Re: [math-fun] Something linguistic
BAIT=BATE and since E=1 already shown, we see I=1, and then RODE=ROAD so A=1 RUDE=RUED proves some kind of commutativity, LOWED=LOAD so W=1, HOUR=OUR so H=1, FLOWER=FLOUR so U=1, SKULL=SCULL and QI=KI and SIC=SICK so K=C=Q=1, GNU=NEW and SIGN=SINE so G=1, PSI=SIGH so P=1, DAMN=DAM so N=1, HYMN=HIM so Y=1, CAN'T=CANT so aprostrophe=1, WRAP=RAP so R=1 , PLUMB=PLUM and CLIME=CLIMB so B=1, PASS=PAS so S=1, HICCOUGH=HICCUP so P=1 (given previous results), WOOD=WOULD so L=1 in view of previous results, WHAT=WATT hence T=1, MUMM=MUM so M=1, GIN=JINN, so G=J so J=1, PASSED=PAST or GUESSED=GUEST (?) would prove D=1 (since T=1) if accepted... oh, just AD=ADD shows that, SOX=SOCKS so X=1 in view of previous results which also is shown by FAUX=FOE, so if I am counting right, this takes care of every letter, plus apostrophe, except that the following letters remain open: V,Z
ANALYSE=ANALYZE so Z=1 since S=1 previously shown. This would make V the only letter not yet shown trivial, albeit if you accepted old lettering where V=U in words, that would suffice... -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
So all we're looking for is two homophones with a different number of Vs? Charles Greathouse Analyst/Programmer Case Western Reserve University On Fri, Sep 25, 2015 at 5:56 PM, Warren D Smith <warren.wds@gmail.com> wrote:
ANALYSE=ANALYZE so Z=1 since S=1 previously shown.
This would make V the only letter not yet shown trivial, albeit if you accepted old lettering where V=U in words, that would suffice...
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Hmm. There's BEVY and BEVVY. —Dan
On Sep 25, 2015, at 3:04 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
So all we're looking for is two homophones with a different number of Vs?
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Fri, Sep 25, 2015 at 5:56 PM, Warren D Smith <warren.wds@gmail.com> wrote:
ANALYSE=ANALYZE so Z=1 since S=1 previously shown.
This would make V the only letter not yet shown trivial, albeit if you accepted old lettering where V=U in words, that would suffice...
If you accept SHIVVER (someone who stabs someone else) then it equals SHIVER. On Fri, Sep 25, 2015 at 6:49 PM, Dan Asimov <asimov@msri.org> wrote:
Hmm. There's BEVY and BEVVY.
—Dan
On Sep 25, 2015, at 3:04 PM, Charles Greathouse < charles.greathouse@case.edu> wrote:
So all we're looking for is two homophones with a different number of Vs?
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Fri, Sep 25, 2015 at 5:56 PM, Warren D Smith <warren.wds@gmail.com> wrote:
ANALYSE=ANALYZE so Z=1 since S=1 previously shown.
This would make V the only letter not yet shown trivial, albeit if you accepted old lettering where V=U in words, that would suffice...
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On Fri, Sep 25, 2015 at 5:51 PM, Warren D Smith <warren.wds@gmail.com> wrote:
WRAP=RAP so R=1 ,
This shows W = 1, but doesn't help us with R.
PASS=PAS so S=1,
PAS isn't a homonym of PASS. Fortunately, it's a homonym of PA, so this still gives is S = 1.
WHAT=WATT hence T=1,
I think WHAT rhymes with NUT, while WATT rhymes with NOT, so these are not homonyms, at least in my dialect. But BUT = BUTT gives T=1. I don't see an F in your list, but RUFF = ROUGH at least gives us F^2 = 1. So I think we still need F, R, and V Andy.Latto@pobox.com
I would love to see some results on how many random relations (of some distribution of length) you have to throw at an n-generator free group before it becomes overwhelmingly likely that the group is (a) finite, (b) trivial. On Fri, Sep 25, 2015 at 6:09 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Sep 25, 2015 at 5:51 PM, Warren D Smith <warren.wds@gmail.com> wrote:
WRAP=RAP so R=1 ,
This shows W = 1, but doesn't help us with R.
PASS=PAS so S=1,
PAS isn't a homonym of PASS. Fortunately, it's a homonym of PA, so this still gives is S = 1.
WHAT=WATT hence T=1,
I think WHAT rhymes with NUT, while WATT rhymes with NOT, so these are not homonyms, at least in my dialect.
But BUT = BUTT gives T=1.
I don't see an F in your list, but RUFF = ROUGH at least gives us F^2 = 1.
