Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.lang.javascript > #24895 > unrolled thread

Bijective basechanger

Started byjonas.thornvall@gmail.com
First post2014-06-17 06:42 -0700
Last post2014-07-22 12:16 -0700
Articles 20 on this page of 60 — 6 participants

Back to article view | Back to comp.lang.javascript

Contents

  Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 06:42 -0700
    Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-17 12:08 -0700
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 13:23 -0700
        Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 13:24 -0700
        Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-17 13:54 -0700
          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 14:07 -0700
            Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 14:15 -0700
              Re: Bijective basechanger John Harris <niam@jghnorth.org.uk.invalid> - 2014-06-23 17:23 +0100
          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 14:23 -0700
          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-17 14:41 -0700
            Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-17 20:28 -0700
              Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 13:04 -0700
                Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-18 16:30 -0700
                  Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 17:45 -0700
                    Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 17:49 -0700
                    Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-19 11:21 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-19 14:44 -0700
                        Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-19 14:48 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-19 14:55 -0700
                        Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-19 19:59 -0700
                          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-21 09:25 -0700
                          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-21 11:15 -0700
                  Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 18:05 -0700
                    Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 18:19 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 18:25 -0700
                    Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-18 18:46 -0700
              Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-23 16:00 -0700
                Re: Bijective basechanger "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-06-23 17:51 -0700
    Re: Bijective basechanger John Harris <niam@jghnorth.org.uk.invalid> - 2014-06-23 17:18 +0100
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-23 11:27 -0700
    Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-06-23 15:29 -0700
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 03:26 -0700
        Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 03:28 -0700
        Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 04:24 -0700
          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 05:36 -0700
          Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-06-24 13:05 -0700
            Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 14:13 -0700
              Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-25 05:36 -0700
              Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-06-25 12:45 -0700
                Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-06-25 13:15 -0700
                  Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-01 02:39 -0700
                    Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-07-03 07:33 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-03 09:54 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-04 02:14 -0700
                        Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-07-05 12:12 -0700
                          Re: Bijective basechanger "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2014-07-05 22:19 +0200
                            Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-07-05 13:50 -0700
                          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-06 11:11 -0700
                            Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-06 14:59 -0700
                              Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-06 15:00 -0700
                              Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-06 15:02 -0700
                          Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-16 23:31 -0700
                      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-07-04 02:28 -0700
                        Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-07-05 12:14 -0700
            Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-25 22:38 -0700
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 03:32 -0700
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 03:44 -0700
      Re: Bijective basechanger jonas.thornvall@gmail.com - 2014-06-24 11:10 -0700
        Re: Bijective basechanger Scott Sauyet <scott.sauyet@gmail.com> - 2014-06-24 11:45 -0700
    Re: Bijective basechanger "Chris M. Thomasson" <no@spam.invalid> - 2014-07-22 12:16 -0700

Page 1 of 3  [1] 2 3  Next page →

#24895 — Bijective basechanger

Fromjonas.thornvall@gmail.com
Date2014-06-17 06:42 -0700
SubjectBijective basechanger
Message-ID<e8847026-67f9-417c-9166-d1368bb9a704@googlegroups.com>
10=A 
100=9A 
1000=99A 
... 
20=1A 
30=2A 
30=3A 
... 
101=A1 
200=19A 
201=1A1 
... 
2000=199A 
2001=19A1 
2010=19AA 
2014=1A14 


1091020134=????????? Brainteaser 

A generic basechanger for anybase is rather easy to accomplish, but i am a bit stumped where to start with doing conversion into bijective bases, i was thinking counting up the base multiples upto the number to convert, and bone it out from there saving the differences. But i just can't recall howto do it. 

Is there an easy obvious way todo this? 
I slightly remember there is something about using pairs, but that may have been for ordinary base conversion. 

Who have the fastest head doing the conversion.

