Try to solve this interesting logical task!!!

Ok guys!
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There are 3 duellists:
John, which can kill enemy(other duellist) with 100% probability,
Garry - 80% probability,
Smit - 50% probability.

They shoots by this sequence:
first - Smit, second - Garry, third - John and then for circle.

What strategy Smit must take, to have maximum chances
to stay living in this duel.

Every duelist has known about possibility of each other.
.......................................
Have ideas. Now I think about it to :rolleyes:

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They shoots by this sequence:
first - Smit, second - Garry, third - John and then for circle.

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What is the meaning of this line? If Smit must face John in the end it doesn't matter anyway since he has 100% probability. I would say, they leave with their back facing each other, while counting loud to 30 and then they turn around and shoot. If Smit knows of his chances and of John's, he'll use foul play, turn around at the count of 1 and shoot John in the back of his head. :D I would go for this. ;)

Hehehehe Simon! ;)

I mean next:

They are all stand face to face ( like in triangle ).
They should shoot by turn.

1) Smit
2) Garry
3) John

Strategy - Smit can shoot in John or Garry or himself :rolleyes: or... etc.
So what strategy has maximum chances to stay living.
Everyone know about everyones probability.

Do U now understood?

dimm, I sent you the answer in a PM.

here is another riddle. Three dwarfs of the Yellowbeard clan are travelling through a tunnel. The tunnel is so narrow that they can only walk one after the other. Suddenly they encounter three dwarfs of the RedBeard clan travelling in the opposite direction. Since they can't move around each other, they decide to climb over each other. While they are discussing this, the dragon that the Yellowbeards wanted to kill appears on the RedBeard side and the dragon for the Redbeards appears at the Yellowbeard side. So they are in a hurry to get to their phat lewt.

How do they do this ?

Scematically it looks like this :

Code:

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X = YellowBeard
O = RedBeard
D = Dragon

DOOO XXXD

note the space in the middle

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Yves, please, one riddle at a time. And even then, two a day is a bit much. Some people are very busy, like getting one of those frustrating rich edit thingies to work.

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Originally posted by Yves M
dimm, I sent you the answer in a PM.

here is another riddle. Three dwarfs of the Yellowbeard clan are travelling through a tunnel. The tunnel is so narrow that they can only walk one after the other. Suddenly they encounter three dwarfs of the RedBeard clan travelling in the opposite direction. Since they can't move around each other, they decide to climb over each other. While they are discussing this, the dragon that the Yellowbeards wanted to kill appears on the RedBeard side and the dragon for the Redbeards appears at the Yellowbeard side. So they are in a hurry to get to their phat lewt.

How do they do this ?

Scematically it looks like this :

Code:

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X = YellowBeard
O = RedBeard
D = Dragon

DOOO XXXD

note the space in the middle

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I'm probably completly out of it, but is that suppose to be easy to explain!!? :) It seems I have trouble to explain the two last riddles. It's like "think i got it, but... how can I... arrggh" :) ;)

JeffB

Jeff, don't worry, if you are not stressed, you understand what the poster is trying to say, you have some time and some paper, all riddles are feasible. :) Unfortunately, for the moment all elements are not present in my case and people just start flooding each other with riddles. AAAAAARGHHHH....

Ok, I'll take out the dwarf and dragon bit and reformulate it schematically only. You have X on the right and O on the left.

• An X can only move to the left, an O can only move to the right.
• An X can jump over a single O if there is a free space behind that O (so that from _OX you get XO_ where _ is a space). Similarly, an O can jump over a single X if there is a space on the right of the X.
• the starting position is OOO_XXX
• the ending position has to be XXX_OOO

You can write the solution as a series of states, so you would write something like
OOO_XXX
OOOX_XX
...
XXX_OOO

Is this clearer ?

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Yves, please, one riddle at a time. And even then, two a day is a bit much. Some people are very busy, like getting one of those frustrating rich edit thingies to work.

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I've been doing that for one year now :D

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I've been doing that for one year now

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Was the picture of the hairy Yves taken from one of those programming moments? ;) :D

It sure was :p

But then again, so is the other one ;)

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Originally posted by Yves M
Ok, I'll take out the dwarf and dragon bit and reformulate it schematically only. You have X on the right and O on the left.

• An X can only move to the left, an O can only move to the right.
• An X can jump over a single O if there is a free space behind that O (so that from _OX you get XO_ where _ is a space). Similarly, an O can jump over a single X if there is a space on the right of the X.
• the starting position is OOO_XXX
• the ending position has to be XXX_OOO

You can write the solution as a series of states, so you would write something like
OOO_XXX
OOOX_XX
...
XXX_OOO

Is this clearer ?



I've been doing that for one year now :D

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What I marked bold is very important. Cannot an X move to the right even if there is the empty space?

YES, now I see the "problem" in here, you know, my first schema was looking like:

Code:

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//Starting:
XXX_OOO

//Some move and it ends like... three dwarfs on the three others heads...
  XXX
  OOO


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You know, if they can walk on top of each others... :D

JeffB

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Originally posted by Gabriel Fleseriu


What I marked bold is very important. Cannot an X move to the right even if there is the empty space?

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Yep, this is very important. X can never move to the right. He wants to get as quickly as he can to the dragon on the left and will bash you with his two meter axe if you tell him to move away from the dragon :p

And Jeff... What do I do with you ? ;)

Yves: are you sure that the problem has a solution? Besides of that one with the dwarves walking on top of each other :)

Yes I'm sure ;) It would even work with 100 dwarves on each side ;) Just give it a little more waste of time, huh I mean thought :p

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 :)

Code:

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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

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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 ;)

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)

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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)

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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.

