Physicist puzzle...
Michael Vang
http://www.highiqsociety.org/common/iqtests/hdtest_printable/06.gif
We've run into some problems with the way the puzzle is set up, like
in how some names are spelled, but what is really killing us is we
cannot get even close to solving it... It is supposed to be one of
those "gotcha" puzzles instead of a brute force one, but so far I am
totally lost...
I don't want an answer, just a small clue... :)
Here is a thread where we have discussed what we have done so far:
http://arstechnica.infopop.net/OpenTopic/page?q=Y&a=tpc&s=50009562&f=6330927813&m=6740938424&p=1
Thanks!
Mike (Xyzzy)
Dogstar
"Michael Vang" <xy...@teamprimerib.com> wrote in message
news:8b0e3a44.02041...@posting.google.com...
I got a feeling of "deja-vu" when I looked at this problem. My feeling is
that there is a "trick" answer, or that the author made a mistake.
I don't have an answer, but I did notice that the names are ordered by the
year they won the Nobel Prize (Glaser, mentioned in the instructions, would
have come just before Feynman). It's possible that the order has nothing to
do with the problem's solution, and that the author just used a list of
Nobel Prize winners to supply the names.
Have you considered that multiple occurrences of the same letter in a name
may only be counted once? Perhaps if this is done, a pattern will emerge.
If the method for determining the value wasn't already given, I would
consider looking for a different way to map values to names (e.g., using
atomic numbers).
Besides Plank/Planck, there is also Block/Bloch and Cherenkov/Cerenkov.
-Dogstar
Patrick Hamlyn
>I got a feeling of "deja-vu" when I looked at this problem. My feeling is
>that there is a "trick" answer, or that the author made a mistake.
>
>I don't have an answer, but I did notice that the names are ordered by the
>year they won the Nobel Prize (Glaser, mentioned in the instructions, would
>have come just before Feynman). It's possible that the order has nothing to
>do with the problem's solution, and that the author just used a list of
>Nobel Prize winners to supply the names.
>
>Have you considered that multiple occurrences of the same letter in a name
>may only be counted once? Perhaps if this is done, a pattern will emerge.
>
>If the method for determining the value wasn't already given, I would
>consider looking for a different way to map values to names (e.g., using
>atomic numbers).
>
>Besides Plank/Planck, there is also Block/Bloch and Cherenkov/Cerenkov.
What are you talking about?? It's just an over-determined system of 25 equations
and 18 unknowns. Solve it by hand in 20 minutes or let Mathematica do it in 2
seconds.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
Frazz
"Patrick Hamlyn" <path@multipro.N_OcomSP_AM.au > schreef in bericht
news:6pnabucfu7np2hvia...@4ax.com...
No, it's not.
F.
Patrick Hamlyn
>"Dogstar" <dog...@microprizes.com> wrote:
>
>
>>I got a feeling of "deja-vu" when I looked at this problem. My feeling is
>>that there is a "trick" answer, or that the author made a mistake.
>>
>>I don't have an answer, but I did notice that the names are ordered by the
>>year they won the Nobel Prize (Glaser, mentioned in the instructions, would
>>have come just before Feynman). It's possible that the order has nothing to
>>do with the problem's solution, and that the author just used a list of
>>Nobel Prize winners to supply the names.
>>
>>Have you considered that multiple occurrences of the same letter in a name
>>may only be counted once? Perhaps if this is done, a pattern will emerge.
>>
>>If the method for determining the value wasn't already given, I would
>>consider looking for a different way to map values to names (e.g., using
>>atomic numbers).
>>
>>Besides Plank/Planck, there is also Block/Bloch and Cherenkov/Cerenkov.
>
>What are you talking about?? It's just an over-determined system of 25 equations
>and 18 unknowns. Solve it by hand in 20 minutes or let Mathematica do it in 2
>seconds.
