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Created May 7, 2015 21:27
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index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>JS Bin</title>
</head>
<body>
<script id="jsbin-javascript">
var sudokuValid1 = '4;1,2,3,4,3,4,1,2,2,3,4,1,4,1,2,3';
var sudokuValid2 = '4;2,3,4,1,4,1,2,3,1,2,3,4,3,4,1,2';
var sudokuInvalid1 = '4;1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4'; // invalid rows
var sudokuInvalid2 = '4;1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4'; // invalid cols
var sudokuInvalid3 = '4;1,2,3,4,2,3,4,1,3,4,1,2,4,1,2,3'; // invalid squares
// looked up a valid solution and entered in the format the challenge specified
var hardModeValid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,4,1';
// below is same as above, just flipped the last two numbers. it's easy to write an invalid sudoku ;)
var hardModeInvalid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,1,4';
// Although in my interview I tried to parse the line into a 2d array, I realized later that wasn't necessary
// The trick is really just to find the right offsets to each row, column, and square
function validateSudoku(sudokuLine) {
// Since we are and-ing all the way down, start with true, any false along the way constitutes a false return
var valid = true;
var parsedSudokuLine = sudokuLine.split(';');
var n = +parsedSudokuLine[0]; // coerce to int
var sudokuArray = parsedSudokuLine[1].split(',');
for(var i=0; i < n; i++) {
valid = valid && validateIteration(sudokuArray, i, n);
// more performant to break if valid ever becomes false
if(!valid) {
break;
}
}
return valid;
}
// validates the i-th row, column, and square simultaneously.
// Originally I had three validation functions but I noticed they were very similar and
// doing them all at once saves time
function validateIteration(sudokuArray, itr, n) {
var rowInitialOffset = itr*n;
var colInitialOffset = itr;
// The square offset derivations were much harder than I anticipated
var squareInitialOffset = (Math.floor(itr / Math.sqrt(n)) * n * Math.sqrt(n)) + ((itr % Math.sqrt(n)) * Math.sqrt(n));
var expectedArray = [];
var actualRowArray = [];
var actualColArray = [];
var actualSquareArray = [];
for(var i=0; i < n; i++) {
expectedArray[i] = i+1;
var rowOffset = i;
var colOffset = i*n;
var squareOffset = (Math.floor(i / Math.sqrt(n)) * n) + (i % Math.sqrt(n));
//console.log('itr ' + itr + ' i ' + i + ' init ' + squareInitialOffset + ' offs ' + squareOffset + ' idx ' + (squareInitialOffset + squareOffset));
actualRowArray[i] = +sudokuArray[rowInitialOffset + rowOffset];
actualColArray[i] = +sudokuArray[colInitialOffset + colOffset];
actualSquareArray[i] = +sudokuArray[squareInitialOffset + squareOffset];
}
var rowValid = arraysIdentical(expectedArray, actualRowArray.sort());
var colValid = arraysIdentical(expectedArray, actualColArray.sort());
var squareValid = arraysIdentical(expectedArray, actualSquareArray.sort());
return rowValid && colValid && squareValid;
}
// Borrowed from Stack Overflow, http://stackoverflow.com/questions/7837456/comparing-two-arrays-in-javascript
function arraysIdentical(a, b) {
var i = a.length;
if (i != b.length) return false;
while (i--) {
if (a[i] !== b[i]) return false;
}
return true;
}
// Found this to make more sense as output, unecessary of course but more fun
function passFail(booleanExpression) {
return booleanExpression ? 'Pass!' : 'Fail :(';
}
console.log('Expect sudokuValid1 to be valid', passFail(validateSudoku(sudokuValid1) === true));
console.log('Expect sudokuValid2 to be valid', passFail(validateSudoku(sudokuValid2) === true));
console.log('Expect sudokuInvalid1 to be invalid', passFail(validateSudoku(sudokuInvalid1) === false));
console.log('Expect sudokuInvalid2 to be invalid', passFail(validateSudoku(sudokuInvalid2) === false));
console.log('Expect sudokuInvalid3 to be invalid', passFail(validateSudoku(sudokuInvalid3) === false));
console.log('Expect hardModeValid to be valid', passFail(validateSudoku(hardModeValid) === true));
