Research on the implementation of teaching reforms and countermeasures of mathematical and intellectual teaching in civil engineering courses

Feature extraction refers to the process of selecting a number of feature words from the image feature space to represent all the features in the feature space based on computer technology, which is the basis of image analysis and image matching. Image features include point features, line features, and surface features, and feature extraction is usually done by feature extraction operators. Multiview image matching is to determine the homonymous image points between images containing the same target based on a specific algorithm, which is the core problem of computer vision and digital photogrammetry, and the accuracy of the result directly affects the subsequent accuracy of airborne triangulation, and is the basis for the subsequent three-dimensional reconstruction and generation of DSM (Digital Surface Model). Image matching is based on gray scale matching and feature matching according to different matching units. The most common application of image feature point extraction is the SIFT operator (Scale Invariant Feature Transform), which is a very famous feature operator for image local feature extraction and matching in the field of computer vision, and it maintains feature invariance for image scale scaling, rotation, luminance change, and affine transformation, and it is suitable for multiview images with target local occlusion, geometric deformation, and clutter scenes.

The SIFT algorithm implementation mainly includes four steps. 1)

Perform keypoint detection in scale space. In the scale space of an image L(x,y,σ)$$L\left( {x,y,\sigma } \right)$$ defined as the original image I(x, y) is convolved with a two-dimensional Gaussian function of variable scale G(x,y,σ)$$G\left( {x,y,\sigma } \right)$$ Convolution operation. 2G(xi,yi,σ)=12πσ2e−x2+y22σ2$$G\left( {{x_i},{y_i},\sigma } \right) = \frac{1}{{2\pi {\sigma^2}}}{e^{ - \frac{{{x^2} + {y^2}}}{{2{\sigma^2}}}}}$$ 3L(x,y,σ)=G(X,Y,σ)*I(x,y)$$L(x,y,\sigma ) = G(X,Y,\sigma )*I(x,y)$$

where X = (x, y, σ)^T is the offset of the keypoint and D is the value of D(x, y, σ) at the keypoint, such that 5∂D(X)∂X=0$$\frac{{\partial D(X)}}{{\partial X}} = 0$$

Obtain the extreme value X^$$\widehat X$$ of X: 6X=∂2D−1∂D∂X2∂X$$\widehat X = \frac{{{\partial^2}{D^{ - 1}}\partial D}}{{\partial {X^2}\partial X}}$$

Corresponding to the extremes, the values of the equations at the extreme points are: 7D(X)=D+12∂DT∂XX$$D(\widehat X) = D + \frac{1}{2}\frac{{\partial {D^T}}}{{\partial X}}\widehat X$$

Where D(X)$$D(\widehat X)$$ is a measure of the contrast value of the feature point, when D(X)<0.03$$D(\widehat X) < 0.03$$, (X)$$(\widehat X)$$ is a non-stable point should be eliminated, and on the contrary, it is retained.

Since DOG will form a strong edge response, resulting in a large principal curvature across the edge region and a small principal curvature in the vertical edge direction, it is necessary to remove all the edge response points, and the principal curvature can be derived from a Hessian matrix of 2*2 for the feature points: 8H=[ Dxx Dxy Dxy Dyy]$$H = \left[ {\begin{array}{*{20}{l}} {{D_{xx}}}&{ {D_{xy}}} \\ {{D_{xy}}}&{ {D_{yy}}} \end{array}} \right]$$

D is the second-order partial derivative, which is derived by interpolating neighboring point sampling points, and the principal curvature of its DOG function is proportional to the eigenvalue of the Hessian matrix, so that ∂ is the largest eigenvalue of the Hessian matrix and β is the smallest eigenvalue of the Hessian matrix, then 9tr(H)=Dxx+Dyy=α+β$$tr(H) = {D_{xx}} + {D_{yy}} = \alpha + \beta$$ 10Det(H)=DxxDyy−(Dxy)2=αβ$$Det(H) = {D_{xx}}{D_{yy}} - {\left( {{D_{xy}}} \right)^2} = \alpha \beta$$

If γ = βγ, then 11tr(H)2Det(H)=(γ+1)2γ$$\frac{{tr{{(H)}^2}}}{{Det(H)}} = \frac{{{{(\gamma + 1)}^2}}}{\gamma }$$

If the two eigenvalues in the above equation are equal then the value of the above equation is the smallest, and as γ increases the value increases, so it can be introduced that the larger the ratio of the two eigenvalues, the larger the gradient value in a certain direction, and the smaller the gradient value in the other direction, and the same as in the case of edge response, in order to eliminate the edge response points and detect whether the principal curvature is under a certain threshold r, only need to compare: 12tr(H)2Det(H)<(γ+1)2γ$$\frac{{tr{{(H)}^2}}}{{Det(H)}} < \frac{{{{(\gamma + 1)}^2}}}{\gamma }$$

Remove the key point that does not satisfy the above equation for the edge response point with lower contrast. 3)

