witzar
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Posts posted by witzar
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Is the road a straight line?
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If you don't like Mr Bayes, you can draw a picture.
Now all you have to do is divide green by (green+red).
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Suppose {a_1, a_2, ..., a_n} contains k numbers <=n. This means that {b_1, b_2, ..., b_n} contains (n-k) numbers <=n. Sequences (a_i) and (b_i) are sorted ascending/descending, so
a_i <= n < b_i for i =1, ..., k
and
a_i > n >= b_i for i = k+1, ..., n.
Therefore
|a_i - b_i| = b_i - a_i for i=1, ..., k
and
|a_i - b_i| = a_i - b_i for i = k+1, ..., n.
This shows that
S = |a_1 - b_1| + ... + |a_n - b_n| = s_1*1 + s_2*2 + ... + s_{2n}*2n
where
s_i = -1 for i=1, ..., n
and
s_i = +1 for i=n+1, ..., 2n.
Therefore
S = -1-2-...-n + (n+1)+(n+2)+...+(n+n) = -(1+2+...+n) + n*n + (1+2+...+n) = n*n. -
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OK. But why all squares of the chessboard are of the same color on the picture?
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A player playing chess blindfolded is not touching any pieces at all.
Nxc6+ Nxc6
Rd5#White have mate in two:
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I never know when to use Q.E.D. vs Without loss of generality, is there a difference or is it just a style thing?
WLOG often starts the proof while QED ends it.
You can decide which one to use on the basis of your progress with the proof
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I. We can construct partition of (n-1) from partition of n by decreasing one of the parts by 1 (if decreased part equals 1 it vanishes). Let's do it for each different part of every partition of n. We will get all partitions of (n-1), but there will be repetitions. Each partition of n produces as many partitions of (n-1) as it has different parts, so we have produced b(n) partitions of (n-1). From the other hand, each partition of (n-1) was produced as many times as it has different parts plus one, therefore:
b(n) = b(n-1) + p(n-1)
This shows, that
b(n) = p(n-1) + p(n-2) + ... + p(0)
II. Let's write all partitions of n in a column. For each partition that has at least one "1" let's rewrite it skipping one "1", right next to original partition. Newly written partitions form second column. Observe, that second column contains each partition of (n-1) exactly once. We repeat the process for second column obtaining partitions of (n-2) in third column, etc.
Not counting first column we have written p(n-1) + p(n-2) + ... + p(0) partitions. From the other hand, each row not counting first partition contains as many partitions as the first partition in the row has ones, so not counting first column we have written a(n) partitions. This proves, that
a(n) = p(n-1) + p(n-2) + ... + p(0) -
For (n+3) points on the circle, number of intersections for chords coming from one of the points equals:
C(n) = 1*n + 2*(n-1) + 3*(n-2) + ... + n*1
Let's call
S(n) = C(n) - C(n-1)
It is easy to see that
S(n) = 1 + 2 + ... + n = n(n+1)/2
So
C(n) = S(n) + S(n-1) + ... + S(1)
It is a known result that
C(n) = n*(n+1)(n+2)/6
Total number of intersections for (n+3) points equals
(n+3)*C(n)/4 = n*(n+1)(n+2)(n+3)/24 -
8 0 1 7 1 1 2 6 2 0 2 6 1 1 2 2 5 3 0 2 5 2 1 1 3 5 1 1 1 3 2 4 4 0 1 4 3 1 1 3 4 2 2 0 2 4 2 1 1 2 3 4 1 1 1 1 4 2 3 3 2 0 2 3 3 1 1 2 2 3 2 1 1 1 3 3 3 1 1 1 1 1 5 2 2 2 2 2 0 1 2 2 2 1 1 2 2 2 2 1 1 1 1 4 2 2 1 1 1 1 1 1 6 2 1 1 1 1 1 1 1 1 8 1 44 42You've just missed one partition:
{3, 2, 2, 1} 1 3So a(8)=b(8)=45.
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I think I can show that
a(n) = p(0) + p(1) + ... + p(n-1)
and
b(n) = p(0) + p(1) + ... + p(n-1),
where p(n) stands for number of partitions of n (assuming p(0)=1).
I accomplish both things by using partitions of n to construct and count partitions of k<n, with two different approaches.
I will provide details in my next post.
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(n-3)^4 + 2(n-3)^3 - (n-3)^2 + 6
(Note: word 'obvious' is used as a joke.)
