Deprecated: Function set_magic_quotes_runtime() is deprecated in /home/otori07org/otori.org/board/common.php on line 88
[phpBB Debug] PHP Notice: in file /includes/session.php on line 821: Cannot modify header information - headers already sent by (output started at /common.php:88)
[phpBB Debug] PHP Notice: in file /includes/session.php on line 821: Cannot modify header information - headers already sent by (output started at /common.php:88)
[phpBB Debug] PHP Notice: in file /includes/session.php on line 821: Cannot modify header information - headers already sent by (output started at /common.php:88)

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 456

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 95

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 95

Deprecated: preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead in /home/otori07org/otori.org/board/includes/bbcode.php on line 95
[phpBB Debug] PHP Notice: in file /includes/functions.php on line 4181: Cannot modify header information - headers already sent by (output started at /common.php:88)
[phpBB Debug] PHP Notice: in file /includes/functions.php on line 4183: Cannot modify header information - headers already sent by (output started at /common.php:88)
[phpBB Debug] PHP Notice: in file /includes/functions.php on line 4184: Cannot modify header information - headers already sent by (output started at /common.php:88)
[phpBB Debug] PHP Notice: in file /includes/functions.php on line 4185: Cannot modify header information - headers already sent by (output started at /common.php:88)
Clan Otori • View topic - Robotic Riddles -- Closed

Robotic Riddles -- Closed

Otori Week returned with a vengeance in 2006. The week-long assortment of mechanical contests and festivities is now archived here for your browsing pleasure.

Moderators: Full Board Control, Otori Week II Organisers

Postby pakwan on Wed Sep 20, 2006 4:14 am

Partial spoiler, then....I hope I am right....I am 99% sure I am, but still...Also, the solution which I submitted to Slashor differs from this in one major respect: the questions presented here require that the "yes" answers point to the number you want; in the solution give to slashor, the questions required that the "no" answers eliminated the numbers that you didn't want, and that the number left would be the desired number.

I apologize for the length of the spoiler. You have been warned.

[spoiler]
The seven questions are all of the form, "Is your number one of the following: _, _, _, _, _, _, _, _?" where each blank stands for one of the integers between 0 and 15.

As stated, these 16 ints can be expressed in binary as no more than 4 bits long. Thus, four of the questions are immediately apparent: you need to ask if there is a 1 in the bit corresponding to 1, 2, 4 or 8.

(1) 1, 3, 5, 7, 9, 11, 13, 15 //for the 1 bit
(2) 2, 3, 6, 7, 10, 11, 14, 15 //for the 2 bit
(3) 4, 5, 6, 7, 12, 13, 14, 15 //for the 4 bit
(4) 8, 9, 10, 11, 12, 13, 14, 15 //for the 8 bit

Notice that if the problem stated that the person could not lie at all, these four questions would suffice. Let 1 represent a "yes" and let 0 represent "no" and place these answers in a row from right to left; the sequence of 4 chars will give you the binary representation of the desired number.

However, the problem is that the person can lie once, and you can not determine from these four questions alone if the person lied and, if so, which answer is false. This is where the Hamming code comes in. A (7,4) Hamming code is a code where every four bits of info is sent as a string of seven bits: four for the actual data, and 3 additional bits as "parity check" bits. I believe that this is the smallest Hamming code possible for correcting single-bit errors in a set of four bits of info.

Thus the other three questions must somehow correspond to these 3 "check" bits. I will not explain the math here on how the following three questions can uniquely determine if an error occurred and, if so, which questions. You can check out this page which I think is at the level of a high school student or maybe 1st-year college student. Nothing harder than knowledge of base 2 and a bit of patience to work through an example step-by-step is required.

(5) 1, 2, 5, 6, 8, 11, 12, 15 //for the 1st check bit
(6) 1, 3, 4, 6, 8, 10, 13, 15 //for the 2nd check bit
(7) 2, 3, 4, 5, 8, 9, 14, 15 //for the 3rd check bit

How are these numbers grouped?
  • The 1st "check" question lists numbers which, when expressed in binary, will yield an odd number when adding up the values in the bits for 8, 2, 1.
  • The 2nd "check" question lists numbers which, when expressed in binary, will yield an odd number when adding up the values in the bits for 8, 4, 1.
  • The 3rd "check" question lists numbers which, when expressed in binary, will yield an odd number when adding up the values in the bits for 8, 4, 2.

Another way of thinking of it is:
  • Question 5 lists numbers for which you would get an odd number of "yes" answers between Questions 4, 2, and 1; that is, either exactly one of these questions or all three question would be answered "yes".
  • Question 6 lists numbers for which you would get an odd number of "yes" answers for Questions 4, 3, and 1.
  • Question 7 lists numbers for which you would get an odd number of "yes" answers for Questions 4, 3, and 2.



That's confusing. Here's an example. The number 11 in binary is (1011). Notice that there is a 0 in the place for the 4 bit, and a 1 for the bits for 8, 2, and 1. Thus, if the person answered these 3 questions truthfully, then you would get a yes for question (5), since 1+1+1 is an odd number, a no for question (6) since 1+0+1 is even, and a no for question (7) since 1+0+1 is an even number. Sorry, that's the best I can do.

OK, so you're probably wondering how you can detect a lie with these questions...again, I won't go into the math and will instead ask you to look at the page linked to earlier in my post.
[/spoiler]
pakwan
Visitor
Visitor
 
Posts: 10
Joined: Wed Aug 30, 2006 11:48 am

Postby Slashor on Sun Sep 24, 2006 5:13 am

NOTE: This is probably the longest post you will ever see or are likely to ever see. Try expanding all the spoiler boxes. I had to use them just to reduce the length of this post.

DirkDiggler and Pakwan Kenobi tied for first place. Interestingly they both failed partially on question 16 by using lemmings instead of clones. That is the fun of changing words to commonish riddles and watching people copy and paste text or something similar. They both scored 18.5 points. OneWingedAngel is runners up for answers on 17 points. There was a total of 21 points available. If you have any queries please get them out fast as I will be distributing prizes soon.

Under the verse side of the competition DirkDiggler won with the most riddles answered correctly in verse. galefrost whilst only answering 6 and getting 5 right was the best answered riddles using verse. Dirk can't win both prizes.

AkravatorTemplar scored adequately by had some very interesting lines of thinking and so gets a very honourable mention. His answers had the most original thought put in and did quite well with his verse as well. Disneeyore1 also had a great set of answers and whilst didn't get a high number right had very good explanation and whilst he didn't use any verse was certainly creative in answering.

To continue I am sorry for taking so long to mark but even with 23 entrants which was roughly one quarter of what I had last year I took at least as long to answer because of the wide variety of good answers and very, very long answers. I also hate RL for not letting me have any free time. I will leave this open for a few days until I get home from camping to see if there are any errors in my marking. Prizes will be distributed upon me getting back. I will try and post the correct answers to everything later today. Look at Pakwan Kenobi's for the best answers though if you want to beat me to the punch. I will probably steal all his text like any good writer would do. Plagiarism is what writers do I assume.


DirkDiggler's entry.

[spoiler]DirkDiggler #881008

My responses are each summarized in a series of Double Dactyl poems, a
verse form whose rules are "complex, unyielding, and moronic."
(http://www.washingtonpost.com/wp-dyn/co ... 00633.html).
Here's an example and summary in one:
Long-short-short, long-short-short
Dactyls in dimeter,
Verse form with choriambs
(Masculine rhyme):
One sentence (two stanzas)
Hexasyllabically
Challenges poets who
Don't have the time. -- Roger L. Robison
(http://www.stinky.com/dactyl/robison.html)


Question 1: Jailhouse Lock

Tiggledy Toggledy
Gaylord the Jailor does
Inebriatedly
latch and unlatch.
Cells get a visit when
Gaylord's pink elephants
signal divisors for
this round do match.

Fictorly Factorly
Number-theoretically
each cell gets toggled once
per divisor.
Fact about factors is
each has a counterpart
pair them together: this
ain't no lie, sir.

Parity Merity
Each perfect square has a
Perfect square root that serves
as its own mate
Ten happy digits found
quite happy occupants
sallied they forth and then
partied til late

(Each Jail cell gets toggled whenever a round number matches a divisor
of the cell number: cell #12 gets toggled on rounds 1,2,3,4,6, and 12,
while cell #75 gets toggled on rounds 1,3,5,15,25, and 75. Most
numbers' divisors come in pairs -- (a, n/a) for number n and divisor a
-- which means that most cells got toggled an even number of times and
thus wound up closed. However, for a perfect square the square root
is its own counterpart divisor, and so cells 1, 4, 9, 16, 25, 36, 49,
64, 81, and 100 (ten in all) were left open.

----------------------------------------------------------------------
Question 2: Greed is good, going first is better

Minimal Maximal
five greedy clannies of
infinite intellect
hash out the loot.
Opportunistical
first-in proposal gives
5, 3, and 1 stuff; the
others, the boot

Simple Majority
rules would require that
for ties (and Balrogs) it's
"Thou shalt not pass."
For the last pairing no
Overlygenerous
actions can help Two: One
gets all the cash.

Not gracious -- Rapacious,
Second-ranked clannie will
happily vote for what-
ever Three's got.
Toss Two a bone, sir
and leave One alone sir --
bloodthirst-a-quenchingly
leaves Three a lot!

Dynamic Programming
Four finds his in'trests are
all-diametrically
opposite three.
Thus he must offer a
pittance to One and a
larger-sized milksop for
Two to be pleased.

Goody bag Goody bag
First mover Five in the
Ol' catbird seat gives a
pair up for One.
One bag for Three plus a
fear of alternatives
(sure-disenfranchisment)
gets the job done.

If it comes down to the last two clannies, #1 will vote down any
proposal put forth by #2: in doing so, #2 gets booted and #1 gets all
the loot (let's call this result 1000, 0, 0, 0, 0). So, #2 *must*
accede to any proposal put forth by #3 which benefits him. (Note that
#3 can put forth a proposal that does benefit #2, and that #2 is
infinitely bloodthirsty. Therefore, #2 will punitively reject anything
#3 puts forth that doesn't leave #2 at least mildly better off: ties
are voted down). This means that if the vote reaches #3, he can put
forth (0, 1, 999, 0, 0); #2 and #3 will vote for it (as otherwise they
each get nothing), and the vote will pass.

If the vote does get to #3's proposal, then 0 and 1 are the most meat
that #2 and #1 can possibly see. This means that #4 has only to
better these paltry amounts and he will have their votes,
disenfranchising #3. If #4 proposes (1, 2, 0, 997, 0):
. #2 must vote for it (otherwise #3 will propose and #2 will only
get 1 goody bag)
. #1 must vote for it (otherwise #3 will propose and #1 will get
nothing)
. #4 is all for it, obviously.

