Challenge Math Challenge - April 2019

[QuantumQuest]

Correct, if the prisoners see 0, 1 or 2 black hats, then the lineup for 0 black hats is needed.

But if there are 5 black hats or more, then the smallest number of black hats any prisoner will see is 4 black hats. So once a prisoner sees 4 black hats or more, he'll know not a single prisoner will think that other prisoners are seeing 0, 1 or 2 black hats. If there's no chance that any prisoner is seeing 0, 1 or 2 black hats, then the lineup for 0 black hats will result in no information, therefore can be skipped. That is, if I see 4 black hats or more, then I know up front that nobody will step up on the 0 black hats lineup.

So I suspect there may be an optimized logic with fewer lineups once a prisoner sees some N number of black hats, but I can't get my head on how that would be.

Let me get things straight. First, it is not known to the prisoners in advance how many black hats will be or who will have it. So, there will be 1 to 11 black hats. It is the warden's decision exactly how many will be and to whom. But any solution must account for all the posibilities and not for specific cases.
Second, the number of line-ups is already decided by the warden and it is an integral part of the test. How can we change the number of line ups?

 

[fbs7]

The size of the triangles added to each iteration is the same as the initial equilateral triangle. On each iteration a new triangle is added on each free edge.

Alright, starting trying my hand at Padawan-Level #3:

Spoiler

The orientation of the triangles is irrelevant; in this example, first set of triangles (call i=0) is brown, second set of triangles (call i=1) is red, i=2 is orange, i=3 is yellow, i=4 is green, i=5 is blue. The number of triangles on each iteration (call $T_i$) is {1, 3, 6, 9, 12, 15, ...}, to it seems that $T_0 = 1, T_i = 3*i$, and the total number of triangles in iteration n is $T_0 + T_1 + ... + T_n = 1 + 3 + 6 + ... + 3*n = 1 + \frac{3}{2}*n*(n+1)$:

But how to prove that - not easy! All kinds of coordinate transformation didn't help much, and trilinear coordinates proved to be a pain to prove anything; after 4 hours on this I'm no closer on finding a proof!

How to prove that each iteration $i$ will add exactly $3$ triangles to the scene... grrr... maddening I can't find that proof! Little tip anyone?

 

[fbs7]

Let me get things straight. First, it is not known to the prisoners in advance neither how many black hats will be nor who will have it. So, there will be 1 to 11 black hats. It is the warden's decision exactly how many will be and to whom. But any solution must account for all the posibilities and not for specific cases.
Second, the number of line-ups is already decided by the warden and it is an integral part of the test. How can we change the number of line ups?

The prisoners will not know in advance, but once a prisoner lines up and sees say 7 black hats, he knows that no other prisoner will see less than 6 black hats; and those that see 6 black hats will know that no other prisoner will see less than 5 black hats. Therefore, those prisoners that see 7 black hats and those that see 6 black hats will all come to the same conclusion, which is that no prisoners are seeing 0 black hats.

Therefore, there's no point in using the 10:00 lineup to test the hypothesis that some prisoner will move up because they see 0 black hats, as that hypothesis is impossible.

So, once the prisoners see 6 black hats or 7 black hats, then they can use the 10:00 lineup to say test a different hypothesis, like that some prisoner is seeing 1 black hat. Therefore the prisoner's action on the 10:00 lineup can be to test a possible hypothesis, not an impossible one.

It's not a matter of changing the number of lineups, but of using the 10:00 lineup to test a different hypothesis and avoid spending time confirming a hypothesis that all prisoners know is impossible once the prisoners see a minimum number of black hats.

 

[QuantumQuest]

The prisoners will not know in advance, but once a prisoner lines up and sees say 7 black hats, he knows that no other prisoner will see less than 6 black hats; and those that see 6 black hats will know that no other prisoner will see less than 5 black hats. Therefore, those prisoners that see 7 black hats and those that see 6 black hats will all come to the same conclusion, which is that no prisoners are seeing 0 black hats.

Therefore, there's no point in using the 10:00 lineup to test the hypothesis that some prisoner will move up because they see 0 black hats, as that hypothesis is impossible.

So, once the prisoners see 6 black hats or 7 black hats, then they can use the 10:00 lineup to say test a different hypothesis, like that some prisoner is seeing 1 black hat. Therefore the prisoner's action on the 10:00 lineup can be to test a possible hypothesis, not an impossible one.

