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Groups > comp.theory > #52765 > unrolled thread
| Started by | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| First post | 2022-06-22 10:40 -0700 |
| Last post | 2022-06-24 21:30 +0100 |
| Articles | 18 on this page of 38 — 5 participants |
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Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-22 10:40 -0700
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-22 10:42 -0700
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-22 10:50 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-22 19:25 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-23 07:03 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-23 23:52 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-23 16:07 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-24 02:28 +0100
Re: Resolved: Even vs. Odd Latin Square Problem Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2022-06-24 01:00 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Paul N <gw7rib@aol.com> - 2022-06-24 04:50 -0700
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-24 08:38 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-24 21:38 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-24 14:04 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Jeff Barnett <jbb@notatt.com> - 2022-06-24 23:06 -0600
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-25 07:22 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Jeff Barnett <jbb@notatt.com> - 2022-06-25 11:25 -0600
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-25 14:06 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Jeff Barnett <jbb@notatt.com> - 2022-06-25 16:40 -0600
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-25 17:21 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-26 01:02 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-25 17:26 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-26 14:44 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-26 07:47 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-26 22:26 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-26 17:57 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-27 02:15 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-26 18:27 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-27 02:54 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-27 08:51 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-27 21:56 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-27 15:56 -0700
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-26 08:03 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-26 21:49 +0100
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-24 21:17 +0100
Re: Resolved: Even vs. Odd Latin Square Problem Paul N <gw7rib@aol.com> - 2022-06-25 04:44 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-24 21:14 +0100
Re: Resolved: Even vs. Odd Latin Square Problem "B.H." <xlt.pjw@gmail.com> - 2022-06-24 08:35 -0700
Re: Resolved: Even vs. Odd Latin Square Problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2022-06-24 21:30 +0100
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-25 17:26 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <18c8fe7f-c245-4fee-b9de-334e723e6965n@googlegroups.com> |
| In reply to | #52985 |
On Saturday, June 25, 2022 at 8:02:04 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Friday, June 24, 2022 at 4:38:27 PM UTC-4, Ben Bacarisse wrote: > > >> I don't want to go into detail about what you wrote unless you ask > >> because I don't want to smear you again and I think pretty much any > >> criticism of you work is a smear in your opinion. But do say if you > >> want to know what I really think. > > > > I don't invite you to smear me; I'm not sure any discussion of my work > > would be informed by an understanding of what a mathematical proof is. > I rather hoped for some clear advice: if I spot another error in > something you post, would you rather I keep quiet about it? And if > you'd rather I pointed it out, how can I do that without offending you? In spite of your record of bad behavior, yes, point it out, but don't be inaccurate and don't smear me, i.e., don't make claims about my mathematical talent that are *untrue*. Since you have a Ph.D., you might want to make sure that you don't disgrace whoever educated you by falsely and intransigently accusing of making errors I haven't made. Clear up your thinking, you have a job and no cognitive disability AFAIK. If you make a mistake, it's OK if you don't insist that someone else is a "crank" or totally wrong based on your own mistakes. Be correct, able to admit to your mistakes, or be silent. Don't over-estimate your abilities, you're not a clear satirist like JSH was. > > Are you really an adult, > Yes. > > representing the UK on the internet, > No. > > or do we need to change those two parameters (adult, UK, to clarify > > for the frivolity quibblers) to describe you? > I am an adult living in the UK. I am not sure why that matters to you, > but that's the truth, for what it's worth. > OK. It's more true than your claims about my math, granted. My life, and thus for now, my reputation as a math thinker, are extremely important to me right now. Getting in my way makes me upset, let's leave it at that. Unlike you, I think, I don't want me to die. -Philip White > -- > Ben.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-26 14:44 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <878rpjv8t7.fsf@bsb.me.uk> |
| In reply to | #52987 |
"B.H." <xlt.pjw@gmail.com> writes: > On Saturday, June 25, 2022 at 8:02:04 PM UTC-4, Ben Bacarisse wrote: >> "B.H." <xlt...@gmail.com> writes: >> >> > On Friday, June 24, 2022 at 4:38:27 PM UTC-4, Ben Bacarisse wrote: >> >> >> I don't want to go into detail about what you wrote unless you ask >> >> because I don't want to smear you again and I think pretty much any >> >> criticism of you work is a smear in your opinion. But do say if you >> >> want to know what I really think. >> > >> > I don't invite you to smear me; I'm not sure any discussion of my work >> > would be informed by an understanding of what a mathematical proof is. >> I rather hoped for some clear advice: if I spot another error in >> something you post, would you rather I keep quiet about it? And if >> you'd rather I pointed it out, how can I do that without offending you? > > In spite of your record of bad behavior, yes, point it out, but don't > be inaccurate and don't smear me, i.e., don't make claims about my > mathematical talent that are *untrue*. I correctly pointed out an error, and you called that a smear. You repeated the mistake, claiming again that the problem you linked to was not an open problem any more. I think it is safer for both of us if I don't point out any further mistakes I see. > Since you have a Ph.D., I don't have a PhD. I said so when you (very politely) called me Dr Bacarisse. While I have examined and supervised several PhDs over the years, but I don't have one myself. > you might want to make sure that you don't disgrace whoever educated > you by falsely and intransigently accusing of making errors I haven't > made. You were wrong about having resolved the open problem you posted about. It's still open. You don't appear to have changed you mind about that. -- Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-26 07:47 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <7ad01d7a-0352-4efe-b81d-bf6c66bb9e08n@googlegroups.com> |
| In reply to | #53006 |
