[math-fun] Don't hire authoritarian programmers
This quick and not at all dirty one liner to draw an orthoscheme, block([o:[0,0,0],a:[1,0,0],b:[0,1,0],c:[0,1,1]],plotfaces(combinations([o,a,b,c],3))) was thwarted by the previous incarnation of this line of code in Macsyma\Macsyma2\share\set2 : /*if not %set=setify(%set) then "Some a&&hole thinks you should be screwed with error(\"First argument to subpowerset must be a set.)",*/ --rwg PS, It can't be news to serious sudokulists, but making the rounds of Mountain View High School is gosper.org/sud5.PNG, a quincunx of overlapping 9x9s with clues in four concentric rings.
At 12:10 AM 5/1/2006, R. William Gosper wrote:
PS, It can't be news to serious sudokulists, but making the rounds of Mountain View High School is http://gosper.org/sud5.PNG, a quincunx of overlapping 9x9s with clues in four concentric rings.
From the NYTimes today, a Sudoku that wasn't solved within the time constraints: 8__75___3 _3__48_2_ 1_______6 34__7___8 79_48__31 2_8____74 5__814__7 _8_327_4_ 4__569__2 Notice that the NYTimes huffs at other newspapers that print Sudoku's, because it doesn't print any itself. ---- I did a Google search on "sudoku" and "direct product", but didn't come up with anything interesting. Since 9=3*3, isn't it possible that at least some Sudoku's are direct product constructions? Aren't at least some Sudoku's the product table of a 9-element group?
At 03:36 PM 5/1/2006, Henry Baker wrote:
I did a Google search on "sudoku" and "direct product", but didn't come up with anything interesting.
Since 9=3*3, isn't it possible that at least some Sudoku's are direct product constructions?
Aren't at least some Sudoku's the product table of a 9-element group?
To answer my own question, the answer is probably "no", at least if group tables are considered. Since each of the subsquares has to have all 9 elements, none of the subsquares can be the product table for a subgroup, because the subgroup must be closed. However, I was able to go backwards -- take the straightforward direct product of Z3xZ3, whose product table is (you'll have to mentally relable i with i-1 to see that this is Z3xZ3): [ 1 2 3 4 5 6 7 8 9 ] [ 2 3 1 5 6 4 8 9 7 ] [ 3 1 2 6 4 5 9 7 8 ] [ 4 5 6 7 8 9 1 2 3 ] [ 5 6 4 8 9 7 2 3 1 ] [ 6 4 5 9 7 8 3 1 2 ] [ 7 8 9 1 2 3 4 5 6 ] [ 8 9 7 2 3 1 5 6 4 ] [ 9 7 8 3 1 2 6 4 5 ] If you then "mix" the columns, by taking the 1st, 4th, 7th, 2nd, 5th, 8th, and then 3rd, 6th, 9th columns (BTW, "mix" is its own inverse), you get: [ 1 4 7 2 5 8 3 6 9 ] [ 2 5 8 3 6 9 1 4 7 ] [ 3 6 9 1 4 7 2 5 8 ] [ 4 7 1 5 8 2 6 9 3 ] [ 5 8 2 6 9 3 4 7 1 ] [ 6 9 3 4 7 1 5 8 2 ] [ 7 1 4 8 2 5 9 3 6 ] [ 8 2 5 9 3 6 7 1 4 ] [ 9 3 6 7 1 4 8 2 5 ] This matrix now satisfies the Sudoku criteria. It also has a very simple structure. I now wonder how many (what % of) sudoku squares can be generated from the direct product of two 3-element groups, followed by some sort of "mix"-ing operation, such as the one above.
Henry:
If you then "mix" the columns, by taking the 1st, 4th, 7th, 2nd, 5th, 8th, and then 3rd, 6th, 9th columns (BTW, "mix" is its own inverse), you get:
[ 1 4 7 2 5 8 3 6 9 ] [ 2 5 8 3 6 9 1 4 7 ] [ 3 6 9 1 4 7 2 5 8 ] [ 4 7 1 5 8 2 6 9 3 ] [ 5 8 2 6 9 3 4 7 1 ] [ 6 9 3 4 7 1 5 8 2 ] [ 7 1 4 8 2 5 9 3 6 ] [ 8 2 5 9 3 6 7 1 4 ] [ 9 3 6 7 1 4 8 2 5 ]
This matrix now satisfies the Sudoku criteria.
It also has a very simple structure.
I now wonder how many (what % of) sudoku squares can be generated from the direct product of two 3-element groups, followed by some sort of "mix"-ing operation, such as the one above.
Up to isomorphism there are two groups of order 9, Z3 x Z3 and Z9. I assume "mix"-ing means doing the usual equivalences: -permuting rows/columns within a band -permuting bands -transposing The group of these equivalences has order 2*6^8=3359232. Your Z3 x Z3 grid has an automorphism group of order 648 (the max possible) so you can get 3359232/648 = 5184 different grids. For the other group of order 9, the cyclic group, you also get a grid with 648 automorphisms, so another 5184 grids. 123 456 789 456 789 123 789 123 456 234 567 891 567 891 234 891 234 567 345 678 912 678 912 345 912 345 678 Gary McGuire
At 04:18 AM 5/17/2006, Gary McGuire wrote:
If you then "mix" the columns, by taking the 1st, 4th, 7th, 2nd, 5th, 8th, and then 3rd, 6th, 9th
columns (BTW, "mix" is its own inverse), you get:
[ 1 4 7 2 5 8 3 6 9 ] [ 2 5 8 3 6 9 1 4 7 ] [ 3 6 9 1 4 7 2 5 8 ] [ 4 7 1 5 8 2 6 9 3 ] [ 5 8 2 6 9 3 4 7 1 ] [ 6 9 3 4 7 1 5 8 2 ] [ 7 1 4 8 2 5 9 3 6 ] [ 8 2 5 9 3 6 7 1 4 ] [ 9 3 6 7 1 4 8 2 5 ]
This matrix now satisfies the Sudoku criteria.
