300 Matching Annotations
  1. Jan 2024
    1. nd 2), then activate the sidebar by clicking the button in the location bar (3).

      yo yo

  2. Jan 2019
  3. Oct 2018
    1. A product under $50 is usually affordable enough that a customer may be ready to buy the first or second time.

      maybe we'll be okay to be deficient here for a while

    2. TODAY

      create urgency/scarcity

  4. Sep 2018
    1. If you are an eCommerce, there will be fewer people who visited the “Check Out” page than the “Shopping Cart”. If you built an audience with your Facebook pixel, you audience size should be different.

      this is basically the point

      --> the people who are checking out are the people you want to target, so that subset is smaller than say all the people mildly interested

    1. I generally stick with drugstore when it comes to eyeliner - although I love black track fluid line by MAC, I've found drugstore options that are just as good. Same goes for mascara. permalinkredditceddit

      MAC is low-end

    1. (like budget, audience, placement, and scheduling) t

      need to know these

    2. Manual to set your own target cost per result.

      which is better? Manual or automatic? Won't fb fuck us on automatic?


    3. conversion pixel or app event.

      what is meant by this? conversion pixel-- a pixel made just for conversion?


    1. About Flexify: Facebook Product Feed

      so essentially it seems that this is about syncing, and considering that a facebook account is something for a little later we can postpone this until further discussion

    2. Dynamically promote your products

      this piece is just an FB ad push, but it basically says that FB is good at timing, devices, relevancy, and scale.

    1. targeting

      this might actually be something propietary, but it is something that can be done by hand regardless

    2. Set up Facebook dynamic ads and retarget shoppers most likely to buy

      from what i understand they're just asking facebook to do the work. they're not actually doing any work themselves

    1. ver 10 months I generated over $55k in sales at around 30% profit margin

      he only made about 20k... + whatever he sold it for

    2. I failed miserably. I built a second store in a different niche and worked tirelessly.

      he worked through a niche, we should maybe try that if this doesn't work out

  5. Jun 2018
    1. Nullable

      what nonterminals can derive epsilon

    2. 7

      not derived from the B, B disappears, goes to epsilon in the follow set of B

      rule 7 makes the capital B disappear

    3. using First()is not sufficien

      soon as you have the idea that capital leters can disappear you find that first is not enough

    4. B →e

      our new rule

    5. Helper Function: Follow( )

      first alone cannot cover all cases

    6. S →Ay

      first terminal expect to see if i have an S on the stack is AyB

    7. first of s that i will see is the begin of file(?)

    8. First

      what is first all about--> use to fill up part of our predict table

    9. match

      have to match-- then you just pop it off

    10. Parsing the Input

      How do we know what to expand next???

      That is the rest of the lecture.

    11. Top-Down Parsing

      summary slide

    12. > S

      S' is non-terminal so expand

    13. →⊦S⫞

      only one rule to expand S'

    14. terminalat

      terminal is a blue letter

    15. →AyB

      replace with AyB

    16. Example

      not like natural deduction only one way

    17. decide which rule to apply next at each step of the derivation

      one of the most important questions

    18. AyB


    19. S



    1. Properties of a Post-Order Traversal

      we do not want ambiguity when evaluating a parse tree


    1. ThePoisso

      big n, small p

    2. r of success

      x is successes in hypergeometric

    3. There is only the one arrangement withxfailures followed by1success. Thisarrangement has probability

      so just x failures followed by 1 success

      same as negative binomial but k = 1 so x+1-1 = x of course where x is the number of failures

    4. different arrangements ofx S’s and(nx)F’sover thentrials.

      how many ways can we order successes over n trials

    5. x+k1x

      how many ways can we order the fucking failures given x+k-1 trials

    6. k1)successes, in any order

      last success is fixed

    7. Xbe the number of failures obtained before thek’th success.

      follows NB if you have S/F with independent trials, with replacement

      k successes x is # of failures until k success

    8. ba+1in order thatbPx=af(x) = 1.

