Solutions to riddles

Just thought it might be a good idea to have a separate thread for the solutions, so no riddles are "spoilt".

(I've solved Klaashaas' riddle, but I think he should post the answer himself! :-) )

Marius

Give us at least two weeks to ponder, please. :)

Prisoners of war riddle:

Replace the colors with the digits 0, 1 and 2. Then, everyone adds all numbers on the prisoners in front of him, and takes modulo 3 of this number (which is always 0, 1 or 2, ofcourse, and thus has a matching color). The prisoner in the back calls out this number. He only has a 1/3 chance to live, I'm afraid :-)

The second guy then subtracts the sum of all prisoners in front of him, modulo 3, from the number the previous one called out. If he gets a negative number, -1 -> 2, -2 -> 1.That is his number/color!

Then, the next person just does the same thing and gets his own number/color, saving everyone except the poor guy in the very rear...

Marius

Painter riddle:

He makes exactly the same path as the first time, poor guy...

Correction to prisoners of war riddle:

Firs guy calls out number A. Second calls out a number as stated. Next, everyone has to take A minus all numbers called out minus the sum of all numbers he can see mod 3, and take this number modulo 3, where modulo 3 of negative numbers is as stated.

Sorry for my mistake in previous post...

Well done Marius!

And as Glenn stated increasing the number of hat colours just maps to calculating in a different modulus :)

Camel riddle:

I'm not sure how the quality of these bananas would be, but this is how I manage to save 500 bananas (or 501, if I don't allow the camel to eat one when we reach our destination, since I no longer need the camel...)

Load up 1000 bananas, go 250 miles, drop 500 bananas, and go back.

Load up 1000 bananas, go 250 miles to first depot, load up another 250 bananas (the same number as the camel has eaten), then go another 250 miles, drop 500 bananas, go back 250 miles to first depot, load up the rest of the bananas (250), and go back to base.

Load up 1000 bananas, travel 500 miles to second depot (first depot is empty now). Load up 500 bananas, and travel to destinataion.

You can use the camel even more efficiently ;-)

OK, I make first a depot of 600 bananas at 200 miles.

Next, I make a depot of 334 bananas at 533 miles.

I now have 200 bananas at 200 miles, and 334 bananas at 533 miles.

Last trip, I will then manage to save 533 bananas (or 534, if I don't let the camel eat when we reach our destination). There's still a banana rotting at the 533 miles depot...

Can I save even more bananas now?

Letting your camel die is not so nice ;)

I think of the man feeding the camel as he walks

so the first depot is as you said on the 200mile mark. The second at 533+1/3 mile from the start.

This way you can save 533+1/3 banana without letting the camel die...

After getting the solution of the prisonner of war riddle - it's a mathematicans solution and it would kill a lot of people because normal gifted enemies of numbers and abstracts thinking would fail. So as the head of the imprissoned army I would judge the one telling the solution - can he shut up? If Yes I would tell him and let him alive - if no I would kill him to save 66,7% of my people. ;) (You have to operate nine cognitive functions without failuere und such a stress, to really save all but one - impossible for untrained people.)

camel riddle - 250 miles first depot - 500 miles second depot - 2500 miles for the whole journey - 500 bananas

Stupid answer to arithemetics riddle:

(1+3)*4 + 6. It's 24 in base 9 - you didn't specify which base we wer to calculate in :-)

Otherwise, I need to use other things than just +-*/. Using other functions it's no problem, but with only these four...?

I looked up the solution on rec.puzzles, and was kicking myself afterwards for not finding it myself. It really is possible with only +, -, *, /. :-)

OK, so then I'll not be able to sleep tonight - I'll just lie awake pondering this. Tomorrow, I'll get it right!!! :-)

how bout 6/[1-(3/4)]

That's it.

Marius' answer to the painter's counter-intuitive trip

is correct. When the ladder tips backwards, Jack's path

is 1/4 of a circle. When the base of the ladder slides

out and the top slides down the house, Jack

FOLLOWS THE *EXACT* SAME PATH AS BEFORE!

