Method of masonry arrangement of secondary structural wall

Abstract

The application provides a brick arrangement method of a secondary structure wall, and can generate an optimal scheme of brick arrangement meeting specification requirements according to boundary and opening information of the secondary structure wall and brick arrangement design information, so that the number of broken bricks is minimized, the workload of manual brick cutting is saved, and material consumption and solid waste are reduced. The application has the advantages that different numbers of openings can be considered to generate multiple sets of optimization schemes, is suitable for different specifications of brick materials, and has the characteristics of strong applicability. Furthermore, the method innovatively adopts five variables to construct a chromosome model of the arrangement scheme, that is, the compliance of the scheme is ensured, and each brick has at most three broken bricks; meanwhile, the height of a guide wall can be optimized to reduce the number of broken bricks at the top.

Description

TECHNICAL FIELD

[0001] The application relates to a brick arrangement method of a secondary structure wall. BACKGROUND

[0002] The secondary structure is a wall for dividing the building space in a large public building, and has a large quantity and wide range, and has a great influence on the construction progress, quality and cost. The construction drawing delivered by the design institute often only gives the basic principle of the brick arrangement of the wall of the secondary structure, such as the distance of the construction column, the distance of the ring beam and the size of the brick, but the design does not provide the brick arrangement drawing of the brick structure. Therefore, the workers often arrange it by themselves according to the actual situation at the construction site, so that the brick materials are cut randomly, a large number of broken bricks are formed, the cost is high, the material waste is large, the aesthetic property is poor and other problems occur frequently.

[0003] Although there are some brick arrangement methods on the market, it is difficult to adapt to different sizes of brick materials and different elevations of door openings and other factors; and cannot give multiple schemes for the construction party to select according to the actual situation, but can only provide a unique scheme, so the practical application is less. SUMMARY

[0004] The application aims to provide a brick arrangement method of a secondary structure wall.

[0005] To solve the above problems, the application provides a brick arrangement method of a secondary structure wall, which comprises the following steps:

[0006] Step 1: setting brick arrangement design information;

[0007] Step 2: establishing a wall BIM model W;

[0008] Step 3: inputting constraint conditions according to the brick arrangement design information and the wall BIM model W;

[0009] Step 4: establishing an initial scheme library GA of the brick arrangement by using a genetic algorithm according to the wall model W, the brick arrangement design information and the constraint conditions, z setting iteration number z=0,

[0010] Step 5: calculating the number of broken bricks of the wall periphery of each scheme ga z in GA i ;

[0011] Step 6: calculating the number of broken bricks of the door opening periphery of each scheme ga z in GA i ;

[0012] Step 7: calculating the fitness value F z of the scheme ga i in GA i according to the number of broken bricks of the wall, Fi F is inversely proportional to the number of broken bricks i =f(s j );

[0013] Step 8: Select N chromosomes from GA i according to the fitness value F z , and put them into the chromosome library GA z+1 ; that is, the elements ga i with high fitness value F i enter GA z+1 with a higher frequency;

[0014] Step 9: Perform chromosome crossover and mutation on the brick arrangement scheme ga z+1 in GA i to form a child chromosome library GA z+2 , increase the iteration number by 1, and assign the value of GA z+2 to GA z ;

[0015] Step 10: When the iteration number z reaches the maximum evolution number Z, go to Step 11; otherwise, go to Step 5.

[0016] Step 11: Select the K groups with the largest resource optimization index F i from the chromosome library GA as the result output.

[0017] Further, in the above method, step 1: setting the masonry arrangement design information includes:

[0018] Step 1.1: inputting masonry size information, including: length bs, width bw, and height bh;

[0019] Step 1.2: inputting mortar joint width t;

[0020] Step 1.3: inputting the length of the dog-tooth d;

[0021] Step 1.4: inputting the upper and lower limits of the guide wall height H1 and H2.

[0022] Further, in the above method, step 2: establishing a wall BIM model W includes:

[0023] Step 2.1: inputting the wall bottom elevation h l , the wall top elevation h t , the wall length L, whether the left side is a dog-tooth c l , and whether the right side is a dog-tooth c r ; that is, W={h l , h t , L, c l , c r}

[0024] Step 2.2: Calculate the odd skin brick length L1 and the even skin brick length L2, the formulas are as follows:

[0025] L1 = L - (c l +c r )*d Formula (1)

[0026] L2 = L; Formula (2)

[0027] Step 2.3: Input the hole information O = {hole o j} on the wall, the general hole o j is rectangular, including the hole bottom elevation ohd j , the hole top elevation oht j , the wall hole width ol j and the distance from the left side of the wall xo j ; that is, the hole o j = (ohd j , oht j , ol j , xo j )

[0028] Further, in the above method, step 3: according to the masonry arrangement design information and the wall BIM model W, input the constraint conditions, including:

[0029] Step 3.1: The first brick length X1, the second brick length X2, the tail brick length X3 of the odd skin brick, and the first brick length Y1, the second brick length Y2, the tail brick length Y3 of the even skin brick are greater than or equal to one-third of the whole brick length, that is:

[0030] bs ≥ X1, X2, X3, Y1, Y2, Y3 ≥ bs / 3 Formula (3)

[0031] Step 3.2: The distance h of the first skin brick from the bottom elevation h l of the wall, that is, the height h of the guide wall, should meet the specification requirements within a certain range, that is, meet the following requirements:

[0032] H1 ≥ h ≥ H2 Formula (4);

[0033] Step 3.3: The staggered length cf1, cf2, cf3, cf4 of the odd skin brick and the even skin brick meets the following requirements:

[0034] Formula (5).

[0035] Further, in the above method, H1 = 200 mm, H2 = 300 mm.

[0036] Furthermore, in the above method, step 4: Based on the wall model W, masonry layout design information, and constraints, a genetic algorithm is used to establish an initial masonry layout scheme library GA. z ,include:

[0037] Step 4.1: Construct the initial chromosome library GA for the bricklaying scheme using the Monte Carlo randomization method. z ={ga i}, i=1-N, where N is the number of chromosomes, and each scheme ga i This includes: the length of the first brick in an odd-numbered set of bricks x 1 i Odd-numbered brick length x 2 i The length Y1 of the first brick in an even-numbered set of bricks i Secondary brick length Y2 i and guide wall height h i Five parameters; namely, scheme ga i =(X1 i X2 i Y1 i Y2 i h i );

[0038] Step 4.2: Calculate ga for each scheme i The length of the tail brick of the odd-numbered bricks x 3 i The length Y3 of the tail brick of an even number of bricks i The calculation formula is as follows:

[0039] X3 i =Mod(L1-X1) i -t-X2 i -t, bs+t) Formula (6);

[0040] Y3 i =Mod(L2-Y1) i -t-Y2 i -t, bs+t) Formula (7);

[0041] Mod() is the function that calculates the remainder;

[0042] Step 4.3: If X3 i Or Y3 i The requirements of formula (3) are not met, so the scheme ga is... j Remove from GA;

[0043] Step 4.4: Calculate ga for each scheme i The staggered joint lengths of odd-numbered and even-numbered tiles are as follows:

[0044] staggered cf1 i =Abs(cl *d+X1 i -Y1 i ) Formula (8);

[0045] staggered cf2 i =Abs(c l *d +X1 i -Y1 i -Y2 i -t) Formula (9);

[0046] staggered cf3 i =Abs(c l *d +X1 i +X2 i -Y1 i -Y2 i ) Formula (10);

[0047] staggered cf4 i =Abs(c r *d +X3 i -Y3 i ) Formula (11);

[0048] If the stitches are misaligned (cf1) i cf2 i cf3 i cf4 i If the requirements of formula (5) are not met, the scheme ga will be changed. j Remove from GA; where Abs() is a function that calculates the absolute value.

