Definition

The following listed the terms defined by this study; these terms would be explained, and their origins of definition and some of their calculation methods would be described also.

3.1.1. Rotation

The “rotation” function defined in this study was: when a 3D object was displayed on a 2D screen, one could rotate the 3D object by a mouse to display it on the screen from multi viewpoint. The students or examinees could see it from multi viewpoint, and all sides could be completely displayed. However, if all six sides of any cube were covered by other cubes, then they couldn’t see it however they used their mouse to rotate it. The example of cube could be rotated by using mouse is as follow:

40 Figure 3-1. Illustration of item rotation

3.1.2. Invisible Cube Number

The cube enumeration test would display a 3D object on a 2D screen, and some cubes would be invisible due to the problem of 2D viewing point. Invisible cube meant the cube that the examinees couldn’t see. “Invisible cube number” meant the total amount of cubes that they couldn’t see; that is, in an item of the test, when subtracting visible cube number from the total cube number then we had “invisible cube number”.

Take the following figure as an example. The total cube number was 8, but when it was displayed on a 2D screen we could only see 6 cubes, and we needed spatial orientation and spatial visualization to display the other two cubes in our minds. These two cubes were invisible cubes, and in this item the “invisible cube number” is 2.

41

Figure 3-2. An example item that the total cube number is 8

3.1.3. Integrity of cubes

According to the literature review, we reconsidered the process of solving cuboid problem from the perspective of layer and analyzed the degree to which each layer of space was filled up by cubes. The Karnaugh map which is the simplification of boolean-algebra was used to calculate the integrity of cubes. While using Boolean theorems to simplify boolean function, we found that it was always difficult to quickly find out which item needed decomposition or combination in the function, and we couldn’t be sure about whether the final result was the most simplified form.

Therefore users expected a step-by-step simplification method with certain rules.

Karnaugh map satisfied the request.

Karnaugh map was developed by an electrical engineer named Karnaugh of Bell Laboratories. It reached the simplification purpose by operating images and using truth table. Karnaugh map with less than 5 elements could be used easily. However, when there were more than 5 elements, it would be too complicated to use it. Hence, we use computer program to do it when needed.

42

The key points of doing simplification using Karnaugh map were described as follows. When there are N variables in Boolean function, Karnaugh map needs to have 2^N squares. For any two adjacent squares, i.e. any two adjacent items, there was only one different character between them. Figure 3-3 to Figure 3-5 represented (a) two-variable Karnaugh map, (b) three-two-variable Karnaugh map, and (c) four-two-variable Karnaugh map respectively.

1. two-variable Karnaugh map

The following was a two-variable Karnaugh map. When any blank of 0, 1, 2, and 3 had the same value with the adjacent one, then they could be combined by the order of 4 blanks, 2 blanks and 1 blank. If all values of 0, 1, 2, and 3 were all the same, then these 4 blanks could be combined to 1 blank and so on.

Figure 3-3. Two-variable Karnaugh map

2. three-variable Karnaugh map

The following was a three-variable Karnaugh map. When any blank of 0, 1, 2, 3, 4, 5, 6, 7 and 8 had the same value with the adjacent one, then they could be combined by the order of 8 blanks, 4 blanks, 2 blanks and 1 blank. If all values of 0, 1, 2, and 3 were the same, then these 4 blanks could be combined to 1 blank and so on. However, 0 and 2 were also adjacent so when the values of 0, 2, 4, and 6 were the same, then they could be combined too.

43 Figure 3-4. Three-variable Karnaugh map

3. four-variable Karnaugh map

The following was a four-variable Karnaugh map. When any blank of 0 ~ 15 (16 blanks) had the same value with the adjacent one, then they could be combined by the order of 16 blanks, 8 blanks, 4 blanks, 2 blanks and 1 blank. If all values of blank 0, 1, 2, and 3 were all the same, then these 4 blanks could be combined to 1 blank and so on.

However, 0 & 2 and 0& 8 were also adjacent so when the values of 0, 2, 8, and 10 were the same, then they could be combined too.

Figure 3-5. Four-variable Karnaugh map

This study adopted the modified Karnaugh map to calculate the integrity of cubes.

In addition to that two adjacent 1 and four adjacent 1 could be eliminated, eight adjacent 1 could eliminate 3 complementary variables and sixteen adjacent 1 could eliminate 4 complementary variables. This study also added a “three adjacent 1, six

44

adjacent 1 and nine adjacent 1 could be eliminated“ rule. The adjacent example is as follow.

The following cubes could be divided into 3 layers and further calculated; they were Layer 1, Layer 2 and Layer 3. For Layer 1, we could find out that the biggest range that we could circle in the first time included the middle 4 blanks, and then we could circle the 2 single cubes on the sides. Hence, “integrity of cubes” of Layer 1 was 3. Next, for layer 2 we could find out that the biggest range that we could circle in the first time included 2 cubes and then we could only circle 1 cube. Because we circled twice, “integrity of cubes” of Layer 2 was 2. Finally, for Layer 3 there was only 1 cube hence we could only circle 1 cube, and “integrity of cubes” of Layer 3 was 1. At last, we added the numbers of these 3 layers and we obtained the total “integrity of cubes”

as 6.

Figure 3-6. An item is divided into 3 layers by horizontal method.

Figure 3-7. Calculate the integrity of cubes.

45

Besides, we could also calculate it by vertical method and the item will be divided into layer A, layer B and layer C. For Layer A, we found out that the biggest range that we could circle included the middle 4 cubes in the first time and we could circle the 2 single cubes on the other side. Hence, “integrity of cubes” of Layer A was 3. Next, for layer B we could find out that the biggest range that we could circle in the first time included 2 cubes and then we could only circle 1 cube. Because we circled twice,

“integrity of cubes” of Layer B was 2. Finally, for Layer C there was only 1 cube hence we could only circle 1 cube, and “integrity of cubes” of Layer 3 was 1. At last, we added the numbers of these 3 layers and we obtained the total “integrity of cubes”

as 6.

Figure 3-8. An item is divided into 3 layers by vertical method.

Figure 3-9. Calculate the integrity of cubes.

46

When the completeness degree of cubes were calculated, circling more independent parts meant that the cube was less completed, and therefore we should try to circle the most adjacent 1 first. If there was any independent cube whose value was 1, then we should circle it independently and it couldn’t be missed. The hypothesis on this factor in this study is that lower integrity value of cubes will reduce item difficulty.