• RajdeepCE

MemberJun 9, 2009

## Brain Blaster : Puzzle

Brain Blaster : Puzzle 1 ->>>>>
This is the famous water jug puzzle of the movie Die Hard 2, but CEans are not for solving such old & easy puzzles. So here it is, the brand new modified water jug puzzle for you all CE,
We have given a barrel full of water with capacity of 20 ltrs. We have two other jugs which can measure 12 ltrs. and 7 ltrs. The puzzle is that how to measure 2 ltr water by using only two jugs?
Let us see, who gets his brain blasted with this puzzle?
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• MemberJun 9, 2009

1. Fill 7 ltr jar (full)
2. empty it in 12 Ltr Jar (Now bigger jar have 5 Ltr capacity remaining)
3. again fill 7 ltr jar and empty it in Bigger jar till its full capacity (7-5 = 2)

Now Smaller jar left with 2 Ltr water.

-CB
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• MemberJun 9, 2009

Great Crazyboy!!!!
You are superfast in solving the puzzle.
Congrats!!!!
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• MemberJun 9, 2009

Now the another quest is same puzzle, but this time you have to measure 3 ltrs.
All the best!!!!!
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• MemberJun 9, 2009

RajdeepCE
Now the another quest is same puzzle, but this time you have to measure 3 ltrs.
All the best!!!!!
1. Fill the 12 ltr jug (full).Now water remaining in 20 ltr jug is 20-12=8 ltr
2. Then fill 7 ltr jug(full) with water from 12 ltr jug. Now free volume in 12 ltr jar is 12-7=5
3. Now again fill the 12 ltr jug(full) from the water remaining in 20 ltr jug
4. When the 12 ltr jug is full, water remaining in 20 ltr jug is 8-5=3 ltr
😁
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• MemberJun 9, 2009

20ltr 12ltr 7ltr
1. 20 0 0
2. 8 12 0
3. 8 5 7
4. 1 12 7
this seems to be wrong result. Correct me if I am wrong.
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• MemberJun 9, 2009

sanandsanu
1. Fill the 12 ltr jug (full).Now water remaining in 20 ltr jug is 20-12=8 ltr
2. Then fill 7 ltr jug(full) with water from 12 ltr jug. Now free volume in 12 ltr jar is 12-7=5
3. Now again fill the 12 ltr jug(full) from the water remaining in 20 ltr jug
4. When the 12 ltr jug is full, water remaining in 20 ltr jug is 8-5=3 ltr
😁
Dude you are wrong in the point 2.

IF we pour water from 12 to 7 then free volume in 12 litres jug will be 7 not 5.
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• MemberJun 9, 2009

My solution is,

1.fill the 12 ltr first ( so the remaining in 20 ltr jug is 8)
2.fill the 7ltr jug using the 12ltr filled jug( so the remaining in 12 ltr jug is 5)
3.fill the 20 ltr jug using 7 ltr jug( so the total in 20 ltr jug is,15)
4.fill the 7ltr jug using 12ltr jug(now the 7ltr jug contains 5ltr)
5.now fill 12ltr jug using 20ltr jug ( therefore the 20ltr jug contains 3ltr, 15-12=3)
20 0 0
8 12 0
8 5 7
15 5 0
15 0 5
3 12 5

Am I right?
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• MemberJun 9, 2009

Congrats for that.!!!
But there is one problem that I forgot to mention that we have to measure 3ltr water in jug not in the barrel of 20ltr.
Anyway now try to get 3ltr any one of the jug.
All the best!!!
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• MemberJun 9, 2009

The another crazy & logical puzzle for you all CEans!!!,
In a garden, there was a total number of 67 roses. Some roses are red and some are yellow. Whenever we randomly pick any two roses, there is atleast one yellow rose in it.
So how many yellow & red roses are there in the garden? Justify the answer with proper logic.
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• MemberJun 9, 2009

1 red and 66 yellow.

