Algorithm to find best dimensions combination - algorithm

I am looking for an algorithm to find the best dimension combination to accomplish a desired result.
Take the following as example:
| A | B | C | y |
|--------|--------|-------|-----|
| dog | house1 | green | 30 |
| dog | house1 | blue | 15 |
| cat | house1 | green | 20 |
| cat | house2 | red | 5 |
| turtle | house3 | green | 50 |
A, B, C are the measured dimensions.
y is the measured result.
If I want to get all combinations of dimensions that accomplish y >= 50 so the results will be:
turtle, house3, green
turtle, any, green
turtle, house3, any
turtle, any, any
any, house3, green
any, house3, any
any, any, green
any, house1, green
any, house1, any
Maybe it's a easy problem but I was trying to figure an optimal solution in terms of O(n) and I didn't found it.

Start with a work queue containing (any, any, ..., any), 0. The elements of this queue will be pairs consisting of a combination and a number of elements on the left that cannot be changed from any (this will make more sense shortly). Until the work queue is empty, remove one element from it and compute the corresponding sum. If it doesn't meet the threshold, then discard it. Otherwise, report it as one of the sought combinations. For each any that can be changed, for each value in that column, enqueue a combination consisting of the current one with any replaced by that value, with the index locking down all previous any values.
Considering an output-sensitive bound, this is within a polynomial factor of optimal (in general, there can be exponentially many combinations).
In Python 3:
def overthreshold(data, threshold):
queue = [(('any',) * len(data[0][0]), 0)]
for combination, begin in queue:
if sum(row[1] for row in data
if all(x in {'any', y}
for x, y in zip(combination, row[0]))) < threshold:
continue
yield combination
for i in range(begin, len(combination)):
if combination[i] == 'any':
queue.extend((combination[:i] + (x,) + combination[i+1:], i + 1)
for x in {row[0][i] for row in data})
def demo():
data = [
(('dog', 'house1', 'green'), 30),
(('dog', 'house1', 'blue'), 15),
(('cat', 'house1', 'green'), 20),
(('cat', 'house2', 'red'), 5),
(('turtle', 'house3', 'green'), 50),
]
for combination in overthreshold(data, 50):
print(combination)

Back here, 8 years later to answer the question using ClickHouse:
WITH table AS (
SELECT 'dog' AS a, 'house1' AS b, 'green' AS c, 30 AS y
UNION ALL
SELECT 'dog' AS a, 'house1' AS b, 'blue' AS c, 15 AS y
UNION ALL
SELECT 'cat' AS a, 'house1' AS b, 'green' AS c, 20 AS y
UNION ALL
SELECT 'cat' AS a, 'house2' AS b, 'red' AS c, 5 AS y
UNION ALL
SELECT 'turtle' AS a, 'house3' AS b, 'green' AS c, 50 AS y
)
SELECT a, b, c, sum(y) y FROM table GROUP BY CUBE(a, b, c)
HAVING y >= 50
FORMAT PrettyCompactMonoBlock;
┌─a──────┬─b──────┬─c─────┬───y─┐
│ turtle │ house3 │ green │ 50 │
│ turtle │ house3 │ │ 50 │
│ turtle │ │ green │ 50 │
│ turtle │ │ │ 50 │
│ │ house3 │ green │ 50 │
│ │ house1 │ green │ 50 │
│ │ house3 │ │ 50 │
│ │ house1 │ │ 65 │
│ │ │ green │ 100 │
│ │ │ │ 120 │
└────────┴────────┴───────┴─────┘

Related

How to dynamically layout atoms in atomic orbitals given a certain set of desired layouts?

