ASF Spoilers Thread

[skyQuake]

a). The kids knew the cookies were Oatmeal?

b). Tilt the drum and pour it until the fluid surface makes a perfect diagonal across the height of the drum

c). The message would take 5 million years to reach the aliens. Although, it's questionable what speed "intergalactic radio message" travels at.

d). A 3 prong plug on the computer vs 2-hole power point from the wall.



I have to say that the questions are not logically watertight enough for an adult... but I'd love it as a kid. I have a friend's 8yr old boy birthday coming up...

Well done. All solved.

Let me know if the kid can answer the questions about displacement (finger in glass of water q), rubber shrinking when heated etc!

 

[bellenuit]

I think the proof is simply that the angular velocity of the swimmer is always greater than the cyclops provided he stays within 1/4 of the radius, and he can afford to swim slightly further out with each zig. The swimmer would have to consult trig tables to calculate exactly what angle is optimum at each zig & zag, but in practice the swimmer would just swim more parallel with the shore until the cyclops reversed.

The zigzags would get shallower as the distance from the centre increased. On the whole I think the zigzag method would be more efficient than the swimming in a circle method (and it would have the benefit of p***ing the cyclops off ).

And many thanks to you & others for providing these puzzles .

Actually the quickest way to get out of the lake safely is to swim in a spiral to a point on the safety circle (the innermost circle with radius A) according to the source article, but the mathematics to prove that is beyond me. Once at the safety zone, one obviously then swims directly to shore.

The spiral makes sense. If you swim directly to the safety circle, the cyclops will have moved to the point directly in line to where you are heading and you then have to swim in a circle until you get 180 degrees ahead of him. Swimming in a spiral means the cyclops will not be at the point directly nearest the point you touch the safety circle, so you will not have to swim as much in a circular direction.

The spiral is like a continuous changing zig, without the zag. Zagging doesn't make sense IMO. Once the cyclops starts running round the lake in anticipation of where you are heading, it never then makes sense for him to change direction, hence causing you to change direction and zag. Once the cyclops has decreased the angular distance between you and him to under 180 degrees (say to 160 degrees), for him to switch direction would suddenly mean the angular separation is now 200 degrees based on the direction he is now heading if you were then to zag, leaving him worse off than if he kept going in the same direction.

 

[bellenuit]

Actually the quickest way to get out of the lake safely is to swim in a spiral to a point on the safety circle (the innermost circle with radius A) according to the source article, but the mathematics to prove that is beyond me. Once at the safety zone, one obviously then swims directly to shore.

I need to change that on re-reading the article. Spiralling is the fastest way to get out, but it doesn't say explicitly that you spiral just to the innermost circle. It just says you spiral to somewhere between the two inner circles and then you still must circle until you are 180 degrees ahead of him before making a direct exit.

 

[cynic]

According to my calculations Denise was born 14th May 2002.

 

[bellenuit]

According to my calculations Denise was born 14th May 2002.

This is my answer too. I reckon it is easier than the original Cheryl's Birthday.

My workings for those who are having difficulty (the number in brackets beside the dates indicate the statement below it that eliminated that particular date)

17 Feb 2001 (3), 16 Mar 2002 (6), 13 Jan 2003 (4), 19 Jan 2004 (5)

13 Mar 2001 (4), 15 Apr 2002 (2), 16 Feb 2003 (7), 18 Feb 2004 (5)

13 Apr 2001 (2), 14 May 2002, 14 Mar 2003 (6), 19 May 2004 (5)

15 May 2001 (3), 12 Jun 2002 (1), 11 Apr 2003 (1), 14 Jul 2004 (5)

17 Jun 2001 (2), 16 Aug 2002 (7), 16 Jul 2003 (7), 18 Aug 2004 (5)

Denise then told Albert, Bernard and Cheryl separately the month, the day and the year of her birthday respectively.

The following conversation ensues:

Albert: I don’t know when Denise’s birthday is, but I know that
Bernard does not know.
1. This rules out 12 Jun 2002 and 11 Apr 2003 because these two DAYs are unique.
2. But the only way Albert can be sure that it is not these two days is because he knows the MONTH is not Apr or Jun. So that also rules out the remaining dates with those months: 15 Apr 2002, 13 Apr 2001 and 17 Jun 2001.

