Bliffoscope Data Analysis Solution Using Javascript
I want to solve Bliffoscope Data Analysis Problem using javascript. I have following question. This is SlimeTorpedo + + +++ +++++++ ++ ++ ++ + ++
Solution 1:
Are you looking for something like this (Fiddle)?
// Create our images: torpedo (object) and background (context)var object = " +\n +\n +++\n +++++++\n ++ ++\n++ + ++\n++ +++ ++\n++ + ++\n ++ ++\n +++++++\n +++",
context = " + + + ++ + +++ + +\n + ++ + + ++++ + + + + + + +++ +++ +\n + + + ++ ++ ++ + ++ + + + + +\n+ ++ + ++ + + + ++ ++ + +\n ++++++ + + + ++ + + + + ++ + + +\n + + + + + ++ + ++ + + + +\n+++ + ++ + + + +++ + + ++ +\n +++++ + + + + + + + +\n + + + + + + + + + + + +\n ++ + + + ++ + + + ++ ";
var c = document.getElementById("test_canvas"),
ctx = c.getContext("2d"),
scale = 10;
// Draw a pixel on canvasfunctiondraw_pixel(x, y, fill_style) {
ctx.fillStyle = fill_style;
ctx.fillRect(x * scale, y * scale, scale, scale);
}
// Receive an array of coordinates, draw pixelsfunctiondraw_image(serialized_image, fill_style) {
for (var i = 0, len = serialized_image.length; i < len; i++) {
draw_pixel(serialized_image[i][0], serialized_image[i][1], fill_style);
}
}
// Receive a text string, turn it into an array of coordinates of filled in pixelsfunctionserialize_map(char_map) {
var x = 0,
y = 0,
c,
map = [];
for (var i = 0, len = char_map.length; i < len; i++) {
c = char_map[i];
if (c == '+') {
map.push([x, y])
}
x += 1;
if (c == '\n') {
x = 0;
y += 1;
}
}
return map;
}
// Find number of intersections between two imagesfunctionarray_intersect() {
var a, d, b, e, h = [],
f = {},
g;
g = arguments.length - 1;
b = arguments[0].length;
for (a = d = 0; a <= g; a++) {
e = arguments[a].length, e < b && (d = a, b = e);
}
for (a = 0; a <= g; a++) {
e = a === d ? 0 : a || d;
b = arguments[e].length;
for (var l = 0; l < b; l++) {
var k = arguments[e][l];
f[k] === a - 1 ? a === g ? (h.push(k), f[k] = 0) : f[k] = a : 0 === a && (f[k] = 0);
}
}
return h;
}
// Translate the coordinates of a serialized imagefunctiontranslate(coords, ix, iy) {
return [coords[0] + ix, coords[1] + iy];
}
// Find in which position the object has more intersections with the backgroundfunctionget_best_position(context, object) {
// Calculate image dimensions
context_width = context.sort(function(a, b) {
return b[0] - a[0];
})[0][0];
context_height = context.sort(function(a, b) {
return b[1] - a[1];
})[0][1];
object_width = object.sort(function(a, b) {
return b[0] - a[0];
})[0][0];
object_height = object.sort(function(a, b) {
return b[1] - a[1];
})[0][1];
// Swipe context, store amount of matches for each patch position
similaritudes = [];
for (var cx = 0; cx < context_width; cx++) {
for (var cy = 0; cy < context_height; cy++) {
translated_object = object.map(function(coords) {
returntranslate(coords, cx, cy);
})
intersection = array_intersect(context, translated_object);
// console.log(translated_object);
similaritudes[intersection.length] = [cx, cy];
}
}
// Return position for which number of matches was greaterreturn similaritudes.slice(-1)[0];
}
// Parse our images from the text stringsvar serialized_context = serialize_map(context);
var serialized_object = serialize_map(object);
// Find best position for our torpedovar best_position = get_best_position(serialized_context, serialized_object);
// Translate torpedo to best position
positioned_object = serialized_object.map(function(coords) {
returntranslate(coords, best_position[0], best_position[1]);
});
// Draw background and torpedo draw_image(serialized_context, "gray");
draw_image(positioned_object, "rgba(0, 255, 0, 0.5)");
<canvasid="test_canvas"width="800"height="120"style="border:1px solid #000000;"></canvas>
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