On 3/8/13, Linda Alvord <[email protected]> wrote:
> ]A=:1 2 3 { 8 32 $ a. NB. Example from definition of a.
> ]AL=:_5}."1 1}."1 A
> ]L=:15}.,AL
> f=: 13 :'((i.#y)~:y i. '':'')#y' NB. Remove the : from list
> f
> ] #~ ([: i. #) ~: ':' i.~ ]
> f L
> 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
Why don't you use this instead?
f=: -.&':'
In any case, this is a nice method, and with some liberal changes, can
be shortened to this.
42}.,4{._32(26&{.)\a.-.':'
(Incidentally, I'm collecting these expressions on
"http://www.jsoftware.com/jwiki/BJonas";.
> Here's where I got lost in your solution:
>
> a.#~2|a.i.&0@/:@,"{'/9@Z`z'
>
> I can't seem to unlock what happens at the heart of it.
> What does this do?
>
> ([: i.&0 /:)
I admit that solution is a bit overcomplicated. It's not what I'd
write in a real program.
Let me explain how it works. This post will be a bit long, sorry, but
this is the easiest way I can explain. I'll show you a much easier
solution at the end though.
I'll show on a shorter range of characters for readability. Suppose
we start from this range of characters.
]alp =: (26{.65}.a.)
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Take this key of six characters which we'll use as endpoints of three
intervals to cut the above.
]key=: 'DHMQTV'
DHMQTV
Suppose we concatenate each letter with the key and sort the resulting
seven character long string.
alp /:~@,"0 1 key
ADHMQTV
BDHMQTV
CDHMQTV
DDHMQTV
DEHMQTV
DFHMQTV
DGHMQTV
DHHMQTV
DHIMQTV
DHJMQTV
DHKMQTV
DHLMQTV
DHMMQTV
DHMNQTV
DHMOQTV
DHMPQTV
DHMQQTV
DHMQRTV
DHMQSTV
DHMQTTV
DHMQTUV
DHMQTVV
DHMQTVW
DHMQTVX
DHMQTVY
DHMQTVZ
As you can see, the single letter we concatenate goes to different
positions in the seven letter string, and jumps to a new position
whenever we reach a letter in the key. For example, ABCD are in the
first position in the first four lines, then EFGH are in the second
position, then IJKLM in the third. The boundaries are exactly DHM
because those are the first three letters of the key.
Now instead of sorting the seven letter strings, we grade them to find
out where the letter we insert goes.
alp /:@,"0 1 key
0 1 2 3 4 5 6
0 1 2 3 4 5 6
0 1 2 3 4 5 6
0 1 2 3 4 5 6
1 0 2 3 4 5 6
1 0 2 3 4 5 6
1 0 2 3 4 5 6
1 0 2 3 4 5 6
1 2 0 3 4 5 6
1 2 0 3 4 5 6
1 2 0 3 4 5 6
1 2 0 3 4 5 6
1 2 0 3 4 5 6
1 2 3 0 4 5 6
1 2 3 0 4 5 6
1 2 3 0 4 5 6
1 2 3 0 4 5 6
1 2 3 4 0 5 6
1 2 3 4 0 5 6
1 2 3 4 0 5 6
1 2 3 4 5 0 6
1 2 3 4 5 0 6
1 2 3 4 5 6 0
1 2 3 4 5 6 0
1 2 3 4 5 6 0
1 2 3 4 5 6 0
As you can see, the position of the zero in this string shows where
the extra letter got inserted. So let's find the position of the
zero, putting the corresponding inserted letter above for clarity.
|:alp;"0 i.&0"1 alp /:@,"0 1 key
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z|
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|0|0|0|0|1|1|1|1|2|2|2|2|2|3|3|3|3|4|4|4|5|5|6|6|6|6|
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
So now the characters up to D get converted to a zero, the characters
up to H get converted to a one, etc. The key string controls where
exactly the numeric result changes. So if we take the letters that
get converted to an odd number, we get exactly the letters
corresponding to the three intervals determined by the key string:
EFGH for the first interval (between D and H), then NOP for the second
interval (from M to P), etc.
alp#~ 2| i.&0"1 alp /:@,"0 1 key
EFGHNOPQUV
In the actual solution to get the alphanumeric characters, I use
'/9@Z`z' as the key because '/9' gives the interval for numeric
characters, '@Z' the one for uppercase letters, etc; and I use a. as
the starting list of letters.
Now it turns out the whole sort thing is a bit superfluous, because
the search primitive gives a much simpler and shorter solution.
alp#~ 2| key I. alp
EFGHNOPQUV
a.#~2|'/9@Z`z'I.a.
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
I wonder why I didn't find that earlier.
--
Ambrus
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