Should I manually apply the production rules to find out the language generated by this grammar? This is tedious, is there any trick/tip to speed up things?
G = {{S, B}, {a, b}, P, S}
P = {S -> aSa | aBa, B -> bB | b}
EDIT: I found Matajon's answer a good one, that is thinking about each language generated by non-terminal symbol and then combine them.
But I'm still stuck when I have to solve some complicated examples like this:
G = {{S, R, T}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, P, S}
P = {S -> A | AS | BR | CT,
R -> AR | BT | C | CS,
T -> AT | B | BS | CR,
A -> 0 | 3 | 6 | 9,
B -> 1 | 4 | 7,
C -> 2 | 5 | 8}
Crazy, isn't it? Taken from past exams (programming languages course).
I don't know any general trick, but usually it helps to think about the language generated from each non-terminal.
In your example language generated from B is obviously L(B) = {b}^+. Then you think about S rules, using the first rule, you can generate sentencial forms {a^n.S.a^n | n >= 1}. If you use second rule on these sentencial forms or on S alone you can generate sentencial forms {a^n.B.a^n | n >= 1}.
Rest is pretty easy, you combine these two things and get L(G) = {a^n.b^+.a^n | n >= 1}
By the way, in the definition of grammar terminals and nonterminals are sets, not tuples. And third component is production rules, not start symbol. So you should write G = {{S, B}, {a, b}, P, S}.
Edit
Actually, there is a way to solve your second example without much thinking just by following something like a cookbook. Because, language generated by your second context-free grammar is in fact regular.
When you substitute rules for A, B and C to first three rules, you get
P' = {S -> 0 | 3 | 6 | 9 | 0S | 3S | 6S | 9S | 1R | 4R | 7R | 2T | 5T | 8T
R -> 0R | 3R | 6R | 9R | 1T | 4T | 7T | 2 | 5 | 8 | 2S | 5S | 8S
T -> 0T | 3T | 6T | 9T | 1 | 4 | 7 | 1S | 4S | 7S | 2R | 5R | 8R}
And P' is regular grammar. Because of that, you can convert it to nondeterministic finite automaton (there is really simple way, look for it) and then convert resulting NFA to the regular expression (this is not so simple but if you follow an algorithm and don't get lost, you should be ok). And it from regular expression it is easy to tell what language it describes.
Also, once you have NFA for this language you can look at it and determine what it does logically (it has something to do with counts of 1,4,7 and 2,5,8 in the word and mod 3 of their difference. Think it through, it is your homework, afterall :-) )
Of course, if you don't context-free grammar generating regular language you can't use this trick. There is no general way to tell what language the grammar generates (language equality problem for CFG's is undecideable), you have to think about every single example and look for similarities and patterns in it's logical structure.
I think you'll just need to apply the production rules.
Related
i want to code a video platform and have a problem and can't think of a solution right now.
I want to divide a rectangle into equal parts
Came up with a solution for a square but i am unable to come up with a solution for different ratios.
Maybe u guys can help me.
example:
BAD GOOD
n=4
________ ________
| | | | | | | |
| | | | | |---|---|
|_|_|_|_| |___|___|
n=2
________ ________
| | | | |
| | | | |
| | | |–––––––|
| | | | |
|___|___| |_______|
Let X be the width of the rectangle and let Y be the height. Let the goal be to divide this rectangle into N rectangles of equal area whose sides are as close to equal as possible.
The solution is not difficult. First, find all factors of the N. Then, write N as a product of two numbers A and B such that A and B are as close as possible (that is, there is no other choice A' and B' such that |A' - B'| < |A - B|). Assume we chose A > B. All we have to do is put A - 1 lines along the long side of the rectangle, and B - 1 lines along the short side.
For example: n = 4, A = 2 and B = 2 is optimal, so you put A - 1 = 1 and B - 1 = 1 lines along each side of the rectangle (as in your GOOD column for n = 4).
For example: n = 21, A = 7 and B = 3 is optional, so you'd put 6 lines along the long edge of the rectangle and 2 lines along the short edge, equally spaced:
_ _ _ _ _ _ _
|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|
Of course, for prime values of N, you are not going to get a very nice solution, but then there is no nice solution in that case. In such cases - where A and B are very very different and the rectangle's dimensions are not similarly different - you might want to choose another solution that doesn't require all rectangles have the same side lengths. You could do better by allowing 2 or 3 kinds of rectangles, or more, into the solution, for instance.
