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Problem statement:
A company sells bowls of various integer sized diameters (inches) and often customers buy a number of these bowls at once.
The company would like to reduce shipping costs by sending the minimum number of packages for an order of bowls to a given customer by finding an optimal nesting of the bowls.
The company has also decided to restrict the nestings with the following limitations:
No more than 3 bowls should be nested in one nesting.
A bowl can be nested inside another if it's smaller but not more than 3 inches smaller than the bowl it's directly nested within.
For example, a customer orders the following bowl sizes:
One 5" bowl
One 8" bowl
Two 11" bowls
One 12" bowl
Two 15" bowls
The follow is a possible (and optimal) nesting:
[15] [15,12,11] [11,8,5]
Is there an algorithm to always provide an optimal nesting?
I've looked through many similar questions here on stackoverflow and googled around, but can't find this exact problem, nor am I able to map any similar problems over to this problem space in a way that solves the problem.
This was actually posted in another forum by a real business owner. A number of the developers tried to help, ultimately finding a heuristic solution that provided an optimal solution most of the time but not always.
I can share the chosen algorithm one of the developers put forward as well as a few approaches I tried myself.
I'm just very curious about this problem and if there is an algorithm that can actually do this, or the best solution will be heuristic. If you can either give an idea of how to approach this, share an algorithm, or send a link to a similar problem that can be mapped to this one, that would be awesome.
This can be solved with dynamic programming in polynomial time.
The idea is that we ONLY care about how many boxes there are total, and how many boxes there are of different top bowl sizes. We don't care about the details beyond that. This is a polynomial amount of state, and so we can track through the calculation and enumerate one arrangement per possible state in a polynomial time. We then reconstruct the minimal packing of bowls into boxes from that arrangement.
class Arrangement:
def __init__(self, next_bowl, prev_arrangement=None):
self.prev_arrangement = prev_arrangement
self.add_rule = None
self.open1 = {}
self.open2 = {}
self.next_bowl = next_bowl
if prev_arrangement is None:
self.boxes = 0
for i in range(next_bowl, next_bowl + 4):
self.open1[i] = 0
self.open2[i] = 0
else:
self.boxes = prev_arrangement.boxes
for i in range(next_bowl, next_bowl + 4):
self.open1[i] = prev_arrangement.open1.get(i, 0)
self.open2[i] = prev_arrangement.open2.get(i, 0)
# This will be tuples of tuples.
def state(self):
open1 = (self.open1[i+self.next_bowl] for i in range(4))
open2 = (self.open2[i+self.next_bowl] for i in range(4))
return (open1, open2)
def next_arrangements(self, bowl):
base_arrangement = Arrangement(bowl, self)
base_arrangement.boxes += 1
base_arrangement.add_rule = ("new",)
old_count = self.open2.get(bowl, 0)
base_arrangement.open2[bowl] = old_count + 1
yield base_arrangement
for i in range(1, 4):
if 0 < self.open1.get(bowl+i, 0):
next_arrangement = Arrangement(bowl, self)
next_arrangement.open1[bowl+i] -= 1
next_arrangement.add_rule = ("open", 1, bowl+i)
yield next_arrangement
if 0 < self.open2.get(bowl+i, 0):
next_arrangement = Arrangement(bowl, self)
next_arrangement.open2[bowl+i] -= 1
next_arrangement.open1[bowl] += 1
next_arrangement.add_rule = ("open", 2, bowl+i)
yield next_arrangement
def find_boxes(self):
items = self._find_boxes()
boxes = items["full"]
for more_boxes in items["open1"].values():
boxes.extend(more_boxes)
for more_boxes in items["open2"].values():
boxes.extend(more_boxes)
return list(reversed(sorted(boxes)))
def _find_boxes(self):
if self.prev_arrangement is None:
return {
"full": [],
"open1": {},
"open2": {},
}
else:
items = self.prev_arrangement._find_boxes()
rule = self.add_rule
if rule[0] == "new":
if self.next_bowl not in items["open2"]:
items["open2"][self.next_bowl] = [[self.next_bowl]]
else:
items["open2"][self.next_bowl].append([self.next_bowl])
elif rule[0] == "open":
if rule[1] == 1:
box = items["open1"][rule[2]].pop()
box.append(self.next_bowl)
items["full"].append(box)
elif rule[1] == 2:
box = items["open2"][rule[2]].pop()
box.append(self.next_bowl)
if self.next_bowl not in items["open1"]:
items["open1"][self.next_bowl] = [box]
else:
items["open1"][self.next_bowl].append(box)
return items
def __str__ (self):
return str(self.boxes) + " open1:" + str(self.open1) + " open2:" + str(self.open2)
def bowl_nesting (bowls):
bowls = list(reversed(sorted(bowls))) # Largest to smallest.
