Hack the elections

Question

You're a professional hacker and your boss has just ordered you to help a candidate win an upcoming election. Your task is to alter the voting machines data to boost the candidate's results.

Voting machines store voting results as two integers : the number of votes for your candidate (v1) and the number of votes for their opponent (v2).

After weeks of research, you have found a security hole in the system and you can increase the value of v1 by an integer x, and decrease the value of v2 by the same x. But there is a constraint, you have to keep the security hash code constant:

• security hash code : (v1 + v2*2) modulo 7

Also, the value for x must be minimal so your changes can go unnoticed.

Your program should accept as input v1 and v2 ; it should output the optimal value for x so v1>v2.

There are some cases for which you cannot hack the results; you don't have to handle them (this might lead to problems with your boss, but that's another story).

Test cases

100,123 --> 14
47,23 --> 0
40,80 --> 21
62,62 --> 7
1134,2145 --> 511

Answer 1

Python 2, 30 bytes

lambda u,t:max(0,(t-u)/14*7+7)

u is our votes, t is their votes.

Answer 2

Python 2, 30 bytes

lambda a,b:max((b-a)/14*7+7,0)

Answer 3

Mathematica, 22 bytes

0//.x_/;2x<=#2-#:>x+7&

Pure function with arguments # and #2. Hits maximum recursion depth if the discrepancy is more than 7*2^16 = 458752.

Explanation

0                       Starting with 0,
 //.                    repeatedly apply the following rule until there is no change:
    x_                    if you see an expression x
      /;                    such that
        2x<=#2-#            2x <= #2-# (equivalently, #+x <= #2-x)
                :>        then replace it with
                  x+7       x+7 (hash is preserved only by multiples of 7)
                     &  End the function definition

Answer 4

Jelly, 9 bytes

IH:7‘×7»0

Try it online!

How it works

IH:7‘×7»0  Main link. Argument: [v1, v2]

I          Increments; compute [v2 - v1].
 H         Halve the result.
  :7       Perform integer division by 7.
    ‘      Increment the quotient.
     ×7    Multiply the result by 7.
       »0  Take the maximum of the product and 0.

Answer 5

Actually, 13 bytes

7;;τ((-\*+0kM

Try it online!

Uses the same max((b-a)/14*7+7,0) formula that xnor and orlp use.

Explanation:

7;;τ((-\*+0kM
7;;            3 copies of 7
   τ           double one of them
    ((-        bring the inputs back to the top, take their difference
       \*+     integer divide by 14, multiply by 7, add 7
          0kM  maximum of that and 0

Answer 6

Groovy, 41 37 bytes

{x,y->[Math.floor((y-x)/14)*7+7,0].max()}

This is an unnamed closure. Thanks to xnor and orlp for the formula and James holderness for pointing out a bug.

The previous solution used intdiv() for integer division but it behaves differently from // used in python.

Try it here!

Answer 7

Haskell, 30 24 bytes

a#b=max 0$div(b-a)14*7+7

An infix operator taking the number of votes of your preferred candidate first. Uses the same logic as the other answers of rounding with /14*7+7.

Answer 8

J, 15 bytes

0>.7+7*14<.@%~-

Kinda interesting, I was working on a problem and I thought I had a solution but as it turns out I was wrong. Oh well. Try it online! Here's the result:

   f =: 0>.7+7*14<.@%~-
   tests =: 123 100 ; 23 47 ; 80 40 ; 62 62 ; 2145 1134
   (,. f/ each) tests
┌─────────┬───┐
│123 100  │14 │
├─────────┼───┤
│23 47    │0  │
├─────────┼───┤
│80 40    │21 │
├─────────┼───┤
│62 62    │7  │
├─────────┼───┤
│2145 1134│511│
└─────────┴───┘

Answer 9

CJam, 13 12 15 bytes

• Saved a byte thanks to Martin Ender.
• Added 3 bytes thanks to Martin Ender.
• Changed ] to [ thanks to ETHproductions.

q~\-Ed/m[)7*0e>

Blatantly stole orlp and xnor's methods.

Input is the two numbers separated by a space: 100 123

Explanation:

q~\-Ed/m])7*0e>
q~\-            e# Input two numbers, swap and subtract them.
    E           e# Push 0xE (15)
     d/m]       e# Float divide and take the floor.
         )7*    e# Increment and multiply by 7.
            0e> e# Max of this and 0.

Answer 10

Excel VBA, ₂₄ 17 Bytes

Immediates window function that takes input from cells A1 and B1 and outputs to the VBE immediates window.

