How to calculate Dilbert error
Johnny
On April 22, 2005, the Dilbert comic strip contained what I think is a
monumental math error, but I'm not sure.
And even if I am right I'm not sure how large that error is.
Please take a look at the part of my web page that starts at
http://barelybad.com/fpdilbert.htm#permutations and let me know whether
I've said anything wrong. Note that there's a link to the Excel
spreadsheet I used, if that's of any help.
Thanks for any and all criticism you can heap on me. I want to get this
right.
--Johnny
johnnyg aattssiiggnn barelybad.com
http://barelybad.com
ni...@fas.harvard.edu
When Dogbert says combination he means pairs of fixes. "Try changing
your monitor settings while in safe mode. Try opening and closing your
CD drive while disconnected from the internet. Try changing your
monitor settings while disconnected from the internet." There are 300
(=25 * 24 / 2) such pairs, so you'd have to try 600 things. If you also
want to try the fixes individually, that's another 25, so 625.
Ari
"Johnny" <bare...@hotmail.com> wrote in message news:70afa$426a8493$407e5bd5$31...@EVERESTKC.NET...
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This post also appears on the alt.math newsgroup.
Please learn to use cross-posting when posting something to more
than one newsgroup.
Aristotle Polonium
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Proginoskes
Dogbert didn't say to try them all in pairs; he said ALL COMBINATIONS.
(I posted a thread about this at alt.comics.dilbert.) That means doing
just one (with 25 ways), doing two of them (300 ways to do two things),
doing three of them, etc.
The answer turns out to be 2^25 - 1 = 33,554,431, the -1 being because
doing nothing will not solve the problem.
If the order matters, then the number of "combinations of fixes" is
P(25,25) + P(25,24), P(25,23) + ... + P(25,1)
= 25! (1/0! + 1/1! + ... + 1/24!)
= 42,163,840,398,198,058,854,693,625.
And, as I noted in the other thread, this doesn't include the
possibility of doing one of the fixes more than once.
--- Christopher Heckman
ni...@fas.harvard.edu
Well, as long as we're speculating on what he meant, maybe you don't
have to restart whenever you finish with one combination. So perhaps
doing the following fixes:
ABCA
counts for doing the combinations ABC and BCA, as well as AB, BC, CA,
A, B, and C. If that's the case, then we only need to try ~25! =
1.551e25 fixes total, which is only 1.533% as many as if we did each
combination individually. Doing it this way, you only have to do ONE
combination of fixes, although that combination is, admittedly,
1.551e25 fixes long.
Patrick Hamlyn
>Hello, helpful people,
>
>On April 22, 2005, the Dilbert comic strip contained what I think is a
>monumental math error, but I'm not sure.
>
>And even if I am right I'm not sure how large that error is.
>
>Please take a look at the part of my web page that starts at
>http://barelybad.com/fpdilbert.htm#permutations and let me know whether
>I've said anything wrong. Note that there's a link to the Excel
>spreadsheet I used, if that's of any help.
>
>Thanks for any and all criticism you can heap on me. I want to get this
>right.
The humour is quite simply 'standard Dogbert fare' in that Dogbert is messing
with Dilbert's brain.
First you should realise that Dogbert is never going to give Dilbert a million
dollars. Clearly Dilbert should realise this too, so it's firstly funny that
Dilbert would go to work either in a bathrobe or naked on such a flimsy premise.
And then secondly it's funny that he would wait until he gets there before
thinking carefully about the wording.
Then it's funny that he would be wrong to be worried about getting the wording
wrong, especially when he has a more obvious problem on his hands - Dilbert is
never going to give him a million dollars.
On the second strip, we should probably calculate how many 'operations' are
required to test every combination of one or more fixes, assuming an operation
is defined as making or undoing a fix.
Example with 2 fixes. 'A' means do fix A, 'a' means undo fix A.
ABa
With 3 fixes, 7 operations are required:
ABaCbAB
Since there are 2^n - 1 combinations, obviously we need at least that many
operations.
But can we always accomplish it? Is there an easy 'generating sequence' to
generate the optimal sequence?
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
Prai Jei
<12vl61hkadjkem14n...@4ax.com>:
> On the second strip, we should probably calculate how many 'operations'
> are required to test every combination of one or more fixes, assuming an
> operation is defined as making or undoing a fix.
>
> Example with 2 fixes. 'A' means do fix A, 'a' means undo fix A.
