Brain Blaster : Puzzle

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

Great Crazyboy!!!!
You are superfast in solving the puzzle.
Congrats!!!!

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

RajdeepCENow 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
😁

I think your answer is wrong cause your explanation leads me as follows,
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.

sanandsanu1. 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.
So your answer is incorrect 😀

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?

Your solution is right, ms_cs.
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!!!

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.

1 red and 66 yellow.

???

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!!!!

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??

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.

Yes SS, you are right!!!
Congrats!!!
@Skipper, nice explanation!!!

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?

Sun 1 - Planet Y
Sun 2 - Planet W
Sun 3 - Planet V
Sun 4 - Planet Z
Sun 5 - Planet X

That's correct ES.
SUN 1 - PLANET Y
SUN 2 - PLANET W
SUN 3 - PLANET V
SUN 4 - PLANET Z
SUN 5 - PLANET X

You both are correct, but ES strikes first.
Congrats ES!!!

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!!!!

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.

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.

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.

RajdeepCEChess 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 !

@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.

@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..
So the answer is wrong..

@Red Hawk, can you please discuss this at original thread.
If you know the answer than post it buddy.

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.

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

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.

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.

@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.

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.😀

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?

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.

@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.

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 😁 😁

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.

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.

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.

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:

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. 😀

I am asking puzzle, not asking how to solve the puzzle.
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.

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!!

RajdeepCEBut 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.

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.

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?)

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.

can there be three queens crossins each other or strictly two only...??

RajdeepCEOk 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. 😀

PS Rest is your wish.

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