[0049] Furthermore, in the above method, step 5: calculate GA z Various schemes in China i The number of broken bricks around the wall includes:

[0050] Step 5.1: Let the number of broken bricks be s. i =0; Calculate the number of brick layers. Where Floor is the floor function; the number of even-numbered tiles P2 = Floor( The number of odd-numbered tiles is P1 = P - P2;

[0051] Step 5.2: Calculate the number of broken bricks s on the left side. il =(Ceil( ) + Ceil ( ))*P1+(Ceil( ) + Ceil ( ))*P2; where Ceil is the floor function; s i =s i + sil ;

[0052] Step 5.3: Calculate the number of broken bricks s on the right side. ir =Ceil( ) * P1 + Ceil ( ) *P2; s i =s i +s ir ;

[0053] Step 5.4: Calculate the number of broken bricks at the top; s it =(Ceil( -P)*Ceil( ); s i =s i +s it .

[0054] Furthermore, in the above method, step 6: calculate GA z Various schemes in China i The number of broken bricks around the opening includes:

[0055] Step 6.1: Traverse each opening o in the opening O j ; Calculate the number of brick courses spanned by the opening, P = Ceil ( The number of odd-numbered tiles is P1 = Floor ( ) + Mod(Floor( ), 2); the number of even-numbered tiles P2 = P - P1;

[0056] Step 6.2: Calculate the number of broken bricks at the bottom, including:

[0057] Step 6.2.1: If <0.5, s i =1+ s i If the condition is met, the process ends; otherwise, proceed to step 6.2.2.

[0058] Step 6.2.2: Calculate the height of the bricks at the bottom of the opening (HOL) i =mod(ohd) j -h i (bh+t)

[0059] Step 6.2.3: If 2t < HOL i <bh / 3, does not conform to specifications, s i =Max, Max is preferably 10000; return to step 6.1; otherwise, proceed to step 6.2.4;

[0060] Step 6.2.4: sod j = Ceil( Fz(Abs(mod(xo j -h i , bh+t)-bh / 2)-bh / 2+2t), where the function Fz(x) is Fz(x)=0 when x≥0; Fz(x)=1 when x<0; s i = s i + sod j ;

[0061] Step 6.3: Calculate the number of broken bricks on the left side, including;

[0062] Step 6.3.1: Calculate the length of odd skin brick on the left side of the hole XOL i = mod(xo j -d-t-X1 i -X2 i , bs+t); the length of even skin brick YOL i = mod(xo j -t-Y1 i -Y2 i , bs+t)

[0063] Step 6.3.2: If 2tXOL i < bh / 3 or 2tYOL i < bh / 3 does not meet the standard, s i = Max, Max is preferably 10000; return to Step 6.1; otherwise, go to Step 6.3.3;

[0064] Step 6.3.3: sol j = P1* Fz(Abs(mod(xo j -d-t-X1 i -X2 i , bs+t)-bs / 2)-bs / 2+2t))+P2* Fz(Abs(mod(xo j -t-Y1 i -Y2 i , bs+t)-bs / 2)-bs / 2+2t); s i = s i + sol j ;

[0065] Step 6.4: Calculate the number of broken bricks on the right side, including;

[0066] Step 6.4.1: Calculate the length of odd skin brick on the right side of the hole XOR i = mod(L-xo j -ol j -d-t-X3 i, bs+t); even skin brick length YOR i = mod(L-xo j -ol j -t-Y3 i , bs+t)

[0067] Step 6.4.2: if 2t < XOR i < bh / 3 or, 2t < YOR i < bh / 3 is not valid, s i = Max, Max is preferably 10000; go back to Step 6.1; otherwise, go to Step 6.4.3;

[0068] Step 6.4.3: sor j = P1* Fz(Abs(mod(L-xo j -ol j -d-t-X3 i , bs+t) - bs / 2) - bs / 2 + 2t) ) + P2* Fz(Abs(mod(L-xo j -ol j -t-Y3 i , bs+t) - bs / 2) - bs / 2 + 2t) ); s i = s i + sor j ;

[0069] Step 6.5: calculate the number of top bricks, including;

[0070] Step 6.5.1 if < 0.5, s i = 1 + s i , end the step; otherwise, go to Step 6.5.2

[0071] Step 6.5.2: calculate the height of the top brick of the opening HOT i = bh + t - mod(oht j -h i , bh + t);

[0072] Step 6.5.3: if 2t < HOT i < bh / 3, not valid, s i = Max, Max is preferably 10000; go back to Step 6.1; otherwise, go to Step 6.5.4;

[0073] Step 6.5.4 sot j = Ceil( ) * Fz(Abs(bh / 2 - mod(oht j -hi ,bh+t))-bh / 2+2t));s i = s i +sot j ;

[0074] Furthermore, in the above method, step 9: for GA z+1 Middle row brick scheme ga i Chromosomal crossover and mutation are performed to form a GA (Genetic Genetic Algebra) library of offspring chromosomes. z+2 ,include:

[0075] Step 9.1: Configure GA z+2 ={};

[0076] Step 9.2: If GA z+1 If there are no elements in GA, the loop ends and proceeds to step 9.7; if GA z+1 There are elements in GA z+1 Two different schemes ga are randomly selected in the middle. i and ga j ; and ga i and ga j From GA z+1 Delete; use a random algorithm in (X1) i X2 i Y1 i Y2 i h i Randomly select a point p in the array and perform an intersection, then proceed to step 9.3;

[0077] Step 9.3: Transfer the paternal chromosome ga i and ga j Chromosome crossing is performed at point p, and the resulting daughter chromosome ga is used. i ', ga j ';

[0078] Step 9.4: Calculate ga according to formulas (6) and (7). i 'of and ga i+1 'X3 and Y3, calculate cf1, cf2, cf3, cf4 according to formulas (8), (9), (10), (11). If X3 or Y3 does not conform to formula (3) or cf1, cf2, cf3, cf4 does not conform to constraint formula (5), proceed to step 9.5; if they all conform, proceed to step 9.6.

[0079] Step 9.5: For ga j ',ga j+1 The method that does not conform to formula (3) or formula (5) is used for mutation processing, that is, in ga i '、 ga i+1 'of (X1)i , X2 i , Y1 i , Y2 i , h i ) are randomly set with any one value selected from the group consisting of formula (3) and formula (4); return to step 9.4;

[0080] Step 9.6: add ga j ' and ga j+1 ' to GA z+2 ; return to step 9.2;

[0081] Step 9.7: GA z = GA z+2 ; set z = z + 1.

[0082] Further, in the above method, K is 10.

