I am just trying to understand the philosophy behind harcoding the particular characters in generating the friendly token. What is the thought process behind this
https://github.com/plataformatec/devise/blob/master/lib/devise.rb#L481
def self.friendly_token(length = 20)
# To calculate real characters, we must perform this operation.
# See SecureRandom.urlsafe_base64
rlength = (length * 3) / 4
SecureRandom.urlsafe_base64(rlength).tr('lIO0', 'sxyz')
end
Here in the above snippet, l,I,O,0 are getting replaced with s, x, y, z respectively. What about the other characters that are getting generated!
Example
SecureRandom.urlsafe_base64(15)
=> "4-6RGWUH1SIsFlXa3C73"
What about R, G, W etc?
The only reason behind that is to remove any characters that are confusing. (hard to distinguish)
Like an O can be confused with 0
Like an l can be confused with I
Specific commit that introduces this explains it in the commit message
https://github.com/plataformatec/devise/commit/6d65c28f1f709722dd86da49241f118813ea7090
Related
I have two problems:
First off, the documentation for tf.keras.datasets.imdb.get_word_index says
Retrieves the dictionary mapping word indices back to words.
While in fact it's the contrary,
print(tf.keras.datasets.imdb.get_word_index())
{'fawn': 34701, 'tsukino': 52006, 'nunnery': 52007
I tried to run this in TensorFlow 2.0
(train_data_raw, train_labels), (test_data_raw, test_labels) = keras.datasets.imdb.load_data()
words2idx = tf.keras.datasets.imdb.get_word_index()
idx2words = {idx:word for word, idx in words2idx.items()}
i = 0
train_ex = [idx2words[x] for x in train_data_raw[0]]
train_ex = ' '.join(train_ex)
print(train_ex)
This result in a nonsense string
the as you with out themselves powerful lets loves their [...]
Shouldn't I get a valid movie review?
I did a bit of digging and found that there are a few "offsets" in the processing which need to be undone in order to return a sensible review language. I modified your line to subtract 3 from the index that appears in the raw sequence (since the default is to start real words with index=3), and also the first character is a dummy marker (set to 1), so the real text starts at position 2 (or index 1 in python).
train_ex = [idx2words[x-3] for x in train_data_raw[0][1:]]
Using the above modification gives me the following for the review you originally selected:
this film was just brilliant casting location scenery story direction everyone's really suited the part they played ...
It appears that some punctuation and capitalization is removed etc, but this seems to return sensible reviews.
I hope this helps.
I know the idea of renaming the variables that is transforming the recurrence to one that you have seen before.
I'm OK with slide until line 4 .. they renamed T(2^m) with S(m) >> this mean they made 2^m = m
So S(m) should be :
S(m)= 2T(m^(0.5)) + m
also m i think we shouldn't leave m as it is, because it here mean 2^m but they in real are not
Could any one explain this to me?
And also how can i know which variables I should use to make it easy to me ?
Everything you're saying is correct up to the point where you claim that since S(m) = T(2m), then m = 2m.
The step of defining S(m) = T(2m) is similar to defining some new function g in terms of an old function f. For example, if you define a new function g(x) = 2f(5x), you're not saying that x = 5x. You're just defining a new function that's evaluated in terms of f.
So let's see what happens from here. We've defined S(m) = T(2m). That means that
S(m) = T(2m)
= 2T(√(2m)) + lg (2m)
We can do some algebraic simplification to see that
S(m) = 2T(2m/2) + m
And, using the connection between T and S, we see that
S(m) = 2S(m/2) + m
Notice that we ended up with the recurrence S(m) = 2S(m/2) + m not by just replacing T with S in the original recurrence, but by doing algebraic substitutions and simplifications.
Once we're here, we can use the master theorem to solve S(m) and get that S(m) = O(m log m), so
T(n) = S(lg n) = O(lg n lg lg n).
As for how you'd come up with this in the first place - that just takes practice. The key insight is that to use the master theorem you need to be shrink the size of the problem down by a constant factor each time, so you need to find a transformation that converts square roots into division by a constant. Square roots are a kind of exponentiation, and logarithms are specifically designed to convert exponentiation into multiplication and division, so it's reasonable to try a log or exponential substitution. Now that you know the trick, I suspect that you'll see it in a lot more places.
You could, as alternative, also just divide the first equation by log(n) to get
T(n)/log(n)=T(sqrt(n))/log(sqrt(n)) + 1
and then just use
S(n) = T(n)/log(n) with S(n) = S(sqrt(n)) + 1
or in a different way
S(k) = T(n^(2^(-k)))/log(n^(2^(-k)))
where then
S(k+1)=S(k)+1
is again a well-known recursive equation.
In an AutoHotkey script, why do dead keys not work with some letters?
As an example, when running AutoHotkey with the following script:
#InstallKeybdHook
EndKeys = {LControl}{RControl}{LAlt}{RAlt}{LShift}{RShift}{LWin}{RWin}{AppsKey}{F1}{F2}{F3}{F4}{F5}{F6}{F7}{F8}{F9}{F10}{F11}{F12}{Left}{Right}{Up}{Down}{Home}{End}{PgUp}{PgDn}{Del}{Ins}{BS}{Capslock}{Numlock}{PrintScreen}{Pause}
<^>!`::
Input, SingleKey, L1, EndKeys
IfInString,SingleKey,a
Send,{U+00E0} ;à
IfInString,SingleKey,e
Send,{U+00E8} ;è
return
return
then pressing the combination of
Alt-Gr & Grave, followed by an 'a', i get à, OK, but
Alt-Gr & Grave, followed by an 'e' does NOT produce è.
