20191117, 14:37  #56 
"Ed Hall"
Dec 2009
Adirondack Mtns
111001110011_{2} Posts 
You guys are making me feel better with your posts. I actually have a high of 916e6 with a poorer implementation of LaurV's concept (arrived at on my own, btw). Unfortunately, you're also encouraging me to take machines away from other projects.
Additional thoughts: I had already decided that there would be situations where swapping two elements could no longer increase the determinant, but my solution was (is currently) to start with an entirely new random matrix. Of course, the success still rests on pure luck at this point. . . Last fiddled with by EdH on 20191117 at 14:41 Reason: Add thoughts. 
20191117, 14:46  #57  
Sep 2017
3^{2}×11 Posts 
Quote:


20191117, 17:43  #58  
"Rashid Naimi"
Oct 2015
Remote to Here/There
11111011100_{2} Posts 
Quote:
I have tried sacrificingfew/manyprawns thousands and thousands of runs with the best LatinSquare solution as well as my own highest solution and gave up on that approach. Back to random swaps with better progress than sacrificing the pawn. Last fiddled with by a1call on 20191117 at 17:56 

20191117, 18:03  #59 
"Rashid Naimi"
Oct 2015
Remote to Here/There
11111011100_{2} Posts 
I think one positive tip (as if I have progressed enough to give them out, which I haven't) is not to discard/reject negative determinants as nothighenough. A detriment will change sign each time you swap a row or column.
Last fiddled with by a1call on 20191117 at 18:04 
20191117, 18:04  #60  
"Ben"
Feb 2007
2^{2}×853 Posts 
Quote:
Best by my search procedure so far is now 925432245. 

20191117, 18:28  #61 
Sep 2017
3^{2}·11 Posts 
I guess this is not for me as it involves luck. I was never lucky. In a previous puzzle, I had to guess the last two digits due to precision, so I tried everything. I got it at 96th trial (out of possible 100).

20191117, 20:43  #62 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·503 Posts 
In the continuing series of tips from someone who hasn't cracked the lower threshold:
* If I had the time I would rewrite the code to shuffle diagonal and parallel to diagonal sets of cells rather than just cells. The logic behind this is that swaging rows and columns does not create any new determinant trials and swapping cells would have wasted efforts equivalents to swapping rows and columns. Swapping diagonals and their parallels should be the most/more efficient brute force approach. Last fiddled with by a1call on 20191117 at 20:45 
20191117, 21:55  #63 
"Ed Hall"
Dec 2009
Adirondack Mtns
3^{3}×137 Posts 
Darn it! I was prepared to call off the search due to total inactivity, which would have let me get back to other things. But, I found that the "total inactivity" was selfinflicted by my poor programming. Upon a small adjustment, I outdid my previous high with 919685927, on the initial test run. Now I guess I'll still let this run awhile. . .
Last fiddled with by EdH on 20191117 at 21:58 Reason: I wanted to use a different word. 
20191118, 22:44  #64 
"Ben"
Feb 2007
2^{2}·853 Posts 
For fun, since progress has pretty much stopped on the 9x9 problem, I thought I'd try to find maximal solutions on smaller problems (https://oeis.org/A301371). Happily, I've rediscovered both a(7) and a(8). a(7) took a couple minutes and a(8) a couple hours. But a(9) is not scaling the same :(

20191120, 03:43  #66  
Jan 2017
2^{3}×11 Posts 
Quote:
Suppose you have a matrix A with determinant d and inverse matrix B. You consider swapping entries at position (r1, c1) and (r2, c2). v=A(r2,c2)A(r1,c1). Let A' be the matrix with those entries swapped (add v to number at (r1,c1), subtract v from number at (r2,c2)). Then det(A') = d * ( (1+v*B(c1,r1))*(1v*(B(c2,r2)) + v*v*B(c2,r1)*B(c1,r2) ) You can derive this by considering the product of B and A'. The determinant of this product is 1/det(A) * det(A'). On the other hand, you can directly calculate the determinant cheaply, since only two rows/columns (depending on whether you multiply from right or left) change from the identity matrix result of A*B, and the remaining zeros of the identity matrix eliminate most determinant terms. 

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