On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Saturday, June 25, 2022 at 8:02:04 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Friday, June 24, 2022 at 4:38:27 PM UTC-4, Ben Bacarisse wrote: > >> > >> >> I don't want to go into detail about what you wrote unless you ask > >> >> because I don't want to smear you again and I think pretty much any > >> >> criticism of you work is a smear in your opinion. But do say if you > >> >> want to know what I really think. > >> > > >> > I don't invite you to smear me; I'm not sure any discussion of my work > >> > would be informed by an understanding of what a mathematical proof is. > >> I rather hoped for some clear advice: if I spot another error in > >> something you post, would you rather I keep quiet about it? And if > >> you'd rather I pointed it out, how can I do that without offending you? > > > > In spite of your record of bad behavior, yes, point it out, but don't > > be inaccurate and don't smear me, i.e., don't make claims about my > > mathematical talent that are *untrue*. > I correctly pointed out an error, and you called that a smear. You > repeated the mistake, claiming again that the problem you linked to was > not an open problem any more. I think it is safer for both of us if I > don't point out any further mistakes I see. You pointing out the error was not a smear, it was a different comment. You don't have to point out anything. > > Since you have a Ph.D., > I don't have a PhD. I said so when you (very politely) called me Dr > Bacarisse. While I have examined and supervised several PhDs over the > years, but I don't have one myself. OK. > > you might want to make sure that you don't disgrace whoever educated > > you by falsely and intransigently accusing of making errors I haven't > > made. > You were wrong about having resolved the open problem you posted about. > It's still open. You don't appear to have changed you mind about that. > I am not wrong at all, it is surprising that you claim to think that that easy proof, which I will not publish, does not exist. I simply didn't publish the proof. > -- > Ben.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-26 22:26 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <87fsjrt8u5.fsf@bsb.me.uk> |
| In reply to | #53007 |
"B.H." <xlt.pjw@gmail.com> writes: > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: >> You were wrong about having resolved the open problem you posted about. >> It's still open. You don't appear to have changed you mind about that. > > I am not wrong at all, it is surprising that you claim to think that > that easy proof, which I will not publish, does not exist. There are counterexamples. There can be no proof that (in your words) "it is actually the case that for every positive integer m, the number of even Latin squares of order m and the number of odd Latin squares of order m are *the same*" because there is (to take just one counterexample) no odd 1x1 Latin square. When I asked about this you said "The odd Latin square of order 1 is simply the square with one cell containing -1" but that is an even Latin square. A list of counterexamples was even included in the link to the problem you said you had resolved. In fact every n for which the counts are currently known is a counterexample. That is why the conjecture is that OLS(n) =/= ELS(n) for all n. > I simply didn't publish the proof. There can't be a proof of your claim to have resolved the conjecture because there are counterexamples. -- Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-26 17:57 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <151e87ad-992b-4eaf-bbfa-6a4b6d75759bn@googlegroups.com> |
| In reply to | #53032 |
On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > > >> You were wrong about having resolved the open problem you posted about. > >> It's still open. You don't appear to have changed you mind about that. > > > > I am not wrong at all, it is surprising that you claim to think that > > that easy proof, which I will not publish, does not exist. > There are counterexamples. There can be no proof that (in your words) > "it is actually the case that for every positive integer m, the number > of even Latin squares of order m and the number of odd Latin squares > of order m are *the same*" > because there is (to take just one counterexample) no odd 1x1 Latin > square. When I asked about this you said > "The odd Latin square of order 1 is simply the square with one cell > containing -1" > but that is an even Latin square. > You finally stated your objection. "A latin square is even if the product of the signs of all of the row and column permutations is 1 and is odd otherwise. " If you treat the same square as the selected row and column, then the "base case" would have to be changed to a 2 x 2 Latin square, where it would work. I am not sharing the rest of the details of the proof with you. My version is defensible, and my original proof was absolutely correct according to my version of what permutations are and whether or not they can be repeated in the case of a 1 x 1 matrix. You vultures are wrong again! Go fly off somewhere else, YOU LOSE THE MATH DEBATE, yet again, right-wing morons! Gosh, I hope no one minds that cathartic insult about IlLoGiCaL right-wing math/CS losers. You lunatics won't admit you were wrong though, that's not how you do things, is it? I bet you don't even know how to finish the proof! -Philip White (philipjwhite@yahoo.com) > A list of counterexamples was even included in the link to the problem > you said you had resolved. In fact every n for which the counts are > currently known is a counterexample. That is why the conjecture is that > OLS(n) =/= ELS(n) for all n. > > I simply didn't publish the proof. > There can't be a proof of your claim to have resolved the conjecture > because there are counterexamples. > > -- > Ben.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-27 02:15 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <877d52uct3.fsf@bsb.me.uk> |
| In reply to | #53036 |
"B.H." <xlt.pjw@gmail.com> writes: > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: >> "B.H." <xlt...@gmail.com> writes: >> >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: >> >> >> You were wrong about having resolved the open problem you posted about. >> >> It's still open. You don't appear to have changed you mind about that. >> > >> > I am not wrong at all, it is surprising that you claim to think that >> > that easy proof, which I will not publish, does not exist. >> There are counterexamples. There can be no proof that (in your words) >> "it is actually the case that for every positive integer m, the number >> of even Latin squares of order m and the number of odd Latin squares >> of order m are *the same*" >> because there is (to take just one counterexample) no odd 1x1 Latin >> square. When I asked about this you said >> "The odd Latin square of order 1 is simply the square with one cell >> containing -1" >> but that is an even Latin square. > > You finally stated your objection. There are two: that the Latin square you gave is even, and that there are counterexamples (included in the page you linked to) to your claimed resolution. > "A latin square is even if the product of the signs of all of the row > and column permutations is 1 and is odd otherwise. " Yes, so your example is an even Latin square, not an odd one. Unless there is (up to isomorphism) exactly one odd Latin square the simple case of n=1 is a counterexample to your claimed resolution. The page you linked to included 7 other counterexamples -- basically every known case is a counterexample. > If you treat the same square as the selected row and column, then the > "base case" would have to be changed to a 2 x 2 Latin square, where it > would work. > > I am not sharing the rest of the details of the proof with you. There can't be a proof of your claim because there are counterexamples. They were even listed in the page you linked to. -- Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-26 18:27 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <e67dccc3-2c3f-4d9b-ae2a-05eb64b6d1can@googlegroups.com> |
| In reply to | #53037 |