I am laying this out as a square:
Typically you see it it in a circle, which is easy you just distribute the electrons evenly.
But in my square case, the outer interior of each square I want to put small squares to layout the electrons according to the Electron shells. I want to do this somewhat dynamically without having to write out each of the 118 atom cases manually.
The desired layouts are like this if there are 8 electrons in a shell:
x x x
x x
x x x
If there are only 7, it should be like this:
x x x
x x
x x
Then these cases:
6:
x - x
x x
x - x
5:
- x -
x x
x - x
4:
- x -
x x
- x -
3:
- x -
- -
x - x
2:
- - -
x x
- - -
1:
- x -
- -
- - -
Then the 18 case is:
x x x x x
x x
x x
x x
x x
x x x x x
x x - x x
x x
x x
x x
x x
x x x x x
x x - x x
x x
x x
x x
x x
x x - x x
x - x - x
x x
x x
x x
x x
x x - x x
x - x - x
x x
x x
x x
x x
x - x - x
I don't have a super-hard-fast desire to layout each configuration in a specific way (even though I started by showing specific configurations). I am mainly looking to figure out some sort of pattern or sort of equation to lay them out in a semi-nice/decent way. How can it be done?
The electron shells are like this:
const SHELLS = `Hydrogen,1
Helium,2
Lithium,2:1
Beryllium,2:2
Boron,2:3
Carbon,2:4
Nitrogen,2:5
Oxygen,2:6
Fluorine,2:7
Neon,2:8
Sodium,2:8:1
Magnesium,2:8:2
Aluminium,2:8:3
Silicon,2:8:4
Phosphorus,2:8:5
Sulfur,2:8:6
Chlorine,2:8:7
Argon,2:8:8
Potassium,2:8:8:1
Calcium,2:8:8:2
Scandium,2:8:9:2
Titanium,2:8:10:2
Vanadium,2:8:11:2
Chromium,2:8:13:1
Manganese,2:8:13:2
Iron,2:8:14:2
Cobalt,2:8:15:2
Nickel,2:8:16:2
Copper,2:8:18:1
Zinc,2:8:18:2
Gallium,2:8:18:3
Germanium,2:8:18:4
Arsenic,2:8:18:5
Selenium,2:8:18:6
Bromine,2:8:18:7
Krypton,2:8:18:8
Rubidium,2:8:18:8:1
Strontium,2:8:18:8:2
Yttrium,2:8:18:9:2
Zirconium,2:8:18:10:2
Niobium,2:8:18:12:1
Molybdenum,2:8:18:13:1
Technetium,2:8:18:13:2
Ruthenium,2:8:18:15:1
Rhodium,2:8:18:16:1
Palladium,2:8:18:18
Silver,2:8:18:18:1
Cadmium,2:8:18:18:2
Indium,2:8:18:18:3
Tin,2:8:18:18:4
Antimony,2:8:18:18:5
Tellurium,2:8:18:18:6
Iodine,2:8:18:18:7
Xenon,2:8:18:18:8
Caesium,2:8:18:18:8:1
Barium,2:8:18:18:8:2
Lanthanum,2:8:18:18:9:2
Cerium,2:8:18:19:9:2
Praseodymium,2:8:18:21:8:2
Neodymium,2:8:18:22:8:2
Promethium,2:8:18:23:8:2
Samarium,2:8:18:24:8:2
Europium,2:8:18:25:8:2
Gadolinium,2:8:18:25:9:2
Terbium,2:8:18:27:8:2
Dysprosium,2:8:18:28:8:2
Holmium,2:8:18:29:8:2
Erbium,2:8:18:30:8:2
Thulium,2:8:18:31:8:2
Ytterbium,2:8:18:32:8:2
Lutetium,2:8:18:32:9:2
Hafnium,2:8:18:32:10:2
Tantalum,2:8:18:32:11:2
Tungsten,2:8:18:32:12:2
Rhenium,2:8:18:32:13:2
Osmium,2:8:18:32:14:2
Iridium,2:8:18:32:15:2
Platinum,2:8:18:32:17:1
Gold,2:8:18:32:18:1
Mercury,2:8:18:32:18:2
Thallium,2:8:18:32:18:3
Lead,2:8:18:32:18:4
Bismuth,2:8:18:32:18:5
Polonium,2:8:18:32:18:6
Astatine,2:8:18:32:18:7
Radon,2:8:18:32:18:8
Francium,2:8:18:32:18:8:1
Radium,2:8:18:32:18:8:2
Actinium,2:8:18:32:18:9:2
Thorium,2:8:18:32:18:10:2
Protactinium,2:8:18:32:20:2
Uranium,2:8:18:32:21:9:2
Neptunium,2:8:18:32:22:9:2
Plutonium,2:8:18:32:24:8:2
Americium,2:8:18:32:25:8:2
Curium,2:8:18:32:25:9:2
Berkelium,2:8:18:32:27:8:2
Californium,2:8:18:32:28:8:2
Einsteinium,2:8:18:32:29:8:2
Fermium,2:8:18:32:30:8:2
Mendelevium,2:8:18:32:31:8:2
Nobelium,2:8:18:32:32:8:2
Lawrencium,2:8:18:32:32:8:3
Rutherfordium,2:8:18:32:32:10:2
Dubnium,2:8:18:32:32:11:2
Seaborgium,2:8:18:32:32:12:2
Bohrium,2:8:18:32:32:13:2
Hassium,2:8:18:32:32:14:2
Meitnerium,2:8:18:32:32:15:2
Darmstadtium,2:8:18:32:32:16:2
Roentgenium,2:8:18:32:32:17:2
Copernicium,2:8:18:32:32:18:2
Nihonium,2:8:18:32:32:18:3
Flerovium,2:8:18:32:32:18:4
Moscovium,2:8:18:32:32:18:5
Livermorium,2:8:18:32:32:18:6
Tennessine,2:8:18:32:32:18:7
Oganesson,2:8:18:32:32:18:8`
.trim()
.split('\n')
.map(x => {
const [a, b] = x.split(',')
const c = b.split(':').map(x => parseInt(x, 10))
return { name: a, shells: c }
})
Is it possible do you think to come up with a simple algorithm for this, or must it be hardcoded?
Some constraints:
The shells have 2, 8, 18, 32, 32 electrons.
The electrons should go into preexisting slots, so there are only 8 slots for the 8, 18 for the 18, etc.. That is, you can't evenly distribute them around the edge.
