The most mentally demanding gift I received for our outflow Xmas final black ;-) is the "Great Minds Aristotle's Number Classic Wooden Puzzle" which yous may purchase at amazon.com. Aristotle must have got non exclusively played amongst that – I think that he invented it. Unless it was invented past times the Chinese because it's also sometimes named the "Lo Shu puzzle". And it may endure just a modern (Chinese?) excogitation inspired past times to a greater extent than primitive piece of occupation of Aristotle's, I am non sure. If yous loathe Sudoku, yous volition loathe this puzzle because it's Sudoku on steroids.
At those times, thinkers must have got played amongst such wooden pieces in addition to marbles a lot. I just verified that upwards to the obvious 12-element symmetry transformations, the solution to this puzzle is unique. This fact seems rather shocking. Why is the disclose of solutions just one? How could he invent it?
OK, what's the puzzle? You have got a large hexagon board amongst 3+4+5+4+3=19 small-scale wooden hexagon pieces inwards it. Those have got numbers betwixt 1 in addition to xix written on them. Your chore is to rearrange the wooden pieces within the large hexagon board inwards such a way that all the fifteen sums inwards the rows are equal to each other.
To endure sure, if yous read the large hexagon every bit 5 rows amongst 3,4,5,4,3 elements, the amount of the 3 pieces inwards the top row must endure X, the amount of the four pieces beneath them must endure X, the amount of the 5 pieces that split the large hexagon to halves must endure X, in addition to and then on. This is v amount rules amongst X on the correct manus side. Because the hexagon may endure sliced inwards 3 mutually similar yet inequivalent ways (which are mutually rotated past times multiples of threescore degrees), at that spot are fifteen amount rules amongst X on the correct manus side.
OK, the kickoff measuring that an intelligent third-grader every bit good every bit your humble correspondent is able to figure out is the value of X. The amount of all pieces is\[
1+2+\dots +19 = \frac{(1+19)\cdot 19}{2} = 190
\] in addition to because the amount of all xix pieces is also the amount of sums of the 5 rows inwards a reading, nosotros may encounter that\[
X = \frac{190}{5} = 38.
\] So the amount inwards each of the fifteen rows has to endure equal to the ugly disclose of 38. Imagine that yous assign xix alphabetic lineament variables to the xix places inwards the puzzle. Those fifteen amount rules plough over yous fifteen weather condition for these xix variables. In fact, exclusively thirteen of these weather condition are independent because "the amount of all pieces is 190" may endure produced every bit a combination of the fifteen amount rules inwards 3 seemingly different ways. (It should truly endure just 19-7=12 independent equations, non 13, I am a combat confused nearly it.)
Spoiler: closed your eyes or plough off your memory.
OK, hither is the unique solution, upwards to symmetries:
I took this moving-picture demo from the Internet. It was created past times individual else in addition to then at that spot is a fault inwards the picture. See the comments for details.
You may encounter – but yous volition kindly forget – that the counterintuitive disclose 5 sits at the center. You could assume that there's some symmetry in addition to then the "middle number" from the laid from 1 to 19, namely 10, would endure inwards the middle. But yous would endure wrong. No such symmetry industrial plant here, partly because of the fact that dissimilar inwards regular Sudoku, nosotros are constraining sums of different numbers of terms. So at that spot is no "Y to 20-Y" counterpart of the "Y to 10-Y" reflection symmetry for the numbers.
Most of the large numbers – all the two-digit numbers – sit down on the boundary of the hexagon. It's because the boundary is where the rows amongst exclusively 3 elements exist, in addition to those must yet create the amount of 38 which agency that the average piece must have got the value of 12.67 over there. The key 7 pieces are the numbers 1-8 amongst the exception of 3 which is the smallest piece that sits at the boundary – on a corner of the hexagon, if yous desire to know it exactly.
How did I banking corporation check that it's the unique solution?
Well, I wrote the whole board inwards price of 7-8 variables inwards Mathematica in addition to tried all possibilities past times creature force. ;-) Well, I used the symmetry grouping amongst 12 elements every bit good every bit some of the inequalities to trim the required fourth dimension to several minutes. Then I was satisfied. You truly don't desire to endeavor all sequences of 7 numbers betwixt 1 in addition to xix because 197 is almost a billion. Even to a greater extent than obviously, yous don't desire to endeavor all permutations of xix pieces because 19! exceeds 1017. So some optimization was done – I could perform a ameliorate optimization but the fourth dimension became manageable.