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1
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If in fact the 7th graders got 8 ponits each, then the possible solutions are 7,8,12,17,or 32
players in the tournament
There are no solutions with 7 or 8 players. To score 8 points you need to have at least 8 opponents.
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So it looks like I have misunderstood the question assuming that both 7-graders scored 8 points each.
Anyway, "my version" of the puzzle was still interesting to solve.
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I made a drawing that shows all possibilities for a blue eyed couple (M=Mother and F=Father) to have a daughter (J=Judy) and a grandChild ( C ) of this daughter with a father of type Bg. Each rectangle represents specific genotype of Father, Mother, Judy and Child. Genotype of the Father can be read on horizontal axis, genotype of Mother is written on vertical axis, genotype of Judy and Child can be read inside of the rectangle. For example rectangle labelled "J:Bg;C:BB" says that Judy is Bg and Child is BB. Area of each rectangle divided by area of the whole square equals probability of the event. Black-hatched rectangles indicate cases that should be excluded due to green eyes of Judy or Child. Blue rectangles indicate cases when Father is of type BB. To answer the question #1 it is sufficient to divide (blue area) by (the area of the whole square minus black-hatched area).
Here is a link to the image:
http://imageshack.us/photo/my-images/545/blueeyesl.jpg/
I'm also trying to attach it, since It looks like I'm not allowed to upload images (why?).
Comments
The logic is impeccable, and the image above looks right with the relative areas. However, I think there might be some arithmetic mistakes in the calculations, since 65/72 is about 90%, and the (blue areas) divided by (the area of the whole square minus black-hatched area) is nearer to 70% than it is to 90%.
(Blue area) equals 46/72 and (the area of the whole square minus black-hatched area) equals 65/72, so my method gives same result as Prime gave: 46/65. There was an arithmetic mistake in my calculations.
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32 players. Divide 8-graders into 2 groups of 15 players. You tie with each player from group 1, lose with each player from group 2, and tie with your friend. This gives you 7.5 + 0 + 0.5 = 8 points. Your friend ties with each player from group 2, loses with each player from group 1 and ties with you, also scoring 8 points. Each game between two 8-graders is a tie, so each 8-grader gets 14.5 points against other 8-graders and 1.5 total against you and your friend, finishing competition with 16 points.
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12 players. Divide 8-graders into 2 groups of 5 players. You beat each player from group 1, and tie with each player from group 2 and with your friend. This gives you 5 + 2.5 + 0.5 = 8 points. Your friend beats each player from group 2 and ties with other players, also scoring 8 points. Each game between two 8-graders is a tie, so each 8-grader gets 4.5 against other 8-graders and 0.5 total against you and your friend, finishing competition with 5 points.
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I made a drawing that shows all possibilities for a blue eyed couple (M=Mother and F=Father) to have a daughter (J=Judy) and a grandChild ( C ) of this daughter with a father of type Bg. Each rectangle represents specific genotype of Father, Mother, Judy and Child. Genotype of the Father can be read on horizontal axis, genotype of Mother is written on vertical axis, genotype of Judy and Child can be read inside of the rectangle. For example rectangle labelled "J:Bg;C:BB" says that Judy is Bg and Child is BB. Area of each rectangle divided by area of the whole square equals probability of the event. Black-hatched rectangles indicate cases that should be excluded due to green eyes of Judy or Child. Blue rectangles indicate cases when Father is of type BB. To answer the question #1 it is sufficient to divide (blue area) by (the area of the whole square minus black-hatched area).
Here is a link to the image:http://imageshack.us/photo/my-images/545/blueeyesl.jpg/
I'm also trying to attach it, since It looks like I'm not allowed to upload images (why?).
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17 players, each game ended in a tie
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[1, 2, 2, 3, 3, 4]
[1, 3, 4, 5, 6, 8] -
(x-25)*100 + (50-x)*(50-x) = 100*x - 2500 + 2500 - 50*x - 50*x + x^2 = x^2
This means that your method works perfectly for every real number x.
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1) 65/72
2) 3/52
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Current relative frequencies of genotypes are not stable.
They will eventually stabilize at values:
BB = 25/64
Bg = 30/64
gg = 9/64
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I would assign values -1, 0, 1 to the doors, then I would agree with Mad Hatter to call value assigned to the correct door by X, and finally I would ask him, "Is 1/X > 0?"
Generation Bee
in New Logic/Math Puzzles
Posted