Re: style question
- To: mathgroup at smc.vnet.net
- Subject: [mg79496] Re: style question
- From: Donald DuBois <donabc at comcast.net>
- Date: Sat, 28 Jul 2007 05:24:46 -0400 (EDT)
>Correction to self. I overlooked the three->dimensionality of m1 and
>m2. Again I'm without Mathematica; still you might try:
>
>matches = MapThread[Equal, {m1, m2}, 3];
>Count[matches, True, {3}]
>
>By the way, your syntax for the first argument of >RandomInteger
>appears to be undocumented.
>
>Vince Virgilio
>
Hello Vince:
I think you were right the first time.
{m1, m2} = RandomInteger[1, {2, 10^5, 2, 2}];
(* First time *)
matches1 = MapThread[Equal, {m1, m2}];
Count[matches1, True]
Out[30]= 6189
(* Second time *)
matches2 = MapThread[Equal, {m1, m2}, 3];
Count[matches2, True, {3}]
Out[31]= 200218
>From probability, one would expect about
6250 matches. This is because {{a,b},{c,d}}
can have 16 different
values since a, b , c and d are either 0 or 1 with
equal probability. So, {{a,b},{c,d}} will match
{{e,f},{g,h}} approx. 1/16-th of the time.
1/16-th of 10^5 = 6250.
Best,
Don