???
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• MemberJun 9, 2009

Yes Differential!!!
You are absolutely right!!!
Here is another blasting puzzle for you all CEans!!!,
In a bookself, there are some books. The books are numbered as 1, 2, 3, 4, 5, & 6. Now the problem is that we have to arrange books in specific order.The order of book is arranged such that,
1>>>> The subtraction of two consequtive book numbers must be greater than or equal to 3.
2>>>> The addition of two consequtive book numbers must be less than or equal to 7.
Go CEans!!! Blast both the puzzle hard!!!!
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• MemberJun 10, 2009

1) 4 1 5 2 6 3

Here, the difference between consecutive numbers is >= 3;

2) 6 1 5 2 3 4

Here, the sum of consecutive numbers is <= 7;
Am I correct??
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• MemberJun 10, 2009

If the law of distribution holds, there must be at least one red rose out of 67. The red and yellow ratio can't be close or you could randomly select 2xred. There must be more yellow than red. At least one must be red.
The puzzle says one is always yellow, allowing for both yellow and at least one out of 67 red. This assumes yellow/red aren't mixed, or there are no part-red, part-yellow ones.
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• MemberJun 10, 2009

Yes SS, you are right!!!
Congrats!!!
@Skipper, nice explanation!!!
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• MemberJun 11, 2009

New reasoning puzzle,
A astronomer discovered 5 suns and their 5 planets in the galaxy. He named 5 sun as Sun 1, Sun 2, Sun 3, Sun 4, and Sun 5. He named planets as Planet V, Planet W, Planet X, Planet Y, Planet Z. He made the list of sun and their respective planet. But he misplaced it. But he remembers the following details:
1. First he discovered the planet of Sun 3, than he found the planet of Sun 4 and than he found Planet W and Planet Y and at last he found the planet of Sun 5.
2. When he discovered Planet X, he already discovered Planet W & Planet Z respectively & also the planet of Sun 1.
3. Planet V was discovered first of all other planets.
Can you arrange all the suns with thier respective planets?
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• MemberJun 11, 2009

Sun 1 - Planet Y
Sun 2 - Planet W
Sun 3 - Planet V
Sun 4 - Planet Z
Sun 5 - Planet X
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• MemberJun 12, 2009

That's correct ES.
SUN 1 - PLANET Y
SUN 2 - PLANET W
SUN 3 - PLANET V
SUN 4 - PLANET Z
SUN 5 - PLANET X
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• MemberJun 12, 2009

You both are correct, but ES strikes first.
Congrats ES!!!
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• MemberJun 12, 2009

Chess Challenger 1:
All you know about 8-queens problem, but here is modified puzzle, "10-queens problem",
You have to arrange 10 queens on a chessboard such that each queen can kill only another one queen.
Come one CEans!!! Kill the queen, but be careful cause you have to kill only one!!!!
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• MemberJun 12, 2009

Is a corollary that the problem means "in one move"? Since once a queen is taken the capturing queen can take be taken, by the queen that's inline with it, as the problem requires? No wait, there should only be pairs that can take each other, so once a queen is captured, its partner holds the independent square with no other queens inline?

In only 1 move per queen all queens can capture at least 1 queen, depending on which one actually moves, so 10 queens, 1 moves and 9 possibilities? Except there should be only one possible capture per move, over 64 squares?

Since each queen can move one square in 8 directions, the next square moved has to have a square it can't move over, then this number of unavailable squares doubles on the next square moved over, for n squares in a single move. Each piece has to be on a single square, at least 1 square from any of the other 9 pieces. It's a 9+1 puzzle in 3 dimensions, 2 dimensions are the x,y distances moved as square 1/64ths of XY the chessboard.

The third Z dimension is the 9+1 pieces = queens. If you remove 1 piece by keeping it "off" the surface as if it's moving, there are 9 pieces to find a position for; repeat with 8+1, then 7+1, until you reach 1+1 which should be trivial. There are then 8 queens "in motion" on 64 squares, and 2 fixed or on 1/32 of the square XY board.

