superprismatic
-
Posts
1267 -
Joined
-
Last visited
-
Days Won
3
Content Type
Profiles
Forums
Events
Gallery
Blogs
Posts posted by superprismatic
-
-
let's say the arrangment is
u
u d
d
if the bartender flips top an bottom, that gives
d
u d
u
and then if he does the other diagonal, that gives.
d
d u
u
I took the statement " Each glass is either right-side-up or up side down" to mean that all the glasses were all in the same orientation to start with. I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid. But, that makes it too easy, I suppose. I haven't yet considered the case where of an arbitrary beginning setup. Thanks for pointing that out, phil.
-
If he always picked two glasses which have one between them (diagonally opposite glasses if the are arranged 90 degrees apart) and turns them both over, then he will always accomplish the task in two moves.
-
here's a picture of superprismatics solution.
Thats a nice picture, phil
That's a nice picture, phil. It's missing the two six-long rows (3rd & 7th). Would you add them?
I'm impressed that you could do such a good job on the two rows closest to the middle row, as
they are each 49 units long whereas the middle row is 50 -- making it difficult to center them.
-
Make 9 rows of cards with each row centered on the ones vertically adjacent to it:
Row 1: 3 cards abutted on their length 5 sides (row has length 21)
Row 2: 5 cards abutted on their length 5 sides (row has length 35)
Row 3: 6 cards abutted on their length 5 sides (row has length 42)
Row 4: 7 cards abutted on their length 5 sides (row has length 49)
Row 5: 10 cards abutted on their length 7 sides (row has length 50)
Row 6: 7 cards abutted on their length 5 sides (row has length 49)
Row 7: 6 cards abutted on their length 5 sides (row has length 42)
Row 8: 5 cards abutted on their length 5 sides (row has length 35)
Row 9: 3 cards abutted on their length 5 sides (row has length 21)
The radius of the smallest circle containing the cards is 25.9326.
I would like to see what you mean.
I wish I could make a picture of it, but I have never done anything like that. I've only used text.
Perhaps a text picture will help. Each 5x7 card is represented by a 5x7 matrix containing 35 instances of a single digit.
First and second rows:
111111122222223333333 111111122222223333333 111111122222223333333 111111122222223333333 111111122222223333333 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888I can't do the third row because the next card would have half of its 7-long side having blank space above it, and the
other half having half of card 4 above it.
-
Make 9 rows of cards with each row centered on the ones vertically adjacent to it:
Row 1: 3 cards abutted on their length 5 sides (row has length 21)
Row 2: 5 cards abutted on their length 5 sides (row has length 35)
Row 3: 6 cards abutted on their length 5 sides (row has length 42)
Row 4: 7 cards abutted on their length 5 sides (row has length 49)
Row 5: 10 cards abutted on their length 7 sides (row has length 50)
Row 6: 7 cards abutted on their length 5 sides (row has length 49)
Row 7: 6 cards abutted on their length 5 sides (row has length 42)
Row 8: 5 cards abutted on their length 5 sides (row has length 35)
Row 9: 3 cards abutted on their length 5 sides (row has length 21)
The radius of the smallest circle containing the cards is 25.9326. -
If x=0, f(x)=1/2;
if x≠0 and 1/x is an integer, f(x)=x/(1+2x);
otherwise, f(x)=x.
Very impressive! Did you just devise this yourself?
I first thought to make room in the range of the function for the extra two (images of 0 and 1)
by moving a subset of the rationals up two spots (as Zeno would have done) thusly:
Going 1/2 way the distance each time, the subset of the rationals is
{1/2,3/4,7/8,15/16,31/32,...}. So, mapping 0 to 1/2 and 1 to 3/4, I just had to
map everything in the subset up two positions forward in the set. Thus, I had
to map 1/2→ 7/8, 3/4→ 15/16, 7/8→ 31/32, 15/16→ 63/64,... I could leave everything
else not in the subset to map to itself. This seemed a bit cumbersome, so I set out
to simplify. At some point I realized than the numerators could all be the same and
I could have used the set {1/2,1/4,1/8,1/16,1/32,...} instead. Then, it hit me that
I didn't need powers of two at all, just increasing denominators, which led me to
use the set {1/2,1/3,1/4,1/5,1/6,...}. From the original Zeno idea to that final
subset of the rationals only took 5 minutes or so, but there was no particular rush
as I noticed that there was no activity on this OP.
-
If x=0, f(x)=1/2;
if x≠0 and 1/x is an integer, f(x)=x/(1+2x);
otherwise, f(x)=x. -
The four numbers (in no particular order) are either
{sqrt(6/5),sqrt(24/5),sqrt(10/3),sqrt(15/2)} or {sqrt(18/5),sqrt(8/5),sqrt(10),sqrt(5/2)}. -
1, 2, 8, 40, 91, 120, 140, 144
-
This algorithm will sort the coins into 3 different piles with
each pile containing coins which all have the same weight.
