Calculate the difficulty of a Code Golf question

Question

Briefing

The difficulty of a Code Golf question can be calculated as such:

$$\text{clamp} \left( \left\lceil \left( \frac v a \div 700 \right) \times 10 \right\rceil, 0, 10 \right)$$

Where \$v\$ is the number of views a question has

and \$a\$ is the number of answers a question has

and \$⌈x⌉\$ is the ceiling operator.

Also:

$$\text{clamp}(x, min, max) = \begin{cases} min & x < min \\ x & min \le x \le max \\ max & max < x \end{cases}$$

This question's current difficulty: ***

Task

Write a program that will take two integers (v and a) and output the difficulty in asterisks (*).

The input can be in the form of an array, a separated string or as separate function arguments

Test Data

Views   Answers Difficulty  Program Output
163     2       2           **
548     22      1           *
1452    24      1           *
1713    37      1           *
4162    32      2           **
3067    15      3           ***
22421   19      10          **********

Example with pseudocode

v: 1713    
a: 37
out = clamp(ceil(((v/a)/700)*10), 0, 10); // evaluates to 1
//program will output '*'

The shortest code in bytes wins! Trailing/ leading spaces are allowed.

Answer 1

JavaScript (ES6), 40 39 bytes

v=>a=>"**********".substring(10-v/a/70)

Because substring provides the required clamping and "ceil"ing behaviour. Edit: Normally I'm too lazy to bother, but because it got 4 upvotes, I've followed @MarsUltor's advice to save 1 byte by currying.

Answer 2

I've been wanting to do this for a while...

HTML + CSS 491 487 485 bytes

-4 bytes thanks to Conor O'Brien
-2 bytes thanks to Releasing Helium Nuclei

Input is taken as the width and height of the page window; width being the number of Views, and height being the number of Answers.

<style>p{overflow:hidden;width:1ch}@media(max-aspect-ratio:70/2){p{width:1ch}}@media(max-aspect-ratio:70/3){p{width:2ch}}@media(max-aspect-ratio:70/4){p{width:3ch}}@media(max-aspect-ratio:70/5){p{width:4ch}}@media(max-aspect-ratio:70/6){p{width:5ch}}@media(max-aspect-ratio:70/7){p{width:6ch}}@media(max-aspect-ratio:70/8){p{width:7ch}}@media(max-aspect-ratio:70/9){p{width:8ch}}@media(max-aspect-ratio:7/1){p{width:9ch}}@media(max-aspect-ratio:70/11){p{width:10ch</style><p>**********

You can try it in your browser by entering

data:text/html,<style>p{overflow:hidden;width:1ch}@media(max-aspect-ratio:70/2){p{width:1ch}}@media(max-aspect-ratio:70/3){p{width:2ch}}@media(max-aspect-ratio:70/4){p{width:3ch}}@media(max-aspect-ratio:70/5){p{width:4ch}}@media(max-aspect-ratio:70/6){p{width:5ch}}@media(max-aspect-ratio:70/7){p{width:6ch}}@media(max-aspect-ratio:70/8){p{width:7ch}}@media(max-aspect-ratio:70/9){p{width:8ch}}@media(max-aspect-ratio:7/1){p{width:9ch}}@media(max-aspect-ratio:70/11){p{width:10ch</style><p>**********

as a url in a new tab.

Answer 3

05AB1E, 11 bytes

/70/î'*T×s£

Explanation

/            # divide v by a
 70/         # divide by 70
    î        # round up, call this n
     '*T×    # push 10 asterisks
         s£  # take n up to 10 asterisk
             # implicitly print

Try it online

Answer 4

Javascript (ES6), 37 36 bytes

v=>a=>"*".repeat((v/=a*70)<9?v+1:10)

