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#56 |
"Ed Hall"
Dec 2009
Adirondack Mtns
3,623 Posts |
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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 2019-11-17 at 14:41 Reason: Add thoughts. |
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#57 | |
Sep 2017
2·72 Posts |
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#58 | |
"Rashid Naimi"
Oct 2015
Remote to Here/There
26·31 Posts |
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I have tried sacrificing-few/many-prawns thousands and thousands of runs with the best Latin-Square 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 2019-11-17 at 17:56 |
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#59 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
198410 Posts |
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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 not-high-enough. A detriment will change sign each time you swap a row or column.
Last fiddled with by a1call on 2019-11-17 at 18:04 |
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#60 | |
"Ben"
Feb 2007
3,371 Posts |
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Best by my search procedure so far is now 925432245. |
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#61 |
Sep 2017
2×72 Posts |
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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).
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#62 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
198410 Posts |
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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 2019-11-17 at 20:45 |
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#63 |
"Ed Hall"
Dec 2009
Adirondack Mtns
70478 Posts |
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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 self-inflicted 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 2019-11-17 at 21:58 Reason: I wanted to use a different word. |
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#64 |
"Ben"
Feb 2007
D2B16 Posts |
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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 :(
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#66 | |
Jan 2017
3·29 Posts |
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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))*(1-v*(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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