      reqt of all probability functions

    9. function(f(x)orX(a)) and thevalue of a function( for examplef(2)orX(a) = 2)

      for a = THH

    10. Proof of (a)
    11. so our sample space has8!5!3!= 56equally possible outcome

      = (8C5)(3C3) = 561 which is also valid

    12. herefore by themultiplication rule there are36(3)56(3)ways to get a number which is even and>3000

      flaw: you cannot make one 4 digit number in p ways and then turn around and make another 4 digit number in q ways and then claim that the there pxq ways to make a single 4 digit number. it doesnt make sense

    13. (4;4)

      only that particular even number is not available, so:

      [3,7] and even is only 4 and 6--> subtract 2

    14. o the probability it is even is theprobability it is either2;4;or6or37

      there are multiple ways to do it and some are simpler than others

      if you have einstellung with one, skip it and move on and you may find a simpler solution

      --> same deal in (b)

    15. Since each of1;2;:::;7isequally likely to be the first digit, we get the probability the number is greater than3000is57

      same deal as in (a)

    16. Case 1:all three letters are the same.

      split into cases when desired/required

    17. Now we can use the same technique to arrange the remaining eight letters. Having placed two of theS’s, there remain eight free boxes, in which we are to place three T’s in83ways, two I’s in52ways,one C in31ways, one A in21ways and finally the remaining S in the last empty box in11way.There are

      Position is unique, not the letters. The letters are indistinguishable.

      i.e: S1=S2=S3 but position 1 != pos2!=pos3

    18. There are9(4)pin numbers that do not contain 1 and so there are10(4)9(4)that do contain a1. Therefor

      the complement of all pins without 1s

      so all pins with 1s which is the same as (9^(3)*2)/(10^(4)) since there are two (2) places for the 1 to go

    19. Since we have a uniform probability model

      every simple event is equally likely

    20. We avoid this complication by starting with the second box




    1. two or more transition leaving the same state can have the same label yet lead to different states

      must also consider/connais that all DFAs are NFAs

    2. e-NFA

      basically your start pos is multiple start pos's

    3. NFA version


    4. (q0,1) U T(q1,1) U T(q2

      yourew moving from any number of these states so of course the final state set is going to be the result of a union

    5. A =

      accepting state

    6. {q2

      intrsxn w accepting states, right?

    7. all the possible subsets of Q

      so here a subset is basically just concatenating instead of putting into a set

    8. q1

      this is a simplistic diagram

      you could have 1 taking you back to q0 from q1 and q2 since you want 2 0s at the end



    1. i-1 times for each i.

      R^(i) = R⋅R^(i-1)--- remember: you already have an R as a "coefficient"

      so: R^(0) = R⋅R^(-1) ??? Wait, then what's R^(-1) which then must exist by this definition?


    2. union of R concatenated with itself any finite number of times

      concatenated to itself i-1 times (for each term in the sum)

    3. superscript

      superscript indicates length

    4. *

      basically like the '*' in regex

    5. R·S = {greyjay, greywhale, bluejay, bluewhale}


    6. Regular Language

      as you will recall, has infinite numbers of valid identifiers

    7. meets the specification

      language spec..

    8. specifyall

      not recognize

    9. Lexical Analysis:

      token analysis



    1. page 43:

      i dont really understand this proof

    2. page 43:

      nonneg: 0 or positive nonpos: 0 or negative

    3. page 37:

      so basically its already known that Span B is a subset of S, basically by definition:

      We already know that all the vectors in B are in S and since S is closed under linear combinations, Span B is a subset of S (i.e: can be written as an LC of the vectors in S). (you cannot go beyond)

    4. the best you can do with n-1 vectors (even if you are touching all 4 dimensions) is to create a hyperplane assuming linear independence

      page 31

    5. thm 5 explanation:

      • we cannot have any other coefficient set which will make the equation 0 besides the trivial solution

      so we find that ci=di for every vector 1<=i<=k in S

    6. the proof makes sense in example 14 since we show that

      • R2 is a subset of Span B and that
      • Span B is a subset of R2
      • B is linearly independent

      page 16

    7. example 17 makes sense since you show that

      • any vector in R2 is a linear combination of the vectors in B by offering a c1, c2 s.t. we have a soln for every x1,x2
      • linear independence is had as we find that the only soln to the problem of x1=x2 =0 is c1=c2=0 (trivial


