Related
The question is as follows:
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
I have written the following code, but I am stuck in terms of how to flag the row as one time to be downward moving, where I increment the start row, but when it's zigzagging back to the top, it should be decremented. I am unable to figure out the logic to make this work without affecting the downward movement. Any help would be appreciated.
const convert = (s, numRows) => {
let startRow = 0
let endRow = numRows - 1
let startColumn = 0
let endColumn = Math.floor((s.length / 2) - 1)
s = s.split('')
let results = []
// to setup the columns
for (let i = 0; i < numRows; i++) {
results.push([])
}
while (startRow <= endRow && startColumn <= endColumn && s.length) {
for (let i = startRow; i <= endRow; i++) {
results[i][startColumn] = s.shift()
}
for (let i = endRow - 1; i >= startRow; i--) {
results[i][startColumn + 1] = s.shift()
startColumn++
}
//this line seems to be the issue
startRow++
}
return results
}
console.log(convert('PAYPALISHIRING', 4))
I rewrote your while loop as follows where I simply walk a "zigzag" pattern! Hopefully, it is simple enough to understand.
let c=0, row=0,col=0, down=0;
while(c<s.length) {
results[row][col]=s[c];
if(down==0) { // moving down
row++;
if(row==numRows) {
down = 1;
col++;
row-=2;
}
} else { // moving up
row--;
col++;
if(row==0) {
down=0;
}
}
c++;
}
Ps. Above code does not handle numRows < 3 so you have to manage them before this loop.
My precalculus is a little rusty, but the logic behind this problem seems like a sine wave. I made a math error somewhere in creating the sin equation that prevents this from working (r never equals c with the current paramaters), but hopefully this will help if this is the direction you choose to go in.
/*If x-axis is position in string, and y-axis is row number...
n=number of rows
Equation for a sin curve: y = A sin(B(x + C)) + D
D=vertical shift (y value of mid point)
D=median of 1 and n
n: median:
1 1
2 1.5
3 2
4 2.5
5 3
6 3.5
7 4
median=(n+1)/2
D=(n+1)/2
A=amplitude (from the mid-point, how high does the curve go)
median + amplitude = number of rows
amplitude = number of rows - median
A=n-D
C=phase shift
Phase shift for a sin curve starting at its lowest point: 3π/2
(so at time 1, row number is 1, and curve goes up from there)
C=3π/2
Period is 2π/B
n p
3 4
4 6
5 8
6 10
period=2(n-1)
2(n-1)=2π/B
B(2(n-1)=2π
B=2π/2(n-1)
B=π/(n-1)
Variables:
s = string
n = number of rows
c = current row number being evaluated
p = position in string
r = row number
*/
var output='';
function convert(s,n) {
D=(n+1)/2
A=n-D
C=(3*Math.PI)/2
B=Math.PI/(n-1)
for (c=1;c<=n;c++) { //loop from 1st row to number of rows
for (p=1;p<=s.length;p++) { //loop from 1st to last character in string
r=A*Math.sin(B*(p+C))+D //calculate the row this character belongs in
if (r==c) { output+= s.charAt(r) } //if the character belongs in this row, add it to the output variable. (minus one because character number 1 is at position 0)
}}
//do something with output here
}
Question
How can i compare two adjacent cells thanks to them coordinates?
Documentation which helped me
I already saw these questions, they helped me but they are different from my case:
question on stackOverflow
question on stackOverflow
question on stackOverflow
Mds documentation to build a dynamic table
Code
I've a dynamically generated table
function tableGenerate(Mytable){
for(var i = 0; i < myTable.length; i++) {
var innerArrayLength = myTable[i].length;
for(var j = 0; j<innerArrayLength; j++){
if(myTable[i][j] === 0){
myTable[i][j]="x";
}else{
myTable[i][j]="y";
};
};
$("#aTable").append("<tr><td>"+ myTable[i].join('</td><td>') + "</td></tr>")
}
}
About the interested cells (two global variables) in actualPosition row and cell have random values
var mainTd = {
name: 'interestedValue',
actualPosition:{
row: 5,
cell: 4
}
};
var otherMainTd = {
actualPosition:{
row: 2,
cell: 3
}
};
The final part of the code works in this way:
I save the position of selectedTd in two differents variables
I create the 2d array directions with the coordinates of near cells relatives to the selectedTd
enter in the first if, compare the two cells. If one of the coordinates are the same, you enter in this last if.