It's not a matter of changing the number of lineups, but of using the 10:00 lineup to test a different hypothesis and avoid spending time confirming a hypothesis that all prisoners know is impossible once the prisoners see a minimum number of black hats.

Do the prisoners know in advance how many black hats are there? What if there is just one? Or two? Or three? Again, I'll state the obvious: there must be the first line-up where if there is just one black hat which the prisoner having it will know before this line-up, then he'll step forward. If none steps forward then we go to the second line-up etc. Prisoners cannot avoid any line-up and at the same time they need to use the outcome of the previous line-up - except the case that there is just one black hat, in order to deduce what is the correct thing to do. I cannot explain this in more simple terms.

 

[QuantumQuest]

and the total number of triangles in iteration n is $T_0 + T_1 + ... + T_n = 1 + 3 + 6 + ... + 3*n = 1 + \frac{3}{2}*n*(n+1)$:

This is not correct.

How to prove that each iteration $i$ will add exactly $3$ triangles to the scene...

Be careful here. Check it again.

 

[fbs7]

Do the prisoners know in advance how many black hats are there? What if there is just one? Or two? Or three? Again, I'll state the obvious: there must be the first line-up where if there is just one black hat which the prisoner having it will know before this line-up, then he'll step forward. If none steps forward then we go to the second line-up etc. Prisoners cannot avoid any line-up and at the same time they need to use the outcome of the previous line-up - except the case that there is just one black hat, in order to deduce what is the correct thing to do. I cannot explain this in more simple terms.

Alright; my last attempt at this: why is this not a solution?

(a) If the prisoner sees 0, 2, 4, 6, 8, 10 black hats then he steps up at 10:00
(b) If the prisoner sees 1, 3, 5, 7, 9, 11 black hats then he steps up at 10:05

 

[QuantumQuest]

Alright; my last attempt at this: why is this not a solution?

(a) If the prisoner sees 0, 2, 4, 6, 8, 10 black hats then he steps up at 10:00
(b) If the prisoner sees 1, 3, 5, 7, 9, 11 black hats then he steps up at 10:05

If you read post #44 carefully it's not difficult to understand it.

 

[fbs7]

I think I'm going to answer my own question. If the prisoners are killed immediately if even one walks up incorrectly even once, then the post above is not a solution, as there's 50% chance they will walk up with the wrong solution by 10:00. But, if they can walk up incorrectly at 10:00, then are not killed, then the walkup at 10:05 will be correct, and that will be a solution.

Therefore I suppose there is a rule to the problem, which is that the prisoners can only walk up once, at some time, that must be the right solution or they will all die immediately.

I didn't understand that rule from the writeups, so I tried to find shorter solutions with a smaller number of walkups, even if some of those walkups could be incorrect. The shortest one I could find in that form is the one above.

So, is it true that if the prisoners walk up incorrectly once they are killed?

 

[SpinFlop]

This isn't true, unfortunately. A continuous function with smaller and smaller "spikes" can be integrable on $[0,\infty)$ and yet fail to have a limit at infinity.

Could you give an example of (or clarify) this? Given that you speak of 'smaller and smaller spikes', this lead me to believe that the limit at infinity is still zero, such as in the classic example of a sine wave in an exponentially decaying envelope:
$$ \lim_{n \rightarrow \infty}e^{-x}sin(x) = 0$$

 

[Infrared]

I mean that the spikes get narrower, not shorter. For example, on the interval $[n,n+1]$ (say for $n>0$), let the graph of $f$ be a triangle with vertices $(n,0),(n+2^{-n},1),(n+2\cdot 2^{-n},0)$ and zero on the rest of the interval.

Then $\int_n^{n+1}f=\frac{1}{2}\cdot 2\cdot 2^{-n}=2^{-n}$, so $\int_1^\infty f=\sum_{n=1}^\infty 2^{-n}=1$

 

[QuantumQuest]

If the prisoners are killed immediately if even one walks up incorrectly even once,...

Isn't this obvious from the wording of the question?

In order to pass the test, all the prisoners with a black hat and only them, will have to step forward during the same line-up. If they succeed, all prisoners will be freed otherwise they will all be executed.

I think that the question states explicitly what is needed to be stated.

 

[Adel Makram]

I think this bit is enough (you don't need the earlier work) since in the last product, the first integral converges iff $\alpha<1$ and the second does if $\alpha>1/2.$

You also seem to have confused which bound went with which variable of integration but as the integrand is symmetric in $x,y$, this is okay!

