user: What would the C source look like for the assembly code provided here?

```asm
<adjust>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  mov    %r14,%rdi
  call   <condread@plt>
  cmp    $0x9,%eax
  jg     L63
  test   %eax,%eax
  jle    L207
  sub    $0x1,%eax
  mov    %r14,0x8(%rsp)
  mov    %r14,%rbp
  lea    0x14(%rsp,%rax,4),%r15
  jmp    L22
  nopl   (%rax)
L105:
  xor    %eax,%eax
  mov    %r14d,%edi
  call   <row@plt>
  cmp    %r12d,%eax
  je     L192
  mov    %r14d,%edi
  xor    %eax,%eax
  call   <row@plt>
  mov    %r12d,%edx
  mov    %rbx,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <rowexchange@plt>
  test   %eax,%eax
  je     L63
L192:
  xor    %eax,%eax
  mov    %r14d,%edi
  call   <column@plt>
  cmp    %r13d,%eax
  je     L0
  mov    %r14d,%edi
  xor    %eax,%eax
  call   <column@plt>
  mov    %r13d,%edx
  mov    %rbx,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <columnexchange@plt>
  test   %eax,%eax
  je     L63
L0:
  add    $0x4,%rbp
  cmp    %rbp,%r15
  je     L226
L22:
  mov    0x0(%rbp),%edi
  xor    %eax,%eax
  call   <row@plt>
  mov    0x0(%rbp),%edi
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <column@plt>
  mov    0x0(%rbp),%edi
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <quadrant@plt>
  movslq %eax,%rdx
  cmpl   $0x1,(%rbx,%rdx,4)
  mov    %rdx,%r14
  lea    0x0(,%rdx,4),%rax
  je     L105
  cmpl   $0x1,0x4(%rbx,%rax,1)
  lea    0x1(%rdx),%edx
  je     L74
  cmpl   $0x1,0x8(%rbx,%rax,1)
  lea    0x2(%r14),%edx
  je     L74
  cmpl   $0x1,0x24(%rbx,%rax,1)
  lea    0x9(%r14),%edx
  je     L74
  cmpl   $0x1,0x28(%rbx,%rax,1)
  lea    0xa(%r14),%edx
  je     L74
  cmpl   $0x1,0x2c(%rbx,%rax,1)
  lea    0xb(%r14),%edx
  je     L74
  cmpl   $0x1,0x48(%rbx,%rax,1)
  lea    0x12(%r14),%edx
  je     L74
  cmpl   $0x1,0x4c(%rbx,%rax,1)
  lea    0x13(%r14),%edx
  je     L74
  add    $0x14,%r14d
  cmpl   $0x1,0x50(%rbx,%rax,1)
  je     L105
  lea    D5(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  jmp    L170
L41:
  add    $0x4,%rcx
  mov    %rcx,0x8(%rsp)
  cmp    %r15,%rcx
  je     L207
L226:
  mov    0x8(%rsp),%rcx
  movslq (%rcx),%rax
  cmpl   $0x1,(%rbx,%rax,4)
  je     L41
  nopl   0x0(%rax)
L63:
  xor    %eax,%eax
L170:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L90
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L207:
  mov    $0x1,%eax
  jmp    L170
L90:
  call   <__stack_chk_fail@plt>
L74:
  mov    %edx,%r14d
  jmp    L105
```

tool_calls: name: read_data
arguments: data_label: D5
type: string

tool: D5: Sudoku invalid.

assistant: ```c
int adjust(sudoku)
int sudoku[];
{
  int condarray[10];
  int a, b, c, i, j, k;
  b = condread(condarray);
  if (b > 9) {
    return (0);
  }
  for (c = 0; c < b; c++) {
    i = row(condarray[c]);
    j = column(condarray[c]);
    k = quadrant(condarray[c]);
    a = k;
    if (sudoku[a] != 1) {
      a = k + 1;
    }
    if (sudoku[a] != 1) {
      a = k + 2;
    }
    if (sudoku[a] != 1) {
      a = k + 9;
    }
    if (sudoku[a] != 1) {
      a = k + 10;
    }
    if (sudoku[a] != 1) {
      a = k + 11;
    }
    if (sudoku[a] != 1) {
      a = k + 18;
    }
    if (sudoku[a] != 1) {
      a = k + 19;
    }
    if (sudoku[a] != 1) {
      a = k + 20;
    }
    if (sudoku[a] != 1) {
      printf("Sudoku invalid.\n");
      return (0);
      printf("Position %d %d %d %d \n", a, row(a), column(a), quadrant(a));
    }
    if (row(a) != i) {
      if (rowexchange(sudoku, row(a), i) == 0) {
        return (0);
      }
    }
    if (column(a) != j) {
      if (columnexchange(sudoku, column(a), j) == 0) {
        return (0);
      }
    }
  }
  for (c = 0; c < b; c++) {
    if (sudoku[condarray[c]] != 1) {
      return (0);
    }
  }
  return (1);
}
```