Physically is possible to simulate such situation on a board, using electronic components.
I got 2 inputs A and B , with 3 possible values for each one (-1,0,1). My final aim is to achieve this following truth table
A | B | result
–1 | –1 | +1
–1 | +1 | 0
0 | 0 | 0
0 | +1 | +1
+1 | –1 | 0
+1 | 0 | +1
+1 | +1 | -1
In pseudo code:
if (A equals B)
result = A * -1
else
result = A + B
Yes it is absolutely possible and this what todays CPUs are using. The so called logic gates.
Of course depending on your project but won't probably need Intel processor to redo your work but much simpler components doing just that. See the above link for example components doing it.
Related
I have a very simple graph with 5 nodes (named n1 - n5), 1 node type (:Node) and 2 relationship types (:r1, :r2). The nodes and relationships are arranged as follows (apologies for the ascii art):
(n1)-[:r1]->(n2)-[:r1]->(n3)
(n1)-[:r2]->(n4)-[:r2]->(n3)
(n1)-[:r1]->(n5)-[:r2]->(n3)
I have a query using a variable length path. I expected to be able restrict the paths returned by describing a specific pattern in the WHERE clause:
MATCH p = (n:Node {name: 'n1'})-[*..2]->()
WHERE (n)-[:r1]->()-[:r1]->()
RETURN p
The problem is that the response returns all possible paths. My question; is it possible to filter the returned paths when specifying a variable length path in a query?
If all relationships or nodes have to adhere to the same predicate, this is easy. You'll need a variable for the path, and you'll need to use all() (or none()) in your WHERE clause to apply the predicate for all relationships or nodes in your path:
MATCH p = (n:Node {name: 'n1'})-[*..2]->()
WHERE all(rel in relationships(p) WHERE type(rel) = 'r1')
RETURN p
That said, when all you want is for all relationships in the var-length path to be of the same type (or types, if you want multiple), that's best done in the pattern itself:
MATCH p = (n:Node {name: 'n1'})-[:r1*..2]->()
RETURN p
For more complicated cases, such as multiple relationship types (where the order of those types matters in the path), or repeating sequences of types or node labels in the path, then alternate approaches are needed. APOC path expanders may help.
EDIT
You mentioned in the comments that your case deals with sequences of relationships of varying lengths. While the APOC path expanders may help, it there are a few restrictions:
The path expanders currently operate on node labels and relationship types, but not properties, so if your expansions rely on predicates on properties, the path expanders won't be able to handle that for you during expansion, that would have to be done by filtering the path expander results after.
There are limits to the relationship sequence support for path expanders. We can define sequences of any length, and can accept multiple relationship types at each step in the sequence, but we don't currently support diverging sequences ((r1 then r2 then r3) or (r2 then r5 then r6)).
If we wanted to do a 3-step sequence of r1 (incoming), r2 (outgoing), then r3 or r4 (with r3 in either direction and r4 outgoing), repeating the sequence up to 3 times we could do so like this:
MATCH (n:Node {name: 'n1'})
CALL apoc.path.expandConfig(n, {relationshipFilter:'<r1, r2>, r3 | r4>', minLevel:1, maxLevel:9) YIELD path
RETURN path
Note that we can provide differing directions per relationship in the filter, or leave off the arrow entirely if we don't care about the direction.
Label filtering is more complex, but I didn't see any need for that present in the examples so far.
Your query return all paths because your WHERE clause (Filter operator) is applied before the VarLengthExpand operator:
+-----------------------+----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| Operator | Estimated Rows | Rows | DB Hits | Page Cache Hits | Page Cache Misses | Page Cache Hit Ratio | Variables | Other |
+-----------------------+----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| +ProduceResults | 0 | 4 | 0 | 0 | 0 | 0.0000 | anon[32], anon[41], n, p | |
| | +----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| +Projection | 0 | 4 | 0 | 0 | 0 | 0.0000 | p -- anon[32], anon[41], n | {p : PathExpression(NodePathStep(Variable(n),MultiRelationshipPathStep(Variable(),OUTGOING,NilPathStep)))} |
| | +----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| +VarLengthExpand(All) | 0 | 4 | 7 | 0 | 0 | 0.0000 | anon[32], anon[41] -- n | (n)-[:*..2]->() |
| | +----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| +Filter | 0 | 1 | 6 | 0 | 0 | 0.0000 | n | n.name = { AUTOSTRING0}; GetDegree(Variable(n),Some(RelTypeName(KNOWS)),OUTGOING) > 0 |
| | +----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
| +NodeByLabelScan | 4 | 4 | 5 | 0 | 0 | 0.0000 | n | :Crew |
+-----------------------+----------------+------+---------+-----------------+-------------------+----------------------+----------------------------+------------------------------------------------------------------------------------------------------------+
This should get you going:
MATCH p = (n:Node {name: 'n1'})-[*..2]->()
WITH n, relationships(p)[0] as rel0, relationships(p)[1] as rel1, p
MATCH (n)-[rel0:r1]->()-[rel1:r1]->()
RETURN p
I have a drug analysis experiment that need to generate a value based on given drug database and set of 1000 random experiments.