Now, as soon as any pile has more than 694 coins in it, we will
know that all of the coins in this pile are gold. We know this
because there are only 694 (=1521-827) non-gold coins.
For ease of reference, number the coins from 1 to 1521.
step 01: let N=1 and put coin 1 into pile A and make piles B and C empty
step 02: re-order the piles so that A has the highest number coin in it; increment N by 1
step 03: weigh coin N against the highest numbered coin in pile A
step 04: if the result is not equal, go to step 08
step 05: add coin N to pile A
step 06: if pile A has 695 coins, stop. This is the pile of gold coins
step 07: go to step 02
step 08: if pile B is empty, put coin N into pile B and go to step 02
step 09: weigh coin N against the highest numbered coin in pile B
step 10: if the result is not equal, go to step 14
step 11: add coin N to pile B
step 12: if pile B has 695 coins, stop. This is the pile of gold coins
step 13: go to step 02
step 14: add coin N to pile C
step 15: if pile C has 695 coins, stop. This is the pile of gold coins
step 16: go to step 02
Suppose coins 1,2, and 3 are gold, silver, and bronze respectively.
- Put coin 1 in pile A.
- Weigh coin 2 against coin 1, not equal. (Both coins have been weighed once)
- Put coin 2 in pile B.
- Re-order piles so coin 2 is in pile A, coin 1 in pile B.
- Weigh coin 3 against coin 2, not equal.
- Weigh coin 3 against coin 1, not equal. (All coins have been weighed twice)
I think this will only work if you can see the result of the third weighing of a coin before the coin magically disappears.
I now see that I have much more of a problem with the algorithm than can be fixed in the simple way I've tried. I must be getting tired.
-
This algorithm will sort the coins into 3 different piles with
each pile containing coins which all have the same weight.
Now, as soon as any pile has more than 694 coins in it, we will
know that all of the coins in this pile are gold. We know this
because there are only 694 (=1521-827) non-gold coins.
For ease of reference, number the coins from 1 to 1521.
step 01: let N=1 and put coin 1 into pile A and make piles B and C empty
step 02: increment N by 1
step 03: weigh coin N against the highest numbered coin in pile A
step 04: if the result is not equal, go to step 08
step 05: add coin N to pile A
step 06: if pile A has 695 coins, stop. This is the pile of gold coins
step 07: go to step 02
step 08: if pile B is empty, put coin N into pile B and go to step 02
step 09: weigh coin N against the highest numbered coin in pile B
step 10: if the result is not equal, go to step 14
step 11: add coin N to pile B
step 12: if pile B has 695 coins, stop. This is the pile of gold coins
step 13: go to step 02
step 14: add coin N to pile C
step 15: if pile C has 695 coins, stop. This is the pile of gold coins
step 16: go to step 02
I think you have a problem, say the coin types were as follows:
0 1 2 3
A B C A
Coin 0 will go to pile A, coin 1 to B and 2 to C, but in the process you have compared coin 0 with 2 coins so you can't compare it against coin 3.
You're right. I had to modify step 2 in my post to fix that problem.
-
This algorithm will sort the coins into 3 different piles with
each pile containing coins which all have the same weight.
Now, as soon as any pile has more than 694 coins in it, we will
know that all of the coins in this pile are gold. We know this
because there are only 694 (=1521-827) non-gold coins.
For ease of reference, number the coins from 1 to 1521.
step 01: let N=1 and put coin 1 into pile A and make piles B and C empty
step 02: re-order the piles so that A and B have the most recent & next most recent additions, respectively; increment N by 1
step 03: weigh coin N against the highest numbered coin in pile A
step 04: if the result is not equal, go to step 08
step 05: add coin N to pile A
step 06: if pile A has 695 coins, stop. This is the pile of gold coins
step 07: go to step 02
step 08: if pile B is empty, put coin N into pile B and go to step 02
step 09: weigh coin N against the highest numbered coin in pile B
step 10: if the result is not equal, go to step 14
step 11: add coin N to pile B
step 12: if pile B has 695 coins, stop. This is the pile of gold coins
step 13: go to step 02
step 14: add coin N to pile C
step 15: if pile C has 695 coins, stop. This is the pile of gold coins
step 16: go to step 02 -
I've exhausted all triangles whose two legs differ by 1 where
the length of the shortest leg is less than or equal to 1010.
I found only 9:
3 4 5
20 21 29
119 120 169
696 697 985
4059 4060 5741
23660 23661 33461
137903 137904 195025
803760 803761 1136689
4684659 4684660 6625109
-
Interesting fact:
Y2+(Y+1)2 becomes (X+1)2-X2 under the change of variable X = Y2+Y.
So, the sum of squares of two integers can be transformed into the difference of squares of two other integers.