I have a character vector that is a string of letters and punctuation. I want to create a data frame where each column is made up of a letter/character from this string.
e.g.
Character string = I WENT TO THE FAIR
Dataframe = | I | | W | E | N | T | | T | O | | T | H | E | | F | A | I | R |
I thought I could do this using a loop with substr, but I can't work out how to get R to write into separate columns, rather than just writing over the previous letter. I'm new to writing loops etc so struggling a bit to get my head around the way in which to compose what I need.
Thanks for any help and advice that you can offer.
Best wishes,
Natalie
This should get that result
string <- "I WENT TO THE FAIR"
df <- as.data.frame(t(as.data.frame(strsplit(string,""))), row.names = "1")
I would like to know how I could eliminate nothing elements in a Julia array (1D) like the one below. It was built from reading a text file with lines with no relevant information mixed with lines with relevant information. "nothing" is type Void and I would like to clean the array of all of it.
nothing
nothing
nothing
nothing
nothing
" -16.3651\t 0.1678\t -4.6997\t -14.0152\t -2.6855\t -16.0294\t -7.8049\t -27.1912\t -5.0354\t -14.5187\t\r\n"
" -16.4490\t -1.0910\t -3.6087\t -12.6724\t -1.5945\t -14.7705\t -7.2174\t -25.2609\t -3.7766\t -14.3509\t\r\n"
" -16.4490\t -2.2659\t -2.4338\t -10.9100\t -0.5875\t -13.6795\t -6.7139\t -22.9950\t -2.9373\t -14.0991\t\r\n"
testvector[testvector.!=nothing] is also a very readable option.
benchmarking can help choose the most efficient code.
How are you reading that file?
You can filter out nothings from an array:
filter(x -> !is(nothing, x), [nothing, 42]) # => Any[42]
But you may want to clean your data first, with a tsv (tab separated values) file like this:
-16.3651 0.1678 -4.6997 -14.0152 -2.6855 -16.0294 -7.8049 -27.1912 -5.0354 -14.5187
-16.4490 -1.0910 -3.6087 -12.6724 -1.5945 -14.7705 -7.2174 -25.2609 -3.7766 -14.3509
-16.4490 -2.2659 -2.4338 -10.9100 -0.5875 -13.6795 -6.7139 -22.9950 -2.9373 -14.0991
Using readdlm:
julia> readdlm("data.tsv")
3x10 Array{Float64,2}:
-16.3651 0.1678 -4.6997 -14.0152 … -27.1912 -5.0354 -14.5187
-16.449 -1.091 -3.6087 -12.6724 -25.2609 -3.7766 -14.3509
-16.449 -2.2659 -2.4338 -10.91 -22.995 -2.9373 -14.0991
Using DataFrmaes.readtable:
julia> df = readtable("data.tsv");
julia> names!(df, [symbol(x) for x in 'A':'J'])
2x10 DataFrames.DataFrame
| Row | A | B | C | D | E | F | G |
|-----|---------|---------|---------|----------|---------|----------|---------|
| 1 | -16.449 | -1.091 | -3.6087 | -12.6724 | -1.5945 | -14.7705 | -7.2174 |
| 2 | -16.449 | -2.2659 | -2.4338 | -10.91 | -0.5875 | -13.6795 | -6.7139 |
| Row | H | I | J |
|-----|----------|---------|----------|
| 1 | -25.2609 | -3.7766 | -14.3509 |
| 2 | -22.995 | -2.9373 | -14.0991 |
one simple way is using filter! function to update your vector like this:
testvector=[fill(nothing,10) ; [1,2,3]];
# =>13-element Array{Any,1}:
# nothing
# nothing
# nothing
# nothing
# nothing
# nothing
# nothing
# nothing
# nothing
# nothing
# 1
# 2
# 3
filter!(x->x!=nothing, testvector)
# => 3-element Array{Any,1}:
# 1
# 2
# 3
thanks #Daniel Arndt
EDIT, Refer to this paragraph from Julia doc:
nothing is a special value that does not print anything at the
interactive prompt. Other than not printing, it is a completely normal
value and you can test for it programmatically.