The original database looks like this where the number in the columns represent the rank for the drug. This is a simplified version of actual database, the actual database will have more Drug and more Gene.
+-------+-------+-------+
| Genes | DrugA | DrugB |
+-------+-------+-------+
| A | 1 | 3 |
| B | 2 | 1 |
| C | 4 | 5 |
| D | 5 | 4 |
| E | 3 | 2 |
+-------+-------+-------+
A score is calculated based on user's input: A and C, using the following formula:
# Compute Function
# ['A','C'] as array input
computeFunction(array) {
# do some stuff with the array ...
}
The formula used will be same for any provided value.
For randomness test, each set of experiment requires the algorithm to provide randomized values of A and C, so both A and C can be having any number from 1 to 5
Now I have two methods of selecting value to generate the 1000 sets for P-Value calculation, but I would need someone to point out if there is one better than another, or if there is any method to compare these two methods.
Method 1
Generate 1000 randomized database based on given database input shown above, meaning all the table should contain different set of value pair.
Example for 1 database from 1000 randomized database:
+-------+-------+-------+
| Genes | DrugA | DrugB |
+-------+-------+-------+
| A | 2 | 3 |
| B | 4 | 4 |
| C | 3 | 2 |
| D | 1 | 5 |
| E | 5 | 1 |
+-------+-------+-------+
Next we perform computeFunction() with new A and C value.
Method 2
Pick any random gene from original database and use it as a newly randomized gene value.
For example, we pick the values from E and B as a new value for A and C.
From original database, E is 3, B is 2.
So, now A is 3, C is 2. Next we perform computeFunction() with new A and C value.
Summary
Since both methods produce completely randomized input, therefore it seems to me that it will produce similar 1000-value outcome. Is there any way I could prove they are similar?
I am trying to expand a data set to include dates outside of the current range.
The data I have ranges from 1992q1 to 2017q1. Each observation exists within a portion of that larger window, for example from 1993q2 to 1997q1.
I need to create quarterly observations for each range to fill the missing time. I have already expanded the existing data into quarters.
What I cannot figure out how to do is add in those missing quarters. For example, country1 may have the dates 1993q2 to 1997q1. I need to add in the missing dates from 1992q1 to 1993q1 and 1997q2 to 2017q1.
A very simple analogue of I want I think is your question is shown by this sandbox dataset.
clear
set obs 10
gen id = cond(_n < 7, 1, 2)
gen qdate = yq(1992, 1) in 1
replace qdate = yq(1992, 3) in 7
bysort id (qdate) : replace qdate = qdate[_n-1] + 1 if missing(qdate)
format qdate %tq
list, sepby(id)
+-------------+
| id qdate |
|-------------|
1. | 1 1992q1 |
2. | 1 1992q2 |
3. | 1 1992q3 |
4. | 1 1992q4 |
5. | 1 1993q1 |
6. | 1 1993q2 |
|-------------|
7. | 2 1992q3 |
8. | 2 1992q4 |
9. | 2 1993q1 |
10. | 2 1993q2 |
+-------------+
fillin id qdate
list, sepby(id)
+-----------------------+
| id qdate _fillin |
|-----------------------|
1. | 1 1992q1 0 |
2. | 1 1992q2 0 |
3. | 1 1992q3 0 |
4. | 1 1992q4 0 |
5. | 1 1993q1 0 |
6. | 1 1993q2 0 |
|-----------------------|
7. | 2 1992q1 1 |
8. | 2 1992q2 1 |
9. | 2 1992q3 0 |
10. | 2 1992q4 0 |
11. | 2 1993q1 0 |
12. | 2 1993q2 0 |
+-----------------------+
So. fillin is a simple way of ensuring that all cross-combinations of identifier and time are present. However, to what benefit? Although not shown in this example, values of other variables spring into existence only as missing values. In some situations, proceeding with interpolation is justified, but usually, you just live with incomplete panels.
How to find solutions like these? One good strategy is to skim through the [D] manual to see what basic data management commands exist.
For a course assignment, we were asked to look up how to convert between various logic gates by creating systems involving two inputs A and B and one output O. The last conversion was from XOR to NOR, but I can't seem to find any answers out there. The logic for each of these is as follows:
XOR
A | B | O
----------
0 | 0 | 0
1 | 0 | 1
0 | 1 | 1
1 | 1 | 0
NOR
A | B | O
----------
0 | 0 | 1
1 | 0 | 0
0 | 1 | 0
1 | 1 | 0
For simplicity, I'll request pseudo-code for a function NOR rather than a diagram. The XOR binary operator is ^. The closest I came was the following function in JavaScript:
function nor(a, b) {
return ((a^1)^(a^b))^((b^0)^(a^b));
}
But this is actually the logic for XNOR:
A | B | O
----------
0 | 0 | 1
1 | 0 | 0
0 | 1 | 0
1 | 1 | 1 (this should be 0 though)
Anyone else who can figure this out?