function compare(selectedTd) {
let tdRow = selectedTd.actualPosition.row;
let tdCell = selectedTd.actualPosition.cell;
let directions = [
[tdRow - 1, tdCell],
[tdRow + 1, tdCell],
[tdRow, tdCell + 1],
[tdRow, tdCell - 1]
]; //these are the TD near the mainTd, the one i need to compare to the others
let tdToCompare = [];
if (selectedTd.name === 'interestedValue') {
tdToCompare = [otherMainTd.actualPosition.row, otherMainTd.actualPosition.cell];
for (let i = 0; i < directions.length; i++) {
if (directions[i] == tdToCompare) {
console.log('you are here');
}
}
} else {
tdToCompare = [mainTd.actualPosition.row, mainTd.actualPosition.cell];
for (let i = 0; i < directions.length; i++) {
if (directions[i] === tdToCompare) {
console.log('you are here');
}
}
}
};
Now the main problem is: I read the coordinates, I store them in the 2 arrays, I can read them but I cannot able to enter in the if statement.
This is what I want to achieve: compare the coordinates of the blackTd with the coordinates of the red-borders td.
Codepen
the interested functions in the codepen are with different names, but the structure is the same that you saw in this post. I changed the original names because I think it could be more clear with general names instead of the names that i choose.
the interested functions are:
function fight(playerInFight) ---> function compare(selectedTd)
function mapGenerate(map) ---> function tableGenerate(MyTable)
mainTd and otherMainTd ---> character and characterTwo
CodepenHere
Update:
Reading your code again I think I figured out the problem. You're comparing array instances instead of their actual values. See this simple example to illustrate the issue:
var a = [1];
var b = [1];
console.log(a===b);
What you'd need to do in your code is this:
if (selectedTd.name === 'interestedValue') {
tdToCompare = [otherMainTd.actualPosition.row, otherMainTd.actualPosition.cell];
for (let i = 0; i < directions.length; i++) {
if (
directions[i][0] === tdToCompare[0] &&
directions[i][1] === tdToCompare[1]
) {
console.log('you are here');
}
}
} else {
tdToCompare = [mainTd.actualPosition.row, mainTd.actualPosition.cell];
for (let i = 0; i < directions.length; i++) {
if (
directions[i][0] === tdToCompare[0] &&
directions[i][1] === tdToCompare[1]
) {
console.log('you are here');
}
}
}
Now it checks if the values, and thus the cells, are matching.
Recommendations:
If I were you, I would write the method a bit different. Below is how I would do it.
function compare(selectedTd) {
const
// Use destructuring assignemnt to get the row and cell. Since these are
// values that won't be changed in the method declare them as "const". Also
// drop the "td" prefix, it doesn't add anything useful to the name.
// See: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment
{ row, cell } = selectedTd.actualPosition,
// Directions can also be a const, it will not be reassigned in the method.
directions = [
[row - 1, cell],
[row + 1, cell],
[row, cell + 1],
[row, cell - 1]
],
// A few things happens in this line:
// - It is a destructuring assignment where the names are changed. In this case
// row and cell are already assigned so it is necessary to give them another name.
// - Don't put the row and cell in an array. You will have to access the actual values
// anyway as you can't compare the array instances.
// - Instead of doing this in the if…else you had, decide here which cell you want to
// look for. It means the rest of the method can be written without wrapping any
// logic in an if…else making it less complex.
{ row: referenceRow, cell: referenceCell } = (selectedTd.name === 'interestedValue')
? otherMainTd.actualPosition
: mainTd.actualPosition,
// Use find instead of a for loop. The find will stop as soon as it finds a match. The
// for loop you had kept evaluating direction items even if the first one was already
// a match.
// The "([row,cell])" is the signature of the callback method for the find. This too is
// a destructuring assignment only this time with one of the arrays of the directions
// array. The first array item will be named "row" and the second item "cell". These
// variable names don't clash with those declared at the top of this method as this
// is a new scope.
// The current directions entry is a match when the row and cell values match.
matchingNeighbor = directions.find(([row, cell]) => row === referenceRow && cell === referenceCell);
// "find" returns undefined when no match was found. So when match is NOT unddefined
// it means directions contained the cell you were looking for.
if (matchingNeighbor !== undefined) {
console.log('you are here');
}
};
const
mainTd = {
name: 'interestedValue',
actualPosition: {
cell: 1,
row: 1
}
},
otherMainTd = {
actualPosition: {
cell: 0,
row: 1
}
};
compare(mainTd);
Orginal answer:
There is quite a bit going on in your question, I hope I understood it properly.
What I've done is create a Grid, you pass this the dimensions and it will create the array for each cell in the grid. Then it returns an object with some methods you can use to interact with the grid. It has the following methods:
cellAtCoordinate: Pass it an X and Y coordinate and it returns the cell.
isSameLocation: Pass it two cells and it checks if the cells are in the same location.
neighborsForCoordinate: Pass it an X and Y coordinate and it returns an array with the cells above, below, to the right, and to the left (if they exist).
