Elementary comparison testing

A test case consists of a logical path from start to end of a process, through one or many decisions. Contradictory situations are deduced from the test case matrix and excluded. The MC/DC approach isolates every condition, neglecting all possible subpath combinations and path coverage.[1]

${\displaystyle T=n+1}$

where

• T is the number of test cases per decision, and
• n the number of conditions.

The decision ${\displaystyle d_{i}}$ consists of a combination of elementary conditions

${\displaystyle {\begin{aligned}\Sigma &=\{0,1\}\\C&=\{c_{0},c_{1},c_{2},c_{3},...,c_{n}\}\end{aligned}}}$

${\displaystyle \epsilon :C\rightarrow \Sigma \times C}$

${\displaystyle D\subseteq C^{*}}$; ${\displaystyle d_{i}\in D}$

The transition function ${\displaystyle \alpha }$ is defined as

${\displaystyle \alpha :D\times \Sigma ^{*}\rightarrow \Sigma \times D}$

Given the transition ${\displaystyle \vdash }$

${\displaystyle \vdash \subseteq (\Sigma \times D\times \Sigma ^{*})\times (\Sigma \times D\times \Sigma ^{*})}$

${\displaystyle S_{j}=(b_{j},d_{m},v_{j})\vdash (b_{j+1},d_{n},v_{j+1})}$
${\displaystyle E_{j}=(a_{j},c_{j})\vdash (a_{j+1},c_{k})}$
${\displaystyle (b_{j+1},d_{n})=\alpha (d_{m},v_{j});(b_{j+1},c_{k})=\epsilon (c_{j});a_{j}\in \Sigma ,}$

the isolated test path ${\displaystyle P_{m}}$ consists of

${\displaystyle {\begin{aligned}P_{m}&=(b_{0},d_{0},v_{0})\vdash ...\vdash (b_{i},d_{i},v_{i})\vdash ^{*}(b_{n},d_{n},v_{n})\\&=(b_{0},c_{0})\vdash ...\vdash (b_{m},c_{m})\vdash ^{*}(b_{n},c_{n})\end{aligned}}}$

${\displaystyle b_{i}\in \Sigma ;c_{m}\in d_{i};v\in C^{*};d_{0}=S;d_{n}=E.}$

A test case graph illustrates all the necessary independent paths (test cases) to cover all isolated conditions. Conditions are represented by nodes and condition values (situations) by edges. All program situations are addressed by an edge. Each situation is connected to one preceding and successive condition. Test cases might overlap due to isolated conditions.

The elementary comparison testing method can be used to determine the number of condition paths by inductive proof.

Figure 2: ECT Inductive Proof Anchor

There are ${\displaystyle r=2^{n}}$ possible condition value combinations

${\displaystyle \forall {i}\in \{1,...,n\}\ c_{i}\mapsto \{0,\ 1\}}$.

When each condition ${\displaystyle c_{i}}$ is isolated, the number of required test cases ${\displaystyle T}$ per decision is:

${\displaystyle T=\log _{2}(r)+1=n+1.}$

Figure 3: ECT Inductive Proof End

${\displaystyle \forall {i}\in \{1,...,n\}}$ there are ${\displaystyle 0<e<i+1}$ edges from parent nodes ${\displaystyle c_{i}}$ and ${\displaystyle s=2}$ edges to child nodes from ${\displaystyle c_{i}}$.

Each individual condition ${\displaystyle c_{i}}$ connects to at least one path

${\displaystyle \forall {i}\in \{1,...,n-1\}\ c_{i}\mapsto \{0,\ 1\}}$

from the maximal possible ${\displaystyle n}$ connecting to ${\displaystyle c_{n}}$ isolating ${\displaystyle c_{n}}$.

All predecessor conditions ${\displaystyle c_{i};\ i<n}$ and respective paths are isolated. Therefore, when one node (condition) is added, the total number of paths, and required test cases, from start to finish increases by:

${\displaystyle {\begin{aligned}T&=n-1+2\\&=n+1.\end{aligned}}}$

Q.E.D.

1. Identify decisions
2. Determine test situations per decision point (Modified Condition / Decision Coverage)
3. Create logical test-case matrix
4. Create physical test-case matrix

Figure 4: ECT Example Control-Flow Graph

Figure 5: ECT Example D2 Conditions

This example shows ETC applied to a holiday booking system. The discount system offers reduced-price vacations. The offered discounts are ${\displaystyle -20\%}$ for members or for expensive vacations, ${\displaystyle -10\%}$ for moderate vacations with workday departures, and ${\displaystyle 0\%}$ otherwise. The example shows the creation of logical and physical test cases for all isolated conditions.