On Sunday, June 26, 2022 at 9:15:22 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > >> > >> >> You were wrong about having resolved the open problem you posted about. > >> >> It's still open. You don't appear to have changed you mind about that. > >> > > >> > I am not wrong at all, it is surprising that you claim to think that > >> > that easy proof, which I will not publish, does not exist. > >> There are counterexamples. There can be no proof that (in your words) > >> "it is actually the case that for every positive integer m, the number > >> of even Latin squares of order m and the number of odd Latin squares > >> of order m are *the same*" > >> because there is (to take just one counterexample) no odd 1x1 Latin > >> square. When I asked about this you said > >> "The odd Latin square of order 1 is simply the square with one cell > >> containing -1" > >> but that is an even Latin square. > > > > You finally stated your objection. > There are two: that the Latin square you gave is even, and that there > are counterexamples (included in the page you linked to) to your claimed > resolution. > > "A latin square is even if the product of the signs of all of the row > > and column permutations is 1 and is odd otherwise. " > Yes, so your example is an even Latin square, not an odd one. Unless > there is (up to isomorphism) exactly one odd Latin square the simple > case of n=1 is a counterexample to your claimed resolution. The page > you linked to included 7 other counterexamples -- basically every known > case is a counterexample. > > If you treat the same square as the selected row and column, then the > > "base case" would have to be changed to a 2 x 2 Latin square, where it > > would work. > > > > I am not sharing the rest of the details of the proof with you. > There can't be a proof of your claim because there are counterexamples. > They were even listed in the page you linked to. > > -- > Ben. I don't know what table you have called master and bowed humbly to, but you are incorrect. I am familiar with the mathematical proof, and your insistently repeated protestations to the contrary are patently incorrect. You ignored my idea about a different permutation definition for the 1 x 1 matrix--where -1 is an odd Latin square because there is only one permutation in this case, and the product, "-1" is taken to be -1 making it odd. You also ignored my 2 x 2 Latin square, opting for empty rhetoric to fool the masses whom you think do not know anyone who knows math, and stating no serious points beyond whining, "but the table, the table!!" Your debating style and efforts to kill me with false claims about math anger me, as you can see. Fortunately, perceptive enough readers will not be fooled by your nonsense; the proof is airtight, and no table from thin air that you cite, that is not even about the version of the conjecture we are discussing, can refute the point. I will be calm in math discussions in public, which I may never have, after I'm out of captivity. Until then, I must point out that you are extremely obnoxious, extremely homicidal and right-wing/pro-slavery, and extremely wrong. -Philip White
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-27 02:54 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <871qvauazi.fsf@bsb.me.uk> |
| In reply to | #53038 |
"B.H." <xlt.pjw@gmail.com> writes: > On Sunday, June 26, 2022 at 9:15:22 PM UTC-4, Ben Bacarisse wrote: >> "B.H." <xlt...@gmail.com> writes: >> >> > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: >> >> "B.H." <xlt...@gmail.com> writes: >> >> >> >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: >> >> >> >> >> You were wrong about having resolved the open problem you posted about. >> >> >> It's still open. You don't appear to have changed you mind about that. >> >> > >> >> > I am not wrong at all, it is surprising that you claim to think that >> >> > that easy proof, which I will not publish, does not exist. >> >> There are counterexamples. There can be no proof that (in your words) >> >> "it is actually the case that for every positive integer m, the number >> >> of even Latin squares of order m and the number of odd Latin squares >> >> of order m are *the same*" >> >> because there is (to take just one counterexample) no odd 1x1 Latin >> >> square. When I asked about this you said >> >> "The odd Latin square of order 1 is simply the square with one cell >> >> containing -1" >> >> but that is an even Latin square. >> > >> > You finally stated your objection. >> >> There are two: that the Latin square you gave is even, and that there >> are counterexamples (included in the page you linked to) to your claimed >> resolution. >> > >> > "A latin square is even if the product of the signs of all of the row >> > and column permutations is 1 and is odd otherwise. " >> >> Yes, so your example is an even Latin square, not an odd one. Unless >> there is (up to isomorphism) exactly one odd Latin square the simple >> case of n=1 is a counterexample to your claimed resolution. The page >> you linked to included 7 other counterexamples -- basically every known >> case is a counterexample. >> >> > If you treat the same square as the selected row and column, then the >> > "base case" would have to be changed to a 2 x 2 Latin square, where it >> > would work. >> > >> > I am not sharing the rest of the details of the proof with you. >> >> There can't be a proof of your claim because there are counterexamples. >> They were even listed in the page you linked to. > > I don't know what table you have called master and bowed humbly to, > but you are incorrect. I am familiar with the mathematical proof, and > your insistently repeated protestations to the contrary are patently > incorrect. All identity permutations are even. Every 1x1 Latin square has 1 even row permutation (the identity permutation) and one even column permutation (the same). Every 1x1 Latin square is even. There are no odd 1x1 Latin squares. > You ignored my idea about a different permutation definition for the 1 > x 1 matrix--where -1 is an odd Latin square because there is only one > permutation in this case, and the product, "-1" is taken to be -1 > making it odd. You claimed to have resolved the conjecture you linked to, not one based on your own notion of what's odd and what's even. I can resolve the twin prime conjecture, provided you accept my alternative definition of "prime" and "twin". > Your debating style and efforts to kill me with false claims about > math anger me, as you can see. I am not trying to kill you. If you feel threatened I will, of course, stop replying. Do you feel safe enough to continue this exchange? -- Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-27 08:51 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <301476fd-94c1-426d-9624-83dedf6ace30n@googlegroups.com> |
| In reply to | #53039 |