start_arrangement = Arrangement(bowls[0])
arrange = {start_arrangement.state(): start_arrangement}
for bowl in bowls:
next_arrange = {}
for state, arrangement in arrange.items():
for next_arrangement in arrangement.next_arrangements(bowl):
state = next_arrangement.state()
if state in next_arrange and next_arrange[state].boxes <= next_arrangement.boxes:
pass # We are not an improvement.
else:
next_arrange[state] = next_arrangement
arrange = next_arrange
min_boxes = len(bowls)
min_box_list = None
for arrangement in arrange.values():
if arrangement.boxes <= min_boxes:
min_boxes = arrangement.boxes
min_box_list = arrangement.find_boxes()
return min_box_list
print(bowl_nesting([15, 15, 12, 11, 11,8,5]))
Now while the above solution works, it is inefficient. Suppose that we have up to k bowls of any given size. The number of combinations of open1[bowl] and open2[bowl] that allows is k choose 2 = k*(k-1)/2). When we consider that our state has 4 sizes in it, that's O(k^8 / 16 possible states. We do that for the number of bowls to get O(n k^8). This doesn't scale well.
We can do better by making the following notes:
In any arrangement with an open2[bowls+3] option, you do not do worse by moving the next bowl out of whatever box you were going to put it in, and putting it there instead.
If there is an open2[bowls+2] option and an open2[bowls+1] option, you never do worse by picking open2[bowls+2].
If there is an open1[bowls+i] option and an open1[bowls+j] option with 1 <= i < j <= 3 then you never do worse picking open1[bowls+i] instead.
This optimization means fewer choices, which speeds you up by a constant. But also you cannot have open2[bowls+3] and also have open2[bowls]. So that O(k^8) becomes O(k^7) states. And adding to the boxes with larger bowls will reduce how much of the potential state space we actually visit. This should lead to a better constant.
Here is this logic with a minor refactor to cleanup the code.
class Arrangement:
def __init__(self, next_bowl, prev_arrangement=None, choice=None, position=None):
self.prev_arrangement = prev_arrangement
self.add_rule = None
self.open1 = {}
self.open2 = {}
self.next_bowl = next_bowl
if prev_arrangement is None:
self.boxes = 0
for i in range(next_bowl, next_bowl + 4):
self.open1[i] = 0
self.open2[i] = 0
else:
self.boxes = prev_arrangement.boxes
for i in range(next_bowl, next_bowl + 4):
self.open1[i] = prev_arrangement.open1.get(i, 0)
self.open2[i] = prev_arrangement.open2.get(i, 0)
if choice is not None:
self.choice(choice, position)
# This will be tuples of tuples.
def state(self):
open1 = (self.open1[i+self.next_bowl] for i in range(4))
open2 = (self.open2[i+self.next_bowl] for i in range(4))
return (open1, open2)
def choice (self, rule, position=None):
self.add_rule = (rule, position)
if rule == "new":
self.boxes += 1
self.open2[self.next_bowl] += 1
elif rule == "open1":
self.open1[position] -= 1
elif rule == "open2":
self.open2[position] -= 1
self.open1[self.next_bowl] += 1
def next_arrangements(self, bowl):
if 0 < self.open2.get(bowl+3, 0):
yield Arrangement(bowl, self, "open2", bowl+3)
else:
yield Arrangement(bowl, self, "new")
for i in [3, 2, 1]:
if 0 < self.open1.get(bowl+i, 0):
yield Arrangement(bowl, self, "open1", bowl+i)
break
for i in [2, 1]:
if 0 < self.open2.get(bowl+i, 0):
yield Arrangement(bowl, self, "open2", bowl+i)
break
def find_boxes(self):
items = self._find_boxes()
boxes = items["full"]
for more_boxes in items["open1"].values():
boxes.extend(more_boxes)
for more_boxes in items["open2"].values():
boxes.extend(more_boxes)
return list(reversed(sorted(boxes)))
def _find_boxes(self):
if self.prev_arrangement is None:
return {
"full": [],
"open1": {},
"open2": {},
}
else:
items = self.prev_arrangement._find_boxes()
rule = self.add_rule
if rule[0] == "new":
if self.next_bowl not in items["open2"]:
items["open2"][self.next_bowl] = [[self.next_bowl]]
else:
items["open2"][self.next_bowl].append([self.next_bowl])
elif rule[0] == "open1":
box = items["open1"][rule[1]].pop()
box.append(self.next_bowl)
items["full"].append(box)
elif rule[0] == "open2":
box = items["open2"][rule[1]].pop()
box.append(self.next_bowl)
if self.next_bowl not in items["open1"]:
items["open1"][self.next_bowl] = [box]
else:
items["open1"][self.next_bowl].append(box)
return items
def bowl_nesting (bowls):
bowls = list(reversed(sorted(bowls))) # Largest to smallest.