console.log('Expect hardModeInvalid to be invalid', passFail(validateSudoku(hardModeInvalid) === false));
</script>
<script id="jsbin-source-javascript" type="text/javascript">var sudokuValid1 = '4;1,2,3,4,3,4,1,2,2,3,4,1,4,1,2,3';
var sudokuValid2 = '4;2,3,4,1,4,1,2,3,1,2,3,4,3,4,1,2';
var sudokuInvalid1 = '4;1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4'; // invalid rows
var sudokuInvalid2 = '4;1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4'; // invalid cols
var sudokuInvalid3 = '4;1,2,3,4,2,3,4,1,3,4,1,2,4,1,2,3'; // invalid squares
// looked up a valid solution and entered in the format the challenge specified
var hardModeValid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,4,1';
// below is same as above, just flipped the last two numbers. it's easy to write an invalid sudoku ;)
var hardModeInvalid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,1,4';
// Although in my interview I tried to parse the line into a 2d array, I realized later that wasn't necessary
// The trick is really just to find the right offsets to each row, column, and square
function validateSudoku(sudokuLine) {
// Since we are and-ing all the way down, start with true, any false along the way constitutes a false return
var valid = true;
var parsedSudokuLine = sudokuLine.split(';');
var n = +parsedSudokuLine[0]; // coerce to int
var sudokuArray = parsedSudokuLine[1].split(',');
for(var i=0; i < n; i++) {
valid = valid && validateIteration(sudokuArray, i, n);
// more performant to break if valid ever becomes false
if(!valid) {
break;
}
}
return valid;
}
// validates the i-th row, column, and square simultaneously.
// Originally I had three validation functions but I noticed they were very similar and
// doing them all at once saves time
function validateIteration(sudokuArray, itr, n) {
var rowInitialOffset = itr*n;
var colInitialOffset = itr;
// The square offset derivations were much harder than I anticipated
var squareInitialOffset = (Math.floor(itr / Math.sqrt(n)) * n * Math.sqrt(n)) + ((itr % Math.sqrt(n)) * Math.sqrt(n));
var expectedArray = [];
var actualRowArray = [];
var actualColArray = [];
var actualSquareArray = [];
for(var i=0; i < n; i++) {
expectedArray[i] = i+1;
var rowOffset = i;
var colOffset = i*n;
var squareOffset = (Math.floor(i / Math.sqrt(n)) * n) + (i % Math.sqrt(n));
//console.log('itr ' + itr + ' i ' + i + ' init ' + squareInitialOffset + ' offs ' + squareOffset + ' idx ' + (squareInitialOffset + squareOffset));
actualRowArray[i] = +sudokuArray[rowInitialOffset + rowOffset];
actualColArray[i] = +sudokuArray[colInitialOffset + colOffset];
actualSquareArray[i] = +sudokuArray[squareInitialOffset + squareOffset];
}
var rowValid = arraysIdentical(expectedArray, actualRowArray.sort());
var colValid = arraysIdentical(expectedArray, actualColArray.sort());
var squareValid = arraysIdentical(expectedArray, actualSquareArray.sort());
return rowValid && colValid && squareValid;
}
// Borrowed from Stack Overflow, http://stackoverflow.com/questions/7837456/comparing-two-arrays-in-javascript
function arraysIdentical(a, b) {
var i = a.length;
if (i != b.length) return false;
while (i--) {
if (a[i] !== b[i]) return false;
}
return true;
}
// Found this to make more sense as output, unecessary of course but more fun
function passFail(booleanExpression) {
return booleanExpression ? 'Pass!' : 'Fail :(';
}
console.log('Expect sudokuValid1 to be valid', passFail(validateSudoku(sudokuValid1) === true));
console.log('Expect sudokuValid2 to be valid', passFail(validateSudoku(sudokuValid2) === true));
console.log('Expect sudokuInvalid1 to be invalid', passFail(validateSudoku(sudokuInvalid1) === false));
console.log('Expect sudokuInvalid2 to be invalid', passFail(validateSudoku(sudokuInvalid2) === false));
console.log('Expect sudokuInvalid3 to be invalid', passFail(validateSudoku(sudokuInvalid3) === false));
console.log('Expect hardModeValid to be valid', passFail(validateSudoku(hardModeValid) === true));
console.log('Expect hardModeInvalid to be invalid', passFail(validateSudoku(hardModeInvalid) === false));</script></body>
</html>
Raw