To see this, suppose we attach the center points of two

identical ladders and put them up flat against this house.

Claim: this sissors-like contraption can fall so that each

ladder follows one of the two described motions and hence,

their center points must follow the same path (since they

are attached).

To see this claim is true, slide one ladder's base away

from the wall by some amount while the top maintains

contact with the wall. This ladder is on the diagonal of

a rectangle. We can always place the other ladder on the other diagonal and in this position, one end abuts the wall/floor corner and the center points of the ladders are together (since the diagonals of a rectangle cross in the middle).

Solution to Glenn's 7 point puzzle:

Well done, Devil! I was sure you would know that. It was a problem I encountered in a math contest, when I was exactly half of my current age. I hadn't solved it.

I knew this figure from the unsolved problem of the chromatic number of the plane. It shows that it is at least 4 (and a different approach shows that it is at most 7).

Scales #1: I figured this out once - took a bit of work to reproduce :-)

There's no simple answer. This is a possible measurement scheme:

Balls numbered A-L. Measurements numbered #1-#3.

(#1) ABCD vs. EFGH {weigh A+B+C+D against E+F+G+H}

if ABCD < EFGH {ABCD is lighter than EFGH}

(#2) ABE vs CDF

if ABE < CDF

(#3) A vs B

if A < B => *A lighter*

if A = B => *F heavier*

if A > B => *B lighter*

if ABE = CDF

(#3) G vs H

if G < H => *H heavier*

if G > H => *G heavier*

if ABE > CDF

(#3) C vs D

if C < D => *C lighter*

if C = D => *E heavier*

if C > D => *D lighter*

if ABCD = EFGH

(#2) IJK vs ABC

if IJK < ABC

(#3) I vs J

if I < J => *I lighter*

if I = J => *K lighter*

if I > J => *J lighter*

if IJK = ABC

(#3) L vs A

if L < A => *L lighter*

if L > A => *L heavier*

if IJK > ABC

(#3) I vs J

if I < J => *J heavier*

if I = J => *K heavier*

if I > J => *I heavier*

if ABCD > EFGH

(#2) ABE vs CDF

if ABE < CDF

(#3) C vs D

if C < D => *D heavier*

if C = D => *E lighter*

if C > D => *C heavier*

if ABE = CDF

(#3) G vs H

if G < H => *G lighter*

if G > H => *H lighter*

if ABE > CDF

(#3) A vs B

if A < B => *B heavier*

if A = B => *F lighter*

if A > B => *A heavier*

Lost the indentation... I hope this is more readable:

Balls numbered A-L. Measurements numbered #1-#3.

(#1) ABCD vs. EFGH

if ABCD < EFGH {ABCD is lighter than EFGH}

...(#2) ABE vs CDF

...if ABE < CDF

......(#3) A vs B

......if A < B => *A lighter*

......if A = B => *F heavier*

......if A > B => *B lighter*

...if ABE = CDF

......(#3) G vs H

......if G < H => *H heavier*

......if G > H => *G heavier*

...if ABE > CDF

......(#3) C vs D

......if C < D => *C lighter*

......if C = D => *E heavier*

......if C > D => *D lighter*

if ABCD = EFGH

...(#2) IJK vs ABC

...if IJK < ABC

......(#3) I vs J

......if I < J => *I lighter*

......if I = J => *K lighter*

......if I > J => *J lighter*

...if IJK = ABC

......(#3) L vs A

......if L < A => *L lighter*

......if L > A => *L heavier*

...if IJK > ABC

......(#3) I vs J

......if I < J => *J heavier*

......if I = J => *K heavier*

......if I > J => *I heavier*

if ABCD > EFGH

...(#2) ABE vs CDF

...if ABE < CDF

......(#3) C vs D

......if C < D => *D heavier*

......if C = D => *E lighter*

......if C > D => *C heavier*

...if ABE = CDF

......(#3) G vs H

......if G < H => *G lighter*

......if G > H => *H lighter*

...if ABE > CDF

......(#3) A vs B

......if A < B => *B heavier*

......if A = B => *F lighter*

......if A > B => *A heavier*

Scales #2:

With one measurement you can discriminate 3 states: scale goes left, goes right, or stays even. With 3 measurements that means 3^3 = 27 states. The #1 riddle has 24 states; 12 balls, each of which can be heavier or lighter.

