Collatz Conjecture in Assembly shortest form

问题

We had an assignment where we had to write the collatz conjecture in 64bit nasm assembly with only 13 commands or less (RET included). Now we are wondering how much you can actually reduce it. We are currently on 9.
Heres the collatz conjecture in pseudo code for reference:

Heres the code we have so far. A few notes:
A tutor of us said we can remove the XOR rax, rax because of some calling convention it's already zero. It doesn't work on my computer though so I've included it here.
I'm aware the two LEA's are probably the most obvious thing to reduce, but we can't think of a way since *6 seems to be the only thing thats literally impossible to do with LEA.

GLOBAL collatz
SECTION .text

collatz:
    XOR rax, rax

    .while:
        SHR rdi, 1
        JNC .even
            LEA rdi, [rdi*2+1]
            LEA rdi, [rdi*2+rdi+1]
        .even:

        INC rax

        CMP rdi, 1
        JA .while
    RET

回答1:

This is somewhat shorter:

collatz:
        or     $-1,%eax
.loop:  inc    %eax
        lea    1(%rdi,%rdi,2),%rsi
        shr    %rdi
        cmovc  %rsi,%rdi
        jnz    .loop
        ret

or, in nasm syntax:

collatz:
        or     eax,-1
.loop:  inc    eax
        lea    rsi,[rdi+rdi*2+1]
        shr    rdi
        cmovc  rdi,rsi
        jnz    .loop
        ret

To understand this code, pay close attention to the carry flag (CF) and zero flag (ZF).

来源：https://stackoverflow.com/questions/42149201/collatz-conjecture-in-assembly-shortest-form