superprismatic
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Posts posted by superprismatic
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Curr3nt and brifri238 both found good adjacency hookups. I think phil1882 got
into a bit of trouble trying to keep the dimension down to two or three. Also,
curr3nt's solution makes for an easy second part solution. I had hoped that
someone would go a bit further and hint at some algorithm for generating these
things. I know of no way to generate a random one -- i.e. an algorithm which
can choose one with probability 1/N where N is the total number of possible
adjacencies. I don't even know how to find N. Thanks for working on it.
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There is only one A which has eight neighbours. There is only one of each of the 26 letters from A to Z, and each letter has to be connected (adjacent) to exactly 8 others.Can the letters be used more than once? If so, are the number of adjacent letters accumulative? (Would the conditions be meet if the first 'A' has 6 adjacent letters and the second 'A' has 2?)
Does the arrangement need to be on a straight plane or can it be on a sperical one? (Could A be next to B and C in the same way that Indiana is next to Kentucky and Illinois?)
The arrangement can be on any manifold with any number of dimensions -- Whence my last paragraph. All that is required is that each of the 26 letters is adjacent to 8 others and "adjacency" is commutative. How this plays out in your head is of no concern. I hope this clarifies things.
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A Boggle-like Challenge
In the game, Boggle, a letter may have at most 8 adjacent letters.
That fact inspired this challenge.
This first part of this challenge is to place letters in such a way
that each letter of the alphabet has precisely eight other different
letters adjacent to it. You must use all 26 letters and, of course,
"adjacent" is a commutative relation. To specify your placement,
all you need to do is list the eight letters adjacent to A, the
eight letters adjacent to B, the eight letters adjacent to C,...,etc.
But remember that, if Q is on A's adjacency list, then A must be
on Q's adjacency list, and this is true for every pair of letters
-- not just A and Q.
The second part of the challenge is to create a cycle of all 26
letters, such that each adjacent pair of letters in the cycle are
adjacent in the sense of the first part of the challenge.
Note that there is no requirement that the graph of adjacent letters
is realizable in a small number of dimensions. So, trying to visualize
such a graph may be hazardous to your mental health!
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I found another 30-face one: 201300110031022220130011003102 but I couldn't find any from n=20 through n=28 after exhausting all for which any die which had i on it, also had n+1-i (where n is even and between 20 and 28). The first one I found was for n=30 and is given above.
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So, after reading your post and looking at the fewest for 3 again. I thought of one configuration to try. Since testing a configuration with my code is fast, I threw it into my code and...
Dice needed are 4-sided, 6-sided, 8-sided, and 12-sided. So 30 total faces.
301210000121033330121000012103
I simply noticed how the best configuration for 3 has two of the best configurations for 2 in it (121), and extrapolated the same way.
I'll still work on code to try and find a shorter one.
Also, will test if I can get a solution with 5 dice the same way...
(Edit: It doesn't work for 5 dice. Which makes sense due to needing one die with a number of faces that is a multiple of 5.)
Nice! I just checked your 30-face solution and it works just fine. I hope you can go lower.
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You do not need to allow a die to have more than one of the same number. You can just increase the value of all faces above it by one to make room for the other.
Also, the relative ordering is all that matters. This means you can simply use the numbers 1 to N, where N is the sum of the number of faces on the dice.
This means that an isomorphic problem would be to find a string of the digits 1 to D, with repititions and where D was the amount of dice in the first problem, with the property if you (uniformly) randomly pick one of each digit and discard the rest, you'll end up with a (uniformly) randomly picked permutation. The length of the string would be N, the sum of the number of faces of the dice from the first problem. An example would be 01101001 is equivalent to two dice with values {1,4,6,7} and {2,3,5,8}.
One thing to notice about these problems is that if you ignore a die (equivalently, all of a given digit in the second problem), you will still be picking uniformly random.
This can lead to insights in the four dice case such as:
1. At least 3 of the dice need an even number of faces.
2. At least 2 of the dice need a number of faces that is a multiple of 3.
3. The product of the number of faces must be a multiple of 72. (The number of permutations of 4 objects (=4!=24) times the extra 3 from (2))
I'm not sure if one of the dice need a number of faces that is a multiple of 4 or not. With prime numbers it is simpler (need at least D+1-p dice with faces that are a multiple of p, where p is any prime number and D is the number of dice)
The fewest total faces is 12. It requires a coin (2 faces), a 4-sided die, and a 6-sided die.
The faces, using the string of digits notation, can be one of the following three (ignoring swapping numbers):
012001100210
012010010210
012100001210
I'm fairly certain the fewest number of faces is greater than 18.
I wrote some code to try and find the fewest, but it ran really slow. I have had a few extra thoughts on optimizations, which should speed things up. It requires a rewrite though.
I thought of limiting the search to palindromes, as the fewest faces for 2 and 3 dice were palindromes. I'll try not to do this unless the code is still far too slow.
Nice solution for 3 dice! I hope you can beat my 4 dice solution of 48 faces. I suspect a solution with fewer than 48 faces exists, but I haven't found one.
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Would you say the winning seat benefited more from 20/21 or from the other two going bust?