[toc] | [next] | [standalone]

#24896

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-17 12:08 -0700
Message-ID<60b6a006-1e9e-4b16-8056-243992faedcb@googlegroups.com>
In reply to#24895
On Tuesday, June 17, 2014 8:42:28 AM UTC-5, jonas.t...@gmail.com wrote:
> 10=A 
> 100=9A
> 1000=99A 
> ... 
> 20=1A 
> 30=2A 
> 30=3A 
> ... 
> 101=A1 
> 200=19A 
> 201=1A1 
> ... 
> 2000=199A 
> 2001=19A1
> 2010=19AA 
> 2014=1A14 
> 1091020134=????????? Brainteaser 
> 
> A generic basechanger for anybase is rather easy to accomplish, but i am a bit stumped where to start with doing conversion into bijective bases, i was thinking counting up the base multiples upto the number to convert, and bone it out from there saving the differences. But i just can't recall howto do it. 
> 
> Is there an easy obvious way todo this? 
> 
> I slightly remember there is something about using pairs, but that may have been for ordinary base conversion. 
> 
> Who have the fastest head doing the conversion.

Not certain if I follow your question completely, so I'm going to give an answer I hope is in the ball park and hope you can correct me where we disconnect:

An integer "i" in base "n" can be represented in the form "i = (k,l)" where
"k = (i - (i mod n)) / n"
and
"l = i mod n"

For example:

253 = ([253 - (253 mod 16)]/16, 253 mod 16)
    = (15, 13)

Converting the pair back into its original form is simply the base expansion
of the pair: i = k*n + l

[toc] | [prev] | [next] | [standalone]

#24897

Fromjonas.thornvall@gmail.com
Date2014-06-17 13:23 -0700
Message-ID<5da0343a-ede6-473a-ae01-58bf34739171@googlegroups.com>
In reply to#24896
Den tisdagen den 17:e juni 2014 kl. 21:08:23 UTC+2 skrev Michael Haufe (TNO):
> On Tuesday, June 17, 2014 8:42:28 AM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 10=A 
> 
> > 100=9A
> 
> > 1000=99A 
> 
> > ... 
> 
> > 20=1A 
> 
> > 30=2A 
> 
> > 30=3A 
> 
> > ... 
> 
> > 101=A1 
> 
> > 200=19A 
> 
> > 201=1A1 
> 
> > ... 
> 
> > 2000=199A 
> 
> > 2001=19A1
> 
> > 2010=19AA 
> 
> > 2014=1A14 
> 
> > 1091020134=????????? Brainteaser 
> 
> > 
> 
> > A generic basechanger for anybase is rather easy to accomplish, but i am a bit stumped where to start with doing conversion into bijective bases, i was thinking counting up the base multiples upto the number to convert, and bone it out from there saving the differences. But i just can't recall howto do it. 
> 
> > 
> 
> > Is there an easy obvious way todo this? 
> 
> > 
> 
> > I slightly remember there is something about using pairs, but that may have been for ordinary base conversion. 
> 
> > 
> 
> > Who have the fastest head doing the conversion.
> 
> 
> 
> Not certain if I follow your question completely, so I'm going to give an answer I hope is in the ball park and hope you can correct me where we disconnect:
> 
> 
> 
> An integer "i" in base "n" can be represented in the form "i = (k,l)" where
> 
> "k = (i - (i mod n)) / n"
> 
> and
> 
> "l = i mod n"
> 
> 
> 
> For example:
> 
> 
> 
> 253 = ([253 - (253 mod 16)]/16, 253 mod 16)
> 
>     = (15, 13)
> 
> 
> 
> Converting the pair back into its original form is simply the base expansion
> 
> of the pair: i = k*n + l
Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?

But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)

[toc] | [prev] | [next] | [standalone]

