Trouble simplifying conditional

user_13082464 0
Viewed 142

I'm having trouble simplifying a conditional statement. It looks like this:
There is an array of arrays like this, dice = [ [a, b], [c, d] ].
a, b, c, and d representing random numbers on a dice 1 > 6.
Every array in the array represents a roll of two dice where the last roll is [c, d] = [2, 5] for example,
and the second-last roll is [a, b] = [3, 6] for example.
Now if you roll 2 x 1 in a row or 2 x 6 in a row something happens. so, a + c || a + d and b + c || b + d may not be 2 or 12.
The conditional below works but I think if you open the door to hell it looks prettier.
dice = last roll = [c, d]
prev = second last roll = [a, b]

if (dice[0] + prev[0] === 2 || dice[1] + prev[1] === 2 || dice[1] + prev[0] === 2 || dice[0] + 
prev[1] === 2 &&
dice[0] + prev[0] === 12 || dice[1] + prev[1] === 12 || dice[1] + prev[0] === 12 || dice[0] + 
prev[1] === 12){
//something happens here
}
if-statement algorithm javascript
6 Answers

I would do something like this:

var combinations = dice.reduce(function(acc, curr) {
    return acc.concat(prev.map(function(curr2) {
        return curr + curr2;
    }));
}, []);

if (combinations.indexOf(2) !== -1 && combinations.indexOf(12) !== -1) {
    //something happens here
}

In ES6 it is a little bit shorter:

const combinations = dice.reduce((acc, curr) => acc.concat(prev.map(curr2 => curr + curr2)), []);

if (combinations.indexOf(2) !== -1 && combinations.indexOf(12) !== -1) {
    //something happens here
}

What this actually does is calculate all combinations (additions) and then check whether one is 2 and one is 12.

user_57945040

I came out with a simpler solution more fitted to your need

const throws1 = [[1, 2], [3, 4]]
const throws2 = [[1, 2], [3, 1]]
const throws3 = [[1, 6], [6, 1]]


const checkCondition = (prev, dice) => [1, 6].every(res => prev.includes(res) && dice.includes(res))

console.log(checkCondition(...throws1))
console.log(checkCondition(...throws2))
console.log(checkCondition(...throws3))

user_143268990

The condition can only happens when both dice and prev contain 1 or 6. So, we can just check that.

let throws = [[5,3],[1,6]];
let prev = throws[0];
let dice = throws[1];

if ((prev.includes(1) && dice.includes(1)) || (prev.includes(6) && dice.includes(6))){
  console.log("2 or 12 occurs")
}
else {
  console.log("2 and 12 do not occur")
}

user_37013930

What about something like the following?:

const one_and_six_found = (roll) => roll.includes(1) && roll.includes(6);
const two_rolls_have_one_and_six = (roll1, roll2) =>
  one_and_six_found(roll1) && one_and_six_found(roll2);

console.log(two_rolls_have_one_and_six( [1, 6], [1, 6] ));
console.log(two_rolls_have_one_and_six( [6, 1], [1, 6] ));
console.log(two_rolls_have_one_and_six( [5, 1], [1, 6] ));

user_144549390

Here's the conditional simplified:

2*(a + b + c + d) == a*b + c*d + 16
user_20347870

I wasn't 100% sure about a couple of these answers so I thought I'd test them:

const orig = ([a, b], [c, d]) =>
  ((a + c === 2 || b + d === 2 || b + c === 2 || a + d === 2) && (a + c === 12 || b + d === 12 || b + c === 12 || a + d === 12));

const A = ([a, b], [c, d]) => 2*(a + b + c + d) == a*b + c*d + 16;
const B = (prev, dice) => [1, 6].every(res => prev.includes(res) && dice.includes(res));
const one_and_six_found = (roll) => roll.includes(1) && roll.includes(6);
const C = (roll1, roll2) => one_and_six_found(roll1) && one_and_six_found(roll2);
const D = (prev, dice) => (prev.includes(1) && dice.includes(1)) || (prev.includes(6) && dice.includes(6));
const E = (prev, dice) => {
  const combinations = dice.reduce((acc, curr) => acc.concat(prev.map(curr2 => curr + curr2)), []);

  return (combinations.indexOf(2) !== -1 && combinations.indexOf(12) !== -1);
};

const test = (fn) => {
  for(let i = 1; i <= 6; i++)
    for(let j = 1; j <= 6; j++)
      for(let k = 1; k <= 6; k++)
        for(let l = 1; l <= 6; l++) 
          if(fn([i, j], [k, l]) !== orig([i, j], [k, l])) {
            console.log('failed on ' + [i, j, k, l].join(', '));
            return false;
          }
  
  return true;
};

console.log('A', test(A) ? 'success' : 'failed');
console.log('B', test(B) ? 'success' : 'failed');
console.log('C', test(C) ? 'success' : 'failed');
console.log('D', test(D) ? 'success' : 'failed');
console.log('E', test(E) ? 'success' : 'failed');

user_144549390
Related