The algorithm should fall out of the backwards induction to 1+1.
1 piece is the redundant state since there are no other queens to take, so 2 queens is where to stop (or start). The problem is either "arrange 5 queens that can't take each other, then 5 more that can only take a single queen from the first 5, and from any other 4 of the 5 extra queens, on a 64 square board".

The 8 degrees of freedom for a queen is constrained if it's at the edge or in corner, A corner position leaves 3 squares to move over, and an edge leaves 5. If a queen is in a corner you can't use the corners for a second queen because the corners are inline (for queens). A corner piece uses up the entire outer set of squares, except for the 6+6 squares out of the range of the corner, and also uses up the diagonal from corner to corner. Another corollary is that each queen will need rank, file and diagonal freedom, and only have 1 other piece along (at least) 3 other possible moves.

I would use 5 black + 5 white pawns as queens, place them as close to the center of the 64 squares as possible, then move them 2 at a time to 'safe' positions. I think each pair will need to be at least 1 square apart, not adjacent, so there's a gap to hide another queen in along a rank, file or diagonal.
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• MemberJun 12, 2009

Hey Skipper, dont make this puzzle so difficult to understand.
It is same as the 8-queens problem, except here you have to arrange 10 queens such that only one queen can be crossed by only one queen. That means that you have to arrange 5 pairs of 2 queens such that they cant cross each other.
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• MemberJun 12, 2009

Yeah, that's what I said... arrange 5 pairs (black&white) without them crossing each others paths except once = traveling salesman problem for 5 places as pairs of queens that capture each other, and no other. A graph traversal problem.

The salesman has to find a route that takes "one queen" of distance, for 5 "towns" on a map with multiple routes between.
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• MemberJun 12, 2009

RajdeepCE
Chess Challenger 1:
All you know about 8-queens problem, but here is modified puzzle, "10-queens problem",

Come one CEans!!! Kill the queen, but be careful cause you have to kill only one!!!!
Man you have not clearly mentioned the size of chessboard. Are we suppose to fit 10 queens on 8x8 chessbaord only 😔 or on 10x10 chessboard ?

[ PS: Also Please comment on my solution for 8 queens problem - #-Link-Snipped-# ]

Thanks !
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• MemberJun 12, 2009

@Shalini, chessboards are always of 8*8 size.
And regarding about your answer of 8-queens puzzle, I am too much busy these days, so I didnt get the chance to check it. I am extremely sorry for that. If it is possible than can you please add image to that answer, so the other CEans can understand.
And also give the solution of this puzzle graphically, so we dont need to clarify our answers.
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• MemberJun 13, 2009

@shalini..
Your answer for the 8 queen puzzle is wrong, because if you arrange like that 4 queens will cross exactly the other four..
For example, 1st row 4th column will strike the 4th row 7th column..
And in the same way all the queens in 2nd and 3rd rows will strike those in 6th and 8th rows respectively..
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• MemberJun 13, 2009

If you know the answer than post it buddy.
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• MemberJun 13, 2009

I think there are 66 yellow roses and 1 red rose. This is because you said whenever we pick any two roses randomly there is atleast 1 yellow. This is only possible if there exists zero probability of picking any two red roses simultaneously. This is only possible if we have one red rose leaving 66 yellow roses.
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• MemberJun 13, 2009

Munguti, its already solved buddy. Anyway the current puzzle is 10-queens problem. Try your mind in it!!!
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• MemberJun 14, 2009

There's a way to solve the 10-queens by using 20 pieces and removing 10. What are the 10 other pieces?
Are you allowed to cheat and use a composite piece?
p.s. has anyone played a variant of chess called ultima, in which the queens are the only actual chess pieces?

pp.s. these questions are a backwards-recursion of the solution.
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• MemberJun 14, 2009