#24898

Fromjonas.thornvall@gmail.com
Date2014-06-17 13:24 -0700
Message-ID<3e123a91-76b7-47d4-b25f-400c4d662718@googlegroups.com>
In reply to#24897
Den tisdagen den 17:e juni 2014 kl. 22:23:15 UTC+2 skrev jonas.t...@gmail.com:
> Den tisdagen den 17:e juni 2014 kl. 21:08:23 UTC+2 skrev Michael Haufe (TNO):
> 
> > On Tuesday, June 17, 2014 8:42:28 AM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > > 10=A 
> 
> > 
> 
> > > 100=9A
> 
> > 
> 
> > > 1000=99A 
> 
> > 
> 
> > > ... 
> 
> > 
> 
> > > 20=1A 
> 
> > 
> 
> > > 30=2A 
> 
> > 
> 
> > > 30=3A 
> 
> > 
> 
> > > ... 
> 
> > 
> 
> > > 101=A1 
> 
> > 
> 
> > > 200=19A 
> 
> > 
> 
> > > 201=1A1 
> 
> > 
> 
> > > ... 
> 
> > 
> 
> > > 2000=199A 
> 
> > 
> 
> > > 2001=19A1
> 
> > 
> 
> > > 2010=19AA 
> 
> > 
> 
> > > 2014=1A14 
> 
> > 
> 
> > > 1091020134=????????? Brainteaser 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > A generic basechanger for anybase is rather easy to accomplish, but i am a bit stumped where to start with doing conversion into bijective bases, i was thinking counting up the base multiples upto the number to convert, and bone it out from there saving the differences. But i just can't recall howto do it. 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Is there an easy obvious way todo this? 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > I slightly remember there is something about using pairs, but that may have been for ordinary base conversion. 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Who have the fastest head doing the conversion.
> 
> > 
> 
> > 
> 
> > 
> 
> > Not certain if I follow your question completely, so I'm going to give an answer I hope is in the ball park and hope you can correct me where we disconnect:
> 
> > 
> 
> > 
> 
> > 
> 
> > An integer "i" in base "n" can be represented in the form "i = (k,l)" where
> 
> > 
> 
> > "k = (i - (i mod n)) / n"
> 
> > 
> 
> > and
> 
> > 
> 
> > "l = i mod n"
> 
> > 
> 
> > 
> 
> > 
> 
> > For example:
> 
> > 
> 
> > 
> 
> > 
> 
> > 253 = ([253 - (253 mod 16)]/16, 253 mod 16)
> 
> > 
> 
> >     = (15, 13)
> 
> > 
> 
> > 
> 
> > 
> 
> > Converting the pair back into its original form is simply the base expansion
> 
> > 
> 
> > of the pair: i = k*n + l
> 
> Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> 
> 
> But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)
Sorry F13=FD

[toc] | [prev] | [next] | [standalone]

#24899

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-17 13:54 -0700
Message-ID<56a7f5f9-25d0-4d19-a545-939eedfdb424@googlegroups.com>
In reply to#24897
On Tuesday, June 17, 2014 3:23:15 PM UTC-5, jonas.t...@gmail.com wrote:

> Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)

If you're using juxtaposition to represent the tuple (which is ambiguous), then
yes, 240 as a base 16 encoding would be F0.

You seem to be using the term "bijective" in a strange way. The above encoding I
presented _is_ bijective.

I don't understand your requirement of a "zeroless encoding"

Another approach is to convert the integer into prime factors, but it will be
quite slow:

864 = (2^5)*(3^3)

Also, there do exist positional number systems that don't use "0", but
honestly I think I want to know the context of your requirement better

[toc] | [prev] | [next] | [standalone]

#24900

Fromjonas.thornvall@gmail.com
Date2014-06-17 14:07 -0700
Message-ID<9b0b48bb-020b-4b5d-bd1e-2033a0891d61@googlegroups.com>
In reply to#24899
Den tisdagen den 17:e juni 2014 kl. 22:54:02 UTC+2 skrev Michael Haufe (TNO):
> On Tuesday, June 17, 2014 3:23:15 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> 
> 
> > Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> > 
> 
> > Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> > 
> 
> > But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)
> 
> 
> 
> If you're using juxtaposition to represent the tuple (which is ambiguous), then
> 
> yes, 240 as a base 16 encoding would be F0.
> 
> 
> 
> You seem to be using the term "bijective" in a strange way. The above encoding I
> 
> presented _is_ bijective.
> 
> 
> 
> I don't understand your requirement of a "zeroless encoding"
> 
> 
> 
> Another approach is to convert the integer into prime factors, but it will be
> 
> quite slow:
> 
> 
> 
> 864 = (2^5)*(3^3)
> 
> 
> 
> Also, there do exist positional number systems that don't use "0", but
> 
> honestly I think I want to know the context of your requirement better

I am not sure what to say about it is an zeroles encoding scheme for the natural numbers. Problem is i do not remember howto approach the problem i Think it must be pairwise from lowest to highest.


