# Paradoxes and Ironies!!!!

like 1,2,4,8,16,32,64,128,256,and the rest of the coins.

THIRU

**Hope, not everybody is aware of this one.**

There are 1000 coins and 10 bags. Fill the bags with these coins in such a way that if I ask you a random number ranging from 1 to 1000, you should be able to give me the bag/s equalling the number of coins asked for. Not a coin more, not a coin less ie. no manipulation of removing or adding according to the number. So figure out how many coins each bag will contain.

There are 1000 coins and 10 bags. Fill the bags with these coins in such a way that if I ask you a random number ranging from 1 to 1000, you should be able to give me the bag/s equalling the number of coins asked for. Not a coin more, not a coin less ie. no manipulation of removing or adding according to the number. So figure out how many coins each bag will contain.

Regarding your fractal problem. I think the solution is mentioned in a book called "The expert's guide to Delphi 3 " . I may have read the chapter on fractal landscapes two years ago, but I dont have even the faintest recollection of it now. So if I make an attempt( without referring to the book , of course), will it be considered a fair win or cheating ?

[This message has been edited by Sahir Shibley (edited May 03, 2001).]

Sheriff

- Instead of dollars and cents, pretend we use dollars and mils, where 1000 mils = one dollar.
- Instead of 20 cents, use a difference of 891 mils.
- Instead of increasing the value of the original check by a factor of 2, use a factor of 10. What's the answer then?

[This message has been edited by Jim Yingst (edited May 03, 2001).]

"I'm not back." - Bill Harding, *Twister*

Sheriff

"I'm not back." - Bill Harding, *Twister*

Sheriff

Uncontrolled vocabularies

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

10 boxes problem: we don�t need weighing at all.

Algoritm:

1) Open all boxes

2) for each box: leave one ball inside, put 9 other on the ground near the box (do not mix balls! )

3) for each heap: put one ball in each of other 9 boxes

4) now all our boxes have the same weight = no need to find �odd box�.

5)

Jim, do not beat me, it was a joke.

Jim, what do you mean by �one weighing�? The scale were loaded with N boxes only once or scale�s value was taken once?

[This message has been edited by Mapraputa Is (edited May 03, 2001).]

Uncontrolled vocabularies

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

Sheriff

Uncontrolled vocabularies

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

Map

**actially read**a puzzle. Before she just tried to solve it. For some reason she decided that there are 10 balls in each box and we should weigh boxed, not balls. Now it's too easy again... Maybe. Or not. Jim, is it important how many balls are there in each box? I mean will it affect your solution somehow?

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

"I'm not back." - Bill Harding, *Twister*

As for the programmer's hats, I still can't come up with a way to save them all, but it does occur to me that even if each programmer is limited to speaking just the single word representing his guess at the colour of his hat, there is still

**tons**of spare bandwidth for extra information. Perhaps when the programmers get together they decide that they will speak the name of their guess, but speak it in English if the hat in front is red, French, if the hat in front is white, and German, if the hat in front is black. If they are limited to using English (boy, this is a tough company to work for), then they can spend a few minutes practicing and speak their guess in a "Russian" accent for a red hat in front, "New York" accent for white hat in front and so on.

Once the guy at the back makes his guess, the programmer in front then knows his colour, and the effect will ripple forward to the front of the row. I still think that the one at the back can only know his colour if he also knows something about the quantities or proportions of the different coloured hats.

The more you look at this, the more it looks like an interesting idea for a "team building excercise". The 100 programmers would have to

*talk to each other*- could this ever happen in real life?

Originally posted by Eager Beaver:

Hi Guys,

Here comes another puzzle.....

A giant truck gets loaded to its max capacity and tips the scale on the weigh bridge at 10 tons. It starts from point A and after traversing a distance of 10 miles arrives at a bridge on river kiev. The board reading 'WARNING : VEHICLES NOT TO EXCEED TOTAL WEIGHT OF 10 TONS. 1 gram (CGS system) MORE AND THE BRIDGE COLLAPSES.' has been ignored by our careless driver. When he is about to get on the bridge a sparrow comes from nowhere and perches atop the truck and stays there while the truck traverses the bridge. Surprise of surprises the bridge remains intact and the truck is happily on its way. How did this miracle happen!!

eagerbeaver.

PS : Hope u all have seen sparrows.

[This message has been edited by Eager Beaver (edited May 02, 2001).]

[This message has been edited by Eager Beaver (edited May 02, 2001).]

Well when it moves 10 Miles, It looses some fuel. So the truck weighs less by the time it reaches the Bridge

Is this right?

Oops . Somebody has already answered this puzzle

[This message has been edited by hemanth kumar (edited May 04, 2001).]

You may use a scale or a spring balance. A traditional balance (one held by the blind lady statue in courts) could also be used and as long as you do not change the number of balls placed on one side of the balance it would be counted as one weighing.