I think all of the conditions below, reach us to the same result
x!=nothing
x!==nothing
!is(x,nothing)
!isa(x,Void)
typeof(x)!=Void
To add to the answers above, it appears:
filter(!isnothing, [nothing, 42])
is a working shorthand for filter(x -> !isnothing(x), [nothing, 42]), and will correctly return 42.
Dear All,
At the end, the code became this:
tmpFile=open(fileName)
tmp=readdlm(tmpFile);
ind=pmap(typeof,tmp[:,1]).!=SubString{ASCIIString}; # if the first column typeof is string, than pmap will return false, else, it return true. This will provide an index of valid/not valid rows.
tmpClean=tmp[ind,:]; # only valid rows will be used
If you may have any suggestion to improve it, I would appreciate it. Thank you for your help.
Apologies for this seemingly minor question, but I can't seem to find the answer anywhere - I'm just coming up to implementing the DAA instruction in my Z80 emulator, and I noticed in the Zilog manual that it is for the purposes of adjusting the accumulator for binary coded decimal arithmetic. It says the instruction is intended to be run right after an addition or subtraction instruction.
My questions are:
what happens if it is run after another instruction?
how does it know what instruction preceeded it?
I realise there is the N flag - but this surely wouldnt definitively indicate that the previous instruction was an addition or subtraction instruction?
Does it just modify the accumulator anyway, based on the conditions set out in the DAA table, regardless of the previous instruction?
Does it just modify the accumulator anyway, based on the conditions set out in the DAA table, regardless of the previous instruction?
Yes. The documentation is only telling you what DAA is intended to be used for. Perhaps you are referring to the table at this link:
--------------------------------------------------------------------------------
| | C Flag | HEX value in | H Flag | HEX value in | Number | C flag|
| Operation | Before | upper digit | Before | lower digit | added | After |
| | DAA | (bit 7-4) | DAA | (bit 3-0) | to byte | DAA |
|------------------------------------------------------------------------------|
| | 0 | 0-9 | 0 | 0-9 | 00 | 0 |
| ADD | 0 | 0-8 | 0 | A-F | 06 | 0 |
| | 0 | 0-9 | 1 | 0-3 | 06 | 0 |
| ADC | 0 | A-F | 0 | 0-9 | 60 | 1 |
| | 0 | 9-F | 0 | A-F | 66 | 1 |
| INC | 0 | A-F | 1 | 0-3 | 66 | 1 |
| | 1 | 0-2 | 0 | 0-9 | 60 | 1 |
| | 1 | 0-2 | 0 | A-F | 66 | 1 |
| | 1 | 0-3 | 1 | 0-3 | 66 | 1 |
|------------------------------------------------------------------------------|
| SUB | 0 | 0-9 | 0 | 0-9 | 00 | 0 |
| SBC | 0 | 0-8 | 1 | 6-F | FA | 0 |
| DEC | 1 | 7-F | 0 | 0-9 | A0 | 1 |
| NEG | 1 | 6-F | 1 | 6-F | 9A | 1 |
|------------------------------------------------------------------------------|
I must say, I've never seen a dafter instruction spec. If you examine the table carefully, you will see that the effect of the instruction depends only on the C and H flags and the value in the accumulator -- it doesn't depend on the previous instruction at all. Also, it doesn't divulge what happens if, for example, C=0, H=1, and the lower digit in the accumulator is 4 or 5. So you will have to execute a NOP in such cases, or generate an error message, or something.
Just wanted to add that the N flag is what they mean when they talk about the previous operation. Additions set N = 0, subtractions set N = 1. Thus the contents of the A register and the C, H and N flags determine the result.
The instruction is intended to support BCD arithmetic but has other uses. Consider this code:
and 15
add a,90h
daa
adc a,40h
daa
It ends converting the lower 4 bits of A register into the ASCII values '0', '1', ... '9', 'A', 'B', ..., 'F'. In other words, a binary to hexadecimal converter.
I found this instruction rather confusing as well, but I found this description of its behavior from z80-heaven to be most helpful.