BASE 10
1=1
10=A 
20=1A          =10+10
30=2A          =20+10
30=3A          =30+10
100=9A         =90+10
101=A1         =100+1
200=19A        =100+90+10
201=1A1        =100+100+1
1000=99A       =900+90+10
2000=199A      =1000+900+90+10
2001=19A1      =1000+900+100+1
2010=19AA      =1000+900+100+10
2014=1A14      =1000+1000+14

And here you have ternary using same principle

BASE3
   Deroless             Ordinary
1 =1 	1		01 
2 =2 	2		02 
3 =3 	3		10 
4 =11 	3+1 		11 
5 =12 	3+2 		12 
6 =13 	3+3 		20 
7 =21 	6+1 		21 
8 =22 	6+2 		22 
9 =23 	6+3 		100 
10=31 	9+1 		101 
11=32 	9+2 		102 
12=33 	9+3 		110 
13=111	9+3+1 		111 
14=112 	9+3+2 		112 
15=113 	9+3+3 		120 
16=121 	9+6+1 		121 
17=122 	9+6+2 		122 
18=123 	9+6+3 		200 
19=131 	9+9+1 		201 
20=132 	9+9+2 		202 
21=133 	9+9+3 		210 
22=211 	18+3+1
23=212 	18+3+2
24=213 	18+3+3
25=221	18+6+1
26=222 	18+6+2
27=223 	18+6+3

[toc] | [prev] | [next] | [standalone]

#24901

Fromjonas.thornvall@gmail.com
Date2014-06-17 14:15 -0700
Message-ID<b030f677-0c59-46a7-b3ef-a9ddefde8fe8@googlegroups.com>
In reply to#24900
Den tisdagen den 17:e juni 2014 kl. 23:07:33 UTC+2 skrev jonas.t...@gmail.com:
> Den tisdagen den 17:e juni 2014 kl. 22:54:02 UTC+2 skrev Michael Haufe (TNO):
> 
> > On Tuesday, June 17, 2014 3:23:15 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > 
> 
> > 
> 
> > > Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> > 
> 
> > > 
> 
> > 
> 
> > > But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)
> 
> > 
> 
> > 
> 
> > 
> 
> > If you're using juxtaposition to represent the tuple (which is ambiguous), then
> 
> > 
> 
> > yes, 240 as a base 16 encoding would be F0.
> 
> > 
> 
> > 
> 
> > 
> 
> > You seem to be using the term "bijective" in a strange way. The above encoding I
> 
> > 
> 
> > presented _is_ bijective.
> 
> > 
> 
> > 
> 
> > 
> 
> > I don't understand your requirement of a "zeroless encoding"
> 
> > 
> 
> > 
> 
> > 
> 
> > Another approach is to convert the integer into prime factors, but it will be
> 
> > 
> 
> > quite slow:
> 
> > 
> 
> > 
> 
> > 
> 
> > 864 = (2^5)*(3^3)
> 
> > 
> 
> > 
> 
> > 
> 
> > Also, there do exist positional number systems that don't use "0", but
> 
> > 
> 
> > honestly I think I want to know the context of your requirement better
> 
> 
> 
> I am not sure what to say about it is an zeroles encoding scheme for the natural numbers. Problem is i do not remember howto approach the problem i Think it must be pairwise from lowest to highest.
> 
> 
> 
> 
> 
> BASE 10
> 
> 1=1
> 
> 10=A 
> 
> 20=1A          =10+10
> 
> 30=2A          =20+10
> 
> 30=3A          =30+10
> 
> 100=9A         =90+10
> 
> 101=A1         =100+1
> 
> 200=19A        =100+90+10
> 
> 201=1A1        =100+100+1
> 
> 1000=99A       =900+90+10
> 
> 2000=199A      =1000+900+90+10
> 
> 2001=19A1      =1000+900+100+1
> 
> 2010=19AA      =1000+900+100+10
> 
> 2014=1A14      =1000+1000+14
> 
> 
> 
> And here you have ternary using same principle
> 
> 
> 
> BASE3
> 
>    Deroless             Ordinary
> 
> 1 =1 	1		01 
> 
> 2 =2 	2		02 
> 
> 3 =3 	3		10 
> 
> 4 =11 	3+1 		11 
> 
> 5 =12 	3+2 		12 
> 
> 6 =13 	3+3 		20 
> 
> 7 =21 	6+1 		21 
> 
> 8 =22 	6+2 		22 
> 
> 9 =23 	6+3 		100 
> 
> 10=31 	9+1 		101 
> 
> 11=32 	9+2 		102 
> 
> 12=33 	9+3 		110 
> 
> 13=111	9+3+1 		111 
> 
> 14=112 	9+3+2 		112 
> 
> 15=113 	9+3+3 		120 
> 
> 16=121 	9+6+1 		121 
> 
> 17=122 	9+6+2 		122 
> 
> 18=123 	9+6+3 		200 
> 
> 19=131 	9+9+1 		201 
> 
> 20=132 	9+9+2 		202 
> 
> 21=133 	9+9+3 		210 
> 
> 22=211 	18+3+1
> 
> 23=212 	18+3+2
> 
> 24=213 	18+3+3
> 
> 25=221	18+6+1
> 
> 26=222 	18+6+2
> 
> 27=223 	18+6+3
Using zeroless base there is no infinity all numbers are recursively constructable?

