tag:blogger.com,1999:blog-4115025577315673827.post5549116047180330801..comments2019-11-07T19:19:08.018+05:30Comments on CSE Blog - quant, math, computer science puzzles: Sam Loyd Puzzle SolvabilityUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4115025577315673827.post-44745069019573787142012-12-24T08:28:37.767+05:302012-12-24T08:28:37.767+05:30Correct solution raghuram. Thanks a ton.Correct solution raghuram. Thanks a ton.Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-58539012751227766902012-11-19T09:02:31.883+05:302012-11-19T09:02:31.883+05:30here is solution.
write the numbers horizontally s...here is solution.<br />write the numbers horizontally so that it represents permutation of numbers 1-15(ignore the blank box).there are two different kind of slides possible.<br />1.horizontal slide<br />2.vertical slide<br />horizontal slide does not change the permutation<br />effect of vertical slide is change the position of a particular element by 3 units .for example in the figure 1 shown if 12 and the blank spot are swapped ,in the permutation position of 12 is 15,and position of 13,15,14 decreases by 1 and rest unchanged.now let us consider the parity of of number of inversions in a permutation.initial permutation has odd number of inversions.final one has even number of permutations.<br />only vertical slide toggles the parity.<br />since the blank occupies same row in both configuration,number of vertical slides are even. that is parity of initial and final permutations can not be different.<br />so not possible raghuram kowdeedhttps://www.blogger.com/profile/10276651805450711325noreply@blogger.com