So #5 just has to make two other players happier than they would be if
#4 were running the show. The proposal (2, 0, 1, 0, 997) is better
for #3 and #1 by one goody bag each than what will happen if the
proposal fails; this proposal passes with votes from #5, #3, and #1.
(Note again that #5 has to sweeten the deal for #3 and #1 over the
alternative: if his proposal only ties then #3 and #1 will
retributively nix it).

----------------------------------------------------------------------

Question 3: Factorial Rumpus

Euclidly Gaussically
two fifty seven bang
has as prime factors those
found in its parts.
Multiplica'tively,
each little integer
puts all its twos and fives
into the pot.

Reasoning logically,
one twenty eight of them
all-turn-a-takingly
toss in a two;
sixty-four more, having
malice of '4'-thought con-
tribute additional
doublings to boot.

So on and so forth, count
binaritistically
each half as many through
two fifty six.
Two fifty five "2"'s the
total arrived at by
brief calculation or
algebra tricks.

Similar mannered-ly
some fifty integers
divisibility
counting by fives.
Ten more, by having a
five squared as factor,
put in a second to
mess up our lives.

Finally, cubically
one twenty fives give us
two contributions (and
now we're home free).
Sixty three fives and a
surfeit of deuces mean
sixty-three 10's are the
most there can be.

(257! ends in 63 0's).

----------------------------------------------------------------------
Question 4: Anno Mirrorabilis

Julian Gregorical
supersymmetrical
dates in an eight digit
padded format
'1' for millenium
yields thirteen hundreds and
thirty-one month days (so
how about that)!

Carefully, Warefully
decimal-formatly
leading a zero gives
extra decades
Numerological
and Horological
Compatibility
each month we check

Tickety Tockety
Check nonuplecticly
thirty days' is all that
September hath;
Reductionisticly,
Palindriomistically,
thus: thirteen eighty on
August the last!

(08/31/1380 -- a later palindrome is possible if 0-padding is not
enforced.)

----------------------------------------------------------------------
#5 The truth, most of the truth, and very little besides the truth?

Exorably XORably,
Boolean Algebra
lets us insert our own
truthiness test.
XNOR your answer with
whether you're lying and
tautologistically
I'll guess the rest!

I would do a binary search using questions of the form
Is the truth value of (the number is greater than or equal to 8) the
same as (you are answering this question truthfully)?
Note that if you are telling the truth, the answer to the question is
identical to that of the range-specifying subquestion. If, however,
you are lying, then the second clause is false. If the first clause
is true, then the truth values are different, and you will answer
'yes' (same as if you were telling the truth). If the first clause is
false, then the truth values are the same, and you will answer 'no'
(also the same as if you were telling the truth).

----------------------------------------------------------------------
#6 is at the end

----------------------------------------------------------------------
#7: Particular Physics


Neutrino Photino
Particle Physicists
Find that invariants
oft do the trick
Axiomatical:
Particle Number and
Gl-Gu diff'rence
values must stick

Tau boson Pi meson
Rule number one implies
absence of gauginos ->
surfeit of p's
Non-irreversible
can't help us anyway
kill all the photinos?
Too many G's!

We have the rule
ga + gl -> 2 ph
and start with
15 ga, 17 gl, 19 ph
The sole interaction conserves the quantities Difference Of G's
D = (gluino-gaugino)
and Total Particle Number
T = (gaugino + gluino + photino)

The universe starts with D=2, which means there will always be unequal
numbers of gauginos and gluinos. For there to be only one particle, we
would need no gauginos and so 2 gluinos -- but T=51 means there would
be 49
photinos. Even if the operation is reversible (2 ph -> ga + gl), for 0
ph
there would be 51 (ga + gl); then D=2 means there would be (25ga, 27gl,
0ph).

It is not possible for there to be only one type of particle after
enough
collisions. (Also I'm assuming that "45" is a typo and T=15+17+19=51)

----------------------------------------------------------------------
#8

Heidegger Heidegger
Free-willer John Stuart
(Utilitarian,
forty three years)
happily sees that a
squared anniversary
there on his pin-up cal-
endar appears!

Pleasantly, Presently
Every few decades do
calendar-riffical
birth years arrive.
Last was the '80 crowd -
they'll all retire when
fifty five, which squared is
two oh two five

In 1849 (the answer), JS Mill was 43, which squared makes 1849.
Similarly the 1892'ers turned forty-four in 44^2=1936, and the 1980
crowd will turn 45 in 45^2=2025.

----------------------------------------------------------------------
#9,#10 at end
----------------------------------------------------------------------
#11 Money -- it's a gas

Secretly Greedily
Three men of Mammon will
game-theoretically
disclose their stash.
Each tells his neighbor a
part of his holdings 'til
each one's fessed up to his
fat stacks of cash.

Multum in Partum, man
Knowing the total means
no man can do more than
debit what's his.
Decompositioning
one from the other an
impossibility
surely it is.

Order the players by player ID. The lowest only talks to the middle,
who only talks to the highest, who only talks to the lowest. Each
player rolls 1d20, waits between zero and that many minutes, and then
-- unless contacted -- starts the process by telling his neighbor an
arbitrary amount of meat between 0 and his total stash. The second
respondent similarly determines an arbitrary amount of meat (between 0
and his total stash), adds that to the total, and sends the result
along to the third respondent, who does the same. Now the we're back
to the initiator, who adds an arbitrary amount of meat between 0 and
his (total, less what he has already disclosed), with the additional
restriction that if the same total comes around twice, and he has not
volunteered all of his meat, he should disclose something. Each
player continues in turn, incrementing the total by some hidden
amount, until all have volunteered their total wealth. (If the same
total circulates for three circuits everyone is all-in). Their
average wealth is then one-third the final total.

This works because each player knows only the total wealth disclosed
so far by the other two players. He has no way of ascribing that
wealth to either player individually: say his neighbor whispers "Two
million" to him, and he put in one million of -- the distribution
could be (1,000,000 and 0), (500,000 and 500,000), or anything else.
All that is known at any point to a player is the amount they
themselves have disclosed and the total that has been collectively
disclosed.

The randomized start is to prevent the no-cycles rule from revealing
information. Imagine that there is one meat in the kingdom, I knew
who started, and I sat second. If the total came around twice as
'zero' I'd know that the player sitting third had the meat. If,
however, I didn't know which of the other two players started I
couldn't know if it had come around fully twice yet. (It's easy to
resolve collisions -- one player will simultaneously send and receive
a message -- and in that case you restart the process).

----------------------------------------------------------------------
#12 Quarters

Googley Moogally,
Otori Riddlers!
Finding the question was
most of the work.
techinterviewdotorg
gave more constraints -- does that
make me a jerk?

Three point one Four one six
Me and al sit at a
circular tabular
play surface, yo.
Each takes turns laying down;
loser is he who's found
zugzwangarifficly
unable to go.

Apposite Opposite
Clever me cleverly
Simply, Symmetricly
game plan adopts.
Right in the center -- not
left too much right too much
superprecisefully
Quarter I drop.

Monkey see Monkey do
Where my opponent goes
theoloditically
position find.
Exactly opposite
each new position my
quite steady hand follows
surveyor's mind.

Complementarially
any position staked
must have a mirror place
where I will go
Thus if the game's to end
his is the losing hand
area finiteness
says it is so!

----------------------------------------------------------------------
#13 Hairy Problem

Headily Hairily,
Beard-wearer Socrates
pigeonhole principled
answers he sought.
"Order your loathers by
Hirsu-itudity
(count up their hairs I mean),
N down to naught.

Chromicly Domicly
Maximum populace
Four eight three, two zero
seven's the peak.
Tonsorialistically,
one more's too hairy or
outlawed explicitly
or not unique.

There cannot be more than 483,207 loathers. Take all your loathers
and stand them in a line, sorted by hirsuteness. The largest populace
will occur when every hair count from 0 to 483,206 is claimed. In
this case, there are a total of 483,207 loathers, with no two having
same haircount, no one having 483,207 hairs, and one more inhabitants
than hairs on the hairiest loather.

Now if Cousin It wants to move to the island, he cannot have more than
483,207 hairs (the new island population) or he'll break rule #3. He
cannot have exactly 483,207 hairs (rule #2), and each hair count from
0 to 483,206 is claimed (rule #1). Unfortunately Cousin It cannot get
an immigrant visa, and the maximum citizens of loathing is 483,207.

#14 Strawbroken Alarm Clock

Heximal-Decimal
Numbers stored digital-
ly have their value
encoded as bits
In this example, two
values are ample: it's
line number nine that is
giving us fits.

A digital line going from the time store to the display is busted
(held at zero). It seems that the clock has an internal value for the
time, stored in binary, representing the number of seconds since
midnight. The internal value for the time is correct, but the ninth
bit is always being reported to the display function as zero.

This means that times up through #000FF (255 seconds, or 00:04:15) are
correct, but that #00100 (256 seconds, or 00:04:16) is being displayed
as #00000 (midnight), #00101 as 00:01:01, and so on. At time #001FF
(511 seconds, or 00:08:31), the displayed time is again #000FF (4:15)
-- but the next value is #00200 (512 seconds, or 00:08:32), which is
rendered properly. The clock will display correct times through
#002FF (767s = 00:12:47); the next tick gives #00300 (00:12:48) which
it will display as #00200 (00:08:32). In general, times with an odd
ninth bit will display as times that are 256 seconds earlier, while
times with a zero ninth bit will display correctl.

----------------------------------------------------------------------
#15

On again Off again,
Intragalactical
Passengers board a fa-
milar row boat.
Cardplayers reckon a
full deck you start with to
start and to end with an
integral note.

Minimal-Maximal
Passengers practical
three quarters (thirty nine)
first disembark.
Seven more board so a
score they set off with --
in cricket and boating a
felicitous mark.

Mutatis mutandis
Fifteen can't stand for us
So at the next stop they
vote with their feet
Qaballa-isticly
seven more join us to
total up twelve: we're the
pride of the fleet!

Lastly not Leastly the
One dozen passengers
disgorge a baseball team
(no DH, please)
Seven more board there, but
self-terminatingly
"nine at the last stop got
off," if you please.

----------------------------------------------------------------------
#16 Oh No More Lemmings! Crazy! Wicked! Havoc!

Chordata Rodentia
Havocs of Lemmings ca-
vort on and off a de-
cameter plank.
Lemmings, like bosons, are
anonymizable:
A null encounter's the same
as a clank.

Crazily Wickedly
Not very quickedly
long walk off short pier
lemming parade
seeing their meetings as
Noninteractingness:
"One hundred minutes" the
answer to grade.

