Understanding/decoding obscure assembly code

Question

Like my previous question, this involves an assignment where a method is called which requires a certain password, the code is hidden and we have to infer the password from the assembly code (I want to avoid hitting . I've completed a few phases so far and I'm getting better at understanding, however this phase has a few aspects I am having trouble with. So far I know the password to this phase is two integers. Backtracing has been my goto method for some of these but is not very helpful for this phase.

1. I understand cltq expands eax(rax) to 4 words, but I'm not sure how this affects calculations, and also unsure what happens if that line is hit multiple times.
2. the phase5+82 -> phase5+65 (a loop?) What values am I trying to start with so ecx(rcx) can pass the final comparison?
3. mov 0x402600(,%rax,4),%eax <- what does this line do exactly? The blank is throwing me off, is a blank = 0?

4. Any other help understanding what is going on and how I should approach figuring out the input would be helpful, I have attempted to convert this back into C code, like previous phases

   0x00000000004010b4 <phase_5+0>:         sub    $0x18,%rsp
   0x00000000004010b8 <phase_5+4>:         lea    0x10(%rsp),%rcx
   0x00000000004010bd <phase_5+9>:         lea    0x14(%rsp),%rdx
   0x00000000004010c2 <phase_5+14>:        mov    $0x4026aa,%esi
   0x00000000004010c7 <phase_5+19>:        mov    $0x0,%eax
   0x00000000004010cc <phase_5+24>:        callq  0x400b80<sscanf@plt>
   0x00000000004010d1 <phase_5+29>:        cmp    $0x1,%eax
   0x00000000004010d4 <phase_5+32>:        jg     0x4010db<phase_5+39>
   0x00000000004010d6 <phase_5+34>:        callq  0x401421(explode_bomb)
   0x00000000004010db <phase_5+39>:        mov    0x14(%rsp),%eax
   0x00000000004010df <phase_5+43>:        and    $0xf,%eax
   0x00000000004010e2 <phase_5+46>:        mov    %eax,0x14(%rsp)
   0x00000000004010e6 <phase_5+50>:        cmp    $0xf,%eax
   0x00000000004010e9 <phase_5+53>:        je     0x40111b <phase_5+103>
   0x00000000004010eb <phase_5+55>:        mov    $0x0,%edx
   0x00000000004010f0 <phase_5+60>:        mov    $0x0,%ecx
   0x00000000004010f5 <phase_5+65>:        add    $0x1,%edx
   0x00000000004010f8 <phase_5+68>:        cltq   
   0x00000000004010fa <phase_5+70>:        mov    0x402600(,%rax,4),%eax
   0x0000000000401101 <phase_5+77>:        add    %eax,%ecx
   0x0000000000401103 <phase_5+79>:        cmp    $0xf,%eax
   0x0000000000401106 <phase_5+82>:        jne    0x4010f5 <phase_5+65>
   0x0000000000401108 <phase_5+84>:        movl   $0xf,0x14(%rsp)
   0x0000000000401110 <phase_5+92>:        cmp    $0xf,%edx
   0x0000000000401113 <phase_5+95>:        jne    0x40111b <phase_5+103>
   0x0000000000401115 <phase_5+97>:        cmp    %ecx,0x10(%rsp)
   0x0000000000401119 <phase_5+101>:       je     0x401120 <phase_5+108>
   0x000000000040111b <phase_5+103>:       callq  0x401421 <explode_bomb>
   0x0000000000401120 <phase_5+108>:       add    $0x18,%rsp
   0x0000000000401124 <phase_5+112>:       retq   
   

Answer 1

The code translates to this:

0x402600: int table[15];
0x4026aa: const char *format;

void func (const char *str)
{
    int a, b, count, sum;

    if (sscanf (str, format, &a, &b) != 2) {
        explode_bomb();
    }

    a = a & 0xF;
    if (a == 0xF) {
        explode_bomb();
    }

    sum = 0;
    count = 0;
    while (a != 0xF) {
        a = table[a];
        sum += a;
        count++;
    }

    if ((count != 0xF) || (sum != b)) {
        explode_bomb ();
    }
}

To answer your specific points:

cltq is used to clear the 4 most significant bytes of rax so as to not interfere with the address calculation in the following instruction. It has no impact on the calculation.

mov 0x402600(,%rax,4),%eax <- what does this line do exactly? The blank is throwing me off, is a blank = 0?

Yes, this is just mov dword eax, [0x402600 + 0 + rax * 4]

Once you have the C equivalent it is easy to figure out how to find a solution.