Yes, it appears so. And the trailing 16s or 64s can be 0 or 1 or 3 or whatever and the answer remains incorrect.

On 12/17/2015 1:40 PM, robert therriault wrote:
Seems to be a combination of having more than 31 leading 2's and the magic 
length of 64 items in the argument :-)

Cheers, bob

     ?. (30#2),34#64
0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 0 0 1 1 54 30 39 63 23 28 
21 14 18 32 21 13 19 23 16 37 16 56 52 14 51 1 22 14 26 63 13 37 54 2 30 16 5 30
    ?. (31#2),33#64
0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 0 0 1 1 0 30 39 63 23 28 21 
14 18 32 21 13 19 23 16 37 16 56 52 14 51 1 22 14 26 63 13 37 54 2 30 16 5 30
    ?. (32#2),32#64
0 1 0 1 1 0 0 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 
1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1
    ?. (33#2),31#64
0 1 0 1 1 0 0 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 
1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1
    ?. (33#2),30#64
1 1 1 0 1 0 0 0 1 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 0 0 1 0 0 52 44 43 35 
59 43 11 11 18 47 3 11 42 49 0 19 27 3 21 20 33 33 49 63 60 42 47 33 33 8

On Dec 17, 2015, at 9:40 AM, 'Pascal Jasmin' via Programming 
<[email protected]> wrote:

?.  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
2 2 2 2 2 2 2 2 2 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16
0 0 0 0 1 1 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 
1 1 1 0 0 1 1 1 8 9 6 8 14 7 12 9 1 15 2 15 3 13 3

?. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
2 2 2 2 2 2 2 2 2 2 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16
0 1 0 1 1 0 0 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 
1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1


one less leading 2 in first example.  47 leading 2s, correct result.  48 
leading 2s and everything is 2.


----- Original Message -----
From: Raul Miller <[email protected]>
To: Programming forum <[email protected]>
Sent: Thursday, December 17, 2015 12:28 PM
Subject: Re: [Jprogramming] bug in ?

Consider
   ?.2 2 2 32
0 1 0 18

Thanks,

--
Raul


On Thu, Dec 17, 2015 at 12:13 PM, 'Pascal Jasmin' via Programming
<[email protected]> wrote:


?. 16 # 2 2 2 16


result is same as

?. 16 # 2 2 2 2

ignores the request for 0-15 range random numbers.


----- Original Message -----
From: Raul Miller <[email protected]>
To: Programming forum <[email protected]>
Sent: Thursday, December 17, 2015 12:10 PM
Subject: Re: [Jprogramming] bug in ?

On Thu, Dec 17, 2015 at 11:49 AM, 'Pascal Jasmin' via Programming
<[email protected]> wrote:
?. 16 # 2 2 2 16
0 1 0 1 1 0 0 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 
1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1


?. 16 # 3 2 2 16
0 2 1 0 2 1 1 1 2 1 0 2 1 2 1 2 1 0 1 1 1 1 1 1 0 0 1 0 0 1 1 0 0 1 1 1 0 1 0 0 
0 1 1 1 1 0 1 0 8 4 14 3 1 6 14 10 15 13 5 6 2 14 0 5

The last 16 numbers in the first example should match the 2nd.
Why do you say that?

Thanks,

--
Raul



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