Bernard: I still don’t know when Denise’s birthday is, but I know that Cheryl still does not know.
3. Since Bernard knows the DAY, this rules out 17 Feb 2001 and 15 May 2001, as these DAYS are unique from what is left over.
4. Because he knows that Cheryl still doesn’t know the birthday, he must know the DAY isn’t 13, because 13 Mar 2001 is the last left over DAY in 2001 meaning Cheryl could know the answer if it were that YEAR. So this eliminates 13 Mar 2001 and also the other DAY with 13: 13 Jan 2003.

Cheryl: I still don’t know when Denise’s birthday is, but I know that Albert still does not know.
5. Since Cheryl knows the YEAR, she must know that it isn’t 2004 because that has a date with a unique left over MONTH, 19 Jan 2004, so if it were 2004, Albert who knows the month could know the birthday. This eliminates all 2004 dates: 19 Jan 2004, 18 Feb 2005, 19 May 2004, 14 Jul 2004 and 18 Aug 2004.

Albert: Now I know when Denise’s birthday is.
6 Since Albert knows the MONTH and knows the answer, then it must be one of those leftover dates that has a unique MONTH. This eliminates the two dates that do not have a unique MONTH: 16 Mar 2002 and 14 Mar 2003.

Bernard: Now I know too.
7. Since Bernard knows the DAY and now knows the birthday, then it must be the 14 May 2002, as the remaining dates have a common DAY, 16. So 16 Feb 2003, 16 Aug 2002 and 16 Jul 2003 are eliminated. The only date left is 14 May 2002, Denise’s birthday.

Cheryl: Me too.

 

[luutzu]

Saudi:

The wise man told them that the Sheik meant to race the camels once it arrives? So the slowest camel will be the one most tired?

 

[luutzu]

Billiard:

1. Put 3 on each side - if it balances, then go to step 2.

2. Put the 6 you just scaled aside, divide the other 6 into 2 group of 3 [say, B1 and B2].

2.1 Take B1 and scale against A1 [any 3 balls, all 6 are identical as they balanced]

If B1 and A1 balanced, then the "defective" one must be in B2...

brb...

 

[bellenuit]

Saudi:

The wise man told them that the Sheik meant to race the camels once it arrives? So the slowest camel will be the one most tired?

This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.

 

[bellenuit]

Billiard:

1. Put 3 on each side - if it balances, then go to step 2.

2. Put the 6 you just scaled aside, divide the other 6 into 2 group of 3 [say, B1 and B2].

2.1 Take B1 and scale against A1 [any 3 balls, all 6 are identical as they balanced]

If B1 and A1 balanced, then the "defective" one must be in B2...

brb...

You have to find the actual ball, not just a group that it is in and you have to determine whether it is heavier or lighter than the others, not just that it is different (defective).

 

[luutzu]

This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.

Yea. I googled the answer.. haha... and didn't agree with their solution
Mine above is bad too... but swapping? That's assuming the other prince allow the swap, and also assumes they've been riding around for days on a few camels, one or two of which happen to belong to other princes.


Haven't finish the billiard yet. Havng too much fun with my new toy at the moment

 

[keithj]

This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.

Not sure what advice the old man gave, but jumping on each others camel & racing to city seems to be the answer. Effectively they are attempting to make the others camel faster.

 

[keithj]

You have 12 billiard balls that look identical in every respect, except one is slightly different in weight (could be heavier or lighter than the other 11). You have a simple balance (one that stays level if both trays contain exactly the same weight, but will tip one way or the other if they are different).

You have to determine by no more than 3 weighings, which ball is different and whether it is heavier or lighter than the others. How can you do it?

This seems to be a complicated method .....

The balls can be in one of 4 states - Unknown, CorrectWeight, CouldBeHeavier, CouldBeLighter
All 12 balls start in the Unknown state.
If they are proven to be the same weight as another ball (or set of balls) they are labeled CorrectWeight & can be eliminated
If they are proven to heavier or lighter than other balls then they are labeled as such.

Empty the balance prior to any weighing.

Label all 12 balls with [B]Unknown[/B].