> ABa
> With 3 fixes, 7 operations are required:
> ABaCbAB
>
> Since there are 2^n - 1 combinations, obviously we need at least that many
> operations.
>
> But can we always accomplish it? Is there an easy 'generating sequence' to
> generate the optimal sequence?
Yes we can. What is required is a "Hamiltonian path" visiting all 2^25
corners of a 25-dimensional hypercube once each, by travelling along a
*subset* of the edges. This is always possible, for any number of
dimensions.
An optimal sequence can be found by solving a 25-disc "Tower of Hanoi"
puzzle, identifying each of the 25 discs with one of the dimensions of the
hypercube. As you move each disc while solving the Hanoi puzzle, your
counterpart moves along the corresponding dimension of the hypercube. (Try
it with a 3-disc Hanoi puzzle and a physical cube.)
(Note that the complete tour of the *edges* of a 25-dimensional hypercube is
not possible. This is possible in even numbers of dimensions only, with the
special exception of 1 dimension. Can the reader see why?)
Exercise for the reader: how many equally optimal sequences are there?
As for the original Dilbert strip, the final panel in each episode of the
"million dollars" storyline is intended to be paradoxical, and the "joke"
is the character eternally wrestling with the paradox and finding no
solution.
--
Paul Townsend
Pair them off into threes
Interchange the alphabetic letter groups to reply
Prai Jei
<1114293460....@g14g2000cwa.googlegroups.com>:
> The answer turns out to be 2^25 - 1 = 33,554,431, the -1 being because
> doing nothing will not solve the problem.
In my own experience I have had several instances where applying a fix has
not solved the problem, but then removing it again *has* solved it.
Illogical perhaps, but that's how M*******t W*****s is.
Johnny
<ni...@fas.harvard.edu> wrote in message
news:1114296798....@f14g2000cwb.googlegroups.com...
Night, your speculation is interesting, and it's one I hadn't thought
of. May I quote you on that page?
--Johnny
johnnyg aattsssiggn b_a_r_e_l_y_b_a_d_DOT_c_o_m
http://barelybad.com
Johnny
"Proginoskes" <progi...@email.msn.com> wrote in message
news:1114293460....@g14g2000cwa.googlegroups.com...
April 24, 2005
Mr. Heckman,
Thanks for your reply, and I did read the thread at alt.comics.dilbert.
As you know, we ended up with the same math result assuming the order of
the fixes makes a difference.
Incidentally, don't you agree it is only reasonable to assume it does?
After all, there's a difference between "Perform Fix #16, then perform
Fix #22" and "Perform Fix #22, then perform Fix #16."
As you might have seen from my spreadsheet, I got only
42,163,840,398,198,100,000,000,000. May I know how you arrived at all
those extra significant digits?
Also, was your second formula, where you multiply the series by 25!,
supposed to run from 1/0! to 1/24! or to 1/25! ?
Perhaps on the same topic, is it always true that P(N,N) = P(N,N-1) ?
For example, my spreadsheet says P(25,25) = P(25,24) =
15,511,210,043,331,000,000,000,000.
I look forward to your response.
--Johnny
======================
Prai Jei
message <1114296798....@f14g2000cwb.googlegroups.com>:
> Well, as long as we're speculating on what he meant, maybe you don't
> have to restart whenever you finish with one combination. So perhaps
> doing the following fixes:
>
> ABCA
>
> counts for doing the combinations ABC and BCA, as well as AB, BC, CA,
> A, B, and C. If that's the case, then we only need to try ~25! =
> 1.551e25 fixes total, which is only 1.533% as many as if we did each
> combination individually. Doing it this way, you only have to do ONE
> combination of fixes, although that combination is, admittedly,
> 1.551e25 fixes long.
That misses (among others) the combination of BC *without* A which may well
be the required fix.
Proginoskes
Using Maple, a symbolic manipulation program which keeps results as
accurate as possible. Mathematica or another mathematics package will
also work.
> Also, was your second formula, where you multiply the series by
> 25!, supposed to run from 1/0! to 1/24! or to 1/25! ?
It's only through 1/24!. This is because if you don't do anything, then
you won't fix anything. Then you add terms for using exactly 1 thing,
exactly 2, etc., getting
P(25,1) + P(25,2) + ... + P(25,25)
= 25! / 24! + 25! / 23! + ... + 25! / 25!
= 25! (1/24! + 1/23! + ... + 1/0!).
