I've been doing past paper questions and keep coming up against these questions that deal with 3 valued logic. My notes mention it but don't give examples that relate to those asked in exams.
I understand the basis that True = 1, False = 0 & Unknown = 1/2 as well as And = Min, Or = Max and Not(x) = 1-x. However I do not know how to apply it to questions such as those below:
In SQL, discuss the possible truth values of the following expression:
R.a > R.b OR R.a <= 0 OR R.b >= 0
Justify your answer.
And:
The phone and age fields of the Owner table might have null values in
them. Considering all possible combinations, show which of the three
truth values might be returned by the expression:
phone = ’141-3304913’ OR age <50 OR age >= 50
Any help in clarifying these for me would be really appreciated :)
I will focus on the concrete example, which is more proper for clarifying things.
Put simply, your logical expression is made of a conjunction of three clauses
C1: phone = '141-3304913'
C2: age < 50
C3: age >= 50
for which tri-boolean logic states that the result is
True, if any clause is true
False, if all clauses are false
Unknown, in all the other cases
Consequently, if the value associated with True is the largest, with False is the smallest, and with Unknown is any intermediate value, then taking the MAX for a conjunction proves correct. Similarly, a disjunction works with the MIN function. Negation works as long as we interpret any value between 0 and 1 (excluded) as Unknown; clearly, if we take 1/2 then the negation function is "stable", but that does not really matter in mathematical terms.
More operatively, the clauses clearly react to the following values (instances) of your phone variable P and your age variable A:
P1 such that P1 = '141-3304913'
P2 such that P2 <> '141-3304913'
P3 such that P3 = NULL
A1 such that A1 < 50
A2 such that A2 >= 50
A3 such that A3 = NULL
In terms of satisfaction of the clauses, we have
P1 -> C1 = 1
P2 -> C1 = 0
P3 -> C1 = 1/2
A1 -> C2 = 1, C3 = 0
A2 -> C2 = 0, C3 = 1
A3 -> C2 = C3 = 1/2
In general there exist 3*3 possible combinations, since each of your two variables takes three possible values:
P1 A1: C1 = 1, C2 = 1, C3 = 0 -> MAX(1,1,0) = 1 -> true
P1 A2: C1 = 1, C2 = 0, C3 = 1 -> MAX(1,0,1) = 1 -> true
P1 A3: C1 = 1, C2 = 1/2, C3 = 1/2 -> MAX(1,1/2,1/2) = 1 -> true
P2 A1: C1 = 0, C2 = 1, C3 = 0 -> MAX(0,1,0) = 1 -> true
P2 A2: C1 = 0, C2 = 0, C3 = 1 -> MAX(0,0,1) = 1 -> true
P2 A3: C1 = 0, C2 = 1/2, C3 = 1/2 -> MAX(0,1/2,1/2) = 1/2 -> unknown
P3 A1: C1 = 1/2, C2 = 1, C3 = 0 -> MAX(1/2,1,0) = 1 -> true
P3 A2: C1 = 1/2, C2 = 0, C3 = 1 -> MAX(1/2,0,1) = 1 -> true
P3 A3: C1 = 1/2, C2 = 1/2, C3 = 1/2 -> MAX(1/2,1/2,1/2) = 1/2 -> unknown
In particular, since C2 and C3 are mutually exclusive, you never get False as a result of the conjunction.
The expression R.a > R.b OR R.a <= 0 OR R.b >= 0 instead presents these cases:
R.a <= 0, R.a > 0, R.a = unknown
R.b >= 0, R.b < 0, R.b = unknown
R.a - R.b > 0, R.a - R.b <= 0, R.a - R.b = unknown
Apparently we have three variables and 27 possible cases, but several related to R.a - R.b can be trivially ruled out.
Related
I have an in-memory table with 3 million records.
Part of the stock data is as follows:
How can I quickly get the last n rows of this table, say 35,000 rows?
Here are two solutions to your problem.