Actually it was 22 unknowns. I typed it into Mathematica, which takes about a
microsecond to inform us there is no solution:
Solve[{r + o + e + n + t + g + e + n == 89, l + o + r + e + n + t + z == 91,
c + u + r + i + e == 67, m + i + c + h + e + l + s + o + n == 95,
l + i + p + p + m + a + n == 123, m + a + r + c + o + n + i == 82,
k + a + m + e + r + l + i + n + g + h == 127, p + l + a + n + k == 77,
s + t + a + r + k == 66, e + i + n + s + t + e + i + n == 109,
b + o + h + r == 43, m + i + l + l + i + k + a + n == 135,
s + i + e + g + b + a + h + n == 114, p + e + r + r + i + n == 78,
r + i + c + h + a + r + d + s + o + n == 115,
h + e + i + s + e + n + b + e + r + g == 122,
s + c + h + r + o + d + i + n + g + e + r == 122,
c + h + a + d + w + i + c + k == 90, a + n + d + e + r + s + o + n == 86,
d + a + v + i + s + s + o + n == 121, f + e + r + m + i == 62,
s + t + e + r + n == 56, b + l + o + c + k == 65,
z + e + r + n + i + k + e == 77,
c + h + e + r + e + n + k + o + v == 81}, {a, b, c, d, e, f, g, h, i, k,
l, m, n, o, p, r, s, t, u, v, w, z}]
Solutions:{}
So either I typed it in wrong or the puzzle setter did. BTW if you remove the
last three equations so it's not overspecified then you get solutions, but not
integral ones. I also tried removing each equation in turn to see if there was
an error in just one of them, but still no joy, so we're looking for multiple
errors, which is about no fun at all.
Al
<SNIP>
> What are you talking about?? It's just an over-determined system of 25
equations
> and 18 unknowns. Solve it by hand in 20 minutes or let Mathematica do it
in 2
> seconds.
> --
There's no more names with Y apart from Feynman, how do I crack that? Or did
I miss something?
-Al
Dirk Haar
Maybe it could not be solved that way (e.g. with Mathematica).
Some example how to code the counting:
o scrabble values
o telephone numbers
o vocals count 3, consonants 5, doubles double or do not count
o alphabet rank summed up
o ..
o ..
.
(I) think about it.
.
DIrk
Dirk Haar
Does it work with corrected names then ?
(e.g. Planck)
Dirk
Gareth Owen
> Does it work with corrected names then ?
> (e.g. Planck)
With planck, bloch and cerenkov: No
With planck, bloch and cherenkov: No
With planck, block and cherenkov: No
With plank, block and cerenkov: No
With plank, block and cherenkov: No
With plank, bloch and cherenkov: No
Any more for any more?
--
Gareth Owen
There ain't no sanity clause
Nick Wedd
<use...@gwowen.freeserve.co.uk> writes
>Dirk Haar <dh...@mideal.de> writes:
>
>> Does it work with corrected names then ?
>> (e.g. Planck)
>
>With planck, bloch and cerenkov: No
>With planck, bloch and cherenkov: No
>With planck, block and cherenkov: No
>With plank, block and cerenkov: No
>With plank, block and cherenkov: No
>With plank, bloch and cherenkov: No
>
>Any more for any more?
If you are going to use correct spellings, then some of the o's want
umlauts on them.
Nick
--
Nick Wedd ni...@maproom.co.uk
Gareth Owen
> >Any more for any more?
>
> If you are going to use correct spellings, then some of the o's want
> umlauts on them.
but then most physicists would treat them as twice differentiated w.r.t time,
which changes the problem considerably. e.g.
ü + u = 0 <=> u = A cos(t) + B sin(t)
--
"Will you kill him in his bed? Stick a dagger in his head?
I would not, could not kill the king. I would not do this evil thing.
I will not wed this girl, you see. So get her to a nunnery."
