Scratch

I'd do it with a formula, rather.

Being ignorant in this subject my way would be count upwards in decimal and add a digit every "base" numbers, but there must be a faster, easier way. Google could help!

Here's a script:

when gf clicked
delete (all v) of [remainders v]
ask [Give a decimal number] and wait
set [number v] to (answer)
ask [Choose a base to convert to] and wait
set [base v] to (answer)
repeat until <(number) = [0]>
	set [rounded down v] to [] //mini script from wiki
	set [count v] to [0]
	repeat until <<(letter (count) of ((number) / (base))) = [.]> or <(count) = (length of ((number) / (base)))>>
		set [rounded down v] to (join (rounded down) (letter (count) of ((number) / (base))))
		change [count v] by (1) //end mini script
	end
	add (item (((number) mod (base)) + (1)) of [in base v]) to [remainders v]
	set [number v] to (rounded down)
end
set [count v] to (length of [remainders v]) //mini script to join digits
set [new num v] to []
repeat (length of [remainders v])
	set [new num v] to (join (new num)(item (count) of [remainders v]))
	change [count v] by (-1)
end
say (join (join([Your number in base](base))[ is ](new num))

Does that work for you?
You'll need a list called "in bases" that has all the characters in the bases. Item 1 of the list should be 0.

BTW, Google helped. XD

Last edited by scimonster (2012-02-02 14:29:01)

I made it work with conversion to binary, but it still isn't working with base 3.

demosthenes wrote:

http://i.imgur.com/QUlJy.gif

Am I seeing a screenshot there 

when green flag clicked
set [base v] to (answer)
ask [Input a number] and wait
set [number v] to (answer)
delete (all v) of [value v]
repeat (10)
  add [0] to [value v]
end
set [difference v] to (number)
repeat until <(difference) = [0]
  set [index v] to [0]
  set [temp v] to [1]
  repeat until <(temp) > (difference)>
    set [temp v] to ((base) * (temp))
    change [index v] by (1)
  end
  replace item (index) of [value v] with (round (((difference) - ((difference) mod ((temp) / (base)))) / ((temp) / (base))))
  set [difference v] to ((difference) - ((item (index) of [value v])*((temp) / (base))))
end

scimonster wrote:

Did you see my post? It works for anything up to base 36. It would be an even shorter script of Scratch has a (round [up/down v] ()) block.

Yes, but it's just too long...

scimonster wrote:

Did you see my post? It works for anything up to base 36. It would be an even shorter script of Scratch has a (round [up/down v] ()) block.

You can just put:

(round ((value) + (0.5)))

and

(round ((value) - (0.5)))

JSO wrote:

scimonster wrote:

Did you see my post? It works for anything up to base 36. It would be an even shorter script of Scratch has a (round [up/down v] ()) block.

Rounding up or down (called 'floor' or 'ceil' in a lot of programming languages) can already be done in Scratch 

The solution is actually fairly simple:

say (round ((value) + (0.5))) //round up
say (round ((value) - (0.5))) //round down

I know what it is in other languages.    Cool, thanks for the workaround. 

RedRocker227 wrote:

scimonster wrote:

Here's a script:

when gf clicked
delete (all v) of [remainders v]
ask [Give a decimal number] and wait
set [number v] to (answer)
ask [Choose a base to convert to] and wait
set [base v] to (answer)
repeat until <(number) = [0]>
	set [rounded down v] to [] //mini script from wiki
	set [count v] to [0]
	repeat until <<(letter (count) of ((number) / (base))) = [.]> or <(count) = (length of ((number) / (base)))>>
		set [rounded down v] to (join (rounded down) (letter (count) of ((number) / (base))))
		change [count v] by (1) //end mini script
	end
	add (item (((number) mod (base)) + (1)) of [in base v]) to [remainders v]
	set [number v] to (rounded down)
end
set [count v] to (length of [remainders v]) //mini script to join digits
set [new num v] to []
repeat (length of [remainders v])
	set [new num v] to (join (new num)(item (count) of [remainders v]))
	change [count v] by (-1)
end
say (join (join([Your number in base](base))[ is ](new num))

Does that work for you?
You'll need a list called "in bases" that has all the characters in the bases. Item 1 of the list should be 0.

BTW, Google helped. XD

How long did it take you to figure that out?! D:

Oh, 5 minutes, once I had the algorithm. 

Here's an updated version of the script:

when gf clicked
delete (all v) of [remainders v]
ask [Give a decimal number] and wait
set [number v] to (answer)
ask [Choose a base to convert to] and wait
set [base v] to (answer)
repeat until <(number) = [0]>
	add (item (((number) mod (base)) + (1)) of [in base v]) to [remainders v]
	set [number v] to (round ((number) - (0.5)))
end
set [count v] to (length of [remainders v]) //mini script to join digits
set [new num v] to []
repeat (length of [remainders v])
	set [new num v] to (join (new num)(item (count) of [remainders v]))
	change [count v] by (-1)
end
say (join(join [Your number in base](base))(join [ is ](new num)))

Last edited by scimonster (2012-02-04 12:54:40)

wilsonbiggs wrote:

I'm doing something similar, building a base conversion block in BYOB... it keeps outputting 0.

base (base =13) 2 digit number (orignum =54) to base 10
set [baseout v] to <<letter (1) of (orignum)> + <(base)*<letter (2) of (orignum)>>>
report baseout

Try using my script.