With all of this out of the way the compare method becomes a little more manageable. Getting the neighbours is now just a single call, the same for the check if two cells match.
Like I said, I hope this is what you were trying to achieve. If I got the problem wrong and something needs some further explaining, please let me know.
/**
* Creates grid with the provided dimensions. The cell at the top left corner
* is at coordinate (0,0). The method returns an object with the following
* three methods:
* - cellAtCoordinate
* - isSameLocation
* - neighborsForCoordinate
*/
function Grid(width, height) {
if (width === 0 || height === 0) {
throw 'Invalid grid size';
}
const
// Create an array, each item will represent a cell. The cells in the
// array are laid out per row.
cells = Array.from(Array(width * height), (value, index) => ({
x: index % width,
y: Math.floor(index / height)
}));
function cellAtCoordinate(x, y) {
// Make sure we don't consider invalid coordinate
if (x >= width || y >= height || x < 0 || y < 0) {
return null;
}
// To get the cell at the coordinate we need to calculate the Y offset
// by multiplying the Y coordinate with the width, these are the cells
// to "skip" in order to get to the right row.
return cells[(y * width) + x];
}
function isSameLocation(cellA, cellB) {
return (
cellA.x === cellB.x &&
cellA.y === cellB.y
);
}
function neighborsForCoordinate(x, y) {
// Make sure we don't consider invalid coordinate
if (x >= width || y >= height || x < 0 || y < 0) {
return null;
}
const
result = [];
// Check if there is a cell above.
if (y > 0) result.push(cellAtCoordinate(x, y - 1));
// Check if there is a cel to the right
if (x < width) result.push(cellAtCoordinate(x + 1, y));
// Check if there is a cell below.
if (y < height) result.push(cellAtCoordinate(x, y + 1));
// Check if there is a cell to the left.
if (x > 0) result.push(cellAtCoordinate(x - 1, y));
return result;
}
return {
cellAtCoordinate,
isSameLocation,
neighborsForCoordinate
}
}
function compareCells(grid, selectedCell) {
const
// Get the neighbors for the selected cell.
neighbors = grid.neighborsForCoordinate(selectedCell.x, selectedCell.y);
compareAgainst = (selectedCell.name === 'interestedValue')
? otherMainTd
: mainTd;
// In the neighbors, find the cell with the same location as the cell
// we want to find.
const
match = neighbors.find(neighbor => grid.isSameLocation(neighbor, compareAgainst));
// When match is NOT undefined it means the compareAgainst cell is
// a neighbor of the selected cell.
if (match !== undefined) {
console.log(`You are there at (${match.x},${match.y})`);
} else {
console.log('You are not there yet');
}
}
// Create a grid which is 3 by 3.
const
myGrid = Grid(3, 3),
// Place the main TD here:
// - | X | -
// - | - | -
// - | - | -
mainTd = {
name: 'interestedValue',
x: 1,
y: 0
},
// Place the other TD here:
// - | - | -
// Y | - | -
// - | - | -
otherMainTd = {
x: 0,
y: 1
};
// Check if the mainTd is in a cell next to the otherMainTd. It is not
// as the neighboring cells are:
// N | X | N
// Y | N | -
// - | - | -
compareCells(myGrid, mainTd);
// Move the mainTd to the center of the grid
// - | - | -
// Y | X | -
// - | - | -
mainTd.y = 1;
// Compare again, now the main TD is next the the other.
// - | N | -
// YN | X | N
// - | N | -
compareCells(myGrid, mainTd);
given a matrix like this one:
1 2 3
4 5 6
7 8 9
which can be represented as a 2 dimensional array:
arr = [[1,2,3], [4,5,6], [7,8,9]];
rotate the array so that it is read diagonally at a 45 degree angle and prints out this:
1
4 2
7 5 3
8 6
9
I spent a while coming up with a solution that I don't even fully intuitively understand, but it works, at least for 3x3 and 4x4 matrices. I was hoping to see more logical and clean implementations.