  6. May 2018
    1. P(BjA)P(A) +P(BjA)P(A)

      prob that b occurs given a + prob that b occurs given a complement

      this just reduces into p(b)

    2. P(A)P(BjA

      P(A)P(B|A) = P(A)P(AnB)/P(A) = P(A)

    3. (AiAjAkAl) +

      notice it keeps going so you would then add back in P(AiAjAkAlAm)

    4. (where the subscripts are all distinct, for examplei < j < k < l).

      clever way to ensure distinctiveness

    5. P(D) = 1PD= 19(4)10(4)

      yes my method works (of choosing one of the four locations) so:

      P(D)=4*9^(4)/10^(4) is right also

      since we can only have one 1, otherwise you would need to use their method of finding the complement

    6. (23) + (32) = 12ways

      makes sense

    7. We need to have more efficient ways of determining the number of outcomes inSor in a compoundevent without having to list them all. Chapter 3 considers ways to do this, and then Chapter 4 developsother ways to manipulate and calculate probabilities

      happens later, chapter 2 is genuinely soft

    8. Solution 1:

      In this solution we simply flesh out the sample space and assign a prob'y dist'n to the events.

      Then we simply count up the number of events which fulfill our criteria and produce the probability.

    9. f2,3,4;:::;A;2~;:::;A|g

      here, the same event is a compound event.

      note that only one of these events will occur in a single trial in both cases, but there are multiple ways in which the event can occur (club)

    10. ={spade, heart, diamond, club}.

      as you can see with this example and the next,: the sample spaces need not be unique in order to produce the same result.

    11. 598

      you do indeed have 5 options for the even digit, and then 9,8,7 for the rest of the digits


    1. Structural recursion

      structural recursion with an accumulator is called accumulative recursion, stores subproblems

    1. Tokens


    2. break up an assembly language instruction into components and classifying these components.

      so that: you can ID errors you can output the appropriate output when desired

    3. lexeme

      basically a fancy way of saying "word"

    4. MSB

      MSB is a 1 so you dupe that and you will get a 2s comp (i.e: signed) -1, i.e.: 0xf

    5. 0x87654321

      0x87654321 has the MSB as 1 which will trigger the different behaviours

      recall that 8= 2^3 and is uniquely represented as the nibble 1000

    6. char(instr>> 24) << char(instr>> 16) << char(instr>> 8) << char(instr);

      printing from MSB to LSB (>>24 comes first!)

    7. least significant byte
      • cuts out MSB --> NOPE:it means what it means, see how char keeps cutting out the MSB until it is equal to the LSB?

      it is saying that these are ALL EQUAL

    8. instr0001


    9. s is 2 = 000102shifted 21 bits left

      consider that a 0 bit shift left would not add any additional 0s to the right.

      1 bit shift left would put 11 zero there.

      26? 26 zeroes

    10. (0x0F)

      number is 15, so we're just going to keep the right nibble on

    11. 3124beyond

      why is beyond on 24 when its just a label??

      ANSWER: You don't want to be going back up to 20 and rerunning it every time. That would be wrong

    12. // y gets 0, and (bar, 0)// gets added to st.

      builtin mechanism where if its not there it will get added

    13. each instruction is exactly 4 bytes long

      aka 32 bits

    14. Big hint:just recognize the proper form and call everything else an error.

      this is a good slide for pseudocoding your soln at least

    15. semantic

      relating to meaning

    16. syntax

      relating to form

    17. 9999999999

      actually, how big can the number be?

