Here's what I get. The line will split but when I paste it back together:
tc =: #:~ 2 #~ [: >./ 1 + [: >. (2 ^. |)`(2 ^.>:)@.(0 <: ])"0
]a=:tc i:6
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
#.a
10 11 12 13 14 15 0 1 2 3 4 5 6
Not good. But your display does not have the first column, so maybe you need
to post the function again.
I tried to remove the first column to match your dislay:
]b=:1}."1 a
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
#.b
2 3 4 5 6 7 0 1 2 3 4 5 6
This is also not working.
Did you really read my post in a reply to a message to Devon? I'll repeat it
here as I am interested in your comments:
]a=:8+i.15
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
]b=:15#15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
]c=:a-b
_7 _6 _5 _4 _3 _2 _1 0 1 2 3 4 5 6 7
f=: 13 :'(x#2)#:y'
f
] #:~ 2 #~ [
The function f converts the lists to binary numbers. The largest number 22
is less than 32 so only 5 digits are require.
]aa=:5 f a
0 1 0 0 0
0 1 0 0 1
0 1 0 1 0
0 1 0 1 1
0 1 1 0 0
0 1 1 0 1
0 1 1 1 0
0 1 1 1 1
1 0 0 0 0
1 0 0 0 1
1 0 0 1 0
1 0 0 1 1
1 0 1 0 0
1 0 1 0 1
1 0 1 1 0
]bb=:5 f b
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
0 1 1 1 1
The result below is the difference in J between the two arrays above. It
is not a conversion using #:
]cc=:aa-bb
0 0 _1 _1 _1
0 0 _1 _1 0
0 0 _1 0 _1
0 0 _1 0 0
0 0 0 _1 _1
0 0 0 _1 0
0 0 0 0 _1
0 0 0 0 0
1 _1 _1 _1 _1
1 _1 _1 _1 0
1 _1 _1 0 _1
1 _1 _1 0 0
1 _1 0 _1 _1
1 _1 0 _1 0
1 _1 0 0 _1
For comparison, the result cc2 is obtained as a J result using #:
]cc2=: 5 f c
1 1 0 0 1
1 1 0 1 0
1 1 0 1 1
1 1 1 0 0
1 1 1 0 1
1 1 1 1 0
1 1 1 1 1
0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 0 1 1
0 0 1 0 0
0 0 1 0 1
0 0 1 1 0
0 0 1 1 1
Or:
]cc3=:#:c
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Both cc2 and cc3 seem implausible because all numbers are known positive
numbers. Now look at cc :
w=: 13 :'(x,y)$|.2^i.y'
w
, $ [: |. 2 ^ [: i. ]
15 w 5
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
16 8 4 2 1
cc*15 w 5
0 0 _4 _2 _1
0 0 _4 _2 0
0 0 _4 0 _1
0 0 _4 0 0
0 0 0 _2 _1
0 0 0 _2 0
0 0 0 0 _1
0 0 0 0 0
16 _8 _4 _2 _1
16 _8 _4 _2 0
16 _8 _4 0 _1
16 _8 _4 0 0
16 _8 0 _2 _1
16 _8 0 _2 0
16 _8 0 0 _1
+/"1 cc*15 w 5
_7 _6 _5 _4 _3 _2 _1 0 1 2 3 4 5 6 7
This is c!
Not only that, J knows it is c.
#.cc
_7 _6 _5 _4 _3 _2 _1 0 1 2 3 4 5 6 7
So what do you think?
Linda
-----Original Message-----
From: programming-boun...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] On Behalf Of Kip Murray
Sent: Friday, December 16, 2011 2:43 AM
To: Programming forum
Subject: Re: [Jprogramming] How #: should have been designed
The verb tc (two's complement) appears to work
tc =: #:~ 2 #~ [: >./ 1 + [: >. (2 ^. |)`(2 ^. >:)@.(0 <: ])"0
twoscomplement i: 6
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
tc i: 6
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
On 12/15/2011 12:52 PM, Kip Murray wrote:
> twoscomplement i: 4 NB. error
> 1 0 0
> 1 0 1
> 1 1 0
> 1 1 1
> 0 0 0
> 0 0 1
> 0 1 0
> 0 1 1
> 1 0 0
>
> rp =: [: }: [: i: 2 ^ <: NB. representable in y bits
>
> rp 3
> _4 _3 _2 _1 0 1 2 3
>
> twoscomplement rp 3 NB. OK
> 1 0 0
> 1 0 1
> 1 1 0
> 1 1 1
> 0 0 0
> 0 0 1
> 0 1 0
> 0 1 1
>
>
> On 12/14/2011 9:13 AM, Raul Miller wrote:
>> The subject line of this thread is arguably wrong -- there are a
>> variety of "good ways" of decomposing integers to binary.
>>
>> That said, it's interesting to think about the various proposals
>> expressed in terms similar to those which could be used to implement
>> monadic #:
>>
>> antibase2=: #:~ 2 #~ 1 + 2<.@^. 1>.>./@,@:|@:<.
>> twoscomplement=: #:~ 2 #~ 1 + 2<.@^. 1 +>./@,@:|@:<.
>> signwithbits=: #:~ 0, 2 #~ 1 + 2<.@^. 1>.>./@,@:|@:<.
>>
>> (In all cases the #: here is dyadic, so these definitions are
>> independent of the definition of monadic #:)
>>
>> antibase2 i: 3
>> 0 1
>> 1 0
>> 1 1
>> 0 0
>> 0 1
>> 1 0
>> 1 1
>> twoscomplement i: 3
>> 1 0 1
>> 1 1 0
>> 1 1 1
>> 0 0 0
>> 0 0 1
>> 0 1 0
>> 0 1 1
>> signwithbits i: 3
>> _1 0 1
>> _1 1 0
>> _1 1 1
>> 0 0 0
>> 0 0 1
>> 0 1 0
>> 0 1 1
>>
>> There's also (* * #:) but that one assumes the antibase2
>> implementation...
>>
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