I’ll use the following simulated data:
def createData(num){
date = take(2022.01.01, num)
time = 00:00:00.000 + take(0..num, num)
a1 = rand(1000, num)
a2 = rand(1000, num)
a3 = rand(1000, num)
a4= rand(1000, num)
a5 = rand(100.0, num)
a6 = rand(100.0, num)
a7 = rand(100.0, num)
a8 = rand(100.0, num)
return table(date, time, a1, a2, a3, a4, a5, a6, a7, a8)
}
num = 3000000
t = createData(num)
Solution 1:
timer{
result = (select * from t order by time desc limit 35000).sortBy!(`time)
}
//Time elapsed: 11.732 ms
Solution 2:
timer{
rowOffset = size(t) - 35000
rowCount = 35000
result = select * from t order by time limit rowOffset, rowCount
}
//Time elapsed: 14.013 ms
I am using rbx.lua. Everything works except for the color changing at the bottom.
Please note that the script is not finished, I just stopped at p1.
--Variables--
local r1 = math.random(100)
local r2 = math.random(100)
local r3 = math.random(100)
local p1 = workspace.Part1
local p2 = workspace.Part2
local p3 = workspace.Part3
local c1 = game.ServerStorage.green
local c2 = game.ServerStorage.yellow
local c3 = game.ServerStorage.red
local grn = NumberRange.new(0, 45)
local ylw = NumberRange.new(46, 75)
local red = NumberRange.new(76, 100)
--Randomizing Y Vector Between 0 and 100--
p1.Size = Vector3.new(4, r1, 4)
p2.Size = Vector3.new(4, r2, 4)
p3.Size = Vector3.new(4, r3, 4)
--Setting up colors--
if p1.Size.Y == grn then
p1.BrickColor = c1
end
if p1.Size.Y == ylw then
p1.BrickColor = c2
end
if p1.Size.Y == red then
p1.BrickColor = c3
end
Try this instead:
Remove the NumberRange variables, and replace the if blocks with this:
if p1.Size.Y <= 45 then
p1.BrickColor = c1
elseif p1.Size.Y <=75 then
p1.BrickColor = c2
else
p1.BrickColor = c3
end
The elseif block won’t be checked unless the p1.Size.Y value is above 45. Same for the else block: it won’t be entered unless the value is above 75.
Hope that helps! You were checking to see if a specific value was equal to (==) a range...not if the value was in the range.
I have a sequence of numbers as follows:
1 , 1, 5, 13, 41, 121, 365, ....
The first two values are:
N(1) = 1 and N(2) = 1
As from 3rd value, N(i) = 2*N(i-1) + 3*N(i-2)
The issue I am facing with is: If I give an argument of p, it should return me the last values of the sequence < p (Using fortran77).
For instance, if p = 90, it should return the value 41.
a = 1
b = 1
while b < p:
c = 2 * b + 3 * a
a = b
b = c
return a
The Fortran equivalent is:
function fct(p) result(a)
integer, intent(in) :: p
integer :: a, b, c
a = 1
b = 1
do while (b < p)
c = 2 * b + 3 * a
a = b
b = c
enddo
end function
program test
integer :: fct
external fct
print *,fct(90)
end program
Assuming you already have the sequence in a variable lst, and p set,
max(filter(lambda x:x<=p, lst))
def get_last_element(p):
n1 = 1
n2 = 1
while True:
if n2 > p:
return n1
n1, n2 = n2, 2*n2 + 3 * n1
print(get_last_element(90))
I wrote a piece of code in Fortran 2003. I defined a type which has memory for two last parts of the sequence.The procedure is a recursive function. The type can be used standalone to get n-th part of the sequence or efficiently placed in a loop to find parts in a row (not necessarily beginning at 1) as it has memory of previous parts. (compiler: gfortran 4.8).
