You are right about that, Raul. Honestly, I may do so - but I am not 100%
sure the solving method will work, so I am grafting it in crudely, with
stacked environments and so forth. If it does work I will do a redesign
and tie things in tighter, for one thing it would allow me to see that the
trial
The exit mechanism, throws, errors and returns are basically your options.
You can combine them, of course. For example, you can catch a throw
and then return. Or you can run a J instance in a separate process and
then exit from that and return from the routine that started the
process. Or maybe y
exit quits the J environment
maybe a return. directive with an appropriate return code would do the job
On Sat, Dec 30, 2017 at 7:48 PM, Nick S wrote:
> Well, exit'' takes J down completely.
>
>
--
For information about J forums
Well, exit'' takes J down completely.
I have read through most of the foreigns again, and through doc, and I
searched the messages from this forum for a bit. Either I can't find the
right search string or it has not been discussed since I have begun
following this. I just can't find an answer.
Still thinking about the puzzle.
Some tools to explore the properties of the conjecture:
NB. pretty print
toca =: '0123456789' {~ ]
tobi =: 3 : '(2#~1+>.2^.>./y)#:y'
todc =: 3 : '(10#~>.10^.>./y)#:y'
todcc =: 3 : '(1+>.10^.>./y)":,.y'
NB. divisor formula from Roger Hui,
http://cod
From http://list.seqfan.eu/pipermail/seqfan/2017-December/018222.html
"
Consider the divisors of 136 (A018299):
1, 2, 4, 8, 17, 34, 68, 136.
136 in binary is 10001000.
Reading off successive bits from the left we have:
1 1
10 2
1004
1000 8
10001 17
10001034
1
The outermost pair of quotes are enclosing the inner pairs.
So you count the pairs and subtract one.
The formula below gives the table but builds it the other way around.
It calculates how many pairs are needed to produce n quotes.
(,.~+:&>:)i.10
2 0
4 1
6 2
8 3
10 4
12 5
14 6
16 7
1
Thanks for all the help on explaining T and U.
Here is my next question:
]V=:((i.8) >:/4#i.8){' '''