jsbin.niyilumocu.js
var sudokuValid1 = '4;1,2,3,4,3,4,1,2,2,3,4,1,4,1,2,3';
var sudokuValid2 = '4;2,3,4,1,4,1,2,3,1,2,3,4,3,4,1,2';
var sudokuInvalid1 = '4;1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4'; // invalid rows
var sudokuInvalid2 = '4;1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4'; // invalid cols
var sudokuInvalid3 = '4;1,2,3,4,2,3,4,1,3,4,1,2,4,1,2,3'; // invalid squares
// looked up a valid solution and entered in the format the challenge specified
var hardModeValid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,4,1';
// below is same as above, just flipped the last two numbers. it's easy to write an invalid sudoku ;)
var hardModeInvalid = '9;3,9,1,2,8,6,5,7,4,4,8,7,3,5,9,1,2,6,6,5,2,7,1,4,8,3,9,8,7,5,4,3,1,6,9,2,2,1,3,9,6,7,4,8,5,9,6,4,5,2,8,7,1,3,1,4,9,6,7,3,2,5,8,5,3,8,1,4,2,9,6,7,7,2,6,8,9,5,3,1,4';
// Although in my interview I tried to parse the line into a 2d array, I realized later that wasn't necessary
// The trick is really just to find the right offsets to each row, column, and square
function validateSudoku(sudokuLine) {
// Since we are and-ing all the way down, start with true, any false along the way constitutes a false return
var valid = true;
var parsedSudokuLine = sudokuLine.split(';');
var n = +parsedSudokuLine[0]; // coerce to int
var sudokuArray = parsedSudokuLine[1].split(',');
for(var i=0; i < n; i++) {
valid = valid && validateIteration(sudokuArray, i, n);
// more performant to break if valid ever becomes false
if(!valid) {
break;
}
}
return valid;
}
// validates the i-th row, column, and square simultaneously.
// Originally I had three validation functions but I noticed they were very similar and
// doing them all at once saves time
function validateIteration(sudokuArray, itr, n) {
var rowInitialOffset = itr*n;
var colInitialOffset = itr;
// The square offset derivations were much harder than I anticipated
var squareInitialOffset = (Math.floor(itr / Math.sqrt(n)) * n * Math.sqrt(n)) + ((itr % Math.sqrt(n)) * Math.sqrt(n));
var expectedArray = [];
var actualRowArray = [];
var actualColArray = [];
var actualSquareArray = [];
for(var i=0; i < n; i++) {
expectedArray[i] = i+1;
var rowOffset = i;
var colOffset = i*n;
var squareOffset = (Math.floor(i / Math.sqrt(n)) * n) + (i % Math.sqrt(n));
//console.log('itr ' + itr + ' i ' + i + ' init ' + squareInitialOffset + ' offs ' + squareOffset + ' idx ' + (squareInitialOffset + squareOffset));
actualRowArray[i] = +sudokuArray[rowInitialOffset + rowOffset];
actualColArray[i] = +sudokuArray[colInitialOffset + colOffset];
actualSquareArray[i] = +sudokuArray[squareInitialOffset + squareOffset];
}
var rowValid = arraysIdentical(expectedArray, actualRowArray.sort());
var colValid = arraysIdentical(expectedArray, actualColArray.sort());
var squareValid = arraysIdentical(expectedArray, actualSquareArray.sort());
return rowValid && colValid && squareValid;
}
// Borrowed from Stack Overflow, http://stackoverflow.com/questions/7837456/comparing-two-arrays-in-javascript
function arraysIdentical(a, b) {
var i = a.length;
if (i != b.length) return false;
while (i--) {
if (a[i] !== b[i]) return false;
}
return true;
}
// Found this to make more sense as output, unecessary of course but more fun
function passFail(booleanExpression) {
return booleanExpression ? 'Pass!' : 'Fail :(';
}
console.log('Expect sudokuValid1 to be valid', passFail(validateSudoku(sudokuValid1) === true));
console.log('Expect sudokuValid2 to be valid', passFail(validateSudoku(sudokuValid2) === true));
console.log('Expect sudokuInvalid1 to be invalid', passFail(validateSudoku(sudokuInvalid1) === false));
console.log('Expect sudokuInvalid2 to be invalid', passFail(validateSudoku(sudokuInvalid2) === false));
console.log('Expect sudokuInvalid3 to be invalid', passFail(validateSudoku(sudokuInvalid3) === false));
console.log('Expect hardModeValid to be valid', passFail(validateSudoku(hardModeValid) === true));
console.log('Expect hardModeInvalid to be invalid', passFail(validateSudoku(hardModeInvalid) === false));
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