Nice work.

Is this function related to Riemann Zeta function in the critical strip?

 

[member 659869]

High School 5:

Spoiler

$$N=6210001000$$
Solution: I started at $N=8100000000$ and kept modifying the number until it stabilized.

 

[Infrared]

Is this function related to Riemann Zeta function in the critical strip?

Not that I'm aware of. Do you have something in mind?

 

[wrobel]

4.4. Let ff be a continuous element of L1((0,∞))L^1((0,\infty)). Suppose that x=x(t)x=x(t) solves the differential equation dxdt=−px+f(t)\frac{dx}{dt}=-px+f(t). Prove that limt→∞x(t)=0\lim_{t\to\infty}x(t)=0.

First it should be clarified what $p$ is. For example if $p=0$ then the assertion is wrong.
If $p=const>0$ then the proof is as follows.

$$x(t)=e^{-pt}x(0)+\int_0^te^{-p(t-s)}f(s)ds.$$ And we have
$$\int_0^te^{-p(t-s)}f(s)ds=\int_0^Te^{-p(t-s)}f(s)ds+\int_T^te^{-p(t-s)}f(s)ds.$$
Since $f\in L^1(\mathbb{R}_+)$ we see that for any $\varepsilon>0$ there exists $T^*>0$ such that for any $T>T^*$ and for any $t>T$ the following estimate holds
$$\Big|\int_T^te^{-p(t-s)}f(s)ds\Big|<\varepsilon.$$
On the other hand, with any fixed $T$ we have
$$\int_0^Te^{-p(t-s)}f(s)ds\to 0$$ as $t\to \infty$
There is no need to assume continuity of $f$.

 

[Master1022]

Here is my attempt at Highschool Problem 1(b). Not sure how rigorous my working needs to be, but here is my attempt.

We know that a minimum path must not include movements in more than one cardinal direction (eg. cannot move left and then later up) because that extra move is redundant and can be avoided by making diagonal moves. Thus, my plan is basically to move diagonally and in only one cardinal direction.

Let us fix our king at some arbitrary origin, and then let us draw lines that extend in the four 'cardinal' directions around the king. Here is a more qualitative description of the method first:
1. Set up boundary lines for your king emanating from his current position
2. For a given destination, evaluate whether it lies on any of the 4 boundary lines
- If it lies on a boundary line, then move in that respective direction and then go back to step 2
3. If not, see which quadrant the destination is in and move diagonally 1 move in that direction - then go back to one and repeat

e.g. path from (0,0) to (4,5): (0,0) --> (1,1) --> (2,2) --> (3,3) --> (4,4) --> (4,5)

Slightly more mathematical, but not much. Let us define a cartesian coordinate system such that the centre of each square is defined by a pair of integer coordinates (x,y) and our King will start at (0,0).

For a given destination (a,b), let us assume that a &gt; b. We will end up moving diagonally b times and then vertically a - b times. Therefore, for destination (a,b), we will only move |a| - |b|+ |b| = |a| times.

This makes sense following on from part (a). Thinking about the possible squares (that the king can land after n moves) as an area of a square (2n + 1)^2 as written by fbs7 for the general n. For us to reach a square in the minimum number of moves (n), we need n such that it is the first square to include our destination point. This will only happen if \textbf{n = number of moves = max(|a|,|b|)} for all n.

How do we know that this method yields a minimum path?
1. There are no redundant moves in two cardinal directions
2. There is not 1 shortest path, can flip around instructions of the 'algorithm' and will generate another version of the path.

Please let me know if this is insufficient (whether or not I have arrived at the correct answer). Many thanks in advance.

 

[wrobel]

5.5. Let f:S5→S3×S2f:S^5\to S^3\times S^2 be a smooth function. Show that ff takes critical values in any set of the form S3×{p}S^3\times\{p\} for p∈S2p\in S^2.

what if the whole sphere is mapped into a single point?

 

[Master1022]

Here is my attempt at Highschool problem 1(c).

Given that we cannot move diagonally, there are squares which we can only reach if the number of moves n is even, and others that can only be reached when n is odd.