Here's my solution:
arr = [[1,2,3,0],[4,5,6,0],[7,8,9,0], [0,0,0,0]];
// arr[i][j];
transform(arr);
function transform(ar) {
// the number of lines in our diagonal matrix will always be rows + columns - 1
var lines = ar.length + ar[0].length - 1;
// the length of the longest line...
var maxLen = ~~(ar.length + ar[0].length)/2;
var start = 1;
var lengths = [];
// this for loop creates an array of the lengths of each line, [1,2,3,2,1] in our case
for (i=0;i<lines; i++) {
lengths.push(start);
if (i+1 < maxLen) {
start++;
} else {
start--;
}
}
// after we make each line, we're going to append it to str
var str = "";
// for every line
for(j=0; j<lengths.length; j++) {
// make a new line
var line = "";
// i tried to do it all in one for loop but wasn't able to (idk if it's possible) so here we use a particular for loop while lengths of the lines are increasing
if (j < maxLen) {
// lengths[j] is equal to the elements in this line, so the for loop will run that many times and create that many elements
for(c=0; c<lengths[j]; c++) {
// if ar[r][c], the pattern here is that r increases along rows (as we add new lines), and decreases along columns. c stays the same as we add rows, and increases across columns
line += ar[lengths[j]-1-c][c] + " ";
// when we've added all the elements we need for this line, add it to str with a line break
if (c == lengths[j]-1) {
line += "\n"; str += line;
}
}
} else {
// when we're headed down or decreasing the length of each line
for (r=0; r<lengths[j]; r++) {
// the pattern here tripped me up, and I had to introduce another changing variable j-maxLen (or the distance from the center). r stays the same as rows increase and decreases across columns. c increases along rows and decreases across columns
line += ar[lengths[j]-r+j-maxLen][j-maxLen+r +1] + " ";
// that's all our elements, add the line to str;
if (r == lengths[j] -1) {
line += "\n"; str += line;
}
}
}
}
console.log(str);
}
The main idea is to partition the original matrix indexed by (i,j) according to i+j.
This is expressed in the code snippet rotated[i+j].push(arr[i][j]) below:
arr = [[1,2,3], [4,5,6], [7,8,9]];
var summax = arr.length + arr[0].length - 1; // max index of diagonal matrix
var rotated = []; // initialize to an empty matrix of the right size
for( var i=0 ; i<summax ; ++i ) rotated.push([]);
// Fill it up by partitioning the original matrix.
for( var j=0 ; j<arr[0].length ; ++j )
for( var i=0 ; i<arr.length ; ++i ) rotated[i+j].push(arr[i][j]);
// Print it out.
for( var i=0 ; i<summax ; ++i ) console.log(rotated[i].join(' '))
Output:
1
4 2
7 5 3
8 6
9
In Ruby
Produces same output:
puts arr.transpose.flatten.group_by.with_index { |_,k|
k.divmod(arr.size).inject(:+) }.values.map { |a| a.join ' ' }
function transform(ar) {
var result = [],
i, x, y, row;
for (i = 0; i < ar.length; i++) {
row = [];
for (x = 0, y = i; y >= 0; x++, y--) {
row.push(ar[y][x]);
}
result.push(row);
}
for (i = 1; i < ar[0].length; i++) {
row = [];
for (x = i, y = ar[0].length - 1; x < ar[0].length; x++, y--) {
row.push(ar[y][x]);
}
result.push(row);
}
return result;
}
This returns the rotated array, to print it out as you go just replace each result.push(row); line with console.log(row.join(" "));.
Here's my approach:
var origMatrix = [[1,2,3,4,5], [4,5,6,7,8], [9,10,11,12,13], [14,15,16,17,18], [19,20,21,22,23]];
var maxSize = origMatrix.length;//Presuming all internal are equal!
var rotatedMatrix = [];
var internalArray;
var keyX,keyY,keyArray;
for(var y=0;y<((maxSize * 2)-1);y++){
internalArray = [];
for(var x=0;x<maxSize;x++){
keyX = x;
keyY = y - x;
if(keyY > -1){
keyArray = origMatrix[keyY];
if(typeof(keyArray) != 'undefined' && typeof(keyArray[keyX]) != 'undefined'){
internalArray.push(keyArray[keyX]);
}
}
}
rotatedMatrix.push(internalArray);
}
//log results
for(var i=0;i<rotatedMatrix.length;i++){
console.log(rotatedMatrix[i]);
}
Here's a JSFiddle of it in action (open the Console to see the results)
The Idea: Walk the Diagonals and Clip
You could use the diagonal enumeration from Cantor, see Cantor pairing function,
which is used to show that the set N x N has the same cardinality as the set N (i.e. natural numbers can be mapped one to one to pairs of natural numbers) and combine it with a condition that one skips those values which lie outside the rectangular matrix.
The Cantor pairing function pi takes two natural numbers i and j, i.e. the pair (i, j) and maps it to a natural number k
pi : |N x |N -> |N : pi(i, j) = k
Use the reverse mapping to get this
pi^-1 : |N -> |N x |N : pi^-1(k) = (i, j)
i.e. one enumerates the cells of the "infinite Matrix" N x N diagonally.
