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  Book of the Day: How Would You Move Mount Fuji? Microsoft's Cult of the Puzzle - How the World's Smartest Company Selects the Most Creative Thinkers
Time: 00:00 EST/05:00 GMT | News Source: ActiveWin.com | Posted By: Robert Stein

Anyone who's interviewed for a job at Microsoft is intimately familiar with questions like the one in this book's title. They've probably also pondered such problems as why are manhole covers round? how do they make M&Ms? what does all the ice in a hockey rink weigh? how many piano tuners are there in the world? Questions like these, which test problem-solving abilities, not specific competencies, are de rigueur at job interviews at Microsoft, other tech firms and on Wall Street. In this hybrid book-it's at once a study of corporate hiring, an assessment of IQ testing's value, a history of interviewing and a puzzle book-science writer Poundstone (Carl Sagan: A Life in the Cosmos) explains the thinking behind this kind of interviewing. In straightforward prose, Poundstone describes the roots of logic questions in interviews (the approach appears to have had its modern beginnings at Shockley Semiconductor Laboratory in 1957), drawing on the history of IQ testing in hiring interviews, psychological studies and interviews with Microsoft ex-interviewers and interviewees, makes a strong case for eliminating standard questions like "What are your strengths and weaknesses?" and replacing them with logic puzzles. Almost half of the book is devoted to an "answer" section, where Poundstone gives possible solutions to the brainteasers. Although it lacks a specific focus, this is a fun, revealing take on an unusual subject.

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#1 By 2332 (65.221.182.2) at 7/11/2003 11:06:12 PM
Here are a few brain teasers for you guys. Please, don't run off an look up the answers on the net. You're just cheating yourself. The 1st one is by far the hardest. The rest are in no particular order.

1.) Continue the following sequence of numbers:

1
11
21
1211
111221
...

2.) There are three light switches in one room, and three light bulbs in another room. The two rooms are seperated by a closed door through which no light can get through. Once you've gone through the door, you cannot return to the previous room. How can you figure out which light switches go with which light bulbs?

3.) A man is looking at a picture of somebody on the wall. He then says "Brothers and sisters I have none, but that man's father is my father's son." Who is the man in the picture?

4.) A man arrives on an island whose inhabitants are quite unique. There are only two groups of people. One is blue. The other is green. One group always lies, and the other always tells the truth. The problem is, the man forgot which was which. While walking through the woods on the island on his way to a town, he arrives at a fork in the road. At the fork stand one blue man, and one green man. Asking only one question of only one man, how can the man figure out which way to go?

I'll post the answers some time tomorrow.

#2 By 7390 (63.211.44.114) at 7/12/2003 12:25:47 AM
#4 ask "Which way is he going to tell me to go?" and go the opposite way.
#3 It is a self portrait.

#3 By 12071 (203.217.72.185) at 7/12/2003 1:59:56 AM
1) (312211)
1 can be said to be one one, i.e. 1 1
11 can be said to be two one's, i.e. 2 1
21 can be said to be one two and one one, i.e. 1 2 1 1
1211 can be said to be one one, one two and two one's, i.e. 1 1 1 2 2 1
111221 can be said to be three one's, two two's and one one, i.e. 3 1 2 2 1 1

2) Turn on one of the switches, leave it on for a while and then turn it off. Turn on a different switch and then go into the room with the lights. The light that is on is controlled by the 2nd switch that you left on and by touching the two other globes you should be able to tell which bulb was switched on by the 1st switch you turned on (i.e. it will be warmer than the other globe).

3) RedHook is correct on this one.

4) RedHook is correct on this one.

#2 I'm sure you probably realise this but you just asked two questions.

Offtopic Slightly: So this is why Microsoft make such buggy software =) They hire all the really intelligent people with great problem solving skills regardless of their ability to code! =)

#4 By 20 (67.9.179.51) at 7/12/2003 2:00:12 AM
*** SPOILER ***

1.) Still thinking...

2.) Old one. Flip on one, wait 5 minutes, flip on the other. Go into the room. The hot one is the first switch, the one that's on but not warm yet is the end, and the off one is the 3rd switch which you didn't touch.

3.) His son.

4.) Another oldy. My favorite answer is, ask the one to ask the other which way is the correct way and repeat it back to you.
if he repeats it back truthfully, he's the truth guy and the other guy is the liar. Go the opposite way.

If he lies, then follow the way the other guy said.

#5 By 20 (67.9.179.51) at 7/12/2003 2:03:08 AM
More of a programming question...

You have a linked list which you wish to traverse, but you cannot be certain that it is not looped. You must traverse the list and process it while checking for loop conditions.

What's the best (least memory, least number of traversals) way to both process the list and watch for loop conditions?

#6 By 20 (67.9.179.51) at 7/12/2003 2:05:19 AM
You have a 3ga bucket and a 5ga bucket and an unlimited supply of water (no other measuring tools and no graduations on the bucket).

What are the steps necessary to end up with 4ga in the 5ga bucket?

#7 By 20 (67.9.179.51) at 7/12/2003 2:07:15 AM
#3, #4: It's not a self-portrait.

The (subject's father) is the (man's father's son).

(man's father's son) = man

(subject's father) = man

If the man is the subject's father, then the subject is his son.

