| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 576.1 | Answer | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Fri Sep 12 1986 16:49 | 17 |
| Judge (To A): Are you the spy?
A: No.
Judge (To B): Are you the spy?
B: No.
(The Judge now knows that C must be the knave.)
A: C is not the spy.
Therefore A must be the spy, and B must be the knight.
-- Barry
|
| 576.2 | no, no | GALLO::JMUNZER | | Fri Sep 12 1986 17:18 | 8 |
| Re .1:
But (no, no) can come from (A = Knight, B = Knave, C = Spy).
So the judge can't conclude from (no, no) that C = Knave.
John
|
| 576.3 | This is the truth | SMURF::DIKE | | Fri Sep 12 1986 17:11 | 27 |
| Re .1: The second question was "Did A tell the truth?"
The following table records the possible identities of A, B, and
C according to the answers that A and B gave to the judge:
Answer
#1 #2 | A | B | C |
-------+--------+--------+--------+
N N | Knight | Knave | Spy | Since the judge already knew
| Knight | Spy | Knave | that C was not the spy, the
| Spy | Knight | Knave | answers must have been NY
-------+--------+--------+--------+ or YY. If they were NY, A
N Y | Knight | Spy | Knave | would be either the knight
| Spy | Knave | Knight | or the spy. Since both are
-------+--------+--------+--------+ allowed to tell the truth,
Y N | Knave | Knight | Spy | the judge would not be able
| Knave | Spy | Knight | to tell who was the spy from
| Spy | Knave | Knight | that extra bit of information.
-------+--------+--------+--------+ If they were YY, A would be
Y Y | Knave | Spy | Knight | either the knave or the spy.
| Spy | Knight | Knave | Since the knave is not allowed
-------+--------+--------+--------+ to tell the truth, A must
be the spy. In addition,
B is the knight and C is the Knave.
Jeff
|
| 576.4 | ...embarrased... | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Fri Sep 12 1986 17:52 | 9 |
| Jeff (.3) is right, I am wrong.
* SIGH *
One of these days I really must learn how to read.
Let me make amends with another problem (to follow)...
|
| 576.5 | Another Question. | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Fri Sep 12 1986 17:54 | 28 |
|
A king of a land consisting only of knights and knaves, one day asked his
Head Royal Bean Counter to compose a report for him detailing how many
subjects he had, how many were knights, and how many were knaves.
After some time here is what his Bean Counter told him:
There are more knights than knaves.
The total number of subjects is a composite number.
The ratio of knaves to knights (knv/knt) is a natural number.
The number of knaves is less then 120.
The total number of subjects is greater than the number of
page table entries in a single page of their computer.
It took the king only a few seconds to figure out exactly how many
subjects of each type he had.
NOW, assuming that the king was a good DEC customer, (I was there on a
residency, supporting their VAXcluster, their only computers), how many
of each kind of subject DID the king have?
-- Barry
|
| 576.6 | Ummm... | SMURF::DIKE | | Fri Sep 12 1986 17:56 | 7 |
| Shoot me if I'm being a dolt, but
If there are more knights than knaves,
Then how can knaves/knights be a natural number?
Jeff
|
| 576.7 | | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Fri Sep 12 1986 18:42 | 1 |
| There is no mistake.
|
| 576.8 | I'd get a new bean counter | 26205::YARBROUGH | | Mon Sep 15 1986 13:01 | 14 |
| Hmmm. Obviously the bean-counter is a knave, so the truth of his
report reads:
knv > kni
knv+kni = prime
knv/kni <> nat no
knv >= 120
knv+kni <= 128
So the number of inhabitants must be 127, the only prime in the
range, but any of the pairs {125,2}, {124,3}, {123,4}, {122,5},
{121,6} or {120,7} satifies the ratio condition. Now if the ratio
condition were actually correct as the B.C. reported it, then
{126,1} would be a unique solution...
No doubt I am missing something.
|
| 576.9 | Hint | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Mon Sep 15 1986 20:48 | 11 |
| Well Lynn, it appears that you have discovered the Head Royal Bean
Counters secret.
So far you are VERY close, but it seems that you have missed one
of the valid non-natural ratios in this range.
I might add that the solution also will explain why the king has
never bothered to get another Head Bean Counter.
-- Barry
|
| 576.10 | Don't see how that helps | 26205::YARBROUGH | | Tue Sep 16 1986 17:57 | 3 |
| Well, I did overlook {127,0} (my aversion to division by zero is
THAT strong) but that is just one more solution to add to the six
I gave above. I don't see a UNIQUE solution.
|
| 576.11 | | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Thu Sep 18 1986 23:33 | 17 |
| ...Sorry, my net link has been down for a couple of days.
Re .10:
Yes, you are right, Lynn. In checking my notes I find that I should
have managed to imply that BOTH the number of knaves, and the total
number of subjects were prime. Oh, well. The correct answer was
supposed to be {127, 0}, and of course (127/0) is not a natural
number (it's not any number at all). I guess this is what I get
for trying to make up a decent problem over lunch.
Oh, wait! Maybe I could say that the narrator was also a knave...,
or maybe a spy...
-- Barry Bv)
|