Since the lemmings are interchangeable and anonymous, the reflection
of two lemmings off each other is indistinguishable from no
interaction at all. That is, they can meet and bounce off; or they
can just high five as they pass, and either way we end up with one
lemming headed left and one headed right at the same speed and time.
The longest trip that such a high-fiving lemming can take is 100m
(end-to-end on the plank), a trip that lasts 100 minutes.

----------------------------------------------------------------------
#17 Plot of land

Higgledy Piggledy
Meat farmer Johnson should
Pythagoristicly
dig in an 'X'
Anything less, and a
vein at an angle will
find the lacuna and
surely will vex

Hewlettly Packardly
Arithametically
here's how to find what the
trench length's to be
(My calculator's re-
verse polish no'tation):
2, enter, 2, square root,
times: less than 3!

----------------------------------------------------------------------
#18 Corsican Yetis

Sleuthily Truthily
Detective Sherlock-like
interrogational
care not bombast
Phrasing it artfully
makes them reveal to me
whether the random's not
next or not last.

Sort the yetis by height (or otherwise arbitrarily order them) from
tallest to shortest, left to right as we see them. The Random yeti
could be tallest (RLT RTL), in the middle (TRL LRT), or shortest (LTR
TLR).

Tallest: "Is the Random Yeti the next-shortest after the Truth Yeti?"
If y, the random yeti must either be tallest or in the middle (RLT
RTL, TRL, LRT) -- not shortest!
Ask the shortest yeti: "Are you the Random Yeti?"
If y, he is the Lying Yeti (RTL or TRL).
Ask him if the tallest Yeti is the Truth Yeti:
If y, we have RTL;
If n, we have TRL with Carson Daly.
If n, he is the Truth Yeti (RLT or LRT).
Ask him if the tallest Yeti is the Random Yeti:
If y, we have RLT;
If n, we have LRT
If n, the random yeti cannot be the middle yeti (RLT RTL TLR LTR).
Ask the middle yeti: "Are you the Random Yeti?"
If y, he is the Lying Yeti (RLT or TLR).
Ask him if the tallest Yeti is the Truth Yeti:
If y, we have RLT;
If n, we have TLR.
If n, he is the Truth Yeti (RTL or LTR).
Ask him if the tallest Yeti is the Random Yeti:
If y, we have RTL;
If n, we have LTR

----------------------------------------------------------------------
#19 Sibling Rivalry

Mile High While Fly
ing to Japan I did
enter the World somewhere
over Taipei
Brief moments later: the
date line, then sister --
Greenwich-Meridianly
the previous day.

I was born on an airplane on Sunday, just before it crossed the date
line -- my younger sister was born moments later on the 'previous'
day.

----------------------------------------------------------------------
#20: Nothing's too good for you

Goldenly Oldenly
This humble puzzler
instantaneously
back in the day.
"Nothing"s my retort to
this hairy chestnut that
blind men saw coming a
mile away.


----------------------------------------------------------------------
#21: Veni Vidi Tottered

Riddling Piddling
Clan of Otori; less
Jeopardy caliber
than, say, "Hee-Haw"!
oscillatorically,
future-historically
paired conjugations name
humble see-saw

----------------------------------------------------------------------
#9


Boisterous Cloister of
Eelymosynarists
some will distribute their
food to their pals
One in particular
gives out his food to those
who are well known to not
do so at all.

Anyone that doesn't get food from that hermit, or Mother Earth.
Obviously all food received from the generous hermit cannot be
transferred back to the generous hermit or indeed to anyone -- else
the hermit wouldn't give it up. But the generous hermit can get food
from a grocery store, or from a food bank, or by growing his own, or
from another hermit who isn't so choosy.

----------------------------------------------------------------------
#10

Diffie and Hellman and
R, S and Adelman,
algamorithmical
answers provide.
Go tell your friend that he
should grow a pair (of keys)
one to keep secret, the
other, plain sight

Cura Te Ipsum We
Take our own medicine
Authenticating with
our private key
Then use his Public key
(off the graffiti wall):
gobbledygookedy
to all but he!

If I understand the problem, my friend and I can be seen communicating,
as
long as the multi can't read the contents of the final message I want
to
send. That is, we can have many plaintext conversations as long as at
the
end I can send a secure encrypted one.

This is exactly the trapdoor public key problem. I tell my friend to
make
a keypair, to keep one secret and to share the other. Any message
encrypted with his public key is immune to decryption by even such as
the
Multi Czar; only my friend knows his private key, and keeps it outside
#6's
ample purview. I can go one step further and first encrypt my message
with
my private key; if my (widely-known) public key unlocks a sensible
message
then only I could have sent it. This final message can have any
combination
of keys and of content: only my friend can read it, and only I could
have
sent it.

----------------------------------------------------------------------
#6

Quackety Yackety
Ducky's in peril; can
math differenti-al
reveal her path?
Involute, tractrix, non
linear cycles and
phantasmagorical
high level math!

Forwards and Backwards the
cat can dissuade the duck
ever from starting not
at center point.
Coupled equations show
sadly for ducky he's
tragic-trajector'ly
stuck in this joint

The cat starts at some point on the periphery, and the duck in the
water
just off that point. The duck will start swimming directly away from
the
cat, who must pick one side of the pond. The duck's path will be such
that
it is always swimming directly away from the cat; the cat moves at 4x
the
speed in a one or the other direction around the outside. Since the
duck's
inst. velocity is optimally directly away from the cat, the line from
cat to
duck is tangent at each point to the duck's path.

This describes a pursuit curve ... yikes. By reversing direction to
match
the duck, the cat can force the duck into the center of the pond. From
that point, there is no escape path which does not complete a circuit;
but
in any such circuit there is a point at which the duck is on the same
side of center as the cat. From there, the cat can force him back to
center.

The duck is trapped.

[/spoiler]

Pakwan Kenobi's answers.

