A boolean vector has been created based on some rule and we need to identify the indices where the rule holds for 2 observations in a row. The following code does that
indices:0101001101b
runs:{0 x\x}"f"$;
where 2=runs indices
Could you please help me understand how the scan operator is used in the definition of the runs function? Appreciate your help.
It's using this special shorthand commonly used in calculating exponential moving averages: https://code.kx.com/q/ref/accumulators/#alternative-syntax
So {0 x\x} is equivalent to:
q){z+x*y}\[0;indices;indices]
0 1 0 1 0 0 1 2 0 1
What this is doing is essentially using the booleans as an on/off switch (via the boolean multiplication) for the rolling sum. It adds (z+) until it hits a negative boolean in which case the rolling sum resets back to zero.
In english: nextValue + [currentValue (starting at 0) * nextValue]
When nextValue is 1, 1 gets added. When nextValue is 0 the result is zero (resetting the rolling sum).
Something like this can achieve the same thing, though no less easy to read at a glance (and using two scans instead of one):
q){s-maxs not[x]*s:sums x}indices
0 1 0 1 0 0 1 2 0 1i
Terry has answered your question about how runs works.
Comparing adjacent items is common. You might prefer to use the prior keyword. Certainly easier to see what it is doing.
q)where (and) prior indices
,7
I modelled the following logic in stella:
(IF "cause" > 0 THEN MONTECARLO("probabilityofconsequence") ELSE 0
But Im not getting the correct syntax on AnyLogic:
(cause > 0) ? (uniform() < probabilityofconsequence) ? 1 : 0 : 0
Any ideas?
Disclaimer:
What stella does is with the Montecarlo function a series of zeros and ones from a Bernoulli distribution based on the probability provided. The probability is the percentage probability of an event happening per DT divided by DT (it is similar too, but not the same as, the percent probability of an event per unit time). The probability value can be either a variable or a constant, but should evaluate to a number between 0 and 100/DT (numbers outside the range will be set to 0 or 100/DT). The expected value of the stream of numbers generated summed over a unit time is equation to probability/100.
MONTECARLO is equivalent to the following logic:
IF (UNIFORM(0,100,<seed>) < probability*DT THEN 1 ELSE 0
the equivalent in anylogic should be:
cause>0 && uniform(0,100) < probability*DT ? 1 : 0
you need to create a variable called DT that is the equal to either the fixed time step that you have chosen in your model configuration, or the value you consider that should be adequate.
Since anylogic depending on how you are running the model, doesn't consider the fixed time step as fixed, you need to define the DT yourself.
No matter what, you are going to get results not exactly equal to stella probably since the time steps are not necessarily the same... but maybe similar enough should satisfy you
implementing isolater logic for the BIDI got me stuck again. Given the following test case.
#Levels: 1 0 1
#Reorder: 0 1 2
R RLI R ; 2 # LTR flag
\____/ \_/ (my run sequences)
Here is my current understanding:
After processing rules X1-X8 you end up with levels 0 0 1 which will convert to two isolated run sequences.
As far as I can tell, rules W1-W7 should not apply to any characters.
Rule N1 however will raise the level of the RLI to 1 as it is an NI between two strong types, the R and the eos of the first sequence.
Rule I1/2 will then push the first R to level 1.
Rule L1 could lower the level of an RLI but does not apply here.
How is that, that the RLI should end up on level 0.
Same problem with similar tests, just invthe opposite direction.
#Levels: 2 1 2
#Reorder: 2 1 0
L LRI L; 4
Sadly http://unicode.org/cldr/utility/bidi.jsp does not implement the latest algorithm including the RLI/LRI.
Any advice welcome :) thank you.
Okay the issue was that Rule N1 should not have been applied since the eos is of type L and not R.
I have missed the Note at X10 stating that: ...and if there is none or the last character of the sequence is an isolate initiator (lacking a matching PDI), with the paragraph embedding level."
