BALLS

Re: BALLS


ooo err

the minimum is 1 the maximum is 6

questions are appreciated

metro17: What do you mean by chances? Are you asking for the minimum number of weighings to guarantee that you've identified the heaviest ball?

If so,
solipsister: How can you identify the heaviest of N in fewer than N-1 weighings?

sum how

if and only if the example is the question

in a perfect world chance is finite

Its an unclear question, because your meaning of chances is not well defined, the capability of the weighing pan is not specified and it is not specified is the user aware of the weights of the balls in the first place.

If the weighing pan reads out total weight upon it, simply keep removing weights until weight goes down by 8lbs. Then best case is one weighing, worse case is seven. Its also seven if you just weigh the first ball versus the second and keep the heaviest, then weight that against the third and keep the heaviest, then weight it and keep the heaviest and so on until you get to the end (7 weightings).


If the weighing mechanism is exotic and forces you to have to sort the balls, then depending on your algorithm (quick sort, heap sort) and how the balls randomly arrive to be sorted, the answer is between order n log2(n) to twice this - so 24 to 48 steps.

fascinating einstein

you only need to weigh 7 balls if you know that the total weight is 36 as in the example so 6 weighings is enough thank you very much

poincare would laugh

Nope - it depends, if it were me and I presume you have a flat surface of constant friction and a gravitational field (else your scales are useless) I'd use the pan of the scales to accelerate each ball as a group in a straight line to a constant velocity and see which travels the longest and which the shortest.

Remember you have momentum and kinetic energy to play with that will differentiate these objects. You could simply collide them to get your answer with no weightings - can't you?

So 0 is likely the minimal correct answer?

you're at sixes and sevens

answers....


I guess 1 and 6 are the right answers....
can you explain please...my explanation is

pan A 1,2,3,4 pan B 5,6,7,8 .............. 1 weigh
1,2 and 3,4 5,6 and 7,8 ............ 2 weighs
2 and 4 6 and 8 ............ 2 weighs
4 and 8.....result is 8 ........... 1 weigh..............total 6 weighs....worst case scenario.....

but 1 is ok if you pick 8 lbs ball against any other ... but you do not know that you have 8 kgs ball...i should have mentioned in the questions all balls are of same shape and size and only differ in weight.....
any more explanations are welcome....

"Balls!" said the Queen: "If i had two, i'd be King!"

And if I had three it would fascinate the life out of the lady who thought she had seen it all.

unreserved apologies n-1 weighings is the minimum for n balls of unknown weight

in my defence i had 3 balls in the air at the time and someone shouted that if you want to kill a clown go for the juggler

after six weighings of balls totalling 36 pounds i'd pick up the last two balls, one in each hand , which could include either the 1 ball or the 8 ball or both and voila

how discerning