start_arrangement = Arrangement(bowls[0])
arrange = {start_arrangement.state(): start_arrangement}
for bowl in bowls:
next_arrange = {}
for state, arrangement in arrange.items():
for next_arrangement in arrangement.next_arrangements(bowl):
state = next_arrangement.state()
if state in next_arrange and next_arrange[state].boxes <= next_arrangement.boxes:
pass # We are not an improvement.
else:
next_arrange[next_arrangement.state()] = next_arrangement
arrange = next_arrange
min_boxes = len(bowls)
min_box_list = None
for arrangement in arrange.values():
if arrangement.boxes <= min_boxes:
min_boxes = arrangement.boxes
min_box_list = arrangement.find_boxes()
return min_box_list
print(bowl_nesting([15, 15, 12, 11, 11,8,5]))
Yes, we can calculate an optimal nesting. As you presented, start with the bowls sorted in reverse order.
15,15,12,11,11,8,5
Assign the minimum number of starting bowls, corresponding to the count of the largest bowl.
[15] [15]
As we iterate element by element, the state we need to keep is the smallest bowl size and count in each container per index visited.
index 0, [(15, 1), (15, 1)]
(The state can be further refined to a multiset of those packages with identical count and smallest bowl size, which would add some complication.)
The choice for any element is which box (or set of boxes with similar state) to add it to or whether to start a new box with it.
index 1, [(15, 1), (12, 2)]
or
index 1, [(15, 1), (15, 1), (12, 1)]
We can explore these branches in an iterative or recursive breadth first search prioritised by the number of elements remaining plus the number of packages in the state, avoiding previously seen states.
We can further prune the search space by avoiding branches with the same or more count of packages than the best we've already seen.
This approach would amount to brute force in the sense of exploring all relevant branches. But hopefully the significant restrictions of package size and bowl size relationship would narrow the search space considerably.
This "Answer" is based on btilly's solution (the accepted answer).
Thank you #btilly for sticking with this and taking the time to revise the algorithm and fix bugs!
Since this was originally set within the context of Google Apps Script, I've rewritten this in Javascript and want to share the JS code with anyone else that might want it.
btilly's improved algorithm does indeed run much quicker than the first. Though the improvement factor depends on the bowls provided I've noticed it running up to 50 times faster in some of my sample sets.
Below is the JS code. Some caveats:
I've kept the same structure and same naming as much as possible in copying over btilly's solution.
There's no guarantee I did not introduce bugs while porting over btilly's code.
I'm not too familiar with many modern/proper JS conventions and also I don't know Python at all, so translating some of the concepts was tough and although I think my code is now bug free, if you spot any bugs, inefficiencies, bad programming ideas, please let me know and I'll update the below code.
I added a count to the state creation to make each state unique, since in my Apps Script implementation the JS runtime kept stringifying the arrays so that two states were sometimes considered the same even if they were not (e.g. the previous arrangement's bowl was the same size as another arrangement's bowl, but not the same bowl - the way two 10" bowls might appear to a 9" bowl for example). This was not needed in Python since the generators were unique based on their memory addresses. If you know a better way to do this in JS, please let me know. Seems a little sloppy the way I did it.
Improved/faster code (Javascript):
class Arrangement2{
constructor(next_bowl, prev_arrangement, choice, position){
this.prev_arrangement = prev_arrangement;
this.add_rule = null;
this.open1 = {};
this.open2 = {};
this.next_bowl = next_bowl;
if (prev_arrangement == null){
this.boxes = 0;
for (let i = next_bowl; i < next_bowl + 4; i++){
this.open1[i] = 0;
this.open2[i] = 0;
}
}
else{
this.boxes = prev_arrangement.boxes;
for (let i = next_bowl; i < next_bowl + 4; i++){
this.open1[i] = prev_arrangement.open1[i] != null ? prev_arrangement.open1[i] : 0;
this.open2[i] = prev_arrangement.open2[i] != null ? prev_arrangement.open2[i] : 0;
}
}
if(choice != null){
this.choice(choice,position);
}
}
state(){
let open1 = {};
let open2 = {};
for(let i = 0; i < 4; i++){
open1[i+this.next_bowl] = this.open1[i+this.next_bowl];