Mathematica commands
First, I modified the shape of the hexagon to i of the 5 x 5 foursquare matrix amongst 3 zeroes inwards the upper correct corner in addition to 3 zeroes inwards the contrary corner of the matrix:
mf[a_, b_, c_, d_, e_, g_, h_, i_] := {{a, b, 38 - a - b, 0, 0},I manually calculated the remaining pieces inwards price of the independent a,b,c,d,e,g,h,i variables in addition to then that the sums I could encounter were equal to 38. OK, the root looks unreadable but inwards the matrix form, the output is a combat to a greater extent than comprehensible:
{c, d, e, 38 - c - d - e, 0},
{38 - a - c, g, h, i, -38 + a + b + c + d + e},
{0, 38 - b - d - g, -38 + b + d + g - c + h + i + b + e + i,
c - i - h, 38 - b - e - i},
{0, 0, -e - h + a + c - i, d + e + h, 38 - a - c - d + i}};
mf[a, b, c, d, e, g, h, i] // MatrixForm
Click at the matrix inwards a higher house to zoom it in. Then I verified whether all the fifteen sums are equal to 38:
m = mf[a, b, c, d, e, g, h, i];The output was
Total[m]
Total[Transpose[m]]
{m[[1, 3]] + m[[2, 4]] + m[[3, 5]],
m[[1, 2]] + m[[2, 3]] + m[[3, 4]] + m[[4, 5]],
m[[1, 1]] + m[[2, 2]] + m[[3, 3]] + m[[4, 4]] + m[[5, 5]],
m[[2, 1]] + m[[3, 2]] + m[[4, 3]] + m[[5, 4]],
m[[3, 1]] + m[[4, 2]] + m[[5, 3]]}
{38, 38, b + d + e + g + h + i, 38, 38}You encounter that many of them are equal to 38 automatically but some of them depend on the variables. All those volition endure equal to 38 assuming\[
{38, 38, b + d + e + g + h + i, 38, 38}
{38, 38, 38, -38 + 2 b + 2 d + 2 e + 2 g + 2 h + 2 i,
76 - b - d - e - g - h - i}
b + d + e + g + h + i = 38
\] Fine. So I decided to substitute \(h=38-b-d-e-g-i\).
nf[a_, b_, c_, d_, e_, g_, i_] :=The matrix "n" exclusively depends on vii variables:
mf[a, b, c, d, e, g, 38 - b - d - e - g - i, i];
n = nf[a, b, c, d, e, g, i];
n // MatrixForm
Total[n]
Total[Transpose[n]]
Also, it's interesting that if yous endeavor to compute the sums of powers of all these elements past times the commands
Total[Total[n]]you volition larn the answers 190, 7220, 274360 which are independent of the variables. If yous idea that the constraints on the "standard deviations of the pieces" etc. volition plough over yous new, nonlinear equations, yous would endure wrong. The fifteen amount rules truly guarantee that all these sums of powers are equal to the sums of powers of the numbers 1...19.
Total[Total[n]*Total[n]]
Total[Total[n]*Total[n]*Total[n]]
Before nosotros calculate, it's useful to write the command
Dynamic[{a, b, c}]which volition dynamically demo us how far the difficult calculation has gotten. Finally, nosotros may run the code that takes several minutes in addition to tries all the possibilities:
For[a = 1, a <= 14, a++,Here I just tried all values for a,b,c,d,e,g,i betwixt 1 in addition to xix except that the intervals were truncated inwards lodge to speed upwards the calculation:
bmax = Min[37 - a - a, 19];
For[b = xix - a, b <= bmax, b++,
cmin = Max[b + 1, xix - a];
cmax = Min[37 - a - a, 19];
For[c = cmin, c <= cmax, c++,
For[d = 1, d <= 19, d++,
emin = Max[1, 39 - b - c - d];
emax = Min[19, 37 - c - d];
For[e = emin, e <= emax, e++,
gmin = Max[1, 39 - b - c - d];
gmax = xix - a - b - c - d;
For[g = gmin, g <= 19, g++,
imin = Min[a + a + c + d - 39, 1, xix - b - d - g];
imax = Max[19, 37 - b - d - g];
For[i = imin, i <= imax, i++,
If[Length[Union[{a, b, c, d, e, g, i}]] < 7, 0,
nsubstitute = nf[a, b, c, d, e, g, i];
uni = Union[Flatten[nsubstitute]];
If[uni == {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19},
Print[MatrixForm[nsubstitute]], 0];
];
]]]]]]];
- I imposed the status that the left upper corner variable "a" is the smallest i amid the half-dozen similar corners. In this way, I guess fixed the symmetry \(\ZZ_6\) rotating the hexagon past times multiples of threescore degrees – which apparently produces equivalent solutions. In this way, the fourth dimension was reduced to one-sixth.
- I imposed the status that the variable "b" is smaller than "c". In this way, I guess fixed the \(\ZZ_2\) symmetry flipping the hexagon amongst abide by to the axis or in addition to then "a". In combination amongst the previous point, I guess fixed the whole symmetry of the hexagon – which is a semidirect production of a \(\ZZ_6\) in addition to a \(\ZZ_2\) in addition to has 12 elements (and it is known every bit the dihedral grouping \(D_6\)). The fourth dimension was reduced past times a cistron of 12 inwards total.
