Bimagic Square of Order 10

Fredrik Jansson

Bimagic Square of Order 10. By @ Fredrik Jansson, 2005. 2) Micke H. 3) @ Christian Boyer, 2006. 4) Micke H. 5) @ Pan Fengchu, 2007. 6) Micke H.

Fredrik Jansson (Turku 1982 -)

I am a physicist, working as a postdoc at CWI in Amsterdam. I work on weather and climate simulations, in particular on coupling local, cloud-resolving atmospheric models with global weather models. From 2013 to 2016 I worked at the Department of Computational Science at the University of Amsterdam. I was working on gene regulatory networks and swarm robotics, in the Swarm Organ project. For my PhD thesis, I studied charge transport in disordered materials. I did this work at the Department of Physics at Åbo Akademi University in Turku, Finland, in collaboration with the Vielteilchen group at Phillipps-Universität Marburg, Germany. I keep a blog about craft and electronics projects, patterns and observations. I am also a somewhat inactive radio amateur. https://user.abo.fi/fjansson/index.php?page=

Story of the first 10th-order bimagic square

Text written by Fredrik Jansson, in February 2005.

I got many of these ideas from Christian Boyer's www.multimagie.com web site, and Walter Trump's story of the 12th-order trimagic square. Thank you! I first listed all bimagic series of order 10, and wrote them in a file. I found 24,643,236 series. This is also the number which was reported on the multimagic series page of www.multimagie.com. A series is represented by a 128-bit number, where bit n is set if the series contains the number n. The 10 rows of a magic square cannot have any number in common, this is quickly checked with the boolean AND function. A set of 10 independent rows are searched for by backtracking. When 10 independent rows are found, the program tries to find a set of 10 columns, such that every column has exactly one number common with every row, and no column has any number in common with another column. This is done by first searching for all series that have exactly one bit in common with every chosen row, and then trying to construct a set of 10 independent columns from these. This too is done by a backtracking search. A quick test to see if two numbers have more than one bit in common is: a AND (a-1) is zero if a has at most one-bit set, because subtracting 1 changes the least-significant set bit to 0, but leaves every other set bit unchanged. If a set of 10 independent columns is found, an at least semi-bimagic square can be constructed from the chosen rows and columns. A diagonal must have one number in common with every row, and one number in common with every column. The series in the list of possible columns already have one number in common with every row, so the program searches through these to find all that have one number in common with every chosen column. The two diagonals must have no numbers in common in an even-order square, so every pair of possible diagonals is tested. If they have no bits in common, these could possibly be the diagonals in our square. There is one additional requirement that almost made me loose hope about finding the square: if row 1 contains the numbers a and b, column 1 contains a and c, column 2 contains b and d, a and d lie on one diagonal, and b and c on the other, then c and d must be in the same row. a .. b .. .. .. c .. d So, when we have chosen the first element on diagonal 1, there is only one row that contains this element, so this must be the first row. There is also only one column that contains this element, so this must be the first column. The first row has exactly one element in common with the second diagonal, this element must be the last element in the first row. This also fixes the position of the column containing that element, it will be the last column. The last column has one element in common with the first diagonal, this element must be in the lower right corner d. The first column has one element in common with the second diagonal, this must be in the lower left corner c. If elements c and d are on different rows, nothing can be done besides using another pair of diagonals. The same check must also be done for the second and second-to-last elements of the diagonals, and so on. I think the key ideas in my program are to use bit-strings to represent the series, and not to consider the order of the rows and columns, or the order of the elements in the rows and columns. This makes the search space considerably smaller. When the program found the solution, it just printed the rows, columns and diagonals used, and I assembled the square by hand. The program was written in C. I had plans to optimize the most timecritical part (searching for a set of 10 columns with no bits in common) with assembler-language, possibly using SSE-instructions that can do calculations on a 128-bit number at once, but the solution was found before I got that done. This project took about a week to program, and the square was found in less than 24h of computing on my Athlon 2800. Fredrik Jansson, Turku, January 2004.

002	019	070	001	066	074	073	060	068	072
058	077	015	003	065	004	067	069	071	076
062	063	082	075	061	059	079	006	005	013
049	018	014	078	098	040	025	096	043	044
094	041	027	042	035	091	021	095	037	022
093	039	023	038	031	090	033	030	029	099
034	100	036	083	045	024	026	028	097	032
008	085	064	057	007	056	080	048	016	084
054	011	086	047	087	012	092	020	050	046
051	052	088	081	010	055	009	053	089	017

024	043	098	086	097	036	032	034	025	030
082	099	100	031	027	035	041	028	029	033
083	052	023	087	061	003	005	076	057	058
038	039	026	019	042	040	095	022	088	096
062	008	063	078	011	070	071	068	002	072
060	007	059	074	010	066	006	080	079	064
016	093	044	037	045	094	053	021	017	085
001	067	018	065	077	056	073	075	069	004
049	050	020	013	046	091	048	092	084	012
090	047	054	015	089	014	081	009	055	051

081	044	041	063	088	003	049	053	001	082
026	038	092	090	025	045	042	062	002	083
096	097	031	046	068	008	022	024	057	056
016	100	009	075	011	071	043	054	065	061
028	048	007	051	034	091	095	059	077	015
013	027	087	014	060	089	055	064	079	017
072	036	052	018	086	047	023	006	066	099
058	010	074	030	084	050	005	094	067	033
080	076	039	098	037	032	078	004	021	040
035	029	073	020	012	069	093	085	070	019

063	075	011	009	056	076	039	059	018	099
057	020	038	060	098	013	048	052	019	100
001	085	026	092	030	090	047	058	040	036
004	005	055	070	093	033	077	079	045	044
074	088	087	014	012	041	037	046	084	022
053	073	050	094	010	067	042	006	086	024
091	043	071	027	051	017	007	096	068	034
065	029	083	049	054	015	095	078	002	035
072	066	081	028	032	089	016	008	082	031
025	021	003	062	069	064	097	023	061	080

089	051	052	088	053	055	010	009	017	081
059	082	062	061	013	006	079	075	063	005
001	002	066	068	072	074	070	073	060	019
042	041	027	022	091	021	035	037	095	094
057	080	064	008	016	085	056	048	007	084
054	020	086	092	011	050	012	087	046	047
078	098	014	040	043	018	044	096	025	049
003	065	004	077	071	058	076	015	067	069
039	030	033	023	090	038	099	031	093	029
083	036	097	026	045	100	024	034	032	028

Micke