[toc] | [prev] | [next] | [standalone]

#24974

FromJohn Harris <niam@jghnorth.org.uk.invalid>
Date2014-06-23 17:23 +0100
Message-ID<v2lgq91rvanhd6t1vplkkvk3bf8fp81enc@4ax.com>
In reply to#24901
On Tue, 17 Jun 2014 14:15:01 -0700 (PDT), jonas.thornvall@gmail.com
wrote:


  <snip>
>Using zeroless base there is no infinity all numbers are recursively constructable?

Natural Numbers are recursively constructable by definition; Real
Numbers are not. In neither is infinity a number.

  John

[toc] | [prev] | [next] | [standalone]

#24902

Fromjonas.thornvall@gmail.com
Date2014-06-17 14:23 -0700
Message-ID<8a732973-4e4f-41e0-b1bd-98311ee3859a@googlegroups.com>
In reply to#24899
Den tisdagen den 17:e juni 2014 kl. 22:54:02 UTC+2 skrev Michael Haufe (TNO):
> On Tuesday, June 17, 2014 3:23:15 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> 
> 
> > Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> > 
> 
> > Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> > 
> 
> > But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)
> 
> 
> 
> If you're using juxtaposition to represent the tuple (which is ambiguous), then
> 
> yes, 240 as a base 16 encoding would be F0.
> 
> 
> 
> You seem to be using the term "bijective" in a strange way. The above encoding I
> 
> presented _is_ bijective.
> 
> 
> 
> I don't understand your requirement of a "zeroless encoding"
> 
> 
> 
> Another approach is to convert the integer into prime factors, but it will be
> 
> quite slow:
> 
> 
> 
> 864 = (2^5)*(3^3)
> 
> 
> 
> Also, there do exist positional number systems that don't use "0", but
> 
> honestly I think I want to know the context of your requirement better

Wiki do call it bijective base-10 system 
http://en.wikipedia.org/wiki/Bijective_numeration
They say it was used in greek by Plato and they guys.

[toc] | [prev] | [next] | [standalone]