Other than that, the general layout should feel somewhat "balanced" (even though that is a fuzzy concept). So if there is just 17, it should take out one from the middle vertically. If there are only 3, it should make them into a triangle sort of thing. I don't see a way out of defining this manually, but I am sure there is a way to do it with some clever perspective.
There might be multiple equally "balanced" ways of creating a layout, so it doesn't matter to me exactly which one is chosen.
It can be simulated just laying out x and - in a monospaced font, so don't need to full Next.js/React/SVG system that I am dealing with currently. Any help would be greatly appreciated, I am stumped.
You could first solve the problem without any actual rendering considerations, but see the orbits as 1-dimensional arrays of bits -- let's say a string of "x" and "-". The inner orbit could have three possibilities:
"--"
"x-"
"xx"
The next orbit would have these:
"--------",
"x-------",
"x---x---",
"x--x-x--",
"x-x-x-x-",
"x-xx-xx-",
"-xxx-xxx",
"xxxx-xxx",
"xxxxxxxx"
To distribute the "x" evenly you would step with fractions of the string length over the desired number of "x". To avoid irregular shapes, you could mirror positions as soon as you find them, and stop generating more when together with the mirrored positions you have them all.
Once you have generated all these bit patterns, we can focus on the format of the squares layout. For that you could define a multiline string that is a template for how you generally want to render it, using specific placeholders for where you want the electron slots to be. So "a" in that string would depict a slot in the inner orbit, "b" in the second one, ...etc. For instance:
┌───────────────────────────────────────────────────┐
│ g │
│ ┌───────────────────────────────────────────┐ │
│ │ f f f f f │ │
│ │ ┌───────────────────────────────────┐ │ │
│ g │ f │ e e e e e e e e e e e │ f │ g │
│ │ │ ┌───────────────────────────┐ │ │ │
│ │ │ │ d d d d d d d d d │ │ │ │
│ │ │ e │ d ┌───────────────────┐ d │ e │ │ │
│ │ │ │ │ c c c c c │ │ │ │ │
│ │ │ │ d │ ┌───────────┐ │ d │ │ │ │
│ │ f │ e │ │ c │ b │ c │ │ e │ f │ │
│ │ │ │ d │ │ b ┌───┐ b │ │ d │ │ │ │
│ │ │ │ │ c │ │ a │ │ c │ │ │ │ │
│ g │ │ e │ d │ │ b │ │ b │ │ d │ e │ │ g │
│ │ │ │ │ c │ │ a │ │ c │ │ │ │ │
│ │ │ │ d │ │ b └───┘ b │ │ d │ │ │ │
│ │ f │ e │ │ c │ b │ c │ │ e │ f │ │
│ │ │ │ d │ └───────────┘ │ d │ │ │ │
│ │ │ │ │ c c c c c │ │ │ │ │
│ │ │ e │ d └───────────────────┘ d │ e │ │ │
│ │ │ │ d d d d d d d d d │ │ │ │
│ │ │ └───────────────────────────┘ │ │ │
│ g │ f │ e e e e e e e e e e e │ f │ g │
│ │ └───────────────────────────────────┘ │ │
│ │ f f f f f │ │
│ └───────────────────────────────────────────┘ │
│ g │
└───────────────────────────────────────────────────┘
I just had some fun with those box drawing characters, but you can of course use an entirely different string. The only requirement is that there are two "a" characters in it, 8 "b", 18 "c", ...etc. All the other characters can be anything.
A little function can find the positions of the slots in the orbits and then use the bit patterns to place the desired character ("x" or "-") at the appropriate slot.
Here is an interactive implementation of that idea:
const templateInput = `
┌───────────────────────────────────────────────────┐
│ g │
│ ┌───────────────────────────────────────────┐ │
│ │ f f f f f │ │
│ │ ┌───────────────────────────────────┐ │ │
│ g │ f │ e e e e e e e e e e e │ f │ g │
│ │ │ ┌───────────────────────────┐ │ │ │
│ │ │ │ d d d d d d d d d │ │ │ │
│ │ │ e │ d ┌───────────────────┐ d │ e │ │ │
│ │ │ │ │ c c c c c │ │ │ │ │
│ │ │ │ d │ ┌───────────┐ │ d │ │ │ │
│ │ f │ e │ │ c │ b │ c │ │ e │ f │ │
│ │ │ │ d │ │ b ┌───┐ b │ │ d │ │ │ │
│ │ │ │ │ c │ │ a │ │ c │ │ │ │ │
│ g │ │ e │ d │ │ b │ │ b │ │ d │ e │ │ g │
│ │ │ │ │ c │ │ a │ │ c │ │ │ │ │
│ │ │ │ d │ │ b └───┘ b │ │ d │ │ │ │
│ │ f │ e │ │ c │ b │ c │ │ e │ f │ │
│ │ │ │ d │ └───────────┘ │ d │ │ │ │
│ │ │ │ │ c c c c c │ │ │ │ │