Sahir,

Try to arrive at a lesser number of weighing. 3 tries is too many!!

Colin,

Forest puzzle....you have once again demonstrated your knack for solving such posers!!

Thiru,

Your answer to (A-X)(B-X)....(Z-X) is correct.

And to all those who have answered the TRUCK/BRIDGE puzzle.....'fuel consumed' was the right answer.

Thanks to all those who have taken a keen interest in this thread.

eagerbeaver.

Sheriff

**The more you look at this, the more it looks like an interesting idea for a "team building excercise". The 100 programmers would have to talk to each other - could this ever happen in real life?**

Frank, you hit the point! Of course, that's the most difficult part. Mapraputa, for example, would say <quote>�yellow with purple polka dots�</quote> because she never pay attention to requirements. Jim, of course, would tell his color without any extra information from those behind

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Here is the real puzzle,

f.y.i...Sherlock & Poirot is still working on it...

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

**Murder in the Black Castle**

It was a dark night, heavy with wind and rain, when three lone knights, strangers to each others, chanced to meet in front of a black and gloomy castle.They were suspicious of each other, but, as was their custom, they approached the castle and sought refuge for the night.

The three knights were greeted by a sour-faced servant who explained that the master had retired for the evening but that their needs would be met. The three starngers were then provided food and shown to separate room.

Sometime during the night

**a murder was commited.**

The crime can be considered somewhat unusual, as

**not only the culprit unknown, but the identity of the victim is also unclear.**

Fortunately, the list of possible culprits and victims can be narrowed to five :

a. the servant

b. the master

c. and the three knights.

Given the following clues, who was the victim and who was the culprit ???

- If the knight in room 1 was culprit, the knight in room 3 was the victim.

- If the knight in room 2 was the victim, the servant was the culprit.

- If the knight in room 3 was the victim, the knight in room 2 was the culprit.

- If the servant was the culprit, the victim was the knight in room 3.

- The servant was not available until the next morning, and was not able to provide an alibi.

- If the knight in room 3 was the culprit, the knight in room 2 was the victim.

- If the knight in room 2 was the culprit, the servant was the victim.

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

enjoy the puzzle & solve the case dude...

stevie

Sheriff

I did spend a whole evening in reclusive contemplation and came up with a way to calculate the number of triangles down to level 4. However i could not formulate expression that could take care of the larger inverted triangles(having more than 5 triangles within).These inverted triangles increase in arithmetic progression.....but have not elegant solution for them....shucks..(:-(. I have given up and looking forward to your post.

eagerbeaver.

Sheriff

For not to count triangles more than once, let's count �basic� one-level triangles separately. Their number on each level is x = x + l*l (if l means level) � it was the easiest part Next go composite triangles in normal direction � these propagate as

level1 0

level2 1

level3 2 + 1

level4 3 + 2 + 1

level5 4 + 3 + 2 + 1

...

if use Y as a number of these triangles, it will give us:

start with n =0; y=0;

then for each level

n = n + l - 1

y = y + n

Then the worst part, inverted triangles

level4 1

level5 2

level6 3 +1

level7 4 +2

level8 5 +3 +1

level9 6 +4 +2

level10 7 +5 +3 + 1

...

if to denote them as Z, then we have a bunch of auxiliary variables:

i = 1 - i

k = k + i

m = m + k

z = z + k

and it�s ugly enough plus I probably already made a mistake somewhere I had a couple of brilliant ideas (like treat each triangle as a half of rhomb or to count a triangle as a function of itself) but they did not work � what a pity...

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

I thought one of you guys would solve it in flash so I did not try. Besides I am in one my lethargic moods. Map, you need to approach it from a different perspective, once you do that it is a piece of cake. How do do you create a fractal . You start with an equilateral triangle, then you join the midpoints of the triangle. Now you have an inverted triangle inside the original triangle. You have divided the original triangle into four triangles. The total number of triangles is 5 including the original triangle. You divide it again you have 4 upright triangles at the base. Try drawing the traingles and you can see the relationship with this problem.

Cheers

Sahir

Here's how i proceeded to analyze it. Here's a table which lists the number of triangles of size 1 to n, for a level n. (In this counting I don't discriminate between normal and inverted triangles. Just the size matters.)

How to read the table:

- Level 2 has 4 size1 triangles, and 1 size2 triangle, giving you T(n) of 5.

- Level 4 has 16 size1 triangles, 7 size2 triangles, 3 size3 triangles, and 1 size4 triangle, giving you T(n) of 27.