-- Green Eggs and Hamlet
Christian Geis
Thank you for this message. I am a Methematica-Newbie and thought I
couldn't even solve such equations since a get {} too. I tried it only
to solve to all 26 letters, thus {a,b,c,d,...,y,z} as the second
variable for Solve[]. Then it should even be underdetermined, but I
don't get a solution neither.
Bye
Mark Brader
> If you are going to use correct spellings, then some of the o's want
> umlauts on them.
Which when using the English alphabet can also be written as E's following
the letter.
--
Mark Brader, Toronto | "This is an excellent opportunity for
m...@vex.net | out-of-context quoting..." --Mike Hardy
Karen Lingel, Penguinist
Dirk Haar wrote in message <3CB59F6F...@mideal.de>...
>Maybe it could not be solved that way (e.g. with Mathematica).
>Some example how to code the counting:
>
>o scrabble values
>o telephone numbers
>o vocals count 3, consonants 5, doubles double or do not count
>o alphabet rank summed up
>o ..
How about something with atom abbreviations, you know Bohr would be Boron
plus Oxygen plus Hydrogen plus, well, R. Or maybe Hr? Or maybe R = 0?
Eh, I'm gettin' nowhere...
-k-
jere7my tho?rpe
<4Itt8.25083$QC1.1...@bgtnsc04-news.ops.worldnet.att.net>,
"Karen Lingel, Penguinist" <karen....@worldnet.att.net.nospam>
wrote:
> How about something with atom abbreviations, you know Bohr would be Boron
> plus Oxygen plus Hydrogen plus, well, R. Or maybe Hr? Or maybe R = 0?
Would anybody know what I was talking about if I mentioned the name
Coroner Sweborg? :)=
----j7y
--
*************************************************************************
jere7my tho?rpe / 734-769-0913 "Homo sum: humani nihil a me
http://homepage.mac.com/jere7my alienum puto." ---Terentius
Mike Williams
But any of those methods, if they are the same for each name, would be
found by solving them as simultaneous equations in Mathematica.
--
Mike Williams
Gentleman of Leisure
Patrick Hamlyn
>Thank you for this message. I am a Methematica-Newbie and thought I
>couldn't even solve such equations since a get {} too. I tried it only
>to solve to all 26 letters, thus {a,b,c,d,...,y,z} as the second
>variable for Solve[]. Then it should even be underdetermined, but I
>don't get a solution neither.
Adding variables after the fact won't 'under-determine' the system. Those
variables don't occur in the equations, so it's overdetermined, period.
Think of it this way: No one set of values assigned to the variables can serve
to make all equations true. Adding other variables can only affect this
situation if you add them to the original equations too.
Angus Walker
Christian Geis
> geis_ch...@hotmail.com (Christian Geis) wrote:
>
>
> >Thank you for this message. I am a Methematica-Newbie and thought I
> >couldn't even solve such equations since a get {} too. I tried it only
> >to solve to all 26 letters, thus {a,b,c,d,...,y,z} as the second
> >variable for Solve[]. Then it should even be underdetermined, but I
> >don't get a solution neither.
>
> Adding variables after the fact won't 'under-determine' the system. Those
> variables don't occur in the equations, so it's overdetermined, period.
>
> Think of it this way: No one set of values assigned to the variables can serve
> to make all equations true. Adding other variables can only affect this
> situation if you add them to the original equations too.
Yes of course it isn't gonna be under-determined (if this word exists,
but I think you now what I mean), but I didn't actually write down all
letters occuring in the system. I just counted the equations and added
26 letters, so on the first view it *seems" to be under-determined.
have a nice day anyway
Patrick Hamlyn
>... so on the first view it *seems" to be under-determined.
OK gotcha.
BTW I found out you can just leave out the second argument to 'Solve', it
assumes you mean all the given variables.
Dirk Haar
Are you right if we look at combined letters or "doubles don't count"?
That would be a coding which would not be included in equations
we've seen in the NG before, where one letter stands for one value.
Dirk