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Interview Question for Software Engineer Intern at Goldman Sachs:
Jul 28, 2009
1 of 1 people found this helpful
Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?
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2 of 12 people found this helpful
Jul 29, 2009
3 times. (2^3 = 8)
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25 of 28 people found this helpful
Jul 29, 2009
Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest.
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4 of 5 people found this helpful
Jul 29, 2009
Brian - this would be correct if you in fact were using a weighing scale, and not a balance scale. The ability to weigh one group against another with a balance scale allows Marty's answer to be a correct answer.
Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer.
However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.
Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer.
However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.
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2 of 3 people found this helpful
Jul 30, 2009
This question is from the book "How to move Mt Fuji".... Marty has already got the right answer.
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0 of 4 people found this helpful
Jul 31, 2009
Actually Bill, by your interpretation of the question the answer is zero, because you could just pick a ball at random. If you get lucky, then you've found the heaviest ball without using the scale at all, thus the least possible amount of times using the scale would be zero.
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1 of 2 people found this helpful
Jul 31, 2009
The answer is 2, as @Marty mentioned. cuz its the worst case scenario which u have to consider, otherwise as @woctaog mentioned it can be zero, u just got lucky picking the first ball....
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0 of 6 people found this helpful
Jul 31, 2009
None- weigh them in your hands.
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0 of 6 people found this helpful
Aug 02, 2009
Assuming that the balls cannot be discerned by physical touch, the answer is 3.
You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.
You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.
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5 of 6 people found this helpful
Aug 04, 2009
2
3a+3b+2 = 8
if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest
if wt(3a) !== wt(3b) then
ignore group of 2
discard lighter group of 3
divide the remaining group of 3 into 2+1
weigh those 2
If == the remaing 1 is the heaviest
if !== the heaviest will be on the scale
3a+3b+2 = 8
if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest
if wt(3a) !== wt(3b) then
ignore group of 2
discard lighter group of 3
divide the remaining group of 3 into 2+1
weigh those 2
If == the remaing 1 is the heaviest
if !== the heaviest will be on the scale
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0 of 3 people found this helpful
Aug 05, 2009
With the systematic approach, the answer is 3.
But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1.
Although the real question is: are the balls truly identical if one is heavier than the rest?
But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1.
Although the real question is: are the balls truly identical if one is heavier than the rest?
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0 of 3 people found this helpful
Aug 05, 2009
just once. Say you are lucky and pick the heavy ball. One use of the scale will reveal your lucky choice
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0 of 2 people found this helpful
Aug 05, 2009
so once, or the creative answer zero if you allow for weighing by hand
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Aug 06, 2009
Without judging by hand: Put 4 balls on one side, and 4 on the other. Take the heavier group and divide again, put 2 balls on one side, and 2 on the other. Take the 2 that were heavier, and put one on each side. You've now found the heaviest ball. This is using the scale 3 times, and will always find the right ball.
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0 of 1 people found this helpful
Aug 06, 2009
None. They are identical. None is heavier.
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4 of 4 people found this helpful
Aug 06, 2009
2 weighings to find the slightly heavier ball.
Step 1.
compare 2 groups of three balls.
Case 1.
if they are both equal in weight, compare the last 2 balls - one will be heavier.
case 2.
If either group of 3 balls is heavier, take 2 balls from the heavier side.
compare 1 ball against the 2nd from the heavy group
result 1. if one ball is heavier than the other, you have found the slightly heavier ball.
result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball.
Easy Shmeezi
Step 1.
compare 2 groups of three balls.
Case 1.
if they are both equal in weight, compare the last 2 balls - one will be heavier.
case 2.
If either group of 3 balls is heavier, take 2 balls from the heavier side.
compare 1 ball against the 2nd from the heavy group
result 1. if one ball is heavier than the other, you have found the slightly heavier ball.
result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball.
Easy Shmeezi
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Aug 06, 2009
Fewest - get lucky and pick the heaviest one. But wait! How would you know it is the heaviest one by just weighing one ball? Your logic is flawed.
Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.
Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.
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0 of 2 people found this helpful
Aug 08, 2009
i think its 3. i would take it like this
OOOO OOOO then
OO OO
then
OO
problem solved. i do this everyday. bye. praise be to allah. thats it.
OOOO OOOO then
OO OO
then
OO
problem solved. i do this everyday. bye. praise be to allah. thats it.
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0 of 1 people found this helpful
Aug 09, 2009
It's 2. Period.
If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.
If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.
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3 of 3 people found this helpful
Aug 09, 2009
The answer is 2.
1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls.
2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one.
3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.
1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls.
2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one.
3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.
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0 of 1 people found this helpful
Aug 17, 2009
2=if all the balls are identical and you pick up the first...weigh it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.
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0 of 1 people found this helpful
Aug 17, 2009
1=if all the balls are identical and you pick up the first...balance it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.
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Aug 19, 2009
Amy is 100% correct for the following reason:
everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one.
Steve
everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one.
Steve
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Aug 21, 2009
First off, take yourself through the process visually and forget square roots, that doesnt apply here and here is why: The question is the Minimum, not the MAXIMUM. BTW, the max would be 7 ( 8-1); you are comparing 2 objects, so 1 ball is eliminated automatically in the first step. Anyway, you have a fulcrom of which you are placing 2 of 8 objects on each end. If by chance you pick the slightly heavier object as one of the two balls, you have in fact, found the slightly heavier one in the first round... btw dont be a smartass with your interviewer, he is looking for smarts not smarmy;)
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Aug 22, 2009
Respectfully, the folks who are answering "3" are mathematically modeling the nature of the balance incorrectly. Performing a measurement on a balance scale is not binary. It is trinary. Each measurement gives you one of three responses: The left is heavier, the right is heavier, or they are equal. So while you do need three binary bits to specify a number from one to eight, you need only two TRINARY-DIGITS
Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements.
Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.
Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements.
Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.
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Aug 24, 2009
English isn't my mother tongue... What is a balance scale? If looking up on Google, I find some devices with two bowls on a bar bearing in the center. Hence, the answer is once (if I'm luck enough to select the heavier ball in teh first measurement) If a balance scale allows to measure only one ball at a time, then it would take two measurements, unless you'd have more information on the weight, which is not listed here, hence doesn't exist in the context of the question^.
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Aug 25, 2009
3 times. Not having looked at the other comments, hopefully, I am the 26th to get this right.
Put the balls 4 and 4 on the scale, Take the heavier side and place those balls 2 and 2 on the scale. Take the heavier side and place them 1 and 1 giving the heaviest ball.
Put the balls 4 and 4 on the scale, Take the heavier side and place those balls 2 and 2 on the scale. Take the heavier side and place them 1 and 1 giving the heaviest ball.
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Aug 25, 2009
OK, now I read the comments and see that the people, like the question are divided into to groups, systematic approach people that say 3 (like I did) and analytic people that say 2. It takes a systematic person (me) a minute to get the answer. I'm guessing it took the analytic 5 minutes just to interpret all the ramifications of the question, i.e. they aren't idenitical if..., do it by hand..., get lucky.
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Aug 25, 2009
minimum is 1 (if lucky - 25% chance by picking 2 balls at random) & max is 2 (using most efficientl process to absolutely determine without luck - 3/3/2 scenario)
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0 of 1 people found this helpful
Aug 25, 2009
While Symantec was busy weighing my balls I took a job with NetApp.... They need to focus on hiring good, capable security engineers, not weighing their balls.
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Mar 17, 2010
The point of these interview questions is to both check your logical brain function and to hear how you think. Most of you are just posting jerk off answers trying to be funny, or you are really dumb. These answer get you nowhere with me in an interview. Think out loud, go down the wrong path back track try another logic path, find the answer. None of this "0 if you use your hands". That is fine if you are interviewing for a job in advertising where creativity is desired, nobody wants you writing code like an 8 year old.
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