│ │ │ e │ d └───────────────────┘ d │ e │ │ │
│ │ │ │ d d d d d d d d d │ │ │ │
│ │ │ └───────────────────────────┘ │ │ │
│ g │ f │ e e e e e e e e e e e │ f │ g │
│ │ └───────────────────────────────────┘ │ │
│ │ f f f f f │ │
│ └───────────────────────────────────────────┘ │
│ g │
└───────────────────────────────────────────────────┘
`.trim();
function parseTemplate(template) {
function shellPattern(length, count) {
const arr = [];
const even = 1 - count % 2;
const symbols = count > length >> 1 && count < length ? "x-" : "-x";
count = symbols[0] == "x" ? length - count : count;
for (let j = 0, step = 0; true; j++) {
const surpass = +((j + 0.5) * count >= step);
const symbol = symbols[surpass];
if (surpass) step += length;
arr[j] = symbol;
if (j) { // Mirror left-right
if (length - j <= j) break;
arr[length - j] = symbol;
}
if (even) { // Mirror top-bottom
if ((length >> 1) - j <= j) break;
arr[(length >> 1) + j] = arr[(length >> 1) - j] = symbol;
}
}
const pat = arr.join("");
// Turn 180° if top cell is not occupied:
return pat[0] == "-" ? pat.slice(length >> 1) + pat.slice(0, length >> 1) : pat;
}
return {
shells: Array.from("abcdefg", ch => {
const forward = [];
const backward = [];
let i = 0;
template.split(/^/gm).forEach((line, y, {length}) => {
const hits = Array.from(line.matchAll(ch), ({index}) => i + index);
if (y * 2 < length) {
backward.push(...hits.slice(0, hits.length >> 1).reverse());
forward.push(...hits.slice(hits.length >> 1));
} else {
backward.push(...hits.slice(0, hits.length >> 1));
forward.push(...hits.slice(hits.length >> 1).reverse());
}
i += line.length;
});
const indices = forward.concat(backward.reverse());
return {
indices,
patterns: ['-'.repeat(indices.length),
...indices.map((_, count) => shellPattern(indices.length, count+1))]
}
}),
template: template.replace(/[a-g]/g, ".")
};
}
function toGrid(model, element) {
const shells = element?.match(/\d+/g)?.map(Number) ?? [];
const arr = [...model.template];
shells.forEach((count, shellNum) => {
const {patterns, indices} = model.shells[shellNum];
Array.from(patterns[count], (ch, i) => arr[indices[i]] = ch);
});
return arr.join("");
}
const model = parseTemplate(templateInput);
// I/O handling
const input = document.querySelector("select");
input.onchange = () =>
document.querySelector("pre").textContent = toGrid(model, input.value);
input.onchange();
pre { font-size: 8px; display: inline-block; width = 50hv; float: left }
<pre></pre>
<select multiple size="15">
<option>Hydrogen,1
<option>Helium,2
<option>Lithium,2:1
<option>Beryllium,2:2
<option>Boron,2:3
<option>Carbon,2:4
<option>Nitrogen,2:5
<option>Oxygen,2:6
<option>Fluorine,2:7
<option>Neon,2:8
<option>Sodium,2:8:1
<option>Magnesium,2:8:2
<option>Aluminium,2:8:3
<option>Silicon,2:8:4
<option>Phosphorus,2:8:5
<option>Sulfur,2:8:6
<option>Chlorine,2:8:7
<option>Argon,2:8:8
<option>Potassium,2:8:8:1
<option>Calcium,2:8:8:2
<option>Scandium,2:8:9:2
<option>Titanium,2:8:10:2
<option>Vanadium,2:8:11:2
<option>Chromium,2:8:13:1
<option>Manganese,2:8:13:2
<option>Iron,2:8:14:2
<option>Cobalt,2:8:15:2
<option>Nickel,2:8:16:2
<option>Copper,2:8:18:1
<option>Zinc,2:8:18:2
<option>Gallium,2:8:18:3
<option>Germanium,2:8:18:4
<option>Arsenic,2:8:18:5
<option>Selenium,2:8:18:6
<option>Bromine,2:8:18:7
<option>Krypton,2:8:18:8
<option>Rubidium,2:8:18:8:1
<option>Strontium,2:8:18:8:2
<option>Yttrium,2:8:18:9:2
<option>Zirconium,2:8:18:10:2
<option>Niobium,2:8:18:12:1
<option>Molybdenum,2:8:18:13:1
<option>Technetium,2:8:18:13:2
<option>Ruthenium,2:8:18:15:1
<option>Rhodium,2:8:18:16:1
<option>Palladium,2:8:18:18
<option>Silver,2:8:18:18:1
<option>Cadmium,2:8:18:18:2
<option>Indium,2:8:18:18:3
<option>Tin,2:8:18:18:4
<option>Antimony,2:8:18:18:5
<option>Tellurium,2:8:18:18:6
<option>Iodine,2:8:18:18:7
<option>Xenon,2:8:18:18:8
<option>Caesium,2:8:18:18:8:1
<option>Barium,2:8:18:18:8:2
<option>Lanthanum,2:8:18:18:9:2
<option>Cerium,2:8:18:19:9:2
<option>Praseodymium,2:8:18:21:8:2
<option>Neodymium,2:8:18:22:8:2
<option>Promethium,2:8:18:23:8:2
<option>Samarium,2:8:18:24:8:2
<option>Europium,2:8:18:25:8:2