Determine the main direction of the key point, generally use the method of the image gradient of the key point to find the stabilizing direction of the local gradient structure, Gaussian pyramid image (x, y) at the gradient size m(x, y), direction θ(x, y) is: 13m(x,y)=L(x+1,y)−L(x−1,y)+(L(x,y+1)−L(x,y−1))2$$m(x,y) = \sqrt {L(x + 1,y) - L(x - 1,y) + {{(L(x,y + 1) - L(x,y - 1))}^2}}$$ 14θ(x,y)=tan−1{(L(x,y+1)−L(x,y−1))L(x+1,y)−L(x−1,y)}$$\theta (x,y) = {\tan^{ - 1}}\left\{ {\frac{{(L(x,y + 1) - L(x,y - 1))}}{{L(x + 1,y) - L(x - 1,y)}}} \right\}$$

Where L is the scale space value of the key point, after the gradient value m(x, y) is calculated, the Gaussian function is used to weight the gradient size of each pixel within the range of the neighboring window (radius meets the scale sampling 3σ), and the gradient direction of the window is counted in the form of a histogram, and the main peak value of the histogram is designated as the main direction of the key point. At this point, the detection of the key points of the image is finished, and the location, scale and direction information of each key point has been obtained, which in turn determines the SIFT feature region. 4)

Use the descriptor for key point characterization, take the key point as the center, select the 8*8-size region in the domain as the sampling window, divide the region into 4*4-size windows, rotate the key point coordinate axis to the main direction to ensure the rotational invariance, and then use the Gaussian weighting method to classify the 8 relative directions between the sampling point and the feature point into the histogram, and use the 4*4-seeded points to describe each key point, and then finally obtain the SIFT feature vector. 4*4*8 a total of 128 dimensional vectors, i.e., SIFT feature vectors, as shown in Fig. 2. The Euclidean distance of the SIFT feature vectors between multiview images is used as the similarity determination measure of the key points.

Figure 2.

Feature vector generated from keypoint domain gradient information

Revit is a widely used 3D modeling software in the field of BIM, with parametric modeling features, which can effectively integrate the design model with the behavioral model as a whole, so that the data and information can achieve the effect of interrelatedness, and to a certain extent, avoid the situation of model error. Its building model has the functions of collaboration, collision analysis, rendering and so on. The BIM data model used in this paper is mainly created through Revit, and through the building construction drawings, the basic building information is obtained to establish a three-dimensional building model. 1)

Elements

Revit interface design defines different functional modules, and commands with similar functions are arranged together by specialty. Element is one of the most basic types of elements in the model, which is the basic unit of the building model and contains various information of the model. Common elements include walls, columns, beams, floor slabs, doors, windows, furniture and electrical appliances, pipelines, etc. When converting the data format of the model, it mainly relies on the information of elements to ensure that the model’s attributes are mapped.

In Revit projects, the elements are divided into three main categories, which are model elements, datum elements, and view elements. Among them, the three categories can be subdivided into five subcategories, as shown in Figure 3. Model elements are divided into main elements and component elements, the main elements include walls, columns, beams, roofs and foundations and other building components, such elements have been set up in the system, you can call them directly, the model is created only on the basis of the original components of the size of the changes can not be individually set on the appearance and attributes; component elements are ancillary parts of the building, dependent on the main building, including doors, windows and indoor furniture, etc., and can not be set up separately. Doors, windows and indoor furniture, etc., can be changed according to the design needs when creating, and can also be imported into the system for direct use by importing the required component types into the external family library. For some indoor furniture, such as tables, cabinets, chairs, beds and other indoor components, they need to be added after the completion of the main building, and are generally imported from the external family library. Benchmark elements include axis network, elevation, reference plane and reference line, etc., which play the role of assisting modeling when creating the model. Axis network and elevation are important parts of building model creation, which play an important role in the positioning of the model, such as layer, and are important references for BIM modeling. The view element mainly includes two parts: detail element and annotation element. Detail element, also called view element, allows real-time viewing of each plan and information table of the model, and different view windows are effectively connected together to view different view surfaces of the model at the same time, and the plan, elevation, and section of the model are relatively independent, so that you can check the visibility, detail, and scale range of the components through the detail element. Annotation elements include dimensional annotation of the model, model notes and symbols and other elements with explanations. When the annotation elements are modified, the information of the components related to them will be automatically updated with the annotation information, which greatly improves the modeling efficiency in the process of model creation and eliminates the need for repetitive modifications, and has the feature of “one-action linkage”. These five types of basic elements constitute the Revit modeling platform for the construction of three-dimensional building models to improve the efficiency and accuracy.

Figure 3.

Detailed classification of primitives

The experiments were conducted using the traditional analytical method and the method of this paper to process the 3D simulation of the experimental building landscape based on BIM technology in civil engineering courses. The time consumption of the traditional method and the method of this paper during the experiment is shown in Fig. 4.

Figure 4.

Comparison of time spent by different methods

From Fig. 4, the best time of traditional parsing method is 32.67ms, while the time consumed by this paper’s method in the process of shaping the 3D simulation map of architectural landscapes in civil engineering courses is as low as 13.78ms and as high as 29.03ms, which is much lower than the traditional parsing method, and it has a high processing efficiency.