On Sunday, June 26, 2022 at 9:54:44 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Sunday, June 26, 2022 at 9:15:22 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: > >> >> "B.H." <xlt...@gmail.com> writes: > >> >> > >> >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > >> >> > >> >> >> You were wrong about having resolved the open problem you posted about. > >> >> >> It's still open. You don't appear to have changed you mind about that. > >> >> > > >> >> > I am not wrong at all, it is surprising that you claim to think that > >> >> > that easy proof, which I will not publish, does not exist. > >> >> There are counterexamples. There can be no proof that (in your words) > >> >> "it is actually the case that for every positive integer m, the number > >> >> of even Latin squares of order m and the number of odd Latin squares > >> >> of order m are *the same*" > >> >> because there is (to take just one counterexample) no odd 1x1 Latin > >> >> square. When I asked about this you said > >> >> "The odd Latin square of order 1 is simply the square with one cell > >> >> containing -1" > >> >> but that is an even Latin square. > >> > > >> > You finally stated your objection. > >> > >> There are two: that the Latin square you gave is even, and that there > >> are counterexamples (included in the page you linked to) to your claimed > >> resolution. > >> > > >> > "A latin square is even if the product of the signs of all of the row > >> > and column permutations is 1 and is odd otherwise. " > >> > >> Yes, so your example is an even Latin square, not an odd one. Unless > >> there is (up to isomorphism) exactly one odd Latin square the simple > >> case of n=1 is a counterexample to your claimed resolution. The page > >> you linked to included 7 other counterexamples -- basically every known > >> case is a counterexample. > >> > >> > If you treat the same square as the selected row and column, then the > >> > "base case" would have to be changed to a 2 x 2 Latin square, where it > >> > would work. > >> > > >> > I am not sharing the rest of the details of the proof with you. > >> > >> There can't be a proof of your claim because there are counterexamples. > >> They were even listed in the page you linked to. > > > > I don't know what table you have called master and bowed humbly to, > > but you are incorrect. I am familiar with the mathematical proof, and > > your insistently repeated protestations to the contrary are patently > > incorrect. > All identity permutations are even. Every 1x1 Latin square has 1 even > row permutation (the identity permutation) and one even column > permutation (the same). Every 1x1 Latin square is even. There are no > odd 1x1 Latin squares. If you re-define permutations in a way that is consistent with non-1x1 Latin squares, the conjecture is provably true. If you don't re-define them, the conjecture I stated is true based on the 2 x 2 Latin square. You are emphasizing the wrong point so you can say something that is true. > > You ignored my idea about a different permutation definition for the 1 > > x 1 matrix--where -1 is an odd Latin square because there is only one > > permutation in this case, and the product, "-1" is taken to be -1 > > making it odd. > You claimed to have resolved the conjecture you linked to, not one based > on your own notion of what's odd and what's even. I can resolve the > twin prime conjecture, provided you accept my alternative definition of > "prime" and "twin". I did, it follows from my own claim. There are different interpretations of the imprecisely phrased conjecture. The conjecture is resolved, I proved it. > > Your debating style and efforts to kill me with false claims about > > math anger me, as you can see. > I am not trying to kill you. If you feel threatened I will, of course, > stop replying. Do you feel safe enough to continue this exchange? > I think you are, though not overtly. I think you should stop replying; I don't like your attitude and don't see whom you think to be believing themselves benefiting from your fake and pointless "service." Why waste everyone's time? If you're trying to be a role-model to bad academic journals that might want to lock me out for political reasons, of course such journals already know how to do that and don't need your eagerly provided "assistance" in figuring out how to bulls---. -Philip White (philipjwhite@yahoo.com) > -- > Ben.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-27 21:56 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <87k091rfjc.fsf@bsb.me.uk> |
| In reply to | #53059 |
"B.H." <xlt.pjw@gmail.com> writes: > On Sunday, June 26, 2022 at 9:54:44 PM UTC-4, Ben Bacarisse wrote: >> "B.H." <xlt...@gmail.com> writes: >> >> > On Sunday, June 26, 2022 at 9:15:22 PM UTC-4, Ben Bacarisse wrote: >> >> "B.H." <xlt...@gmail.com> writes: >> >> >> >> > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: >> >> >> "B.H." <xlt...@gmail.com> writes: >> >> >> >> >> >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: >> >> >> >> >> >> >> You were wrong about having resolved the open problem you posted about. >> >> >> >> It's still open. You don't appear to have changed you mind about that. >> >> >> > >> >> >> > I am not wrong at all, it is surprising that you claim to think that >> >> >> > that easy proof, which I will not publish, does not exist. >> >> >> There are counterexamples. There can be no proof that (in your words) >> >> >> "it is actually the case that for every positive integer m, the number >> >> >> of even Latin squares of order m and the number of odd Latin squares >> >> >> of order m are *the same*" >> >> >> because there is (to take just one counterexample) no odd 1x1 Latin >> >> >> square. When I asked about this you said >> >> >> "The odd Latin square of order 1 is simply the square with one cell >> >> >> containing -1" >> >> >> but that is an even Latin square. >> >> > >> >> > You finally stated your objection. >> >> >> >> There are two: that the Latin square you gave is even, and that there >> >> are counterexamples (included in the page you linked to) to your claimed >> >> resolution. >> >> > >> >> > "A latin square is even if the product of the signs of all of the row >> >> > and column permutations is 1 and is odd otherwise. " >> >> >> >> Yes, so your example is an even Latin square, not an odd one. Unless >> >> there is (up to isomorphism) exactly one odd Latin square the simple >> >> case of n=1 is a counterexample to your claimed resolution. The page >> >> you linked to included 7 other counterexamples -- basically every known >> >> case is a counterexample. >> >> >> >> > If you treat the same square as the selected row and column, then the >> >> > "base case" would have to be changed to a 2 x 2 Latin square, where it >> >> > would work. >> >> > >> >> > I am not sharing the rest of the details of the proof with you. >> >> >> >> There can't be a proof of your claim because there are counterexamples. >> >> They were even listed in the page you linked to. >> > >> > I don't know what table you have called master and bowed humbly to, >> > but you are incorrect. I am familiar with the mathematical proof, and >> > your insistently repeated protestations to the contrary are patently >> > incorrect. >> All identity permutations are even. Every 1x1 Latin square has 1 even >> row permutation (the identity permutation) and one even column >> permutation (the same). Every 1x1 Latin square is even. There are no >> odd 1x1 Latin squares. > > If you re-define permutations in a way that is consistent with non-1x1 > Latin squares, And, apparently, you have redefined the very notion of counting Latin squares since you believe there are at least two 1x1 Latin squares. So you redefined how you count the squares, and you redefined what even and odd squares are and you proved something about these new counts. That unseen proof can't be about the conjecture you linked to because that conjecture counts equivalence classes of squares so that the actual n symbols used don't matter. > the conjecture is provably true. Some other, as yet unstated, conjecture is provably true. You would need to state, at least, how you counting Latin squares since you count more 1c1 squares than anyone else. The conjecture you linked to can not be resolved in the way you claimed (all the counts being the same) because there are counterexamples. > If you don't re-define them, the conjecture I stated is true based on > the 2 x 2 Latin square. Quite possibly. I am only talking about the claim to have resolved the conjecture you linked to. > You are emphasizing the wrong point so you > can say something that is true. Unfortunately that is in the very nature of pointing out an error. I could go look for some true things you've said, but I don't even know how you are counting squares (you never said) so I would have to guess what you say you have proved. >> > You ignored my idea about a different permutation definition for the 1 >> > x 1 matrix--where -1 is an odd Latin square because there is only one >> > permutation