Firstly, apologies for not using the correct terminology here. I don't actually know the correct terms, and thus have failed miserably to find a solution. Please accept my example as the question, and I'll update the question accordingly if somebody can enlighten me (or delete and read the actual solution where it exists).
So far my research has only yielded comparisons of single bits within the whole, rather than showing all.
Given a set of integers:
ELEMENT
-----------
Bricks 1
Plaster 2
Cement 4
Concrete 8
I have a result set that provides how these materials are used:
MIXTURE ELEMENTS
----------------------
MixtureFoo 3
MixtureBar 7
MixtureBaz 11
I need to show the final set of mixtures, but with each constituent element listed that is used in the respective mixture:
MIXTURE ELEMENTS ELEMENT
------------------------------
MixtureFoo 3 1
MixtureFoo 3 2
MixtureBar 7 1
MixtureBar 7 2
MixtureBar 7 4
MixtureBaz 11 1
MixtureBaz 11 2
MixtureBaz 11 8
You could use bitwise operations:
SELECT *
FROM t
JOIN ELEMENT e
ON t.ELEMENTS & e.w = e.w
ORDER BY MIXTURE, w;
db<>fiddle demo
Related
I'm setting up a morphological table that will have to go through potentially a couple hundred items, so it's desirable for this process to not be done by hand.
Here's a small summary of the situation:
fin
eng
op
fli
A
2
4
6
8
B
1
3
5
4
C
1
2
3
5
D
1
4
7
2
The first column holds named ranges A through D which have associated values from the 4 categories in row 1.
In a second table we create configurations based on which features are selected, something like this:
Config 1
Config 2
A
B
C
D
What I'm looking for is a formula that would read for each configuration which named range is selected, add the score for each category and return it in a simple array. Something like
Config 1 {3,6,9,13}, Config 2 {2,7,12,6}
So far I've found that the Indirect formula works exactly the way I want but I have to manually input each range. Something like:
=INDIRECT(A1)+INDIRECT(A2)
I've played around with different permutations of sum functions but instead of returning the arrays it returns the sum of the first values.
=SUM(INDIRECT(A1:A2))
Amy suggestion would be welcome.
I know this would probably be much simpler with code but this study needs to be done in excel..
I'm not sure if this answers your question as it doesn't use named ranges, but you could try something like this:
=MMULT(SEQUENCE(1,4,1,0),$B$2:$E$5*COUNTIF(INDEX($H$2:$I$3,0,ROW()-ROW($A$7)+1),$A$2:$A$5))
I am trying to write a program that can generate permutations using the Johnson-Trotter method with varying number of elements. I am still confused on how to get the permutations exactly. For 5 elements, I can only get this far and then I get stuck. I am not asking for all of them just a few more so I can get the pattern down.
1 2 3 4 5
1 2 3 5 4
1 2 5 3 4
1 5 2 3 4
5 1 2 3 4
I found the following page:
http://introcs.cs.princeton.edu/java/23recursion/JohnsonTrotter.java.html
Try running this Java program step by step, it will surely be of great help...
I have used the Johnson-Trotter algorithm in a card-melding game. The princeton.edu link mentioned above is widely quoted, but pretty confusing. And the code I've seen is either recursive (yuck) or inefficient. I rewrote it as an iterator, and it works great in my game. See my other post here: https://stackoverflow.com/a/28241384/4266886 with (very short) code. Let me know if I can help further.
Given a goal state
int final[3][3]={{1,2,3},
{4,5,6},
{7,8,9}};
and a random initial state, I want to sort my array as final only by shifting rows (right or left) and columns (up and down) of my table
7 8 4 by shifting to the right the first row it will become 4 7 8
2 1 9 2 1 9
6 5 3 6 5 3
So I want to use a* search and I'm trying to find a good heuristic.
I've already tried with misplaced array elements.
Any suggestions?
I view this as an algebraic problem. You are given a group of permutation which is generated by 6 cycles (3 rows and 3 columns) and you want to find some more moves which help you to get to any permutation.
First advice: not all permutations are possible! Since every shift is an even permutation (a 3-cycle is the composition of two transpositions) only even permutations are possible. Hence you will not find any solution to a configuration where all is in place but two swapped numbers as in (2,1,3),(4,5,6),(7,8,9).