The first six non-trivial examples for the problem at hand are:
83-73 = 169 = 132 = (72-62)2
1053-1043 = 32761 = 1812 = (912 - 902)2
14563 - 14553 = 6355441 = 25212 = (12612 - 12602)2
202733 - 202723 = 1232922769 = 351132 = (175572 - 175562)2
2823603 - 2823593 = 239180661721 = 4890612 = (2445312 - 2445302)2
39327613 - 39327603 = 46399815451081 = 68117412 = (34058712 - 34058702)2 -
5
4 9
7 11 2
8 1 12 10
6 14 15 3 13 -
I don't understand your example. It seems to me that
C is joined to E is joined to A is joined to F is joined to D is joined to B,
but B and C aren't joined. So, it seems that player 2 didn't win because
a loop wasn't made! Can you clear up my misunderstanding?
-
let x = sqrt(2+sqrt(2+sqrt(2+sqrt(2+...
then
...
I hope you're not intimating that -1 is also a solution!
-
A 1 in row i and column j of this matrix indicates
that teacher i is assigned to committee j.11110000000000000000 00001111000000000000 00000000111100000000 00000000000011110000 00000000000000001111 10001000100010000000 01000100010000000100 00100010000000100010 00010000000100010001 00000001001001001000 10000100001000010000 00001000010000100001 00010000100001000010 00100001000010000100 01000010000100001000
-
Consider N strands of spaghetti. Some of these strands may comprise original strands tied end-to-end. Now, it is easy to see that
there are N ways to pick a pair of strand ends which, when tied together would cause a loop to form. Since there are 2NC2 ways
to pick a pair of ends, it follows that the probability of randomly picking ends which would result in a loop is N/2NC2 which
is equal to 1/(2N-1). So, with probability 1/(2N-1), we will form a loop AND decrease the number of strands by 1. With probability 1-(1/(2N-1)),
we will not form a loop but we will decrease the number of strands by 1. Let E(N) be the expected number of loops formed in this way.
Then, E(N) = (1/(2N-1))*(1+E(N-1)) + (1-(1/(2N-1)))*E(N-1) = 1/(2N-1) + E(N-1). Using this recursive formula, we can telescope out to
E(N) = 1/(2N-1) + 1/(2N-3) + ... + E(1). But E(1)=1. So, E(N) = 1/(2N-1) + 1/(2N-3) + ... + 1/1. This says that the expected number of
loops formed is the sum of reciprocals of all odd integers from 1 to 2N-1. -
If a deck of 52 cards is cut exactly in half (top half and bottom half),
then the top half can be partitioned into T segments (keeping the order
of the cards) and, similarly, the bottom half can be partitioned into B
segments. If T=B then there are two ways to interleave these partitions
of cards -- dropping segments of cards alternately from T and B, beginning
with T (the first way) or beginning with B (the second way). If T=B+1,
then there is only one way to interleave the segments -- one must begin
with T. If T+1=B, one must begin with B. Any other relationship between
T and B make them impossible to interleave. We can count the number of
partitions of 26 in all possible arrangements, they are:
#partitions of 26; #of ways to make partitions of this number respecting order
1 1
2 25
3 300
4 2300
5 12650
6 53130
7 177100
8 480700
9 1081575
10 2042975
11 3268760
12 4457400
13 5200300
14 5200300
15 4457400
16 3268760
17 2042975
18 1081575
19 480700
20 177100
21 53130
22 12650
23 2300
24 300
25 25
26 1
We can now count all possible T,B pairs where T and B are at most one
apart. This gives us a grand total of 374,369,872,911,804 possible
riffle shuffles of 52 cards. Considering that 52! is the enormous
80658175170943878571660636856403766975289505440883277824000000000000,
I can't see how to show that there is some number, N, of riffle shuffles
such that one can always produce any particular permutation of the 52!
by applying at most N riffle shuffles to the identity permutation.
-
That answer works. Is there a shorter answer, one that uses less numbers?1, 1, 1, 5
I don't think so. 5,1,1,1 also works. I think these are the only solutions
There are two others: 1,5,1,1 and 1,1,5,1.
But is there a three number solution?
No
-
If there are N pieces of spaghetti, the expected number of loops formed is the sum of reciprocals of all odd integers from 1 to 2N-1.
-
That answer works. Is there a shorter answer, one that uses less numbers?1, 1, 1, 5
I don't think so. 5,1,1,1 also works. I think these are the only solutions
There are two others: 1,5,1,1 and 1,1,5,1.
-
What's a "random shuffle"? If it's the same as a random permutation, then phil1882 must be correct. So, it's not that. Please define "random shuffle".
blind bartender
in New Logic/Math Puzzles
Posted
But I assumed that the bartender had to change at least one glass on his first move. Then the spin would cause a problem for him.