[spoiler]> All but two riddles are answered. I know, it's sad. May I be
> obliterated by a shit-ton of mechs like Heihachi Mishima was.
>
> (1) On the nth round, a door's state will be changed iff n is a
> divisor of the number on the door. As only square numbers have an odd
> number of divisors, only the doors with square numbers on them will
be
> open, since an open door implies an odd number of state changes.
>
> Therefore, only 10 doors out of the 100 will be open after the jailor
> decides to call it quits.
>
> (2) Proceed by induction on p, the number of pirates. For the sake of
> the inductive proof, let a "majority" be defined as 50% or greater.
>
> If p=1, this is easy.
>
> If p=2, then this is also easy, as the higher-ranked pirate will
> propose to take all 1000 bags and get 50% of the vote (his own).
>
> If p=3, then call the pirates P1, P2, and P3, with P1 making the
first
> proposal. He only offers P3 1 bag, and P3 must vote for this, because
> if P3 votes no, P2 and P3 will constitute the majority, P1 will get
> banished, and then P2 will take all 1000 bags. Thus P1 gets 999, P2
> gets 0, and P3 gets 1.
>
> If p=4, then call the pirates P1, P2, P3, and P4, with P1 making the
> proposal. He only offers P3 1 bag, and P3 must vote for this, because
> if P3 votes no, P2, P3, and P4 will constitute the majority, P1 will
> get banished, and P2 will propose the plan outlined in the previous
> step of the proof, leading to P2 getting 999, P3 getting 0, and P4
> getting 1. So, with this plan, P1 gets 999, P2 gets nothing, P3 gets
> 1, and P4 gets nothing.
>
> If p=5, then call the pirates P1 through P5, with P1 making the
> proposal. He only offers P3 and P5 1 bag each, keeping 998 bags. P3
> and P5 must vote yes, because otherwise P1 will get banished, and P2
> will propose the plan outlined in the previous step, which would
leave
> P3 and P5 with nothing. Thus, they will accept P1's plan, P1 will get
> 998, P2 will get 0, P3 will get 1, P4 will get 0, and P5 will get 1.
>
> (3) To determine the number of trailing zeroes at the end of 257!,
> observe the prime factorization of 257!. It is quickly noticeable
that
> there are fewer integers between 1 and 257 that are divisble by 5
than
> are divisible by 2. Thus the factorization into prime powers of 257!
> will be of the form 2^n * 5^m * k, where n > m and where k is the
> integer resulting from multiplying all the other factors together.
The
> problem is reduced to finding the value of m, as 2^n * 5^m is equal
to
> 2^(n-m) * 10^m.
>
> There are 51 multiples of 5 between 1 and 257. Of these, 2 are
> divisible by 125 = 5^3 (i.e., 125 and 250). Of the remaining 49, 8
are
> divisible by 25 = 5^2 (i.e., 25, 50, 75, 100, 150, 175, 200, and
225).
> The remaining 41 multiples of 5 between 1 and 257 are not divisible
by
> any power of 5 higher than 5^1. When all of these multiples of 5 are
> multiplied together, their product will be divisible by 5^(2*3 + 8*2
+
> 41*1) = 5^63, and no higher power of 5 will be a divisor of this
> product. Thus, 257! will have 63 trailing zeroes.
>
> (4) August 31, 1380 (08/31/1380) is the most recent palindromic date
> from the second millenium AD. The first two digits of the year must
be
> the reverse of a number for a valid day of a month. Thus any year
> between 1400 and 1999 will not work as there is no 41st, 51st, 61st,
> 71st, 81st, or 91st day of a month. Since the range 1300-1399 is the
> most recent range that works, we must look for months with 31 days,
> namely January, March, May, July, August, October, and December. By
> brute force, it is easy to verify that August gives the date with the
> most recent year.
>
> (5) This can be solved with usage of the Hamming code, a code that
can
> detect and correct single-bit errors. Since integers between 0 and 15
> can be represented with only 4 bits, and since 3 bits will be
> necessary to detect errors and correct them, seven questions are
> necessary. The error correction is important because the person being
> asked is allowed to lie no more than once. The seven questions are
all
> of the form, "Is your number one of the following: _, _, _, _, _, _,
> _, or _?" where the blanks represent a set of 8 integers. The seven
> sets of integers are as follows:
>
> 0, 1, 4, 5, 8, 9, 12, 13
> 0, 1, 2, 3, 8, 9, 10, 11
> 0, 1, 2, 3, 4, 5, 6, 7
> 0, 2, 5, 7, 9, 11, 12, 14
> 0, 2, 4, 6, 8, 10, 12, 14
> 0, 3, 4, 7, 9, 10, 13, 14
> 0, 3, 5, 6, 8, 11, 13, 14
>
> Write down the integers 0-15 inclusive on a piece of paper. Ask each
> question. For each question, if the person answers no, place an x
> under each number contained in the question; if the person answers
> yes, place an x under each number that was NOT contained in the
> question.
>
> After asking the seven questions, there should be one one number with
> no x's underneath; that is the desired number. If the person being
> asked the questions lied, then there will be one number with only 1 x
> underneath; that will be the desired number
>
> (6) Let n be a arbitrarily small but positive real number. The duck
> swims first to the center of the pond, then swims out to a distance
of
> (R/4 - n) from the center of a pond and swims in a circle until the
> duck is ready to begin, where R is the radius of the pond.
>
> The duck then swims around this circle until it is diametrically
> opposed to the cat (i.e., the cat and the duck are on a diameter of
> the pond, and the distance between the cat and the duck is (5R/4 -
n).
>
> The duck then swims the distance n along this diameter AWAY from the
cat.
> If the cat doesn't move, repeat the previous step. If the cat does
> move, then the duck must immediately move in a direction
PERPENDICULAR
> to the direction in which the duck traveled the distance n and
> OPPOSITE the direction in which the cat ran.
>
> (example: say the cat is at the easternmost side of the pond, and the
> duck is on the diameter heading towards the westernmost edge of the
> pond. After swimming the distance n westward, if the cat moves
> counter-clockwise, the duck should immediately turn SOUTH and start
> swimming as fast as possible. If, on the other hand, the cat decided
> to run clockwise, the duck should immediately turn NORTH and start
> swimming as fast as possible.)
>
> The duck is now swimming for a point that is more than halfway around
> the pond from where the cat started. If the cat slows down or changes
> direction such that the cat and the duck are on another diameter of
> the pond, the duck should immediately stop heading towards the edge
of
> the pond, and instead starts swimming around in a circle at this new
> radius. The radius of this new circle is guaranteed to be greater
than
> the previous radius, (R/4 - n) because the duck swam at least n
> farther away from the center of the pond. Along this new circle, the
> duck then swims around until the cat and duck are diametrically
> opposed again, and then starts the process over.
>
> Since the duck never gets closer to the center of the pond, and since
> the cat can never catch the duck until the duck gets to the edge, the
> duck will eventually win.
>
> (7) The difference between the numbers of photinos and gluinos is
> 19-17=2. The difference between the numbers of photinos and gauginos
> is 19-15=4. The difference between the number of gauginos and gluinos
> is 17-15=2. Every time two different particles collide, they
disappear
> and form two new particles of the other kind. This would cause one of
> the three differences to remain the same, another difference to
> increase by 3, and the last difference to decrease by 3. Say a gluino
> and photino collide, forming 2 gauginos. Then the three differences
> are now 18-16=2, 18-17=1, and 16-17=-1. However, if all of the
> particles were going to become of only one kind, then the differences
> would be 51-0=0, 51-0=0, and 0-0=0. As the initial differences were
2,
> 4, and 2, and since the differences are either increased by 3 or
> decreased by 3, there is no possible way for one of the differences
to
> become 0. Therefore, it is impossible to end up with one kind of
> particle.
>
> (8) John Stuart Mill was born in 1806. In 1936, people born in 1892
> were 44 years old, which is the square root of 1936. In 2025, people
> borin in 1980 will be 45 years old, which is the square root of 2025.
> Since John Stuart Mill was born in the 19th century, there is only
one
> year in the 19th century that is a square number: 43^2 = 1849. Thus,
> in the year 1849, Mill would have been 43 years old, which puts his
> year of birth at 1806.
>
> (9) No one gives that hermit food....unless you want to talk about
his
> friend the barber who shaves every man in his village that doesn't
> shave himself. ;) The hermit either keeps food for himself (as a
> person can not "give" himself food) or someone sells him food
(selling
> is not the same as "giving" food for free).
>
> (10) Encrypt your kmail and send it along with a plaintext message
> instructing the recipient to encrypt the message with another
> algorithm and send it back doubly encrypted. Decrypt your level of
encryption of the
> doubly-encrypted message, and then send it again. The recipient can
> now decrypt his own encryption and read the message unencrypted
> completely.
>
> (11) I'm not sure how to do this one. If you extend the contest, I
> will try this one...
>
> (12) This game has a winning strategy (it can not end in a tie) as it
> is stated that you really want to go first. Suppose it is player 2
> that has the winning strategy. Then, in particular, there is a
winning
> reply to the owner placing the first quarter in a certain spot. But
> then the owner should just make that winning reply his first move,
> contradicting the premise that player 2 can always win. Thus, it must
> be player 1 that has the winning strategy.
>
> (13) There are at most 483,207 inhabitants in the Kingdom, because
> there are 483,207 nonnegative integers less than 483,207 (i.e., 0
> through 483,206). If there is an additional inhabitant, then either
> the first statement is violated (there will be two people with the
> same number of hairs), the second statement is violated (there will
be
> a person with 483,207 hairs), or the third statement is violated (the
> population will be less than or equal to the number of hairs on
> someone's head).
>
> (14) 12:04:15 is 255 seconds after midnight, which is 0111 1111 in
> binary. 256 seconds after midnight, the clock should read 12:04:16,
> but if that eight bit from the right can't be turned on, then 1000
000
> (i.e., 256 in binary) will read as 0000 0000, which means that it
will
> be read as 0 seconds after midnight.
>
> When it is 12:08:31, which is 511 seconds after midnight, the bit
> counter should be 0000 1111 1111 (511 in binary), but instead reads
> 0000 0111 111 (255 in binary) which is equivalent to 12:04:15 on the
> clock. However, at 12:08:32, the bit counter will be 0001 0000 0000
> (512 in binary), and the eighth bit from the left won't be used again
> for another 255 seconds, and so the clock will display the correct
> time for the next 255 seconds...until 768 seconds after midnight,
> which will be displayed as 12:08:32 but will actually be 12:12:48.
>
> (15) Nine passengers got off at that last stop.
>
> Let 64x be the number of people on the boat before the Castle.
>
> 64x becomes 16x which becomes (16x+7) people who leave the Castle.
> (16x+7) becomes (4x + 7/4) which becomes (4x + 35/4) who leave this
stop.
> (4x + 35/4) becomes (x + 35/16) which becomes (x + 147/16) who leave
this stop.
>
> Since we will assume that the number of passengers at any time is a
> positive integer (ahem...), we need to find the smallest x such that
> 64x is an integer and such that (x + 147/16) is also an integer. Let
x
> = A/64, where A is an integer. Then:
>
> A/64 + 147/16 = A/64 + 588/64 must be an integer. The smallest
> multiple of 64 that is greater than 588 is 640, which means A = 52.
> Substituting 52/64 in for x, we find that 12 people arrived at the
> last stop, and 9 got off the bus.
>
> (16) It takes at most 100 minutes for the last lemming to drop off
the
> edge. Proceed by induction on n, the number of lemmings. For n=1, put
> the lemming at one end, and have him walk towards the other end. For
> n=2, if the lemmings face the same direction, then it takes at most
100 minutes for the second lemming to drop off (the second lemming must
be at one end and facing the other). If the 2 lemming face each other,
then the farthest either one will walk before bumping into the other
one is 50, and this is only if the lemmings started on opposite ends of
the plank. Let lemming 1 be at D1 and let lemming 2 be at D2, where D1
and D2 are between 0 and 100; without loss of generality, let D1<D2>
> (17) Couldn't get this one either...boo. I had an idea, but it didn't
> seem to work...because I don't see how digging on 3 sides is enough
> info to determine the meat vein location. Call the square ABCD, and
dig two semi-circles such that AB and CD are diameters. If only one hit
on the vein takes place, the vein is tangent to the two semi-circles.
Otherwise, you have two points to determine the location of the
vein....Of course, the total distance dug is PI*(length of one side), which is
more than 3*(length of side)...
>
> (18) Line them up next to each other, and call them A, B, and C. Ask
> A, "Is B more likely to tell the truth than C is?"
>
> If A answers yes, then there are 4 possibilities: TRL, LRT, RLT, or
> RTL, where T stands for the pure truth-teller, L is the pure liar,
and
> R is the random answerer. Notice that in all four of these
> configurations, C is **not** the random guy. Thus your last two
> questions go to this guy.
>
> If A answers no, then the 4 possibilities are: TLR, LTR, RLT, or RLT.
> Here, B is the non-random guy so he will get the last two questions.
>
> The second question should be, "Do you know who A is?" If he says
yes,
> then C (or B, whoever you asked) is a truth teller. Your third
> question will be, "Can A tell the truth ever?" If the answer is yes,
> then A is the random guy, as liars can't tell the truth. If the
answer
> is no, then A is the liar.
>
> If, on the other hand, the second question is answered no, then C (or
> B, whoever you asked) is the pure liar. The third question will then
> be, "Can A ever lie?" If the answer is yes, then A is the
> truth-teller. If the answer is no, A is the random guy.
>
> (19) I've got two possibilities for this one....but in both cases,
the
> baby and his sibling are actually twins. Fraternal or identical, it
> doesn't matter.
>
> A baby is born after the spring equinox (say Easter Sunday). His
> sibling is then born a few minutes later, which would make him
> younger. However, before the winter solstice, the first baby dies,
> thus preventing him from "experiencing winter". As dead people don't
> age, the remaining baby eventually gets older than the baby born
> first.
>
> OR
>
> A baby is born on an air-conditioned airplane after the spring
> equinox. The plane must cross the International Date Line immediately
> after the baby is born, but before the second baby is born. The
second
> baby is born after the first, but because of the IDL, the birthday
> will be one day before the birthday of the first baby.
>
> (20) Nothing existed before the universe, etc.
>
> I almost considered putting a blank entry for this riddle, and then I
> realized I might not get credit for that "answer". :)
>
> (21) see-saw ("...present tense, yet past tense, too...").
>
> Thanks again. Hope you haven't torn out your hair...but seeing as you
> are a CS student, you've seen math beyond calculus, I think. :)
>

[/spoiler]

OneWingedAngel's answers

[spoiler]Haha... I did not know that it was open for another day... this is my
final copy.


*** I am very sorry for my answers for #5, #6, and #18. ***

1) 10 prisoners find their doors open.

2) The solution is that 5 will propose 998 coins for himself and 1
coin
each for 1 and 3.

3) 61 zeros.

4) 08/31/1380

5) With 7 bits, you can have 4 of information (a number between
0 and 15) and 3 of redundance. If at most one of the 7 bits is changed
you can find out which was by the break in the redundance.

1. Is the least significant bit (bit 0) set?
2. Is bit 1 set?
3. Is bit 2 set?
4. Is bit 3 set?
5. Of bits 0, 1, and 2, are an odd number set?
6. Of bits 0, 1, and 3, are an odd number set?
7. Of bits 0, 2, and 3, are an odd number set?