In order to have a more comprehensive understanding of the teaching effect of Mathematical Intelligence, we evaluate the learning effect of students’ homework results, stage test results, accompanying test results and final test results in combination with the “Satisfaction Questionnaire for Mathematical Intelligence Teaching in Civil Engineering Courses” research.

In the reform of Numerical Intelligence Teaching, teachers will have accompanying test questions before, during and after each class. The results of each chapter test were summarized, and the specific results before, during and after class are shown in Figures 5, 6 and 7, respectively.

Figure 5.

The distribution of completion and correct rate of each course before class

Figure 6.

The distribution of completion and correct rate of each course between class

Figure 7.

The distribution of completion and correct rate of each course after class

In terms of the completion rate of test questions, the average completion rate in class is about 90%, which is significantly greater than before and after class, mainly because the teacher sets aside time in class to let students complete the test and explain the answers, so that the class participation is the highest; The accuracy rate is highly correlated with the content of the chapter and the form of the questions, and the accuracy rate of the non-computational chapters is the highest, and the correct rate of the exercises after class is significantly higher than that before and during class, and the correct rate value between the class and the pre-class is not much different, mainly because the pre-class assessment questions are relatively simple, and each test question follows the corresponding knowledge point explanation. The difficulty of the practice questions in the class is relatively high, coupled with the short calculation time, and the engagement of some students is not high, so that the accuracy rate is the same as that of the pre-class test. After learning the knowledge points in the pre-class class, most of the students can complete the after-class test questions well, and the accuracy rate is higher. For example, in the more difficult chapters 5 and 6, the correct rates in the middle of the class were 43.74% and 48.63%, respectively, and after classroom teaching and post-course consolidation, the correct rates in the post-course test increased to 59.82% and 53.61%.

In addition, for the same knowledge points and difficulty, the scores in the midterm test were generally lower than those in the chapter test, mainly because no systematic review was organized before the midterm test and some students had forgotten the more difficult knowledge points. In contrast, before the final test, teachers organized a systematic review of all the knowledge points in the textbook and conducted a series of lectures on the difficult and important points, so that the students were able to integrate what they had learned and performed better in the final results.

In order to compare the effectiveness of teaching and learning, the two traditional civil engineering teaching classes without teaching reforms were used as the basic control group, recorded as Class 1 and Class 2, and the civil engineering class that implemented the teaching reforms of Math and Wisdom was used as the experimental class, recorded as Class 3.

Table 1 shows the basic descriptive analysis of the performance of the three classes. According to Table 1, it can be seen that comparing the scores of the two basic control classes, Class 1 (69.73±22.432) and Class 2 (70.54±19.524), the scores of the experimental class (Class 3) (81.21±16.706) have been substantially improved, with the average scores increased by more than 10 points, and the overall class scores have reached a moderate to high level.

In order to further determine the significance of the difference in performance between classes, post hoc multiple comparisons were conducted using the LSD method, and Table 2 shows the results of the LSD multiple comparisons.

The results show that the performance sig. of the experimental class (class 3) and the traditional teaching class 1 and 2 are 0.000 and 0.001 respectively, which are less than 0.005, indicating that there is a significant difference between both. Reflects that due to the differences in teaching methods, means and so on, led to carry out teaching reform and the implementation of traditional teaching, two different teaching modes brought about by the significant difference in student performance.

Teaching innovation and reform, also includes the reform of the assessment content. When designing the examination questions, in addition to including the examination of basic theoretical knowledge, should also include the examination of civil engineering knowledge application ability. Independent samples t-tests were conducted on the grades of the traditional teaching control class 1 and the experimental class 3 with the same examination paper to further analyze the differences in the degree of mastery of various aspects of the students’ knowledge.

Because the judgment question is only 5 points, it is difficult to distinguish the difference between the two classes, so the judgment question and the multiple choice questions are combined into objective questions, and fill in the blanks, calculations, applications, respectively, as the dependent variable of the independent samples t-test, the results of which are shown in Table 3.

According to the data in Table 3, students in the experimental class scored higher than the control class in objective questions, subjective questions, and total scores, which indicates that the teaching effect after the implementation of the teaching reform is significantly better than that of the traditional teaching method, and that the penetration of a variety of teaching methods has led to a more pronounced improvement in both theoretical and applied knowledge. Meanwhile, the results also show that in the objective and fill-in-the-blank sections, there is no significant difference between the experimental class and the control class (P>0.05). While in the calculation questions, there is a significant difference (P<0.05); in the application questions, there is a highly significant difference (P<0.01). Traditional teaching pays more attention to students’ solid mastery of theoretical knowledge, so in terms of objective and fill-in-the-blank questions that test basic theoretical knowledge, the performance of traditional teaching classes is not significantly inferior to that of experimental classes. However, for the examination of knowledge application ability, due to the penetration of a variety of teaching methods, classroom teaching from teacher-led, gradually transitioned to student-centered, which makes the students more fully join the teaching process, hands-on practice of the initiative is more intense, so the experimental class students in the application of the ability of the class is significantly better than the traditional teaching class, with a very significant difference. Through the above fundamental change in the teaching mode, the overall improvement of learning achievement was finally achieved.