Re: puzzle question
- From: "M.A.Fajjal" <h2maf@xxxxxxxxx>
- Date: Mon, 02 Jan 2006 06:56:17 EST
> On Mon, 02 Jan 2006 05:28:30 EST, "M.A.Fajjal"
> <h2maf@xxxxxxxxx>
> wrote:
>
> >> On Mon, 02 Jan 2006 03:36:35 EST, "M.A.Fajjal"
> >> <h2maf@xxxxxxxxx>
> >> wrote:
> >>
> >> >> On Sun, 01 Jan 2006 22:06:34 EST, "M.A.Fajjal"
> >> >> <h2maf@xxxxxxxxx>
> >> >> wrote:
> >> >>
> >> >> >> On Sun, 01 Jan 2006 13:31:29 EST,
> "M.A.Fajjal"
> >> >> >> <h2maf@xxxxxxxxx>
> >> >> >> wrote:
> >> >> >>
> >> >> >> >> On Sun, 01 Jan 2006 02:02:08 -0500, quasi
> >> >> >> >> <quasi@xxxxxxxx> wrote:
> >> >> >> >>
> >> >> >> >> >On 31 Dec 2005 22:46:54 -0800, "mathbee"
> >> >> >> >> <buntygill@xxxxxxxxx> wrote:
> >> >> >> >> >
> >> >> >> >> >>Given a stick of length L, what is the
> >> chance
> >> >> >> that
> >> >> >> >> you can break it
> >> >> >> >> >>into 3 pieces that can form a triangle?
> >> >> >> >> >>
> >> >> >> >> >>My answer is 1/8 (many other people got
> >> the
> >> >> same
> >> >> >> >> answer) but the
> >> >> >> >> >>interviewer said it is 1/3.
> >> >> >> >> >>
> >> >> >> >> >>Any ideas?
> >> >> >> >> >
> >> >> >> >> >It depends on the assumptions for the
> how
> >> the
> >> >> >> >> breaking is done.
> >> >> >> >> >
> >> >> >> >> >Assuming the 2 breakpoints are chosen
> >> >> >> independently
> >> >> >> >> and with a uniform
> >> >> >> >> >distribution, I don't get 1/8 or 1/3 --
> I
> >> get
> >> >> a
> >> >> >> >> probability of 1/4.
> >> >> >> >> >
> >> >> >> >> >quasi
> >> >> >> >>
> >> >> >> >> Ok, quick challenge problem ...
> >> >> >> >>
> >> >> >> >> With the same assumptions, what is the
> >> >> probability
> >> >> >> >> that you can form
> >> >> >> >> an acute triangle?
> >> >> >> >>
> >> >> >> >let the angls of triangle are a, b, c
> >> >> >> >
> >> >> >> >a+b+c=180
> >> >> >> >
> >> >> >> >assume the angles are integers
> >> >> >> >
> >> >> >> >Acute triangle cases = 44/179
> >> >> >> >
> >> >> >> >Probability of acute triangle for the same
> >> case
> >> >> >> assumptions = 1/4 * 44/179 = 11/179
> >> >> >>
> >> >> >> As pointed out in another reply, there no
> >> simple
> >> >> way
> >> >> >> to convert the
> >> >> >> problem from a problem of choosing 2 random
> >> break
> >> >> >> points to a problem
> >> >> >> of choosing angles.
> >> >> >>
> >> >> >> But even if we change the problem to a
> problem
> >> of
> >> >> >> choosing angles, I
> >> >> >> don't see how you got 44/179. What do you
> mean
> >> by
> >> >> >> "Acute triangle
> >> >> >> cases"?
> >> >> >>
> >> >> >> And why are you multiplying by 44/179 by
> 1/4?
> >> >> What's
> >> >> >> the logic?
