How to create a magic square if we know a date. Eg-22-04-2014 The first column should have 22 2nd-04 3rd-20 and 4th -14. I believe ramanujan created the same thing for his birthday but I don't know the method he used. Please help. Is there any method to do it? I tried doing it using the fact that the row diagonal 2*2 square sum should all be equal to the total sum of the first row in this case 22+4+20+14=60
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Please be more specific : Which entries should the magic square have beside the given ones ? – Peter Aug 13 '14 at 21:36
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Let the magic square be $\begin{bmatrix}a & b & c & d \\ x_1 & x_2 & x_3 & x_4 \\ x_5 & x_6 & x_7 & x_8 \\ x_9 & x_{10} & x_{11} & x_{12}\end{bmatrix}$ where $a,b,c,d$ represents the date.
The conditions that the rows, columns, and diagonals sum to $a+b+c+d$ gives us $9$ linear equations for $12$ variables. The resulting system has rank $8$.
Therefore, any solution can be obtained by taking a particular solution and adding a linear combination of $12-8 = 4$ "basis vectors" for the nullspace of the system.
The matrix $\begin{bmatrix}a & b & c & d \\ c & d & a & b \\ d & c & b & a \\ b & a & d & c\end{bmatrix}$ is a magic square with the correct top row.
The following $4$ matrices form a basis of the nullspace of the system (all rows, cols, diags sum to $0$):
$\begin{bmatrix}0 & 0 & 0 & 0\\ 2 & -1 & -1 & 0 \\ -2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 1 & 0 & 0 & -1 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 1 & 0 & -1 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 2 & -1 & 0 & -1 \\ -1 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1\end{bmatrix}$.
Take the first matrix and add any linear combination of the above four matrices to get a solution.
JimmyK4542
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I think there should not be any number repeated also. Please see the ramanujan birth date magic square. – geek101 Aug 14 '14 at 06:22
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Here is a solution in Perl, which makes no claim to optimization but does produce results. The algorithm uses a mixing strategy to obtain balanced squares. For today the fourteenth of August 2014 we get the following squares (initial segment shown)
$ ./md-shuffle.pl 14-08-2014 56 014 008 020 014 022 023 006 005 008 024 003 021 012 001 027 016 014 008 020 014 022 023 005 006 001 018 010 027 019 007 021 009 014 008 020 014 022 023 005 006 007 024 004 021 013 001 027 015 014 008 020 014 022 023 003 008 005 024 006 021 015 001 027 013
For Xmas eve we obtain
$ ./md-shuffle.pl 24-12-2014 70 024 012 020 014 009 021 022 018 007 004 023 036 030 033 005 002 024 012 020 014 009 021 007 033 022 034 008 006 015 003 035 017 024 012 020 014 009 021 005 035 022 036 008 004 015 001 037 017
Euler's birthday was 15-04-1707:
$ ./md-shuffle.pl 15-04-1707 43 015 004 017 007 004 018 014 007 011 009 002 021 013 012 010 008 015 004 017 007 004 010 009 020 005 008 016 014 019 021 001 002
Gauss's birthday was 30-04-1777:
$ ./md-shuffle.pl 30-04-1777 128 030 004 017 077 089 014 013 012 007 036 065 020 002 074 033 019 030 004 017 077 089 014 013 012 008 037 064 019 001 073 034 020 030 004 017 077 089 014 013 012 003 032 069 024 006 078 029 015
This is the code which it may be a useful exercise to study and improve. As I mentioned it really does admit improvement in several spots but demonstrates proof of concept.
#! /usr/bin/perl -w
#
sub search {
my ($sq, $sum, $sofar, $seen) = @_;
if($sofar == 16){
for(my $row=0; $row<4; $row++){
for(my $col=0; $col<4; $col++){
printf "%03d ", $sq->[$row][$col];
}
print "\n";
}
print "\n";
return;
}
my $loc_col = $sofar % 4;
my $loc_row = ($sofar - $loc_col) / 4;
my $tries = [(1..$sum)];
for(my $pos=$sum-1; $pos>=0; $pos--){
my $targ = int rand($pos+1);
my $tmp;
$tmp = $tries->[$targ];
$tries->[$targ] = $tries->[$pos];
$tries->[$pos] = $tmp;
}
my $ind = 0;
while($ind < scalar(@$tries)){
my $nxt = $tries->[$ind];
if(!exists($seen->{$nxt})){
$seen->{$nxt} = 1;
$sq->[$loc_row][$loc_col] = $nxt;
my $no_admit = undef; my ($empty, $sval);
for(my $row=0; $row<4; $row++){
$empty = 0; $sval = 0;
for(my $col=0; $col<4; $col++){
if($sq->[$row][$col] == -1){
$empty++;
}
else{
$sval += $sq->[$row][$col];
}
}
if(($empty == 0 && $sval != $sum) ||
($empty > 0 && $sval >= $sum)){
$no_admit = 1;
}
}
for(my $col=0; $col<4; $col++){
$empty = 0; $sval = 0;
for(my $row=0; $row<4; $row++){
if($sq->[$row][$col] == -1){
$empty++;
}
else{
$sval += $sq->[$row][$col];
}
}
if(($empty == 0 && $sval != $sum) ||
($empty > 0 && $sval >= $sum)){
$no_admit = 1;
}
}
$empty = 0; $sval = 0;
for(my $diag=0; $diag<4; $diag++){
if($sq->[$diag][$diag] == -1){
$empty++;
}
else{
$sval += $sq->[$diag][$diag];
}
}
if(($empty == 0 && $sval != $sum) ||
($empty > 0 && $sval >= $sum)){
$no_admit = 1;
}
$empty = 0; $sval = 0;
for(my $diag=0; $diag<4; $diag++){
if($sq->[$diag][3-$diag] == -1){
$empty++;
}
else{
$sval += $sq->[$diag][3-$diag];
}
}
if(($empty == 0 && $sval != $sum) ||
($empty > 0 && $sval >= $sum)){
$no_admit = 1;
}
if(!defined($no_admit)){
search($sq, $sum, $sofar+1, $seen);
}
$sq->[$loc_row][$loc_col] = -1;
delete $seen->{$nxt};
}
$ind++;
}
}
MAIN: {
my $date = shift || '31-12-2014';
die "not a date" if
$date !~ /^(\d\d)-(\d\d)-(\d\d)(\d\d)$/;
my $first = [$1, $2, $3, $4];
my $sum = $first->[0]+$first->[1]
+$first->[2]+$first->[3];
print "$sum\n";
my $sq = []; push @$sq, $first;
for(my $row=1; $row<4; $row++){
push @$sq, [-1, -1, -1, -1];
}
my $seen = {
$first->[0] => 1, $first->[1] => 1,
$first->[2] => 1, $first->[3] => 1
};
search($sq, $sum, 4, $seen);
}
Marko Riedel
- 61,317
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I wanted it to be done physically as a sum and not by the computer but I am impressed. I guess we can do it in matlab also then. Ramanujan did not do it with the help of a computer. – geek101 Aug 14 '14 at 06:16