#24903

Fromjonas.thornvall@gmail.com
Date2014-06-17 14:41 -0700
Message-ID<f3dd2e6b-b193-4990-8f5f-191f8609c4bd@googlegroups.com>
In reply to#24899
Den tisdagen den 17:e juni 2014 kl. 22:54:02 UTC+2 skrev Michael Haufe (TNO):
> On Tuesday, June 17, 2014 3:23:15 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> 
> 
> > Ok base 16 but lets 10=A,B...,C...,D...,E...,F,*G*=16 so  253=F13 
> 
> > 
> 
> > Well using base 16 if we would use your suggeste method to convert 240 it would end up be F0?
> 
> > 
> 
> > But I want a zeroless encoding (bijective?) EG=(14*16)+(16*1)
> 
> 
> 
> If you're using juxtaposition to represent the tuple (which is ambiguous), then
> 
> yes, 240 as a base 16 encoding would be F0.
> 
> 
> 
> You seem to be using the term "bijective" in a strange way. The above encoding I
> 
> presented _is_ bijective.
> 
> 
> 
> I don't understand your requirement of a "zeroless encoding"
> 
> 
> 
> Another approach is to convert the integer into prime factors, but it will be
> 
> quite slow:
> 
> 
> 
> 864 = (2^5)*(3^3)
> 
> 
> 
> Also, there do exist positional number systems that don't use "0", but
> 
> honestly I think I want to know the context of your requirement better
According to wiki it is a The bijective base-10 system, used by the old greeks Plato and they guys.

http://en.wikipedia.org/wiki/Bijective_numeration

[toc] | [prev] | [next] | [standalone]

#24906

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-17 20:28 -0700
Message-ID<5bf41f47-caa2-405f-8c13-dc67498702f9@googlegroups.com>
In reply to#24903
On Tuesday, June 17, 2014 4:41:47 PM UTC-5, jonas.t...@gmail.com wrote:
> According to wiki it is a The bijective base-10 system, used by the old greeks Plato and they guys.
> 
> http://en.wikipedia.org/wiki/Bijective_numeration

Ah, I vaguely recall this from a planetmath.org article from a few years ago.

A cursory Google search yielded the following which could be a decent starting point to create a JavaScript implementation:

http://math.stackexchange.com/questions/172808/finding-a-bijective-function-from-integer-to-words

[toc] | [prev] | [next] | [standalone]

#24910

Fromjonas.thornvall@gmail.com
Date2014-06-18 13:04 -0700
Message-ID<89156a84-6fa7-4b79-9a4c-f53686792543@googlegroups.com>
In reply to#24906
Den onsdagen den 18:e juni 2014 kl. 05:28:03 UTC+2 skrev Michael Haufe (TNO):
> On Tuesday, June 17, 2014 4:41:47 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > According to wiki it is a The bijective base-10 system, used by the old greeks Plato and they guys.
> 
> > 
> 
> > http://en.wikipedia.org/wiki/Bijective_numeration
> 
> 
> 
> Ah, I vaguely recall this from a planetmath.org article from a few years ago.
> 
> 
> 
> A cursory Google search yielded the following which could be a decent starting point to create a JavaScript implementation:
> 
> 
> 
> http://math.stackexchange.com/questions/172808/finding-a-bijective-function-from-integer-to-words

The guys at rec.puzzles really clever.
Here is what they had to say.

Andrew B
Isn't it just A8AA1A134; convert each (non-leading) 0 into an A and subtract one from the number to its left, iterating if necessary? 

Simple and elegant.

[toc] | [prev] | [next] | [standalone]

#24932

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-18 16:30 -0700
Message-ID<e9444b01-8627-462c-94bd-1190f99d9665@googlegroups.com>
In reply to#24910
On Wednesday, June 18, 2014 3:04:08 PM UTC-5, jonas.t...@gmail.com wrote:
>
> The guys at rec.puzzles really clever.
> Here is what they had to say.
> 
> Andrew B
> 
> Isn't it just A8AA1A134; convert each (non-leading) 0 into an A and subtract one from the number to its left, iterating if necessary? 
> 
> Simple and elegant.

It makes an assumption about the base used which makes me skeptical of its applicability beyond this specific case.

[toc] | [prev] | [next] | [standalone]

#24937

Fromjonas.thornvall@gmail.com
Date2014-06-18 17:45 -0700
Message-ID<ad574b14-98d3-4ca2-93e5-757bdb7fe8cc@googlegroups.com>
In reply to#24932
Den torsdagen den 19:e juni 2014 kl. 01:30:09 UTC+2 skrev Michael Haufe (TNO):
> On Wednesday, June 18, 2014 3:04:08 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> >
> 
> > The guys at rec.puzzles really clever.
> 
> > Here is what they had to say.
> 
> > 
> 
> > Andrew B
> 
> > 
> 
> > Isn't it just A8AA1A134; convert each (non-leading) 0 into an A and subtract one from the number to its left, iterating if necessary? 
> 
> > 
> 
> > Simple and elegant.
> 
> 
> 
> It makes an assumption about the base used which makes me skeptical of its applicability beyond this specific case.

