Binary Data Compression.
I think this code:
:D→B
:0 // this will make sure that if I is greater than the highest index, 0 will be returned
:If D:Then //log(0) would give domain error, and 0d=0b.
:For(A,1+int(log(D)/log(2)),I,-1) //A will run through the indexes, from high to low
stopping at I(the one we needed). log(D)/log(2) = base 2 log of D
:0
:If 2^A≥B: Then
:B-2^A→B
:1
End:End:End
Could be shortened to this for most scenarios:
int(2fPart(.5D/2^I
For example, to test bit 2 of the value 18:
.5*18/2^2
9/4
2.25
int(2fPart(2.25
int(2*.25
0
and 18=10010b, which fits. Testing bit 4:
.5*18/2^4
9/16=.5625
int(2fPart(.5625
int(1.125
1
I am not sure how much accuracy it will retain for very large numbers, though, so you will probably need to do a check if I>32 or something.
EDIT: I often use this short routine for testing bits when I don't want to actually convert a number to binary. For example, with the routine I posted for raising A^K mod M, I used this to test the bits in K.
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