Pseudocode

if days > 15 or price > 1000 or member then
    return −0.2
else if (days > 8 and days ≤ 15 or price ≥ 500 and price ≤ 1000) and workday then
    return −0.1
else
    return 0.0

Factors

• Number of days: ${\displaystyle <8;\ 8-15;\ >15}$
• Price (euros): ${\displaystyle <500;\ 500-1000;\ >1000}$
• Membership card: none; silver; gold; platinum
• Departure date: workday; weekend; holiday

${\displaystyle T=3\times 3\times 4\times 3=108}$ possible combinations (test cases).

Step 1: Decisions

Table 1: Example D1 MC/DC

		Outcome
Decision D1		1			0
Conditions		c1	c2	c3	c1	c2	c3
c1	${\displaystyle {\text{days}}>15}$	1	0	0	0	0	0
c2	${\displaystyle {\text{price}}>1000}$	0	1	0	0	0	0
c3	${\displaystyle {\text{member}}}$	0	0	1	0	0	0

${\displaystyle {\begin{aligned}d_{1}&={days}>15\ {\text{or}}\ {\text{price}}>1000\ {\text{Eur}}\ {\text{or}}\ {\text{member}}\\c_{1}&={\text{days}}>15\\c_{2}&={\text{price}}>1000\\c_{3}&={\text{member}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}d_{2}&=(8<{\text{days}}<15\ {\text{or}}\ 500<{\text{price}}<1000\ {\text{Eur}})\ {\text{and}}\ {\text{workday}}\\c_{4}&=8<{\text{days}}<15\\c_{5}&=500<{\text{price}}<1000\ {\text{Eur}}\\c_{6}&={\text{workday}}\\\end{aligned}}}$

Step 2: MC/DC Matrix

Table 2: Example D2 MC/DC

		Outcome
Decision D2		1			0
Conditions		c4	c5	c6	c4	c5	c6
c4	${\displaystyle 8<{\text{days}}<15}$	1	0	1	0	0	1
c5	${\displaystyle 500<{\text{price}}<1000}$	0	1	1	0	0	1
c6	${\displaystyle {\text{workday}}}$	1	0	1	1	0	0

The highlighted diagonals in the MC/DC Matrix are describing the isolated conditions:

${\displaystyle (c_{i},c_{i})\mapsto \{1,0\}}$

all duplicate situations are regarded as proven and removed.

Step 3: Logical test-Case matrix

Table 3: Example Logical Test Case Matrix

Situation ${\displaystyle S_{j}}$	${\displaystyle T_{1}}$	${\displaystyle T_{2}}$	${\displaystyle T_{3}}$	${\displaystyle T_{4}}$	${\displaystyle T_{5}}$	${\displaystyle T_{6}}$	${\displaystyle T_{7}}$
${\displaystyle \alpha (d_{1},\mathbf {1} 00)\mapsto (1,E)}$	x
${\displaystyle \alpha (d_{1},\mathbf {0} 00)\mapsto (0,d_{2})}$		x			x	x	x
${\displaystyle \alpha (d_{1},0\mathbf {1} 0)\mapsto (1,E)}$			x
${\displaystyle \alpha (d_{1},00\mathbf {1} )\mapsto (1,E)}$				x
${\displaystyle \alpha (d_{2},\mathbf {1} 01)\mapsto (1,E)}$		x
${\displaystyle \alpha (d_{2},\mathbf {0} 01)\mapsto (1,E)}$						x
${\displaystyle \alpha (d_{2},0\mathbf {1} 1)\mapsto (1,E)}$					x
${\displaystyle \alpha (d_{2},11\mathbf {0} )\mapsto (0,E)}$							x

Test cases are formed by tracing decision paths. For every decision ${\displaystyle d_{i};\ 0<i<n+1}$ a succeeding and preceding subpath is searched until every connected path has a start ${\displaystyle S}$ and an end ${\displaystyle E}$:

${\displaystyle {\begin{aligned}T_{1}&=(d_{1},100)\vdash (1,E)\\T_{2}&=(d_{1},000)\vdash (0,d_{2},100)\vdash (1,E)\\T_{3}&=(d_{1},010)\vdash (1,E)\\\vdots \\T_{n+1}\end{aligned}}}$

Step 4: Physical test-case matrix

Table 4: Example Physical Test Cases

Factor\Test Case	${\displaystyle T_{1}}$	${\displaystyle T_{2}}$	${\displaystyle T_{3}}$	${\displaystyle T_{4}}$	${\displaystyle T_{5}}$	${\displaystyle T_{6}}$	${\displaystyle T_{7}}$
days	16	14			8	8	8
price			1100			600
departure							sa
member				silver
Result
0							0
-10		1			1	1
-20	1		1	1

Physical test cases are created from logical test cases by filling in actual value representations and their respective results.

Test-case graph

Figure 6: ECT Example Test Case Graph

In the example test case graph, all test cases and their isolated conditions are marked by colors and the remaining paths are implicitly passed.

• Multiple condition coverage
• Control-flow graph
• Decision to decision path