Saved 1 byte by currying, thanks to TheLethalCoder

let F=
v=>a=>"*".repeat((v/=a*70)<9?v+1:10)

console.log("Test #1:", F(163)(2))    // **
console.log("Test #2:", F(548)(22))   // *
console.log("Test #3:", F(1452)(24))  // *
console.log("Test #4:", F(1713)(37))  // *
console.log("Test #5:", F(4162)(32))  // **
console.log("Test #6:", F(3067)(15))  // ***
console.log("Test #7:", F(22421)(19)) // **********

Answer 5

Mathematica, 38 35 bytes

StringRepeat["*",10,⌈#/#2/70⌉]&

Thanks to @MartinEnder for 3 bytes

Answer 6

CJam, 18 15 14 bytes

Saved 1 byte thanks to Peter Taylor and 3 bytes thanks to Adnan

'*A*q~d/70/m]<

Try it online

'*A*            e# Push "**********"
    q~d/        e# Get the input and divide the two numbers
        70/     e# Divide by 70
           m]   e# Ceil, yielding x
             <  e# Slice the string, taking the first x elements

Answer 7

C#, 68 49 48 bytes

v=>a=>"**********".Substring((int)(10-v/a/70d));

This is the C# version of this excellent answer by Neil.

Saved another 19 bytes thanks to Neil

Answer 8

Java 8, 57 bytes

Uses a lambda to save bytes, performs the calculation and substrings to return the answer.

(v,a)->"**********".substring(Math.max(0,(700*a-v)/70/a))

---

Here is my class for testing it.

public class DifficultyCalculator{
    static interface h{ String f(int v, int a);}
    static void g(h H){
        System.out.print(H.f(163,2));System.out.println("\t**");
        System.out.print(H.f(548,22));System.out.println("\t*");
        System.out.print(H.f(1452,24));System.out.println("\t*");
        System.out.print(H.f(1713,37));System.out.println("\t*");
        System.out.print(H.f(4162,32));System.out.println("\t**");
        System.out.print(H.f(3067,15));System.out.println("\t***");
        System.out.print(H.f(22421,19));System.out.println("\t**********");
    }
    public static void main(String[] args) {
        g( // 70
            (v,a)->"**********".substring(java.lang.Math.max(0,(int)(10-v/70d/a)))
        );
    }
}

---

Update

• -3 [16-08-19] Utilized integer division
• -10 [16-08-18] Removed unnecessary import, thanks to @OlivierGrégoire!
• -18 [16-08-17] Return string instead of print

Answer 9

MATL, 12 bytes

/70/0:9>42*c

Try it online!

Explanation

The rounding up and clamping are done simultaneously as follows: the number x = v/a/70 is compared against each element of the array [0 1 ... 9]. The numbers of that array that are exceeded by x will become asterisks, and the rest will be spaces.

/      % Take the two numbers implicitly. Divide. (Example: for inputs 3067, 15
       % we get 204.47)
70/    % Divide by 70 (we get 2.92)
0:9    % Push array [0 1  ... 9]
>      % See which of those are exceeded by the previous number (2.92 exceeds
       % 0, 1 and 2, so we get [1 1 1 0 ... 0]). This does the rounding up
       % and the clamping
42*    % Multiply by 42, which is the ASCII code of '*' (we get [42 42 42 0 ... 0])
       % Char 0 will be displayed as space
c      % Convert to char. Implicitly display

Answer 10

Python2, 32 bytes

saved 3 + 2 bytes and corrected off by one error thanks to Leaky Nun

lambda v,a:('*'*10)[:~-v/a/70+1]

similar to Neils answer. Uses the fact that Python2 does integer division.

Answer 11

EXCEL, 29 bytes

If you count Excel as a representation of VBA Excel, then you can use

=REPT("*",MIN(1+v/(70*a),10))

where v and a are the name of reference cells.

Answer 12

Haskell, 35 bytes

v#a=[1..min(ceiling$v/a/70)10]>>"*"

[1..min(ceiling$v/a/70)10] creates a range from 1 to the computed difficulty (an empty list for difficulty 0). a>>b repeats the list b length a often.

Answer 13

Pyth, 17 13 bytes

4 bytes in credit to Luis Mendo for his algorithm.

