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Topic: partial algorithm to dropping eggs (Read 4875 times) 

summersolstice
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partial algorithm to dropping eggs
« on: Oct 5^{th}, 2012, 1:12pm » 
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hello I'm new here (first post) I saw a simliar problem to this on another site (with snookerballs instead of eggs) and actually figured out multiple ball puzzles though like I said it the title its incomplete atm but I thought I'd share the progress there exists some (very complicated) formula that I think might be expressed as a continued product, for now I'll write it as f(e,d) = F e is number of eggs d is number of drops F is the maximum number of floors that you can investigate now if you have F and e you put them into the formula and find d, (however it might need trial and error but it doesn't matter, because its a whole number) what you need is the smallest integer value for d so that the result is greater than/equal to the number of floors you have (i.e. if you lowered d by one it would be less than the number of floors) now once you find d you can call this the initial d everytime you drop an egg, d (the current value) goes down by one if it breaks then e also goes down by one. at each step you have 3 values current value of e (number of eggs left) current value of d (number of drops left) lowest it didn't break (this doesn't have to be the previous drop just the last time it didn't break) I'll call this g take e1 and d1 and put them into the formula add g then add 1 now go the the floor whose number correponds to the result and do this on every step and this means that on or before d drops and breaking no more than e eggs you'll find the correct floor I'll be interested to hear what you think now


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