[0083] Compared with the prior art, the application can generate an optimal scheme of masonry arrangement according to the boundary and opening information of the secondary structure wall and the design information of the masonry arrangement, so as to meet the specification requirements, minimize the number of broken bricks, save the workload of manual brick cutting, and reduce material consumption and solid waste. The application has the advantages of considering different numbers of openings, generating multiple optimization schemes, being suitable for different specifications of masonry materials, and having strong applicability. Furthermore, the method innovatively adopts five variables to construct a chromosome model of the arrangement scheme, which guarantees the compliance of the scheme and ensures that each brick has at most three broken bricks, while supporting the reduction of the number of top broken bricks by optimizing the height of the guide wall. The method can adapt to the intelligent arrangement of masonry of the secondary structure wall with different masonry sizes and different elevation door openings, generate multiple optimized masonry arrangement schemes that meet the specification requirements according to various complex and variable site conditions, assist workers in quickly determining the masonry arrangement scheme, improve construction quality and efficiency, reduce masonry material consumption and cost, and ensure the aesthetics and compliance of the wall. BRIEF DESCRIPTION OF DRAWINGS

[0084] Figure 1 is a flowchart of a masonry arrangement method of a secondary structure wall according to an embodiment of the application;

[0085] Figure 2 is a schematic diagram of a wall model according to an embodiment of the application;

[0086] Figure 3 is a schematic diagram of a chromosome code according to an embodiment of the application;

[0087] Figure 4 is a schematic diagram of a chromosome crossing process according to an embodiment of the application. EMBODIMENT

[0088] To make the above-mentioned objects, features and advantages of the present invention more apparent and understandable, the present invention will be further described in detail below with reference to the accompanying drawings and specific embodiments.

[0089] like Figure 1 As shown, the present invention provides a method for arranging masonry in a secondary structural wall, comprising:

[0090] Step 1: Set the masonry layout design information, which includes the following steps:

[0091] Step 1.1: Input the masonry dimensions, including: length bs, width bw, and height bh;

[0092] Step 1.2: Input the mortar joint width t;

[0093] Step 1.3: Input the length d of the toothed joint;

[0094] Step 1.4: Input the upper and lower limits of the guide wall height, H1 and H2.

[0095] Step 2: Create the wall BIM model W, which includes the following steps:

[0096] Step 2.1: Enter the elevation h of the bottom surface of the wall. l The top surface elevation of the wall is h t The wall length is L, and whether the left side has a toothed joint is c. l (1 for yes, 0 for no), is the right side a toothed joint? r (1 for yes, 0 for no); that is, W={h} l h t L,c l c r};

[0097] Step 2.2: Calculate the length L1 of odd-numbered tiles and the length L2 of even-numbered tiles, using the following formula:

[0098] L1=L - (c l +c r Formula (1)*d

[0099] L2=L; Formula (2)

[0100] Step 2.3: Input the wall opening information O = {opening o} j}, generally the entrance o j It is rectangular, including the bottom elevation of the opening (ohd). j The elevation of the top surface of the opening is oht j Wall opening width ol j and the distance xo from the left side of the wall j ; that is, the entrance to the cave. j =(ohd j ohtj ol j xo j );

[0101] Step 3: According to the specifications, based on the masonry layout design information and the wall BIM model W, input the constraints, specifically including the following:

[0102] Step 3.1: The lengths of the first brick (X1), the second brick (X2), and the last brick (X3) of odd-numbered brick courses, and the lengths of the first brick (Y1), the second brick (Y2), and the last brick (Y3) of even-numbered brick courses, are greater than or equal to one-third of the total brick length, i.e.:

[0103] bs≥X1, X2, X3, Y1, Y2, Y3≥bs / 3 Formula (3)

[0104] Step 3.2: The elevation h of the first layer of bricks from the bottom surface of the wall l The distance h, i.e., the height h of the guide wall, should meet the requirements of the specification and be within a certain range, that is, satisfy the following requirements:

[0105] H1≥h≥H2 formula (4);

[0106] Preferably, H1=200mm, H2=300mm;

[0107] Step 3.3: The staggered joint lengths cf1, cf2, cf3, and cf4 of odd-numbered and even-numbered tiles must meet the following requirements:

[0108] Formula (5);

[0109] Step 4: Based on the wall model W, masonry layout design information, and constraints, a genetic algorithm is used to establish an initial masonry layout scheme library GA. z Set the iteration number z=0.

[0110] Step 4.1: Set the iteration number z=0, and use the Monte Carlo randomization method to construct the initial chromosome library GA for the brick-laying scheme. z ={ga i}, i=1-N, where N is the number of chromosomes, and each scheme ga i This includes: the length of the first brick in an odd-numbered set of bricks x 1 i Odd-numbered brick length x 2 i The length Y1 of the first brick in an even-numbered set of bricks i Secondary brick length Y2 i and guide wall height h i Five parameters; namely, scheme ga i =(X1 i X2 i Y1 i Y2i , h i );

[0111] Step 4.2: Calculate the tail brick length X3 of odd skin brick and Y3 of even skin brick for each scheme ga i i and even skin brick i , the calculation formula is as follows:

[0112] X3 i =Mod(L1-X1 i -t-X2 i -t,bs+t) formula (6)

[0113] Y3 i =Mod(L2-Y1 i -t-Y2 i -t,bs+t) formula (7)

[0114] Where Mod() is a function to calculate the remainder;

[0115] Step 4.3: If X3 i or Y3 i does not meet the requirements of formula (3), delete the scheme ga j from GA;

[0116] Step 4.4: Calculate the misalignment length of odd skin brick and even skin brick for each scheme ga i , as follows:

[0117] Misalignment cf1 i =Abs(c l *d+X1 i -Y1 i ) formula (8)

[0118] Misalignment cf2 i =Abs(c l *d+X1 i -Y1 i -Y2 i -t) formula (9)

[0119] Misalignment cf3 i =Abs(c l *d+X1 i +X2 i -Y1 i -Y2 i ); formula (10)

[0120] Misalignment cf4 i =Abs(c r *d+X3 i -Y3​i ); formula (11)

[0121] If the stitches are misaligned (cf1) i cf2 i cf3 i cf4 i If the requirements of formula (5) are not met, the scheme ga will be changed. j Remove from GA; where Abs() is a function that calculates the absolute value.

[0122] Step 5: Calculate GA z Various schemes in China i The number of broken bricks around the wall is calculated using the following method:

[0123] Step 5.1: Let the number of broken bricks be s. i =0; Calculate the number of brick layers.

[0124] Where Floor is the floor function; the number of even-numbered tiles P2 = Floor( The number of odd-numbered tiles is P1 = P - P2;

[0125] Step 5.2: Calculate the number of broken bricks s on the left side. il =(Ceil( ) + Ceil ( ))*P1+(Ceil( ) + Ceil ( ))*P2; where Ceil is the floor function; s i =s i + s il ;

[0126] Step 5.3: Calculate the number of broken bricks s on the right side. ir =Ceil( ) * P1 + Ceil ( ) *P2; s i =s i +s ir ;

[0127] Step 5.4: Calculate the number of broken bricks at the top; s it =(Ceil( -P)*Ceil( ); s i =s i +s it ;

[0128] Step 6: Calculate GA z Various schemes in China i The number of broken bricks around the opening;

[0129] Step 6.1: Traverse each hole o in hole openings O j ; calculate the number of odd skin bricks P1 = Floor(P / 2) + Mod(Floor(P / 2), 2); and the number of even skin bricks P2 = P - P1.