I am trying to create a tic-tac-toe program as a mental exercise and I have the board states stored as booleans like so:
http://i.imgur.com/xBiuoAO.png
I would like to simplify this boolean expression...
(a&b&c) | (d&e&f) | (g&h&i) | (a&d&g) | (b&e&h) | (c&f&i) | (a&e&i) | (g&e&c)
My first thoughts were to use a Karnaugh Map but there were no solvers online that supported 9 variables.
and heres the question:
First of all, how would I know if a boolean condition is already as simple as possible?
and second: What is the above boolean condition simplified?
2. Simplified condition:
The original expression
a&b&c|d&e&f|g&h&i|a&d&g|b&e&h|c&f&i|a&e&i|g&e&c
can be simplified to the following, knowing that & is more prioritary than |
e&(d&f|b&h|a&i|g&c)|a&(b&c|d&g)|i&(g&h|c&f)
which is 4 chars shorter, performs in the worst case 18 & and | evaluations (the original one counted 23)
There is no shorter boolean formula (see point below). If you switch to matrices, maybe you can find another solution.
1. Making sure we got the smallest formula
Normally, it is very hard to find the smallest formula. See this recent paper if you are more interested. But in our case, there is a simple proof.
We will reason about a formula being the smallest with respect to the formula size, where for a variable a, size(a)=1, for a boolean operation size(A&B) = size(A|B) = size(A) + 1 + size(B), and for negation size(!A) = size(A) (thus we can suppose that we have Negation Normal Form at no cost).
With respect to that size, our formula has size 37.
The proof that you cannot do better consists in first remarking that there are 8 rows to check, and that there is always a pair of letter distinguishing 2 different rows. Since we can regroup these 8 checks in no less than 3 conjuncts with the remaining variable, the number of variables in the final formula should be at least 8*2+3 = 19, from which we can deduce the minimal tree size.
Detailed proof
Let us suppose that a given formula F is the smallest and in NNF format.
F cannot contain negated variables like !a. For that, remark that F should be monotonic, that is, if it returns "true" (there is a winning row), then changing one of the variables from false to true should not change that result. According to Wikipedia, F can be written without negation. Even better, we can prove that we can remove the negation. Following this answer, we could convert back and from DNF format, removing negated variables in the middle or replacing them by true.
F cannot contain a sub-tree like a disjunction of two variables a|b.
For this formula to be useful and not exchangeable with either a or b, it would mean that there are contradicting assignments such that for example
F[a|b] = true and F[a] = false, therefore that a = false and b = true because of monotonicity. Also, in this case, turning b to false makes the whole formula false because false = F[a] = F[a|false] >= F[a|b](b = false).
Therefore there is a row passing by b which is the cause of the truth, and it cannot go through a, hence for example e = true and h = true.
And the checking of this row passes by the expression a|b for testing b. However, it means that with a,e,h being true and all other set to false, F is still true, which contradicts the purpose of the formula.
Every subtree looking like a&b checks a unique row. So the last letter should appear just above the corresponding disjunction (a&b|...)&{c somewhere for sure here}, or this leaf is useless and either a or b can be removed safely. Indeed, suppose that c does not appear above, and the game is where a&b&c is true and all other variables are false. Then the expression where c is supposed to be above returns false, so a&b will be always useless. So there is a shorter expression by removing a&b.
There are 8 independent branches, so there is at least 8 subtrees of type a&b. We cannot regroup them using a disjunction of 2 conjunctions since a, f and h never share the same rows, so there must be 3 outer variables. 8*2+3 makes 19 variables appear in the final formula.
A tree with 19 variables cannot have less than 18 operators, so in total the size have to be at least 19+18 = 37.
You can have variants of the above formula.
QED.
One option is doing the Karnaugh map manually. Since you have 9 variables, that makes for a 2^4 by 2^5 grid, which is rather large, and by the looks of the equation, probably not very interesting either.