> >> >> >>
> >> >> >The probability to form any triangle is 1/4
> >> >> >
> >> >> >let the angls of the formed triangle are a, b,
> c
> >> >> >
> >> >> >a+b+c=180
> >> >> >
> >> >> >assume the angles are integers
> >> >> >
> >> >> >possible angles = 179*178/2 = 15931
> >> >> >
> >> >> >Acute triangle where 1=<a<90 and 1=<b<90 and
> >> 1=<c<90
> >> >> >
> >> >> >Acute triangle cases = 3916 (Counting by
> simple
> >> >> program)
> >> >> >
> >> >> >Probability of acute triangle for the same
> case =
> >> >> 1/4 * 3916/15931 = 11/179
> >> >> >
> >> >> >The code for the program in VB6 shall be as
> >> follows
> >> >> >
> >> >> >
> >> >> >Private Sub Command1_Click()
> >> >> >Open "D:\puzzle.txt" For Output As #1
> >> >> >Print #1, Time
> >> >> >k = 0
> >> >> >u = 0
> >> >> >For a = 1 To 180
> >> >> > For b = 1 To 180
> >> >> > For c = 1 To 180
> >> >> > DoEvents
> >> >> > If a + b + c = 180 Then
> >> >> > k = k + 1
> >> >> > If a < 90 and b < 90 and c < 90 Then
> >> >> > u = u + 1
> >> >> > End If
> >> >> > End If
> >> >> > Next
> >> >> > Next
> >> >> > Next
> >> >> > Print #1, u & " " & k
> >> >> > Print #1, Time
> >> >> > End
> >> >> >
> >> >> >End Sub
> >> >> >
> >> >> >The results
> >> >> >19:02:50
> >> >> >3916 15931
> >> >> >19:02:58
> >> >>
> >> >> Ok, thanks for explaining it -- at least now I
> >> >> understand your logic.
> >> >>
> >> >> The answer is wrong for at least 5 reasons:
> >> >>
> >> >> (1) You are assuming integer angles. Hence,
> even
> >> if
> >> >> the rest was
> >> >> correct, the result would only be approximate.
>
> >> >>
> >> >> (2) You are assuming the first angle is
> uniformly
> >> >> distributed.
> >> >> However, the set of possible triangles
> obtainable
> >> >> from 2 random break
> >> >> points does not imply a uniform distribution on
> >> the
> >> >> angles.
> >> >>
> >> >> (3) You are assuming the 2nd angle must be
> >> different
> >> >> than the first
> >> >> (as shown by your choice of the 2nd factor as
> >> 178).
> >> >> There's no reason
> >> >> to assume that the 2nd angle must be different.
> >> >>
> >> >> (4) You are assuming that, except for being not
> >> >> equal, the value of
> >> >> the 2nd angle is independent of the value of
> the
> >> >> first. This is
> >> >> clearly false. An easy way to see this is to
> note
> >> >> that the first angle
> >> >> has taken up some amount of the 180 total, so
> the
> >> >> bigger the first
> >> >> angle, the less choices for the 2nd.
> >> >>
> >> >> Of all the errors so far, this is the worst
> one,
> >> but
> >> >> there's an even
> >> >> worse error here -- namely, your calculation
> >> >> 179*178/2 includes pairs
> >> >> of angles whose sum is already greater than 180
> --
> >> >> your pair sums
> >> >> range from 3 degrees to 357 degrees.
> >> >>
> >> >> (5) Your counting loop counts the same triple
> >> again
> >> >> in a permuted
> >> >> order, but your calculation 179*178/2 is
> >> inconsistent
> >> >> with this
> >> >> assumption since you are dividing by 2.
> >> >>
> >> >> But even here, there is a worse error. To see
> the
> >> >> error dramatically,
> >> >> take out the test for acute and count again.
> >> You'll
> >> >> get a count of
> >> >> 180^3, but by your logic, the total should
> equal
> >> >> 179*178/2. However,
> >> >> 180^3 and 179*178/2 are not even close.
> >> >>
> >> >> The errors here, especially (2), (4), (5),
> totally
> >> >> invalidate any
> >> >> result you might obtain.
> >> >>
> >> >
> >> >The triangle has been already formed with the
> >> probability of 1/4.
> >> >Now, starting new problem, what the probability
> of
> >> cute triangle. All angle values are possible.
> >>
> >> Once a triangle has been formed, all angles are
> not
> >> possible.
> >>
> >> Obviously, you don't understand the experiment.
> >>
> >> The break points are being chosen at random, not
> the
> >> angles.
> >>
> >> >If fraction of angles is allowed for a, b, c
> >> (uniform distributed),
> >>
> >> As pointed out above, it's the break points that
> are
> >> uniformly
> >> distributed.
> >>
> >> The fact that the break points are uniformly
> >> distributed doesn't
> >> automatically imply that the angles are uniformly
> >> distributed.