When this instruction is executed, the A register is BCD corrected using the contents of the flags. The exact process is the following: if the least significant four bits of A contain a non-BCD digit (i. e. it is greater than 9) or the H flag is set, then $06 is added to the register. Then the four most significant bits are checked. If this more significant digit also happens to be greater than 9 or the C flag is set, then $60 is added.
This provides a simple pattern for the instruction:
if the lower 4 bits form a number greater than 9 or H is set, add $06 to the accumulator
if the upper 4 bits form a number greater than 9 or C is set, add $60 to the accumulator
Also, while DAA is intended to be run after an addition or subtraction, it can be run at any time.
This is code in production, implementing DAA correctly and passes the zexall/zexdoc/z80test Z80 opcode test suits.
Based on The Undocumented Z80 Documented, pag 17-18.
void daa()
{
int t;
t=0;
// 4 T states
T(4);
if(flags.H || ((A & 0xF) > 9) )
t++;
if(flags.C || (A > 0x99) )
{
t += 2;
flags.C = 1;
}
// builds final H flag
if (flags.N && !flags.H)
flags.H=0;
else
{
if (flags.N && flags.H)
flags.H = (((A & 0x0F)) < 6);
else
flags.H = ((A & 0x0F) >= 0x0A);
}
switch(t)
{
case 1:
A += (flags.N)?0xFA:0x06; // -6:6
break;
case 2:
A += (flags.N)?0xA0:0x60; // -0x60:0x60
break;
case 3:
A += (flags.N)?0x9A:0x66; // -0x66:0x66
break;
}
flags.S = (A & BIT_7);
flags.Z = !A;
flags.P = parity(A);
flags.X = A & BIT_5;
flags.Y = A & BIT_3;
}
For visualising the DAA interactions, for debugging purposes, I have written a small Z80 assembly program, that can be run in an actual ZX Spectrum or in an emulation that emulates accurately DAA: https://github.com/ruyrybeyro/daatable
As how it behaves, got a table of flags N,C,H and register A before and after DAA produced with the aforementioned assembly program: https://github.com/ruyrybeyro/daatable/blob/master/daaoutput.txt
I understand how to use a breadth first search and A* in a tree structure, but given the following graph, how would it be implemented? In other words, how would the search traverse the graph? S is the start state
Graph Here
It's exactly the same as doing it in a tree. You just need to somehow keep track of which nodes you've already visited so that you don't end up going in circles.
Basically, you treat a graph the same way that you'd treat a tree, except you need to keep track of nodes you've already visited. That's fine for BFS. On top of that, in the case of A*, consider what you'd do when you revisit a node but have found a cheaper route to it.
Paint the graph - Recursively search each node and mark the nodes you visited as dirty. Only recurse when the graph is not dirty.
If memory is not an issue, copy the graph and instead of marking the nodes, remove them from the copy graph.
It's weighted graph. Do you want to find shortest paths or just traverse it?
If you want just traversing, here it is:
1) there is only S in the queue
2) we are adding C and A in the queue, only they are reachable from S directly (with one edge)
3) D, G2 - from C
4) B, E - from A
5) G1 - from D (G2 is already in the queue)
6) there no outgoing edge from G2
7) there's no adjacent nodes of B which aren't already in the queue
So here's the order how nodes where added in the queue: S, C, A, D, G2, B, E, G1
I don't know how helpful you will find this, but here's a complete solution coded in the functional language J (available for free from jsoftware.com).
First, it's probably simplest to work directly from a representation of the graph you show in your picture. I represent this as a (# nodes) x (# nodes) table with a number at (i,j) for the value of the link between node-i and node-j. Also, along the diagonal I've put the number associated with each node itself.
So, I enter this as follows - don't worry too much about the unfamiliar notation, you'll soon see what the result looks like:
grph=: <;.1&>TAB,&.><;._2 ] 0 : 0
A B C D E G1 G2 S
A 2 1 8 2
B 1 1 1 4 2
C 3 1 5
D 1 5 2
E 6 9 7
G1 0
G2 0
S 2 3 5
)
So, I've assigned the variable "grph" as a 9x9 table where the first row and first column are the labels "A"-"E", "G1", "G2", and "S"; I've used tabs to delimit items so this could be cut-and-pasted to or from a spreadsheet as needed.