I should stipulate that I'm forwarding, without proof a conjecture that, since each piece in chess is a kind of composite or 'semi-piece', and the board trivially demonstrates a proof of the 2-color problem (it's a symmetrical square 8x8 matrix) it's a binary space

The pawns are bivectors, queens are pawns promoted to be able to move n squares and take 1 of 3 ways from a corner, whereas a pawn moves 1 squares times 1 direction, but takes 1 of 2 diagonal ways from a square not at the edge (not along a or h files). The pawn files but takes diagonally, the queen ranks and files, taking in either direction so has 2x2(+/-2) symmetry, which is 'positive' for a1 and 'negative' for h8.

Rooks are promoted from pawns by having 1 file (as 2 pieces) and 1 rank movement x2 taking directions; in a corner it has 2x(+/-2) symmetry. Corner pawns have the lowest or 1x1, pawns on inner files b-g have 1x2, etc. You induce queens and rooks by promoting pawns. Promote 2 pawns into a 2-color piece that can occupy 1x2 squares. Recolor a chess board by coloring it either all white or all black.
Place the black/white semi-pawn on the corner so it occupies about 1/64th of the area, recolor the 1/64th according to the angle, white-black makes, so the square is 1/2 black and 1/2 white. There are 4 possible ways to do this.

I conjecture that this procedure will allow you to replace the semi-pawns with semi-queens and resolve the 10-queens problem.

p.s. I goofed with the ultima version I know about, the king is the only piece that retains the chess function, queens take in a double move, first moving next to a piece, then moving away from it on a subsequent move -which the queen's owner can delay - in any free direction, so it can't take more than one opponent's piece. Queens are 'takers', bishops I think take in the usual way, but you can't take bishops with bishops. Pawns are passive locators, if you line up an opponent's piece squarely with a couple of pawns you can take it, and your opponent can subsequently take both pawns if you don't take a located piece. I can't remember all of it.
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• MemberJun 14, 2009

@Skipper, can we have the answer please. Cause I am tired reading such long discriptions which are of no use as per me.
So just make it simple, solve puzzle & less discussion.
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• MemberJun 14, 2009

I am trying to make it simple. Have you tried placing 10 pawns on a chessboard and treating them like queens that can only move 1 square?
Do you know what decomposition is? How to decompose a composition (say, by Ravel or Rachmaninoff)?

Alternate that with decomposing a game of chess and determining what the rules are, then altering the rules, then the board. Start with 1 square that can be colored 4 ways with 2 colors, like the corners of a finished board.

Simplfication #1 cut out the center 4 squares of a finished board (make it out of paper first).

Now fold up the remainder into 4 sides of a cube. Each 2x2 set of squares can hold a single "semi-pawn" which is a degenerate state for a queen (= a crippled queen that has to hop 1 square, or, 1 sq' at a time) the square and the square(s) queen alternate over the folded regions. Simply think in 4 dimensions as you do this.😀
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• MemberJun 14, 2009

A side-puzzle, which I conjecture is still within the domain of RajdeepCE's original 10-queens problem (which is equivalent to finding a solution for switching 10 signals so they don't interfere, over 5x2-input gates, and 5 levels - this is tricky since you need to interconnect 5 sets of binary switches so each original input pair (a,a'), (b,b'), ..., (e,e') is preserved at the outputs so, how many switches are needed for each successive level?)

The question is: if 100 people have two pawns, one white and one black; another 100 have two black or two white pawns, then if you select 1 pawn from each of the people in the first lot, what probability is there of selecting two black or two white in sequence? What's the probability with the second lot of people?

If you select 2 pawns from each set, what changes and why? Why do you get a different distribution, which are the mixed states?
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• MemberJun 18, 2009

OK, mixed states are (obviously) white & black; all white or all black are pure states.

