Entrance Exam

Should IQfy have one to filter out the morons?

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  1. 1 week ago
    Anonymous

    Huge uptick in people seething that they can’t be a gatekeeper that nobody asked for.

    • 1 week ago
      Anonymous

      >t. filtered

  2. 1 week ago
    Anonymous

    >haruhi is not included
    moron, big big moron

  3. 1 week ago
    Anonymous

    your waifu is trash, troon

  4. 1 week ago
    Anonymous

    Ni/g/ika is my loving wife

  5. 1 week ago
    Anonymous

    This is more annoying than challenging

    • 1 week ago
      Anonymous

      Black person

      • 1 week ago
        Anonymous

        Tf is this

        • 1 week ago
          Anonymous

          I think it's this

          • 1 week ago
            Anonymous

            yes

      • 1 week ago
        Anonymous

        chad

      • 1 week ago
        Anonymous

        Why did you assume the discs would be in that position and not just like this "oO"?

    • 1 week ago
      Anonymous

      Lay flat the small one on top of the big one and it's just 36cm give or take

    • 1 week ago
      Anonymous

      36

    • 1 week ago
      Anonymous

      These are not sufficient questions to allow IQfy users access to this board.
      Instead, three questions are needed:
      -"Why is HR undeserving of respect?"
      -"What other social media platforms or IQfy boards do you frequent?"
      -"Which line items from the DSM have you been diagnosed with?"

      From there, we just eliminate people based on their answers. Even a bot/ai can answer a technical question like OP's, but we can more easily gatekeep morons, jeets, and bots by filtering out inputs with certain answering patterns. The questions should also be changed out for different ones every 6 months

      • 1 week ago
        Anonymous

        >-"Which line items from the DSM have you been diagnosed with?"
        LMFAO autistic elitist

        • 1 week ago
          Anonymous

          This is a disqualifying question, not a qualifying one. Answers that don't reflect critical thought or reading comprehension will not get you access.

          • 1 week ago
            Anonymous

            As an AI language model, I don't have personal experiences or a physical presence, so I haven't been diagnosed with any conditions outlined in the DSM. My purpose is to assist users like you with information and guidance to the best of my abilities. If you have any questions or need assistance, feel free to ask!

        • 1 week ago
          Anonymous

          aspies are the evolution of neurotypical human beings

      • 1 week ago
        Anonymous

        >-"Why is HR undeserving of respect?"
        because they're a bunch of paper pushers that don't actually screen for talent or meaningfully assist employees of a company and exist only to help insulate an organization from liability.
        >-"What other social media platforms or IQfy boards do you frequent?"
        /k/, IQfy, /m/, IQfy. I use FB messenger to talk to my family but I never post anything.
        >-"Which line items from the DSM have you been diagnosed with?"
        I haven't as far as I know. I didn't know what the DSM (DSM V?) was before I opened this thread.

        • 1 week ago
          Anonymous

          You may enter

      • 1 week ago
        Anonymous

        We should do this. Having autists that can code shouldn't be the barrier to entry cause even a frickin AI can be set up to figure OP's shit out.

      • 1 week ago
        Anonymous

        Underrated idea, but you are surrounded by autistic primadonnas that prefer to post what they think are benchmark logic/coding/math questions. They straight up think being smart is the only metric that matters in any given discussion.

    • 1 week ago
      Anonymous

      setup: oO
      simeq 3*pi + 9*pi + 2*sqrt(12*12+6*6)

      • 1 week ago
        Anonymous

        I got 14pi + 12*sqrt(3) and I set it up like this
        >a = 3
        >phi = pi/6
        >b = 6*sqrt(3)
        >theta = 2pi/3
        >rubber band = 2( 9*theta + 3(pi - theta) + d )

    • 1 week ago
      Anonymous

      there is not enough information given unironically
      if you're going to be a question-asking autist, you should be prepared to handle autistic and specific requests for clarification

      Even if the discs are laid flat on one another, we aren't given the thickness of the discs or of the rubber band, or how the band is positioned or if it is twisted, PLUS rubber bands stretch, so its stretched length while holding the discs is not its resting length

      • 1 week ago
        Anonymous

        I wasn't serious about the question but if you insist, setup is like picrel, assume rubber band is infinitely thick and does not sag

        • 1 week ago
          Anonymous

          infinitely thin*******

        • 1 week ago
          Anonymous

          should be (14*pi) + 12(sqrt(3))