<option>Gadolinium,2:8:18:25:9:2
<option>Terbium,2:8:18:27:8:2
<option>Dysprosium,2:8:18:28:8:2
<option>Holmium,2:8:18:29:8:2
<option>Erbium,2:8:18:30:8:2
<option>Thulium,2:8:18:31:8:2
<option>Ytterbium,2:8:18:32:8:2
<option>Lutetium,2:8:18:32:9:2
<option>Hafnium,2:8:18:32:10:2
<option>Tantalum,2:8:18:32:11:2
<option>Tungsten,2:8:18:32:12:2
<option>Rhenium,2:8:18:32:13:2
<option>Osmium,2:8:18:32:14:2
<option>Iridium,2:8:18:32:15:2
<option>Platinum,2:8:18:32:17:1
<option>Gold,2:8:18:32:18:1
<option>Mercury,2:8:18:32:18:2
<option>Thallium,2:8:18:32:18:3
<option>Lead,2:8:18:32:18:4
<option>Bismuth,2:8:18:32:18:5
<option>Polonium,2:8:18:32:18:6
<option>Astatine,2:8:18:32:18:7
<option>Radon,2:8:18:32:18:8
<option>Francium,2:8:18:32:18:8:1
<option>Radium,2:8:18:32:18:8:2
<option>Actinium,2:8:18:32:18:9:2
<option>Thorium,2:8:18:32:18:10:2
<option>Protactinium,2:8:18:32:20:2
<option>Uranium,2:8:18:32:21:9:2
<option>Neptunium,2:8:18:32:22:9:2
<option>Plutonium,2:8:18:32:24:8:2
<option>Americium,2:8:18:32:25:8:2
<option>Curium,2:8:18:32:25:9:2
<option>Berkelium,2:8:18:32:27:8:2
<option>Californium,2:8:18:32:28:8:2
<option>Einsteinium,2:8:18:32:29:8:2
<option>Fermium,2:8:18:32:30:8:2
<option>Mendelevium,2:8:18:32:31:8:2
<option>Nobelium,2:8:18:32:32:8:2
<option>Lawrencium,2:8:18:32:32:8:3
<option>Rutherfordium,2:8:18:32:32:10:2
<option>Dubnium,2:8:18:32:32:11:2
<option>Seaborgium,2:8:18:32:32:12:2
<option>Bohrium,2:8:18:32:32:13:2
<option>Hassium,2:8:18:32:32:14:2
<option>Meitnerium,2:8:18:32:32:15:2
<option>Darmstadtium,2:8:18:32:32:16:2
<option>Roentgenium,2:8:18:32:32:17:2
<option>Copernicium,2:8:18:32:32:18:2
<option>Nihonium,2:8:18:32:32:18:3
<option>Flerovium,2:8:18:32:32:18:4
<option>Moscovium,2:8:18:32:32:18:5
<option>Livermorium,2:8:18:32:32:18:6
<option>Tennessine,2:8:18:32:32:18:7
<option>Oganesson,2:8:18:32:32:18:8
</select>
Working out the ideal positions
Let's take for example the orbit with 8 slots and 6 electrons to populate on it. Let's define a unit of measure such that the orbit has a circumference of 8 units (so we take a slot-to-slot distance on the circumference as unit of measure). If we for a moment forget about the slots, then a perfect distribution would be to have a distance (on the circumference) of 8 / 6 between every consecutive pair of electrons, because the sum of these 6 distances would be 8, i.e. making a full circle.
To avoid that the limited floating point precision gives us less accurate results, we could redefine the unit of measure by multiplying the numerator and denominator by 6 (the electron count), so the circumference is actually 8*6 units long, and each step is 8 (the distance between two consecutive electrons).
This is why in the loop of shellPattern you see the step variable increase with length (which is the number of slots in the orbit), giving us the distance (on the circumference) from the "home" position (at 0) to each electron. To translate this unit of measure back to the original unit of measure, we would divide by the electron count (the variable count). But instead of dividing, we multiply the index of the slot by count so we can avoid the floating point issues of a division. This product gives us the distance of the slot from the home position expressed in the new unit of measure. Every time this product passes over the current step, we should "place" the electron in the corresponding slot. This is where the rounding gets done, because we ignore the overrun of the product (the part that is more than step).
The + 0.5 is to make sure that the grid is horizontally mirrored in such a way that the home position is at the exact top. We want to "collapse" the perfect calculated position into a slot index. So we don't want the left picture, but the right picture (the shaded areas represent ranges that would collapse to the slots that are positioned at the center of them):
The difference is that additional 0.5*count which represents a half slot section. Note that 0.5 poses no problem for floating point: it has a perfect representation for 0.5