- I reduced several intervals for the 7 variables from 1...19 to a narrower interval past times imposing the status that some of the entries inwards the matrix that are expressed every bit complicated sums or differences of variables (and/or addition minus 38) are betwixt 1 in addition to 19, too.
- Before I verified whether all numbers 1...19 are represented inwards the matrix, I checked whether the numbers a,b,c,d,e,g,i are all different from each other, in addition to if they are not, I avoided the verification of the total matrix. I don't know whether this two-step decomposition of the chore sped upwards the calculation substantially.
\left(
\begin{array}{ccccc}
3 & 17 & eighteen & \circ & \circ \\
xix & 7 & 1 & eleven & \circ \\
xvi & 2 & 5 & 6 & nine \\
\circ & 12 & four & 8 & fourteen \\
\circ & \circ & 10 & thirteen & fifteen \\
\end{array}
\right)
\] Not bad. Note that 3 is the smallest disclose amid the 6 corner numbers 3,18,9,15,10,16 spell its horizontal vecino 17 is smaller than the vertical vecino 19, every bit nosotros required. For the sake of readability, I replaced 0 inwards the corners amongst \(\circ\) – I did in addition to then inwards our linear algebra textbook to a greater extent than than ii decades ago, too.
I am non certain whether I would endure able to solve – allow lonely invent – the puzzle without the help of a computer. To holler back some features of the diagram – similar 3 inwards the corner, 5 inwards the center, fifteen on the contrary side of 3, "3,17,18" on i edge, or something similar that – seems to endure the exclusively viable way for me to endure able to solve the puzzle inwards the future, without the help of a computer.
Now, a deeper point. This is of class recreational mathematics but fifty-fifty though the solution is unique, the solution looks remarkably "symmetry-breaking" in addition to "non-obvious". The 12-15 amount rules underdetermine the xix variables but the status that each disclose betwixt 1 in addition to xix is represented just i time happens to just compensate for this shortage of conditions.
In some sense, the regular Sudoku looks similar the families of \(U(N)\) groups. This Aristotle's Sudoku is to a greater extent than similar to an especial group. It doesn't seem to follow whatever "easy template" or "trivial logic" but it yet happens to be and, inwards fact, endure the unique solution to some rather natural conditions. Exceptional, unexpected, unique solutions to weather condition play a rather of import operate inwards mathematics, grouping theory, theoretical physics, grand unification, in addition to string theory, amid related corners of physics in addition to mathematics.
With a lot of creature force, i may solve this Aristotle's Sudoku puzzle – in addition to also detect finite sporadic groups such every bit the monster grouping or the especial \(E_8\) Lie group. But because of the uniqueness or close uniqueness of these structures, i is tempted to think that at that spot must ever be a truly simple, nearly spiritual explanation of the beingness of these structures in addition to solutions. The \(E_8\) is the guess grouping on the boundary of M-theory's spacetime (and hence inwards a heterotic string theory, too). We know that it works, why it works, numerous anomaly cancellation weather condition in addition to other expert tidings that the grouping obeys. But isn't at that spot some "really simple" explanation why such a grouping exists in addition to sits on the M-theory's boundaries?
Similarly, why does the monster grouping appear every bit the discrete grouping inwards the CFT dual to the pure AdS3 gravity? And if I brand a bound from these physically justifiable questions, I may inquire a to a greater extent than speculative one: Is at that spot a solution to (or compactification of) string theory that depends on the beingness of this solution to Aristotle's Sudoku puzzle?
Bonus: Finally, I realized why at that spot are just 12 independent "sum rule" equations. There are fifteen to start with. But at that spot are 3 ways of combining the rows to larn "the amount of xix pieces is 190" which reduces fifteen to 13. But at that spot is i to a greater extent than redundance. There are 6 centers of the edges, 19,17,11,14,13,12. It is no coincidence that if yous split them to 2 triangles – 19,11,13 in addition to 17,14,12 – the sums of both groups are equal (it's non slow to encounter why it has to endure 43). The combination 19+11+13-17-14-12 may truly endure obtained inwards ii ways. You either amount the border rows amongst alternating signs; or yous amount the "second rows" amongst alternating signs (one large internal triangle minus the contrary one).
My electrical flow way of remembering the correct setup is: a 3 inwards a corner in addition to 17,19 every bit neighbors. 5 at the center. fifteen inwards the corner against 3 – similar if the solution wants to state that 3*5 = 15. 43 every bit the ii sums of the triangles at centre of the edges. nine inwards some other corner, the exclusively other 1-digit piece inwards a corner. There are ii possibilities where to house it. With these pieces, the completion of the chore is almost straightforward – at most, i tries ii possibilities nearly twice.
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