An Approach
lWrite iterative pseudo-code,
then construct recursive equivalent.
then construct recursive equivalent.
lL = É list to be converted É;
Result = 0;
while( L != [ ] )
{
Result = 2*Result + first(L);
L = rest(L);
}
É answer is in Result ...
Result = 0;
while( L != [ ] )
{
Result = 2*Result + first(L);
L = rest(L);
}
É answer is in Result ...
lDefining
fromBinary3(L, Result):
lfromBinary3([],
Result) => Result;
lfromBinary3([F
| R], Result) =>
fromBinary3(R, 2*Result+F);
fromBinary3(R, 2*Result+F);
lfromBinary(L)
= fromBinary3(L, 0);
lfromBinary3([1,
0, 0, 1, 0, 1], 0) ==>
lfromBinary3([0,
0, 1, 0, 1], 1) ==>
lfromBinary3([0,
1, 0, 1], 2) ==>
lfromBinary3([1,
0, 1], 9) ==>
lfromBinary3([0,
1], 9) ==>
lfromBinary3([1],
18) ==>
lfromBinary3([],
37) ==>
l37