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Ed H
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Joined: 30 Nov 2007
Posts: 138
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 Posted: 17 Oct 2008 12:00:10 am Post subject: |
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The program takes the digits of a number from a list, and the base of the number from a variable, and returns the number.
Inputs:
B - the base of the input number
Ans - a list containing the digits of the input number in reverse order
Output:
Ans - the value of the number inputted
Example:
With 5 in B, and {1,2,3,4} in Ans, return 586 = 1 + 2*5 + 3*25 + 4*125.
Goal: 20 bytes
Last edited by Guest on 11 Dec 2008 03:22:29 pm; edited 1 time in total |
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bfr
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Joined: 13 Feb 2006
Posts: 108
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| Posted: 17 Oct 2008 08:42:19 pm Post subject: |
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20 bytes:
Quote:
sum(AnsB‾¹B^(cumSum(Ans/Ans
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Ed H
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Joined: 30 Nov 2007
Posts: 138
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| Posted: 17 Oct 2008 09:48:04 pm Post subject: |
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This is what I had:
Quote: [font="Courier New"]:sum(Ans/BB^cumSum(Ans or 1 |
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calc84maniac
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Joined: 22 Jan 2007
Posts: 770
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| Posted: 18 Oct 2008 08:37:48 am Post subject: |
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bfr wrote: 20 bytes:
Quote:
sum(AnsB‾¹B^(cumSum(Ans/Ans
[post="128005"]<{POST_SNAPBACK}>[/post]
That would give an error if any of the digits are zero.  |
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Ed H
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Joined: 30 Nov 2007
Posts: 138
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| Posted: 18 Oct 2008 10:33:13 am Post subject: |
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| Good catch, calcmaniac84. |
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elfprince13
Retired
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Joined: 11 Apr 2005
Posts: 3500
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| Posted: 18 Oct 2008 12:23:05 pm Post subject: |
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17 bytes, assuming everything is a 1 byte token.
Quote:
"B^(n-1->u
sum(Ansu(1,dim(Ans
Last edited by Guest on 18 Oct 2008 12:30:08 pm; edited 1 time in total |
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bfr
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Joined: 13 Feb 2006
Posts: 108
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| Posted: 18 Oct 2008 12:26:55 pm Post subject: |
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That returns a list, btw
EDIT: Nevermind, your new code doesn't return a list anymore. That still doesn't give the right answer, though, because you're doing: [digit]*B*1+[digit]*B*2... etc. instead of [digit]*B*0+[digit]*B*1
Without cumSum(, something like that would be a pretty nice solution though
Last edited by Guest on 18 Oct 2008 12:32:31 pm; edited 1 time in total |
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elfprince13
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| Posted: 18 Oct 2008 12:31:16 pm Post subject: |
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my wireless pooped out right in the middle of fixing that with an edit  or I would have beat you to it. |
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Ed H
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Joined: 30 Nov 2007
Posts: 138
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| Posted: 18 Oct 2008 12:33:40 pm Post subject: |
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elfprince13 wrote: 14 bytes, assuming everything is a 1 byte token.
Quote:
"B^n->u
sum(Ansu(1,dim(Ans
[post="128033"]<{POST_SNAPBACK}>[/post]
That returns the answer times B. Somewhere, you need to include a B-1. Also, the way we usually count program sizes is by taking its size from the Mem Mgmt menu and subtracting the length of the program title, so your program takes 27 bytes, by that standard.
Last edited by Guest on 22 Jul 2010 12:29:22 pm; edited 1 time in total |
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elfprince13
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| Posted: 18 Oct 2008 10:50:11 pm Post subject: |
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haha, yeah, I just didn't have a calculator or emulator with me so I was doing that in my head, and I don't know how many bytes all the tokens I used are. I really just felt a need to exploit that cool u/v/w() thing DarkerLine posted.
also, you missed my edit with the fixed version. |
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Ed H
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Joined: 30 Nov 2007
Posts: 138
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| Posted: 20 Oct 2008 06:33:36 pm Post subject: |
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Using the sequence functions in that way only really helps if you're using seq( multiple times with the same expression but different start and end values. It is very cool, though - a useful substitute for seq( in sum situations.
However, in this situation, seq( is more efficient than the sequence functions. You could do
Quote: :sum(Ans/BB^seq(X,X,1,dim(Ans and that would be 27 bytes, compared to 30 with your program.
The real point of this challenge was to use [font="Courier new"]cumSum(Ans or 1 as a replacement for seq(X,X,1,dim(Ans, which gives the smallest size at 20 bytes.
Last edited by Guest on 22 Jul 2010 12:28:36 pm; edited 1 time in total |
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DarkerLine
ceci n'est pas une |
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Posts: 8328
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| Posted: 20 Oct 2008 11:03:08 pm Post subject: |
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| In general, cumSum(Ans=Ans) is more effective than cumSum(Ans or 1); although it's the same size, it won't fail for complex inputs. I suppose it doesn't really matter here, though. |
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TI-Basic Brain Teasers => TI-BASIC
