I'm developing a rudimentary word finder app with sql and ruby where I have an array of letters to find available words. It's easier to make the query by narrowing down what alphabetic letters aren't in the array. For ex.
alphabet= %w{a b c d e f g h i j k l m n o p q r s t u v w x y z}
available_letters = %w{p k z l p m t l n g g r u a r t n d z w a l m n e}
I can then subtract from alphabet, letters to exclude from my search and end up with an sql query like the one below.
select * from words
where word not like '%b%' and word not like '%c%' and word not like '%f%'.....
This gives me all the available words with a combination of all available letters. It does not narrow them down by the number of times that letter occurs. So if I only have one "E", I would like the query to narrow down words that only contain one e. I'm not sure if this can be done with an sql query or whether I will need to use a procedure. Anyone know a good way of solving this?
You will probably want to look into faster ways of implementing this, but to answer your question, you can exclude words with more than one "e" with not like '%e%e%'.
Related
I have some data that looks like this, and identifies pairs that are related:
From_ID To_ID
A C
B C
D E
E D (note this is the same pair as above, in a different order)
E F
A F
G H
Using the logic of 'if x is paired with y, and y is paired with z, then x is paired with z', how can I run an SQL query to return all members of a group?
So for the table above I would like a set of results that identifies or returns two groups: 'A, B, C, D, E, F' and 'G, H', not fussy about how this is done.
It feels like some kind of iterative query but I really have no idea where to start with this so any pointers would be appreciated.
edit: could be run in SQL Developer or HiveQL.
I have a question regarding this question:
L= empty where the alphabet is {a,b}
how to create a grammar for this ? how can be the production rule ?
thanks in advance
A grammar G is an ordered 4-tuple {S, N, E, e, P} where:
N is a set of non-terminal symbols
E is a set of terminal symbols
N and E are disjoint
E is a superset of the alphabet of L(G)
e is the empty string
P is a set of ordered pairs of elements of (N U E U e); that is, P is a subset of (N U E U e) X (N U E U e)*.
S, the start symbol, is in N
A derivation in G is a sequence of elements of (N U E U e)* such that:
The first element is S
Adjacent elements w[i] and w[i+1] can be written as w[i] = uxv and w[i+1] = uyv such that (x, y) is in P
If there is a derivation in G whose last element is a string w[n] over (E U e)*, we say G generates w[n]; that is, w[n] is in L(G).
Now, we want to define a grammar G such that L(G) is the empty set. We fix the alphabet E = {a, b}. We must still define:
N, the set of nonterminals
S, the start symbol
P, the productions
We might as well take S as our start symbol. So N contains at least S; N is a superset of {S}. We will only add more nonterminals if we determine we need them. Let us turn our attention to the condition that L(G) is empty.
If L(G) is empty, that means there is no derivation in G that leads to a string of only terminal symbols. We can accomplish this easily be ensuring all our productions produce at least one nonterminal with any terminal. Or produce no terminals at all. So the following grammars would all work:
S := S
or
S := aSb
or
S := aXb | XXSSX
X := aabbXbbaaS
etc. All of these grammars have L(G) empty since none of them can derive a string of nonterminals.
I'm looking at some exercices that reguards FSAs and my teacher is giving and odd solution for the language
I would have solved that language with a Not-Deterministic TM cause you have to remember n for a, b and c.
This is the given solution
Is this solution incorrect?
The example above does work and therefore is correct.
It is clear that for n = 1 the FSA posted above does work.
Now let's condiser the case n > 1, we have to have the same amount of a, b and c between every possible string composed of {a,b,c}. This string can be seen as the same string with n = 1 and the other n - 1 repetitions can be put in the {a,b,c}* group of strings, hence this FSA can accept that language correctly.
**{a^i b^j c^k d^m | i+j=k+m | i<m}**
The grammar should allow the language in order abbccd not cbbcda. First should be the a's then b's and so on.
I know that you must "count" the number of a's and b's you are adding to make sure there are an equivalent number of c's and d's. I just can't seem to figure out how to make sure there are more c's than a's in the language. I appreciate any help anyone can give. I've been working on this for many hours now.
Edit:
The grammar should be Context Free
I have only got these two currently because all others turned out to be very wrong:
S -> C A D
| B
B -> C B D
|
C -> a
| b
D -> c
| d
and
S -> a S d
| A
A -> b A c
|
(which is close but doesn't satisfy the i < k part)
EDIT: This is for when i < k, not i < m. OP changed the problem, but I figure this answer may still be useful.
This is not a context free grammar, and this can be proven with the pumping lemma which states that if the grammar is context free, there exists an integer p > 0, such that all strings in the language of length >= p can be split into a substring uvwxy, where len(vx) >= 1, len(vwx) <= p, and uvnwxny is a member of the language for all n >= 0.
Suppose that a value of p exists. We can create a string such that:
k = i + 1
j = m + 1
j > p
k > p
v and x cannot contain more than one type of character or be both on the left side or both on the right side, because then raising them to powers would break the grammar immediately. They cannot be the same character as each other, because then multiplying them would break the rule that i + j = k + m. v cannot be a if x is d, because then w contains the bs and cs, which makes len(vwx) > p. By the same reasoning, v cannot be as if x is cs, and v cannot be bs if x is ds. The only remaining option is bs and cs, but setting n to 0 would make i >= k and j >= m, breaking the grammar.
Therefore, it is not a context free grammar.
There has to be at least one d because i < m, so there has to be a b somewhere to offset it. T and V guarantee this criterion before moving to S, the accepted state.
T ::= bd | bTd
U ::= bc | bUc
V ::= bUd | bVd
S ::= T | V | aSd
I'm not sure how to express this problem, so my apologies if it's already been addressed.
I have business rules summarized as a table of outputs given two inputs. For each of five possible value on one axis, and each of five values on another axis, there is a single output. There are ten distinct possibilities in these 25 cells, so it's not the case that each input pair has a unique output.
I have encoded these rules in TSQL with nested CASE statements, but it's hard to debug and modify. In C# I might use an array literal. I'm wondering if there's an academic topic which relates to converting logical rules to matrices and vice versa.
As an example, one could translate this trivial matrix:
A B C
-- -- -- --
X 1 1 0
Y 0 1 0
...into rules like so:
if B OR (A and X) then 1 else 0
...or, in verbose SQL:
CASE WHEN FieldABC = 'B' THEN 1
WHEN FieldABX = 'A' AND FieldXY = 'X' THEN 1
ELSE 0
I'm looking for a good approach for larger matrices, especially one I can use in SQL (MS SQL 2K8, if it matters). Any suggestions? Is there a term for this type of translation, with which I should search?
Sounds like a lookup into a 5x5 grid of data. The inputs on axis and the output in each cell:
Y=1 Y=2 Y=3 Y=4 Y=5
x=1 A A D B A
x=2 B A A B B
x=3 C B B B B
x=4 C C C D D
x=5 C C C C C
You can store this in a table of x,y,outvalue triplets and then just do a look up on that table.
SELECT OUTVALUE FROM BUSINESS_RULES WHERE X = #X and Y = #Y;