*\*htS[0T.EccFQ70
*\*s>LccFQ70T

Test suite.

Answer 14

Pyke, 13 9 bytes

/70/\**T<

Try it here!

Explanation:

/         -    num_1 / num_2
 70/      -   ^ / 70
    \**   -  "*" * ^
       T< - ^[:10]

Answer 15

C#, 97 89 87 77 42 41 bytes

v=>a=>new string('*',(v/=a*70)<9?v+1:10);

Saved 10 bytes thanks to Adám

Saved a few bytes thanks to Arnauld

Answer 16

Perl, 35 32 bytes

say"*"x(10-($-=10-pop()/70/pop))

Use -E to activate say and give the arguments in the reverse order:

perl -E 'say"*"x(10-($-=10-pop()/70/pop))' 2 163

If arguments on STDIN are allowed the following is 29 bytes:

(echo 163; echo 2) | perl -pe '$_="*"x(10-($-=10-$_/70/<>))'

Answer 17

R, 68, 50 52 bytes

f=function(v,a)cat(rep("*",1+min(v/a/70,10)),sep="")

rep implicitly places a min on number of 0.

Thanks to @plannapus and @Anastasiya-Romanova秀 for spotting my error.

Answer 18

Javascript ES6, 48 bytes

a=>b=>"*".repeat(Math.ceil(Math.min(a/b/70,10)))

Answer 19

m4, 136 135 bytes

define(r,`ifelse($1,0,,eval($1>9),1,*`r(9)',*`r(decr($1))')')define(f,`r(ifelse(eval($1%($2*70)),0,eval($1/$2/70),eval($1/$2/70+1)))')

Defines a macro f which takes v and a, and expands to the correct output. Most of the program is an implementation of ceiling.

Answer 20

C, 54, 51, 50, 49 bytes

Assuming that v is positive or zero and a positive, the x < min clamping case is never met, since there is no way the result of the ceiling operation can be negative. Additionally, integer maths on non-negative values always yields the floor of the result, so we add 1 to get the ceiling.

This solution requires a write function, works on Linux at least.

F(v,a){write(1,"**********",(v/=a*70)>9?10:v+1);}

Test main:

int main() {
  F(163, 2);
  putchar('\n');
  F(548, 22);
  putchar('\n');
  F(1452, 24);
  putchar('\n');
  F(1713, 37);
  putchar('\n');
  F(4162, 32);
  putchar('\n');
  F(3067, 15);
  putchar('\n');
  F(22421, 19);
  putchar('\n');
}

Answer 21

javascript: 82 73 bytes

 (v,a)=>console.log("*".repeat(Math.min(Math.max(0,Math.ceil(v/a/70),10)))

• saved some bytes after Adám pointed out I overlooked the /700*10=/70, and the removal of parens

Answer 22

Dyalog APL, 15 bytes

'*'⍴⍨10⌊⌈⎕÷70×⎕

'*'⍴⍨ the character repeated this many times:
10⌊ min(10,...
⎕÷ input divided by
70× seventy times
⎕ input

TryAPL online!

Answer 23

Jellyfish, 18 bytes

P
#'*
mM/%i
10 %70

Takes input in the format [a v]. Try it online!

Explanation

• % is reciprocal, so %70 is 1/70.
• i is input, as a two-element array.
• /% with inputs i and %70 reduces the array i by flipped division with initial value %70. In other words, it computes v/(a/(1/70)), which is equal to v / (70*a).
• M takes the ceiling of this value, and m takes the maximum of that and 10.
• #'* repeats the literal * character that many times.
• P prints the result without quotes.

Answer 24

MATLAB, 34 33 bytes

Because I like this challange so much, here is one for MATLAB (outputs trailing whitespaces):

@(v,a)[(ceil(v/a/70)>0:9)*42,'']

Inspired by @Luis Mendo's answer. Thanks to @pajonk for saving one byte.

