DejMar
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Your rules indicate the only operations permitted are concatenation ||, subtraction -, and division /. Is that correct?
[spoiler=Or are these special operations and mathematical notations also permitted?]exponetiation (visually, does not require any special symbols when using simple superscription. The character symbols ^ and ** are only used when superscripting is not an option), tetration (the same is true for this operation, when superscripting the operation is alternately represented by ^^ or Knuth's double arrows. The other notation I inquire is whether one can use base notation.
I am assuming you are using a modified interpretation of the concatenation operator, allowing it to concatenate the result of an operation and not simply two numerals. The operator precedence of the concatenation operator, in regards to the other operators, is last. Does this still hold true for this puzzle? -
There does not appear to be a generic formula. If there were, I believe the Circle Packing article in Wolfram would have provided such. The article indicates that diameter of the circle inscribed with N unit-circles generally increases, but not always and sometimes the diameter of the next circle is actually smaller, with each unit-circle packed.
Where N is the number of unit-circles packed within a circle of diameter D.
N D
1 1
2 2
3 1 + 2/3*SQRT(3) = ~ 2.15470053837925
4 1 + SQRT(2) = ~ 2.41421356237309
5 1 + SQRT(1 + 1/SQRT(5)) = ~ 2.20300191001509
6 3
7 3
8 1 + csc(PI/7) ~= 3.30476487096249
9 1 + SQRT(2*(2 + SQRT(2))) = ~ 3.61312592975275
10 ~ 3.82...
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One important assumption not to make is that the walls are all connected. There can be islands.
That is, you won't necessarily get through the maze, nor explore all of its area,
simply by keeping a wall on the robot's left or right.
Having referenced the Wikipedia article on "Maze solving algorithm", a possible algorithm to use is the "Pledge algorithm." By tracking the angular sum of the turns made and the current heading, a 'robot' could avoid being trapped in a loop. As the objective of the 'robot' is not to find an exit, but all the flags, the algorithm would need be modified. An important inclusion to the algorithm would be to "map" the grid, so as not to miss any passages or locations. An incorporation of an aspect to "Trémaux's algorithm" could be used to register the visited locations and note any, as of yet, unexplored junctions.
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N Number of permutations of length 26 with longest subsequence of length N
1 1
2 18367353072151 OEIS A001453
3 32159889269079495426 OEIS A001454
4 73852382382545858737126 OEIS A001455
The total, 73884542290182291314704, subtracted from 26!, 403291461126605635584000000, then divided by 26!
equals 0.99981679616502718947142164987304, which is the probability that 5 letters in proper alphabetical order will occur if that alphabet is written down in a random sequence. -
Is the task then to provide pseudo-code for an algorithm that correctly "maps" the maze using a virtual 'robot' as an exploration tool? And, though it must be preferred, is a most efficient algorithm - with the restraints given - required for the task?
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Within the spoiler of my prior post, when I give the counts for the maximum length of a subsequence they are for that length and do not include the counts for lengths smaller. To be correct, each count should include the counts of the lesser length, but as indicated many, as yet, are unknown.
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Here it is: https://en.m.wikipedia.org/wiki/Erdős–Szekeres_theoremThis problem reminds me of another unsolvable BMAD puzzle involving the task of sorting DVDs using insertion sort.
I think back then, Yoruichi-San, pointed us to a Wikipedia article that showed how to calculate an upper bound.
Surprisingly, I think this might hold the key to this question.
The theorem says that, for any given r and s, any sequence of at least (r-1)(s-1)+1 orderable elements has an ascending subsequence of length r or a descending subsequence of length s. In our case, we have 26 orderable elements, which suggests that every string contains an ascending or descending subsequence of length 6. Symmetry tells us that the probability of just an ascending subsequence of length 6 is at least 50%.
You could be right in that the theorem holds a key, but I am unsure on how to apply such for this problem. As to symmetry, it may apply to the number of ascending versus the number of descending subsequences, but not to the length of subsequences alone.
Where the maximum length of a subsequence is 1, the count is 1.
Where the maximum length of a subsequence is 2, the count is equal to the 26th Catalan number - 1, which, if I factored and calculated correctly, is 18,367,353,072,151.
(Probability for a length of at least 3 would then be (26! - 18367353072152) / 26!.
Where the maximum length is 25, the count is equal to 25^2, or 625. (Probability = (625+1) / 26!)
Where the maximum length of a subsequence is 26, the count is 1. (Probability = 1 / 26!)
I do not, as yet, know the answer for the maximum length of a subsequence of 5.
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Sorry about that gavinksong. I didn't mean to attribute the quote incorrectly. The quoting structure is more complex than it should be, yet I will be more careful, as you request. I didn't remove any brackets. I used the quote tool on the toolbar, but, apparantly, it has changed with the upgrade(s) to the site. Unlike the Spoiler tag, it didn't provide anywhere for a reference to the quote -- except within the quote. i just picked the wrong name.
And though the difference between 0 and 1 may seem to be trivial, it is not so trivial. -
I apologize to gavinksong, he is correct. I guess lack of sleep had my mind thinking a bit off.
Consider the regulation required a minimum age of 66 years and 9 months for the bus pass. Let the younger reach this minimum on December 31 of the second year of Bolandia's minimum age schedule. The following day, January 1 of the third year, the minimum age would be 67. The maximum age the elder could be and not be able to apply for the bus pass at this time, would be an age of 66 years, 11 months and a day short of another month. Thus the maximum age differences between the elder and younger is (B) just under 3 months (66yr 11mo ~30da - 66yr 9mo 0 da = 0yr 2mo ~30da).
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gavinksong, on 23 Oct 2014 - 12:58 AM, said:
"the function would have to be 0 for n=1"
Why? The least number of apples Alice can place in a room is 0, not 1. 0 is a non-negative integer, and thus a valid option for Alice.
"each room can have any non-negative integer number of apples"
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This is incorrect because the minimum requirement only advances three months at a time. It does not suddenly jump forward by a year.Let the younger person apply for a bus pass on the same day he reaches 66, and this day happens immediately before the day the minimum age becomes 67. On that same day, let the older person be two days shy of 67. On the subsequent day, the older 66-year-old would not be permitted to apply for a bus pass, as the minimum age would be 67, and he would still be shy by 1 day. Thus, the answer is D, the younger would be just under 1 year younger and have a free bus pass, while the elder could not get one (until the next day).
However, by the same reasoning, a person can be younger by just under three months. The answer is B.
As a person does not necessarily apply for the free bus pass when he reaches retirement age, it matters not the interval of advancement, but at what point (day) it changes. The 3-month intervals only represent a cycle. The unit (day) prior to the point of change is the point open to a given minimum age, and then is closed to that age the day following. As age is measured in years -- not 3 month intervals, the maximum length between two ages of the same age would be under 1 year. What occurs is that this maximum span of age for two 66-year-olds, though not the same two 66-year-olds as they all age during the intervals, happens 4 times in the two year cycle.
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Let the younger person apply for a bus pass on the same day he reaches 66, and this day happens immediately before the day the minimum age becomes 67. On that same day, let the older person be two days shy of 67. On the subsequent day, the older 66-year-old would not be permitted to apply for a bus pass, as the minimum age would be 67, and he would still be shy by 1 day. Thus, the answer is D, the younger would be just under 1 year younger and have a free bus pass, while the elder could not get one (until the next day).
Six sixes to make 1,000 (modified version)
in New Logic/Math Puzzles
Posted · Edited by DejMar