So counting up k and applying the inverse function will give the proper pair of indices (i, j) for printing the rotated matrix.
Example:
0->(0, 0) 2->(0, 1) | 5->(0, 2) 9->(0, 3) . .
1->(1, 0) 4->(1, 1) | 8->(1, 2)
3->(2, 0) 7->(2, 2) |
---------------------+ <- clipping for 3 x 2 matrix
6->(3, 0)
.
.
Calculation of the Inverse Cantor Pair Function
For input k these formulas give the pair (i, j):
w = floor((sqrt(8*k + 1) - 1) / 2)
t = (w*w + w) / 2
j = k - t
i = w - j
See the link given above for a derivation.
Resulting Algorithm
Given a m x n matrix A: i from [0, .., m - 1] enumerates the rows, and j from [0, .., n - 1] enumerates the columns
Start at k = 0
calculate the corresponding index pair (i, j)
print the matrix value A[i, j] if the indices i and j lie within your matrix dimensions m and n
print a new line, once your i hit the top of the matrix, i.e. if i == 0
increment k
continue with step 2 until you arrived at the index pair (i, j) = (n - 1, n - 1)
Sample Implementation in JavaScript
Note: I tried this out in the MongoDB shell, using its print() function.
Helper functions
function sprint(k) {
var s = '' + k;
while (s.length < 3) {
s = ' ' + s;
}
return s;
}
function print_matrix(a) {
var m = a.row_size;
var n = a.column_size;
for (var i = 0; i < m; i++) {
var s = '';
for (var j = 0; j < n; j++) {
s += sprint(a.value[i][j]);
}
print(s);
}
}
The Inverse of the Cantor pairing function
// inverse of the Cantor pair function
function pi_inv(k) {
var w = Math.floor((Math.sqrt(8*k + 1) - 1) / 2);
var t = (w*w + w) /2;
var j = k - t;
var i = w -j;
return [i, j];
}
The algorithm
// "rotate" matrix a
function rot(a) {
var m = a.row_size;
var n = a.column_size;
var i_max = m - 1;
var j_max = n - 1;
var k = 0;
var s = '';
do {
var ij = pi_inv(k);
var i = ij[0];
var j = ij[1];
if ((i <= i_max) && (j <= j_max)) {
s += sprint(a.value[i][j]);
}
if (i == 0) {
print(s);
s = '';
}
k += 1
} while ((i != i_max) || (j != j_max));
print(s);
}
Example usage
// example
var a = {
row_size: 4,
column_size: 4,
value: [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]
};
print('in:');
print_matrix(a);
print('out:');
rot(a);
Output for 4x4 Matrix
in:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
out:
1
5 2
9 6 3
13 10 7 4
14 11 8
15 12
16
This method works for any m x n Matrix, e.g. 4 x 5:
in:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
out:
1
6 2
11 7 3
16 12 8 4
17 13 9 5
18 14 10
19 15
20
or 4 x 3:
in:
1 2 3
4 5 6
7 8 9
10 11 12
out:
1
4 2
7 5 3
10 8 6
11 9
12
Maybe i am just not that good enough in math, but I am having a problem in converting a number into pure alphabetical Bijective Hexavigesimal just like how Microsoft Excel/OpenOffice Calc do it.
Here is a version of my code but did not give me the output i needed:
var toHexvg = function(a){
var x='';
var let="_abcdefghijklmnopqrstuvwxyz";
var len=let.length;
var b=a;
var cnt=0;
var y = Array();
do{
a=(a-(a%len))/len;
cnt++;
}while(a!=0)
a=b;
var vnt=0;
do{
b+=Math.pow((len),vnt)*Math.floor(a/Math.pow((len),vnt+1));
vnt++;
}while(vnt!=cnt)
var c=b;
do{
y.unshift( c%len );
c=(c-(c%len))/len;
}while(c!=0)
for(var i in y)x+=let[y[i]];
return x;
}
The best output of my efforts can get is: a b c d ... y z ba bb bc - though not the actual code above. The intended output is suppose to be a b c ... y z aa ab ac ... zz aaa aab aac ... zzzzz aaaaaa aaaaab, you get the picture.
Basically, my problem is more on doing the ''math'' rather than the function. Ultimately my question is: How to do the Math in Hexavigesimal conversion, till a [supposed] infinity, just like Microsoft Excel.
And if possible, a source code, thank you in advance.
Okay, here's my attempt, assuming you want the sequence to be start with "a" (representing 0) and going:
a, b, c, ..., y, z, aa, ab, ac, ..., zy, zz, aaa, aab, ...
