Crazy Tronners :: View topic

P4 · Posted: Wed Apr 11, 2007 9:37 am

I got some math tasks too Wink

1)
Using only addition (+) write number 28 with five 2's and number 1000 with eight 8's.

2)
We got 3 identical bottles. One of them consider 2/3 liters of water and rest is empty. How to measure exactly 1/2 liters of water?

3)
How to write number 1 using all decimal integers (0-9)?

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wrtlprnft · Posted: Wed Apr 11, 2007 9:48 am

- 22+2+2+2 = 22+6 = 28
- 888+88+8+8+8 = 1000
1. Distribute the water into two bottles so they contain the same amount of water, that gives you {1/3; 1/3; 0}
2. Distribute the water of one bottle into the empty one so they contain the same amount of water, this gives you {2/6; 1/6; 1/6}
3. Now you only have to add the 2/6l and one 1/6l bottle to get 3/6l=1/2l of water
(9-8)+(7-6)+(5-4)-(3-2)-(1-0) = 9-8+7-6+5-4-3+2-1+0 = 1
EDIT: Alternative: 0*23456789 + 1

Last edited by wrtlprnft on Wed Apr 11, 2007 10:11 am; edited 1 time in total
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wrtlprnft · Posted: Wed Apr 11, 2007 10:04 am

Got another riddle, too, and this one isn't easy Razz

N people, standing in a straight line, get different colored hats put onto their heads (at random).
They can only see the hats of the people in front of them, not their own (ALL hats in front of them, so the last guy sees all hats but his)
They know what different colors the hats can have, say red, green and blue
They do not know how many hats of which color they are, they just know that everybody gets one
Before they get their hats put on they can decide on a strategy to use
Once they get their hats put on they're not allowed to say anything but one color (so only �red�, �green�, or �blue� in this case).
Once they said that color they have to be silent
Everybody hears everything
The goal is to have as many people say their own hat's color as possible

What is that strategy? Hint: The optimal strategy still works if you change the number of people or the number of colors.

Last edited by wrtlprnft on Wed Apr 11, 2007 10:28 am; edited 1 time in total
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Voodoo · Posted: Wed Apr 11, 2007 10:11 am

Have no idea ! Wink

Well let's start by the last one who have nobody behind him :
-He says the color of the guy hat in front of him,
- the next one of course now know is color hat, so he says it,
- next one don't know so, he says the color of the guy in front of him,
- etc ...

N/2 people will give the good answer ...

That's the only strategy I can think of ... Dunno if it is the optimal one Wink

wrtlprnft · Posted: Wed Apr 11, 2007 10:15 am

So N/2 people will say the correct color for sure, and of the other N/2 people, on average 1/3 will be lucky and say their own color simply because they've got the same color as the guy in front of them.

That'd make for 2/3 of all saying their own color.

Not bad, but the optimal strategy results in N-1 people saying the correct color and the remaining one having a 1/3 chance Smile

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owned · Posted: Wed Apr 11, 2007 11:17 am

wrtlprnft wrote:

So N/2 people will say the correct color for sure, and of the other N/2 people, on average 1/3 will be lucky and say their own color simply because they've got the same color as the guy in front of them.

That'd make for 2/3 of all saying their own color.

Not bad, but the optimal strategy results in N-1 people saying the correct color and the remaining one having a 1/3 chance Smile

are they aloud to say something that isn't a color?

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ed · Posted: Wed Apr 11, 2007 11:20 am

wrtlprnft wrote:

Once they said that color they have to be silent

wrtlprnft · Posted: Wed Apr 11, 2007 11:21 am

No. They're only allowed to say �possible� colors (so they can't say purple with a shade of red or something like that).

Anyways, here's the solution:

Code:

#include <iostream>
#include <stdlib.h>

#define NUM_COLORS 3
#define NUM_PEOPLE 10

class Hat {
public:
typedef enum {
   COLOR_RED = 0,
   COLOR_GREEN = 1,
   COLOR_BLUE = 2
} color;
color m_color;
Hat(void) {
   m_color = (color)(random() % NUM_COLORS);
}
Hat(color theColor) : m_color(theColor) {}
void Print(std::ostream &s) const {
   switch (m_color) {
      case COLOR_RED : s << " red"; break;
      case COLOR_GREEN: s << "green"; break;
      case COLOR_BLUE : s << " blue"; break;
      default: s << "BUG!";
   }
}
};

std::ostream &operator<< (std::ostream &s, Hat const &hat) {
hat.Print(s);
return s;
}

class Guy {
int m_SaidColors;
int m_HeardFirstColor;
Hat::color m_FirstColor;
public:
Hat m_hat;
Guy *m_next;
Guy() : m_SaidColors(0), m_HeardFirstColor(false) {}
void HearAColor(Hat::color color) {
   if(!m_HeardFirstColor) {
      m_HeardFirstColor = true;
      m_FirstColor = color;
   } else {
      m_SaidColors += color;
      m_SaidColors %= NUM_COLORS;
   }
}
Hat::color SayYourColor() {
   int colorsInFront = 0;
   for(Guy *guy = m_next; guy != 0; guy = guy->m_next) {
      colorsInFront += guy->m_hat.m_color;
      colorsInFront %= NUM_COLORS;
   }
   if(!m_HeardFirstColor) {
      return (Hat::color)(colorsInFront);
   }
   int totalColors = (m_SaidColors + colorsInFront) % NUM_COLORS;
   int myColor = m_FirstColor - totalColors;
   if(myColor < 0) myColor += NUM_COLORS;
   return (Hat::color)(myColor);
}
};

int main(int ac, char **av) {
srand(time(0));
Guy guys[NUM_PEOPLE];
guys[NUM_PEOPLE - 1].m_next = 0;
for(int i = NUM_PEOPLE -2; i>=0; --i) {
   guys[i].m_next = guys + i + 1;
}

std::cout << "Colors :";
for(Guy *guy = guys; guy != 0; guy = guy->m_next) {
   std::cout << " " << guy->m_hat;
}
std::cout << std::endl;

std::cout << "Guesses:";
for(Guy *guy = guys; guy != 0; guy = guy->m_next) {
   Hat::color color = guy->SayYourColor();
   for(Guy *guy2 = guy->m_next; guy2 != 0; guy2 = guy2->m_next) {
      guy2->HearAColor(color);
   }
   std::cout << " " << Hat(color);
}
std::cout << std::endl;
}

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P4 · Posted: Wed Apr 11, 2007 12:56 pm

pretty nice one wrtl
this one is much easier

There's 5 houses. Each one has the front doors in a different coulour. Each one is occupied by person different nationalty who owns different animal and who likes different drink. Each person eats different food.
The Autralian lives in house with red doors.
The Italian owns a dog.
A coffe is drunk in the house with green doors.
The Ukrainian drinks tea.
When you're looking on the ecri doored house the house with green doors is on right.
The eater of mushrooms owns snails (slugs)
Apples are eaten in the house with yellow doors.
A milk is drunk in the middle house.
The Norway guy lives it he 1st house from left.
An onion eater lives in neighbourhoods with the guy who ows fox.
Apples are eaten in the house who neighbours with the horse owner house.
Cookies eater drinks orange juice.
The Japanese eats bananas.
The Norway guy lives near the house with blue doors.

Who drinks water and who owns Zebra?

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P4

owned · Posted: Wed Apr 11, 2007 7:58 pm

10 = 50

bob says it's true
and after he told them why, everyone believed him. Why?

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kyle · Posted: Wed Apr 11, 2007 8:04 pm

10 is base 50

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In theory it should work in reality who knows if it will work.

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owned · Posted: Wed Apr 11, 2007 10:49 pm

that isn't wat I was thinking.

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Tadd wrote:

The C in CT stands for "crazy". And in our forum I think 1/4 of our posts are ironic, funny, offtopic or to piss someone off.

kyle · Posted: Wed Apr 11, 2007 11:10 pm

still works

_________________

In theory it should work in reality who knows if it will work.

MB53: I am 26 years old how can i be a teen ? Very Happy

owned: You got the heart of a teen Smile

Gonzap: and the innocence of a little child

SidneY · Posted: Fri Apr 20, 2007 3:20 am

I got one!

Please fill in the next row in the pattern:

1
11
21
1211
111221
312211
13112221

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@ Xevi: " Smile

"
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@ Kyle: "Profound."
@ Voodoo: "I will, thanks."
@ Ed: "No Sig?!"
@ P4: "Only 5 songs in the top 10?"

wrtlprnft · Posted: Fri Apr 20, 2007 5:36 am

1113213211

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