open2[i+this.next_bowl] = this.open2[i+this.next_bowl];
}
var toReturn = [];
//Used to make each state unique, without this the algorithm may not always find the best solution
Arrangement2.count++;
toReturn.push(Arrangement2.count);
toReturn.push(open1);
toReturn.push(open2);
return toReturn;
}
choice(rule, position){
this.add_rule = [rule, position];
if( rule == "new" ){
this.boxes += 1;
this.open2[this.next_bowl] += 1;
}
else if( rule == "open1" ){
this.open1[position] -= 1;
}
else if( rule == "open2" ){
this.open2[position] -= 1;
this.open1[this.next_bowl] += 1;
}
}
* next_arrangements (bowl){
if( 0 < (this.open2[bowl+3] != null ? this.open2[bowl+3] : 0)){
yield new Arrangement2(bowl, this, "open2", bowl + 3);
}
else{
yield new Arrangement2(bowl, this, "new", null);
for(let i = 3; i > 0; i--){
if (this.open1[bowl+i] != null ? this.open1[bowl+i] : 0){
yield new Arrangement2(bowl, this, "open1", bowl+i);
break ;
}
}
for(let i = 2; i > 0; i--){
if (this.open2[bowl+i] != null ? this.open2[bowl+i] : 0){
yield new Arrangement2(bowl, this, "open2", bowl+i);
break ;
}
}
}
}
find_boxes(){
let items = this._find_boxes();
let boxes = items["full"];
for (const [key, more_boxes] of Object.entries(items["open1"])) {
boxes = boxes.concat(more_boxes);
}
for (const [key, more_boxes] of Object.entries(items["open2"])) {
boxes = boxes.concat(more_boxes);
}
//Max --> Min (i.e [ 12, 12, 11, 11, 10, 7, 7, 7 ])
boxes.sort(function(a, b){return b - a});
return boxes; //boxes.sort().reverse(); //list(reversed(sorted(boxes)));
}
_find_boxes(){
if (this.prev_arrangement == null){
return {
"full": [],
"open1": {},
"open2": {},
}
}
else{
let items = this.prev_arrangement._find_boxes();
let rule = this.add_rule;
if (rule[0] == "new"){
if (!(this.next_bowl in items["open2"])){
items["open2"][this.next_bowl] = [[this.next_bowl]];
}
else{
items["open2"][this.next_bowl].push([this.next_bowl]);
}
}
else if( rule[0] == "open1"){
let box = items["open1"][rule[1]].pop();
box.push(this.next_bowl);
items["full"].push(box);
}
else if( rule[0] == "open2"){
let box = items["open2"][rule[1]].pop();
box.push(this.next_bowl);
if (!(this.next_bowl in items["open1"])){
items["open1"][this.next_bowl] = [box];
}
else{
items["open1"][this.next_bowl].push(box);
}
}
return items;
}
}
__str__(){
return this.next_bowl + " " + JSON.stringify(this.boxes) + " open1:" + JSON.stringify(this.open1) + " open2:" + JSON.stringify(this.open2);
}
}
allStates_nesting_improved = function (bowls){
//Used to make each state unique, without this the algorithm may not always find the best solution
Arrangement2.count = 0;
//Max --> Min (i.e [ 12, 12, 11, 11, 10, 7, 7, 7 ])
bowls.sort(function(a, b){return b - a});
let start_arrangement = new Arrangement2(bowls[0], null);
let returnObj = start_arrangement.state();
let arrange = {[returnObj]:start_arrangement};
for (const [key, bowl] of Object.entries(bowls) ) {
let next_arrange = {};
for (let [state, arrangement] of Object.entries(arrange) ) {
let next_arrangements = arrangement.next_arrangements(bowl);
let next_arrangement = next_arrangements.next();
while(next_arrangement.value != undefined){
next_arrangement = next_arrangement.value;
let state = next_arrangement.state();
let nextArrange_state = next_arrange[state];
if ( next_arrange[state] != undefined && (nextArrange_state === state) && next_arrange[state].boxes <= next_arrangement.boxes){
continue ; // # We are not an improvement.
}
else{
next_arrange[next_arrangement.state()] = next_arrangement;
}
next_arrangement = next_arrangements.next();
}
}
arrange = next_arrange;
}
let min_boxes = bowls.length;
let min_box_list = null;
for (const [key, arrangement] of Object.entries(arrange) ) {
if (arrangement.boxes <= min_boxes){
min_boxes = arrangement.boxes;
min_box_list = arrangement.find_boxes();
}
}
console.log(min_box_list);
return min_box_list;
}
Original code (Javascript):
class Arrangement1{
constructor(next_bowl, prev_arrangement){
this.prev_arrangement = prev_arrangement;
this.add_rule = null;
this.open1 = {};
this.open2 = {};
this.next_bowl = next_bowl;
if (prev_arrangement == null){
this.boxes = 0;
for (let i = next_bowl; i < next_bowl + 4; i++){
this.open1[i] = 0;
this.open2[i] = 0;
}
}
else{
this.boxes = prev_arrangement.boxes;
for (let i = next_bowl; i < next_bowl + 4; i++){
this.open1[i] = prev_arrangement.open1[i] != null ? prev_arrangement.open1[i] : 0;
this.open2[i] = prev_arrangement.open2[i] != null ? prev_arrangement.open2[i] : 0;
}
}
}
state(){