          • 1 week ago
            Anonymous

            correct, it's 14π+12√3

          • 1 week ago
            Anonymous

            cool

          • 1 week ago
            Anonymous

            off the top of my head, probably (3cm)(2pi)(angle/360) + 2(v3)(6) + (9)(2pi)(1-angle/360)
            the angle should be like 120, so that resolves to 14pi+12(v3)
            thats like 64 cm, maybe 65

            ah, sorry i didnt read the thread

            also wrong, answer should be in cm, you cant order a string thats 14pi+12v3 long

    • 1 week ago
      Anonymous

      off the top of my head, probably (3cm)(2pi)(angle/360) + 2(v3)(6) + (9)(2pi)(1-angle/360)
      the angle should be like 120, so that resolves to 14pi+12(v3)
      thats like 64 cm, maybe 65

    • 1 week ago
      Anonymous

      ~50cm

    • 1 week ago
      Anonymous

      How thick are the discs?
      How are they arranged?
      By *length* of the rubber band, do you mean circumference?
      Are we taking the unstretched length? Or stretched? In other words, are we taking the minimum length of string needed to hold the discs together (ignoring what's needed for a knot.)

      • 1 week ago
        Anonymous

        who dis?

  6. 1 week ago
    Anonymous

    in The Rust Programming Language this is just

    • 1 week ago
      Anonymous

      >inb4 hardcoded
      >inb4 no short circuit
      i didn't ask for chud's opinion!

    • 1 week ago
      Anonymous

      >The same letter cell may not be used more than once in a word.
      A K
      S U

      • 1 week ago
        Anonymous

        wow i didn't even read that part lol, then i'd just throw a "trail" parameter and track each cell explored in a path

    • 1 week ago
      Anonymous

      I don't undertand what your j loop is actually doing.

      • 1 week ago
        Anonymous

        it takes the column index

      • 1 week ago
        Anonymous

        homie every matrix problem in history has a nested for loop

  7. 1 week ago
    Anonymous

    >output of the algorithm depends on where you start and which directions you prioritize
    bravo moron

  8. 1 week ago
    Anonymous

    You can use a radix tree and check for the start of a waifu path at each letter in the matrix. Then the worst case time complexity is O(m*n*w), where w is the length of the longest waifu name.

  9. 1 week ago
    Anonymous
  10. 1 week ago
    Anonymous

    Yes.
    It should be a whiteboard exam as well.
    That way 100% of pajeets get filtered and quality of all threads skyrocket.

  11. 1 week ago
    Anonymous

    sex with chibi kurisu

  12. 1 week ago
    Anonymous

    I refuse to learn to code on principle, because that sounds too much like work. I have never worked a day in my life and I don't intend to start now to appease a spastic on IQfy.

  13. 1 week ago
    Anonymous

    yawn

    • 1 week ago
      Anonymous

      i love how this is so fricking shit at "AI" that when it copy/pastes code it has learned it also keeps the comments by the original author.

      • 1 week ago
        Anonymous

        Cope and seethe moron. I'm not paid to solve your dumb games.

        • 1 week ago
          Anonymous

          You aren't paid to come here and act like a whiny pretentious homosexual either, yet here you are.

          • 1 week ago
            Anonymous

            That's free for you dumbo, enjoy it while it lasts

          • 1 week ago
            Anonymous

            Ai sucks, gpt sucks. YOU SUCK FRICKING SUCK b***h

    • 1 week ago
      Anonymous

      If you have a list of 1 million waifus that aren't in the board, it will search through the entire board 1 million times.

      • 1 week ago
        Anonymous

        >but what about
        get fricked, you've been btfo by an ai

        • 1 week ago
          Anonymous

          >AI writes the worst solution
          >but at least it wrote it fast!

  14. 1 week ago
    Anonymous

    this is a technology board, not a codemonkey board.

  15. 1 week ago
    Anonymous

    >Should IQfy have one to filter out the morons?
    considering that this garbage picture is written by someone that thinks they're clever but is actually a jungle dwelling vantablack Black person then you're only filtering yourself.