No it doesn't, i recall it to be the correct solution.

[toc] | [prev] | [next] | [standalone]

#24938

Fromjonas.thornvall@gmail.com
Date2014-06-18 17:49 -0700
Message-ID<80d440d6-9eba-49cd-ae80-c35480a56e83@googlegroups.com>
In reply to#24937
Den torsdagen den 19:e juni 2014 kl. 02:45:21 UTC+2 skrev jonas.t...@gmail.com:
> Den torsdagen den 19:e juni 2014 kl. 01:30:09 UTC+2 skrev Michael Haufe (TNO):
> 
> > On Wednesday, June 18, 2014 3:04:08 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > >
> 
> > 
> 
> > > The guys at rec.puzzles really clever.
> 
> > 
> 
> > > Here is what they had to say.
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Andrew B
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Isn't it just A8AA1A134; convert each (non-leading) 0 into an A and subtract one from the number to its left, iterating if necessary? 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Simple and elegant.
> 
> > 
> 
> > 
> 
> > 
> 
> > It makes an assumption about the base used which makes me skeptical of its applicability beyond this specific case.
> 
> 
> 
> No it doesn't, i recall it to be the correct solution.

The idea is sound in any base just have to take care of leading and reoccuring zeros, and this is simply done by look one step forward beyond each zero.

[toc] | [prev] | [next] | [standalone]

#24945

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-19 11:21 -0700
Message-ID<efb887c8-ee9c-4024-8b22-d951ba26056c@googlegroups.com>
In reply to#24937
On Wednesday, June 18, 2014 7:45:21 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> No it doesn't, i recall it to be the correct solution.

To what question? What base? 16? There is no '0' in this encoding, nor a substituted letter for it so the answer to unquoted question I'm assuming you asked doesn't seem to make sense. It's position based, so a hex encoding (contra your mapping above) would instead be:

0 - ε
1 - 1
2 - 2
3 - 3
4 - 4
5 - 5
6 - 6
7 - 7
8 - 8
9 - 9
10 - A
11 - B
12 - C
13 - D
14 - E
15 - F
16 - 11  //NOT 'G'
...

[toc] | [prev] | [next] | [standalone]

#24946

Fromjonas.thornvall@gmail.com
Date2014-06-19 14:44 -0700
Message-ID<c267c834-0f19-4dc0-a576-7a3ffecb0e46@googlegroups.com>
In reply to#24945
Den torsdagen den 19:e juni 2014 kl. 20:21:42 UTC+2 skrev Michael Haufe (TNO):
> On Wednesday, June 18, 2014 7:45:21 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > No it doesn't, i recall it to be the correct solution.
> 
> 
> 
> To what question? What base? 16? There is no '0' in this encoding, nor a substituted letter for it so the answer to unquoted question I'm assuming you asked doesn't seem to make sense. It's position based, so a hex encoding (contra your mapping above) would instead be:
> 
> 
> 
> 0 - ε
> 
> 1 - 1
> 
> 2 - 2
> 
> 3 - 3
> 
> 4 - 4
> 
> 5 - 5
> 
> 6 - 6
> 
> 7 - 7
> 
> 8 - 8
> 
> 9 - 9
> 
> 10 - A
> 
> 11 - B
> 
> 12 - C
> 
> 13 - D
> 
> 14 - E
> 
> 15 - F
> 
> 16 - 11  //NOT 'G'
> 
> ...

No using the encoding scheme on HEX would give G your 21 is Another encoding scheme.