So I think we still need F, R, and V
Andy.Latto@pobox.com
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That depends on what the definition of is overwhelmingly is. —Dan
On Sep 25, 2015, at 3:48 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I would love to see some results on how many random relations (of some distribution of length) you have to throw at an n-generator free group before it becomes overwhelmingly likely that the group is (a) finite, (b) trivial.
On Fri, Sep 25, 2015 at 6:09 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Sep 25, 2015 at 5:51 PM, Warren D Smith <warren.wds@gmail.com> wrote:
WRAP=RAP so R=1 ,
This shows W = 1, but doesn't help us with R.
PASS=PAS so S=1,
PAS isn't a homonym of PASS. Fortunately, it's a homonym of PA, so this still gives is S = 1.
WHAT=WATT hence T=1,
I think WHAT rhymes with NUT, while WATT rhymes with NOT, so these are not homonyms, at least in my dialect.
But BUT = BUTT gives T=1.
I don't see an F in your list, but RUFF = ROUGH at least gives us F^2 = 1.
So I think we still need F, R, and V
Andy.Latto@pobox.com
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I was leaving the definition of "overwhelmingly" up to anyone who knew any results. My intuition is that there is a "phase change" phenomenon, where below a certain critical threshhold almost no groups collapse, while above it almost all do; if that is true, then the position of the halfway point will answer my question. But anything about this general question will interest me. On Fri, Sep 25, 2015 at 6:52 PM, Dan Asimov <asimov@msri.org> wrote:
That depends on what the definition of is overwhelmingly is.
—Dan
So I think we still need F, R, and V
Andy.Latto@pobox.com
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On Sep 25, 2015, at 3:48 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I would love to see some results on how many random relations (of some distribution of length) you have to throw at an n-generator free group before it becomes overwhelmingly likely that the group is (a) finite, (b) trivial.
On Fri, Sep 25, 2015 at 6:09 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Sep 25, 2015 at 5:51 PM, Warren D Smith <warren.wds@gmail.com> wrote:
WRAP=RAP so R=1 ,
This shows W = 1, but doesn't help us with R.
PASS=PAS so S=1,
PAS isn't a homonym of PASS. Fortunately, it's a homonym of PA, so this still gives is S = 1.
WHAT=WATT hence T=1,
I think WHAT rhymes with NUT, while WATT rhymes with NOT, so these are not homonyms, at least in my dialect.
But BUT = BUTT gives T=1.
I don't see an F in your list, but RUFF = ROUGH at least gives us F^2 =
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Okay, that's a plausible hypothesis. (Btw, I sent a strikethrough across the first "is" — which I guess is not ASCII.) Now to answer that we should come up with at least one concrete definition of "a random relation". So we need a positive probability distribution on Z^26. An easy one is this: For any subset X of Z^26, let Prob(X) := Sum_{x in X} rho(cube(x)) , where cube(x) is the unit cube centered at x and aligned with the axes, and rho is the probability (defined on Borel subsets of R^26) given by the standard normal distribution on R^26. (Or, choose your scale factor to taste.) —Dan
On Sep 25, 2015, at 3:55 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I was leaving the definition of "overwhelmingly" up to anyone who knew any results. My intuition is that there is a "phase change" phenomenon, where below a certain critical threshhold almost no groups collapse, while above it almost all do; if that is true, then the position of the halfway point will answer my question. But anything about this general question will interest me.
On Fri, Sep 25, 2015 at 6:52 PM, Dan Asimov <asimov@msri.org> wrote:
That depends on what the definition of is overwhelmingly is.
—Dan
So I think we still need F, R, and V
Andy.Latto@pobox.com
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On Sep 25, 2015, at 3:48 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I would love to see some results on how many random relations (of some distribution of length) you have to throw at an n-generator free group before it becomes overwhelmingly likely that the group is (a) finite, (b) trivial.
On Fri, Sep 25, 2015 at 6:09 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Sep 25, 2015 at 5:51 PM, Warren D Smith <warren.wds@gmail.com> wrote:
WRAP=RAP so R=1 ,
This shows W = 1, but doesn't help us with R.
PASS=PAS so S=1,
PAS isn't a homonym of PASS. Fortunately, it's a homonym of PA, so this still gives is S = 1.
WHAT=WATT hence T=1,
I think WHAT rhymes with NUT, while WATT rhymes with NOT, so these are not homonyms, at least in my dialect.
But BUT = BUTT gives T=1.
I don't see an F in your list, but RUFF = ROUGH at least gives us F^2 =
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participants (6)
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Allan Wechsler -
Andy Latto -
Charles Greathouse -
Dan Asimov -
Jon Ziegler -
Warren D Smith