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

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 8

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 9

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 10

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 11

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 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

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 20

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 21

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 22

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 23

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 24

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 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

PowerShell v2+, 47 bytes

-join(('*'*11)[1..($args[0]/$args[1]/70+.499)])

Somewhat a port of @Neil's JavaScript answer.

Takes input $args and divides them, then divides that by 70, and adds .499. Since PowerShell does banker's rounding, this is effectively ceil to two decimal points of precision. If additional precision is required, tack on as many additional 9s as required.

Along with the 1.., this forms a range index into a string. The string is '*'*11, i.e. '***********'. That results in a char-array, so we -join it together back into a string. That string is left on the pipeline and output is implicit. Like Neil's answer, this effectively "clamps" the output to be between 1 and 10 stars.

Test Suite

PS C:\Tools\Scripts\golfing> @(@(163,2), @(548,22), @(1452,24), @(1713,37), @(4162,32), @(3067,15), @(22421,19))|%{($_-join', ')+" -> " +(.\difficulty-of-a-question $_[0] $_[1])}
163, 2 -> **
548, 22 -> *
1452, 24 -> *
1713, 37 -> *
4162, 32 -> **
3067, 15 -> ***
22421, 19 -> **********

Answer 29

Python 3, 69 68 bytes

I didn't want to copy the Python 2 answer, so mine is slightly longer.

from math import*
x=lambda v,a:print(max(0,min(ceil(v/a/70),10))*'*')

Saved 1 byte thanks to Program man

Answer 30

Actually, 14 bytes

:70a\\u9ukm'**

Try it online!

Takes advantage of the fact that 0 views and 0 answers is impossible, and thus ceil(v/a) > 0.

Explanation:

:70a\\u9ukm'**
:70             push 70 ([70 a v])
   a            invert stack ([v a 70])
    \\          integer division twice ([v//a//70])
      u         add 1 ([v//a//70 + 1])
       9uk      push 10, make stack into list ([[v//a//70+1, 10]])
          m     minimum of list
           '**  push "*", repeat