This works and hopefully makes some sense. The funky line is there because it mathematically makes more sense for 0 to be represented by the empty string and then "a" would be 1, etc.
alpha = "abcdefghijklmnopqrstuvwxyz";
function hex(a) {
// First figure out how many digits there are.
a += 1; // This line is funky
c = 0;
var x = 1;
while (a >= x) {
c++;
a -= x;
x *= 26;
}
// Now you can do normal base conversion.
var s = "";
for (var i = 0; i < c; i++) {
s = alpha.charAt(a % 26) + s;
a = Math.floor(a/26);
}
return s;
}
However, if you're planning to simply print them out in order, there are far more efficient methods. For example, using recursion and/or prefixes and stuff.
Although #user826788 has already posted a working code (which is even a third quicker), I'll post my own work, that I did before finding the posts here (as i didnt know the word "hexavigesimal"). However it also includes the function for the other way round. Note that I use a = 1 as I use it to convert the starting list element from
aa) first
ab) second
to
<ol type="a" start="27">
<li>first</li>
<li>second</li>
</ol>
:
function linum2int(input) {
input = input.replace(/[^A-Za-z]/, '');
output = 0;
for (i = 0; i < input.length; i++) {
output = output * 26 + parseInt(input.substr(i, 1), 26 + 10) - 9;
}
console.log('linum', output);
return output;
}
function int2linum(input) {
var zeros = 0;
var next = input;
var generation = 0;
while (next >= 27) {
next = (next - 1) / 26 - (next - 1) % 26 / 26;
zeros += next * Math.pow(27, generation);
generation++;
}
output = (input + zeros).toString(27).replace(/./g, function ($0) {
return '_abcdefghijklmnopqrstuvwxyz'.charAt(parseInt($0, 27));
});
return output;
}
linum2int("aa"); // 27
int2linum(27); // "aa"
You could accomplish this with recursion, like this:
const toBijective = n => (n > 26 ? toBijective(Math.floor((n - 1) / 26)) : "") + ((n % 26 || 26) + 9).toString(36);
// Parsing is not recursive
const parseBijective = str => str.split("").reverse().reduce((acc, x, i) => acc + ((parseInt(x, 36) - 9) * (26 ** i)), 0);
toBijective(1) // "a"
toBijective(27) // "aa"
toBijective(703) // "aaa"
toBijective(18279) // "aaaa"
toBijective(127341046141) // "overflow"
parseBijective("Overflow") // 127341046141
I don't understand how to work it out from a formula, but I fooled around with it for a while and came up with the following algorithm to literally count up to the requested column number:
var getAlpha = (function() {
var alphas = [null, "a"],
highest = [1];
return function(decNum) {
if (alphas[decNum])
return alphas[decNum];
var d,
next,
carry,
i = alphas.length;
for(; i <= decNum; i++) {
next = "";
carry = true;
for(d = 0; d < highest.length; d++){
if (carry) {
if (highest[d] === 26) {
highest[d] = 1;
} else {
highest[d]++;
carry = false;
}
}
next = String.fromCharCode(
highest[d] + 96)
+ next;
}
if (carry) {
highest.push(1);
next = "a" + next;
}
alphas[i] = next;
}
return alphas[decNum];
};
})();
alert(getAlpha(27)); // "aa"
alert(getAlpha(100000)); // "eqxd"
Demo: http://jsfiddle.net/6SE2f/1/
The highest array holds the current highest number with an array element per "digit" (element 0 is the least significant "digit").
When I started the above it seemed a good idea to cache each value once calculated, to save time if the same value was requested again, but in practice (with Chrome) it only took about 3 seconds to calculate the 1,000,000th value (bdwgn) and about 20 seconds to calculate the 10,000,000th value (uvxxk). With the caching removed it took about 14 seconds to the 10,000,000th value.
Just finished writing this code earlier tonight, and I found this question while on a quest to figure out what to name the damn thing. Here it is (in case anybody feels like using it):
/**
* Convert an integer to bijective hexavigesimal notation (alphabetic base-26).
*
* #param {Number} int - A positive integer above zero
* #return {String} The number's value expressed in uppercased bijective base-26
*/
function bijectiveBase26(int){
const sequence = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const length = sequence.length;
if(int <= 0) return int;
if(int <= length) return sequence[int - 1];
let index = (int % length) || length;
let result = [sequence[index - 1]];
while((int = Math.floor((int - 1) / length)) > 0){
index = (int % length) || length;
result.push(sequence[index - 1]);
}
return result.reverse().join("")
}
I had to solve this same problem today for work. My solution is written in Elixir and uses recursion, but I explain the thinking in plain English.