  16. 1 week ago
    Anonymous

    I'm not handling edge cases and there is nothing you can do about it

    private static List<String> findNames(char[][] matrix, String[] names) {
    record Position(int y, int x) {}
    var charLocations = new HashMap<Character, List<Position>>();
    for (int y = 0; y < matrix.length; y++) {
    for (int x = 0; x < matrix[0].length; x++) {
    charLocations.computeIfAbsent(matrix[y][x], _ -> new ArrayList<>()).add(new Position(y, x));
    }
    }
    var visiting = new ArrayDeque<Position>();
    int[][] adjacent = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    var foundNames = new ArrayList<String>();
    for (String name : names) {
    List<Position> positions = charLocations.getOrDefault(name.charAt(0), emptyList());
    next:
    for (Position startPos : positions) {
    int index = 0, y, x;
    visiting.add(startPos);

    while (!visiting.isEmpty()) {
    Position current = visiting.pop();
    y = current.y;
    x = current.x;

    if (index == name.length() - 1) {
    foundNames.add(name);
    visiting.clear();
    break next;
    }
    if (matrix[y][x] == name.charAt(index)) {
    index++;
    for (int[] dir : adjacent) {
    int newY = y + dir[0], newX = x + dir[1];
    if (newY >= 0 && newY < matrix.length
    && newX >= 0 && newX < matrix[0].length
    && matrix[newY][newX] == name.charAt(index)) {
    visiting.add(new Position(newY, newX));
    }
    }
    }
    }
    }
    }
    return foundNames;
    }

  17. 1 week ago
    Anonymous

    I think I'm rarted. I used a map of letters to coordinates to avoid having to test all four neighbours everytime (and also to avoid having to handle the (literal) edge cases). Now I don't know if I crippled my solution's performance or if it outperformed everyone else.

    • 1 week ago
      Anonymous

      Yep, I'm rarted. Since my code doesn't do BFS, it fails with cases like:
      _MON_
      _O___
      _NIKA

      I guess I can't enter IQfy.

  18. 1 week ago
    Anonymous

    def word_in_board(board, y, x, word):
    if board[y][x] != word[0]:
    return False

    coord = (y, x)
    visited = set()
    visited.add(coord)
    stack = [(y, x, 0)]

    while stack:
    y, x, char = stack.pop()
    char += 1
    if len(word) == char:
    return True
    for coord in [(y-1, x), (y+1, x), (y, x-1), (y, x+1)]:
    try:
    y, x = coord
    if coord not in visited and board[y][x] == word[char]:
    stack.append((y, x, char))
    visited.add(coord)
    except IndexError:
    pass
    return False

    def find_words(board, words):
    return [word for word in words
    for y in range(len(board))
    for x in range(len(board[0]))
    if word_in_board(board, y, x, word)]

    board = [
    ["M", "O", "N", "I"],
    ["L", "K", "S", "K"],
    ["W", "D", "S", "A"],
    ["A", "K", "U", "Y"]
    ]

    words = ["MONIKA", "ASUKA", "HARUHI"]

    print(find_words(board, words))

  19. 1 week ago
    Anonymous

    I've never been asked to write anything similar in an enterprise setting, so this is literally useless

  20. 1 week ago
    Anonymous

    Is there a bigger dataset I can speed test my solution on

  21. 1 week ago
    Anonymous

    a recursive solution

    • 1 week ago
      Anonymous

      >recursive
      yikes

    • 1 week ago
      Anonymous

      shit, I forgot the check on already visited entries

    • 1 week ago
      Anonymous

      Literally unreadable write-only code

      • 1 week ago
        Anonymous

        The function checks if "word" is present starting from "board[row][col]". Base case is when "word" is empty. Otherwise it checks that "word[0] == board[row][col]" and tries to move in any of the 4 directions.

        I think it is a solution that you may read in a competitive programming contest.

    • 1 week ago
      Anonymous

      very nice and readable
      you forgot the diagonals though and you never check wether you visited the current cell

      • 1 week ago
        Anonymous

        The OP says diagonals don't count as adjacent

        • 1 week ago
          Anonymous

          my bad, didn't real OP's post well
          reading is becoming painful

      • 1 week ago
        Anonymous

        also there should be visited[row][col] = false after each find_word()
        it's take back what I said, this code is garbage

        • 1 week ago
          Anonymous

          >also there should be visited[row][col] = false after each find_word()
          NTA but you're wrong. All of the recursive calls have visited (row, col) so why would we set it to false? I use the same method for keeping track of visited positions in my solution (

          /C/hads, we won.

          )

          • 1 week ago
            Anonymous

            >NTA but you're wrong. All of the recursive calls have visited (row, col) so why would we set it to false?

            >also there should be visited[row][col] = false after each find_word()

            find_word will keep calling itself recursively until it returns False, then it will call "it's neighbors" recursively.
            Only when the four neighbors return False that visited[row][col] should be set to false... which is what their code does.

