Riddles, brain puzzles and mathematical problems

[BAndrew]

Holy stromboli this is a tough one. BUT I'M SO CLOSE.

I understand very well what you mean. When I tried to solve this puzzle at first I managed to exchange the positions of 3 horses, but the 4th could just not be moved! Anyway, I am happy to hear you have progress. Keep it up.

Yes the riddle really is a tough one. I wanted to start hot with a tough riddle

[Statyk]

Same... asdjlasjdfa My brain is going "WHAT THE HELL MAN? You want me to WORK? PFFF"

EDIT: I'll come back on it later, as of right now, I give up.

[Nice]

i'd try to come out with a solution but im afraid i'm going to pull my muscle

[Statyk]

i'd try to come out with a solution but im afraid i'm going to pull my muscle

[Froge]

Solution:


I created an algorithm for this. If anyone wants to know what it was, just tell me.

Let B1 be the knight on 6, B2 be the knight on 1, W2 be the knight on 7, W1 be the knight on 5

B2 --> 4 --> 10 --> 2 --> 8
W2 --> 1 --> 4 --> 10 --> 2
W1 --> 7 --> 1 --> 4 --> 10
B1 --> 4 --> 1 --> 7 --> 5
W1 --> 4 --> 1 --> 7
W2 --> 10 --> 4 --> 1
B2 --> 2 --> 10 --> 4 --> 6
W2 --> 4 --> 10 --> 2
W1 --> 1 --> 4 --> 10
B2 --> 4 --> 1 --> 7
W1 --> 4 --> 6
W2 --> 10 --> 4 --> 1

I'll post my next question once i think of one.

[BAndrew]

Solution:

I created an algorithm for this. If anyone wants to know what it was, just tell me.

Let B1 be the knight on 6, B2 be the knight on 1, W2 be the knight on 7, W1 be the knight on 5

B2 --> 4 --> 10 --> 2 --> 8
W2 --> 1 --> 4 --> 10 --> 2
W1 --> 7 --> 1 --> 4 --> 10
B1 --> 4 --> 1 --> 7 --> 5
W1 --> 4 --> 1 --> 7
W2 --> 10 --> 4 --> 1
B2 --> 2 --> 10 --> 4 --> 6
W2 --> 4 --> 10 --> 2
W1 --> 1 --> 4 --> 10
B2 --> 4 --> 1 --> 7
W1 --> 4 --> 6
W2 --> 10 --> 4 --> 1

I'll post my next question once i think of one.

Congratulations! You are the first person to solve this riddle! I am interested in seeing an algorithm for this.

Looking forward to your riddle also

I updated my first post so people know that the riddle is solved. When you post your riddle I will update my post again.

[Froge]

I lied. It's not an algorithm, more like a mapping.

Spoiler below!

Move W1, W2, B2 to everything below 4, move B1 to 5, move both whites back to stuff before 4, place B2 at 6, move all the whites out to stuff below 4, move B2 beside B1, place whites wherever at 1 or 6 to finish.

Next question:

This crudely drawn circuit has 3 switches on top and 2 at bottom. The probability that any switch is open is 1/3.

Find the probability that a current can travel across this circuit.

[BAndrew]

My first guess was

P(Σ)= P(A) + P(B) + P( C ) + P(D) + P(E) =
1/3 + 1/3 + 1/3 + 1/3 + 1/3 = 5/3

But I doubt it's that simple.

ERM NO It's a possibility over 100%

Another thought was

(1/3) * (1/3) (1/3) * (1/3) * (1/3) = 1/243 which is approximately 0,004115 ~ 0,4115%
which makes more sense.
But I haven't been taught about probabilities at school so I am not sure.

[Froge]

Just to clarify for those who do not know anything about circuits: the word in the question is across, not around.

Edit: Far from correct, BAndrew.

[BAndrew]

Just to clarify for those who do not know anything about circuits: the word in the quesiton is across, not around.

Edit: Far from correct, BAndrew.

oops Didn't see across there.

---

Just to clarify for those who do not know anything about circuits: the word in the question is across, not around.

So either both the three upper or both the two lower switches have to work? (or every switch of course)
Is that how it goes?

In that case I think it will be
(1/3 * 1/3 * 1/3) + (1/3 * 1/3) - (1/3 * 1/3 * 1/3 * 1/3 * 1/3) =
1/27 + 1/9 - 1/243 = 9/243 + 27/243 - 1/243 = 35/243 =~ 0.144 =~ 14,4%