Jump to content
BrainDen.com - Brain Teasers

Anza Power

Members
 View Profile  See their activity

  • Posts

    79

  • Joined

  • Last visited

  • Days Won

    12

 Content Type 

Posts posted by Anza Power

  1. Mailing incongruent cubes to a friend

    in New Logic/Math Puzzles

    Posted · Edited by Anza Power

    1st, the problem can be reduced to 2D, if you just look at the bottom row, then it's a problem of fitting a rectangle using several squares of different sizes.

    Imagine that someone tried to do it in 2D, take a look at the squares touching the bottom side of the rectangle, they may look something like this:

    ####

    #########

    #########

    #########

    Now look at the smallest of these squares (in this example the green one) I say you cannot fill the row directly above it, because you cannot use another square of the same size, so you'll have to use smaller squares, but then reapply this proof recursivly until you get to the smallest square of size 1x1 with a 1x1 gap above it and you are stuck.

  6. dice problem

    in New Logic/Math Puzzles

    Posted · Edited by Anza Power

    It seems that this puzzle is NP-hard and not NP-complete (if I understand this correctly), as there is not an efficient way to prove that the solution is minimal, without an exhaustive search for a better one. (Though a solution can be proved to be valid easily enough, just not minimal.)

    Actually it's O(1).

    Your comment got me thinking, I couldn't say that it's in NP because there is no n in the problem, We have 130 as a starting point and we want to see if any of the binome(129,8) dice combinations have 120 different sum...

    Unless you are referring to the problem with "N-sided Dice", then yeah that's crazy exponential...

  7. A boy is late for school

    in New Logic/Math Puzzles

    Posted

    Is the arrival time continuous or discrete (whole minutes)?

    Can we assume the distribution is uniform?

    Continuous time is the interesting case so let's look at that with uniform distribution:

    Let t=0 be 7:58, so the bus arrives anywhere in time [0,4], the kid arrives anywhere in [-3,3].

    50% chance the kid arrives in segment [-3,0] and in this case his chance of catching the bus is 1.

    50% chance the kid arrives in segment [0,+3] at time t1, then he would catch the bus iff it arrives in the interval [t1,4], so his chance is (4-t1)/4 = 1-t1/4, so on average that's the integral from 0 to 3 of 1-x/4 divided by 3 for normalization, which is (3-9/8)/3 = 1-3/8 = 5/8

    So the final probability is (1+5/8)/2 = 13/16

  8. [0,1] to (0,1)

    in New Logic/Math Puzzles

    Posted · Edited by Anza Power

    Are there conditions other than being one-to-one and onto?

    No other conditions, although if you want something really interesting try having a continuous function...

    continuous image of compact space has to be compact.

    Oh, that's even simpler than what I had in mind:

    Since the function is continuous and reversible, it has to be monotonic, assume it's monotonically rising, by definition of "continuous" lim

    x->1f(x) = f(1), but f(1) is somewhere in the open segment (0,1), so let's say f(1)=1-epsilon, but since f is monotonically rising that means there is nothing mapped to the segment (1-epsilon,1)

  9. An irrational "aha!" puzzle

    in New Logic/Math Puzzles

    Posted

    Golden ratio satisfies

    g=1/(g-1)

    Assume that it is rational, then we can write g=p/q where gcd(p,q)=1

    Now we have p/q=1/(p/q-1), after moving some stuff around you end up with p2-pq=q2 -> p2-q2=pq -> (p+q)(p-q)=pq.

    p and q are natural numbers, easy to see that p is not 1, let m be some prime divisor of p, since gcd(p,q)=1 then a does not divide q, on the RHS of the equation we have a but on the LHS we have multiplication of two things that do not divide a so LHS cannot divide a

  11. building a bridge

    in New Logic/Math Puzzles

    Posted

    Imagine that the river is on the x axis and A and B are some dots below and above it, since the river width is consistent you can assume we already paid for the bridge and eliminate the river, as in make it's width 0, and now the shortest distance is a straight line

    Well, except that we are given the river as is, with a fixed width and a bridge perpendicular to its flow.

    We do not know whether the line joining the cities lines up with the bridge - a very special case with a trivial solution.

    We also do not know that the cities are equidistant from the river - a fact that bears on bridge placement if is not parallel to the joining line.

    I'm curious to know the two different ways BMAD has to optimize the bridge placement.

    You might've misunderstood me a little bit, I should have detailed a bit more, let the river be of width 1 unit, align the axis such that the x axis is parallel with and centered the river, in other words the river would be the area -0.5<y<0.5, now "removing" the river means collapsing it and so every point that was below it now goes up by 0.5 units and every point above it goes doen 0.5 units...

  14. Sharing a sandwhich

    in New Logic/Math Puzzles

    Posted

    You mean show that for any shape there exists a direction such that if you slice through it you get two equal halves?

    Well then just put the shape at point (0,0,0) and imagine slicing it parallel to the yz axis at point x, let v1(x) be the volume of the first half and v2(x) the volume of the second, if x is at the far left then v1(x)=0 and v2(x)=1, if x is at the far right then v1(x)=1 and v2(x)=0, all that's left to show is that v2 and v2 are continous and then simply apply the mean value theorem and you know there exists some x where v1(x)=v2(x)=0.6

  18. Mensa Math/Logic Problems...How to Solve?

    in New Logic/Math Puzzles

    Posted

    So many, don't have the energy to go through them all...

    The "find the missing number" type of questions don't usually have a strategy, you just stare at them and try to draw connections, look for patterns like for example are the numbers rising or falling, usually the answer should be very simple like taking a number and summing it's two digits and stuff like that.

    The sum of every 4 circles in a line is 21

    Take the numbers in the left and right squares and reverse them then sum them, so for first row 31+12 becomes 14+21 which is 35

    http://en.wikipedia.org/wiki/Binomial_coefficient

    Also the chosen answer is wrong, it should be 120

    We simply skip prime numbers.

  19. Four Numbers

    in New Logic/Math Puzzles

    Posted

    If we multiply all 6 pairs we will get (a*b*c*d)^3.

    Assume that the missing product was a*b, then among the 5 given products we see (a*c)*(b*d)=(a*d)*(b*c), if we look at the numbers obviously 5 is out of the equation because there is nothing to balance it so the only remaining match is 2*6=3*4, so a*b*c*d=2*6=12.

    So the product of all 6 pairs should be 12^3, the product of the given 5 is 720, so the missing one is 12/5=2.4

  23. Making a fair choice

    in New Logic/Math Puzzles

    Posted

    Aha, ok then here's another idea:

    First everyone sends everyone their preference, and they all email each other to make sure they have the same information.

    Now each one of the three generates a random 1024 bit number, call it x1 x2 x3, but first before revealing those numbers they share a cryprographically secure hash of these numbers (say SHA-512), and then after they make sure everyone know all 3 hashes they share the actual x1 x2 x3, they calculate x1+x2+x3 (mod 2^1024) and treat that as a uniform sampled random number, you can then use it to pick a restaurant

×
×
  • Create New...