> >>
> >> In fact, the angles are _not_ uniformly
> distributed.
> >
> >Why?
>
> Because it's not the angles that are being chosen --
> it's the break
> points. The angles (and also the sides) are forced
> once the break
> points are chosen.
>
> The break points are assumed independent and
> uniformly distributed in
> the interval [0,1].
>
> But the angles are not uniformly distributed (in
> [0,180]) and also not
> independent of each other.
>
> Similarly, the sides are also not uniformly
> distributed and also not
> independent of each other.
>
> If you want to try a discrete experiment to
> approximate the correct
> answer, revise your program as follows:
>
> (1) Assume the length of the stick is 100 units.
>
> (2) Keep 2 counters:
>
> * A count of the total number of trials
> * A count of the number trials which resulted in
> acute triangles
>
> Initialize both counters to 0.
>
> (2) For all pairs x,y of integers with 0 < x < y <
> 100, use x,y as
> break points. For each pair (x,y) increment the count
> of the number of
> trials.
>
> (3) Let a=x, b=y-x, c=100-y
>
> (4) Reject (a,b,c) unless the triangle inequalities
> hold:
>
> a+b>c, b+c>a, c+a>b
>
> (5) Use the law of cosines to determine if the
> triangle with sides
> a,b,c is acute. Reject (a,b,c) if the triangle is not
> acute. If the
> triangle is acute, increment the count of the number
> of acute
> triangles.
>
> (6) Once all pairs (x,y) have been tested, output the
> value of the
> ratio (acute count)/(number of trials).
>
> If you implement a program using the above design,
> the result you get
> will be approximately correct since it's based on
> choosing break
> points, not angles (the angles and side are both
> forced).
>
> Note, the result you get by the above method will
> only be an
> approximation since, in the actual problem, the
> assumption is that the
> break points can vary continuously.
>
> To get the exact answer, a standard approach is:
>
> (1) View the break points x,y as corresponding to a
> point (x,y) of the
> unit square S. Thus, the unit square can be viewed as
> the sample
> space. Since we are assuming that x,y are independent
> and uniformly
> distributed in [0,1], it follows that the point (x,y)
> is uniformly
> distributed in S.
>
> (2) Let D be the region of S such that, if x,y were
> used as break
> points for a stick of length 1, the resulting 3
> segments could be used
> to form an acute triangle. Since the area of S is 1,
> the probability
> that a random point of S is in D is area(D)/area(S),
> and since
> area(S)=1, the probability is just area(D).
>
> (3) By symmetry, it suffices to consider the
> subregion D1 of D for
> which x<y. After finding the area of D1, we can then
> simply multiply
> by 2 to get the area of D.
>
> (4) Find the equations of the boundary curves for D1
> and draw the
> region D1.
>
> (5) Calculate the area of D1. For many such problems
> of this type, the
> area of the desired region can be calculated by
> simple geometry,
> however for this problem, calculus will be required.
> Once you see the
> region D1, it will be clear why.
>
> If you follow the plan outlined above, you'll get the
> exact answer:
>
> 3*ln(2) - 2
>
> which is approximately:
>
> 0.079441542
>
> quasi
I think your answar is correct
I have got a near value by the following program
Private Sub Command1_Click()
Open "D:\Sys\K\Paper quintic\7_.txt" For Output As #1
k = 0
u = 0
v = 0
For x = 1 To 1000
For y = x + 1 To 1000
DoEvents
L1 = x
L2 = y - x
L3 = 1000 - y
k = k + 1
If L1 + L2 > L3 And L2 + L3 > L1 And L1 + L3 > L2 Then
u = u + 1
a = ((L2 ^ 2 + L3 ^ 2 - L1 ^ 2) / (2 * L2 * L3))
b = ((L1 ^ 2 + L3 ^ 2 - L2 ^ 2) / (2 * L1 * L3))
c = ((L2 ^ 2 + L1 ^ 2 - L3 ^ 2) / (2 * L2 * L1))
If a > 0 And b > 0 And c > 0 Then
v = v + 1
End If
End If
Next
Next
Print #1, k & " " & u & " " & v
End
End Sub
results
499500 124251 39765
Probability of cute triangle = 39765/499500 = 0.07960960960960
.
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