I hadn't realized that so many people had sent me PMs already ;) Sorry for the delay on the answers. So far there are two correct (and by the way optimal) answers from Goodz13 and JeffB. Sorry Jeff, but Goodz sent his PM a full minute earlier ;)
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I hadn't realized that so many people had sent me PMs already ;) Sorry for the delay on the answers. So far there are two correct (and by the way optimal) answers from Goodz13 and JeffB. Sorry Jeff, but Goodz sent his PM a full minute earlier ;)
"The last man on earth sat alone in his room. Suddenly there was a knock at the door!"
Can you change one word in the first sentence to make the man's isolation before the knock at the door more complete?
pm me with answers. I will post the answer in a few days.
JeffB is requesting answer and explanation to that &?$#%? 80 coins riddle... :D
JeffB
Actually, for simplicity say we have 81 coins.
Then we can divide this neatly into three piles of 27 coins. We can weigh two of these piles against each other. If they weigh as much, the third remaining pile contains the false coin, otherwise the lighter pile contains the coin. Now we have 27 coins where we can use the same procedure again.
We have one operation to select a pile of 27 out of 81 coins.
We have one operation to select a pile of 9 out of 27 coins.
We have one operation to select a pile of 3 out of 9 coins.
We have one operation to select a pile of 1 out of 3 coins.
These are 4 operations. With 80 coins we use the same procedure but always use one coin less in the pile that is not used on the balance.
It took me a couple of seconds to realize 80 is close to 81 which is 3 to the fourth power. Exactly 81 coins would be a bit suspicious.
In free time visit this:
http://www.loveis.ru/games/pearl2/index.html
and play this logical game.
(Main site by russion but game on english. Look at center of page for big black square with man).
Enjoy!!! :D
And what about common strategy for this game. Is it possible :confused:
I trying to find it now.
Ok, here is the solution to the dwarf battles :)
So there are 16 moves. The first correct answer was by Goodz and JeffB (1 minute apart) Then a correct one by dimm_coder. Gabriel was getting closer and closer, but he kept on going backwards ;)Code:01 OOO_XXX
02 OOOX_XX
03 OO_XOXX
04 O_OXOXX
05 OXO_OXX
06 OXOXO_X
07 OXOXOX_
08 OXOX_XO
09 OX_XOXO
10 _XOXOXO
11 X_OXOXO
12 XXO_OXO
13 XXOXO_O
14 XXOX_OO
15 XX_XOOO
16 XXX_OOO
Actually Jeff raised an interesting idea. It *seems* that as soon as you make a move which is not in the line with the optimal solution, you get stuck. So every valid solution will be in 16 moves (and there are actually only two valid solutions, one starting OOOX_XX and the other OO_OXXX)
Actually, it do not seems, it is. Any non-conform move will prevent the dwarf to pass one on top of another, but that is only if they CAN'T step back. I've just done a small VB program (for fun), and it is easy to try and see that there is only 2 solution, but it's still an interesting "mathematical" riddle.Quote:
Originally posted by Yves M
Actually Jeff raised an interesting idea. It *seems* that as soon as you make a move which is not in the line with the optimal solution, you get stuck. So every valid solution will be in 16 moves (and there are actually only two valid solutions, one starting OOOX_XX and the other OO_OXXX)
Oh and by the way, there is 15 moves... since the first is not really a move... :)
Which riddle left now... the man alone, ... any others??! :)
JeffB
Here's the intended answer:Quote:
"The last man on earth sat alone in his room. Suddenly there was a knock at the door!"
"The last person on earth..."
thereby eliminating the possibility that his wife or daughter or mom or someone was knocking on the door.
I got answers from Gabriel Fleseriu and JeffB at exactly the same time! Only, JeffB's answer:
"The last man on universe..."
doesn't make sense grammatically in English.
Gabriel is the only one who sent the correct answer, stating it as:
"The last human on earth..."
which is equally valid.
Here's a new problem to have fun with:
Name something money can't buy.
I have one specific answer in mind, but I'm sure there are really hundreds. pm me.
;) :) :) :D :cool:
Oh thanks to "tell others" :D :DQuote:
Originally posted by solarflare
Only, JeffB's answer:
"The last man on universe..."
doesn't make sense grammatically in English.
What exactly doesn't make sense? Is universe not a word in english?? :confused: in universe? of the universe, oh yes, it is of the, it is what Altavista Translation software just tell me... nevermind :D :D
JeffB - Thanks to not read that post :)
I believe it would be "the last man in the universe..."Quote:
Originally posted by JeffB
What exactly doesn't make sense?
cout << endl;
;) :)
Incorrect, get your Chrisitianity straight before you try to speak.Quote:
Originally posted by Simon666
Thus, if there is a God, you will in the best case (Christianity) send 0.67*10.000.000.000 people = 6.700.000.000 people to h*ll, just because they do not recognize or haven't been told about God.
Figured the riddle without reading the entire 11 or 12 pages of this. 3 hats, 2 white, 5 men, 4 have black hats, doesn't add up.
Temple,Quote:
Originally posted by temple
Incorrect, get your Chrisitianity straight before you try to speak.
religious discussions are likely to produce more heat than light. This comment of yours is rather on the rude side, in my oppinion. It is essential to understand that it is perfectly ok to fight an idea in the thread of the discussion, but it's not allowed to start a flamewar against anyone. I thought I state this clearly before it happens.
That were 3 men, 5 hats (3 black and 2 white).Quote:
Figured the riddle without reading the entire 11 or 12 pages of this. 3 hats, 2 white, 5 men, 4 have black hats, doesn't add up.
Ok, we really need a new riddle here ;)
I'll do a variation on the coins problem.
You have 12 coins, one of which is a false coin and has a different weight from the other ones. You don't know whether it's lighter or heavier. You can only use scales to weigh them (so you'll know whether one pile is heavier or lighter than the other one, or whether they have the same weight).
Now, by using the scales at most three times, find out which coin is the false one and whether it's lighter or heavier than the other ones.
As always, if you think you have the solution, send a private message to me. If you want to discuss the problem, you can do it here ;)
Thou have to send a message to Yves M. because thou want to answer a riddle. Thou have a headache by trying to explain the answer and the following message (see attachement) appears as you finally click "Send message button", what to do?
JeffB:D