Look at the answers to questions 5, 6, and 7. See if they're
consistent with the answers to questions 1, 2, 3, and 4.

If exactly one of 5, 6, and 7 is inconsistent with 1, 2, 3, 4, that's
the question that was answered incorrectly. 1, 2, 3, and 4 were all
answered correctly, and you can calculate the number.

If more than one of 5, 6, and 7 is inconsistent with 1, 2, 3, or 4,
whichever bit is common to all of the inconsistent "parity" questions
was answered incorrectly. (That is, if all three of 5, 6, 7 are
inconsistent, question 1 was answered incorrectly. If only 5 and 6 are
inconsistent, question 2 was answered incorrectly, and so on)


6) Define our distance scale so that the radius of the circle is 1,
and define time so that the duck's speed is also 1 and cats speed is
4. The duck can swim out to a point 1/4 - x from the center (x is any
small positive number). The duck can swim around this circle faster
than the cat can run around the pond because its circumference is less
than 1/4 the circumference of the pond. This means he can position
himself at the opposite side of this subcircle from the cat by racing
around this inner circle faster than the cat can keep up (he could do
this more efficiently but I'm just saying it can be done). Once he is
opposite the cat, then his distance to the edge is 3/4+x, while the
cat will have to run a distance of pi to make it to that point. As
long as we pick x <pi>.>) escape as long as the cat's speed is not about 4.6 times greater than the duck's speed. As the cat is only 4 times as fast, the duck can reach the perimeter.
Interesting Alternatives:
Note that your puzzle didn't mention that the duck couldn't fly and didn't give a fly speed. A duck should be able to fly faster than it can swim and thus provide a way to escape or reach the perimeter of the pond before the cat.
If the pond was small enough (maybe 1 square foot), then I would imagine that the cat could just reach its paw out and slash the duck before the poor duck could even move. How cruel.

7. Verse:
Tag decided to take some gauginos
And mix them with a few gluinos
But soon there transpired
that which was not desired.
Nothing was left but photinos.

Explanation:
It isn't clear whether gauginos and gluinos can mix freely, for example, if a gluino + gluino collision results in photinos. If they can, then after enough collisions there will be only photinos. Only 16 collisions would need to occur between the two G particles. Example: 15 gaugino + 15 gluinos, 1 gluino + 1 gluino.
If they cannot mix, then it may still be possible for all to become photinos. If a photino + (gaugino or gluino) collision results in photinos, then all may change into photinos.
However, if the only coupling that can occur is gaugino + gluino, then there will be at least 2 gluinos remaining.

8. Verse:
Eighteen-oh-six is a time I hold dear.
John Stuart could shed his first tear.
Comedy show he can't host
But I think he can boast
Of a square in his fourty-third year.

Explanation:
The boast is that the current year is the square of the person's age. So, people born in 1892 were 44 in 1936 and 44^2 = 1936. John Stuart Mill was born in 1806 and in 1849 he could have boasted about 43^2 = 1849.

10. Verse:
To get past the Czar on the fly,
Public key encrypt you might try.
Message in the chat pane,
Or something not in the game
To do better than this is a lie.

Explanation:
Encrypting the message once or a thousand times doesn't matter. The message must be "secure", which presumably means encrypted because an unencrypted message will be "found out" allowing him to "exact retribution." Now, the easiest way of exchanging the message is clearly to do so in person, over an instant-messenger, etc. Alternatively, the decryption key could be exchanged over these non-kmail mediums and then the friend would have the key and the kmail containing the message could be sent. Both of these tactics, although valid, seem to defeat the "spirit" of the riddle.
A better idea is to use public key cryptography. Ask the friend to send you his or her public key (getting one first obviously). Then the message can be encrypted using that key and only the friend could read it. Given that this is how symmetric keys are sent in real systems, this is probably the "best" solution (cryptographically, at least).

11. Verse:

Explanation:
The wording "How can they find out their average wealth without disclosing their own wealth?" is ambiguous. One possibility is that disclosing their own wealth means to find the average wealth of the 3 richest players without telling others. In this case, they just tell each other and calculate the average. Another possibility is that "their average wealth" means each individual's average wealth. Here, the average wealth of the individual is simply the individual's wealth.
One alternate idea is that "without disclosing their own wealth" means that absolutely nothing or no one can see or know that player's wealth except that player. This is absurd because it means that it is impossible to find the average wealth.
The conclusion I've come to is that this disclosure simply means that the rich player cannot tell another person their wealth.
Given this assumption, I forsee 2 options:
First, ask Jick or some other member of Asymmetric to calculate the average wealth. They have not disclosed anything and have found out the average.
Secondly, use a program that calculates an average after three players enter their money and that has no memory and doesn't output to another source. This would allow each player to not "disclose" their wealth to beggars.

12. Verse:

Explanation:
What is not revealed is that they are placing down quarters of a pie and that the last quarter of the pie to be placed down must be eaten. The table I am playing on has only enough room for one pie and thus I will place the first and third quarters. This will leave the owner to place the last quarter, forcing him to eat the pie. Knowing that eating a lot of pies makes one sick and sleepy (plus having added some sleeping draught) I will force him to eat many pies. When he falls asleep I will steal his keys and sneak inside. Victory!

13. Verse:
Jick and his crew can't be glad.
Exact hair limit on people is bad.
The server's exploding.
My page is still loading.
A hairy hippy invasion gone mad.

Explanation:
There can be up to 483207 inhabitants of the kingdom (a filthy lie!). Reasoning is as follows:
All inhabitants have unique number of hairs. So, let inhabitant #1 have 0 hairs, #2 have 1 hair, ..., #483207 have 483206. No one can have 483207 so the fewest hairs #483208 could have is 483208, which is not more than the hairs on the head of any one inhabitant. Thus, the maximum number of inhabitants is 483207.

15. Verse:

Explanation:
A minimum number of passengers is 0. Then, 7 people got on. The next stop saw 5 people leave then 7 get on, for a total of 9. The next stop saw 6 people leave then 7 get on for a total of 10 people. How many got off at the last stop? It depends on whether "last" refers to the previous stop or a final stop. 3/4 people got off at the previous stop (or 6) and we don't know how many got off at the "last" so we can only say we know of 0 that got off.

16. Verse:

Explanation:
The clones are likely to be facing "either" end of the plank, meaning the plank has only 2 ends. Since the plank was "thin" let's say the thin part was maybe 1/3 meter, making the plank 1/3 by 100 meters. Place the lengthwise ends (one every 100 meters) on two sides of the ravine. Now you have a bridge across a ravine, but still satisfying the "either end" (2 ends) of the plank requirement. Now, anywhere you place the clones, the end of the plank they face is up to 1/3 of a meter away. Within the first second of moving, all clones will have fallen off the plank. Since we don't know the fall speed, the longest time that will elapse until all lemmings are dead is 1 second or rather 1/3 of a second.

17. Verse:
There once was a meat farmer, O'Brian,
Who dug for big veins with his lion.
He'd dig sides of three
But then farmer McGee
Said diagonals are all that need tryin'.

Explanation:
Two diagonal cuts (from corner to corner) through his land will find any meat veins. One diagonal cut is enough to find all horizontal and vertical veins that might run exist. However, diagonal veins may not be found with only one. This reduces the number of total feet (meters, whatever) that must be dug to guarantee the vein is found.

18. Verse:
To get to the truth of the matter,
Arrange yetis up by who's fatter.
Ask if liar's position to random
Is immediate, are they in tandem?
And you'll know where isn't the latter.

Then ask non-rand if he's rand
and you'll know which side he has manned.
Query the yeti again (oh brother)
Give the identity of another.
Finally, you know the whole band.

Explanation:
First, note that there are multiple solutions to this. Abbreviations are as follows: R - random yeti, T - truth-only yeti, L - lying-only yeti
Arrange them in a horizontal line. Let #1 be leftmost, #2 be middle, and #3 be rightmost yeti.
1) Ask #1, "Is L to the immediate left of R?"
If yes, the possible positions are (based on who you asked):
T: TLR L:LTR R:RLT RTL
We know now that R cannot be #2.

If no, the positions can be:
T: TRL L:LRT R:RLT RTL
We know now that R cannot be #3.

2) If 1 was yes, ask #2, "Is #2 R?"
If yes, possible positions are: RLT TLR
#2 is L.

If no, possible positions are: RTL LTR
#2 is T.

3) Ask #2, "Is #1 R?"
Since we are asking a yeti we have identified (in question 2), we can determine #1's identity from this answer. Yes means #1 is R if #2 is T, etc. Obviously, the final yeti can only have one identity then. Thus, we know each yeti.

If the answer to question 1 was No, then it is trivial to construct alternative questions. Simply, replace #2 in all further questions by #3 and you identify the yetis.


20. Verse:
Ponder a bit and you'll find,
This riddle to rich is quite kind.
A classical ode
Hints are sowed.
If you can't answer nothing you're blind.

Explanation:
I'm glad this was included in the contest. It is indeed a classic. The answer is "nothing."

21. Verse:
In a game all children pursue.
Weights on each end they accrue.
Lose one pair,
Flip into the air.
The see-saw launching was due.

Explanation:
The answer is "see-saw."
[/spoiler]

Disneeyore1

[spoiler]Great contest quite fun. I attached a word file in case this
gets garbled and so that you can see the picture I got for
#17. Thanks again. Email if you've questions and good luck
checking them all.



1a) Creative Answer: All of them. If one prisoner was free
and saw the jailor collapsed on the floor, wouldn’t he
free everyone else?

1b) Real Answer: 10

1c) Rationale: Following this pattern, the jailor only
leaves perfect squares open because perfect squares are the
only numbers with an odd number of divisors. Thus the only
cells left open are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

2a) Creative Answer: 1000. Leader 5 takes all the bags and
runs. I mean 1000 bags must be worth a mention on the looter
website.

2b) Real Answer: 998 (or 997 if he’s playing it safe).
Either 1 bag to Leaders 2 and 4 and the rest to himself or 1
bag to Leader 4, 2 bags to Leader 2, and 997 bags to
himself.

2c) Rationale: Let’s work backwards.
If there was only 1 leader, he would get 1000 bags.

If there were 2 leaders. Leader 1 knows he can get 1000 bags
so he will vote down whatever proposal Leader 2 gives. Thus
Leader 2 is screwed and expects nothing. Remember 50% of the
vote is still not a majority.

If there were 3 leaders. Leader 1 still wants all 1000 and
Leader 2 will take anything .So Leader 3 can offer 2 a
single bag and he would come out ahead. Thus Leader 3 can
secure 2/3 of the vote and 999 bags of goodies for himself.