This is a mnemonic (classical) way to see quantum states in QIS and switching of wavefunctions. In a bipartite state, there are exactly 4 states available = {00,01,10,11} and in QIS these are probability amplitudes. The problem is separability - how do you switch (phase-shift) the pure (00) or (11) as inputs? All you can do with these is invert them, you can't change a pure 2-input state into a mixed state without losing the state = erasing one of the inputs.

This is important to the Boolean logic approach in computation using quantum states.

Back to the 10-queens algorithm. Is it a question of finding an adjacency graph, for the 64 'occupied or not' squares for the queens? Each pair needs a directed path which doesn't intersect any other pair's path and is one move long, or p(1), a polynomial => Tn = T(1); the steps in the induction of the adjacency graph are "the problem" since you have to exclude T(n+1), T(n+2), ...; there can be only one move.

Is there more than one solution, and is there a solution by first placing two queens, then two more untll there are 5 pairs only 1 move apart and all on different paths? There must be a polynomial for n=1 moves, for m = 5 pairs (placements in the 8x8 graph). The solution should look like: P(1,5) = p(1,1).p(1,2).p(1,3).p(1,4).p(1,5); 5 degrees for pairs and one for moves, means factorial 5 steps in the induction...! There has to be a way to exclude certain paths so you don't have to traverse the board to find a location for the next pair, then to alter the paths excluded, by moving the previous pair, so you fit all 5 pairs by moving all previous positions to "make room". Alternatively start with one pair and add pairs until re-positioning is needed, then recurse backwards over the board each time.

The result and the induction steps differ by up to 5! I think... this puzzle isn't that easy but I'm sure there's either a way to do it or show it can't be done.
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• MemberJun 18, 2009

@Skipper, if puzzle was easy, I wouldn't post it. And one more thing buddy, give it a try instead of messing with algorithms. I am sure that it can be solved by backtracking method.
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• MemberJun 18, 2009

1]first fill the 7 ltr jug and pour it into the 12 ltr jug........

2]then again fill 7 ltr jug again and pour it till the 12 ltr jug until it gets filled,,,,,,

3]after 12 ltr jug has been fill 2 ltr will be remaining in the 7 ltr jug....😎 😁 ;-)

12-7=5
7-5=2 😁 😁
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• MemberJun 20, 2009

Here's my solution to the ongoing brain-numbing chess prob in two subsets:
1) solve the 8-queens problem
2a) move three of the queens so they're next to 3 of 5 others, and the remaining two are on 'safe' squares
2b) add 2 more queens as in 2a

For extra marks, show this is equivalent to finding 10 independent pairs of algebraic states in a 8-dimensional vector space, which is equivalent to a byte-oriented (mod 2) STN or switching net.
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• MemberJun 20, 2009

Instead of giving solution, can we have the image of the solution, Skipper?
It will be easy to understand the other CEans if you post the image.
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• MemberJun 20, 2009

Ok, but it's the image of any 8-queens solution, that satisfies the second condition. If it doesn't satisfy 2a it fails.

I look at it like this: there are queens a..h in the input, which is |a,b,...,h> , and axb, cxd; also ab, cd and a+b = b+a, etc are the possible interactions, so that for 8x8 squares each queen has a1, a2, ..., ai possible cross or 'dot' products, at least one of which is excluded as a row or column, or product = diagonal cover of the board.
The input has to pass the first 'parse' so that only those that have a 6-queens solution as pairs that can remove each other (ab = -ab) plus two independent c,d queens, to get to the next filter which starts with 7-queens.
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• MemberJun 20, 2009

I just simply said to post the solution in image format. I dont told you to discuss the puzzle here.😡
Next time you better dont solve the puzzle ( I mean dont discuss about how to solve it, how to blah blah...). I am tired of reading your long descriptive methods. Short & simple, if you know the answer than post the answer and if you dont than no need to discuss how to blah blah😡😡😡.
Here is the solution,

Thank you skipper for not posting in my thread next time:evil:
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• MemberJun 21, 2009