<pre><big>

1 2 3 4 5 6 7 8 9 10 TOTAL

1 1 1

2 4 1 5

3 9 3 1 13

4 16 7 3 1 27

5 25 13 6 3 1 48

6 36 21 11 6 3 1 78

7 49 31 18 10 6 3 1 118

8 64 43 27 16 10 6 3 1 170

9 81 57 38 24 15 10 6 3 1 235

10 100 73 51 34 22 15 10 6 3 1 315

</big></pre>

And some more samples:<pre><big>

n T(n)

11 411

12 525

13 658

14 812

15 988</big></pre>

Do you recognize some patterns? One simple pattern is, for a level n, number of triangles of size1 is n<sup>2</sup> (That is, n*n). That's cute right? Can you prove (geometrically) why that's true

When you recognize more patterns, try to put that into an equation. So you would first arrive at a summation (sigma) equation. Then solve this to get the final answer!

Pourquoi voulez-vous mon nom?

Sheriff

enjoy!

*One simple pattern is, for a level n, number of triangles of size1 is n2 (That is, n*n). That's cute right? Can you prove (geometrically) why that's true*

well, lets see a composite triangle as a half of rhomb. Number of triangles of size1 is twice as many as rhombes of the same size, but our �big� triangle includes only half of rhomb, so 2 * � * n <SUP>2</SUP>

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

Sheriff

x (i,j) = f (i, j)

where f (i, j) is:

if j = 1, f (i, j) = i * i;

else

k = i - 2*j + 2;

if k < 0, f (i, j) = x(i-1, j-1) <br /> if k > 0, f (i, j) = x(i-1, j-1) - k

if produces correct result, except it is more obscure than Ugly Three-step Algorithm

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

The complete equation, for even, and odd values of 'n' is:

T(n) = floor [n * (2 * n + 1) * (n + 2) / 8]

(I can hear you thinking, duh, why didn't I think of using floor for odd numbers )

Could you briefly describe how you arrived at this solution?

Map, your recursive equation looks right. How are the values of i and j related to n, the level of the triangle... I was wondering.

Pourquoi voulez-vous mon nom?

Sheriff

Now consider inverted triangles of size k. This is a bit harder to see, but again look at possible locations for just one corner, and you can see it's a triangular area with side length n-2k+1. Which means that the number of points in that triangle, and hence the number of inverted triangles of length k, is (n-2k+1)(n-2k+2)/2. But this formula only works for 2k <= n. Well, actually it's OK for 2k = n+1 and 2k = n+2 as well, since those yield zero, but after that it gives meaningless results. There are zero inverted triangles with size k > n/2 - this can be easily verified graphically. At this point I decided to only worry about even values of n for simplicity - obviously the odd values would follow a similar formula, but who needed the extra work?

So, to get the total number of triangles we need to sum two series:

sum(k=1...n) [ (n-2k)(n-2k+1)/2 ] + sum(k=1...n/2) [ (n-2k+1)(n-2k+2)/2 ]

In order to combine these, I needed both series to use the same set of indices, so I rewrote the first one by defining u = 2k:

sum(k=1...n) [ (n-2k)(n-2k+1)/2 ] = sum(u=1...n/2) [ (n-u+2)(n-u+3)/2 + (n-u+1)(n-u+2)/2 ]

The first term handles the odd values of k; the second handles the even ones. Once that's done, rename u to k -- we can call the index anything we want, and now it has the same range of values as the k in the inverted triangle formula, so we can combine them. The total number of triangles is now

sum(k=1...n/2) [ (n-2k+2)(n-2k+3)/2 + (n-2k+1)(n-2k+1)/2 + (n-2k+1)(n-2k+2)/2 ]

What follows after this is a lot of messy algebra. Eventually I got it to the form

1/2 * sum(k=1...n/2) [(3n^2 + 11n + 10) - (12n + 22)k + 12k^2]

This groups into three sums. The first is just a constant, added N = n/2 times. The second is a constant times 1+2+3+...N, which we know the formula for. And the last is a constant times 1^2 + 2^2 + 3^3 +...+ N^2, which I looked up rather than derive, and that evaluates to N(N+1)(2N+1)/6. Substitute these formulas in, do some more algebra, and it eventually simplifies to n(n+2)(2n+1)/8

"I'm not back." - Bill Harding, *Twister*

Sheriff

*How are the values of i and j related to n, the level of the triangle*

i = level, j = size

To print your �level� X �size� table:

x (level,size) = f (level, size)

where f (level, size) is:

if size = 1, f (level, size) = level * level;

else

k = level - 2*size + 2;

if k < 0, f (level, size) = f(level-1, size-1) <br /> if k > 0, f (level, size) = f(level-1, size-1) - k

But Jim already expressed all this as one formula

[This message has been edited by Mapraputa Is (edited May 10, 2001).]

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Map, your recursive solution is cool, quite elegant, actually!

Pourquoi voulez-vous mon nom?

Sheriff

**How many beans can you put into an empty sack**

One.

Jim, see - Map also can solve something

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet

Sheriff

Sheriff

"I try my best to make *all* my posts nice, even when I feel upset" -- Philippe Maquet