Created
September 20, 2017 12:39
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var A = [ | |
["1", "0", "0"], | |
["1", "0", "1"], | |
["1", "1", "0"], | |
["0", "0", "0"], | |
["0", "0", "1"], | |
["0", "1", "0"], | |
["0", "1", "1"] | |
]; | |
var i = 0; | |
var chainSize = A[0].length; | |
//работам пока не закончится цепочка или пока не найдём решение | |
while(i<chainSize && !haveSameNumberOfZerosAndOnes(A, i)){ | |
var mode = mostElement(A, i); //нашли самые популярный | |
A = A.filter(function(element){ | |
return element[i]===mode; //оставили в массиве только элементы с ним | |
}); | |
i++; | |
} | |
if (haveSameNumberOfZerosAndOnes(A, i)){ | |
console.log("Found", A); | |
}else{ | |
console.log("Chain is over") | |
} | |
//функция проверяет одинаковое ли число 1 и 0 на позиции idx в цепочках массива | |
function haveSameNumberOfZerosAndOnes(arr, idx){ | |
var zeros = 0; | |
var ones = 0; | |
for (var j=0; j<arr.length; j++){ | |
if (arr[j][idx]==="1") ones++; | |
if (arr[j][idx]==="0") zeros++; | |
} | |
return zeros === ones; | |
} | |
//функция ищет наиболее популярное число на позиции idx в цепочках массива | |
function mostElement(arr, idx){ | |
var zeros = 0; | |
var ones = 0; | |
for (var j=0; j<arr.length; j++){ | |
if (arr[j][idx]==="1") ones++; | |
if (arr[j][idx]==="0") zeros++; | |
} | |
if (ones>zeros){ | |
return "1"; | |
} else { | |
return "0"; | |
} | |
} |
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