For any n moves, we know that the furthest squares that can be reached are all the squares whose coordinates (a,b) are such that |a| + |b| = n. Drawing this out on a cartesian coordinate system draws out a rotated square, which makes sense from intuition. So we have dealt with the outermost squares, but now we need to deal with all squares within our region. So when n is, for example, even (and greater than 0), it can only move in increments on 2 moves. Therefore, it will always remain on a square with the colour that it started with (helpful to picture a chess-board as in the question). For an odd n, the king will be forced to end on a square that is of a different colour to the one that it initially began in. The diagonal movement was able to act as effectively 2 moves, thus allowing the king to move to a square of the same colour.

Thus, the possible inner squares will be all the smaller rotated squares for any n_i that is smaller than n and returns the same result: remainder(n,2) = remainder(n_i ,2) (not worrying about n = 0 at the moment). It can reach these smaller squares just by oscillating back and forth, between adjacent squares, if necessary.

Getting to some maths:
Splitting this into: for even n = 0:
the number of squares = 1

for even n > 0:
each new 'layer' will have 4n squares of possible destinations.

Thus, the total number of possible squares = 1 + \sum_{n=2}^n 4 n, with the latter part forming the sum of an arithmetic sequence (a = 2, d = 2, \lambda = n/2). This can thus be re-written as 1 + 4 \times \frac{\lambda}{2}(2 + 2 \lambda) = \mathbf{ 1 + 4 \lambda(1 + \lambda)}

For odd n > 0:
Following on from the same logic, basically, everything is the same as even numbers, except that it won't be able to return to its original square, so it won't have the extra +1. However, the formula for the arithmetic sequence needs to be redone:
let \mu = \frac{n+1}{2}
number of squares = 4\sum_{n=1}^n n = 4 * \frac{\mu}{2}(2(1) + 2(\mu - 1)) = 4 \mu ^ 2 = \mathbf{(n+1)^2}

In conclusion, for n moves, \lambda = n/2, the king can move:
if n = 0: 1 square
if n > 0 and even: 1 + 4 \lambda(1 + \lambda) squares
if n > 0 and odd: (n+1)^2 squares

 

[Infrared]

@wrobel

Apologies, I forgot the crucial condition that $f$ is onto!

 

[fresh_42]

@wrobel

Apologies, I forgot the crucial condition that $f$ is onto!

Edited.

 

[QuantumQuest]

Let us fix our king at some arbitrary origin, and then let us draw lines that extend in the four 'cardinal' directions around the king. Here is a more qualitative description of the method first:
1. Set up boundary lines for your king emanating from his current position
2. For a given destination, evaluate whether it lies on any of the 4 boundary lines
- If it lies on a boundary line, then move in that respective direction and then go back to step 2
3. If not, see which quadrant the destination is in and move diagonally 1 move in that direction - then go back to one and repeat

I see a good effort here but I can't visualize it in the exact way needed.

Let us define a cartesian coordinate system such that the centre of each square is defined by a pair of integer coordinates $(x,y)$ and our King will start at $(0,0)$ .

For a given destination $(a,b)$ , let us assume that $a > b$ . We will end up moving diagonally b times and then vertically $a - b$ times. Therefore, for destination $(a,b)$ , we will only move $|a| - |b|+ |b| = |a|$ times.

This makes sense following on from part (a). Thinking about the possible squares (that the king can land after n moves) as an area of a square $(2n + 1)^2$ as written by fbs7 for the general n. For us to reach a square in the minimum number of moves $(n)$ , we need $n$ such that it is the first square to include our destination point. This will only happen if n = number of $\textbf{n = number of moves = max(|a|,|b|)}$ for all n.

I don't ask for a rigorous way of describing the minimum number of moves required to reach a destination. I just ask for the optimal strategy for this, described as a number of steps. So, moving in the diagonal direction as you say, is the first step. Now, from what you write I see that you're on the right way for the next step but how can we describe it in simple words i.e. next, what is the move in order to reach the destination, independently of whether it is on the diagonal or not?

 

[Master1022]

I see a good effort here but I can't visualize it in the exact way needed.

Thank you for your response. I have attached a quick sketch to try to help visualise my thinking.

Edit: Please let me know if this is not legible

 

[QuantumQuest]

Here is my attempt at Highschool problem 1(c)...

For even $n \gt 0$ the number of squares is not correct. Check it again.

 

[QuantumQuest]

I have attached a quick sketch to try to help visualise my thinking.

Yes. So, again to my question: You're moving diagonally and if the destination is not on the diagonal, what is the next move?

 

[Master1022]

For even $n \gt 0$ the number of squares is not correct. Check it again.