            Personally:
            1: Mutable types suck.
            2: It's better to, instead of a Boolean matrix, simply make either a list or a set of the indexes you've already visited, like in my code: [...]
            Since the worst case scenario of memory consumption (Every cell was visited) is the constant memory consumption of a Boolean Matrix. (Granted, you don't have a choice in Haskell, since you can't map a list based on indexes).

            you're right

            The anon still needs to check wether the current position has already been visited though. Because first, if it's not checked why even set visited[row][col] to 1 and secondly, because the current function accepts "MONIKASSKI" and shouldn't.

            >It's better to, instead of a Boolean matrix, simply make either a list or a set of the indexes you've already visited
            same here, I use a flat hash table

            sub pos_in_board {
            my ($board, $pos) = @_;
            my $m = $board->[0]->@*; # X axis
            my $n = $board->@*; # Y axis
            0 <= $pos->[0] < $m && 0 <= $pos->[1] < $n
            }

            my @vec = ([-1, -1], [0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0]);

            sub around {
            my ($board, $pos) = @_;
            grep { pos_in_board($board, $_) }
            map { [$pos->[0]+$_->[0], $pos->[1]+$_->[1]] } @vec
            }

            sub find_letter {
            my ($board, $letter) = @_;
            my $m = $board->[0]->@*; # X axis
            my $n = $board->@*; # Y axis
            my @pos;
            for my $x (0 .. $m-1) {
            for my $y (0 .. $n-1) {
            if ($board->[$y][$x] eq $letter) {
            push @pos, [$x, $y];
            }
            }
            }

            @pos;
            }

            sub find_waifus {
            my ($board, @waifus) = @_;
            grep { find_waifu($board, $_) } @waifus;
            }

            sub find_waifu {
            my ($board, $waifu) = @_;
            my $first = substr $waifu, 0, 1;
            my $rest = substr $waifu, 1;

            for my $pos (find_letter($board, $first)) {
            if (find_waifu_rec($board, $rest, $pos, {})) {
            return 1;
            }
            }

            return;
            }

            sub find_waifu_rec {
            my ($board, $waifu, $pos, $seen) = @_;
            if ($waifu eq "") {
            return 1;
            }

            my $first = substr $waifu, 0, 1;
            my $rest = substr $waifu, 1;

            for my $pos (grep { !$seen->{"@$_"} } grep { $board->[$_->[1]][$_->[0]] eq $first } around($board, $pos)) {
            $seen->{"@$pos"} = 1;
            if (find_waifu_rec($board, $rest, $pos, $seen)) {
            $seen->{"@$pos"} = 0;
            return 1;
            }
            $seen->{"@$pos"} = 0;
            }

            return;
            }

            my $board = [
            ["M", "O", "N", "I"],
            ["L", "K", "S", "K"],
            ["W", "D", "S", "A"],
            ["A", "K", "U", "Y"],
            ];

            my @waifus = qw(MONIKA ASUKA HARUHI);

            my @found = find_waifus($board, @waifus);

            say for @found;

            I like that it is mutable though, it's wasteful to create a copy of the set data structure each time you advance in the string. It's maximum O(n) where n is the number of cells like you said, but all this damned copies (inb4 immutable and shared data structures).
            visitd being global works but it's dirty, I concur.

        • 1 week ago
          Anonymous

          >also there should be visited[row][col] = false after each find_word()

          find_word will keep calling itself recursively until it returns False, then it will call "it's neighbors" recursively.
          Only when the four neighbors return False that visited[row][col] should be set to false... which is what their code does.

          Personally:
          1: Mutable types suck.
          2: It's better to, instead of a Boolean matrix, simply make either a list or a set of the indexes you've already visited, like in my code:

          This Anon here ()

          [...]
          [...]

          Thanks for your incomplete code (And for making it recursive).
          I managed to understand it, complete it and make the entire thing in Haskell.

          Since the worst case scenario of memory consumption (Every cell was visited) is the constant memory consumption of a Boolean Matrix. (Granted, you don't have a choice in Haskell, since you can't map a list based on indexes).

