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| Jun-26-08 | | SetNoEscapeOn: <WannaBe>
I like your questions, and they all seem to matter, especially #2. The answer to #5 is probably: because it is in Microsoft's book of recommended interview questions :) |
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| Jun-26-08 | | SetNoEscapeOn: But, let's assume that they meant that he flashlight must always be "held" by those crossing the bridge, and that if two people are on the bridge, then they must be walking "abreast"- i.e the flashlight must always be "with" anyone while they are on the bridge, in the same location <while they are moving>. I think this is probably what the questioner intended... I don't see the answer yet, but I would guess that those are the rules. |
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Jun-26-08
 | | al wazir: <WannaBe>: There are two solutions: Time THIS SIDE OTHER SIDE
0------ABCD
2------CD--------AB
3------ACD-------B
13-----A---------BCD
15-----AB--------CD
17---------------ABCD
Time THIS SIDE OTHER SIDE
0------ABCD
2------CD--------AB
4------BCD-------A
14-----B---------ACD
15-----AB--------CD
17---------------ABCD
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Jun-26-08
 | | WannaBe: Thanks, <al wazir>! =) |
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Jun-26-08
 | | OhioChessFan: Another way to address my last problem:
Take P = S+a, Q = S+b, R = S+c with a>b>c then PS = QR implies S = bc/(a-b-c) so clearly a >= b+c+1, and the inequality holds if [ (P-S)^2 - 4S - 8 ] [a - b - c] > 0, i.e. if
a^2 (a-b-c) - 4bc - 8(a-b-c) > 0. This is increasing in a when a>b+c so we can substitute b+c+1 for a without affecting the validity of the inequality. This gives (b+c+1)^2 - 4bc - 8 = (b-c)^2 + 2(b+c) -7.
But c>=1, b>=2 and b-c>=1 so that (b-c)^2 + 2(b+c) - 7 >= 0 |
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| Jun-26-08 | | SetNoEscapeOn: <al wazir>
Elegant... I actually peeked at the answer on another site, something that I knew I would regret (and of course do). I made it more complicated than it had to be. I kept thinking that somehow the answer had to lie with leaving somebody standing on the bridge halfway (this just works against the goal though). My other issue is that somehow I kept thinking that A always had to be the "sole runner", not realizing that it's quicker to have B do it one of the trips... Very well done. |
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Jun-26-08
| | al wazir: <OCF>: Interesting. My proof may be equivalent to a restatement of your; I'm not sure. Both arguments lead to the same conclusion, namely (P-S)^2 ≥ 4S+8, not (P-S)^2 > 4S+8. Is there any choice of P, Q, R, S for which *equality* holds? I haven't found one. <WannaBe, SetNoEscapeOn>: Full disclosure: I've seen this problem before. But all "crossing" problems are solved by essentially the same trick -- someone has to make the reverse trip. |
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Jun-26-08
| | OhioChessFan: <<OCF>: Interesting. My proof may be equivalent to a restatement of your; I'm not sure.> LOL, I gave up trying to decide. Very similar defining of terms. |
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| Jun-27-08 | | Once: Love the bridge problem. I didn't quite get it, but I was close. Here's my logic - finished off by <Al Wazir> giving us the answer. The "obvious" solution does not work. If we use A to make all the return trips, we have A+D = 10
A returns 11
A+C = 16
A returns = 17
And poor A and B are stuck on the wrong side of the bridge with the flash light out of juice. So let's think laterally. Clearly C and D need to cross at the same time. That way they take the time of the slowest (10 minutes) rather than their combined time (15 minutes). But C and D cannot cross on the first attempt (because there would be no-one to take the flashlight back to the beginning). And they cannot cross on the last attempt (because there would have been no-one to bring the torch back to them from the previous crossing). So C and D have to cross in one of the middle trips. That's as far as I got last night, but Al Wazir finished it off for me. The first move has to be A and B (2 minutes), because we need to leave C and D where they are. Then A returns (3 minutes). Then the longest trip of all (C and D) - 13 minutes so far. Then B goes back to pick A up (another 2 minutes there and 2 minutes back) = 17 minutes. Very glad that I didn't get this in a job interview! Incidentally, I did have a job interview yesterday, but luckily no questions as hard as this one. |
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| Jun-27-08 | | kellmano: <WannaBe: I don't have the solution to this one, I've been pondering this one for a while...