[toc] | [prev] | [next] | [standalone]

#24947

Fromjonas.thornvall@gmail.com
Date2014-06-19 14:48 -0700
Message-ID<f635c1a2-508b-4eb6-aeae-07729bf76a5f@googlegroups.com>
In reply to#24946
Den torsdagen den 19:e juni 2014 kl. 23:44:43 UTC+2 skrev jonas.t...@gmail.com:
> Den torsdagen den 19:e juni 2014 kl. 20:21:42 UTC+2 skrev Michael Haufe (TNO):
> 
> > On Wednesday, June 18, 2014 7:45:21 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > > 
> 
> > 
> 
> > > No it doesn't, i recall it to be the correct solution.
> 
> > 
> 
> > 
> 
> > 
> 
> > To what question? What base? 16? There is no '0' in this encoding, nor a substituted letter for it so the answer to unquoted question I'm assuming you asked doesn't seem to make sense. It's position based, so a hex encoding (contra your mapping above) would instead be:
> 
> > 
> 
> > 
> 
> > 
> 
> > 0 - ε
> 
> > 
> 
> > 1 - 1
> 
> > 
> 
> > 2 - 2
> 
> > 
> 
> > 3 - 3
> 
> > 
> 
> > 4 - 4
> 
> > 
> 
> > 5 - 5
> 
> > 
> 
> > 6 - 6
> 
> > 
> 
> > 7 - 7
> 
> > 
> 
> > 8 - 8
> 
> > 
> 
> > 9 - 9
> 
> > 
> 
> > 10 - A
> 
> > 
> 
> > 11 - B
> 
> > 
> 
> > 12 - C
> 
> > 
> 
> > 13 - D
> 
> > 
> 
> > 14 - E
> 
> > 
> 
> > 15 - F
> 
> > 
> 
> > 16 - 11  //NOT 'G'
> 
> > 
> 
> > ...
> 
> 
> 
> No using the encoding scheme on HEX would give G your 21 is Another encoding scheme.
Sorry yout 11 isn't same encoding scheme i Think you  try coding base 15.

[toc] | [prev] | [next] | [standalone]

#24948

Fromjonas.thornvall@gmail.com
Date2014-06-19 14:55 -0700
Message-ID<37079c88-aac9-4633-bfbc-dbabf328109b@googlegroups.com>
In reply to#24945
Den torsdagen den 19:e juni 2014 kl. 20:21:42 UTC+2 skrev Michael Haufe (TNO):
> On Wednesday, June 18, 2014 7:45:21 PM UTC-5, jonas.t...@gmail.com wrote:
> 
> > 
> 
> > No it doesn't, i recall it to be the correct solution.
> 
> 
> 
> To what question? What base? 16? There is no '0' in this encoding, nor a substituted letter for it so the answer to unquoted question I'm assuming you asked doesn't seem to make sense. It's position based, so a hex encoding (contra your mapping above) would instead be:
> 
> 
> 
> 0 - ε
> 
> 1 - 1
> 
> 2 - 2
> 
> 3 - 3
> 
> 4 - 4
> 
> 5 - 5
> 
> 6 - 6
> 
> 7 - 7
> 
> 8 - 8
> 
> 9 - 9
> 
> 10 - A
> 
> 11 - B
> 
> 12 - C
> 
> 13 - D
> 
> 14 - E
> 
> 15 - F
> 
> 16 - 11  //NOT 'G'
> 
> ...
You do realise the numerals for a base should ideally be unique. But it works as well giving commaseparated Powers for the positions.

And your bijective/zeroless HEX encoding only have 15 unique numerals.

[toc] | [prev] | [next] | [standalone]

#24951

From"Michael Haufe (TNO)" <tno@thenewobjective.com>
Date2014-06-19 19:59 -0700
Message-ID<72f14f01-bb4a-43ca-bf53-766527f95307@googlegroups.com>
In reply to#24948
On Thursday, June 19, 2014 4:55:20 PM UTC-5, jonas.t...@gmail.com wrote:

> And your bijective/zeroless HEX encoding only have 15 unique numerals.

You're correct. The alphabet for base k should be of magnitude k

[toc] | [prev] | [next] | [standalone]

Page 1 of 3  [1] 2 3  Next page →

Back to top | Article view | comp.lang.javascript

csiph-web