Skip to main content

Pebble Placement Puzzle 1

Source: AUSTMS Gazette 35


There are several pebbles placed on an n × n chessboard, such that each pebble is inside a square and no two pebbles share the same square.

Perry decides to play the following game. At each turn, he moves one of the pebbles to an empty neighboring square. After a while, Perry notices that every pebble has passed through every square of the chessboard exactly once and has come back to its original position.

Prove that there was a moment when no pebble was on its original position.


  1. Assume there are M stones in n x n board (M < nxn). WLOG assume that M is the last stone which will be moved from its original place(whenever it may be moved).
    Since the rest M-1 stones have not yet occupied the "Original Position" of M, neither one has returned to their respective original positions.
    All the M-1 stones are not in their respective original positions.
    Thus when M is moved the current positions of all the pebbles will be different from their original positions.

  2. For the first case, If we assume that the number of empty cells are more than the number of pebbles then after the first round, none of the pebbles will be in their original position

    For the second case, when the number of free cells are less than the number of pebbles, even if there is only one empty space left, that space will traverse throughout until all the pebbles have been moved. At this point none of the pebbles are in their original position

  3. Let, "x" be the first element to visit all the squares exactly once and return to it's original position. The claim is that no other stone can be in it's original position in the state when x is about to enter it's original position. If y is in it's original position just before x is about to enter, then "y" never left it's original position, since "x" is the first element to return. If "y" never left, "x" could not have visited y's cell, thus a contradiction. So no stone is in it's original position


Post a Comment

Popular posts from this blog

Asking a girl out

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this article though :P

Most of my friends already read an article that I wrote more than an year back - "Speak Up"

Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it!

Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis.

@Consultants, correct me if I am wrong in my estimates. :P

Why is there a need to be selective?

From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once every week right …

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

1) Calculate P{X>Y}
2) What's the expected value of X
3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

Fraction Brainteaser

Sent to me by Gaurav Sinha

Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions. In total, Siddhant scores 25 out of 34. 

Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34.

Note that
a) Vaibhav scores more than Siddhant
b) Siddhant score better than Vaibhav in both individual topics - 5/6 > 20/25 and 20/28 > 6/9

How is it possible?