//Used to make each state unique, without this the algorithm may not always find the best solution
Arrangement1.count++;
let open1 = {};
let open2 = {};
for(let i = 0; i < 4; i++){
open1[i+this.next_bowl] = this.open1[i+this.next_bowl];
open2[i+this.next_bowl] = this.open2[i+this.next_bowl];
}
var toReturn = [];
toReturn.push(Arrangement1.count);
toReturn.push(open1);
toReturn.push(open2);
return toReturn;
}
* next_arrangements (bowl){
let base_arrangement = new Arrangement1(bowl, this);
base_arrangement.boxes += 1;
base_arrangement.add_rule = ["new"];
let old_count = this.open2[bowl] != null ? this.open2[bowl] : 0;
base_arrangement.open2[bowl] = old_count + 1;
yield base_arrangement;
for(let i = 1; i < 4; i++){
if (0 < (this.open1[bowl+i] != null ? this.open1[bowl+i] : 0)){
let next_arrangement = new Arrangement1(bowl, this);
next_arrangement.open1[bowl+i] -= 1;
next_arrangement.add_rule = ["open", 1, bowl+i];
yield next_arrangement;
}
if (0 < (this.open2[bowl+i] != null ? this.open2[bowl+i] : 0)){
let next_arrangement = new Arrangement1(bowl, this);
next_arrangement.open2[bowl+i] -= 1;
next_arrangement.open1[bowl] += 1;
next_arrangement.add_rule = ["open", 2, bowl+i];
yield next_arrangement;
}
}
}
find_boxes(){
let items = this._find_boxes();
let boxes = items["full"];
for (const [key, more_boxes] of Object.entries(items["open1"])) {
boxes = boxes.concat(more_boxes);
}
for (const [key, more_boxes] of Object.entries(items["open2"])) {
boxes = boxes.concat(more_boxes);
}
//Max --> Min (i.e [ 12, 12, 11, 11, 10, 7, 7, 7 ])
boxes.sort(function(a, b){return b - a});
return boxes;
}
_find_boxes(){
if (this.prev_arrangement == null){
return {
"full": [],
"open1": {},
"open2": {},
}
}
else{
let items = this.prev_arrangement._find_boxes();
let rule = this.add_rule;
if (rule[0] == "new"){
if (!(this.next_bowl in items["open2"])){
items["open2"][this.next_bowl] = [[this.next_bowl]];
}
else{
items["open2"][this.next_bowl].push([this.next_bowl]);
}
}
else if( rule[0] == "open"){
if (rule[1] == 1){
let box = items["open1"][rule[2]].pop();
box.push(this.next_bowl);
items["full"].push(box);
}
else if( rule[1] == 2){
let box = items["open2"][rule[2]].pop();
box.push(this.next_bowl);
if (!(this.next_bowl in items["open1"])){
items["open1"][this.next_bowl] = [box];
}
else{
items["open1"][this.next_bowl].push(box);
}
}
}
return items;
}
}
__str__(){
return this.next_bowl + " " + JSON.stringify(this.boxes) + " open1:" + JSON.stringify(this.open1) + " open2:" + JSON.stringify(this.open2);
}
}
allStates_nesting = function (bowls){
//Used to make each state unique, without this the algorithm may not always find the best solution
Arrangement1.count = 0;
//Max --> Min (i.e [ 12, 12, 11, 11, 10, 7, 7, 7 ])
bowls.sort(function(a, b){return b - a});
let start_arrangement = new Arrangement1(bowls[0], null);
let returnObj = start_arrangement.state();
let arrange = {[returnObj]:start_arrangement};
for (const [key, bowl] of Object.entries(bowls) ) {
let next_arrange = {};
for (let [state, arrangement] of Object.entries(arrange) ) {
let next_arrangements = arrangement.next_arrangements(bowl);
let next_arrangement = next_arrangements.next();
while(next_arrangement.value != undefined){
next_arrangement = next_arrangement.value;
let state = next_arrangement.state();
let nextArrange_state = next_arrange[state];
if ( next_arrange[state] != undefined && (nextArrange_state === state) && next_arrange[state].boxes <= next_arrangement.boxes){
continue ; // # We are not an improvement.
}
else{
next_arrange[state] = next_arrangement;
}
next_arrangement = next_arrangements.next();
}
}
arrange = next_arrange;
}
let min_boxes = bowls.length;
let min_box_list = null;
for (const [key, arrangement] of Object.entries(arrange) ) {
if (arrangement.boxes <= min_boxes){
min_boxes = arrangement.boxes;
min_box_list = arrangement.find_boxes();
}
}
return min_box_list;
}
See it in action
Here is a link to a spreadsheet testbed with 3 algorithms:
Algorithm 1: A heuristic algorithm another developer provided (runs fast but doesn't always find the optimal solution and ignores some of the requirements in some of its solutions for simplicity's sake)
Algorithm 2: btilly's revised algorithm (faster)
Algorithm 3: btilly's first attempt
Bowl Nesting Spreadsheet
Feel free to make a copy and modify the code and/or add your own algorithm to compare it with the others. (The orange "Run" button won't work since the spreadsheet is in "Viewer" mode. You'll need to make a copy to run it).