Here are some example transformations:
0 -> "A", 1 -> "B", 2 -> "C", 3 -> "D", ..
25 -> "Z", 26 -> "AA", 27 -> "AB", ...
At first glance it might seem like a normal 26-base counting system
but unfortunately it is not so simple.
The "problem" becomes clear when you realize:
A = 0
AA = 26
This is at odds with a normal counting system, where "0" does not behave
as "1" when it is in a decimal place other than then unit.
To understand the algorithm, consider a simpler but equivalent base-2 system:
A = 0
B = 1
AA = 2
AB = 3
BA = 4
BB = 5
AAA = 6
In a normal binary counting system we can determine the "value" of decimal places by
taking increasing powers of 2 (1, 2, 4, 8, 16) and the value of a binary number is
calculated by multiplying each digit by that digit place's value.
e.g. 10101 = 1 * (2 ^ 4) + 0 * (2 ^ 3) + 1 * (2 ^ 2) + 0 * (2 ^ 1) + 1 * (2 ^ 0) = 21
In our more complicated AB system, we can see by inspection that the decimal place values are:
1, 2, 6, 14, 30, 62
The pattern reveals itself to be (previous_unit_place_value + 1) * 2.
As such, to get the next lower unit place value, we divide by 2 and subtract 1.
This can be extended to a base-26 system. Simply divide by 26 and subtract 1.
Now a formula for transforming a normal base-10 number to special base-26 is apparent.
Say the input is x.
Create an accumulator list l.
If x is less than 26, set l = [x | l] and go to step 5. Otherwise, continue.
Divide x by 2. The floored result is d and the remainder is r.
Push the remainder as head on an accumulator list. i.e. l = [r | l]
Go to step 2 with with (d - 1) as input, e.g. x = d - 1
Convert """ all elements of l to their corresponding chars. 0 -> A, etc.
So, finally, here is my answer, written in Elixir:
defmodule BijectiveHexavigesimal do
def to_az_string(number, base \\ 26) do
number
|> to_list(base)
|> Enum.map(&to_char/1)
|> to_string()
end
def to_09_integer(string, base \\ 26) do
string
|> String.to_charlist()
|> Enum.reverse()
|> Enum.reduce({0, nil}, fn
char, {_total, nil} ->
{to_integer(char), 1}
char, {total, previous_place_value} ->
char_value = to_integer(char + 1)
place_value = previous_place_value * base
new_total = total + char_value * place_value
{new_total, place_value}
end)
|> elem(0)
end
def to_list(number, base, acc \\ []) do
if number < base do
[number | acc]
else
to_list(div(number, base) - 1, base, [rem(number, base) | acc])
end
end
defp to_char(x), do: x + 65
end
You use it simply as BijectiveHexavigesimal.to_az_string(420). It also accepts on optional "base" arg.
I know the OP asked about Javascript but I wanted to provide an Elixir solution for posterity.
I have published these functions in npm package here:
https://www.npmjs.com/package/#gkucmierz/utils
Converting bijective numeration to number both ways (also BigInt version is included).
https://github.com/gkucmierz/utils/blob/main/src/bijective-numeration.mjs
I am working on looping through a few grid items and was hoping someone could help me find a way to figure out how to get to the items that I need and possibly I will better understand how to access the items in the grid. So here is the grid.
[0] [1] [2] [3] [4]
[5] [6] [7] [8] [9]
[10] [11] [12] [13] [14]
[15] [16] [17] [18] [19]
[20] [21] [22] [23] [24]
This is basically a 5x5 grid, however it could be any size but I just used this for the example. There are two ways I would like to loop through this. The first one being in this order:
0,1,2,3,4,9,14,19,24,23,22,21,20,15,10,5,6,7,8,13,18,17,16,11,12
Basically all that is doing is going around the outside starting from the top left. The next way I would want to loop through that is through the same exact values except in reverse order (basically inside out instead of outside in) and actually thinking about this now I could just loop through the first method backwards. If anyone can help me with this it would be great. I really want to learn more on how to loop through items in crazy arrangements like this.
This function
function n(i, w, h)
{
if (i < w)
return i;
if (i < w + h-1)
return w-1 + (i-w+1)*w;
if (i < w + h-1 + w-1)
return w-1 + (h-1)*w - (i - (w + h-1 - 1));
if (i < w + h-1 + w-1 + h-2)
return (h - (i - (w + w-1 + h-1 - 2)))*w;
var r = n(i - (w-1)*2 - (h-1)*2, w-2, h-2);
var x = r % (w-2);
var y = Math.floor(r / (w-2));
return (y+1)*w + (x+1);
}
accepts as input
i: Index of the item you're looking for
w: Width of the grid
h: Height of the grid
and returns the corresponding element of the grid assuming that clock-wise spiral traversal.