By inspection, it doesn't look like a Karnaugh map will give you any useful information (Karnaugh maps basically reduce expressions such as ((!a)&b) | (a&b) into b), so in that sense of simplification, your expression is already as simple as it can get. But if you want to reduce the number of computations, you can factor out a few variables using the distributivity of the AND operators over ORs.
The best way to think of this is how a person would think of it. No person would say to themselves, "a and b and c, or if d and e and f," etc. They would say "Any three in a row, horizontally, vertically, or diagonally."
Also, instead of doing eight checks (3 rows, 3 columns, and 2 diagonals), you can do just four checks (three rows and one diagonal), then rotate the board 90 degrees, then do the same checks again.
Here's what you end up with. These functions all assume that the board is a three-by-three matrix of booleans, where true represents a winning symbol, and false represents a not-winning symbol.
def win?(board)
winning_row_or_diagonal?(board) ||
winning_row_or_diagonal?(rotate_90(board))
end
def winning_row_or_diagonal?(board)
winning_row?(board) || winning_diagonal?(board)
end
def winning_row?(board)
3.times.any? do |row_number|
three_in_a_row?(board, row_number, 0, 1, 0)
end
end
def winning_diagonal?(board)
three_in_a_row?(board, 0, 0, 1, 1)
end
def three_in_a_row?(board, x, y, delta_x, delta_y)
3.times.all? do |i|
board[x + i * delta_x][y + i * deltay]
end
end
def rotate_90(board)
board.transpose.map(&:reverse)
end
The matrix rotate is from here: https://stackoverflow.com/a/3571501/238886
Although this code is quite a bit more verbose, each function is clear in its intent. Rather than a long boolean expresion, the code now expresses the rules of tic-tac-toe.
You know it's a simple as possible when there are no common sub-terms to extract (e.g. if you had "a&b" in two different trios).
You know your tic tac toe solution must already be as simple as possible because any pair of boxes can belong to at most only one winning line (only one straight line can pass through two given points), so (a & b) can't be reused in any other win you're checking for.
(Also, "simple" can mean a lot of things; specifying what you mean may help you answer your own question. )
I have two vectors in Matlab, z and beta. Vector z is a 1x17:
1 0.430742139435890 0.257372971229541 0.0965909090909091 0.694329541928697 0 0.394960106863064 0 0.100000000000000 1 0.264704325268675 0.387774594078319 0.269207605609567 0.472226643323253 0.750000000000000 0.513121013402805 0.697062571025173
... and beta is a 17x1:
6.55269487769363e+26
0
0
-56.3867588816768
-2.21310778926413
0
57.0726052009847
0
3.47223691057151e+27
-1.00249317882651e+27
3.38202232046686
1.16425987969027
0.229504956512063
-0.314243264212449
-0.257394312588330
0.498644243389556
-0.852510642195370
I'm dealing with some singularity issues, and I noticed that if I want to compute the dot product of z*beta, I potentially get 2 different solutions. If I use the * command, z*beta = 18.5045. If I write a loop to compute the dot product (below), I get a solution of 0.7287.
summation=0;
for i=1:17
addition=z(1,i)*beta(i);
summation=summation+addition;
end
Any idea what's going on here?
Here's a link to the data: https://dl.dropboxusercontent.com/u/16594701/data.zip
The problem here is that addition of floating point numbers is not associative. When summing a sequence of numbers of comparable magnitude, this is not usually a problem. However, in your sequence, most numbers are around 1 or 10, while several entries have magnitude 10^26 or 10^27. Numerical problems are almost unavoidable in this situation.
The wikipedia page http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems shows a worked example where (a + b) + c is not equal to a + (b + c), i.e. demonstrating that the order in which you add up floating point numbers does matter.
I would guess that this is a homework assignment designed to illustrate these exact issues. If not, I'd ask what the data represents to suss out the appropriate approach. It would probably be much more productive to find out why such large numbers are being produced in the first place than trying to make sense of the dot product that includes them.