If there were 4 leaders. Leader 4 runs into the same issue
that Leader 2 had. Leaders 1 and 3 really can’t be bought
as they have pretty good deals so far. The best Leader 4 can
get is half the vote (by offering Leader 2 a bag or two).
Thus Leader 4 is screwed. Remember 50% of the vote is still
not a majority.

Finally we have all 5 leaders. Leaders 1 and 3 still want
the maximum and will vote down any scheme Leader 5 proposes.
But Leader 5 ignores them. Leader 4 is happy with a single
bag. Leader 2 breaks even with 1 bag and comes out ahead
with two bags. That leaves 998 or 997 bags left for Leader
5. Leader 5 just has to decide whether Leader 2 will accept
an equitable amount now or if he needs a definitively larger
amount.

3a) Creative Answer: 5. “257!” is a big number!!!! See
there that was definitely a 4 exclamation sentence. That
makes five exclamations total. And it isn’t that hard to
demote those periods to zeros…

3b) Real Answer: 51

3c) Rationale: A zero appears at the end after every time
the number is multiplied by a factor of 10 or multiplied by
a number ending in 2 and by a number ending in 5. Thus you
get 2 zeros for every 10 units multiplied. 2 * 25 = 50.
Plus you have both a 2 and 5 in those last seven numbers so
you get to add one more. 50 + 1 = 51.
4a) October 26, 2005- On White Wednesday this caused the
Great Time Catastrophe. At first glance this isn’t a
palindrome and is after the date asked for. But trust me,
it’s true.
4b) Real Answer: December 31, 1321

5a) Creative Answer: Is it 7? Really? Why? Why? Why? In bed?
Ok I love ya bye bye.
5b) Real Answer:
1. Is it greater than 7
2. Is it odd
3. Is it 0, 1, 2, 3, 8, 9, 10, 11
4. Is it 0, 1, 4, 5, 8, 9 12, 13

5. Is 1, 2, 3 true?
6 Is 1, 2, 4 true?
7 Is 1, 3, 4 true?
5c) Rationale: Steps 1-4 solve the number. Steps 5-7 verify
all of the information. If 5-7 are all Yes, then there is no
lying. If there is one no in 5-7, that’s where someone
lied. If there is two or three no’s, then you can pinpoint
which of 1-4 is the lie.
6a) Creative Answer: Yes. It is a duck. It has wings. It can
fly. Unless the cat had wings too. But then it would be a
Kitty Hawk…
6b) Real Answer: No

6c) Rationale: The best case scenario for the duck is for it
to be at the midpoint of the circle and to swims to the
opposite point of where the cat was located. This means the
duck must go the radius of the circle. The cat must run half
the circumference of the circle or 2pi r/2 or the radius
times pi. Canceling out the radiuses and realizing that the
cat is 4 times faster, we see that 1> pi/4. Thus the cat
will always be there before the duck.

7a) Creative Answer: Tagliatelle’s head explodes once he
realizes that 45 != 51.

7b) Real Answer: No

7c) Rationale: To have only one particle, you must at some
point have two particles equate to each other. Given that
x+y=2z, y+z=2x, and x+z=2y, you form a circular logic.
So if you try and shift them around you either spin on
particle significantly away from the middle or oscilate from
15, 17, 19 to 16, 17, 18 and back again.

8a) Creative Answer: John Stuart was born November 28,
1962. He now hosts the Daily Show.

8b) Real Answer: 1806

8c) Rationale: These are all dates where people can reach
an age that is the square root of their year. John Stuart
Mill turned 43 which was his own square of 1806.

9a) Creative Answer: Otori does. Where did you think all
those buffs went?

9b) Real Answer: Either you can reason it as that one hermit
buys his food (thus it is not free or given) or that he
receives it from a source from outside of the hermitage (and
thus does not fall under its jurisdiction or rule system.)

10a) Creative Answer: You tell them in person or outside of
KOL. If the eye sees that, then I’d be Really worried in
the bathroom.

10b) Real Answer: Set a pattern within the encryptions. IE
First letters spell something, or Morse Code using time, or
even a pattern of the Fibonacci Sequence. Just make it overt
enough to be understood but not a blatant answer that
MultiCzar could interpret.

11a) Creative Answer: All three players realize that Jick
actually has all of their meat totals. In an effort to
protext their secret, they all decide to meet and kill Jick.
After an interesting episode involving Badgers on a
Submersible Vehicle, the dirty deed is done. Then they all
realize that no more updates will be forthcoming and that
their hard-earned meat is meaningless. Their average meat
total then becomes moot.

11b) Real Answer: Each player kmails one player a random
percent of their wealth and kmails the other 100- that
random percent of their wealth. Each player then adds up
those two numbers and all present their sum. The three new
sums can then be averaged to find their meat average.

12a) Creative Answer: You accidentally roll a quarter
towards the card game. As the owner chase off after it, you
pocket the money and walk inside.

12b) Real Answer: Going first always gives you the
initiative you can prove how to win from the end and work
backwards to always find the winning number. IE: A quarter
can block anywhere from 1 to 4 spots. If there are four
spots open, you can place it in the middle and win. So if
the owner is given 5 openings to start with, you win. So
long as the sum total of your play and the owners totals 9,
you can reach this point by ensuring on you first move that
that the number of openings available to the owner on his
more = 9n +5 where n is any whole number.

13a) Creative Answer: 1. With any larger number, it would be
impossible for people to pull out their hair whenever nerfs
get rolled out.

13b) Real Answer: 483,207

13c) Rationale: The first player can have a bald head. Every
nth player after that can have n-1 hairs. If none can have
483,207 hairs, then their can be no 483,208 inhabitant.

14a) Because the machines are taking over. My gosh, you’d
think with the robotic takeover that you’d notice…

14b) The 9th wire for the binary code appears faulty.
That’s why it keeps resetting. The counter counts normally
but clock shows time with 9th bit always unset.

15a) Creative Answer: One chicken. The washing machine ate
my SOCK and that was all that was left in the quantum egg.

15b) Real Answer: 52

15c) Rationale: 52 *.25 = 13 +7 = 20 *.25 = 5 +7 =
12 *.25 = 3
Anything smaller starts using fractions of people.

16a) Creative Answer: That depends. If they played a turn
then its an hour. Otherwise it’s permanent. That’s long
enough to even bother Anakin.

16b) Real Answer: 100 minutes.

16c) Rationale: Eventually you can make the intuitive leap
that when two clones bump, it is just the same as if the
passed each other. So long as one clone is on the edge of
the plank facing the other end, it doesn’t matter if you
have 1 or 100 clones. Regardless of the number of clones,
the last one will walk 100 meters in 100 minutes and then
fall.

17a) Creative Answer: He goes out and get a leprechaun with
a meat detector. Look ma, no digging at all!

17b) Real Answer: This will be a picture as it is difficult
to explain. Think 2sqrt(2 times side length)


18a) Creative answer: Why did you leave the Icy Peak? Where
would you take the Ice Queen if you went on a date? Where
did you get all those penguin skins?

18b) Real Answer: Ask two of them: Considering the other two
yetis, would one always tell the truth and the other always
lie? Both the truth and the liar cannot answer. Thus the one
that spoke or the third yeti is the switcher. Then ask any
non-switching yeti: Does 2 + 2 equal 4. This should easily
determine the nature of that yeti and thus the whole group
is revealed.

19a) Creative Answer: Babes on a Plane!! You know very well
what I meant; you’ve a very dirty mind.

19b Real Answer: Two babies were born on a moving object
that was crossing time zones in a hemisphere that was not
experiencing winter. The first could be born at 1:05 and the
next would be born at 12:06, because he crossed a time zone.

20a) Creative Answer: Otori of course. You are old,
powerful, and help the needy.

20b) Real Answer: Nothing. Fun fact- 75%+ of 1st graders
answer this correctly but less than 10% of college student
get it correct.

21a) Creative Answer: Mr A mall prices

21b) A see-saw.

[/spoiler]
Last edited by Slashor on Sun Sep 24, 2006 6:04 am, edited 1 time in total.
Image
User avatar
Slashor
Friend
Friend
 
Posts: 4123
Joined: Thu Mar 24, 2005 8:25 am

Postby Slashor on Sun Sep 24, 2006 5:53 am

Here are my final set of answers. I do know some answers were easily googeable but I wanted it like that so some people would have a chance of at least getting one answer correct. Suprisingly some people didn't get all those easy ones correct. Next time I am just going to have riddles with small answers so I can make this fast. Reading different logic patterns whilst fun takes way too long. If there are still queries please raise them. I also did generally take the best answers I found and put them in for each question. In some cases there are multiple answers or variants on the same answer and I did not list all of them although I highlighted the case in some riddles.

[spoiler]1) There is a circular jail with 100 cells numbered 1-100. Each cell has an inmate and the door is locked. One night the jailor gets drunk and starts running around the jail in circles. In his first round he opens each door. In his second round he visits every 2nd door (2, 4, 6…) and shuts the door. In the 3rd round he visits every 3rd door (3, 6, 9…) and if the door is shut he opens it, if it is open he shuts it. This continues for 100 rounds and exhausted the jailor falls down. How many prisoners found their doors open after 100 rounds?

Answer:
On the nth round, a door's state will be changed if and only if n is a divisor of the number on the door. As only square numbers have an odd number of divisors, only the doors with square numbers on them will be open, since an open door implies an odd number of state changes.

Therefore, only 10 doors out of the 100 will be open after the jailor decides to call it quits.

2) Otori loots 1000 goodie bags from another clan. The bags have to be split among the 5 leaders: 1, 2, 3, 4, and 5 in order of rank. The leaders have the following important characteristics: infinitely smart, bloodthirsty, greedy and stunningly attractive. Starting with leader #5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the leader is thrown into /newbie. A proposal is accepted if and only if a majority of the leaders agrees on it. What proposal should leader 5 make?

Answer:
This riddle had one logical error in that if there is a tie it was not stated who would win that. As a result there could be two different answers. Either way an answer that involved two players getting either 1 or 2 goodie bags whilst you get 997 or 998 bags would win. Technically it would have to be 1 and 2 goodie bags split as they are bloodthirsty but I was lazy in marking and missed the problem in answers before I could effectively change my marks to that person. The answer below is the most correct one by DirkDiggler and recognises that issue with this riddle.

If it comes down to the last two clannies, #1 will vote down any proposal put forth by #2: in doing so, #2 gets booted and #1 gets all the loot (let's call this result 1000, 0, 0, 0, 0). So, #2 *must* accede to any proposal put forth by #3 which benefits him. (Note that #3 can put forth a proposal that does benefit #2, and that #2 is infinitely bloodthirsty. Therefore, #2 will punitively reject anything #3 puts forth that doesn't leave #2 at least mildly better off: ties are voted down). This means that if the vote reaches #3, he can put forth (0, 1, 999, 0, 0); #2 and #3 will vote for it (as otherwise they each get nothing), and the vote will pass.

