cyclomatic complexity of the pseudo-code

Question

Cyclomatic complexity is number of test cases that are necessary to achieve thorough test coverage of a particular module.

Consider the following pseudo-code :

If (A > B) and (C > D) then
    A = A + 1
    B = B + 1
Endif

I think only two test cases are required here, one for true condition another for false condition, so cyclomatic complexity should be 2 but answer is 3(not sure why).

Someone please help me understand why the answer is 3 instead of 2.

Answer 1

The code will be translated by compilers to something like the following pseudo-assemley code:

A - B
branch if greater then 0 to end
C - D
branch if greater then 0 to end
inc A
inc B
end:
...

Now, in order to make all possible choices in all branches, you need 3 test cases:

1. A <= B (branch in first branch instruction)
2. C <= D (branch in second branch instruction)
3. A > B and C > D (don't branch in 2nd branch instruction and get to the increasing instructions)

Note that for alternative definition of "coverage" - "All instructions coverage", 1 test case can be enough (with the above compiled code), because even if you don't branch, the "branch if.." instruction is still executed, so for A>B and C>D test case, you actually go through all instructions.

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P.S. assuming here:

Cyclomatic complexity is number of test cases that are necessary to achieve thorough test coverage of a particular module.

Answer 2

Cyclomatic complexity directly measures the number of linearly independent paths through a program's source code.

If (A > B) and (C > D) then
    A = A + 1
    B = B + 1
Endif

Case 1.

If (true) and (true) then
    // Execute some code
Endif

Case 2.

If (true) and (false) then
    // Don't execute code
Endif

Case 3.

If (false) and (2nd part won't be executed) then
    // Don't execute code
Endif

So Cyclomatic complexity will be 3.