in this case, and the product, "-1" is taken to be -1 >> > making it odd. >> You claimed to have resolved the conjecture you linked to, not one based >> on your own notion of what's odd and what's even. I can resolve the >> twin prime conjecture, provided you accept my alternative definition of >> "prime" and "twin". > > I did, it follows from my own claim. There are different > interpretations of the imprecisely phrased conjecture. The conjecture > is resolved, I proved it. You should have know that you were not addressing the open problem given in the link, since that page clearly shows that there is only one 1x1 Latin square. Since your notion of counting differs from that used to state the conjecture, how could you think you had resolved it? But, in fact, there is nothing imprecise about the conjecture. You didn't know what some of the terms meant (even and odd Latin squares, how to count distinct squares), but guessing does not mean the wording was imprecise. Not all conjectures are explained in the most basic terms. For some, you have to know how the "terms of art" are defined. >> > Your debating style and efforts to kill me with false claims about >> > math anger me, as you can see. >> >> I am not trying to kill you. If you feel threatened I will, of course, >> stop replying. Do you feel safe enough to continue this exchange? > > I think you are, though not overtly. That's a shocking thing to say. > I think you should stop replying; OK. I read this only after typing the above, so I'll send it but you won't hear anything more from me on this topic. Would you rather I don't reply to any of your technical posts? I am very happy to refrain. I have no desire to cause you any distress. -- Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-27 15:56 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <c036bbba-4114-431e-9073-f3a314f304een@googlegroups.com> |
| In reply to | #53092 |
On Monday, June 27, 2022 at 4:56:59 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Sunday, June 26, 2022 at 9:54:44 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Sunday, June 26, 2022 at 9:15:22 PM UTC-4, Ben Bacarisse wrote: > >> >> "B.H." <xlt...@gmail.com> writes: > >> >> > >> >> > On Sunday, June 26, 2022 at 5:26:29 PM UTC-4, Ben Bacarisse wrote: > >> >> >> "B.H." <xlt...@gmail.com> writes: > >> >> >> > >> >> >> > On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > >> >> >> > >> >> >> >> You were wrong about having resolved the open problem you posted about. > >> >> >> >> It's still open. You don't appear to have changed you mind about that. > >> >> >> > > >> >> >> > I am not wrong at all, it is surprising that you claim to think that > >> >> >> > that easy proof, which I will not publish, does not exist. > >> >> >> There are counterexamples. There can be no proof that (in your words) > >> >> >> "it is actually the case that for every positive integer m, the number > >> >> >> of even Latin squares of order m and the number of odd Latin squares > >> >> >> of order m are *the same*" > >> >> >> because there is (to take just one counterexample) no odd 1x1 Latin > >> >> >> square. When I asked about this you said > >> >> >> "The odd Latin square of order 1 is simply the square with one cell > >> >> >> containing -1" > >> >> >> but that is an even Latin square. > >> >> > > >> >> > You finally stated your objection. > >> >> > >> >> There are two: that the Latin square you gave is even, and that there > >> >> are counterexamples (included in the page you linked to) to your claimed > >> >> resolution. > >> >> > > >> >> > "A latin square is even if the product of the signs of all of the row > >> >> > and column permutations is 1 and is odd otherwise. " > >> >> > >> >> Yes, so your example is an even Latin square, not an odd one. Unless > >> >> there is (up to isomorphism) exactly one odd Latin square the simple > >> >> case of n=1 is a counterexample to your claimed resolution. The page > >> >> you linked to included 7 other counterexamples -- basically every known > >> >> case is a counterexample. > >> >> > >> >> > If you treat the same square as the selected row and column, then the > >> >> > "base case" would have to be changed to a 2 x 2 Latin square, where it > >> >> > would work. > >> >> > > >> >> > I am not sharing the rest of the details of the proof with you. > >> >> > >> >> There can't be a proof of your claim because there are counterexamples. > >> >> They were even listed in the page you linked to. > >> > > >> > I don't know what table you have called master and bowed humbly to, > >> > but you are incorrect. I am familiar with the mathematical proof, and > >> > your insistently repeated protestations to the contrary are patently > >> > incorrect. > >> All identity permutations are even. Every 1x1 Latin square has 1 even > >> row permutation (the identity permutation) and one even column > >> permutation (the same). Every 1x1 Latin square is even. There are no > >> odd 1x1 Latin squares. > > > > If you re-define permutations in a way that is consistent with non-1x1 > > Latin squares, > And, apparently, you have redefined the very notion of counting Latin > squares since you believe there are at least two 1x1 Latin squares. > > So you redefined how you count the squares, and you redefined what even > and odd squares are and you proved something about these new counts. > That unseen proof can't be about the conjecture you linked to because > that conjecture counts equivalence classes of squares so that the actual > n symbols used don't matter. > > the conjecture is provably true. > Some other, as yet unstated, conjecture is provably true. You would > need to state, at least, how you counting Latin squares since you count > more 1c1 squares than anyone else. > > The conjecture you linked to can not be resolved in the way you claimed > (all the counts being the same) because there are counterexamples. > > If you don't re-define them, the conjecture I stated is true based on > > the 2 x 2 Latin square. > Quite possibly. I am only talking about the claim to have resolved the > conjecture you linked to. > > You are emphasizing the wrong point so you > > can say something that is true. > Unfortunately that is in the very nature of pointing out an error. I > could go look for some true things you've said, but I don't even know > how you are counting squares (you never said) so I would have to guess > what you say you have proved. > >> > You ignored my idea about a different permutation definition for the 1 > >> > x 1 matrix--where -1 is an odd Latin square because there is only one > >> > permutation in this case, and the product, "-1" is taken to be -1 > >> > making it odd. > >> You claimed to have resolved the conjecture you linked to, not one based > >> on your own notion of what's odd and what's even. I can resolve the > >> twin prime conjecture, provided you accept my alternative definition of > >> "prime" and "twin". > > > > I did, it follows from my own claim. There are different > > interpretations of the imprecisely phrased conjecture. The conjecture > > is resolved, I proved it. > You should have know that you were not addressing the open problem given > in the link, since that page clearly shows that there is only one 1x1 > Latin square. Since your notion of counting differs from that used to > state the conjecture, how could you think you had resolved it? > > But, in fact, there is nothing imprecise about the conjecture. You > didn't know what some of the terms meant (even and odd Latin squares, > how to count distinct squares), but guessing does not mean the wording > was imprecise. Not all conjectures are explained in the most basic > terms. For some, you have to know how the "terms of art" are defined. > >> > Your debating style and efforts to kill me with false claims about > >> > math anger me, as you can see. > >> > >> I am not trying to kill you. If you feel threatened