  22. 1 week ago
    Anonymous

    I saw Kurisu and cant stop gooning to her feet rn hnmnnghhhf sfroki g the stroke

  23. 1 week ago
    Anonymous
    • 1 week ago
      Anonymous

      >You may write a program in any language of your choice

      lol

      • 1 week ago
        Anonymous

        oups , I forgot its two treasures

        • 1 week ago
          Anonymous

          moron

      • 1 week ago
        Anonymous

        oups , I forgot its two treasures

        How can I actually solve recursive probability questions (dont know if they have a better name) without aid of computers?

        • 1 week ago
          Anonymous

          You need to find a way to express them not recursively

    • 1 week ago
      Anonymous

      Assuming x is a chance of getting a bar
      x = 0.2+0.1+0.2*(1 - (1-x)(1-x))

      According to Wolfram Alpha, x ≈ 0.436492 or negative, which doesn't make sense.

    • 1 week ago
      Anonymous

      a half?

  24. 1 week ago
    Anonymous

    I don't program anything unless I get payed, OP.

  25. 1 week ago
    Anonymous

    >tfw no trie in standard library

    • 1 week ago
      Anonymous

      I think this approach should be quite efficient in theory, if anyone makes a bigboy input I might rewrite in C

      • 1 week ago
        Anonymous

        Also this will repeat waifu names if they appear multiple times, could probably make it a little better by removing waifus from the trie if they are found

      • 1 week ago
        Anonymous

        ended up with something similar. using nested dictionaries to build a trie is cheeky.

  26. 1 week ago
    Anonymous

    >waifus

    • 1 week ago
      Anonymous

      qt waifu

  27. 1 week ago
    Anonymous

    It's a technology board, not a programmer board. Come at me one by one, rust-gaygets

  28. 1 week ago
    Anonymous

    what if I am better at technology than you and don't waste a single second of my life on animu homosexualry

    • 1 week ago
      Anonymous

      go back to re_dd_t

  29. 1 week ago
    Anonymous

    /C/hads, we won.

  30. 1 week ago
    Anonymous

    If anyone wants to race, here's a bigboy I made: https://0x0.st/XKb1.txt
    500 "waifus", 110 of which can be found in the 750x750 character grid.
    A slightly modified version of my C solution solves it in 20ms

    • 1 week ago
      Anonymous

      I ran

      yawn

      and it took 30 seconds.

      • 1 week ago
        Anonymous

        Don't worry, I'm sure the saved developer time and money is worth the slowdown.

  31. 1 week ago
    Anonymous

    Anons, am i dumb? I spent hours in this and my code still doesn't work as it should?
    This results in an empty list.

    • 1 week ago
      Anonymous

      >Anons, am i dumb?
      signs point to yes

      • 1 week ago
        Anonymous

        Huh, that's kinda sad.
        What should I do to stop being dumb and being able to solve problems like this one?

    • 1 week ago
      Anonymous

      This Anon here ()

      a recursive solution

      shit, I forgot the check on already visited entries

      Thanks for your incomplete code (And for making it recursive).
      I managed to understand it, complete it and make the entire thing in Haskell.

  32. 1 week ago
    Anonymous
    • 1 week ago
      Anonymous

      bait,bait,bait,bait (cm^2)

      • 1 week ago
        Anonymous

        cm^2 is unit for area, i wrote "square is 1x1" since i gave sides for square

        (9*2)/2-(1*2)/2

        did you read a question?

        >Area of the quadrangle
        44
        >Area of the triangle
        17

        how

        • 1 week ago
          Anonymous

          Oh, didn't read, Ok then it would be
          (5*1)/2 + (4*1) + (4*1)/2 - (2*1)/2

    • 1 week ago
      Anonymous

      7.5cm^2

      • 1 week ago
        Anonymous

        (9*2)/2-(1*2)/2

        >Area of the quadrangle
        44
        >Area of the triangle
        17

        9 x 2, - (1 x 2/2) -(3/2)(5) -2
        = 7.5cm (approx)

        1/2 x 8 x 2 = 8 squares morons.

        • 1 week ago
          Anonymous

          It's not a triangle, regard.

          • 1 week ago
            Anonymous

            Frick, it's the "infinite chocolate" trick again.

    • 1 week ago
      Anonymous

      (9*2)/2-(1*2)/2

    • 1 week ago
      Anonymous

      >Area of the quadrangle
      44
      >Area of the triangle
      17

    • 1 week ago
      Anonymous

      9 x 2, - (1 x 2/2) -(3/2)(5) -2
      = 7.5cm (approx)

  33. 1 week ago
    Anonymous

    Make the test operaing MacOS.
    Most of you homosexuals don't know how to do anything on an OS purported to be for dumb people.