The type is defined in mymod.f90 file as
module mymod
implicit none
type seq_t
integer :: saved_i = 0, saved_val_i = 0, saved_val_i_1 = 0
contains
procedure :: getpart => getpart_seq
end type
contains
recursive function getpart_seq(this,i) result(r)
class(seq_t) :: this
integer, intent(in) :: i
integer :: r,r_1,r_2
if (i.eq.1.or.i.eq.2) then
r = 1
elseif(i.eq.this%saved_i) then
r = this%saved_val_i
elseif(i.eq.this%saved_i-1) then
r = this%saved_val_i_1
else
r_1 = this%getpart(i-1)
r_2 = this%getpart(i-2)
r = 2*r_1 + 3*r_2
this%saved_val_i_1 = r_1
end if
this%saved_i = i
this%saved_val_i = r
end function getpart_seq
end module mymod
The main program for the requested case is
program main
use mymod
implicit none
type (seq_t) :: seq
integer :: i,p,tmp_new,tmp_old,ans
! Set the threshold here
p = 90
! loop over parts of the sequence
i = 0
do
i = i + 1
tmp_new = seq%getpart(i)
print*,tmp_new
if (tmp_new>p) then
ans = tmp_old
exit
end if
tmp_old = tmp_new
end do
print*,"The last part of sequence less then",p," is equal to",ans
end program
The outcome is
1
1
5
13
41
121
The last part of sequence less then 90 is equal to 41.
I am aware of this post :- Locking mechanism of Mifare Classic 1K
However, it is really not clear - how a value like FF 07 80 FF is calculated in this string:
D3 F7 D3 F7 D3 F7 FF 07 80 FF 00 00 00 00 00 00
This means that the blocks can be read with key A and written with Key B but does not allow inc/dec.
How should the access bits look like if I have to support increment and decrement operations. I understand that C1, C2 and C3 must be 1,1,0 how does this reflect to the byte 6, 7 and 8.
Any help would be highly appreciate.
The access bits FF 07 80 translate to
C1 = 0x0 => C1_3 = 0, C1_2 = 0, C1_1 = 0, C1_0 = 0
C2 = 0x0 => C2_3 = 0, C2_2 = 0, C2_1 = 0, C2_0 = 0
C3 = 0x8 => C3_3 = 1, C3_2 = 0, C3_1 = 0, C3_0 = 0
So the sector trailer can be read and written using key A only (Cx_3 = 0 0 1). All operations (read, write, increment, decrement, etc) can be performed on the data blocks using key A only (Cx_{0,1,2} = 0 0 0, key B is disabled due to the access conditions of the trailer block).
If you want be able to read all blocks with key A, write with key B, perform value block increments with key B and perform value block decrement, etc. with keys A and B, you could use access conditions like this:
sector trailer write with key B only: Cx_3 = 0 1 1
data/value blocks: read/decrement with key A, write/increment with key B: Cx_{0,1,2} = 1 1 0
C1_3 = 0, C1_2 = 1, C1_1 = 1, C1_0 = 1 => C1 = 0x7
C2_3 = 1, C2_2 = 1, C2_1 = 1, C2_0 = 1 => C2 = 0xF
C3_3 = 1, C3_2 = 0, C3_1 = 0, C3_0 = 0 => C3 = 0x8
This leads to the access bits 08 77 8F. Hence, you sector trailer could look like this (with key A = D3F7D3F7D3F7 and key B = 000000000000):
D3F7D3F7D3F7 08778F FF 000000000000
I'm working on a VBA script that is to work through an extensive list of email addresses and flag the ones that are suspected of being wrong.
I'd like to refine the routine by adding a function that would spot typos in common domain names such as gmail, hotmail, msn, skynet, etc. I'll have a list of these common display names in an array.
The string function would see if the inputted string looks similar but is not the same as an element in the array, and return true as boolean if it is the case.
Idea is to spot erroneous entries such as: homtail, mns, slynet, hotmal, yahooo, etc.