[0130] Step 6.2: Calculate the number of lower broken bricks, including:

[0131] Step 6.2.1: If <0.5, s i =1+ s i , end the step; otherwise, go to Step 6.2.2.

[0132] Step 6.2.2: Calculate the hole bottom brick height HOL i =mod(ohd j -h i , bh+t);

[0133] Step 6.2.3: If 2t i bh / 3, it is not in compliance, s i =Max, Max is preferably 10000; return to Step 6.1; otherwise, go to Step 6.2.4.

[0134] Step 6.2.4: sod j = Ceil( ) *Fz(Abs(mod(ohd j -h i , bh+t) - bh / 2) - bh / 2 + 2t), where the function Fz(x) is Fz(x)=0 when x≥0; and Fz(x)=1 when x<0; and s i = s i +sod j

[0135] Step 6.3: Calculate the number of left broken bricks, including:

[0136] Step 6.3.1: Calculate the hole left odd skin brick length XOL i = mod(xo j -d-t-X1 i -X2 i , bs+t); and the even skin brick length YOL i =mod(xo j -t-Y1 i -Y2 i ​​​​(bs+t)

[0137] Step 6.3.2: If 2t < XOL i <bh / 3 or, 2t<YOL i <bh / 3 does not conform to the specifications, s i =Max, Max is preferably 10000; return to step 6.1; otherwise, proceed to step 6.3.3;

[0138] Step 6.3.3: sol j =P1* Fz(Abs(mod(xo j -dt-X1 i -X2 i , bs+t)-bs / 2)-bs / 2+2t))+P2* Fz(Abs(mod(xo j -t-Y1 i -Y2 i , bs+t)-bs / 2)-bs / 2+2t));s i =s i + sol j ;

[0139] Step 6.4: Calculate the number of broken bricks on the right side, including;

[0140] Step 6.4.1: Calculate the XOR of the length of the odd-numbered bricks on the right side of the opening. i = mod(L-xo) j -ol j -dt-X3 i (bs+t); even-numbered brick length YOR i = mod(L-xo) j -ol j -t-Y3 i (bs+t)

[0141] Step 6.4.2: If 2t < XOR i <bh / 3 or, 2t<YOR i <bh / 3 does not conform to the specifications, s i =Max, Max is preferably 10000; return to step 6.1; otherwise, proceed to step 6.4.3;

[0142] Step 6.4.3: sor j = P1* Fz(Abs(mod(L-xo j -ol j -dt-X3 i , bs+t)-bs / 2)-bs / 2+2t))+P2* Fz(Abs(mod(L-xoj -ol j -t-Y3 i , bs+t)-bs / 2)-bs / 2+2t));s i =s i + sor j ;

[0143] Step 6.5: Calculate the number of broken bricks at the top, including;

[0144] Step 6.5.1 If <0.5, s i =1+ s i If the condition is met, the process ends; otherwise, proceed to step 6.5.2.

[0145] Step 6.5.2: Calculate the height of the brick at the top of the opening (HOT) i =bh+t-mod(oht) j -h i (bh+t)

[0146] Step 6.5.3: If 2t < HOT i <bh / 3, does not conform to specifications, s i =Max, Max is preferably 10000; return to step 6.1; otherwise, proceed to step 6.5.4;

[0147] Step 6.5.4 sot j = Ceil( )*Fz(Abs(bh / 2-mod(oht) j -h i ,bh+t))-bh / 2+2t));s i = s i +sot j ;

[0148] Step 7: Calculate GA based on the number of broken bricks in the wall. z Chinese solution ga i fitness value F i F i F is inversely proportional to the number of broken bricks. i =f(s) j For example, F i = ;like Figure 2 As shown, for ga1, F i = =0.013;

[0149] Step 8: As Figure 4 As shown, a random competitive selection algorithm is used, based on the fitness value F. i , from GAz Screening N chromosomes into chromosome library GA z+1 ; that is, fitness value F i High element ga i Enter GA with higher frequency z+1 ;

[0150] Step 9: Chromosome crossover and mutation are performed on GA z+1 Brick laying scheme ga i to form offspring chromosome library GA z+2 , iteration number z is added by 1; specifically comprising the following steps:

[0151] Step 9.1: Set GA z+2 ={};

[0152] Step 9.2: If there is no element in GA z+1 , the loop ends and step 9.7 is entered; if there is an element in GA z+1 , two different schemes ga z+1 and ga i are randomly selected from GA j ; and ga i and ga j are deleted from GA z+1 ; a point p is randomly selected in (X1 i , X2 i , Y1 i , Y2 i , h i ) using a random algorithm for crossover, and step 9.3 is entered;

[0153] Step 9.3: Parent chromosomes ga i and ga j are crossed at point p, and child chromosomes ga i ', ga j ' are generated after crossing;

[0154] Step 9.4: X3 and Y3 of ga i ' and ga i+1 ' are calculated according to formulas (6) and (7), and cf1, cf2, cf3, cf4 are calculated according to formulas (8), (9), (10), (11); if X3 or Y3 does not meet formula (3) or cf1, cf2, cf3, cf4 does not meet constraint formula (5), step 8.5 is entered; if all meet, step 9.6 is entered;

[0155] Step 9.5: The method in ga j ', ga j+1 that does not meet formula (3) or formula (5) is subjected to mutation treatment, that is, ga i', ga i+1 ' of formula (3) and formula (4); return to step 9.4; i ', X2 i ', Y1 i ', Y2 i ', h i ' is randomly set; return to step 9.4;

[0156] Step 9.6: add ga j ' to GA j+1 '; return to step 9.1;

[0157] Step 9.7: GA z = GA z+2 ; set the iteration number z = z + 1;

[0158] Step 10: when the iteration number z reaches the maximum evolution number Z, go to step 11; otherwise, go to step 5;

[0159] Step 11: select the K groups of resource optimization indexes F i ' from the chromosome library GA l with the largest value as the result output as the selectable optimal arrangement scheme, and K can be selected as 10.

[0160] In summary, the present application can generate an optimal masonry arrangement scheme that meets the specification requirements according to the boundary and opening information of the secondary structure wall and the masonry arrangement design information, so that the number of broken bricks is minimized, the workload of manual brick cutting is saved, and material consumption and solid waste are reduced. The present application has the advantages of considering different numbers of openings, generating multiple optimization schemes, being suitable for different specifications of masonry materials, and having strong applicability. Moreover, the method innovatively uses 5 variables to construct a chromosome model of the arrangement scheme, which not only guarantees the compliance of the scheme, but also ensures that each brick has at most 3 broken bricks, while supporting the reduction of the number of top broken bricks by optimizing the height of the guide wall. The method can adapt to different masonry sizes and different elevation door openings, and can generate multiple optimized masonry arrangement schemes that meet the specification requirements according to various complex and variable site conditions, thereby assisting workers in quickly determining the masonry arrangement scheme, improving construction quality and efficiency, reducing masonry material consumption and cost, and ensuring the aesthetics and compliance of the wall.