Next, we really want the slots to be filled in a way that the rounding is done symmetrically, so that the left and right side look the same. This is what the first if block does: whenever we have determined whether a slot is to be filled or left empty, we do exactly the same thing at the mirrored slot. If it turns out the slot number of the mirror is less than the current slot number, then we know we have "crossed" over and can stop the loop.
In case the number of electrons is even we also want the top and bottom half of the distribution to be mirrored. That is what the second if block does, using the same principle. In this case we only need to do one quarter of the total circumference as the rest is derived by mirroring.
I don't know Next.js/React/SVG, so here is python.
At least the matrix indices should be correct regardless of the language.
In the code I built a square matrix representing the ascii drawing. In the matrix, 0 represents an empty ascii space; 1 represents an empty electron spot; 2 represents an actual electron. Substitute 0-> , 1->-, 2->x to get your ascii drawing.
First I wrote a function make_layer that builds a layer as a 1d array. For instance, you can build the third layer of manganese by calling make_layer(3, 18, 13) because it's layer number 3, which holds a total of 18 spots, but only 13 electrons. The logic for this function is that we can distribute k items 0..k-1 evenly among n spots 0..n-1 by placing item j at spot floor(j * n / k). We do this twice: first we place k = n_electrons electrons among n = layer_size spots; then we place these k = layer_size spots inside a blank string of length n = ascii_layer_size.
Then I wrote a function add_layer_to_matrix that wraps a layer around inside a square matrix. This requires juggling with indices. I added a diagram below to explain visually.
And finally, function draw_square declares a square matrix of the appropriate size, and iterates on the layers to build them and wrap them around.
data_as_string = '''Hydrogen,1
Helium,2
Lithium,2:1
Beryllium,2:2
Berkelium,2:8:18:32:27:8:2
Oganesson,2:8:18:32:32:18:8'''
data_as_dict = {row[0]: list(map(int,row[1].split(':'))) for line in data_as_string.split('\n') if len(line) >= 1 and (row:=line.strip().split(','))}
# {'Hydrogen': [1], 'Helium': [2], 'Lithium': [2, 1], 'Beryllium': [2, 2], 'Berkelium': [2, 8, 18, 32, 27, 8, 2], 'Oganesson': [2, 8, 18, 32, 32, 18, 8]}
import numpy as np
def make_layer(radius, layer_size, n_electrons):
square_side_length = 2 * radius + 1
electrons_in_layer = np.ones(layer_size, dtype=int)
electrons_in_layer[(np.arange(n_electrons) * layer_size) // n_electrons] = 2
ascii_layer_size = (square_side_length - 1) * 4
layer = np.zeros(ascii_layer_size, dtype=int)
layer[(np.arange(layer_size) * ascii_layer_size) // layer_size] = electrons_in_layer
return layer
def add_layer_to_matrix(matrix, layer, R, r):
i = R - r
l = 2 * r + 1
L = 2 * R + 1
matrix[i, R:L-1-i] = layer[:r]
matrix[i:L-1-i, L-1-i] = layer[r:3*r]
matrix[L-1-i, i+1:L-1-i+1] = layer[5*r-1:3*r-1:-1]
matrix[i+1:L-1-i+1, i] = layer[7*r-1:5*r-1:-1]
matrix[i, i:R] = layer[7*r:]
def draw_square(layer_list):
max_radius = len(layer_list)
square_side_length = 2 * max_radius + 1
square_matrix = np.zeros((square_side_length, square_side_length), dtype=int)
for (layer_size, (radius, n_electrons)) in zip((2,8,18,32,32,18,8), enumerate(layer_list, start=1)):
layer = make_layer(radius, layer_size, n_electrons)
add_layer_to_matrix(square_matrix, layer, max_radius, radius)
return square_matrix
for element_name in ('Hydrogen', 'Beryllium', 'Berkelium'):
mat = draw_square(data_as_dict[element_name])
ascii = '\n'.join(''.join(' -x'[i] for i in row) for row in mat)
print(element_name, data_as_dict[element_name])
print(ascii)
print()
Output:
Hydrogen [1]
x
-
Beryllium [2, 2]
- x -
x
- -
x
- x -
Berkelium [2, 8, 18, 32, 27, 8, 2]
- x -
- - x - x
xxx- xxxx x
x xxxxxxxxxx-
xxxx xxx x-
xxxx x xxxxx
--x x xx
- xxxx xxxx -
xx x xx-
xxxxx x xxxx
xx xxx xxxx
-xxxxxxxxxx x
x -xxx xxx-
x - x - -
- x -
The arithmetic to wrap a layer around in the matrix follows this diagram:

Idea for algorithm to arrange balls of different weights into boxes

The problem
Suppose I have some balls and box types:
Each ball has a different weight.
Each type of box has its min and max capacity, and a penalty when used.
There are unlimited number of boxes for each type.
How can I arrange the balls into the least boxes such that:
The total weight of the balls in each box is within its min and max capacity.
The total penalty of the used boxes is minimized.
There may be multiple solutions. However, the accepted solution is where the total weight of the balls in each box is nearest to its max capacity.
Example
For example, there are 5 balls of weight 31, 14, 13, 12, 7 respectively, and 3 box types:
type│ min │ max │penalty
────┼─────┼─────┼───────
A │ 11 │ 20 │ 1
B │ 21 │ 30 │ 1
C │ 31 │ 40 │ 5
The possible combinations are:
boxTypes│ 31 │ 14 │ 13 │ 12 │ 7 │ penalty
────────┼────┼────┼────┼────┼────┼─────────
ABC │ C │ B │ B │ A │ C │ 7
BBC │ C │ B1 │ B2 │ B2 │ B1 │ 7
CC │ C1 │ C2 │ C2 │ C2 │ C1 │ 10
ACC │ C1 │ C2 │ C2 │ A │ C2 │ 11
and many other unlisted possibilities where the set of box types are the same or the penalty is just too high.
Notice that there are 2 solutions with the same penalty. However, considering the third condition:
boxTypes │ box1 │ box2 │ box3 │ shortfall
─────────┼──────┼──────┼──────┼──────────────────────────────────
ABC │ 12 │ 27 │ 38 │ (20-12) + (30-27) + (40-38) = 13
BBC │ 21 │ 25 │ 31 │ (30-25) + (30-25) + (40-31) = 19
The ABC box combination is chosen due to filling the most capacity of the boxes.
My code
I am currently recursively generating all combinations of the balls, and check whether there is a set of box that fits the ball groups.
I am able to improve the performance by:
Early halt when a group weight is out of the maximum capacity (40 in this example)
Limit the number of boxes (2 - 3 instead of 5, i.e. 1 box for each ball)
However, my solution still cannot handle more than 15 balls.
Is there a better algorithm other than bruteforce to solve this problem?

Multiple arrays Clickhouse

Problem:
Count distinct values in an array filtered by another array on same row (and agg higher).
Explanation:
Using this data:
In the Size D70, there are 5 pcs available (hqsize), but shops requests 15. By using the column accumulatedNeed, the 5 first stores in the column shops should receive items (since every store request 1 pcs). That is [4098,4101,4109,4076,4080].
It could also be that the values in accumulatedNeed would be [1,4,5,5,5,...,15], where shop 1 request 1 pcs, shop2 3 pcs, etc. Then only 3 stores would get.
In the size E75 there is enough stock, so every shop will receive (10 shops):
Now i want the distinct list of shops from D70 & E75, which would be be final result:
[4098,4101,4109,4076,4080,4062,4063,4067,4072,4075,4056,4058,4059,4061] (14 unique stores) (4109 is only counted once)
Wanted result:
[4098,4101,4109,4076,4080,4062,4063,4067,4072,4075,4056,4058,4059,4061]. (14 unique stores)
I'm totally open to structure the data otherwise if better.
The reason why it can't be precalculated is that the result depends on which shops that are filtered on.
Additional issue
The answer below from Vdimir is good and I've used it as basics for the final solution, but the solution does not cover (partial fullfillment).
If the stock number is in the runningNeed array we are all goodt, but remainers are not handled.
If you got:
select 5 as stock,[2,2,3,3] as need, [1,2,3,4] as shops, arrayCumSum(need) as runningNeed,arrayMap(x -> (x <= stock), runningNeed) as mask
You will get:
This is not correct since the 3rd shop should have 1 from stock (5-2-2 = 1)
I can't seem to get my head around how to make an array with "stock given", which in this case would be [2,2,1,0]
I use this query to create table with data similar to your screenshot:
CREATE TABLE t
(
Size String,
hqsize Int,
accumulatedNeed Array(Int),
shops Array(Int)
) engine = Memory;
INSERT INTO t VALUES ('D70', 5, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15], [4098,4101,4109,4076,4080,4083,4062,4063,4067,4072,4075,4056,4057,4058,4059]),('E75', 43, [1,2,3,4,5,6,7,8,9,10], [4109,4062,4063,4067,4072,4075,4056,4058,4059,4061]);
Find which shops that can receive enough items:
SELECT arrayMap(x -> (x <= hqsize), accumulatedNeed) as mask FROM t;
┌─mask────────────────────────────┐
│ [1,1,1,1,1,0,0,0,0,0,0,0,0,0,0] │
│ [1,1,1,1,1,1,1,1,1,1] │
└─────────────────────────────────┘
Filter not fulfilled shops according to this mask:
Note that shops and accumulatedNeed have to have equals sizes.
SELECT arrayFilter((x,y) -> y, shops, mask) as fulfilled_shops, arrayMap(x -> (x <= hqsize), accumulatedNeed) as mask FROM t;
┌─fulfilled_shops─────────────────────────────────────┬─mask────────────────────────────┐
│ [4098,4101,4109,4076,4080] │ [1,1,1,1,1,0,0,0,0,0,0,0,0,0,0] │
│ [4109,4062,4063,4067,4072,4075,4056,4058,4059,4061] │ [1,1,1,1,1,1,1,1,1,1] │
└─────────────────────────────────────────────────────┴─────────────────────────────────┘
Then you can create table with all distinct shops:
SELECT DISTINCT arrayJoin(fulfilled_shops) as shops FROM (
SELECT arrayMap(x -> (x <= hqsize), accumulatedNeed) as mask, arrayFilter((x,y) -> y, shops, mask) as fulfilled_shops FROM t
);
┌─shops─┐
│ 4098 │
│ 4101 │
│ 4109 │
│ 4076 │
│ 4080 │
│ 4062 │
│ 4063 │
│ 4067 │
│ 4072 │
│ 4075 │
│ 4056 │
│ 4058 │
│ 4059 │
│ 4061 │
└───────┘
14 rows in set. Elapsed: 0.049 sec.
Or if you need single array group it back:
SELECT groupArrayDistinct(arrayJoin(fulfilled_shops)) as shops FROM (
SELECT arrayMap(x -> (x <= hqsize), accumulatedNeed) as mask, arrayFilter((x,y) -> y, shops, mask) as fulfilled_shops FROM t
);
┌─shops───────────────────────────────────────────────────────────────────┐
│ [4080,4076,4101,4075,4056,4061,4062,4063,4109,4058,4067,4059,4072,4098] │
└─────────────────────────────────────────────────────────────────────────┘
If you need data only from D70 & E75 you can filter extra rows from table with WHERE before.