Not looking for a script per se, looking for inspiration of how to tackle this problem...
a fuzzy comarison is what you need - there is code here that will compare two strings, and give you a score from 0 to 1 depending on how close they are. It will be up to you to decide how close they are to do automatic substitution.
example results:
server text fuzzy score
------- -------- -----------
hotmail hotmale 0.7619048
hotmail hot 0.4285714
hotmail notmail 0.8571429
hotmail NotEvenClose 0.1944444
hotmail hotmail 1
hotmail yellow 0.0952381
hotmail homtail 0.7142857
The the source code has been released under GNU Lesser GPL
in case of link rot, here's the code:
Public Function Fuzzy(ByVal s1 As String, ByVal s2 As String) As Single
Dim i As Integer, j As Integer, k As Integer, d1 As Integer, d2 As Integer, p As Integer
Dim c As String, a1 As String, a2 As String, f As Single, o As Single, w As Single
'
' ******* INPUT STRINGS CLEANSING *******
'
s1 = UCase(s1) 'input strings are converted to uppercase
d1 = Len(s1)
j = 1
For i = 1 To d1
c = Mid(s1, i, 1)
Select Case c
Case "0" To "9", "A" To "Z" 'filter the allowable characters
a1 = a1 & c 'a1 is what remains from s1 after filtering
j = j + 1
End Select
Next
If j = 1 Then Exit Function 'if s1 is empty after filtering
d1 = j - 1
s2 = UCase(s2)
d2 = Len(s2)
j = 1
For i = 1 To d2
c = Mid(s2, i, 1)
Select Case c
Case "0" To "9", "A" To "Z"
a2 = a2 & c
j = j + 1
End Select
Next
If j = 1 Then Exit Function
d2 = j - 1
k = d1
If d2 < d1 Then 'to prevent doubling the code below s1 must be made the shortest string,
'so we swap the variables
k = d2
d2 = d1
d1 = k
s1 = a2
s2 = a1
a1 = s1
a2 = s2
Else
s1 = a1
s2 = a2
End If
If k = 1 Then 'degenerate case, where the shortest string is just one character
If InStr(1, s2, s1, vbBinaryCompare) > 0 Then
Fuzzy = 1 / d2
Else
Fuzzy = 0
End If
Else '******* MAIN LOGIC HERE *******
i = 1
f = 0
o = 0
Do 'count the identical characters in s1 and s2 ("frequency analysis")
p = InStr(1, s2, Mid(s1, i, 1), vbBinaryCompare)
'search the character at position i from s1 in s2
If p > 0 Then 'found a matching character, at position p in s2
f = f + 1 'increment the frequency counter
s2 = Left(s2, p - 1) & "~" & Mid(s2, p + 1)
'replace the found character with one outside the allowable list
'(I used tilde here), to prevent re-finding
Do 'check the order of characters
If i >= k Then Exit Do 'no more characters to search
If Mid(s2, p + 1, 1) = Mid(s1, i + 1, 1) Then
'test if the next character is the same in the two strings
f = f + 1 'increment the frequency counter
o = o + 1 'increment the order counter
i = i + 1
p = p + 1
Else
Exit Do
End If
Loop
End If
If i >= k Then Exit Do
i = i + 1
Loop
If o > 0 Then o = o + 1 'if we got at least one match, adjust the order counter
'because two characters are required to define "order"
finish:
w = 2 'Weight of characters order match against characters frequency match;
'feel free to experiment, to get best matching results with your data.
'If only frequency is important, you can get rid of the second Do...Loop
'to significantly accelerate the code.
'By altering a bit the code above and the equation below you may get rid
'of the frequency parameter, since the order counter increments only for
'identical characters which are in the same order.
'However, I usually keep both parameters, since they offer maximum flexibility
'with a variety of data, and both should be maintained for this project
Fuzzy = (w * o + f) / (w + 1) / d2
End If
End Function
What you want to do is called Hamming codes (or hamming distance) -
try this