Ha ha ! Rajdeep also get angry. 😀 Chill man. All people never think in same direction. May be his way of thinking was much better than yours but you were not even trying to understand his points or may be he was not able to express his answer well. 😀
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• MemberJun 21, 2009

I am tolerating this guy ever since he made his first post here. At that time I just thought that he will post the answer in next post. But I was wrong. Look at his all the post and especially last two post. I am consequently telling him to post the answer but still he was discussing blah blah😡😡. Don't he have any sense that I am asking for solution, not for blah blah😡.
I am tired of this kind of person who makes needless discussions😡.
What to do now ? I am thinking of posting the variant of this puzzle. But now I am scared of this guy😁.
I will cool down my mind and will post at next time.
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• MemberJun 21, 2009

Hey Rajdeep buddy,
Now don't get too tensed. Just forget it, ok?
And by the way, I too was a bit irritated by his replies. But no qualms.

And then, as for this problem, is this the only solution or does it have any other solutions?

And continue posting such puzzles. There are people who are eager for them!!
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• MemberJun 21, 2009

RajdeepCE
But now I am scared of this guy😁.
ROFL ! Chill ,Grant him excuse and keep on posting. 😀

PS: For God sake don't make his fun. He at least tried your puzzle (as compared to rest) though the way he tried didn't please you.
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• MemberJun 21, 2009

Ok.
Here is the another variant of this puzzle,
Arrange 14 queens such that each queen crosses other 2 queens.
For reference, check the image on the top. In that image, each queen is crossed by only one queen. Similarly, here you have to do same job with 14 queens and place each queen such that it can be crossed by 2 othe queens.
Please, post only the solution of puzzle.😁. If you dont have image, than no problem, just post the solution by raw:column format. I will post the image after varifying the answer.
@SS, I manage to find only 1 solution. May be you can try to find another one.
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• MemberJun 21, 2009

Notice, Raj ol' buddy, I wasn't interested in "a" solution but in how to get to a solution. For instance, and just to be that little more annoying: can you say if the solution you posted is unique? Is it the only one possible? How many other solutions are there, and what formula will generalize all of them?

Hint: I'm more interested in numbers and where they come from, so what number of solutions corresponds to a solved 5-pairs of queens problem? Another detail: when you hand in homework, you get a mark for the correct answer, and you get more marks for a correct proof of your answer. In exams the answer is worth less than the work you do getting to it. (at least, that was my experience; sciencey types can be so picky someytimes, huh?)
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• MemberJun 21, 2009

Ok skipper, thats fine. But dont discuss here please. If you know the answer or found the algorithms behind it, than post it. Otherwise just wait & watch for others to answer.
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• MemberJun 22, 2009

can there be three queens crossins each other or strictly two only...??
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• MemberJun 22, 2009

RajdeepCE
Ok skipper, thats fine. But dont discuss here please. If you know the answer or found the algorithms behind it, than post it. Otherwise just wait & watch for others to answer.
Hey Rajdeep, I guess there is no harm if someone is trying to find out the generic solution for such puzzles. What is the objective of posting the puzzles here then if you are simply asking and giving answers without telling the strategy of solving them right? If you will tell the strategy of solving them alongwith asking, it would really benefit all of us from next time. 😀

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• MemberJun 22, 2009

well first of all fill completely the 7 lt barrel n put the water in 12 lt barrel..then again fill the 7lt barrel n fill the remaining part of 12lt barrel wath the 7lt barrel...the remaining water in the 7lt jug is 2 litre....hence the problem
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• MemberJun 22, 2009

he is right.if we pour water from 12 ltr to 7 ltr then 12-7=5.
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• MemberJun 22, 2009

Hey buddy, current puzzle is to arrange 14 queen problem.
@Sayeed, the following image will guide to solve the puzzle,

Hope that everyone is cleared with puzzle.
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• MemberJun 24, 2009

Where is the " Vishvanathan Anand of CE "?
I am waiting for this puzzle to be solved.
Should I proceed with the another puzzle or post the solution of this puzzle first?
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• MemberJun 24, 2009