Answer 25

dc, 110 108 104 98 bytes

This was a doozy since slicing isn't a thing. Also, dc doesn't manipulate strings. I just really was waiting for a string one that would be < 5 hours of coding. On the plus side, I finally started writing down common constructs, like for loops. Also had to formulate rounding/ceiling, so thanks for that.

[42P]sd[dsi[li0!=dli1-dsi0!=L]dsLx]sl[Isi[li0!=dli1-dsi0!=L]dsLx]sg[1+]saIk/70*0k1~0!=adI>ldI!>gIP

Invoked in bash:

echo 'v a (above)'|dc
# Wholly:
>> echo '163 2 [42P]sd[dsi[li0!=dli1-dsi0!=L]dsLx]sl[Isi[li0!=dli1-dsi0!=L]dsLx]sg[1+]saIk/70*0k1~0!=adI>ldI!>gIP'|dc
# outputs:
**
>> 

Replacing (above) with the code, and v and a with their respective counterparts above. The single quotes are important (otherwise you get bash's history stuff).

---

Explained:

[42P]sd   # Here we store a macro in register d to print 1 * without a newline

[dsi[li0!=dli1-dsi0!=L]dsLx]sl # Store the "less than" case, a for loop which
                        # uses the top-of the stack as it's number of iterations.
[Isi[li0!=dli1-dsi0!=L]dsLx]sg # Store the "greater than" case. It's the above,
                        # but it puts 10 on the stack to use instead.

[1+]sa # Store a macro to add 1 to whatever is the top-of-stack.


Ik # Set precision at non-zero to allow decimal division

/70* # Divide the top two of the stack, v/a; multiply by 70 (`/700*10` == `/70`)
             # dc is postfix and stack-based, so operators come after operands.

0k1~0!=a     # This is a ceiling function.
|> 0k  # set precision to 0 to perform integer division
|> 1~  # push the quotient of integer division by 1, and then the remainder. (r is top)
|> 0!=a # If the top-of-stack (decimal part) is not 0, add 1 to the quotient

dI>ldI!>g # Conditional statement
|> dI>l  # (d)uplicate the top, push 10 on. If 10 > the old top, execute the `l`ess-than
          # case, which loops top-of-stack times.
|> dI!>g # Complement of the above, using the `g`reater-than to loop 10 times.

IP # print a newline

This is probably more golf-able, but I was trying to get it finished to avoid premature optimization.

• 2 bytes saved by duplicating-saving instead of saving-loading
• 4 bytes saved dividing by 70
• 6 bytes from daniero's suggestions (non-strings, ASCII nums instead; 10 => I)

Answer 26

Haskell, 35 bytes

This solution is as completely different from Laikonis answer as it gets for something this trivial. Yet the score (for now) is exactly the same.

v%a=take(ceiling$v/a/70)[0..9]>>"*" 

This produces ten stars, then shaves off some. Easy to extend to arbitrary difficulty with an infinite list.

I did manage to shave off one more byte. But while all test cases work, this shouldn't be correct in general.

v%a=take(1+div v(a*70))[0..9]>>"*"

Answer 27

TI-Basic, 39 bytes

Prompt V,A
sub("**********",1,max(0,min(10,int(V/A/70)+1

Answer 28

Ruby, 47 35 bytes

f=->(v,a){'*'*((v/=a*70)<9?v+1:10)}

Called with f.call(v,a)

Saved 12 bytes thanks to user3334690, now the function implicitly returns the result.

Answer 29

Jelly, 10 bytes

÷÷70Ċ«⁵”*x

Try it online!

How it works

÷÷70Ċ«⁵”*x - Main link. Takes v on the left and a on the right
÷          - v ÷ a
 ÷70       - v ÷ a ÷ 70
    Ċ      - ⌈v ÷ a ÷ 70⌉
     «⁵    - min(⌈v ÷ a ÷ 70⌉, 10)
       ”*x - Repeat "*" min(⌈v ÷ a ÷ 70⌉, 10) times

Answer 30

Vyxal, 9 bytes

/70/ṙ›\**

Try it Online!