Out of 1,000,000,000 simulated games:
Seat 1 won 342,546,409 times, 281,000,589 of which were because seat 1 got 20/21, and 61,545,820 of which were because Seats 2 and 3 both went over 21. In this simulation, Seat 2 won 318,668,459 times and Seat 3 won 338,785,132 times.
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Would you say the winning seat benefited more from 20/21 or from the other two going bust?
I don't know, but I can modify my program to count those things. I have an appointment soon.
When I return from that, I'll fix up the program. Perhaps knowing such things may make it easier
to analyse the problem analytically.
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My 1,000,000,000 game simulation gave the following probabilities of winning:
Seat 1: 0.342568700
Seat 2: 0.318651372
Seat 3: 0.338779928
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d1 2 5 11 16
d2 1 7 12 14
d3 4 6 9 15
d4 3 8 10 13
No, that can't work because there are 24 permutations of 4 things and your dice can produce 256 different results, but 256 results can't be split evenly amongst 24 permutations. Some permutations would have to get more dice results than others.
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If we have any 4-tuple of distinct real
numbers, there is a simple way to have
this determine a permutation on 4 things:
Just replace each number with its ranking
amongst the 4 numbers. For example,
suppose I had the 4-tuple, (36,95,1,18).
By replacing each number with its rank,
I get the permutation (2,1,4,3).
I would like to be able to use 4 fair
dice to generate a permutation in this
way. The dice would give me the 4-tuple
(each die would have its own spot in the
tuple) and I would use the ranks to
determine a permutation. Of course, In
order to insure that I get 4 distinct
numbers, no die can have a number which
is on any other die. But it may be the
case that a paricular die has two or more
faces having the same value. The number
of faces on the dice may be any positive
integer (I'm assuming that fair dice can
always be made this way). It is not
necessary that all of the dice have the
same number of faces.
Can you construct a set of 4 dice which
can produce all 24 permutations of 4
things, each with probability 1/24 ?
I have several such sets with 12 faces
on each die. Can you find a set with
fewer total faces?
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Josef Kieffer
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You can't delete posts, only moderators can. But, I think you may want to leave it here so that others may enjoy working on it. You might even offer hints and/or answer questions for people who may have trouble with it. Thanks for posting.Ive figured it out
How can i delete the post ? -
I can prove that it can be done in ceiling(log2n)+ceiling(log2m)-1. My conjecture is that this is minimal.
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Thank you SP,
I think you will have to re-check your answer..
What I have learned is that in counting of years, 1AD starts forward from the birth of Jesus, the year 1BC starts backwards from the same instant.
Therefore when we say an event to have happened 1year after 1BC, it actually have happened in the year 2AD.
I would like to have comments of other members also.....!
I used the Wikipedia article entitled "Anno Domini" as the source of my information. There's a section of the page about going from 1BC to 1AD with no year 0 between them.
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5 AD. From year Michael was born till Samuel died there are 95 years. Of these 95 years, their collective lifespan covers 80. This leaves a gap of 15 years when neither lived. So Samuel would have been born 15 years after Michel dies - 5 AD
because there was no year 0. The year after 1BC is 1AD. So, 15 years after 10BC is 6AD.
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There is a logical solution.
Yep. I didn't think things through enough.
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If I happen to point to the jack, then your response will convey no information to me since that response depends only on a process (the random yes-no generator) which is completely hidden from me. This is, therefore, a case where I could not learn anything about where an ace may be. I conclude, then, that any answer to this puzzle must depend on trick wording or some such chicanery.
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Find 11 common English words, each of
which is a subsequence of the string
PLUFIEMDATNGORPAPNLCHGE. The words
are all related by their meanings.
As an example, IMPALE is a subsequence
since its letters occur, in sequence,
from left to right in the string
PLUFIEMDATNGORPAPNLCHGE. Unfortunately,
IMPALE isn't one of the words in the
answer to this puzzle.
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Arrr, mesmells a rat................Methinks Simpson's Paradox is rearing its ugly head!
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I had to write a little program to get:
This 55-long sequence: 42 1318 2259 2339 5355 5709 9393 11299 12230 12641 14000 14289 17990 18498 18638 19218 21861 24351 25846 27141 28105 33535 35597 37723 38410 38833 38994 39295 45392 46268 46480 47318 47877 48856 50628 56087 57163 59481 60292 60868 65793 67511 67635 70066 75993 78289 79572 85182 87577 92362 94360 94963 95425 97491 97684
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24,214
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Nice, but do you have any reason to believe there is no 8-long?1 2 4 6 24 144 288 312 311
is my personal "best fit".
If you look at the end of my 9-long, you'll see that it ends in what
are almost "wasted" moves:
1 2 3 9 18 324 315 312 311
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You guys are much better than me at doing this. I fooled around with
311 for some time, and I can't do it with an 8-long program (including
the first 1). But eventually I got a 9-long which ended in 311. Can
you guys do it in an 8-long, like you did with the previous numbers?
A Boggle-like Challenge
in New Logic/Math Puzzles
Posted
Perhaps going down to a smaller alphabet will help
shed light on this problem. I wrote a little program
to generate all possible ways of making adjacencies on
an alphabet of size 8 with each letter having 3 adjacencies.
The number of these I got was 19,355.