above.


EDIT: oops, started typing this before replies above…

Here, done it in scratchblocks:

((( [floor v] of (((i)-(1))/(27)) ) * (3)) + ( (([floor v] of (((i)-(1))/(3)) ) mod (3)) + (1) ) )

Hope I got all the brackets in the right places!

Lol, i can't believe this response being so quickly, obviously you are a great mathematician.
Greetings.

So now I'm curious - why would you want that odd looking output sequence?

Paddle2See wrote:

So now I'm curious - why would you want that odd looking output sequence?

me too

Paddle2See wrote:

So now I'm curious - why would you want that odd looking output sequence?

When wrapped around in nines it looks like this:

1 1 1 2 2 2 3 3 3 /n
1 1 1 2 2 2 3 3 3 /n
1 1 1 2 2 2 3 3 3 /n
4 4 4 5 5 5 6 6 6 /n
4 4 4 5 5 5 6 6 6 /n
4 4 4 5 5 5 6 6 6 /n
7 7 7 8 8 8 9 9 9 /n
7 7 7 8 8 8 9 9 9 /n
7 7 7 8 8 8 9 9 9 /n

It's for a Sudoku solver - I can take a number's position in a list and know what box it's in.

https://scratch.mit.edu/projects/119524351/

Now released:
https://scratch.mit.edu/projects/119524351/