The implementation simply checks if we're on the top side (i<w), on the downward right side (i<w+h-1) and so on and for these cases it computes the cell element explicitly.
If we complete one single trip around the spiral then it calls recursively itself to find the element in the inner (w-2)*(h-2) grid and then extracts and adjusts the two coordinates considering the original grid size.
This is much faster for big grids than just iterating and emulating the spiral walk, for example computing n(123121, 1234, 3012) is immediate while the complete grid has 3712808 elements and a walk of 123121 steps would require quite a long time.
You just need a way to represent the traversal pattern.
Given an NxM matrix (e.g. 5x5), the pattern is
GENERAL 5x5
------- -------
N right 5
M-1 down 4
N-1 left 4
M-2 up 3
N-2 right 3
M-3 down 2
N-3 left 2
M-4 up 1
N-4 right 1
This says move 5 to the right, 4 down, 4 left, 3 up, 3 right, 2 down, 2 left, 1 up, 1 right. The step size shifts after each two iterations.
So, you can track the current "step-size" and the current direction while decrementing N, M until you reach the end.
IMPORTANT: make sure to write down the pattern for a non-square matrix to see if the same pattern applies.
Here's the "walking" method. Less efficient, but it works.
var arr = new Array();
for(var n=0; n<25; n++) arr.push(n);
var coords = new Array();
var x = 0;
var y = 0;
for(var i=0; i<arr.length; i++) {
if( x > 4 ) {
x = 0;
y++;
}
coords[i] = {'x': x, 'y': y};
x++;
}
// okay, coords contain the coordinates of each item in arr
// need to move along the perimeter until a collision, then turn.
// start at 0,0 and move east.
var dir = 0; // 0=east, 1=south, 2=west, 3=north.
var curPos = {'x': 0, 'y': 0};
var resultList = new Array();
for(var x=0; x<arr.length; x++) {
// record the current position in results
var resultIndex = indexOfCoords(curPos, coords);
if(resultIndex > -1) {
resultList[x] = arr[resultIndex];
}
else {
resultList[x] = null;
}
// move the cursor to a valid position
var tempCurPos = movePos(curPos, dir);
var outOfBounds = isOutOfBounds(tempCurPos, coords);
var itemAtTempPos = arr[indexOfCoords(tempCurPos, coords)];
var posInList = resultList.indexOf( itemAtTempPos );
if(outOfBounds || posInList > -1) {
dir++;
if(dir > 3) dir=0;
curPos = movePos(curPos, dir);
}
else {
curPos = tempCurPos;
}
}
/* int indexOfCoords
*
* Searches coordList for a match to myCoords. If none is found, returns -1;
*/
function indexOfCoords(myCoords, coordsList) {
for(var i=0; i<coordsList.length; i++) {
if(myCoords.x == coordsList[i].x && myCoords.y == coordsList[i].y) return i;
}
return -1;
}
/* obj movePos
*
* Alters currentPosition by incrementing it 1 in the direction provided.
* Valid directions are 0=east, 1=south, 2=west, 3=north
* Returns the resulting coords as an object with x, y.
*/
function movePos(currentPosition, direction) {
var newPosition = {'x':currentPosition.x, 'y':currentPosition.y};
if(direction == 0) {
newPosition.x++;
}
else if(direction == 1) {
newPosition.y++;
}
else if(direction == 2) {
newPosition.x--;
}
else if(direction == 3) {
newPosition.y--;
}
return newPosition;
}
/* bool isOutOfBounds
*
* Compares the x and y coords of a given position to the min/max coords in coordList.
* Returns true if the provided position is outside the boundaries of coordsList.
*
* NOTE: This is just one, lazy way of doing this. There are others.
*/
function isOutOfBounds(position, coordsList) {
// get min/max
var minx=0, miny=0, maxx=0, maxy=0;
for(var i=0; i<coordsList.length; i++) {
if(coordsList[i].x > maxx) maxx = coordsList[i].x;
else if(coordsList[i].x < minx) minx = coordsList[i].x;
if(coordsList[i].y > maxy) maxy = coordsList[i].y;
else if(coordsList[i].y < miny) miny = coordsList[i].y;
}
if(position.x < minx || position.x > maxx || position.y < miny || position.y > maxy) return true;
else return false;
}
This will move through the grid in a direction until it hits the bounds or an item already in the results array, then turn clockwise. It's pretty rudimentary, but I think it would do the job. You could reverse it pretty simply.
Here's a working example: http://www.imakewidgets.com/test/walkmatrix.html