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

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"The last man on earth sat alone in his room. Suddenly there was a knock at the door!"

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Here's the intended answer:
"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:

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Originally posted by solarflare
Only, JeffB's answer:
"The last man on universe..."
doesn't make sense grammatically in English.

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Oh thanks to "tell others" :D :D

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 :)

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Originally posted by JeffB
What exactly doesn't make sense?

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I believe it would be "the last man in the universe..."

cout << endl;

;) :)

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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.

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Incorrect, get your Chrisitianity straight before you try to speak.

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.

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Originally posted by temple
Incorrect, get your Chrisitianity straight before you try to speak.

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Temple,

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.

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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.

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That were 3 men, 5 hats (3 black and 2 white).

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

You take one character off your message :p

Interesting answer by the way. Unfortunately 3*3 is not 12 though ;)

Here's the intended answer to my previous puzzle:

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Originally posted by solarflare
Name something money can't buy.

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poverty

by the way, no correct answers on this one... :(:(:(

All these coin problems where you can only use balance scales... hmmm... You know, in my country, most post offices have real scales which weigh things in grams or ounces or whatever instead of against other things. Also, it doesn't cost anything to use, so you can weigh things like coins as many times as you want. Maybe I'm just special to have access to such an extraordinary tool, I don't know. But I guess these coin problems are just too unrealistic and impractical. The dwarves and dragons problem, now there's something that is likely to happen in the real world ;).

:D :cool: :cool: :p

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But I guess these coin problems are just too unrealistic and impractical. The dwarves and dragons problem, now there's something that is likely to happen in the real world

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I couldn't have put it better ;) Ok, let's rephrase my coin problem then.

You have twelve dwarfs. They go kill a dragon and get the phat loot. They decide to divide it up equally, but one of the dwarfs is a bit of a joker and took a different share. Noone knows which one it is though. So since the dwarfs are all of the same clan (the Yellowbeards), they are all the same model number and hence all weigh the same.

They don't have access to any high technology so they decide to figure out who is the joker by weighing themselves using a fallen tree. They realize that the tree won't be able to withstand more than three weighings, so they have to be very careful on the procedure. Luckily a wandering elf is kind enough to provide them with a solution that will find out which dwarf is the joker and whether he took more or less gold.

What did the elf suggest ?

Are the dwarfs going to kill the elf instead of trying to figure out who is the joker ?

WOW Yves M., how interesting!

Was that elf a drow? :) :D

JeffB

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Originally posted by JeffB
WOW Yves M., how interesting!

Was that elf a drow? :) :D

JeffB

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That depends on whether you believe that his solution actually works or is just an elaborate plan to kill the dwarfs ;)

The elf gave an incorrect solution! The dwarves didn't realize this until after their third weighing, and went on the tree branch a fourth time. It broke, of course, and they plummeted 4 feet (long distance for a dwarf) into the murky swamp below. The elf was going to then kill the dwarves and eat them, but the dragon came back to life! It ate the elf as the dwarves came out of the swamp. The dwarves slayed the dragon again and got double the gold! Yes, that's right, the dragon's gold reserves regenerated when it was resuscitated. So the dragon is dead, the lying bastarrd elf is dead, and the dwarves have double the gold! Oh, and by the way, the jokester dwarf confessed. The end!

Since the subject of this thread is "Try to solve this interesting logical task" I thought I'd add another interesting logical task...

Whats wrong with this picture?

pm answers to me!

;) ;) ;) ;) ;) ;) ;) :cool: :cool: :D

Not exactly a difficult logical task ;)

Hum and your solution to the dwarfs riddle second edition... what can I say... There is another outcome ;)

[edit:
Wow, I didn't expect this one to fall as quickly ;) solarflare PMed me his solution to the 12 coin problem and it's correct.]

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Originally posted by Yves M
Not exactly a difficult logical task ;)

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It's funny, though! It scared the cr*p out of me the first time I saw it... my speakers were pretty loud...

Quote:

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Originally posted by Yves M
Wow, I didn't expect this one to fall as quickly ;) solarflare PMed me his solution to the 12 coin problem and it's correct.

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Actually, it was pretty difficult for me... at first I expected the solution to be symmetrical, but it's not! Very odd.

:D :cool: ;) :p :p :o

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Originally posted by solarflare

It's funny, though! It scared the cr*p out of me the first time I saw it... my speakers were pretty loud...

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Well, I didn't have the sound on since I was listening to music ;)

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Originally posted by solarflare
Actually, it was pretty difficult for me... at first I expected the solution to be symmetrical, but it's not! Very odd.

:D :cool: ;) :p :p :o

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Yep, it's much more interesting than the one where the coin is lighter. The solution is "sort of" symmetric, but it's true that you have to think about it a bit before you get it.

Actually I got the riddle off some other forum some years ago. After I posted it here I had to spend a quarter of an hour finding a solution, since I wasn't sure anymore that it was even feasible :p

Yves, I just read your "new" riddle, but the solution (or the method used) is EXACTLY the same as the one I already posted here to the previous coin problem.

Hum, are you quite sure, Mr. Simon ? The point being that in my riddle you don't know whether the false coin (resp. the dwarf joker) is lighter or heavier. Have you tried it ?

I can tell you that the solution is quite a bit different from the one to your problem ;)

That makes it different indeed.

Good luck, Simon :)

Hum, nobody else has tried the 12 coin riddle ?

No time, seems impossible anyway.

Plus I got hit by a car which may have damaged my brain. Have I mentioned yet I like gurls and chicken chop?

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Originally posted by Simon666
Plus I got hit by a car which may have damaged my brain. Have I mentioned yet I like gurls and chicken chop?

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Hum... even if you'vew got hit by a car, you could still be considerate with other members of the board ;)