Second advice. If r is a row shift and c is a coumn shift, compute the action of rcr'c' where r' and c' are the inverse shifts. This "commutator" is again a cycle of 3 elements but this time they are not in a row or column. By choosing different r and c you get a lot of 3-cycles which can be used in the third advice.
Third advice. Consider the region of numbers which are already in their final position. Apply 3-cycles to the complement of this set to reduce it, until you get to a solution.
I have a large vector of numbers, say 500 numbers. I would like a program to detect patterns (reoccurrence in this case) in such vector based on following rules:
A sequence of numbers is a pattern if:
The size of the sequence is between 3 and 20 numbers.
The RELATIVE positions of the numbers in sequence is repeated at
least one other time in a vector. So let's say if I have a sequence
(1,4,3) and then (3,6,5) somewhere else in the vector then (1,4,3) is
a pattern. (as well as (2,5,4), (3,6,5) etc.)
The sequences can't intersect. So, a vector (1,2,3,4,5) does not
contain patterns (1,2,3) and (3,4,5)(we can't use the same number for
both sequences). However, (1,2,3,3,4,5) does contain a pattern
(1,2,3) (or (3,4,5))
A subset A of a pattern B is a pattern ONLY IF A appears somewhere
else outside B. So, a vector (1,2,3,4,7,8,9,2,3,4,5) would contain
patterns (1,2,3,4) and (1,2,3), because (1,2,3,4) is repeated (in a
form of (2,3,4,5)) and (1,2,3) is repeated (in a form (7,8,9)).
However, if the vector was (1,2,3,4,2,3,4,5) the only pattern will
be (1,2,3,4), because (1,2,3) appeares only in context of (1,2,3,4).
I'd like to know several things:
First of all I hope the rules don't go against each other. I made them myself so there might be a clash somewhere that I didn't notice, please let me know if you do notice it.
Secondly, how would one implement such system in the most efficient way? Maybe someone can point out towards some particular literature on the subject? I could go number by number starting with searching a sequence repetition for all subsets of 3, then 4,5 and till 20. But that seems to be not very efficient..
I am interested in implementation of such system in C, but any general guidance is very welcome.
Thank you in advance!
Just a couple of observations:
If you're interested in relative values, then your first step should be to calculate the differences between adjacent elements of the vector, e.g.:
Original numbers:
1 4 3 2 5 1 1 3 6 5 6 2 5 4 4 4 1 4 3 2
********* ********* ********* *********
Difference values:
3 -1 -1 3 -4 0 2 3 -1 1 4 3 -1 -3 0 -3 3 -1 -1
****** ****** ****** ******
Once you've done that, you could use an autocorrelation method to look for repeated patterns in the data. This can be computed in O(n log n) time, and possibly even faster if you're only concerned with exact matches.
I'm having trouble with my n-puzzle solver. Thought it was working, but it turns out it is solving insoluble puzzles. I've tried to trace it, but that's a lot of tracing and so far I see no cheating. I think I understand the algorithm for determining solubility, and my implementation agrees with the odd/even parity of some examples from the web... that is to say, when I count up the number of tiles after a given tile that are smaller than it, for every tile, and then add the row index of the blank tile, I get the same odd or even number as others have gotten.
So a thought that has occurred to me. In my model of, say, the 8-puzzle, my solution state is:
_ 1 2
3 4 5
6 7 8
Rather than
1 2 3
8 _ 4
7 6 5
Or
1 2 3
4 5 6
7 8 _
As it is in some other formulations. Could this be affecting which puzzles are soluble and which are not?
Thanks!
z.
In general, yes: If a configuration is solvable to the standard solution, it will not be solvable to an unsolvable configuration.
In particular, it depends on the exact configuration you're using as a solution. You will need to check to see if you can solve from that configuration to the standard one.
EDIT: This of it this way:
Let A be the standard solution.
Let B be your preferred solution.
Let C be your starting configuration.
If you can get from A to B, and you can get from C to A, then you can get from C to B.
But if you can't get from A to B, and you can get from C to A, then you can't get from C to B.