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Re: Boolean 0/1 function
*To*: MathGroup at yoda.physics.unc.edu
*Subject*: Re: Boolean 0/1 function
*From*: villegas
*Date*: Tue, 3 Nov 92 08:07:55 CST
Joseph McWilliams asked how to define a Boolean function which
returns 1 or 0 (instead of the Mathematica Booleans 'True' and
'False'):
> I am trying to recursively define a piecewise function in the
> following manner:
>
> ========================
> s=0
> For[i=0,i<=(n-1), i++,
> s = s + f[y,x[i]]*Indicator[y,x[i],x[i+1]]
> ]
> ========================
>
> I would like Indicator[y,x[i],x[i+1]] to equal 1 if x[i]<= y <=x[i+1]
> and 0 otherwise.
You could define a function which maps the symbol 'True' to 1, and 'False'
to 0:
BooleanBit[True] = 1
BooleanBit[False] = 0
Then apply attach this function to any expression whose output you expect
to be 'True' or 'False'. For instance, the inequality you wrote:
s=0
For[i=0,i<=(n-1), i++,
s = s + f[y,x[i]] * BooleanBit[ x[i] <= y <= x[i+1] ]
]
If you wished, you could define BooleanBit to return some flag for cases
where a test returns neither 'True' nor 'False', which could happen if
you ever had symbols involved in the comparison instead of numbers.
An alternative way to code this computation, more Mathematica-esque
in that it is structural and functional, could go something like:
(* Initialization of values *)
xlist = {x1, x2, ..., xn} (where any 'xi' stands for the value of 'x[i]'
in your code)
y = <whatever> (the value you had for 'y')
(* Computation *)
fvalues = Map[f[y, #]&, Drop[xlist, -1] ]
consecutivepairs = Partition[xlist, 2, 1]
boolvalues = Apply[BooleanBit[#1 <= y <= #2]&, consecutivepairs, {1}]
total = Apply[Plus, fvalues * boolvalues]
The 'fvalues' could also be without using any pure functions, with a
structural command:
fvalues = Thread[ f[y, xlist] ]
as long as 'y' isn't a list itself (if it's a number, say).
Robby Villegas
Technical Support
Wolfram Research
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