The issue is not related to grave (`), the same thing happens with any other dead keys (like circumflex, acute, macron etc.)
In my particular case, the letters not working are: e y s d k n. Could it have something to do with the keyboard layout? (I am using a UK English). Any ways of approaching the issue to ensure the dead keys will work?
Thank you!
In my particular case, the letters not working are: e y s d k n
Try reorganizing these letters. I find this very hillarious indeed. Please insert any expression of laughter yourself, for it would not be welcomed on stackoverflow if I did.
You forgot to include your %'s. It should be
Input, SingleKey, L1, %EndKeys%
Otherwise, only e, n, d, k, y, s will be recognized as EndKeys
I'm trying to understand this algorithm the DFA minimization algorithm at http://www.cs.umd.edu/class/fall2009/cmsc330/lectures/discussion2.pdf where it says:
while until there is no change in the table contents:
For each pair of states (p,q) and each character a in the alphabet:
if Distinct(p,q) is empty and Distinct(δ(p,a), δ(q,a)) is not empty:
set distinct(p,q) to be x
The bit I don't understand is "Distinct(δ(p,a), δ(q,a))" I think I understand the transition function where δ(p,a) = whatever state is reached from p with input a. but with the following DFA:
http://i.stack.imgur.com/arZ8O.png
resulting in this table:
imgur.com/Vg38ZDN.png
shouldn't (c,b) also be marked as an x since distinct(δ(b,0), δ(c,0)) is not empty (d) ?
Distinct(δ(p,a), δ(q,a)) will only be non-empty if δ(p,a) and δ(q,a) are distinct. In your example, δ(b,0) and δ(c,0) are both d. Distinct(d, d) is empty since it doesn't make sense for d to be distinct with itself. Since Distinct(d, d) is empty, we don't mark Distinct(c, b).
In general, Distinct(p, p) where p is a state will always be empty. Better yet, we don't consider it because it doesn't make sense.
Given:
I have no idea what the accepted language is.
From looking at it you can get several end results:
1.) bb
2.) ab(a,b)
3.) bbab(a, b)
4.) bbaaa
How to write regular expression for a DFA
In any automata, the purpose of state is like memory element. A state stores some information in automate like ON-OFF fan switch.
A Deterministic-Finite-Automata(DFA) called finite automata because finite amount of memory present in the form of states. For any Regular Language(RL) a DFA is always possible.
Let's see what information stored in the DFA (refer my colorful figure).
(note: In my explanation any number means zero or more times and Λ is null symbol)
State-1: is START state and information stored in it is even number of a has been come. And ZERO b.
Regular Expression(RE) for this state is = (aa)*.
State-4: Odd number of a has been come. And ZERO b.
Regular Expression for this state is = (aa)*a.
Figure: a BLUE states = EVEN number of a, and RED states = ODD number of a has been come.
NOTICE: Once first b has been come, move can't back to state-1 and state-4.
State-5: comes after Yellow b. Yellow b means b after odd numbers of a.
Once you gets b after odd numbers of a(at state-5) every thing is acceptable because there is self a loop for (b,a) at state-5.
You can write for state-5 : Yellow-b followed-by any string of a, b that is = Yellow-b (a + b)*
State-6: Just to differentiate whether odd a or even.
State-2: comes after even a then b then any number of b. = (aa)* bb*
State-3: comes after state-2 then first a then there is a loop via state-6.
We can write for state-3 comes = state-2 a (aa)* = (aa)*bb* a (aa)*
Because in our DFA, we have three final states so language accepted by DFA is union (+ in RE) of three RL (or three RE).
So the language accepted by the DFA is corresponding to three accepting states-2,3,5, And we can write like:
State-2 + state-3 + state-5
(aa)*bb* + (aa)*bb* a (aa)* + Yellow-b (a + b)*
I forgot to explain how Yellow-b comes?
ANSWER: Yellow-b is a b after state-4 or state-3. And we can write like:
Yellow-b = ( state-4 + state-3 ) b = ( (aa)*a + (aa)*bb* a (aa)* ) b
[ANSWER]
(aa)*bb* + (aa)*bb* a (aa)* + ( (aa)*a + (aa)*bb* a (aa)* ) b (a + b)*
English Description of Language: DFA accepts union of three languages
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's,
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's, FOLLOWED BY ODD NUMBERs OF a's.
A PREFIX STRING OF a AND b WITH ODD NUMBER OF a's, FOLLOWED BY b, FOLLOWED BY ANY STRING OF a AND b AND Λ.
English Description is complex but this the only way to describe the language. You can improve it by first convert given DFA into minimized DFA then write RE and description.
Also, there is a Derivative Method to find RE from a given Transition Graph using Arden's Theorem. I have explained here how to write a regular expression for a DFA using Arden's theorem. The transition graph must first be converted into a standard form without the null-move and single start state. But I prefer to learn Theory of computation by analysis instead of using the Mathematical derivation approach.
I guess this question isn't relevant anymore :) and it's probably better to guide you through it then just stating the answer, but I think I got a basic expression that covers it (it's probably minimizable), so i'll just write it down for future searchers
(aa)*b(b)* // for stoping at 2
U
(aa)*b(b)*a(aa)* // for stoping at 3
U
(aa)*b(b)*a(aa)*b((a)*(b)*)* // for stoping at 5 via 3
U
a(aa)*b((a)*(b)*)* // for stoping at 5 via 4
The examples (1 - 4) that you give there are not the language accepted by the DFA. They are merely strings that belong to the language that the DFA accepts. Therefore, they all fall in the same language.
If you want to figure out the regular expression that defines that DFA, you will need to do something called k-path induction, and you can read up on it here.