> Perhaps on the same topic, is it always true that P(N,N) = P(N,N-1) ?
> For example, my spreadsheet says P(25,25) = P(25,24) =
> 15,511,210,043,331,000,000,000,000.
>
> I look forward to your response.
Yes, it is. P(n,r) = n! / (n-r)!, so
P(n,n) = n! / 0! = n!, and
P(n,n-1) = n! / 1! = n!.
--- Christopher Heckman
ni...@fas.harvard.edu
Johnny wrote:
> <ni...@fas.harvard.edu> wrote in message
> > Well, as long as we're speculating on what he meant, maybe you
don't
> > have to restart whenever you finish with one combination. So
perhaps
> > doing the following fixes:
> >
> > ABCA
> >
> > counts for doing the combinations ABC and BCA, as well as AB, BC,
CA,
> > A, B, and C. If that's the case, then we only need to try ~25! =
> > 1.551e25 fixes total, which is only 1.533% as many as if we did
each
> > combination individually. Doing it this way, you only have to do
ONE
> > combination of fixes, although that combination is, admittedly,
> > 1.551e25 fixes long.
>
> Night, your speculation is interesting, and it's one I hadn't thought
> of. May I quote you on that page?
Actually, you probably shouldn't. I realize now I made a mistake. I was
thinking after the first 24 fixes, you could keep doing a new length-25
combination with (almost) every new fix you added to the sequence. But
this is definitely not the case. I don't know what the correct number
is.
ni...@fas.harvard.edu
> > Well, as long as we're speculating on what he meant, maybe you
> > don't have to restart whenever you finish with one combination.
> > So perhaps doing the following fixes:
> >
> > ABCA
> >
> > counts for doing the combinations ABC and BCA, as well as AB,
> > BC, CA, A, B, and C.
>
> which may well be the required fix.
> --
> Paul Townsend
That depends on how you imagine a single fix. If you imagine it as a
state (like removing the disk drive) then you're right. But if you
imagine it as a task (like jiggling the cable), then only the sequence
matters, and "BC without A" doesn't make any sense.
Johnny
======================
April 25, 2005
Mr. Heckman,
Thanks again for your help, and for making my spreadsheet make sense
again. I take it 0! = 1, not 0. (Eeps!)
On that Web page I'd be happy to credit you or the alt.comics.dilbert
group or whatever you like with supplying all those extra digits if you
want me to. Also, I have been unable to find a way to verify that they
are correct short of buying Maple or Mathematica. Are you sure they
are, i.e., might you have mis-typed?
--Johnny
johnnyg aaattsiggn _b_a_r_e_l_y_b_a_d_ DOT *c*o*m*
http://barelybad.com
======================
Proginoskes
Yes. Mainly to make formulas for C(n,r) and P(n,r) turn out right when
r = n.
> On that Web page I'd be happy to credit you or the
> alt.comics.dilbert group or whatever you like with supplying
> all those extra digits if you want me to.
Okay, go ahead. If you want to link to my personal webpage, it's
http://math.asu.edu/~checkman/ .
> Also, I have been unable to find a way to verify that they
> are correct short of buying Maple or Mathematica. Are you
> sure they are, i.e., might you have mis-typed?
I copied-and-pasted the output from Maple to Firefox. (I use Google
Groups for posting.) There's no chance of a typo, but MAYBE a digit got
changed when it got sent. 8-)
--- Christopher Heckman
Johnny
April 26, 2005
Well, if a digit got changed then everyone who tries Dogbert's advice
will be misled about how long the whole thing will take. I hope no one
sues me over any such error, and I hope you know I will be counter-suing
you if that happens. Or Maple. Or Firefox. Or Google.
With that out of the way, let me say that I have changed my Dilbert page
to show a link to your site, which is fascinating. In addition, I've
added a link elsewhere on my site to your What Day Is Today? page.
Thanks again.
--Johnny
===================
Proginoskes
Johnny wrote:
>
> With that out of the way, let me say that I have changed
> my Dilbert page to show a link to your site,
Thanks.
> which is fascinating. In addition, I've
> added a link elsewhere on my site to your What Day Is
> Today? page.
I got the idea for that page from Robert Anton Wilson's "How to live
eleven days in 24 hours". Wilson is best known as the co-author of the
Illuminatus! trilogy, and has written lots of books and articles about
breaking out of "reality tunnels."
--- Christopher Heckman