You have 8 identically shaped objects, however, one of them weights a little more (or less) than the other 7, now, with a balance, can you determine the one that is more/less with just 2 weighings?> I'm pretty sure that's impossible. If you weigh two versus two and the scales stay level, then you will have one use of the scale and four objects to choose from, which is impossible. If you weigh three versus three, and the scale tilts, it will be impossible to establish which of the six objects is different. Here is a similar, but definitely solvable puzzle. There are twelve coins, one of which is fake and a slightly different weight to the others. You have three uses of a pair of weighing scales to establish which is fake. This is a very tricky puzzle. There are at least two solutions, one very elegant, and one really rather vulgar. I should like the elegant one. |
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Jun-27-08
 | | SwitchingQuylthulg: <kellmano> Here's a solution I came up with (don't know if it's the elegant one or the vulgar one - it looks elegant enough because you can do the same weighings regardless of previous weighings' results, but I won't mind if you call it vulgar). Name the 12 coins from A to L. First weighing: ABCD vs. EFGH, second weighing: ABCE vs. DIJK; and third weighing, ADHI vs. CGJL Depending on which coin is fake, the scale will have tilted as follows: A -> L-L-L or R-R-R
B -> L-L-n or R-R-n
C -> L-L-R or R-R-L
D -> L-R-L or R-L-R
E -> L-R-n or R-L-n
F -> L-n-n or R-n-n
G -> L-n-L or R-n-R
H -> L-n-R or R-n-L
I -> n-L-R or n-R-L
J -> n-L-L or n-R-R
K -> n-L-n or n-R-n
L -> n-n-R or n-n-L
All of which, as far as I can see, completely unambiguously tell you which coin is fake. |
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| Jun-27-08 | | kellmano: <SwitchingQuylthulg> No offence but that was the one i consider a little vulgar. :-) It's a bit exhaustive. The other solution involves varying subsequent choices depending on the outcome of each weighing. See if you can work that out. By the way, what does your name mean? |
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| Jun-27-08 | | Once: I think the scales problem only works if know whether the odd object is heavier or lighter than the others. Then <Thorski>'s solution works. Let's assume that the odd object is heavier than the others. Divide the eight objects into two groups of three and one group of two. Weigh the two groups of three against each other. If the two sets of three are identical, then the odd one must be in the group of two. Weight them against each other to find it. If the two sets of three are different, take the heavier one. Pick two objects from this group and set the other aside. Weigh the first two against each other. If one is heavier than the other, then it is the odd one out. If the two are identical, then the odd one out is the one you set aside. <Kellmano's> problem is similar, but I think you can solve it without knowing whether the fake is heavier or lighter than the others. Divide the 12 coins into four groups of three. Put two of these groups aside and compare the other two. If they are the same, then the fake is in the two groups that you set aside. Take one of the groups that you set aside and one of the groups that you have already weighed and compare them. You will end up with one group of three after two weighings. Then use the solution as for the eight - put one aside and weigh the other two. |
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Jun-27-08
| | SwitchingQuylthulg: <Once> I don't think your solution works since you don't know if the fake is heavier or lighter than other coins. You can only work out which of three coins is a fake with one weighing if you have additional information (such as whether the fake is heavier or lighter than other coins.) Otherwise, you might end up knowing that one of two coins is heavier than the other but with no way of telling which one is the supposed weight... I didn't get much sleep this night so I might have missed something trivial :) |
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| Jun-27-08 | | Once: <SwitchingQuylthulg> You might be right, but my head is starting to spin at this point! I thought I had covered this by using the second weighing to compare one of the unweighed groups with one of the groups that I had already weighed. If we know that the weighed group is the "right" weight for three coins, then the difference between this and the unweighed group tells us whether the fake is heavier or lighter. Then we use the third weighing to narrow down which of the three remaining coins is lighter or heavier. I think the flaw is my approach is that you don't compare the unweighed fourth group with anything until the third weighing. Might need to think again on this one! |
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Jun-27-08
| | al wazir: <OCF>: The answer to my question (<Is there any choice of P, Q, R, S for which *equality* holds?>) is yes. If P = 6, Q = 4, R = 3, and S = 2, then (P-S)^2 = 4S+8. I don't know how I failed to notice that before. |
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| Jun-27-08 | | SetNoEscapeOn: <kellmano>
That other puzzle with eight objects is not impossible to solve, unless you mean that it must be made clear whether the "oddball" is heavier or lighter than the others (the problem cannot just state that it is either or). |
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Jun-28-08
| | OhioChessFan: Change the position of only one number in the sequence given below in order that the amended sequence corresponds to a palindromic sequence. 1, 4, 9, 6, 2, 1, 5, 10, 4
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| Jun-28-08 | | arifattar: Move 1 from the start of the sequence to the end of the sequence, and you have a palindrome. |
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Jun-29-08
| | al wazir: <OCF>: IIVIXVIIIVXIV --> IVIXVIIIVXIVI (That's almost certainly what <arifattar> meant, so he should get the credit for solving it first.) I liked your math problem better. This one smacks of subterfuge. |
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Jun-29-08
| | al wazir: Since <chessgames.com> solvers like trick problems and live all over the globe, here's a tricky geography quiz: 1. Mexico and Canada are the two nearest neighbors of the U.S. What country is next? 2. (a) Which is farther west, Reno or Los Angeles? (b) Washington, D.C., or Lima, Peru? 3. (a) Which is farther north, Naples or New York? (b) Rome or Chicago? (c) Nice, France, or Toronto, Canada? 4. (a) Do you go east or west in passing through the Panama Canal from the Pacific to the Atlantic Ocean? (b) What about passing from Lake Huron to Lake Erie? (c) From Lake Erie to Lake Ontario? 5. What U.S. state is (a) farthest north? (b) farthest west? (c) farthest east? (d) farthest south? |
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Jun-29-08
| | OhioChessFan: I'll try <aw> without looking any up. I'll ignore the tricky aspect and just put what I really think. 1. Russia
2. a. Have to pick the obvious LA
b. DC
3. a. NY b. Chicago c. Nice
4. a. north b. west c. east
5. a Alaska b. Alaska c Maine d. Hawaii |
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Jun-29-08
| | al wazir: <OCF>: Not too bad. Almost half right. |
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| Jun-29-08 | | dalbertz: I remember my 13 year old son telling me about #5. Alaska is also the farthest east, because the string of Alaskan islands stretch across the international date line. :) |
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Jun-30-08
| | OhioChessFan: Imagine that. A cooked Stumper. In fact, putting the 2 at the end,
1, 4, 9, 6, 1, 5, 10, 4, 2
IIVIXVIIVXIVII
is what I was looking for.
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