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>> At this point, they figured out how the device actually works. It
>> isn't an oscilator, it's a radio receiver.
>
> Interesting. I read about that, or something similar, where the device
> had a block of four or five transitors wired together but not connected
> to anything else. When they removed the transistors, it stopped working.
>
> I wonder if it's the same device.
That does, in fact, sound vaguely familiar...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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On Wed, 20 Jan 2010 08:56:25 +0100, scott wrote:
> http://www.glassdoor.com/blog/top-oddball-interview-questions-2009
>
> A lot of them seem like genuinely useful questions to ask candidates,
> one of our lecturers at University would regularly ask us questions like
> #7, 11 or 19 - it's interesting to see what assumptions and reasoning
> people make to get an estimate.
I agree that some do see useful. I like #19, probably partly because I
know how I'd answer: "Roughly one per space where a light bulb can be
fitted to the fixture." (This would allow for some of the fixtures to be
empty).
Jim
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On Wed, 20 Jan 2010 08:11:28 -0800, Darren New wrote:
> Warp wrote:
>> I have hard time parsing that question.
>
> It *is* messed up. The usual question is something like
>
> You have a race two miles long. For the first mile, one car goes 60MPH
> and the other goes 30MPH. How fast does the other car have to go to
> finish the race at the same time as the one going 60MPH?
Even then it doesn't parse correctly, because they've given you the
speeds for both cars - so asking how fast the slower car should go
doesn't make sense, because you've been told how fast it is going.
And I'm not sure that discarding the length of the track allows the
question to be answered even then. I assume they're asking about when
the two cars would cross the finish line at the same time, not when they
would finish at the same time (since the slower car would clearly be laps
behind the faster car). Assuming both are measured at a running start
(ie, at speed), one car going half the speed of another should cross the
finish line of a closed track at the same time every 2 laps. I think
that's the answer they're going for.
Jim
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On Wed, 20 Jan 2010 09:15:50 -0500, Warp wrote:
> However, from a practical point of view those pragmatic answers are
> much, much better than any convoluted answer you could come up with.
> Why? Because if you come up with a convoluted answer it shows that you
> tend to come up with convoluted solutions to simple problems, rather
> than doing the practical thing and just consult a manual or other source
> which *already* knows the answer, which would be the best thing to do in
> a practical situation in the job.
Except that often times in the real world the answers aren't so
straightforward, and some creativity is being figured. The answer to the
question isn't important, it's how you got there. I do agree that on
their own they're not useful, but in conjunction with other questions
(both "interesting" questions and "non-interesting" questions) can tell a
lot about the candidate.
Jim
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Jim Henderson wrote:
> On Wed, 20 Jan 2010 08:11:28 -0800, Darren New wrote:
>
>> Warp wrote:
>>> I have hard time parsing that question.
>> It *is* messed up. The usual question is something like
>>
>> You have a race two miles long. For the first mile, one car goes 60MPH
>> and the other goes 30MPH. How fast does the other car have to go to
>> finish the race at the same time as the one going 60MPH?
>
> Even then it doesn't parse correctly, because they've given you the
> speeds for both cars - so asking how fast the slower car should go
> doesn't make sense, because you've been told how fast it is going.
The race is two miles long. The slow car goes at a speed for one mile.
What's the speed in the second mile?
(I guess I left out that the fast car keeps going the same speed.)
> And I'm not sure that discarding the length of the track allows the
> question to be answered even then.
No, that would eliminate the ability to answer.
Anyway, that's an easy question to answer. A slightly better version is this:
You want to average 60MPH over two minutes. You go 30MPH for the first mile.
How fast do you have to go to average 60MPH for two minues?
Second alternate question: you want to average 60 MPH. You drive 10 miles at
30MPH. How many miles do you need to drive at 90MPH to average 60MPH?
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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scott wrote:
> http://www.glassdoor.com/blog/top-oddball-interview-questions-2009
>
> A lot of them seem like genuinely useful questions to ask candidates,
Google had a lot of these kinds of questions. Stuff like "how can you
determine how big a machine's cache is?"
I couldn't figure out if they wanted a program to measure the timing of
something assuming different cache sizes and look for the step in the
timing, or "look it up in the x86 instruction set manual".
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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scott wrote:
> http://www.glassdoor.com/blog/top-oddball-interview-questions-2009
As for delightfully fun puzzles I got at an interview, I got this one:
You have an infinitely long line. Two robots are dropped on parachutes onto
the line at random, and each drops its parachute (leaving it behind) where
it lands. Your task is to crash the two robots into each other.