If the vote does get to #3's proposal, then 0 and 1 are the most meat that #2 and #1 can possibly see. This means that #4 has only to better these paltry amounts and he will have their votes, disenfranchising #3. If #4 proposes (1, 2, 0, 997, 0): . #2 must vote for it (otherwise #3 will propose and #2 will only get 1 goody bag). #1 must vote for it (otherwise #3 will propose and #1 will get nothing). #4 is all for it, obviously.

So #5 just has to make two other players happier than they would be if #4 were running the show. The proposal (2, 0, 1, 0, 997) is better for #3 and #1 by one goody bag each than what will happen if the proposal fails; this proposal passes with votes from #5, #3, and #1. (Note again that #5 has to sweeten the deal for #3 and #1 over the alternative: if his proposal only ties then #3 and #1 will retributively nix it).


3) 257! contains how many trailing zeroes?

Answer:
To determine the number of trailing zeroes at the end of 257!, observe the prime factorization of 257!. It is quickly noticeable that there are fewer integers between 1 and 257 that are divisible by 5 than are divisible by 2. Thus the factorization into prime powers of 257! will be of the form 2^n * 5^m * k, where n m and where k is the integer resulting from multiplying all the other factors together. The problem is reduced to finding the value of m, as 2^n * 5^m is equal to 2^(n-m) * 10^m.

There are 51 multiples of 5 between 1 and 257. Of these, 2 are divisible by 125 = 5^3 (i.e., 125 and 250). Of the remaining 49, 8 are divisible by 25 = 5^2 (i.e., 25, 50, 75, 100, 150, 175, 200, and 225). The remaining 41 multiples of 5 between 1 and 257 are not divisible by any power of 5 higher than 5^1. When all of these multiples of 5 are multiplied together, their product will be divisible by 5^(2*3 + 8*2 + 41*1) = 5^63, and no higher power of 5 will be a divisor of this product. Thus, 257! will have 63 trailing zeroes.

4) "October 2, 2001" in MMDDYYYY format is a palindrome. When was the last date before October 2, 2001 that is also a palindrome?

Answer:

August 31, 1380 (08/31/1380) is the most recent palindromic date from the second millennium AD. The first two digits of the year must be the reverse of a number for a valid day of a month. Thus any year between 1400 and 1999 will not work as there is no 41st, 51st, 61st, 71st, 81st, or 91st day of a month. Since the range 1300-1399 is the most recent range that works, we must look for months with 31 days, namely January, March, May, July, August, October, and December. By brute force, it is easy to verify that August gives the date with the most recent year.

5) I am thinking of an integer between 0 and 15 inclusive. To figure out what number I'm thinking of, you can ask me 7 yes-or-no questions. The questions must be independent of each other, their answers, and the order in which they are answered. When you ask me your seven questions, I am allowed to lie about at most one of the answers. What seven questions can you ask to determine n? (NOTE: You are not actually asking me questions but sending what questions you would ask.)

Answer:

This can be solved with usage of the Hamming code, a code that can detect and correct single-bit errors. Since integers between 0 and 15 can be represented with only 4 bits, and since 3 bits will be necessary to detect errors and correct them, seven questions are necessary. The error correction is important because the person being asked is allowed to lie no more than once. The seven questions are all of the form, "Is your number one of the following: _, _, _, _, _, _, _, or _?" where the blanks represent a set of 8 integers. The seven sets of integers are as follows:

0, 1, 4, 5, 8, 9, 12, 13
0, 1, 2, 3, 8, 9, 10, 11
0, 1, 2, 3, 4, 5, 6, 7
0, 2, 5, 7, 9, 11, 12, 14
0, 2, 4, 6, 8, 10, 12, 14
0, 3, 4, 7, 9, 10, 13, 14
0, 3, 5, 6, 8, 11, 13, 14

Write down the integers 0-15 inclusive on a piece of paper. Ask each question. For each question, if the person answers no, place an x under each number contained in the question; if the person answers yes, place an x under each number that was NOT contained in the question.

After asking the seven questions, there should be one number with no x's underneath; that is the desired number. If the person being asked the questions lied, then there will be one number with only 1 x underneath; that will be the desired number.

6) A cat chases a duck into a circular pond. The cat can run four times as fast as the duck can swim, but cannot enter the water. Can the duck get to the perimeter of the pond without the cat being on top of him?

Answer:

Let n be an arbitrarily small but positive real number. The duck swims first to the centre of the pond, and then swims out to a distance of (R/4 - n) from the centre of a pond and swims in a circle until the duck is ready to begin, where R is the radius of the pond.

The duck then swims around this circle until it is diametrically opposed to the cat (i.e., the cat and the duck are on a diameter of the pond, and the distance between the cat and the duck is (5R/4 - n).

The duck then swims the distance n along this diameter AWAY from the cat. If the cat doesn't move, repeat the previous step. If the cat does move, then the duck must immediately move in a direction PERPENDICULAR to the direction in which the duck travelled the distance n and OPPOSITE the direction in which the cat ran.

(Example: say the cat is at the easternmost side of the pond, and the duck is on the diameter heading towards the westernmost edge of the pond. After swimming the distance n westward, if the cat moves counter-clockwise, the duck should immediately turn SOUTH and start swimming as fast as possible. If, on the other hand, the cat decided to run clockwise, the duck should immediately turn NORTH and start swimming as fast as possible.)

The duck is now swimming for a point that is more than halfway around the pond from where the cat started. If the cat slows down or changes direction such that the cat and the duck are on another diameter of the pond, the duck should immediately stop heading towards the edge of the pond, and instead starts swimming around in a circle at this new radius. The radius of this new circle is guaranteed to be greater than the previous radius, (R/4 - n) because the duck swam at least n farther away from the centre of the pond. Along this new circle, the duck then swims around until the cat and duck are diametrically opposed again, and then starts the process over.

Since the duck never gets closer to the centre of the pond, and since the cat can never catch the duck until the duck gets to the edge, the duck will eventually win.

7) Tagliatelle created a mini-universe that was made up of 51 (45 was improperly stated) particles of three different types: 15 are gaugino’s, 17 are gluino’s and 19 are photino’s. If two particles (gaugino and gluino) collide the both change to the third (photino). Is it possible for there to be only one type of particle after enough collisions?

Answer:

The difference between the numbers of photinos and gluinos is 19-17=2. The difference between the numbers of photinos and gauginos is 19-15=4. The difference between the number of gauginos and gluinos is 17-15=2. Every time two different particles collide, they disappear and form two new particles of the other kind. This would cause one of the three differences to remain the same, another difference to increase by 3, and the last difference to decrease by 3. Say a gluino and photino collide, forming 2 gauginos. Then the three differences are now 18-16=2, 18-17=1, and 16-17=-1. However, if all of the particles were going to become of only one kind, then the differences would be 51-0=0, 51-0=0, and 0-0=0. As the initial differences were 2, 4, and 2, and since the differences are either increased by 3 or decreased by 3, there is no possible way for one of the differences to become 0. Therefore, it is impossible to end up with one kind of particle.

8) Back in the year 1936, people born in 1892 were able to make an unusual mathematical boast, a boast that people born in 1980 will be able to make at some time during the 21st century. John Stuart Mill, the English philosopher and economist, would also have been able to make the same boast, had he noticed it. Given that he was born in the 19th century, can you tell me which year?

Answer:


1806. In 1936, people born in 1892 had ages (44) that were the square root of the current year. For Mill, this happens in 1849. For the people in 1980, it happens in 2025.

9) Somewhere in the big mountains there is a very special cult of hermits. Within this cult there is one hermit who gives away free food. This hermit gives food to everyone who doesn’t give out food. Who gives that hermit food?

Answer:

No one gives that hermit food....unless you want to talk about his friend the barber who shaves every man in his village that doesn't shave himself. ;) The hermit either keeps food for himself (as a person can not "give" himself food) or someone sells him food (selling is not the same as "giving" food for free).

Pakwan Kenobi again was the only one to get this one properly. I did make a massive blunder in this riddle but he was the only one who at least stated where I originally got this riddle from to modify. Look up the barber paradox for a proper version of this riddle.

10) You want to send a kmail to a friend without the Multi Czar finding out. You have found a way to encrypt your message multiple times and have the password key to unlock each encryption method. However, your friend does not have any of the keys and if you send a message without encryption the Czar will find out and enact retribution on you. How can you get the message securely to your friend?

Answer:

Encrypting the message once or a thousand times doesn't matter. The message must be "secure", which presumably means encrypted because an unencrypted message will be "found out" allowing him to "exact retribution." Now, the easiest way of exchanging the message is clearly to do so in person, over an instant-messenger, etc. Alternatively, the decryption key could be exchanged over these non-kmail mediums and then the friend would have the key and the kmail containing the message could be sent. Both of these tactics, although valid, seem to defeat the "spirit" of the riddle.
A better idea is to use public key cryptography. Ask the friend to send you his or her public key (getting one first obviously). Then the message can be encrypted using that key and only the friend could read it. Given that this is how symmetric keys are sent in real systems, this is probably the "best" solution (cryptographically, at least).

Another way to say it is as below:

Encrypt your kmail and send it along with a plaintext message instructing the recipient to encrypt the message with another algorithm and send it back doubly encrypted. Decrypt your level of encryption of the doubly-encrypted message, and then send it again. The recipient can now decrypt his own encryption and read the message unencrypted completely.

11) The three richest players in the kingdom would like to know their average wealth. However they are secretive and will not tell each other their total wealth for fear of newbies begz0ring them to death. How can they find out their average wealth without disclosing their own wealth?

Answer:

There were several different answers that made it in to this as this was one of the questions I didn’t require a set out answer as there are many ways to do it. The best answer I found was this one below.

Order the players by player ID. The lowest only talks to the middle, who only talks to the highest, who only talks to the lowest. Each player rolls 1d20, waits between zero and that many minutes, and then -- unless contacted -- starts the process by telling his neighbour an arbitrary amount of meat between 0 and his total stash. The second respondent similarly determines an arbitrary amount of meat (between 0 and his total stash), adds that to the total, and sends the result along to the third respondent, who does the same. Now that we're back to the initiator, who adds an arbitrary amount of meat between 0 and his (total, less what he has already disclosed), with the additional restriction that if the same total comes around twice, and he has not volunteered all of his meat, he should disclose something. Each player continues in turn, incrementing the total by some hidden amount, until all have volunteered their total wealth. (If the same total circulates for three circuits everyone is all-in). Their average wealth is then one-third the final total.