Still, You can't do the trick with 13 balls (26 states), because you can only weigh 4 against 4 balls in the first measurement. If you measure 5 against 5, and one group is heavier than the other, you're left with 10 states for the second measurement, which is impossible (you've got 2 measurements left, which is a maximum of 3^2=9 states).

For 4 measurements, there's a theoretical maximum of 3^4=81 states; but because of the above, you can't weigh more than 12 against 12 the first time. That gives a maximum of 3*12 = 36 balls (72 states).

Scales #1 is correct. But you can measure more balls than 36 with 4 weighings! :-)

River riddle:

The easiest way would be to swim 100m in one direction, turn 90 degrees, and swim 100m again.

Assuming that he is at the very edge of the river (and still doesn't feel it!), and happens to start out up- or downstream, that would guarantee a safe arrival within 200m.

I haven't looked too hard at the problem, but I guess a cirkular curve would land him as well. In that case, the maximum distance would be (2*100m*pi)/4, or approx 157m.

Are there still better ways?

That doesn't work; suppose he starts near the shore, starts swimming at a 45 deg angle for 100 m, turns 90 deg towards the shore he came from. He'd end up at the same distance from the shore, only 141 m further along the river...

Scales#2: you may be able to discriminate 13 balls with 3 measurements if you have a reference ball. That would make it possible to handle 38 balls with 4 measurements. You weigh 13 against 13. If the scale is even, the problem reduces to the 12 balls left over, else you use one of the 12 left-over balls as a reference. Haven't worked out a measuring plan for 13 balls yet, though.

You can do even better than 38, Marc! :-)

River riddle: if he swims in a circle with r=50m, worst case scenario is that he needs to swim 3/4 of the circle (~240m), right? I'm thinking about starting to swim 1/4 of a circle, and if not reached the shore, change tactic. Need to ponder a bit more about this one. Thanks for the nice riddle, btw!:)

I am glad you liked it klaashaas :-)

worst case scenario is that he needs to swim 3/4 of the circle (~240m), right?

Wrong! He needs to swim the whole circle (314m).

If he swims a circle of 100m, he has to swim a little less than a quarter of the circle. -

But look at Abigail in the riddle post. ;)

Quarter of a circle with radie 50m doesnt do the trick. The height at the smallest part is to small.

Also 240m is a bit too long ;)

U=2r*pi

r=100.000.....1

1/4U ~ 132m

Every shorter r may lead to a longer distance.

Sorry, maybe 133m (calculating without paper or calculator)

In the 12-ball puzzle, there are solutions where the

weighings do not depend on the results of previous weighings.

For example: (number the balls from 1 through 12)

<br/>             Left Pan    Right Pan<br/>             ========    =========<br/>Weighing 1   4 8 10 11   1 2 5  7<br/>Weighing 2   2 4  7 12   3 5 6 11<br/>Weighing 3   5 6 10 12   7 8 9 11<br/><br/>

Now start with a sum of 0 and perform the weighings, observing which side is heavy, if any.

<br/>             Left Heavy   Right Heavy<br/>             ==========   ===========<br/>Weighing 1   Subtract 1   Add 1<br/>Weighing 2   Subtract 3   Add 3<br/>Weighing 3   Subtract 9   Add 9.<br/><br/>

The absolute value of the sum is the number of the bad

ball and it is light or heavy depending upon its appearance

on the light or heavy side of a weighing.

It's also possible to use letters instead of numbers.

The cutest solution of this form is in the rec.puzzles

archive.

<br/><br/>Ma Do vs. Like<br/>Me To vs. Find<br/>Fake  vs. Coin<br/><br/>

With this method, it is not as simple to determine the

bad ball (or Coin as in an alternate version), but it sure is easy to remember the weighings; just remember the

sentence "Ma Do Like Me to Find Fake Coin"

Coin problem:

I'm able to do it with 9 weightings. Is it possible to do it in less?