Each robot runs the identical program starting at the beginning, and the
parachutes are identical, and they are in all ways identical other than
where they land, which is random.
You have the following instructions available for programming the robots:
Move the robot one step left.
Move the robot one step right.
Move the instruction pointer to a particular instruction (i.e., "goto")
Skip the following instruction if you're standing on a parachute.
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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On Wed, 20 Jan 2010 16:27:06 -0800, Darren New wrote:
> Jim Henderson wrote:
>> On Wed, 20 Jan 2010 08:11:28 -0800, Darren New wrote:
>>
>>> Warp wrote:
>>>> I have hard time parsing that question.
>>> It *is* messed up. The usual question is something like
>>>
>>> You have a race two miles long. For the first mile, one car goes 60MPH
>>> and the other goes 30MPH. How fast does the other car have to go to
>>> finish the race at the same time as the one going 60MPH?
>>
>> Even then it doesn't parse correctly, because they've given you the
>> speeds for both cars - so asking how fast the slower car should go
>> doesn't make sense, because you've been told how fast it is going.
>
> The race is two miles long. The slow car goes at a speed for one mile.
> What's the speed in the second mile?
That makes more sense - but fairly easy to answer algebraically.
> (I guess I left out that the fast car keeps going the same speed.)
>
>> And I'm not sure that discarding the length of the track allows the
>> question to be answered even then.
>
> No, that would eliminate the ability to answer.
How so? If you know the distances and speeds, then the time at which
both cars reaches the end of the distance becomes a solvable problem.
> Anyway, that's an easy question to answer. A slightly better version is
> this:
>
> You want to average 60MPH over two minutes. You go 30MPH for the first
> mile. How fast do you have to go to average 60MPH for two minues?
>
> Second alternate question: you want to average 60 MPH. You drive 10
> miles at 30MPH. How many miles do you need to drive at 90MPH to average
> 60MPH?
I agree, that question formulation would be easier to answer.
Jim
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Darren New <dne### [at] san rrcom> wrote:
> Jim Henderson wrote:
> > On Wed, 20 Jan 2010 08:11:28 -0800, Darren New wrote:
> >
> >> Warp wrote:
> >>> I have hard time parsing that question.
> >> It *is* messed up. The usual question is something like
> >>
> >> You have a race two miles long. For the first mile, one car goes 60MPH
> >> and the other goes 30MPH. How fast does the other car have to go to
> >> finish the race at the same time as the one going 60MPH?
> >
> > Even then it doesn't parse correctly, because they've given you the
> > speeds for both cars - so asking how fast the slower car should go
> > doesn't make sense, because you've been told how fast it is going.
> The race is two miles long. The slow car goes at a speed for one mile.
> What's the speed in the second mile?
I understood it to mean "when the faster car hits the one mile mark,
how fast should the slower car go from that point forward in order to
catch up and get to the finish line at the same time as the faster car"
(which is an interesting question).
However, if it really meant "when the *slower* car hits the one mile
mark, how fast must it go from that point forward to catch up" it becomes
a trick question (because the faster car has already reached the finish
line by that point).
> Anyway, that's an easy question to answer. A slightly better version is this:
> You want to average 60MPH over two minutes. You go 30MPH for the first mile.
> How fast do you have to go to average 60MPH for two minues?
> Second alternate question: you want to average 60 MPH. You drive 10 miles at
> 30MPH. How many miles do you need to drive at 90MPH to average 60MPH?
I like this version of two things going at different speeds more (I don't
remember the *exact* setup, so I'll just invent the numbers):
A man and a dog are walking home, which is at a distance of 1 mile.
The man walks at 3 mph. The dog start running at 10 mph towards home.
When it reaches it, it turns around and runs back to the man (keeping
the 10 mph velocity at all times), then turns around and runs to home,
and so on, until the man reaches home. What's the total distance that
the dog ran?
The problem sounds like you would need to calculate an infinite sum,
but there's actually a much simpler solution.
(There's also an anecdote of this problem having been given to some
mathematician (don't remember name), and him giving the answer almost
immediately. The person who posed the problem then remarked that "ah,
you figured out pretty fast that you don't need an infinite sum to get
the answer, but that there's a much simpler way", to which the mathematician
responded "what? I *did* use an infinite sum".)
--
- Warp
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On Wed, 20 Jan 2010 16:33:44 -0800, Darren New wrote:
[...]
Now my brain is going to be busy for the next month..... ;-)
Jim
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