I will, of course, > >> stop replying. Do you feel safe enough to continue this exchange? > > > > I think you are, though not overtly. > That's a shocking thing to say. > > I think you should stop replying; > OK. I read this only after typing the above, so I'll send it but you > won't hear anything more from me on this topic. Would you rather I > don't reply to any of your technical posts? I am very happy to refrain. > I have no desire to cause you any distress. > I already advised you to stop replying, you are lying, there is no error, and all you do is fake attemptedly persuasive sounding rebuttals of my correct mathematical arguments to try to sway a few fools who might believe you...what a waste of your time. I think you should not reply to my posts, I already said that, I don't see why you replied again. I don't have time to wade through the numerous paragraphs of your newly created fake-rebuttal-of-math bologna; no matter how many absurd arguments I rebut, more absurd lies are generated from thin air. It damages you more than it does me; I might lose a handful of supporters who don't get you, you have bet whatever reputation as an honest/virtuous man you had left and lost. That's the end of this discussion for me too; I won the debate decisively. -Philip White > -- > Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-26 08:03 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <25511ed2-0dd0-4942-8922-462cdd7db6d6n@googlegroups.com> |
| In reply to | #53006 |
On Sunday, June 26, 2022 at 9:44:07 AM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Saturday, June 25, 2022 at 8:02:04 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Friday, June 24, 2022 at 4:38:27 PM UTC-4, Ben Bacarisse wrote: > >> > >> >> I don't want to go into detail about what you wrote unless you ask > >> >> because I don't want to smear you again and I think pretty much any > >> >> criticism of you work is a smear in your opinion. But do say if you > >> >> want to know what I really think. > >> > > >> > I don't invite you to smear me; I'm not sure any discussion of my work > >> > would be informed by an understanding of what a mathematical proof is. > >> I rather hoped for some clear advice: if I spot another error in > >> something you post, would you rather I keep quiet about it? And if > >> you'd rather I pointed it out, how can I do that without offending you? > > > > In spite of your record of bad behavior, yes, point it out, but don't > > be inaccurate and don't smear me, i.e., don't make claims about my > > mathematical talent that are *untrue*. > I correctly pointed out an error, and you called that a smear. You > repeated the mistake, claiming again that the problem you linked to was > not an open problem any more. I think it is safer for both of us if I > don't point out any further mistakes I see. > > Since you have a Ph.D., > I don't have a PhD. I said so when you (very politely) called me Dr > Bacarisse. While I have examined and supervised several PhDs over the > years, but I don't have one myself. > > you might want to make sure that you don't disgrace whoever educated > > you by falsely and intransigently accusing of making errors I haven't > > made. > You were wrong about having resolved the open problem you posted about. > It's still open. You don't appear to have changed you mind about that. > > -- > Ben. By the way, you can squirm and argue and maneuver all day, but your dream of trafficking the next Einstein is already dead. I have recovered. -Philip White
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-26 21:49 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <87letjtak3.fsf@bsb.me.uk> |
| In reply to | #53009 |
"B.H." <xlt.pjw@gmail.com> writes: > By the way, you can squirm and argue and maneuver all day, but your > dream of trafficking the next Einstein is already dead. I have > recovered. That's a nasty accusation. I know why you says such things, and Usenet is largely unread, so I consider it harmless here, but you could get into a lot of trouble saying things like in other contexts. -- Ben.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-24 21:17 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <87wnd5x1d8.fsf@bsb.me.uk> |
| In reply to | #52876 |
Paul N <gw7rib@aol.com> writes: > On Friday, June 24, 2022 at 9:00:10 AM UTC+1, malcolm.ar...@gmail.com wrote: >> On Friday, 24 June 2022 at 02:28:34 UTC+1, Ben Bacarisse wrote: >> > "B.H." <xlt...@gmail.com> writes: >> > >> > > On Thursday, June 23, 2022 at 6:52:30 PM UTC-4, Ben Bacarisse wrote: >> > >> "B.H." <xlt...@gmail.com> writes: >> > >> >> > >> > On Wednesday, June 22, 2022 at 2:25:32 PM UTC-4, Ben Bacarisse wrote: >> > >> >> "B.H." <xlt...@gmail.com> writes: >> > >> >> >> > >> >> > Hi everyone, >> > >> >> > >> > >> >> > I just found an easy proof that the answer to this problem: >> > >> >> > >> > >> >> > http://www.openproblemgarden.org/op/even_vs_odd_latin_squares >> > >> >> > >> > >> >> > ...is that the conjecture is false, assuming I understand it correctly. >> > >> >> > >> > >> >> > According to my understanding, it is actually the case that for every >> > >> >> > positive integer m, the number of even Latin squares of order m and >> > >> >> > the number of odd Latin squares of order m are *the same*. >> > >> >> There is one even latin square of order 1. What is the odd latin square >> > >> >> of order one? How could there possibly be such a thing? >> > >> >> >> > >> >> The number of even and odd latin squares is known for small n and they >> > >> >> are (so far) all different. >> > >> >> >> > >> > >> > >> > Since there are infinitely many integers, I decided to count Latin >> > >> > squares by sign. That is, each Latin square cell is either a "1" if >> > >> > its sign is negative, or a "0" if its sign is positive. >> > >> > >> > >> > The odd Latin square of order 1 is simply the square with one cell >> > >> > containing -1. >> > >> > >> > >> > I resolved the conjecture >> > >> So you were addressing some other conjecture of your own. >> > > >> > > It's close enough. >> > For what? >> > > The same proof logic resolves all versions of the conjecture, and the >> > > purpose is to understand the logic of the proof and what it can imply, >> > > not just to understand one single fact about Latin squares. >> > The version in the link you posted remains an open problem. >> > >> If the terms >> > >> used in the page you linked to have their usual meanings, it's clear >> > >> that the number of even Latin squares of order m and the number of odd >> > >> Latin squares of order m can not possibly be the same. >> > > >> > > What are the usual meanings? The terms are vague. Whether or not you >> > > have a math degree, you might want to clarify yourself. >> > You announced a solution without knowing what even and odd mean in the >> > context of Latin squares? >> > >> I had a look at the link and couldn't work out what the problem was. >> Something to do with the product of the signs, but it doesn't actually >> tell you which square is multiplied with which (It can't be simply >> "multiply all squares" as the wording implies, because that's just a >> parity problem.) >> It's understandable that someone would have misunderstood the problem >> if relying on that link. > > I too would be interested to know what the problem is about. I don't > think I have ever heard of the "sign" of a permutation. It's common, when dealing with parity, to make even/odd to 1/-1. That way parities can be combined with multiplication, just as permutations can be factored and multiplied. > If it simply > means whether the permutation can be made up of an odd or even number > of pair switches then the results don't seem consistent with those > given on the page. See my answer to MM on what the problem is about. Do you still see an inconsistency? If so, maybe someone here can help of you explain what you think is inconsistent. -- Ben.
