我将如何找到具有静态角元素的 4x4 矩阵的所有可能排列？

Question

到目前为止，我一直在使用 python 来生成矩阵排列来寻找幻方。所以到目前为止我一直在做的（对于 3x3 矩阵）是我使用 itertools.permutations 找到集合 {1,2,3,4,5,6,7,8,9} 的所有可能排列，存储它们作为一个列表，做我的计算并打印我的结果。

现在我想找出 4 阶幻方。因为找到所有排列意味着 16！可能性，我想通过在角落放置可能的元素来提高效率，例如 1, 16 在对角线的一个角上，4, 13 在对角线的两个角上。

那么我如何找到集合 {1,2....16} 的排列，其中一些元素没有移动是我的问题

Answer 1

只需将放置的数字从排列集中拉出来。然后将它们插入到生成的排列中的适当位置。

对于您的示例，您将取出 1、16、4、13。对于每个排列，在 (2、3、5、6、7、8、9、10、11、12、14、15) 上排列, 在您预先选择的位置插入 1、16、4、13 以放置它们。

Answer 2

#Rajiv Ravishankar
#rravisha
#21-301 Assignment #1, Qns 4

from numpy import matrix
import itertools

def eligCheck(m):
    #We know certain properties of magic squares that can help identify if a 3x3 matrix is a magic square or not
    #These properties include the following checks listed below.
    #
    #
    #The main purpose of this function is to check is a 3x3 matrix is a magic square without having to add all the 
    #rows, columns and diagonals.
    flag=0
    #Check 1 if the matrix is indeed 4x4
    if (len(m)==4 and len(m[0])==4 and len(m[1])==4 and len(m[2])==4):
        flag=flag+1
    #Check 2 if the 2nd diagonal adds up
    if (m[0][3] + m[1][2] + m[2][1] + m[3][0] == 34):
        flag=flag+1
    #Checks 2 if the first diagonal adds up 
    if (m[0][0] + m[1][1] + m[2][2] + m[3][3] == 34):
        flag=flag+1
    #To save resources and increase efficency, only if all three checks return true will we add the rows and columns to check.      
    if (flag==3):
        return True
    else:
        return False

def elementAdder(m):
    #This function is to be called only AFTER eligCheck() returns TRUE for a given matrix.  Since a 4x4 matrix that satisfies the checks 
    #in eligCheck() does not mean that it is a magic square, we add each row, each column and both diagonals an see if the sum
    #is equal to 15.  Splitting into two function save processing power.
    #
    #
    #Checking if all rows add up to 15
    flag=0
    #Check 1 if row 1 adds up
    if (m[0][0]+m[0][1]+m[0][2]+m[0][3] == 34):
        flag=flag+1
    else:
        return False
    #Check 2 if row 2 adds up   
    if (m[1][0]+m[1][1]+m[1][2]+m[1][3] == 34):
        flag=flag+1
    else:
        return False    
    #Check 3 if row 3 adds up
    if (m[2][0]+m[2][1]+m[2][2]+m[2][3] == 34):
        flag=flag+1
    else:
        return False
    #Check if row 4 adds up
    if (m[3][0]+m[3][1]+m[3][2]+m[3][3] == 34):
        flag=flag+1
    else:
        return False    
    #Check 4 if column 1 adds up    
    if (m[0][0]+m[1][0]+m[2][0]+m[3][0] == 34):
        flag=flag+1
    else:
        return False
    #Check 5 if column 2 adds up    
    if (m[0][1]+m[1][1]+m[2][1]+m[3][1] == 34):
        flag=flag+1
    else:
        return False
    #Check 6 if column 3 adds up
    if (m[0][2]+m[1][2]+m[2][2]+m[3][2] == 34):
        flag=flag+1
    else:
        return False
    #Check 7 if column 4 adds up    
    if (m[0][3]+m[1][3]+m[2][3]+m[3][3] == 34):
        flag=flag+1
    else:
        return False
    #Note that diagonal checks have already been verified in eligCheck() represents the diagonal from left to right

    #The strategy here is to set flag as zero initially before the additiong checks and then run each check one after the other.  If a
    #check fails, the matrix is not a magic square.  For every check that passes, flag is incremented by 1.  Therefore, at the end of 
    #all the check, if flag == 8, it is added proof that the matrix is a magic square.  This step is redundant as the program has been 
    #written to stop checks as soon as a failed check is encountered as all checks need to be true for a magic square.
    if flag==8:
        print m
        return True
    else:
        print "**** FLAG ERROR: elementAdder(): Line 84 ***" 
        print m

def dimensionScaler(n, lst):
    #This function converts and returns a 1-D list to a 2-D list based on the order.  #Square matrixes only.
    #n is the order here and lst is a 1-D list.
    i=0
    j=0
    x=0
    #mat = [[]*n for x in xrange(n)]
    mat=[]
    for i in range (0,n):
        mat.append([])
        for j in range (0,n):
            if (j*n+i<len(lst)):
                mat[i].append(lst[i*n+j])
    return mat

#mtrx=[]

def matrixGen():
#Brute forcing all possible 4x4 matrices according to the previous method will require 16!*32*16 bits or 1.07e6 GB of memory to be allocated in the RAM (impossible today)./, we 
#use an alternative methos to solve this problem.
#
#
#We know that for the sums of the diagonals will be 34 in magic squares of order 4, so we can make some assumtions of the corner element values 
#and also the middle 4 elements.  That is, the values of the diagonals.
#The strategy here is to assign one set of opposite corner elements as say 1 and 16 and the second as 13 and 4
#The remaining elements can be brute forced for combinations untill 5 magic squares are found.
    setPerms=itertools.permutations([2,3,5,6,7,8,9,10,11,12,14,15],12)
    final=[0]*16
    count=0
    #print final
    for i in setPerms:
        perm=list(i)
        setCorners=list(itertools.permutations([1,4,13,16],4))


        for j in range(0,len(setCorners)):
            final[0]=setCorners[j][0]
            final[1]=perm[0]
            final[2]=perm[1]
            final[3]=setCorners[j][1]
            final[4]=perm[2]
            final[5]=perm[3]
            final[6]=perm[4]
            final[7]=perm[5]
            final[8]=perm[6]
            final[9]=perm[7]
            final[10]=perm[8]
            final[11]=perm[9]
            final[12]=setCorners[j][2]
            final[13]=perm[10]
            final[14]=perm[11]
            final[15]=setCorners[j][3]
            if eligCheck(dimensionScaler(4,final))==True:
                elementAdder(dimensionScaler(4,final))

matrixGen()