Does 14 queens mean 14 groups of moves over m,n squares? That would be an interesting link to the number of ways to arrange 14 things, right?, or is it 8 things?
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• MemberJun 24, 2009

No skipper. We have 14 queens(things). And we have to arrange it.
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• MemberJun 24, 2009

Ha, I got this 14 queens, 3 at a time in a 8x8 matrix.
14 = 3+3 + 3+3 +?(+)2 the last 2 have to interact with only 1 other queen; then forwards it's: place 3 in a row or column (or triangle), place 3 more the same way so they don't intersect, repeat the first 2 steps, halt if the last 2 can't be placed as in the first step.
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• MemberSep 3, 2009

66 yellow and 1 red.
for any 2 roses selected there will be atleast 1 yellow.
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• MemberSep 4, 2009

Here is solution for n queens(here n=14) for any n n ≥ 4 or n = 1:

1.Divide n by 12. Remember the remainder (n is 8 for the eight queens puzzle).
2.Write a list of the even numbers from 2 to n in order.
3.If the remainder is 3 or 9, move 2 to the end of the list.
4.Append the odd numbers from 1 to n in order, but, if the remainder is 8, switch pairs (i.e. 3, 1, 7, 5, 11, 9, …).
5.If the remainder is 2, switch the places of 1 and 3, then move 5 to the end of the list.
6.If the remainder is 3 or 9, move 1 and 3 to the end of the list.
7.Place the first-column queen in the row with the first number in the list.
8.place the second-column queen in the row with the second number in the list, etc.

14 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 3, 1, 7, 9, 11, 13, 5.

Solving the problem requires O(n!) time. 😁
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• MemberSep 7, 2009

1) out of 20 lits pour into 7 lits jug
2) empty 7 lits jug into 12 lits jug
3) refill 7 lits jug from 20 lits barrel (in which 13 lits is remaining)
4) pour water from 7 lits jug into 12 lits jug (you can fill 5 lits into it)
5) remaining water in 7 lits jug = 2 lits
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• MemberOct 13, 2009

raajdeep
u get angry very soon. try to chill man 😛

itchap
the solution which u have provided is for 8x8 chessboard or 14x14
as far as i can make out it works for 14x14 and not for 8x8
please correct me if i m wrong
how one should solve this kind of puzzle please tell me
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• MemberOct 15, 2009

RajdeepCE
Brain Blaster : Puzzle 1 ->>>>>
This is the famous water jug puzzle of the movie Die Hard 2, but CEans are not for solving such old & easy puzzles. So here it is, the brand new modified water jug puzzle for you all CE,

ans--
first take 7 liter in small jug and put into 12 lit. jug.. do it again
n u ll get 2 liter remaining in small jug....
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• MemberDec 20, 2009

fill the 12lt jug completely and pour it into 7lt jug until it is filled.now the big jug contains 5 lt.

now empty the 7lt jug completely and pour the 5lt water left in the big jug into the small jug....so now the small jug has 5lt water and 2lt volume is free...

now fill the big jug completely and use it to fill the free volume in small jug...so the big jug has now 10 lt...

now empty the 7lt jug and fill it with the 10lt water in the big jug..hence only 7lt is used and 3lt is left in the big jug.....
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• MemberDec 20, 2009

1)fill 12lt jug fully and pour it into 7lt jug...so 5lt is left in the big jug ..now empty the 7lt jug

2)now pour the remaining 5 lt water in the big jug into the 7lt jug so tha it has now(7-5=2)lt left in it as free volume

3)now fill the 12lt jug completely and use it to fill the remaining 2lt space in the small jug

..so now the big jug has(12-2=10)lt...now empty the 7lt jug and refill it using the water in the big jug(having 10lt)...so the big jug now has (10-7=3)lt...i.e. 3 lt left...so we can now measure 3lts
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• MemberAug 7, 2010

great cean
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