?([A1-B1]\14)*7+7

Subroutine Version, 43 Bytes

takes input b, c as variant\integer and prints to the VBE immediates window

Sub a(b,c):Debug.?Int((c-b)/14)*7+7:End Sub

Answer 11

Julia 0.5, 26 bytes

v\w=max(fld(w-v,14)*7+7,0)

Try it online!

Answer 12

PHP, 41 39 bytes

    <?=7*max(0,1+($argv[2]-$argv[1])/14|0);

takes input from command line arguments; run with -r.

7 5 extra bytes just to handle $a>$b :-/

Answer 13

Japt, 14 bytes

V-U /2+7 f7 w0

Run it here!

Thank you ETHproductions for shaving off 3 bytes!

Answer 14

05AB1E, 9 bytes

-14÷>7*0M

Try it online!

Explanation

-          # push difference of inputs
 14÷       # integer divide by 14
    >      # increment
     7*    # times 7
       0   # push 0
        M  # take max

Or a corresponding function with same byte-count operating on a number-pair

Î¥14÷>7*M

Try it online!

Answer 15

Dyalog APL, 14 bytes

Takes v1 as right argument and v2 as left argument.

0⌈7×1+(⌊14÷⍨-)

0 ⌈ the maximum of zero and

7 × seven times

1 + (...) one plus...

 ⌊ the floor of

 14 ÷⍨ a fourteenth of

 - the difference (between the arguments)

TryAPL online!

Answer 16

Befunge, 19 bytes

777+:&&\-+\/*:0`*.@

Try it online!

This relies on a slightly different formula to that used by orlp and xnor, since the Befunge reference interpreter has different rounding rules to Python. Befunge also doesn't have the luxury of a max operation.

The basic calculation looks like this:

x = (v2 - v1 + 14)/14*7
x = x * (x > 0)

Examining the code in more detail:

7                     Push 7                                      [7]
 77+:                 Push 14 twice.                              [7,14,14]
     &&               Read v1 and v2 from stdin.                  [7,14,14,v1,v2]
       \-             Swap the values and subtract.               [7,14,14,v2-v1]
         +            Add the 14 that was pushed earlier.         [7,14,14+v2-v1]
          \/          Swap the second 14 to the top and divide.   [7,(14+v2-v1)/14]
            *         Multiply by the 7 that was pushed earlier.  [7*(14+v2-v1)/14 => x]
             :        Make a copy of the result                   [x,x]
              0`      Test if it's greater than 0.                [x,x>0]
                *     Multiply this with the original result.     [x*(x>0)]
                 .@   Output and exit.

Answer 17

Go, 36 bytes

func(a,b int)int{return(b-a)/14*7+7}

Try it online!

Answer 18

JavaScript (ES6), 31 bytes

(a,b,c=(b-a)/14|0)=>c>0?c*7+7:0

f=(a,b,c=(b-a)/14|0)=>c>0?c*7+7:0
document.write(f(1134,2145))

Answer 19

Java 8, 31 bytes

(a,b)->b<a?0:(a=(b-a)/2)+7-a%7;

This is a lambda expression assignable to IntBinaryOperator.

a is your candidate's votes, b is your opponent's.

java rounds down for division with positive integers, so +7-a%7 is used to bump up the value to the next multiple of 7.

Answer 20

Ruby, 26 27 bytes

->a,b{[(b-a)/14*7+7,0].max}

Basically the same as xnor's and orlp's Python solution, with a twist (no need to add 7, because of negative modulo, saves 1 byte in ruby, don't know about python)

No twist, the twist was just a bad case of cognitive dissonance. Forget it. Really. :-)

Answer 21

Scala, 31 bytes

(a,b)=>Math.max((b-a)/14*7+7,0)

The ternary version is 2 bytes longer

Answer 22

Noodel, 16 bytes

⁻÷14ɲL×7⁺7ḋɲl⁺÷2

Pulled equation from xor and orlp answers, but since Noodel does not have a max capability had to work around that.

Try it:)

How it works

⁻÷14ɲL×7⁺7       # The equation...
⁻                # v2 - v1
 ÷14             # Pops off the difference, then pushes on the (v2 - v1)/14
    ɲL           # Applies lowercase which for numbers is the floor function.
      ×7         # Multiplies that by seven.
        ⁺7       # Then increments it by seven.

          ḋɲl⁺÷2 # To relate with the other answers, this takes the max between the value and zero.
          ḋ      # Duplicates what is on the top of the stack (which is the value just calculated).
           ɲl    # Pops off the number and pushes on the magnitude (abs value).
             ⁺   # Add the abs to itself producing zero if the number came out negative (which means we are already winning).
              ÷2 # Divides the result by two, which will either be zero or the correct offset.

Answer 23

Pyth, 16 bytes

MtS+0[+7*7/-HG14

Try it here!