Diberikan daftar koordinat 2d (x, y), tentukan berapa banyak kuadrat satuan (panjang tepi 1 unit) yang dapat dibentuk menggunakan koordinat.
- Input akan berupa larik 0 atau lebih pasangan koordinat:
misalnya dalam JavaScript:numofsq([[0,0], [1,0], [1,1], [0,1]]) - Tidak ada koordinat duplikat dalam input
- Urutan input akan acak (koordinat 2d acak).
- Bentuk koordinat: [koordinat x, koordinat y] (duh)
- Urutan koordinat: [0,0], [1,0], [1,1], [0,1] dan [0,0], [0,1], [1,1], [1,0 ] menunjukkan satuan satuan yang sama (dihitung sekali saja)
- Koordinat dapat berupa bilangan bulat negatif atau positif (duh)
- Fungsi hanya akan mengembalikan jumlah kuadrat unit yang mungkin, 0 atau lebih.
Kasus uji:
Input Coordinates Pairs Expected Output
[0,0], [0,1], [1,1], [1,0], [0,2], [1,2] 2
[0,0], [0,1], [1,1], [1,0], [0,2], [1,2], [2,2], [2,1] 3
[0,0], [0,1], [1,1], [1,0], [0,2], [1,2], [2,2], [2,1], [2,0] 4
[0,0], [0,1], [1,1], [1,0], [0,2], [1,2], [2,2], [2,1], [2,0], [9,9] 4
Peringatan Spoiler: Hal-hal solusi di sini-seterusnya [JS]
Pendekatan non-Golf, cepat dan kotor, brute-force (termasuk untuk memberikan beberapa arahan).
//cartesian distance function
function dist(a, b) {
if (b === undefined) {
b = [0, 0];
}
return Math.sqrt((b[0] - a[0]) * (b[0] - a[0]) + (b[1] - a[1]) * (b[1] - a[1]));
}
//accepts 4 coordinate pairs and checks if they form a unit square
//this could be optimized by matching x,y coordinates of the 4 coordinates
function isUnitSquare(a) {
var r = a.slice(),
d = [],
c = [],
i,
j = 0,
counter = 0;
for (i = 1; i < 4; i++) {
if (dist(a[0], a[i]) === 1) {
d.push(a[i]);
r[i] = undefined;
counter++;
}
}
r[0] = undefined;
c = d.concat(r.filter(function(a) {return undefined !== a}));
if (dist(c[0], c[1]) === 1) {j++};
if (dist(c[1], c[2]) === 1) {j++};
if (dist(c[2], c[0]) === 1) {j++};
return counter === 2 && j === 2;
}
//a powerset function (from rosetta code)
//however, we will need only "sets of 4 coordinates"
//and not all possible length combinations (sets of 3 coords or
//sets of 5 coords not required). Also, order doesn't matter.
function powerset(ary) {
var ps = [[]];
for (var i=0; i < ary.length; i++) {
for (var j = 0, len = ps.length; j < len; j++) {
ps.push(ps[j].concat(ary[i]));
}
}
return ps;
}
//so to capture only sets of 4 coordinates, we do
var res = powerset([[0,0], [0,1], [1,1], [1,0], [0,2], [1,2], [2,2], [2,1], [2,0]])
.filter(function (a) {return a.length === 8;});
//and a little bit of hoopla to have a nice set of sets of 4 coordinates.
//(Dizzy yet? Wait for the generalized, 3D, cube of any edge length version ;))
var squareQuads = res.map(function(ary) {
return ary.join().match(/\d\,\d/g)
.map(function(e) {
return [+e.split(',')[0], +e.split(',')[1]];
});
});
//Finally, the last bit
var howManyUnitSquares = 0;
squareQuads.map(function(quad) {
if (isUnitSquare(quad)) {
howManyUnitSquares++;
}
});
console.log(howManyUnitSquares);
//Cleaning up and putting those in-line stuff into a function
function howManySquares(r) { //r = [[x,y], [x,y], [x,y], [x,y], ......];
var res = powerset(r)
.filter(function (a) {return a.length === 8;});
var squareQuads = res.map(function(ary) {
return ary.join().match(/\d\,\d/g)
.map(function(e) {
return [+e.split(',')[0], +e.split(',')[1]];
});
});
var answer = 0;
squareQuads.map(function(quad) {
if (isUnitSquare(quad)) {
answer++;
}
});
return answer;
}
@JohannesKuhn Tidak, panjang tepi mereka bukan kesatuan, tetapi sqrt (2), jadi mereka bukan kotak satuan.
Tetapi
—
Johannes Kuhn
[-2,0],[0,-2],[2,0],[0,2]memiliki panjang tepi 2. Kotak?
[-1,0],[0,-1],[1,0],[0,1]kotak?