tag:blogger.com,1999:blog-4115025577315673827.post3669194663878306165..comments2019-05-18T16:40:30.584+05:30Comments on CSE Blog - quant, math, computer science puzzles: Buying in Rocket Ships and Selling in Fire SaleUnknownnoreply@blogger.comBlogger9125tag:blogger.com,1999:blog-4115025577315673827.post-46229410082243912582016-02-07T00:51:22.616+05:302016-02-07T00:51:22.616+05:30what happens if the cost of each of the items are ...what happens if the cost of each of the items are different ??<br />My hypothesis is that we can, while buying, make a grid similar to what has been suggested by John and buy the most expensive item each month, deleting that row and column every month. In the limiting case of all items costing the same, the solution reduces to that of John.Manish Guptahttps://www.blogger.com/profile/17501086330737919407noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-49276474152978646492016-02-07T00:49:14.029+05:302016-02-07T00:49:14.029+05:30what happens if the cost of each of the items are ...what happens if the cost of each of the items are different ??<br />My hypothesis is that we can, while buying, make a grid similar to what has been suggested by John and buy the most expensive item each month, deleting that row and column every month. In the limiting case of all items costing the same, the solution reduces to that of John.Manish Guptahttps://www.blogger.com/profile/17501086330737919407noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-44350845337319881272016-01-07T13:34:20.293+05:302016-01-07T13:34:20.293+05:30For selling:
I think we should calculate decrease...For selling:<br /><br />I think we should calculate decrease in price at each time step for each remaining item and sell the one which might depreciate the most in value if we do not sell it immediately. This way we are greedily minimizing the total loss incurred.<br /><br />Proof doesn't seem to be obvious, but it works on toy cases I constructed. Has anyone proved this or has a link to a solution?Ashishhttps://www.blogger.com/profile/16792135802098470087noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-61152999469081921892015-12-04T21:36:41.733+05:302015-12-04T21:36:41.733+05:30just a thought, is there an optimized way, even if...just a thought, is there an optimized way, even if initial prices of items differ ? like p,q,r,...instead of all being 100. i guess we should seek the sum of the entire column instead of just looking for r_i<br />Shankarhttps://www.blogger.com/profile/16014872260680603611noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-91901209654391536742015-11-08T22:27:16.996+05:302015-11-08T22:27:16.996+05:30How to solve the second part apart from an exponen...How to solve the second part apart from an exponential method<br />Parikshit Khannahttps://www.blogger.com/profile/12953083405737499517noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-48256237805318151682015-07-28T10:29:27.433+05:302015-07-28T10:29:27.433+05:30For the selling case, consider the following rates...For the selling case, consider the following rates( in absolute terms ) : <br />99, 99, 1, 0.5<br /><br />If you sell them in order of decreasing rates, you will get the following returns :<br />100/(1) + 100/(100) + 100/(2*2) + 100/(1.5*1.5*1.5 ) = 155.63<br /><br />This is not the best selling sequence. A better sequence is : ( 99, 1, 0.5, 99 )<br />100/(1) + 100/(2) + 100/(1.5*1.5) + 100/(100*100*100 ) = 194.44Ankush Jainhttps://www.blogger.com/profile/14341091026907249096noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-27529030089700578722015-07-28T02:53:31.983+05:302015-07-28T02:53:31.983+05:30For the selling case, I think, we'd need to se...For the selling case, I think, we'd need to select in the order of increasing r_i. Since, we're trying to maximise the sum in that case. Correct?gjhttps://www.blogger.com/profile/12510270934879404415noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-54474717561969727872015-07-12T00:11:05.436+05:302015-07-12T00:11:05.436+05:30it is correct, it is a greedy solution. the soluti...it is correct, it is a greedy solution. the solution is simple, i think it is the correctness proof that requires some clever argumentsparthahttps://www.blogger.com/profile/16617572727465811884noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-37200114274888261342015-06-19T22:28:09.857+05:302015-06-19T22:28:09.857+05:30Are all items priced at $100? If they are, what...Are all items priced at $100? If they are, what's wrong with buying (and selling) the one with highest r_i first (and then in order of decreasing r_i)? <br /><br />Put more formally, considering n items, I would use a grid:<br />1,(1+r1),(1+r1)^2,.....(1+r1)^n-1<br />.<br />.<br />1,(1+rn),(1+rn)^2,.....(1+rn)^n-1<br /><br />The question is now to select one no. from each row and column to minimise the sum. It can be proven that this happens when you select in order of decreasing r_i.<br /><br />I feel like I'm missing something big here.. What part of above is incorrect?John Mitchelhttps://www.blogger.com/profile/09581635974186122859noreply@blogger.com