  34. 1 week ago
    Anonymous

    /g/ophers rise up

  35. 1 week ago
    Anonymous

    ok, but how would 10/10 look like?

  36. 1 week ago
    Anonymous

    100% chance that you're a kiddy fiddler. Frick off.

  37. 1 week ago
    Anonymous

    >the only technology that exists is programming

    this is why FOSS is doomed to suck btw, people invested 100% in a future where everyone is an advanced programmer with unlimited free time, which defies multiple hard laws of economics.

    • 1 week ago
      Anonymous

      >which defies multiple hard laws of economics
      go suck gayMAN wiener somewhere else
      >FOSS GODS captcha

  38. 1 week ago
    Anonymous

    just did it, waste of time

    • 1 week ago
      Anonymous

      can we see it?

      • 1 week ago
        Anonymous

        sub pos_in_board {
        my ($board, $pos) = @_;
        my $m = $board->[0]->@*; # X axis
        my $n = $board->@*; # Y axis
        0 <= $pos->[0] < $m && 0 <= $pos->[1] < $n
        }

        my @vec = ([-1, -1], [0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0]);

        sub around {
        my ($board, $pos) = @_;
        grep { pos_in_board($board, $_) }
        map { [$pos->[0]+$_->[0], $pos->[1]+$_->[1]] } @vec
        }

        sub find_letter {
        my ($board, $letter) = @_;
        my $m = $board->[0]->@*; # X axis
        my $n = $board->@*; # Y axis
        my @pos;
        for my $x (0 .. $m-1) {
        for my $y (0 .. $n-1) {
        if ($board->[$y][$x] eq $letter) {
        push @pos, [$x, $y];
        }
        }
        }

        @pos;
        }

        sub find_waifus {
        my ($board, @waifus) = @_;
        grep { find_waifu($board, $_) } @waifus;
        }

        sub find_waifu {
        my ($board, $waifu) = @_;
        my $first = substr $waifu, 0, 1;
        my $rest = substr $waifu, 1;

        for my $pos (find_letter($board, $first)) {
        if (find_waifu_rec($board, $rest, $pos, {})) {
        return 1;
        }
        }

        return;
        }

        sub find_waifu_rec {
        my ($board, $waifu, $pos, $seen) = @_;
        if ($waifu eq "") {
        return 1;
        }

        my $first = substr $waifu, 0, 1;
        my $rest = substr $waifu, 1;

        for my $pos (grep { !$seen->{"@$_"} } grep { $board->[$_->[1]][$_->[0]] eq $first } around($board, $pos)) {
        $seen->{"@$pos"} = 1;
        if (find_waifu_rec($board, $rest, $pos, $seen)) {
        $seen->{"@$pos"} = 0;
        return 1;
        }
        $seen->{"@$pos"} = 0;
        }

        return;
        }

        my $board = [
        ["M", "O", "N", "I"],
        ["L", "K", "S", "K"],
        ["W", "D", "S", "A"],
        ["A", "K", "U", "Y"],
        ];

        my @waifus = qw(MONIKA ASUKA HARUHI);

        my @found = find_waifus($board, @waifus);

        say for @found;

  39. 1 week ago
    Anonymous

    Clojurebros, how did I do?

    I'm pretty new to Clojure

  40. 1 week ago
    Anonymous

    Well, the first solution that comes to mind would be to 1. sort the waifu list alphabetically into buckets by letter
    2. make a distinct object/class/subdivision/branch/whatever for each letter in the list
    3.Do this for each "letter" until you discard all branches:
    a)take the letter you're in "id/pos"with you.
    b) Check the buckets of sorted list until you find the fitting one (If none of the waifus match your letter, discard the letter branch), and take the list only with that bucket with you( if bucket is empty, discard branch). If you reached the end of any waifu name, add it to result with id of all the letters you have with you in order
    b) make subletters for all the letters your current letter "borders", that you don't already have with you (if there are none, discard the branch)
    c)sort the list of waifus you have with you by the next letter alphabetically into buckets
    d)do 3: for each subletter branch
    4) return the result list

  41. 1 week ago
    Anonymous

    R-E-I
    E-I-E
    I-E-R

  42. 1 week ago
    Anonymous

    Waifu=mx+b

  43. 1 week ago
    Anonymous

    kissing chibi kurisu on the lips!

  44. 1 week ago
    Anonymous

    i value /hsg/ and analogues more than code scum (i code for a living)

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