Now, I'll check the size of my table and display it:
$grph
9 9
grph
+---+--+--+--+--+--+---+---+--+
| | A| B| C| D| E| G1| G2| S|
+---+--+--+--+--+--+---+---+--+
| A | 2| 1| | | 8| | | 2|
+---+--+--+--+--+--+---+---+--+
| B | | 1| 1| 1| | 4 | | 2|
+---+--+--+--+--+--+---+---+--+
| C | | 3| 1| | | | 5 | |
+---+--+--+--+--+--+---+---+--+
| D | | | | 1| | 5 | 2 | |
+---+--+--+--+--+--+---+---+--+
| E | | | | | 6| 9 | 7 | |
+---+--+--+--+--+--+---+---+--+
| G1| | | | | | 0 | | |
+---+--+--+--+--+--+---+---+--+
| G2| | | | | | | 0 | |
+---+--+--+--+--+--+---+---+--+
| S | 2| | 3| | | | | 5|
+---+--+--+--+--+--+---+---+--+
It looks OK and it's easy to compare this to the picture of the graph to check it.
Now I'll drop the first row and column so we're left only with numbers (as boxed literals),
and remove any extraneous tab characters.
grn=. TAB-.~&.>}.}."1 grph
You can see I assign this result to the variable "grn".
Next, I'll replace any empty cells with "_" - which represents infinity - then convert the literals to numeric representation (re-assigning the result to the same name "grn"):
grn=. ".&>(0=#&>grn)}grn,:<'_'
Finally, I'll move the last column and row to the beginning since this is the one for "S" and it's supposed to be first. I'll also display the result to confirm that it looks correct.
]grn=. _1|."1]_1|.grn NB. "S" goes first.
5 2 _ 3 _ _ _ _
2 2 1 _ _ 8 _ _
2 _ 1 1 1 _ 4 _
_ _ 3 1 _ _ _ 5
_ _ _ _ 1 _ 5 2
_ _ _ _ _ 6 9 7
_ _ _ _ _ _ 0 _
_ _ _ _ _ _ _ 0
So, now that I have a simple 8x8 table of numbers representing the graph, it's a simple matter to traverse it.
Here's a simple J function, called "traverseGraph", to read this table, traverse the graph it represents, and return two results: the indexes (0-based origin) of the nodes visited, and the values of the points and edges in the order visited.
traverseGraph=: 3 : 0
pts=. ,_-.~,ix{y [ nxt=. ix=. ,0
while. 0~:#nxt=. ~.ix-.~;([:I._~:])&.><"1 nxt{y do.
ix=. ix,nxt [ pts=. pts,_-.~,nxt{y
end.
ix;pts
)
We start by initializing three variables: the list of indexes "ix" (to zero, since we want to begin in the zeroth row of the table), a variable "nxt" to point to the next group of nodes (initially the same as the starting node), and the list of point values "pts" (starting as the 0th row of our input table, known internally as "y", with all the infinite values removed.)
In the "while." loop, we continue as long as there's more than zero "nxt" values resulting from pulling the current row out of the table and removing any nodes (in "ix") we've already visited. Inside the loop, we accumulate the next set of indexes onto the end of "nxt" and the point values onto "pts". At the end, we return the indexes and point values as our (two-element) result.
We run it like this - it displays the result by default:
traverseGraph grn
+---------------+---------------------------------------------+
|0 1 3 2 5 7 4 6|5 2 3 2 2 1 8 3 1 5 2 1 1 1 4 6 9 7 0 1 5 2 0|
+---------------+---------------------------------------------+
So, the first box contains the indexes starting with "0" and ending with "6". The second boxed item is the vector of point values in the order we accumulated them. I don't know what you do with these, so I just show them.
We can use the indexes to display the node names like this:
0 1 3 2 5 7 4 6{(<"0'SABCDE'),'G1';'G2'
+-+-+-+-+-+--+-+--+
|S|A|C|B|E|G2|D|G1|
+-+-+-+-+-+--+-+--+
I don't know how useful you'll find this but it does outline a simple solution to your problem.