Clickhouse: runningAccumulate() does not work as I expect

Say, we have a table testint.
SELECT *
FROM testint
┌─f1─┬─f2─┐
│ 2 │ 3 │
│ 2 │ 3 │
│ 4 │ 5 │
│ 4 │ 5 │
│ 6 │ 7 │
│ 6 │ 7 │
└────┴────┘
We try to query runningAccumulate() with sumState().
SELECT runningAccumulate(col)
FROM
(
SELECT sumState(f1) AS col
FROM testint
GROUP BY f1
)
┌─runningAccumulate(col)─┐
│ 8 │
│ 12 │
│ 24 │
└────────────────────────┘
Why is the first row in the response 8, and not 4? If we are grouping by f1, the first row seems to be 4 (we do sum the first 2 and the second 2 in the column f1).
For accumulate-functions the order of elements is important, so just add ORDER BY to fix it:
SELECT runningAccumulate(col)
FROM
(
SELECT sumState(f1) AS col
FROM testint
GROUP BY f1
ORDER BY f1 ASC /* <-- */
)
You got the result [8, 12, 24] for input data [8, 4, 12] when should be used the ordered input - [4, 8, 12].

How to use results of [with totals] modifier

We have modifier [with totals] that can summarize values across all rows and get the total result with key value=0 or null or smth like this
The problem is that I don't understand how I can use these values in the next calculations
Maybe I'm using the wrong format
select processing_date,count(*)
from `telegram.message`
where processing_date>='2019-05-01'
group by processing_date with totals
The documentation says that
You can use WITH TOTALS in subqueries, including subqueries in the
JOIN clause (in this case, the respective total values are combined).
Example subqueries in the JOIN (CH tests scripts in github):
SELECT k, s1, s2
FROM
(
SELECT intDiv(number, 3) AS k, sum(number) AS s1
FROM
(
SELECT *
FROM system.numbers
LIMIT 10
)
GROUP BY k WITH TOTALS
)
ANY LEFT JOIN
(
SELECT intDiv(number, 4) AS k, sum(number) AS s2
FROM
(
SELECT *
FROM system.numbers
LIMIT 10
)
GROUP BY k WITH TOTALS
) USING (k)
ORDER BY k ASC
/* Result:
┌─k─┬─s1─┬─s2─┐
│ 0 │ 3 │ 6 │
│ 1 │ 12 │ 22 │
│ 2 │ 21 │ 17 │
│ 3 │ 9 │ 0 │
└───┴────┴────┘
Totals:
┌─k─┬─s1─┬─s2─┐
│ 0 │ 45 │ 45 │
└───┴────┴────┘
*/
As a workaround, you can combine results of several totals using client libraries.
Using "with rollup" instead of "with totals" decides problems with format

Resources