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| From | Paul N <gw7rib@aol.com> |
|---|---|
| Date | 2022-06-25 04:44 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <07dc5d3d-98c0-431c-a2a0-21cec0e282een@googlegroups.com> |
| In reply to | #52903 |
On Friday, June 24, 2022 at 9:17:25 PM UTC+1, Ben Bacarisse wrote: > Paul N <gw7...@aol.com> writes: > > > On Friday, June 24, 2022 at 9:00:10 AM UTC+1, malcolm.ar...@gmail.com wrote: > >> On Friday, 24 June 2022 at 02:28:34 UTC+1, Ben Bacarisse wrote: > >> > "B.H." <xlt...@gmail.com> writes: > >> > > >> > > On Thursday, June 23, 2022 at 6:52:30 PM UTC-4, Ben Bacarisse wrote: > >> > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > >> > >> > On Wednesday, June 22, 2022 at 2:25:32 PM UTC-4, Ben Bacarisse wrote: > >> > >> >> "B.H." <xlt...@gmail.com> writes: > >> > >> >> > >> > >> >> > Hi everyone, > >> > >> >> > > >> > >> >> > I just found an easy proof that the answer to this problem: > >> > >> >> > > >> > >> >> > http://www.openproblemgarden.org/op/even_vs_odd_latin_squares > >> > >> >> > > >> > >> >> > ...is that the conjecture is false, assuming I understand it correctly. > >> > >> >> > > >> > >> >> > According to my understanding, it is actually the case that for every > >> > >> >> > positive integer m, the number of even Latin squares of order m and > >> > >> >> > the number of odd Latin squares of order m are *the same*. > >> > >> >> There is one even latin square of order 1. What is the odd latin square > >> > >> >> of order one? How could there possibly be such a thing? > >> > >> >> > >> > >> >> The number of even and odd latin squares is known for small n and they > >> > >> >> are (so far) all different. > >> > >> >> > >> > >> > > >> > >> > Since there are infinitely many integers, I decided to count Latin > >> > >> > squares by sign. That is, each Latin square cell is either a "1" if > >> > >> > its sign is negative, or a "0" if its sign is positive. > >> > >> > > >> > >> > The odd Latin square of order 1 is simply the square with one cell > >> > >> > containing -1. > >> > >> > > >> > >> > I resolved the conjecture > >> > >> So you were addressing some other conjecture of your own. > >> > > > >> > > It's close enough. > >> > For what? > >> > > The same proof logic resolves all versions of the conjecture, and the > >> > > purpose is to understand the logic of the proof and what it can imply, > >> > > not just to understand one single fact about Latin squares. > >> > The version in the link you posted remains an open problem. > >> > >> If the terms > >> > >> used in the page you linked to have their usual meanings, it's clear > >> > >> that the number of even Latin squares of order m and the number of odd > >> > >> Latin squares of order m can not possibly be the same. > >> > > > >> > > What are the usual meanings? The terms are vague. Whether or not you > >> > > have a math degree, you might want to clarify yourself. > >> > You announced a solution without knowing what even and odd mean in the > >> > context of Latin squares? > >> > > >> I had a look at the link and couldn't work out what the problem was. > >> Something to do with the product of the signs, but it doesn't actually > >> tell you which square is multiplied with which (It can't be simply > >> "multiply all squares" as the wording implies, because that's just a > >> parity problem.) > >> It's understandable that someone would have misunderstood the problem > >> if relying on that link. > > > > I too would be interested to know what the problem is about. I don't > > think I have ever heard of the "sign" of a permutation. > It's common, when dealing with parity, to make even/odd to 1/-1. That > way parities can be combined with multiplication, just as permutations > can be factored and multiplied. > > If it simply > > means whether the permutation can be made up of an odd or even number > > of pair switches then the results don't seem consistent with those > > given on the page. > See my answer to MM on what the problem is about. Do you still see an > inconsistency? If so, maybe someone here can help of you explain what > you think is inconsistent. Hi Ben Thanks for your help. I was mis-reading it, hence my mistake. A 2x2 Latin square has one row of (1 2) which is even and one of (2 1) which is odd so I thought the square as a whole was odd. In fact you multiply all the rows AND all the columns so it is actually even. Paul.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-24 21:14 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <8735ftyg2l.fsf@bsb.me.uk> |
| In reply to | #52875 |
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
> On Friday, 24 June 2022 at 02:28:34 UTC+1, Ben Bacarisse wrote:
>> "B.H." <xlt...@gmail.com> writes:
>>
>> > On Thursday, June 23, 2022 at 6:52:30 PM UTC-4, Ben Bacarisse wrote:
>> >> "B.H." <xlt...@gmail.com> writes:
>> >>
>> >> > On Wednesday, June 22, 2022 at 2:25:32 PM UTC-4, Ben Bacarisse wrote:
>> >> >> "B.H." <xlt...@gmail.com> writes:
>> >> >>
>> >> >> > Hi everyone,
>> >> >> >
>> >> >> > I just found an easy proof that the answer to this problem:
>> >> >> >
>> >> >> > http://www.openproblemgarden.org/op/even_vs_odd_latin_squares
>> >> >> >
>> >> >> > ...is that the conjecture is false, assuming I understand it correctly.
>> >> >> >
>> >> >> > According to my understanding, it is actually the case that for every
>> >> >> > positive integer m, the number of even Latin squares of order m and
>> >> >> > the number of odd Latin squares of order m are *the same*.
>> >> >> There is one even latin square of order 1. What is the odd latin square
>> >> >> of order one? How could there possibly be such a thing?
>> >> >>
>> >> >> The number of even and odd latin squares is known for small n and they
>> >> >> are (so far) all different.
>> >> >>
>> >> >
>> >> > Since there are infinitely many integers, I decided to count Latin
>> >> > squares by sign. That is, each Latin square cell is either a "1" if
>> >> > its sign is negative, or a "0" if its sign is positive.
>> >> >
>> >> > The odd Latin square of order 1 is simply the square with one cell
>> >> > containing -1.
>> >> >
>> >> > I resolved the conjecture
>> >> So you were addressing some other conjecture of your own.
>> >
>> > It's close enough.
>> For what?
>> > The same proof logic resolves all versions of the conjecture, and the
>> > purpose is to understand the logic of the proof and what it can imply,
>> > not just to understand one single fact about Latin squares.
>> The version in the link you posted remains an open problem.
>> >> If the terms
>> >> used in the page you linked to have their usual meanings, it's clear
>> >> that the number of even Latin squares of order m and the number of odd
>> >> Latin squares of order m can not possibly be the same.
>> >
>> > What are the usual meanings? The terms are vague. Whether or not you
>> > have a math degree, you might want to clarify yourself.
>> You announced a solution without knowing what even and odd mean in the
>> context of Latin squares?
>>
> I had a look at the link and couldn't work out what the problem was.
> Something to do with the product of the signs, but it doesn't actually
> tell you which square is multiplied with which (It can't be simply
> "multiply all squares" as the wording implies, because that's just a
> parity problem.)
> It's understandable that someone would have misunderstood the problem
> if relying on that link.
Sure. I would hope, though, that someone tackling a hard open problem
would make sure they know what the terms mean ahead of time. A lot of
effort could be wasted otherwise.
For reference... Permutations are even or odd depending whether they
can be factored into an even or an odd number of swaps.