    18. a kind

      a what?

      maybe they just mean some "type", basically

    19. 10

      note that the scanner is providing the output as a decimal value

    20. errors

      to stop program early, i suppose

    21. << 2 =1 0

      fall off is lost



    1. R

      sas university is also an option


    1. 16 Unnecessary Complications

      rethink the belows' use

    2. ature that all of the names are in the scope of all the value-expressions. It canbe used to define rec

      idgi but sounds important for recursive+mutually-recursive

    3. Before the implicitbeginin alambdaor function definition, Racket allows internaldefines.This is an alternative to immediately using one of the constructs in the next section (internal defi-nitions a

      idgi but sounds important

    4. Racket also provides one-armedwhenandunless


    5. ote that the values of top-level expressions are printed after evaluation if they are not<void>.If you

      ote that the values of top-level expressions are printed after evaluation if they are not<void>.If you</void>

    6. Racket Reference

      referring to a resrouce listed above (in ss 1)

    1. S

      if you got an 'a' then you go to the empty set

    2. he setof states {1, 3}

      at the same time

    3. A string is accepted if at least one pathleads to an accepting state.

      if you end up in q1 and q2 then you will accept (at least 1)

    4. Differs

      not a single answer but a set of answers

    5. q0

      you are here AND in q1


    1. state d(state, ci)

      update state based on input

      can think of as just a big lookup table

    2. d

      state symbol pair

    3. (S, b)

      in S and see a b go to b

      in state b see an e then go to be,,

      and so on

    4. A=

      can be empty

    5. S

      subset of states

    6. deterministic

      as stated before, know where im going next, for every step, i know here im going to next EX;; b + e = be-- > we know this, non-dfa would not know this

    7. 0

      beginning in an accepting state

    8. more


    9. *

      Kleene Star

    10. a

      if input stopped here ---> in an nonaccepting or error state -------> not a sucessful termination of the program, will JUST hang

    11. ,

      a or b

    12. ab

      accepts this specific string only

    13. nothing

      no concentric circle state, accepts {} (nothing, epsilon)

    14. Err

      idea of error state

    15. Transi

      equivalent of arrows in prev diagram

    16. at

      why is sleepy not an accepting state?

      psbl ans: can't stay in "sleepy"

    17. accepting state

      goal is to end up in an accepting state

      as many as you want

      BUT only one start state

    18. deterministic

      meaning every move is determined



    1. Recognizing A Regular Language

      May 24th, 2018

    2. infinite

      obviously going to be infinite

    3. .e

      ^what the star means>>

    4. ε

      means the empty string

    5. infinitee

      most prog langs

    6. bababa

      combinations of our set

    7. { a, b}

      irl it would be ascii

    8. Input:intmaxEntry

      easy to figure out whther say, youve been given a valid register (finite)

      but this is an infinite sized langauge --> so we must specify and recognize them

    9. machine code to be able to convert an assembly language

      how can i break strings into a series of tokens that i understand (?what?)


    1. capricious

      easily swung this way or that emotional given to sudden and unaccountable changes of mood or behaviour

    1. update

      why is the order reversed?

      why do we update then load & restore then update?

    2. in our MIPS simulator we use $30

      stack pointer location


    1. archetypes

      stereotype is generally one dimensional and negative archetype is the source from which copies are made

      Stereotype is a "conceptual model" created by abstracting the key features of current examples. ---> abstracting means extracting so an investment banker stereotype may have the abstracted traits of selfishness and vanity

    2. asetiological

      origin stories

    3. Mythology is for me a medium of exploration

      author perspective

    1. myths which we will be looking at in this course willcontain elements from

      contain elements across these boundaries we have just created

    2. Perceptions ofhistorical truth are inherently subjective


    3. Troy in the Troad

      might be based in reality

    1. complements

      completes, vs compliment which is just a nice thing to say

  7. Apr 2018
    1. 1

      1 X is funct-field determined

    2. , the main control unit generates the ALUOp bits, which then are used as input to the ALU control that generates the actual signals to control the ALU unit


    3. add $t1,$t2,$t3, which reads $t2 and $t3 and writes $t1

      class 1 of instructions

    4. and implement a nondelayed beq instruction.

      idk about irl/ in the courseontes but this is imp/good to know

    5. data memory must be written on store instructions

      data memory is the fucking data which is not the fucking register

    6. while the value written will be available to a read in a subsequent clock cycle


    7. Zero

      used for branch instructions, ex: beq

    8. Readdata 2

      32 bits wide

    9. Readdata 1

      32bits wide

    10. adder to increment the PC to the address of the next instruction. Th is adder, which is combinational

      wired to be a perma-adder

    11. ombinational logic

      alu or and for instance