My shot at this using alternate:
// %0 contains [0,0,1,0]
// %1 contains [A,B,C,D]
F("pick_reduce_4") %0, %1;
pick (%0, %1) %2;
alternate (%2, %2) %3 %4,
recur (%3, %4, %3) %5;
alternate (%5, %5) %6 %7,
recur (%6, %7, %6) %8;
extract (%8, w(0ul)) %9
ret (%9);
Some example runs:
input: %0 = [1,0,0,0] %1 = [a,b,c,d]
%2 = [a,,,]
%3 = [a,a,,] %4=[,,,]
%5 = [a,a,,]
%6 = [a,a,a,a] %7=[,,,]
%8 = [a,a,a,a]
%9 = a
input: %0 = [0,1,0,0] %1 = [a,b,c,d]
%2 = [,b,,]
%3 = [,,b,b] %4=[,,,]
%5 = [,,b,b]
%6 = [,,,] %7=[b,b,b,b]
%8 = [b,b,b,b]
%9 = b
input: %0 = [0,0,1,0] %1 = [a,b,c,d]
%2 = [,,c,]
%3 = [,,,] %4=[c,c,,]
%5 = [c,c,,]
%6 = [c,c,c,c] %7=[,,,]
%8 = [c,c,c,c]
%9 = c
input: %0 = [0,0,0,1] %1 = [a,b,c,d]
%2 = [,,,d]
%3 = [,,,] %4=[,,d,d]
%5 = [,,d,d]
%6 = [,,,] %7=[d,d,d,d]
%8 = [d,d,d,d]
%9 = d
Shuffle is to reorder a vector? If I understand it correctly:
shuffle ([a,b,c,d], [3,2,1,0]) = [d,c,b,a]
shuffle ([a,b,c,d], [1,1,1,1]) = [b,b,b,b]
shuffle ([a,b,c,d], [0, None, None, None]) = [a, None, None, None]
Then perhaps we can use the left operation to find the index of the selected element, this should work for vectors of any size and it would be O(1):
F("pick_reduce") %0, %1;
left (%0) %2;
// this just builds a vector the same width as the data vector
// the first element is the index of the element we want
// the rest do not matter and can be junk
inject (%1, %2, 0) %3;
shuffle (%1, %3) %4;
extract (%4, 0) %5,
ret (%5);
If I understand correctly, the left operation will return the amount of leading zeroes. So if %0 = [0,0,1,0], then %2 will be 2.
Then I create a vector with that result as the first element in %3 and use that to shuffle the input vector, the first element in the result will be the element at position 2.
Is this correct Ivan?