#8 By 12071 (203.217.72.185) at 7/12/2003 3:29:25 AM
#7 Fill up the 3ga bucket and pour the contents into the 5ga bucket. (3ga is empty, 5ga has 3 ga of water in it)
Full up the 3ga bucket again and pour 2ga into the 5ga bucket (i.e. as much as the 5gz can fit = 2ga). The 3ga bucket now has 1ga of water and the 5ga bucket is full.
Empty the 5gz bucket and then fill it with the 1gz of water that's in the 3ga bucket.
Fill up the 3ga bucket for the last time and pour the whole amount into the 5ga bucket, leaving the 3ga bucket empty and the 5ga bucket with exactly 4ga of water (as long as you didn't spill too much of it in the process =))

#8 You are quite correct, it is his son in the picture.... I rushed into answering that one a bit quickly!

#9 Just admit that you're wrong because you asked two questions.

This post was edited by chris_kabuki on Saturday, July 12, 2003 at 03:33.

#9 By 7390 (63.211.44.114) at 7/12/2003 11:42:36 AM
#8, I also was quick to answer. You are right it is the man's son.

#10 By 2332 (65.221.182.2) at 7/12/2003 3:17:39 PM
Correct answers:

1.)
-------------
1
11
21
1211
111221
312211
13112221
...
-------------
Each number is the previous number "spelled" out. So "11" is "one 1". "21" is "two 1's". "1211" is "one 2, one 1". Etc.

2.) Using the heat of the light bulbs is correct.

3.) The person in the picture is the son of the man talking.

4.) There are actually two possible ways to solve this problem. The first way is the one that I'm sure some of you looked up on the net, and is the most common answer. You simply ask one of the men what the other one would say, then do the opposite. My answer (and the answer I was hoping somebody would come up with so I could be more confident that you didn't cheat) was you could ask either of the men what they would say if you were to ask them which way to go. The liar would lie about his lie, which would be the truth. The honest man would tell the truth anyway.


#4 - chris_kabuki - Did you cheat on this one, or had you heard it before. I was unable to figure this out on my own... and very few people could get this in less than a couple of days, much less a couple of hours. Come on. Be honest. :-)

#11 By 2332 (65.221.182.2) at 7/12/2003 3:27:44 PM
#12 - That solution doesn't work because a linked list doesn't have to loop back to the very beginning. It could loop back to someplace in the middle. For instance:

A->B->C->D->B

If you just stored B, you would not detect the loop and would be stuck.

The best solution to this is to add a marker bit to each node in the list. Before you begin your iterations, clear all the marker bits. Each time you visit a node, you check the marker bit. If it's set, you've detected a loop. If it's not set, set it, and continue.

This would require a single bit for each node in the list. The bits don't have to be stored on the actual objects in the list. Instead, you could store them in a sufficently large variable outside of the list. You then use bitwise operations to change the appropriate bit.

This only encounters problems if items are added to or removed from the list *during* your iteration. If this is the case, you must store the bit on the actual linked objects.

Another option would be to use a link counter. Each time an object in the list is referenced as a forward pointer, it gets a counter incremented. If you detect a counter greater than 0 anywhere past the first index, you've got a loop.

There are lots of solutions to this... obviously. :-)

#12 By 12071 (203.217.72.185) at 7/13/2003 2:12:42 AM
#13 I've encountered that question in the past which is why I knew the answer - although I have to admit that the first time I saw it I wasn't quite sure about it as I though it was just some sequence of 1 and 2's. I've done plenty of these sorts of puzzles in the past, the most recent lot was for the reverse-engineering academy - an online course where you get to practise different skills including these sorts of puzzles. It was quite a lot of fun - but as a result you do get to see the same questions or at least very similar questions repeated all over the place.

#13 By 20 (67.9.179.51) at 7/13/2003 2:38:02 AM
#14: That answer requires more memory as n approaches maximum, so that's not a great solution.

Also, there is an assumption that the list is not changing (or is locked or something).

The best solution I've heard and heard repeated several times is this:

Maintain two pointers. One point is the current node pointer (which you would have anyhow during a regular traversal) and maintain another pointer.

Increment pointer #1 node by node as you go through the list.

Increment pointer #2 by two nodes as you go.

If pointer #2 ever == pointer #1, you have a loop.

If pointer #1 reaches the end, you do not have a loop.



#14 By 135 (208.50.204.91) at 7/13/2003 12:10:11 PM
To move Mount Fuji...

I would redraw the map. Defining the point where we want Mount Fuji to be as Mount Fuji, and then renaming the old Mount Fuji to Mount 'Formerly known as Fuji'.

That is the easiest way.

BTW... Chris Sells has some stuff on his website regarding Microsoft interviews:
http://www.sellsbrothers.com/fun/msiview/

Oh, and I came across this while looking up that info... (this is funny)

http://bbspot.com/News/2003/01/os_quiz.php

I'm classified as AmigaOS. :)

#15 By 16302 (64.110.237.213) at 7/13/2003 4:54:14 PM
#2 - another solution
open the door, and while remaining in the room, test each light switch. The puzzle did not say that the door had to remain closed; just if you go through it, you cannot return.

#16 By 2332 (65.221.182.2) at 7/13/2003 7:56:21 PM
#18 - That was my mistake. It should have said just that. Otherwise, it would have been a dumb puzzle.

#17 By 20 (67.9.179.51) at 7/13/2003 11:43:45 PM
#18,19: I always heard that the room with the lights was upstairs and you can only go up the stairs once and you can't see the lights or light eminating from the room from the switches

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