[0161] Specifically, the masonry intelligent arrangement method for a secondary structure wall comprises the following steps:

[0162] Step 1: set the initial information, which specifically comprises the following steps:

[0163] Step 1.1: input the masonry size information, including length bs, width bw, and height bh; for example,Figure 2 As shown, the common brickwork bs=600mm, bw=240mm; bh=240mm;

[0164] Step 1.2: input mortar joint width t; as shown, the general mortar joint width is: 10mm; Figure 2

[0165] Step 1.3: input the length of the rabbet d; as shown, the general rabbet length is: 60mm; Figure 2

[0166] Step 2: establish a wall model W, specifically including the following steps:

[0167] Step 2.1: input the wall bottom elevation h l , wall top elevation h t , wall length L, whether the left side is a rabbet c l (1 for yes, 0 for no), whether the right side is a rabbet c r (1 for yes, 0 for no); that is, W={h l , h t , L, c l , c r}; as shown, c l =1, c r =1; W={0, 3900, 6600, 1, 0}; Figure 2

[0168] Step 2.2: calculate the length of odd skin bricks L1 and the length of even skin bricks L2, the formula is as follows:

[0169] L1=L-(c l +c r )*d Formula (1)

[0170] L2=L; Formula (2)

[0171] As shown, L2=3300; L1=6600-60=6540mm; Figure 2

[0172] Step 2.3: input the hole information on the wall O={hole o j}, the general hole o j is rectangular, including hole bottom elevation ohd j , hole top elevation oht j , wall hole width ol j and distance from the left side of the wall xo j ; that is, hole o j =(ohd j , oht j , ol j ​​​​xo j );

[0173] like Figure 2 As shown, O={ o 1， o2}, where o2 = (1400, 3500, 2000, 2000);

[0174] Step 3: According to the specification requirements, input the constraints, specifically including the following:

[0175] Step 3.1 The lengths of the first brick (X1), the second brick (X2), and the last brick (X3) of odd-numbered brick courses, and the lengths of the first brick (Y1), the second brick (Y2), and the last brick (Y3) of even-numbered brick courses, are greater than or equal to one-third of the total brick length, i.e.:

[0176] bs≥X1, X2, X3, Y1, Y2, Y3≥bs / 3 Formula (3)

[0177] like Figure 2 As shown, X1=600, X2=540, X3=520, Y1=360, Y2=540, Y3=300;

[0178] Step 3.2 The elevation h of the first layer of bricks from the bottom of the wall l The distance h, i.e., the height h of the guide wall, should meet the requirements of the specification and be within a certain range, that is, satisfy the following requirements:

[0179] H1≥h≥H2 formula (4);

[0180] Preferably, H1=200mm, H2=300mm;

[0181] like Figure 2 As shown, H1 = 300 mm;

[0182] Step 3.3 The staggered joint lengths cf1, cf2, cf3, and cf4 of odd-numbered and even-numbered tiles must meet the following requirements:

[0183] Formula (5);

[0184] like Figure 2 As shown;

[0185] Step 4: Based on the wall model W, the initial input information, and the specification requirements, a genetic algorithm is used to establish the masonry layout scheme GA. z Set the iteration algebra z=0;

[0186] Step 4.1: Construct the initial chromosome library GA for the bricklaying scheme using the Monte Carlo randomization method. z ={ga i}, i = 1-N, N is the number of chromosomes. Where each scheme ga i , the first brick length of odd skin brick X1 i , the second brick length of odd skin brick X2 i , the first brick length of even skin brick Y1 i , the second brick length of even skin brick Y2 i and the height of guide wall h i Five parameters; that is, the scheme ga i = (X1 i , X2 i , Y1 i , Y2 i , h i );

[0187] As shown in Figure 2 and Figure 2 , N = 100; GA z = {ga i}, i = 1-100; randomly generated scheme includes ga1= (600, 540, 360, 540, 300); ga2= (540, 540, 320, 540, 200);

[0188] Step 4.2: Calculate the tail brick length X3 i of the odd skin brick and the tail brick length Y3 i of the even skin brick of each scheme ga i , the calculation formula is as follows:

[0189] X3 i = Mod (L1-X1 i -t-X2 i -t, bs+t) formula (6)

[0190] Y3 i = Mod (L2-Y1 i -t - Y2 i -t, bs+t) formula (7)

[0191] Where, Mod() is the function of calculating the remainder;

[0192] For ga1, X31=Mod(6540-600-540,610)= 520; Y31=Mod(6600-360-540,610)= 210;

[0193] Step 4.3 If X3 i or Y3 i does not meet the requirements of formula (3), the scheme ga j is deleted from GA;

[0194] AsFigure 3 As shown, analysis found that X31 and Y31 are greater than bs / 3=200, meeting the requirements;

[0195] Step 4.4 Calculate each scheme ga i The staggered length of odd skin bricks and even skin bricks is as follows:

[0196] Staggered cf1 i =Abs(c l *d+X1 i -Y1 i ) Formula (8)

[0197] Staggered cf2 i =Abs(c l *d+X1 i -Y1 i -Y2 i -t) Formula (9)

[0198] Staggered cf3 i =Abs(c l *d+X1 i +X2 i -Y1 i -Y2 i ) Formula (10)

[0199] Staggered cf4 i =Abs(c r *d+X3 i -Y3 i ) Formula (11)

[0200] If the staggered cf1 i , cf2 i , cf3 i , cf4 i does not meet the requirements of formula (5), the scheme ga j is deleted from GA; wherein Abs() is an absolute value function.

[0201] As Figure 2 shown, for ga1, cf11= Abs(60+600-360)=300; cf21= Abs(60+600-360-540-10)=250; cf13= Abs(60+600+540-360-540)=300; cf24= Abs(520-210)=310; it can be seen that, cf11, cf21, cf31, cf41 meet the requirements of formula (5);

[0202] Step 5: Calculate GA z Number of broken tiles around the perimeter of the wall body according to each scheme ga i , the specific calculation method is as follows:

[0203] Step 5.1: Set the number of broken tiles s i =0; Calculate the number of whole tile skins P=Floor( ), wherein Floor is the floor function; the number of even skin tiles P2=Foor( ); the number of odd skin tiles P1=P-P2;

[0204] As shown in Figure 2 , for ga1, P=Floor( )=14, P2=Foor( )=7, P1=7;

[0205] Step 5.2: Calculate the number of left broken tiles s il =(Ceil( )+ Ceil( )) *P1+ (Ceil( )+ Ceil( )) *P2; wherein Ceil is the ceiling function; s i =s i + s il ;

[0206] As shown in Figure 2 , for ga1, s jl =(Ceil( )+ Ceil( )) *7+ (Ceil( )+Ceil( )) *7=21; s1=21;

[0207] Step 5.3: Calculate the number of right broken tiles s ir =Ceil( ) *P1+Ceil( ) *P2; s i =s i +s ir ;

[0208] As shown in Figure 2 , for ga1, s jr =Ceil( ) *7+Ceil( ) *7=14; s1=21+14=35;

[0209] Step 5.4: Calculate the number of top broken tiles; sit =(Ceil( -P)*Ceil( ); s i =s i +s it ;

[0210] like Figure 2 As shown, for ga1, s jt =Ceil( ) = 19; s1 = 35 + 19 = 54;

[0211] Step 6: Calculate GA z Various schemes in China i The number of broken bricks around the opening;

[0212] Step 6.1: Traverse each opening o in the opening O j ; Calculate the number of brick courses spanned by the opening, P = Ceil ( The number of odd-numbered tiles is P1 = Floor ( ) + Mod(Floor( ), 2); the number of even-numbered tiles P2 = P - P1;

[0213] like Figure 2 As shown, for ga1, taking o2 as an example, calculate the number of brick courses spanned by the opening P=Ceil( =9; the number of odd-numbered tiles P1 = 4 + Mod(Floor( ), 2) = 5; the number of even-numbered tiles P2 = 4;

[0214] Step 6.2: Calculate the number of broken bricks at the bottom, including:

[0215] Step 6.2.1: If <0.5, s i =1+ s i If the condition is met, the process ends; otherwise, proceed to step 6.2.2.