I am sorry Jupp but your solutions has flaws. First, from what you write I think you mean a circle of radius 100m. Then you say that all the guy needs is to swim a quarter of this circle. A quarter of this circle is about 157m and not 133m. But he needs to swim half of the circle which is 314m, rather long.

Nick Wedd mentioned in the other thread, a solution of 231m.

I found one that is 400/sqrt(3) = 230,94 m. I think Nick meant this one. This is how: Go straight for 200/sqrt(3) meters then turn 60 degrees and go another 200/sqrt(3) meters. In other words, make the two sides of an equilateral triangle (a triangle with 3 equal sides) that its height is 100 meters (or perhaps a little bigger, 100,00...01 meters).

I meant turn 120 degrees.

This is a very good solution close to the optimal one :-)

I think getting closer can be done with some reasoning and even finding the optimal path is possible without too much mathematics, although proving that this is optimal is very hard...

Glenn, great solution!

This way, you can also solve the 4 weighings case (it's 39 balls).

If you continue the pattern, it looks like with N weighings, you can solve (3^N-3)/2 balls.

4 weighings (try to find a mnemonic for *this* one :-) ):

#1: [2 7 8 13 14 19 20 25 26 31 32 37 38] vs. [1 4 5 10 11 16 17 22 23 28 29 34 35]

#2: [3 6 11 13 14 16 21 24 29 31 32 34 39] vs. [2 4 5 7 12 15 20 22 23 25 30 33 38]

#3: [5 7 11 13 14 16 18 20 22 33 35 37 39] vs. [6 8 9 10 12 15 17 19 21 32 34 36 38]

#4: [15 17 19 21 23 25 27 29 31 33 35 37 39] vs. [14 16 18 20 22 24 26 28 30 32 34 36 38]

. Left Heavy Right Heavy

Weighing 1 Subtract 1 Add 1

Weighing 2 Subtract 3 Add 3

Weighing 3 Subtract 9 Add 9

Weighing 4 Subtract 27 Add 27

39 balls is right, and I think your solution is more elegant than mine. However, there's something wrong with your general formula, I think...

As for the guy in the river, I too found the triangle giving 231 meter. I haven't found a better solution yet, and if I keep scratching my head over this one, I'll soon go bald...

Ypercup

happy birthday first.

Second: You're right, it's ~ 158m.

You wrong about the halfcircle.

If the radius is 100m, the swimmer is in distance to the shore of the river 0,01m and starts to swim with one arm (the wrong one!) a perfect circle, he has completed a right angle after 1/4 of the circle from mid point to starting point and to actual point. The mid point is in 0,01m distance to the shore, the swimmer is 100,01m in distance of the straight shore in an right angle.

Well, Jupp, but suppose he starts out not 90 degrees from the shor, but 89 degrees? Then he needs to complete the halfcircle...

Marius,

if he is nearer to the goal (that's Your proposal) than the distance is shorter.

The smimmer does't start in a degree distance, he starts in a meter distance.

The soccer problem.

There is the triangel goal line, running line, shot line. The angle goal line, running line is a right one. So he has to shot, when the two other angles are equal.

And because he is perfect and the goal keeper knows this, he will only score, if the goal keeper is k.o. before. ;)

How do I make drawings here? It would be much easier to explain then. However. He HAS to comlete a halfcircle. And not only does he start at a specific point, he also start in a specific direction, which can be given as an angle from the shoreline. Suppose he starts 0.01 m from the shore. He starts out 89 degrees from the shore (nearly perpendicular to it). If he then swims in a circular motion with radius 100 meters, he has come ALMOST to the other shore after a quarter of a circle. The next quarter of the circle takes him back to the first shore again.

Another way of seeing it: Suppose he starts out 0.01 meters from the shore, and starts out swimming PARALLELL to the shore. Then, obviously, he has to make a halfcircle if the radius is 100 meters.

To include a drawing, put it on a web page and type .