To make a copy go to
File -> Make a copy.
Once you have your own copy, you can click the "Run" button or go to the code by clicking
Extensions -> Apps Script
You can then modify and/or add your own algorithm to the mix.
You'll also have to authorize the script to run as with all Apps Script scripts.
If you're worried about authorizing it, of course check out the code before clicking run to make sure there isn't anything nefarious in there.
I am developing an application to store the relationship data of customers and accounts (w.r.t banking domain). Typically in banks a customer can have an account which is a sole account or have a joint account with another customer.
Eg 1: Customer C1 has a sole account A1.
Eg 2: Customer C1 and C2 have a joint account JA1 where C1 is primary holder and C2 is non-primary holder.
I am looking for an algorithm that will generate all possible combinations of relationships for a given number of customers and accounts.
For example: if the number of customers = 2 and number of accounts = 2, then the algorithm should generate the below entries.
Combination #1:
C1-A1-Primary
C1-A2-Primary
C2-A1-Non-Primary
C2-A2-Non-Primary
Combination #2:
C1-A1-Primary
C1-A2-Non-Primary
C2-A1-Non-Primary
C2-A2-Primary
Combination #3:
C1-A1-Non-Primary
C1-A2-Primary
C2-A1-Primary
C2-A2-Non-Primary
Combination #4:
C1-A1-Non-Primary
C1-A2-Non-Primary
C2-A1-Primary
C2-A2-Primary
Combination #5:
C1-A1-Sole
C1-A2-Primary
C2-A2-Non-Primary
Combination #6:
C1-A1-Sole
C1-A2-Non-Primary
C2-A2-Primary
Combination #7:
C1-A2-Sole
C1-A1-Primary
C2-A1-Non-Primary
Combination #8:
C1-A2-Sole
C1-A1-Non-Primary
C2-A1-Primary
Edit: This is not the complete list of combinations - but the algorithm is supposed to generate all those.
You have 2 problems to solve here:
Get all possible account types for N customers. You can do it like this:
const allAccounts = [];
for (let i = 1; i <= customersNumber; i++) {
allAccounts.push(C${i}-Sole);
for (let j = 1; j <= customersNumber; j++) {
if (i === j) continue;
allAccounts.push(C${i}-Primary C${j}-NonPrimary);
}
}
For 2 customers the result will be:
[
"C1-Sole",
"C1-Primary C2-NonPrimary",
"C2-Sole",
"C2-Primary C1-NonPrimary"
]
Get all possible combinations of length r (with repetitions) from this array. We want to exclude two types of combinations here:
ones that have 2 or more sole accounts for the same customer.
ones that are not connected (have no common customers, if I got you right)
// checks if two accounts are connected
function connected(customers1, customers2) {
return customers1.filter(cu => customers2.includes(cu)).length > 0;
}
// checks if acc1 and acc2 are the same Sole account
function sameSoleAccount(acc1, acc2) {
return acc1.type === 'Sole' && acc1 === acc2;
}
function printAccount(i, a) {
const c = a.customers;
return a.type === 'Sole' ? `${c[0]}-A${i}-Sole` : `${c[0]}-A${i}-Primary ${c[1]}-A${i}-NonPrimary`;
}
function combination(chosen, arr, index, r) {
if (index === r) {
const combination = chosen.map((c, i) => printAccount(i + 1, arr[c])).join(', ');
console.log(combination);
return;
}
for (let i = 0; i < arr.length; i++) {
if (chosen.length === 0 ||
chosen.some(ch => !sameSoleAccount(arr[ch], arr[i])
&& connected(arr[ch].customers, arr[i].customers))) {
const copy = chosen.slice();
copy[index] = i;
combination(copy, arr, index + 1, r);
}
}
}
function allPossibleCombinations(accountsNumber, customersNumber) {
const allAccounts = [];
for (let i = 1; i <= customersNumber; i++) {
allAccounts.push({customers: [`C${i}`], type: 'Sole'});
for (let j = 1; j <= customersNumber; j++) {
if (i === j) continue;
allAccounts.push({customers: [`C${i}`, `C${j}`], type: 'Joint'});
}
}
console.log(`All possible combinations for ${customersNumber} customers and ${accountsNumber} accounts: `);
combination([], allAccounts, 0, accountsNumber);
}
allPossibleCombinations(2, 2);
If you have limited relationship between accounts and customers:
1) create dict with:
dMapCustomer = {<nCustomerId>: [<nAccountId1>, <nAccountId2>]}
2) for each customer create all possible pairs, it is just
lCustomerPairs = [(nCustomerId, nAccountId1), (nCustomerId, nAccountId2), ...]