This works because each player knows only the total wealth disclosed so far by the other two players. He has no way of ascribing that wealth to either player individually: say his neighbour whispers "Two million" to him, and he put in one million of -- the distribution could be (1,000,000 and 0), (500,000 and 500,000), or anything else. All that is known at any point to a player is the amount they themselves have disclosed and the total that has been collectively disclosed.

The randomized start is to prevent the no-cycles rule from revealing information. Imagine that there is one meat in the kingdom, I knew who started, and I sat second. If the total came around twice as 'zero' I'd know that the player sitting third had the meat. If, however, I didn't know which of the other two players started I couldn't know if it had come around fully twice yet. (It's easy to resolve collisions -- one player will simultaneously send and receive a message -- and in that case you restart the process).

12) You finally ascend again for the umpteenth time but there was a problem, you must face a new challenge. The owner of the Vahalladay Inn will not let you inside until you beat him in a game. He gives you a huge pile of quarters and tells you that you and he will take turns putting quarters down without overlapping and resting face down. You are just about to begin when the owner says he will start first. You interject immediately and ask if you can go first as that will guarantee you victory. Explain how this is so.

Answer:


13) In the Kingdom of Loathing, the following facts are true:

- No two inhabitants have exactly the same number of hairs.
- No inhabitant has exactly 483,207 hairs.
- There are more inhabitants than there are hairs on the head of any one inhabitant.

What is the largest possible number of inhabitants of the kingdom?

Answer:

There are at most 483,207 inhabitants in the Kingdom, because there are 483,207 nonnegative integers less than 483,207 (i.e., 0 through 483,206). If there is an additional inhabitant, then either the first statement is violated (there will be two people with the same number of hairs), the second statement is violated (there will
be a person with 483,207 hairs), or the third statement is violated (the population will be less than or equal to the number of hairs on someone's head).

14) My fancy new digital alarm clock is broken! The time 'jumps' around.

When I reset it, it reads 12:00:00. Then it runs as it should, but after 12:04:15 it resets back to 12:00:00. It counts up to 12:04:15 again and then it jumps to ... 12:08:32! Do you know what's wrong with my alarm clock?

Answer:


The wire for the 9th bit is broken. The numbers in the circuitry are represented in binary (base-2). An 8 bit counter counts from 0 to 255 and if increased beyond that it will reset to 0. At exactly 12:04:15, 255 seconds have elapsed. The next second forces the 8-bit counter to reset to 0 (12:00:00 on the clock). Since a 12 hour clock has 16 bits, however, the problem must occur after the 8th bit. Since the clock jumps ahead to 12:08:32, the 10th bit must be working and receiving data from the 9th bit, meaning the problem lies somewhere with the connection with the 9th bit.

15) The intragalactic rowboat had a minimal number of passengers. When it arrived at the Castle, 3/4 of the passengers got out, and 7 people got on. At the next two stops, the same thing happened. How many got off at the last stop?

Answer:

Let 64x be the number of people on the boat before the Castle.

64x becomes 16x which becomes (16x+7) people who leave the Castle. (16x+7) becomes (4x + 7/4) which becomes (4x + 35/4) who leave this stop. (4x + 35/4) becomes (x + 35/16) which becomes (x + 147/16) who leave this stop.

Since we will assume that the number of passengers at any time is a positive integer (ahem...), we need to find the smallest x such that 64x is an integer and such that (x + 147/16) is also an integer. Let x = A/64, where A is an integer. Then:

A/64 + 147/16 = A/64 + 588/64 must be an integer. The smallest multiple of 64 that is greater than 588 is 640, which means A = 52. Substituting 52/64 in for x, we find that 12 people arrived at the last stop, and 9 got off the bus.

16) Somewhere in the big mountains, 20 YaYaMcG clones are planning a group suicide. Each of the clones will be placed in a random position along a thin, 100 meter long plank of wood which is suspended above a large ravine. Each clone is equally likely to be facing either end of the plank. When the bell rings the lemmings all walk forward at a slow speed of 1 meter per minute. If a clone bumps into another clone, they both reverse directions. When a clone reaches the end she falls to her death. What is the longest time that must elapse till all the lemmings have died?

Answer:

Several n00bs used lemmings instead for this which cost them.

It takes at most 100 minutes for the last lemming to drop off the edge. Proceed by induction on n, the number of lemmings. For n=1, put the lemming at one end, and have him walk towards the other end. For n=2, if the lemmings face the same direction, then it takes at most 100 minutes for the second lemming to drop off (the second lemming must be at one end and facing the other). If the 2 lemming face each other, then the farthest either one will walk before bumping into the other one is 50, and this is only if the lemmings started on opposite ends of the plank. Let lemming 1 be at D1 and let lemming 2 be at D2, where D1 and D2 are between 0 and 100; without loss of generality, let D1<D

17) A meat farmer has a square plot of land. One day he finds out that a long meat vein passes under his land. It is a well known fact that meat veins run in straight lines parallel to the ground. He decides that if he digs on 3 sides he will find the location on the vein and harvest it. His friend tells him he is on the right track but he can do better. What solution does the farmer’s friend have in mind?

Answer:

There were a few answers for this one as well. If it was well thought out I gave points.

I figure that if you make an "x" shape, you'll hit it at some point, also. However, if you want to dig up the least land, you want a shape that consists of a straight line of some length running through the centre of the plot and then ending symmetrically at two points. Then, dig from the four corners to those endpoints of the line, two corners to an endpoint. The result looks a bit like >-<but>-< with 0.42 on the horizontal section is the best solution I can think of at the moment.

18) Of three yetis, one yeti always tells the truth, one always tells lies, and one answers yes or no randomly. Each yeti knows which yeti is who. You may ask three yes/no question to determine who is who. If you ask the same question to more than one person you must count it as question used for each person whom you ask. What three questions should you ask?

Answer:

I really couldn’t have been stuffed marking this one so if it looked good you got a mark. In fact I am only assuming this answer is correct because it looks all scary with the steps and everything.

Truth Lying Lying Random
Yeti Yeti Yeti
----- ----- ------
I A B C
II A C B
III B A C
IV B C A
V C A B
VI C B A


Step 1: Ask A, "Is B more likely to tell the truth than C?" If yes go to step 2, if no go to step 5.
Step 2: Ask C, "Are you the random guy?" If yes go to step 3, if no go to step 4.
Step 3: Ask C, "Is A the truth guy?" If yes then scenario IV, if no then scenario II.
Step 4: Ask C, "Is A the lying guy?" If yes then scenario V, if no then scenario VI.
Step 5: Ask B, "Are you the random guy?" If yes then step 6, if no then step 7.
Step 6: Ask B, "Is A the truth guy?" If yes then scenario VI, if no then scenario I.
Step 7: Ask B, "Is A the lying guy?" If yes then scenario III, if no then scenario IV.

19) My younger sibling is older than me and I have never experienced winter.

Why is this so?

Answer:

My only original riddle and it stumped a lot of people. Several answers had something involving time zones and planes which would account for the sibling being older and several somehow accounted for not experiencing winter but not many people put that together or even did it well. This haiku answer was the only answer that was actually what I was thinking although a few others got it correct.

Born on February 29
Change hemispheres every fall
Sibling higher age


20) I existed before the universe.
I am more powerful than God.
I am older than time itself.
Rich people want me, yet poor people have me.

What am I?

Answer:

“I almost considered putting a blank entry for this riddle, and then I
realized I might not get credit for that "answer". :)” - Pakwan Kenobi

21) What goes up, but at the same time comes down, up to the sky and down to the ground, my name present tense yet past tense too.

Answer:

See-saw
[/spoiler]
Image
User avatar
Slashor
Friend
Friend
 
Posts: 4123
Joined: Thu Mar 24, 2005 8:25 am

Postby pakwan on Sun Sep 24, 2006 12:07 pm

I hated lemmings anyway. ;) They always cropped up at the most inopportune times. I wish there had been a button available so that those lemmings would just scream, "Oh, no!" in unison and then explode leaving no trace behind.

Congrats to all those who won prizes, even if it turns out that due to miscount I am not one of them. I had fun answering the riddles. Also, a big thank you to Slashor for slogging through a ton of entries that must have all looked the same after a while. I sympathize completely with the deal about school and work taking up more time than you would want.

Oh, and I am :oops: since Slashor credited me with a few of his answers listed above.

Even though I'm not a member of Otori, I'm glad no one has flamed me for trying to sound like a smart-ass on these forums. Thanks, I'll be lurking more here in the future.
pakwan
Visitor
Visitor
 
Posts: 10
Joined: Wed Aug 30, 2006 11:48 am

Postby Tagliatelle on Sun Sep 24, 2006 3:47 pm

Even though I'm not a member of Otori, I'm glad no one has flamed me for trying to sound like a smart-ass on these forums.


On the contrary, it was a pleasure hearing your views on the riddles. I look forward to seeing you about the place!
Image
User avatar
Tagliatelle
Shogun
Shogun
 
Posts: 3028
Joined: Fri Mar 18, 2005 3:05 am

Postby Shadari on Wed Sep 27, 2006 4:13 am

I still dont get the sibling one....

Explain in non-haiku?
Shadari
Visitor
Visitor
 
Posts: 7
Joined: Wed Aug 30, 2006 12:42 pm

Postby Slashor on Wed Sep 27, 2006 8:58 am

19) My younger sibling is older than me and I have never experienced winter.

Why is this so?

Answer:

My only original riddle and it stumped a lot of people. Several answers had something involving time zones and planes which would account for the sibling being older and several somehow accounted for not experiencing winter but not many people put that together or even did it well. This haiku answer was the only answer that was actually what I was thinking although a few others got it correct.

Born on February 29
Change hemispheres every fall
Sibling higher age


As you were born on the 29th of February your birtday only comes around once every four years which accounts for a sibling having a higher age than you. To never experience winter you just have to swap hemispheres every Autumn. I can explain it in greater depth but hopefully that is enough.
Image
User avatar
Slashor
Friend
Friend
 
Posts: 4123
Joined: Thu Mar 24, 2005 8:25 am

Postby Slashor on Mon Oct 02, 2006 1:38 am

All my major prizes are sent out now. I will be sending out consolation prizes to everyone who didn't get one as soon as I can't a friend online.
Image
User avatar
Slashor
Friend
Friend
 
Posts: 4123
Joined: Thu Mar 24, 2005 8:25 am

Previous

Return to Otori Week II: The Rise of the Machines

Who is online

Users browsing this forum: No registered users and 1 guest

cron