[0216] like Figure 2 As shown, for ga1, If the value is >0.5, proceed to step 6.2.2;

[0217] Step 6.2.2: Calculate the height of the bricks at the bottom of the opening (HOL) i =mod(ohd) j -h i (bh+t)

[0218] like Figure 2 As shown, for ga1, HOL i =mod(ohd)j -h i , bh + t) = mod(1400 - 300, 250) = 100;

[0219] Step 6.2.3: If 2t < HOL i < bh / 3, not in specification, s i = Max, Max is preferably 10000; go to Step 6.1; otherwise, go to Step 6.2.4;

[0220] As Figure 2 shown, for ga1, HOL i = 100 > 240 / 3 = 80, in specification, go to Step 6.2.4;

[0221] Step 6.2.4: sod j = Ceil ) * Fz(Abs(mod(ohd j -h i , bh + t) - bh / 2) - bh / 2 + 2t)), where function Fz(x), Fz(x) = 0 when x ≥ 0; Fz(x) = 1 when x < 0; s i = s i + sod j ;

[0222] As Figure 2 shown, for ga1, sod2= Ceil ) * Fz(Abs(mod(ohd j -h i , bh + t) - bh / 2) - bh / 2 + 2t)) = 4 * Fz(Abs(mod(1400 - 300, 250) - 250 / 2) - 250 / 2 + 20) = 4,

[0223] s i = s i + sod j = 58;

[0224] Step 6.3: Calculate the number of left-side broken bricks, including;

[0225] Step 6.3.1: Calculate the length of odd-numbered skin bricks on the left side of the opening XOL i = mod(xo j - d - t - X1 i - X2 i , bs + t); the length of even-numbered skin bricks YOL i = mod(xo j - t - Y1 i - Y2 ibs+t);

[0226] As Figure 2 shown, for ga1, XOL i = mod(810, 610) = 200; YOL i = mod(2000 -10-360-540, 610) = 480;

[0227] Step 6.3.2: If 2t < XOL i < bh / 3 or, 2t < YOL i < bh / 3 is not in specification, s i = Max, Max is preferably 10000; go to Step 6.1; otherwise, go to Step 6.3.3;

[0228] As Figure 2 shown, for ga1, XOL i and YOL i are in specification, go to Step 6.3.3;

[0229] Step 6.3.3: sol j = P1* Fz(Abs(mod(xo j - d - t - X1 i - X2 i , bs+t) -bs / 2) -bs / 2+2t)) + P2* Fz(Abs(mod(xo j - t - Y1 i - Y2 i , bs+t) -bs / 2) -bs / 2+2t));

[0230] s i = s i + sol j ;

[0231] As Figure 2 shown, for ga1, sol j = 5* Fz(Abs(mod(2000-60-10-600-540, 610) -bs / 2) -bs / 2+2t)) + 4* Fz(Abs(mod(2000 -10-360-540, 610) -bs / 2) -bs / 2+2t)) = 9; s i = s i + sol j = 58+9 = 67;

[0232] Step 6.4: Calculate the number of right side broken bricks, including;

[0233] Step 6.4.1: Calculate the XOR of the length of the odd-numbered bricks on the right side of the opening. i = mod(L-xo) j -ol j -dt-X3 i (bs+t); even-numbered brick length YOR i = mod(L-xo) j -ol j -t-Y3 i (bs+t)

[0234] like Figure 2 As shown, for ga1, XOR i = mod (3860, 610) = 200; YOR i = mod(4180, 610) = 520;

[0235] Step 6.4.2: If 2t < XOR i <bh / 3 or, 2t<YOR i <bh / 3 does not conform to the specifications, s i =Max, Max is preferably 10000; return to step 6.1; otherwise, proceed to step 6.4.3;

[0236] like Figure 2 As shown, for ga1, XOR i and YOR i If it meets the specifications, proceed to step 6.4.3;

[0237] Step 6.4.3: sor j = P1* Fz(Abs(mod(L-xo j -ol j -dt-X3 i , bs+t)-bs / 2)-bs / 2+2t))+P2* Fz(Abs(mod(L-xo j -ol j -t-Y3 i , bs+t)-bs / 2)-bs / 2+2t));s i =s i + sor j ;

[0238] like Figure 2 As shown, for ga1, sor j= 5* Fz(Abs(mod(6600-2000-200 -60-10-520, 610)-300)-300+20))+4* Fz(Abs(mod(6600-2000-200 -10-210, 610)-300)-300+20))=9; s i =67+9=76;

[0239] Step 6.5: Calculate the number of top bricks, including;

[0240] Step 6.5.1 If <0.5, s i =1+ s i , end step; otherwise, go to Step 6.5.2

[0241] As shown in Figure 2 , for ga1, >0.5, go to Step 6.5.2;

[0242] Step 6.5.2: Calculate the height of the top brick at the opening, HOT i =bh+t-mod(oht j -h i , bh+t);

[0243] As shown in Figure 2 , for ga1, HOT i = 250-mod(3170, 250)=80;

[0244] Step 6.5.3: If 2t i < bh / 3, not in compliance, s i =Max; return to Step 6.1; otherwise, go to Step 6.5.4;

[0245] As shown in Figure 2 , for ga1, HOT i >80, in compliance with the requirements, go to Step 6.5.4;

[0246] Step 6.5.4 sot j = Ceil( )*Fz(Abs(bh / 2-mod(oht j -h i , bh+t)) -bh / 2+2t)); i = s i +sot j ;

[0247] As shown in Figure 4 , for ga1, sot j= Ceil( )*Fz(Abs(mod(3500-300,610)-120)-120+2t));s i =76+3=79;

[0248] Step 7: Calculate GA based on the number of broken bricks in the wall. z Chinese solution ga i fitness value F i F i F is inversely proportional to the number of broken bricks. i =f(s) j For example, F i = ;like Figure 4 As shown, for ga1, F i = =0.013;

[0249] Step 8: Use a random competitive selection algorithm based on the fitness value F. i , from GA z N chromosomes were selected and placed into the chromosome library GA. z+1 That is, the fitness value F i High element ga i Entering GA at a higher frequency z+1 ;

[0250] like ​ As shown, two ga1s are added to the GA. z+1 The probability of ga2 is low; only 1 will be added to GA. z+1 middle;

[0251] Step 9: For GA z+1 Middle row brick scheme ga i Chromosomal crossover and mutation are performed to form a new generation of chromosome library GA. z+2 Specifically, it includes the following steps:

[0252] Step 9.1: Configure GA z+2 ={};