3) concat all pairs from step 2.
l = []
for nCustomer in lCustomer:
l += lCustomerPairs
If any account can be linked with any customer, then just:
lAccounts = [1,2,3]
lCustomers = [4,5,6]
list(product(lCustomers, lCustomers)) # all possible pairs of account and customer
Function product generates all possible pairs from two lists:
def product(l1, l2):
pools = [tuple(pool) for pool in [l1, l2]]
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
I've been working with this variation of dynamic programming to solve a knapsack problem:
KnapsackItem = Struct.new(:name, :cost, :value)
KnapsackProblem = Struct.new(:items, :max_cost)
def dynamic_programming_knapsack(problem)
num_items = problem.items.size
items = problem.items
max_cost = problem.max_cost
cost_matrix = zeros(num_items, max_cost+1)
num_items.times do |i|
(max_cost + 1).times do |j|
if(items[i].cost > j)
cost_matrix[i][j] = cost_matrix[i-1][j]
else
cost_matrix[i][j] = [cost_matrix[i-1][j], items[i].value + cost_matrix[i-1][j-items[i].cost]].max
end
end
end
cost_matrix
end
def get_used_items(problem, cost_matrix)
i = cost_matrix.size - 1
currentCost = cost_matrix[0].size - 1
marked = Array.new(cost_matrix.size, 0)
while(i >= 0 && currentCost >= 0)
if(i == 0 && cost_matrix[i][currentCost] > 0 ) || (cost_matrix[i][currentCost] != cost_matrix[i-1][currentCost])
marked[i] = 1
currentCost -= problem.items[i].cost
end
i -= 1
end
marked
end
This has worked great for the structure above where you simply provide a name, cost and value. Items can be created like the following:
items = [
KnapsackItem.new('david lee', 8000, 30) ,
KnapsackItem.new('kevin love', 12000, 50),
KnapsackItem.new('kemba walker', 7300, 10),
KnapsackItem.new('jrue holiday', 12300, 30),
KnapsackItem.new('stephen curry', 10300, 80),
KnapsackItem.new('lebron james', 5300, 90),
KnapsackItem.new('kevin durant', 2300, 30),
KnapsackItem.new('russell westbrook', 9300, 30),
KnapsackItem.new('kevin martin', 8300, 15),
KnapsackItem.new('steve nash', 4300, 15),
KnapsackItem.new('kyle lowry', 6300, 20),
KnapsackItem.new('monta ellis', 8300, 30),
KnapsackItem.new('dirk nowitzki', 7300, 25),
KnapsackItem.new('david lee', 9500, 35),
KnapsackItem.new('klay thompson', 6800, 28)
]
problem = KnapsackProblem.new(items, 65000)
Now, the problem I'm having is that I need to add a position for each of these players and I have to let the knapsack algorithm know that it still needs to maximize value across all players, except there is a new restriction and that restriction is each player has a position and each position can only be selected a certain amount of times. Some positions can be selected twice, others once. Items would ideally become this:
KnapsackItem = Struct.new(:name, :cost, :position, :value)
Positions would have a restriction such as the following:
PositionLimits = Struct.new(:position, :max)
Limits would be instantiated perhaps like the following:
limits = [Struct.new('PG', 2), Struct.new('C', 1), Struct.new('SF', 2), Struct.new('PF', 2), Struct.new('Util', 2)]
What makes this a little more tricky is every player can be in the Util position. If we want to disable the Util position, we will just set the 2 to 0.
Our original items array would look something like the following:
items = [
KnapsackItem.new('david lee', 'PF', 8000, 30) ,
KnapsackItem.new('kevin love', 'C', 12000, 50),
KnapsackItem.new('kemba walker', 'PG', 7300, 10),
... etc ...
]
How can position restrictions be added to the knapsack algorithm in order to still retain max value for the provided player pool provided?
There are some efficient libraries available in ruby which could suit your task , Its clear that you are looking for some constrain based optimization , there are some libraries in ruby which are a opensource so, free to use , Just include them in you project. All you need to do is generate Linear programming model objective function out of your constrains and library's optimizer would generate Solution which satisfy all your constrains , or says no solution exists if nothing can be concluded out of the given constrains .
Some such libraries available in ruby are
RGLPK
OPL
LP Solve
OPL follows the LP syntax similar to IBM CPLEX , which is widely used Optimization software, So you could get good references on how to model the LP using this , Moreover this is build on top of the RGLPK.
As I understand, the additional constraint that you are specifying is as following:
There shall be a set of elements, out which only at most k (k = 1 or
2) elements can be selected in the solution. There shall be multiple
such sets.
There are two approaches that come to my mind, neither of which are efficient enough.
Approach 1:
Divide the elements into groups of positions. So if there are 5 positions, then each element shall be assigned to one of 5 groups.