Now let s(p) = 1 if p is an even permutation and -1 is p is an odd
permutation. The parity of a Latin square, L, is given by
S(L) = (Product{k=1,n} s(r_k)) . (Product{k=1,n} s(c_k))
where r_k and c_k are the permutations present in the kth row and kth
column of L. Like s(p), S(L) is either 1 or -1 indicating even and odd
parity respectively.
--
Ben.
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| From | "B.H." <xlt.pjw@gmail.com> |
|---|---|
| Date | 2022-06-24 08:35 -0700 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <12d37df2-f9b9-48cb-931d-abb32828ed8cn@googlegroups.com> |
| In reply to | #52864 |
On Thursday, June 23, 2022 at 9:28:34 PM UTC-4, Ben Bacarisse wrote: > "B.H." <xlt...@gmail.com> writes: > > > On Thursday, June 23, 2022 at 6:52:30 PM UTC-4, Ben Bacarisse wrote: > >> "B.H." <xlt...@gmail.com> writes: > >> > >> > On Wednesday, June 22, 2022 at 2:25:32 PM UTC-4, Ben Bacarisse wrote: > >> >> "B.H." <xlt...@gmail.com> writes: > >> >> > >> >> > Hi everyone, > >> >> > > >> >> > I just found an easy proof that the answer to this problem: > >> >> > > >> >> > http://www.openproblemgarden.org/op/even_vs_odd_latin_squares > >> >> > > >> >> > ...is that the conjecture is false, assuming I understand it correctly. > >> >> > > >> >> > According to my understanding, it is actually the case that for every > >> >> > positive integer m, the number of even Latin squares of order m and > >> >> > the number of odd Latin squares of order m are *the same*. > >> >> There is one even latin square of order 1. What is the odd latin square > >> >> of order one? How could there possibly be such a thing? > >> >> > >> >> The number of even and odd latin squares is known for small n and they > >> >> are (so far) all different. > >> >> > >> > > >> > Since there are infinitely many integers, I decided to count Latin > >> > squares by sign. That is, each Latin square cell is either a "1" if > >> > its sign is negative, or a "0" if its sign is positive. > >> > > >> > The odd Latin square of order 1 is simply the square with one cell > >> > containing -1. > >> > > >> > I resolved the conjecture > >> So you were addressing some other conjecture of your own. > > > > It's close enough. > For what? It is close enough to provide a clear solution to whatever conjecture you likely think it is. > > The same proof logic resolves all versions of the conjecture, and the > > purpose is to understand the logic of the proof and what it can imply, > > not just to understand one single fact about Latin squares. > The version in the link you posted remains an open problem. I disagree. The proof is useless, I might publish it if you would agree to acknowledge that it is correct and not just auto-smear me. > >> If the terms > >> used in the page you linked to have their usual meanings, it's clear > >> that the number of even Latin squares of order m and the number of odd > >> Latin squares of order m can not possibly be the same. > > > > What are the usual meanings? The terms are vague. Whether or not you > > have a math degree, you might want to clarify yourself. > You announced a solution without knowing what even and odd mean in the > context of Latin squares? I don't have time to read academic papers, I'm not being paid. I read the definitions on OPG and Wikipedia. Do you know what they are? I stated my understanding, why do you think it's I who have some serious explaining to do, Dr. Bacarisse? > > -- > Ben. At least you're getting some more attention paid to my threads. -Philip White (philipjwhite@yahoo.com)
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-06-24 21:30 +0100 |
| Subject | Re: Resolved: Even vs. Odd Latin Square Problem |
| Message-ID | <87r13dx0s4.fsf@bsb.me.uk> |
| In reply to | #52883 |
"B.H." <xlt.pjw@gmail.com> writes: > On Thursday, June 23, 2022 at 9:28:34 PM UTC-4, Ben Bacarisse wrote: >> "B.H." <xlt...@gmail.com> writes: >> >> > On Thursday, June 23, 2022 at 6:52:30 PM UTC-4, Ben Bacarisse wrote: >> >> "B.H." <xlt...@gmail.com> writes: >> >> >> >> > On Wednesday, June 22, 2022 at 2:25:32 PM UTC-4, Ben Bacarisse wrote: >> >> >> "B.H." <xlt...@gmail.com> writes: >> >> >> >> >> >> > Hi everyone, >> >> >> > >> >> >> > I just found an easy proof that the answer to this problem: >> >> >> > >> >> >> > http://www.openproblemgarden.org/op/even_vs_odd_latin_squares >> >> >> > >> >> >> > ...is that the conjecture is false, assuming I understand it correctly. >> >> >> > >> >> >> > According to my understanding, it is actually the case that for every >> >> >> > positive integer m, the number of even Latin squares of order m and >> >> >> > the number of odd Latin squares of order m are *the same*. >> >> >> There is one even latin square of order 1. What is the odd latin square >> >> >> of order one? How could there possibly be such a thing? >> >> >> >> >> >> The number of even and odd latin squares is known for small n and they >> >> >> are (so far) all different. >> >> >> >> >> > >> >> > Since there are infinitely many integers, I decided to count Latin >> >> > squares by sign. That is, each Latin square cell is either a "1" if >> >> > its sign is negative, or a "0" if its sign is positive. >> >> > >> >> > The odd Latin square of order 1 is simply the square with one cell >> >> > containing -1. >> >> > >> >> > I resolved the conjecture >> >> So you were addressing some other conjecture of your own. >> > >> > It's close enough. >> For what? > > It is close enough to provide a clear solution to whatever conjecture you likely think it is. > >> > The same proof logic resolves all versions of the conjecture, and the >> > purpose is to understand the logic of the proof and what it can imply, >> > not just to understand one single fact about Latin squares. >> The version in the link you posted remains an open problem. > > I disagree. So far there is no proof one way or the other. That makes it an open problem. What you posted about before was about another problem (which may or not have been open at the time) based on what you interpreted even/odd as meaning for Latin squares. It had to be about another problem, because the number of even and odd Latin squares is know for small n. > The proof is useless, I might publish it if you would agree to > acknowledge that it is correct and not just auto-smear me. I did not intend to smear you, automatically or otherwise. And I have no wish to add to your worries. I will happily not read what you write if that would be preferable for you. Please let me know. >> You announced a solution without knowing what even and odd mean in the >> context of Latin squares? > > I don't have time to read academic papers, I'm not being paid. I read > the definitions on OPG and Wikipedia. > > Do you know what they are? I stated my understanding, why do you > think it's I who have some serious explaining to do, Dr. Bacarisse? I have posted a reply to Malcolm with what I think is the correct meaning for odd and even in the context of Latin squares. (I do not have a PhD -- I am plain Mr -- but I prefer the more chatty Ben). -- Ben.
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