Weigh 1.
Put 4 on each side
If they balance then
    Label those eight on the balance as C (for Correct weight). We now have 4 Unknowns.
    Weigh 2.
         Put 3 (of the 4 remaining) Unknowns on one side the balance & 3 of the C on the other.
         If they balance then
                label the 3 Unknowns on the balance as C. 
                So we now have 1 Unknowns but we don;t know if it's heavier or lighter, so...
                Weight 3a.
                    Put the only remaining Unknowns of the balance opposite one C, so we can find out if it's heavier or lighter
         else if the C and Unknown is heavier then
            label the Unknown on the heavier side as [B]UBCBH [/B](Unknown but could be heavier)
            label the 2 Unknowns on the lighter side as [B]UBCBL [/B](Unknown but could be lighter)
            Weigh 3b
                    Put one UBCBH and one UBCBL on one side and 2 C on the other
                    if they balance then
                            the UBCBL that is not on the scales is the lighter than the other 11
                    if the side with one UBCBH and one UBCBL is heavier then the UBCBH is heavier than the other 11
                    else UBCBL is lighter than the other 11
         else if the 2 Unknowns are heavier then
                label them both UBCBH
                Weight 3c.
                    Label the 2 Unknowns that aren't on the balance as C
                    Put both the UBCBH on opposite sides of the balance
                    One will be heavier than the other 11


As they don't balance then
        label the 4 not on the balance as Correct
        label the heavier 4 Unknowns as UBCBH and the other 4 UBCBL
        Weigh 2a.
        Put 2 UBCBH and one UBCBL on the LHS of the balance, and 1 UBCBH and 2 UBCBL on the RHS
        and leave one of each aside.
        if they balance then
            Label all 6 on the balance as C
            Weight 3d.
            Put the remaining UBCBL on the balance opposite a C
            if they balance then the remaining one (ie not on the balance)  is heavier than the other 11
            else the one on the balance (labeled UBCBL) is lighter than the other 11
        else they don't balance
            Label the 2 that aren't on the balance as C
            if the LHS is heavier (it contains 2 UBCBH and only one UBCBL) then
                there are 2 possibilities (either the lighter one by itself is lighter than all others, or one of the 2 UBCBH is heavier than all others)
                Label the UBCBL on the LHS as C
                Label the 2 UBCBH on the RHS as C
                We are left with 3 balls - 1 UBCBL and 2 UBCBH
                Weight 3e.
                    Put one of the UBCBH and the only remaining UBCBL opposite 2 C
                    if they balance then the UBCBH is heavier than the other 1
                    else if the 2 Cs are heavier the the UBCBL on the balance is lighter than the other 11
                    else the UBCBH on the balance is heavier than the other 11
            else the RHS must be heavier then
                just do the inverse of Weight 3e.

Is there an easier way ?

 

[bellenuit]

Not sure what advice the old man gave, but jumping on each others camel & racing to city seems to be the answer. Effectively they are attempting to make the others camel faster.

Yes, that was the advice. Swap camels.

 

[cynic]

Scenario 1:

Weigh balls 1-4 against 5-8

If even weigh balls 1-3 against 9-11

If even weigh ball 12 (which is oddball) against 1 to determine relative weight by direction of tilt.

Scenario 2:

Weigh balls 1-4 against 5-8

If even weigh balls 1-3 against 9-11

If uneven 9-11 contains oddball and relative weight is now known from direction of tilt.

Weigh ball 9 against 10

If even 11 is the oddball otherwise it can be identified as either 9 or 10 by the direction of the tilt.

Scenario 3:

Weigh balls 1-4 against 5-8

If uneven weigh balls 4-7 against 8-11

If even 1-3 contains oddball and relative weight is now known from direction of tilt in first weighing.

Weigh ball 1 against 2

If even ball 3 is the oddball otherwise it can be identified as either 1 or 2 from the direction of the tilt.

Scenario 4:

Weigh balls 1-4 against 5-8

If uneven weigh balls 4-7 against 8 - 11

if uneven and tilt unchanged then oddball is either 4 or 8.

Weigh ball 4 against 1.

If uneven then oddball is 4 otherwise it is 8.

Relative weight can now be deduced from the direction of the tilt in the second weighing.

Scenario 5:

Weigh balls 1-4 against 5-8

If uneven weigh balls 4-7 against 8 - 11

If uneven and tilt changed then oddball is in 5-7 and relative weight is now known by direction of tilt.

Weigh ball 5 against 6.

If even ball 7 is the oddball otherwise it can be identified as either 5 or 6 from the direction of the tilt.

 

[bellenuit]

Both cynic and keithj seem correct. I can't see any flaw in their answers.

 

[cynic]

3211000

 

[cynic]

3211000

42101000

 

[cynic]

42101000

521001000

 

[cynic]

521001000

6210001000

 

[cynic]

6210001000

C210 0000 0000 1000