It also has a very simple structure.
I now wonder how many (what % of) sudoku squares can be generated from the direct product of two 3-element groups, followed by some sort of "mix"-ing operation, such as the one above.
Up to isomorphism there are two groups of order 9, Z3 x Z3 and Z9.
Don't you mean the only _commutative_ groups? Aren't there a lot more non-commutative groups of order 9? (I'm very rusty on my group theory...) Latin square operation tables only require "solvability" of equations, but not necessarily the associativity or identities of a group operation. But I'm willing to look at only group operation tables for the moment.
I assume "mix"-ing means doing the usual equivalences: -permuting rows/columns within a band -permuting bands -transposing
No, "mix"-ing does _not_ preserve sudoku-ness. This is a different operation that can create sudoku-ness out of a non-sudoku.
The group of these equivalences has order 2*6^8=3359232. Your Z3 x Z3 grid has an automorphism group of order 648 (the max possible) so you can get 3359232/648 = 5184 different grids. For the other group of order 9, the cyclic group, you also get a grid with 648 automorphisms, so another 5184 grids.
123 456 789 456 789 123 789 123 456
234 567 891 567 891 234 891 234 567
345 678 912 678 912 345 912 345 678
Gary McGuire
Groups of order P and P^2 are always commutative. The only P^2 groups are (mod P^2) and Zp x Zp. There are 5 groups of order P^3, 3 commutative Z(p^3), Zp x Z(p^2), Zp x Zp x Zp, and two non-commutative. When P=2, the noncomms are D4 (group of the square) and the quaternions +-(1,i,j,k). When P>2, the two noncomms are different from the P=2 case, but I don't remember the rules. I think one has every element (except E) of order P, while the other has some order P^2 elements. For order P^4, there are 14 groups when P=2, a few more when P>2. All have 5 commutatives, based on partitions of 4 as 4,31,22,211,1111. It would be interesting to look at Sudoku Edits: small changes that preserve S-ness. The simplest is to find a sub-pattern like A B B A where the left pair lies in a box, and the right pair in (another) box. Then the exchange to B A A B produces another Sudoku. (A puzzle with this completed pattern will be ambiguous unless one of the A or B cells is a starting clue.) A fancier edit might use 6 or 8 or 9 cells. Is there a small set of edits that makes the Sudokus into a connected graph? (Similar questions have been asked about magic squares.) Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Henry Baker Sent: Wed 5/17/2006 10:36 AM To: Gary McGuire Cc: math-fun Subject: Re: [math-fun] Sudokus from direct products At 04:18 AM 5/17/2006, Gary McGuire wrote:
If you then "mix" the columns, by taking the 1st, 4th, 7th, 2nd, 5th, 8th, and then 3rd, 6th, 9th
columns (BTW, "mix" is its own inverse), you get:
[ 1 4 7 2 5 8 3 6 9 ] [ 2 5 8 3 6 9 1 4 7 ] [ 3 6 9 1 4 7 2 5 8 ] [ 4 7 1 5 8 2 6 9 3 ] [ 5 8 2 6 9 3 4 7 1 ] [ 6 9 3 4 7 1 5 8 2 ] [ 7 1 4 8 2 5 9 3 6 ] [ 8 2 5 9 3 6 7 1 4 ] [ 9 3 6 7 1 4 8 2 5 ]
This matrix now satisfies the Sudoku criteria.
It also has a very simple structure.
I now wonder how many (what % of) sudoku squares can be generated from the direct product of two 3-element groups, followed by some sort of "mix"-ing operation, such as the one above.
Up to isomorphism there are two groups of order 9, Z3 x Z3 and Z9.
Don't you mean the only _commutative_ groups? Aren't there a lot more non-commutative groups of order 9? (I'm very rusty on my group theory...) Latin square operation tables only require "solvability" of equations, but not necessarily the associativity or identities of a group operation. But I'm willing to look at only group operation tables for the moment.
I assume "mix"-ing means doing the usual equivalences: -permuting rows/columns within a band -permuting bands -transposing
No, "mix"-ing does _not_ preserve sudoku-ness. This is a different operation that can create sudoku-ness out of a non-sudoku.
The group of these equivalences has order 2*6^8=3359232. Your Z3 x Z3 grid has an automorphism group of order 648 (the max possible) so you can get 3359232/648 = 5184 different grids. For the other group of order 9, the cyclic group, you also get a grid with 648 automorphisms, so another 5184 grids.
123 456 789 456 789 123 789 123 456
234 567 891 567 891 234 891 234 567
345 678 912 678 912 345 912 345 678
Gary McGuire
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participants (4)
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Gary McGuire -
Henry Baker -
R. William Gosper -
Schroeppel, Richard