[0253] Step 9.2: If GA z+1 If there are no elements in GA, the loop ends and proceeds to step 9.7; if GA z+1 There are elements in GA z+1 Two different schemes ga are randomly selected in the middle. i and ga j ; and ga i and ga j From GA z+1 Delete; use a random algorithm in (X1) i X2 i Y1i Y2 i h i Randomly select a point p in the array and perform an intersection, then proceed to step 9.2;

[0254] like ​ As shown, the intersection point p=3;

[0255] Step 9.3: Transfer the paternal chromosome ga i and ga j Chromosome crossing is performed at point p, and the resulting daughter chromosome ga is used. i ', ga j ';

[0256] like ​ As shown, the generated ga i ', ga j '; if ga i = (600, 540, 360, 540, 300); ga j = (540, 540, 320, 540, 200); then ga i '= (600, 540, 360, 540, 200), ga j = (540, 540, 320, 540, 300);

[0257] Step 9.4: Calculate ga according to formulas (6) and (7). i 'of and ga i+1 'X3 and Y3, calculate cf1, cf2, cf3, cf4 according to formulas (8), (9), (10), (11). If X3 or Y3 does not conform to formula (3) or cf1, cf2, cf3, cf4 does not conform to constraint formula (5), proceed to step 8.5; if they all conform, proceed to step 9.6.

[0258] with ga i Taking (600, 540, 360, 540, 200) as an example, X3=520 and Y3=210, which meet the requirements; cf1, cf2, cf3, and cf4 remain unchanged, which also meets the requirements.

[0259] Step 9.5: For ga j ',ga j+1 The method that does not conform to formula (3) or formula (5) is used for mutation processing, that is, in ga i '、 ga i+1 'of (X1) i X2 i Y1 i Y2 i h i) still selects one value randomly under the premise of satisfying formula (3) and formula (4); return to step 9.4;

[0260] ga i ’= (600, 540, 360, 540, 200) as an example, randomly mutate X1 between 【200, 600】, for example, ga i ’= (500, 540, 360, 540, 200);

[0261] Step 9.6: add ga j ’ and ga j+1 ’ to GA z+2 ; return to step 9.1;

[0262] Step 9.7: GA z = GA z+2 ; set z = z + 1;

[0263] Step 9: when z reaches the maximum evolution times Z, enter step 10; otherwise, enter step 5;

[0264] Step 10: select the K groups of resource optimization indexes F i with the largest value from the chromosome library GA as the result output, as the selectable optimal arrangement scheme, K can be selected as 10.

[0265] Each embodiment in the specification is described in a progressive manner, and each embodiment focuses on the difference from other embodiments. The same and similar parts between each embodiment can be referred to each other.

[0266] The skilled person can further realize that the units and algorithm steps of each example described in combination with the embodiments disclosed in the present text can be realized in electronic hardware, computer software or a combination of both. In order to clearly show the interchangeability of hardware and software, the components and steps of each example have been described in the above description. Whether the functions are executed in hardware or software depends on the specific application and design constraints of the technical solution. The skilled person can use different methods to implement the described functions for each specific application, but such implementation should not be considered beyond the scope of the present application.

[0267] Obviously, those skilled in the art can make various modifications and variations to the application without departing from the spirit and scope of the application. Thus, if these modifications and variations of the application fall within the scope of the claims of the application and their equivalents, the application also intends to include these modifications and variations.

Claims

1. A method of masonry construction of a secondary structural wall, characterized in that, Comprising: Step 1: setting the masonry arrangement design information; Step 2: establishing a wall BIM model W; Step 3: inputting constraint conditions according to the masonry arrangement design information and the wall BIM model W; Step 4: According to the wall model W, the masonry arrangement design information and the constraint conditions, an initial scheme library GA of the masonry arrangement is established by using a genetic algorithm z ; set iteration number z = 0; Step 5: Calculate GA z The wall body perimeter of each scheme ga i The number of broken bricks of the wall body perimeter; Step 6: Calculate GA z the hole perimeter in each scheme ga i the number of broken bricks Step 7: Calculate GA from the number of broken bricks of the wall z The fitness value F of the middle scheme ga i i , F i is inversely proportional to the number of broken bricks, F i = f(s j );​ Step 8: Using random tournament selection algorithm, select N chromosomes from GA based on fitness value F i , from GA z , filter N chromosomes into GA z+1 ; i.e. elements ga i with high fitness value F i enter GA z+1 with higher frequency; Step 9: GA z+1 Mid row brick scheme ga i Perform chromosome crossover and mutation to form a child chromosome pool GA z+2 , add 1 to the iteration number z; replace GA z+2 with GA z ; Step 10: when the iteration number z reaches the maximum evolution number Z, entering step 11; Otherwise, enter step 5; Step 11: Selecting resource optimization index F from chromosome library GA i The largest K group is output as a result; Step 1: setting the masonry arrangement design information, comprising: Step 1.1: inputting masonry size information, including: length bs, width bw, and height bh; Step 1.2: inputting the mortar joint width t; Step 1.3: inputting the length of the jagged groove d; Step 1.4: inputting the upper and lower limits H1 and H2 of the guide wall height; Step 2: establishing a wall BIM model W, comprising: Step 2.1: input the bottom elevation h of the wall l , the top elevation h of the wall t , the length L of the wall, whether the left side is a zigzag c l , whether the right side is a zigzag c r ; that is, W = {h l , h t , L, c l , c r} Step 2.2: calculating the length of the odd skin brick L1 and the length of the even skin brick L2, the formula is as follows: L1 = L - (c l + c r )*d Equation (1); L2 = L Formula (2); Step 2.3: input the hole information on the wall O = {hole o j}, the general hole o j is rectangular, including the hole bottom elevation ohd j , the hole top elevation oht j , the wall hole width ol j and the distance from the left side of the wall xo j ; that is, the hole o j = (ohd j , oht j , ol j , xo j ).

2. The method of claim 1, wherein the secondary structural wall is a masonry wall. Step 3: inputting constraint conditions according to the masonry arrangement design information and the wall BIM model W, comprising: Step 3.1: the length of the first brick X1, the length of the second brick X2, the length of the tail brick X3 of the odd skin brick, and the length of the first brick Y1, the length of the second brick Y2, the length of the tail brick Y3 of the even skin brick are greater than or equal to one-third of the length of the whole brick, that is: bs≥X1, X2, X3, Y1, Y2, Y3≥bs / 3 Formula (3) Step 3.2: Distance of first skin brick from the bottom level of the wall h l The distance h, i.e. the height h of the guide wall, should be in accordance with the requirements of the specifications, within a certain range, i.e. to meet the following requirements: H1≥h≥H2 Formula (4); Step 3.3: the staggered joint length cf1, cf2, cf3, cf4 of the odd skin brick and the even skin brick meets the following requirements:

3. The method of claim 2, wherein the secondary structural wall is a masonry wall. Step 4: According to the wall model W, the masonry arrangement design information and the constraint conditions, an initial scheme library GA of the masonry arrangement is established by using a genetic algorithm z , comprising: Step 4.1: Constructing initial chromosome library of brick arrangement scheme GA using Monte Carlo random method z = {ga i}, i = 1 - N, N is the number of chromosomes, wherein each scheme ga i includes: the length of the first brick of odd skin brick X1 i , the length of the second brick of odd skin brick X2 i , the length of the first brick of even skin brick Y1 i , the length of the second brick of even skin brick Y2 i and the height of the guide wall h i five parameters; that is, the scheme ga i = (X1 i , X2 i , Y1 i , Y2 i , h i ); Set the iteration number z = 0; Step 4.2: Calculate the length of the tail brick X3 for each scheme ga i and the length of the tail brick Y3 for even skin bricks i i The calculation formula is as follows:​ X3 i = Mod(L1 - X1 i - t - X2 i - t, bs + t) Equation (6) Y3 i = Mod(L2 - Y1 i - t - Y2 i - t, bs + t) Equation (7) Where, Mod() is the function of calculating the remainder; Step 4.3: If X3 i or Y3 i does not satisfy the requirement of equation (3), the scheme ga j is deleted from GA; Step 4.4: Calculate the odd and even skin brick misalignment length for each scheme ga i of the odd and even skin bricks as follows: Staggered cf1 i = Abs(c l * d + X1 i - Y1 i ) Equation (8); Staggered cf2 i = Abs(c l * d + X1 i - Y1 i - Y2 i t) Equation (9); Staggered cf3 i = Abs(c l * d + X1 i + X2 i - Y1 i - Y2 i ) Equation (10); Staggered cf4 i = Abs(c r * d + X3 i - Y3 i ) Equation (11); if misaligned cf1 i , cf2 i , cf3 i , cf4 i does not meet the requirements of equation (5), the scheme ga j is deleted from GA; wherein Abs() is a function for calculating absolute value.

4. The method of claim 3, wherein the secondary structural wall is a masonry wall. Step 5: Calculate GA z The wall body perimeter broken brick quantity of each scheme ga i The wall body perimeter broken brick quantity, comprising: Step 5.1: Set the number of broken bricks s i = 0; Calculate the number of whole tiles Where Floor is the floor function; the number of even tile skins The number of odd tile skins P1 = P - P2; Step 5.2: Calculate the number of left-over tiles where Ceil is the ceiling function; s i = s i + s il ; Step 5.3: Calculate the number of right-side broken bricks s i = s i + s ir ; Step 5.4: Calculate the number of top broken bricks s i = s i + s it .

5. The method of claim 4, wherein the secondary structural wall is a masonry wall. Step 6: Calculate GA z each of the schemes ga i the number of broken tiles around the opening of the hole, comprising: Step 6.1: Traverse each hole o in the hole opening O j ; Calculate the number of odd skin bricks Number of odd skin bricks Number of even skin bricks P2 = P - P1; Step 6.2: calculating the number of lower broken bricks, comprising: Step 6.2.1 : If End Step; else, go to Step 6.2.2; Step 6.2.2: Calculate the height of the bottom brick of the opening hole H0L i = mod(ohd j - h i , bh + t); Step 6.2.3: if 2t < HOL i < bh / 3, not in specification, s i = Max, Max is preferably 10000; go to Step 6.1; otherwise, go to Step 6.2.4; Step 6.2.4: where the function Fz(x) is Fz(x) = 0 when x > 0; Fz(x) = 1 when x < 0; s i = s i + sod j ; Step 6.3: calculating the number of left broken bricks, comprising; Step 6.3.1: Calculate odd skin brick length XOL on left side of opening i = mod(xo j - d - t - X1 i - X2 i , bs + t); even skin brick length YOL i = mod(xo j - t - Y1 i - Y2 i , bs + t); Step 6.3.2: if 2t < XOL i < bh / 3 or, 2t < YOL i < bh / 3 is not in spec, s i = Max, Max is preferably 10000; go to Step 6.1; else, go to Step 6.3.3; Step 6.3.3: sol j = P1 * Fz(Abs(mod(xo j - d - t - X1 i - X2 i , bs + t) - bs / 2) - bs / 2 + 2t)) + P2 * Fz(Abs(mod(xo j - t - Y1 i - Y2 i , bs + t) - bs / 2) - bs / 2 + 2t)); s i = s i + sol j ; Step 6.4: calculating the number of right broken bricks, comprising; Step 6.4.1: Calculate odd skin brick length XOR right side of opening i = mod(L - xo j - ol j - d - t - X3i,bs + t) i = mod(L - xo j - ol j - t - Y3 i ,bs + t) Step 6.4.2: if 2t < XOR i < bh / 3 or, 2t < YOR i < bh / 3 is not in specification, s i = Max, Max is preferably 10000; go to Step 6.1; otherwise, go to Step 6.4.3; Step 6.4.3: sor j = P1 * Fz(Abs(mod(L-xo j -ol j -d-t-X3 i , bs+t)-bs / 2)-bs / 2+2t))+P2*Fz(Abs(mod(L-xo j -ol j -t-Y3 i , bs+t)-bs / 2)-bs / 2+2t)); s i = s i +sor j ; Step 6.5: calculating the number of top broken bricks, comprising; Step 6.5.1 if End Step; else, go to Step 6.5.2 Step 6.5.2: Calculate the height of the top of the opening brick, HOT i = bh + t - mod(oht j - h i , bh + t); Step 6.5.3: If 2t < HOT i < bh / 3, not in spec, s i = Max; go to Step 6.1; else, go to Step 6.5.4; Step 6.5.4 6. The method of claim 5, wherein the secondary structural wall is a concrete wall. Step 9: GA z+1 Mid row brick scheme GA i Performing chromosome crossover and mutation to form a child chromosome pool GA z+2 Incrementing the iteration number z by 1 comprises: Step 9.1: Set GA z+2 = {}; Step 9.2: If GA z+1 has no elements, the loop ends and goes to Step 9.7; if GA z+1 has elements, two different solutions ga z+1 and ga i are randomly selected from GA j ; ga i and ga j are deleted from GA z+1 ; a random algorithm is used to randomly select a point p in (X1 i , X2 i , Y1 i , Y2 i , h i ) for crossover, and goes to Step 9.3; Step 9.3: Chromosome crossing is performed at point p, and the child chromosome ga i and ga j is generated after crossing at point p. i ', ga j '; Step 9.4: Calculate ga according to formula (6) and (7) i X3and Y3of the previous step, calculate cf1, cf2, cf3, cf4 according to formula (8), (9), (10), (11), if X3or Y3does not comply with formula (3) or cf1, cf2, cf3, cf4 does not comply with constraint formula (5) go to Step 8.5; if all comply, go to Step 9.6 i+1 X3and Y3of the previous step, calculate cf1, cf2, cf3, cf4 according to formula (8), (9), (10), (11), if X3or Y3does not comply with formula (3) or cf1, cf2, cf3, cf4 does not comply with constraint formula (5) go to Step 8.5; if all comply, go to Step 9.6 Step 9.5: For ga j ',ga j+1 The method that does not conform to formula (3) or formula (5) is used for variation processing, that is, in ga i '、ga i+1 'of(X1) i X2 i Y1 i Y2 i h i Choose any one value from the given information and set it randomly while satisfying formulas (3) and (4); return to step 9.4; Step 9.6: Add ga j ’, ga j+1 ’ to GA z+2 ; go to step 9.1 ; Step 9.7: Set z = z + 1. z = GA z+2 ; Set z = z + 1.

7. The method of claim 1, wherein the secondary structural wall is a masonry wall. The K is 10.

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