Thank you for your response. Is the answer not (n+1)^2, which is equivalent to what I have written? I thought it would be this as it will be the total number of squares of the same colour as the starting square within our region of furthest possible squares.

 

[Master1022]

Yes. So, again to my question: You're moving diagonally and if the destination is not on the diagonal, what is the next move?

Thanks for your response. Maybe my steps weren't clear enough. However, to address your question, do we just move in one cardinal direction that will most quickly put us on a similar diagonal with the destination? However, isn't this method the same as the one as I outlined, with the steps just reversed?

 

[QuantumQuest]

However, to address your question, do we just move in one cardinal direction that will most quickly put us on a similar diagonal with the destination? However, isn't this method the same as the one as I outlined, with the steps just reversed?

Let's take it in a simple way. We are moving along a diagonal - whichever it is. If we meet our target (=destination) we're done. If not , there is a simple way to state what you do according to where is the destination relatively to the point on the diagonal that you are. That's all I ask. The way you describe it is as taking another diagonal etc. Is this the optimal way?

 

[QuantumQuest]

Is the answer not $(n+1)^2$ , which is equivalent to what I have written? I thought it would be this as it will be the total number of squares of the same colour as the starting square within our region of furthest possible squares.

You start with the premise that

each new 'layer' will have $4n$ squares of possible destinations.

For instance take $n = 2$. Then what you write means that starting from, say, a black square, the possible squares of the new layer, as you call it, will be $8$. Now I ask, is this premise correct?

EDIT: As a hint, if you take the next positive even integer i.e. $n = 4$ then what I ask above may be more evident.

 

[Master1022]

You start with the premise that
For instance take $n = 2$. Then what you write means that starting from, say, a black square, the possible squares of the new layer, as you call it, will be $8$. Now I ask, is this premise correct?

EDIT: As a hint, if you take the next positive even integer i.e. $n = 4$ then what I ask above may be more evident.

Thank you for your response. This is what I found for n = 2 and n = 4. Are you saying that I am missing some squares?

The outer layers seem to have 4n squares in them, so I will keep looking

 

[QuantumQuest]

@Master1022 as I check again High School problem $1 c$ I see that you mean that the total sum is $1$ plus a new layer (=possible positions) each time and the series is on even numbers, so you finally mean $(n + 1)^2$ but the way is written is more cumbersome than it needs to. So, what is the difference from $n$ being odd?

 

[Master1022]

@Master1022 as I check again High School problem $1 c$ I see that you mean that the total sum is $1$ plus a new layer (=possible positions) each time and the series is on even numbers, so you finally mean $(n + 1)^2$ but the way is written is more cumbersome than it needs to. So, what is the difference from $n$ being odd?

I realized after posting that they were the same, but not when I was working through the mathematics.

 

[Master1022]

Let's take it in a simple way. We are moving along a diagonal - whichever it is. If we meet our target (=destination) we're done. If not , there is a simple way to state what you do according to where is the destination relatively to the point on the diagonal that you are. That's all I ask. The way you describe it is as taking another diagonal etc. Is this the optimal way?

Thank you for your response. I fail to see how my method doesn't get a king/piece to the destination square in an optimal number of moves. In response to your suggestion, I will keep thinking. At first, I thought about closest approach (i.e. you move diagonally until the direction between you and the destination is perpendicular to the diagonal you move on, but that only works if destination square is of the same colour as the diagonal). Would you be able to explain what is redundant in my method which is basically "move diagonally towards the destination until you are below/above/left/right it, then move straight towards it"?

 

[QuantumQuest]

I realized after posting that they were the same, but not when I was working through the mathematics.

So, in both cases it is $(n + 1)^2$ squares and you get the credit for it.

As a simpler way to analyze and state the whole thing, you can say that starting from a certain square whether white or black, for each new $n$, we take into account all the squares of the same color ($n$ even) or the opposite color ($n$ odd) that are on the "border" - created by these squares, plus the squares of the same color (with the ones on the border) inside it.

 

[QuantumQuest]

Would you be able to explain what is redundant in my method which is basically "move diagonally towards the destination until you are below/above/left/right it, then move straight towards it"?

The way you state the steps in post #66 - taking also into account the diagram in your post #72, is basically correct but you can state it in a simpler manner - as you do in the above quote i.e. Move diagonally towards the destination until the king reaches the row or column that includes the destination and then move in a straight line. Describing it this way, it is obvious that if king is already on the same row or column with the destination then there will be zero diagonal moves.