tag:blogger.com,1999:blog-4115025577315673827.post719452637311127746..comments2019-05-18T16:40:30.584+05:30Comments on CSE Blog - quant, math, computer science puzzles: Pebble Placement Puzzle 1Unknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4115025577315673827.post-90069082488698002662014-12-14T15:45:41.875+05:302014-12-14T15:45:41.875+05:30good one!!good one!!ashwani pandeyhttps://www.blogger.com/profile/09629580972154099734noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-61308579422996765942014-12-01T13:11:03.099+05:302014-12-01T13:11:03.099+05:30Let, "x" be the first element to visit a...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 positionparikshit sharmahttps://www.blogger.com/profile/13260882434065119099noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-91438156105778614302014-11-07T23:06:25.177+05:302014-11-07T23:06:25.177+05:30For the first case, If we assume that the number o...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<br /><br />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 positionBlue swordhttps://www.blogger.com/profile/14333877783200238339noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-45731014482200499602014-11-07T14:58:54.394+05:302014-11-07T14:58:54.394+05:30Assume there are M stones in n x n board (M < n...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).<br />Since the rest M-1 stones have not yet occupied the "Original Position" of M, neither one has returned to their respective original positions.<br />All the M-1 stones are not in their respective original positions.<br />Thus when M is moved the current positions of all the pebbles will be different from their original positions.Aditya Tirumalahttps://www.blogger.com/profile/05892898666205305657noreply@blogger.com