如何查找具有静态角点元素的4x4矩阵的所有可能排列？

Question

So far I have been using python to generate permutations of matrices for finding magic squares. So what I have been doing so far (for 3x3 matrices) is that I find all the possible permutations of the set {1,2,3,4,5,6,7,8,9} using itertools.permutations, store them as a list and do my calculations and print my results.

Now I want to find out magic squares of order 4. Since finding all permutations means 16! possibilities, I want to increase efficiency by placing likely elements in the corner, for example 1, 16 on diagonal one corners and 4, 13 on diagonal two corners.

So how would I find permutations of set {1,2....16} where some elements are not moved is my question

Answer 1

Just pull the placed numbers out of the permutation set. Then insert them into their proper position in the generated permutations.

For your example you'd take out 1, 16, 4, 13. Permute on (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15), for each permutation, insert 1, 16, 4, 13 where you have pre-selected to place them.

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()