EDIT To summarize, make a NOR operator / function of two arguments using A, B, the constants 0 and 1 as needed, and the only operator allowed is the XOR operator (^).
It is not possible. For instance, you can check
a XOR b XOR 1
to achieve equivalence, but you cannot exclude AND from it, because XOR is not a universal gate.
Not going to give complete solution to homework. But I was curious and so:
You write you have A B and 0 as inputs, but you use a 1 as input in your code.
If you are not allowed to to use 1, you might want to consider what A XOR 0 is.
Your edit looks like 1 is allowed, in which case as far as I can see the solution is to use the direct input in one side of a gate, and a combined input as the other. Like ((a) ^ (a ^ b)).
EDIT Posted the reply without a finished solution so you had something to work on. Just noting it is possible there is no solution.
Assume, that we have large file, which contains descriptions of the cells of two matrices (A and B):
+---------------------------------+
| i | j | value | matrix |
+---------------------------------+
| 1 | 1 | 10 | A |
| 1 | 2 | 20 | A |
| | | | |
| ... | ... | ... | ... |
| | | | |
| 1 | 1 | 5 | B |
| 1 | 2 | 7 | B |
| | | | |
| ... | ... | ... | ... |
| | | | |
+---------------------------------+
And we want to calculate the product of this matrixes: C = A x B
By definition: C_i_j = sum( A_i_k * B_k_j )
And here is a two-step MapReduce algorithm, for calculation of this product (I will provide a pseudocode):
First step:
function Map (input is a single row of the file from above):
i = row[0]
j = row[1]
value = row[2]
matrix = row[3]
if(matrix == 'A')
emit(i, {j, value, 'A'})
else
emit(j, {i, value, 'B'})
Complexity of this Map function is O(1)
function Reduce(Key, List of tuples from the Map function):
Matrix_A_tuples =
filter( List of tuples from the Map function, where matrix == 'A' )
Matrix_B_tuples =
filter( List of tuples from the Map function, where matrix == 'B' )
for each tuple_A from Matrix_A_tuples
i = tuple_A[0]
value_A = tuple_A[1]
for each tuple_B from Matrix_B_tuples
j = tuple_B[0]
value_B = tuple_B[1]
emit({i, j}, {value_A * value_b, 'C'})
Complexity of this Reduce function is O(N^2)
After the first step we will get something like the following file (which contains O(N^3) lines):
+---------------------------------+
| i | j | value | matrix |
+---------------------------------+
| 1 | 1 | 50 | C |
| 1 | 1 | 45 | C |
| | | | |
| ... | ... | ... | ... |
| | | | |
| 2 | 2 | 70 | C |
| 2 | 2 | 17 | C |
| | | | |
| ... | ... | ... | ... |
| | | | |
+---------------------------------+
So, all we have to do - just sum the values, from lines, which contains the same values i and j.
Second step:
function Map (input is a single row of the file, which produced in first step):
i = row[0]
j = row[1]
value = row[2]
emit({i, j}, value)
function Reduce(Key, List of values from the Map function)
i = Key[0]
j = Key[1]
result = 0;
for each Value from List of values from the Map function
result += Value
emit({i, j}, result)
After the second step we will get the file, which contains cells of the matrix C.
So the question is:
Taking into account, that there are multiple number of instances in MapReduce cluster - which is the most correct way to estimate complexity of the provided algorithm?
The first one, which comes to mind is such:
When we assume that number of instances in the MapReduce cluster is K.
And, because of the number of lines - from file, which produced after the first step is O(N^3) - the overall complexity can be estimated as O((N^3)/K).
But this estimation doesn't take into account many details: such as network bandwidth between instances of MapReduce cluster, ability to distribute data between distances - and perform most of the calculations locally etc.
So, I would like to know which is the best approach for estimation of efficiency of the provided MapReduce algorithm, and does it make sense to use Big-O notation to estimate efficiency of MapReduce algorithms at all?
as you said the Big-O estimates the computation complexity, and does not take into consideration the networking issues such(bandwidth, congestion, delay...)
If you want to calculate how much efficient the communication between instances, in this case you need other networking metrics...
However, I want to tell you something, if your file is not big enough, you will not see an improvement in term of execution speed. This is because the MapReduce works efficiently only with BIG data. Moreover, your code has two steps, that means two jobs. MapReduce, from one job to another, takes time to upload the file and start the job again. This can affect slightly the performance.
I think you can calculate the efficiently in term of speed and time as the MapReduce approach is for sure faster when it comes to big data. This is if we compared it to the sequential algorithms.
Moreover, efficiency can be with regards to the fault-tolerance. This is because MapReduce will manage to handle failures by itself. So, no need for the programmers to handle instance failure or networking failures..