Iterate (or recur) through all the combinations by selecting 1 (or 2) element from each group and checking the total value and cost. There are ways in which you can fathom some combinations. For example, in a group if there are two elements in which one gives more value at lesser cost, then the other can be rejected from all solutions.
Approach 2:
Mixed Integer Linear Programming Approach.
Formulate the problem as follows:
Maximize summation (ViXi) {i = 1 to N}
where Vi is value and
Xi is a 1/0 variable denoting presence/absence of an element from the solution.
Subject to constraints:
summation (ciXi) <= C_MAX {total cost}
And for each group:
summation (Xj) <= 1 (or 2 depending on position)
All Xi = 0 or 1.
And then you will have to find a solver to solve the above MILP.
This problem is similar to a constraint vehicle routing problem. You can try a heuristic like the saving algorithm from Clarke&Wright. You can also try a brute-force algorithm with less players.
Considering players have Five positions your knapsack problem would be:-
Knpsk(W,N,PG,C,SF,PF,Util) = max(Knpsk(W-Cost[N],N-1,...)+Value[N],Knpsk(W,N-1,PG,C,SF,PF,Util),Knpsk(W-Cost[N],N-1,PG,C,SF,PF,Util-1)+Value[N])
if(Pos[N]=="PG") then Knpsk(W-Cost[N],N-1,....) = Knpsk(W-Cost[N],N-1,PG-1,....)
if(Pos[N]=="C") then Knpsk(W-Cost[N],N-1,....) = Knpsk(W-Cost[N],N-1,PG,C-1....)
so on...
PG,C,SF,PF,Util are current position capacities
W is current knapsack capacity
N number of items available
Dynamic Programming can be used as before using 7-D table and as in your case the values of positions are small it will slow down algorithm by factor of 16 which is great for n-p complete problem
Following is dynamic programming solution in JAVA:
public class KnapsackSolver {
HashMap CostMatrix;
// Maximum capacities for positions
int posCapacity[] = {2,1,2,2,2};
// Total positions
String[] positions = {"PG","C","SF","PF","util"};
ArrayList playerSet = new ArrayList<player>();
public ArrayList solutionSet;
public int bestCost;
class player {
int value;
int cost;
int pos;
String name;
public player(int value,int cost,int pos,String name) {
this.value = value;
this.cost = cost;
this.pos = pos;
this.name = name;
}
public String toString() {
return("'"+name+"'"+", "+value+", "+cost+", "+positions[pos]);
}
}
// Used to add player to list of available players
void additem(String name,int cost,int value,String pos) {
int i;
for(i=0;i<positions.length;i++) {
if(pos.equals(positions[i]))
break;
}
playerSet.add(new player(value,cost,i,name));
}
// Converts subproblem data to string for hashing
public String encode(int Capacity,int Totalitems,int[] positions) {
String Data = Capacity+","+Totalitems;
for(int i=0;i<positions.length;i++) {
Data = Data + "," + positions[i];
}
return(Data);
}
// Check if subproblem is in hash tables
int isDone(int capacity,int players,int[] positions) {
String k = encode(capacity,players,positions);
if(CostMatrix.containsKey(k)) {
//System.out.println("Key found: "+k+" "+(Integer)CostMatrix.get(k));
return((Integer)CostMatrix.get(k));
}
return(-1);
}
// Adds subproblem added hash table
void addEncode(int capacity,int players,int[] positions,int value) {
String k = encode(capacity,players,positions);
CostMatrix.put(k, value);
}
boolean checkvalid(int capacity,int players) {
return(!(capacity<1||players<0));
}
// Solve the Knapsack recursively with Hash look up
int solve(int capacity,int players,int[] posCapacity) {
// Check if sub problem is valid
if(checkvalid(capacity,players)) {
//System.out.println("Processing: "+encode(capacity,players,posCapacity));
player current = (player)playerSet.get(players);
int sum1 = 0,sum2 = 0,sum3 = 0;
int temp = isDone(capacity,players-1,posCapacity);
// Donot add player
if(temp>-1) {
sum1 = temp;
}
else sum1 = solve(capacity,players-1,posCapacity);
//check if current player can be added to knapsack
if(capacity>=current.cost) {
posCapacity[posCapacity.length-1]--;
temp = isDone(capacity-current.cost,players-1,posCapacity);
posCapacity[posCapacity.length-1]++;
// Add player to util
if(posCapacity[posCapacity.length-1]>0) {
if(temp>-1) {
sum2 = temp+current.value;
}
else {
posCapacity[posCapacity.length-1]--;
sum2 = solve(capacity-current.cost,players-1,posCapacity)+current.value;
posCapacity[posCapacity.length-1]++;
}
}
// Add player at its position
int i = current.pos;
if(posCapacity[i]>0) {
posCapacity[i]--;
temp = isDone(capacity-current.cost,players-1,posCapacity);
posCapacity[i]++;
if(temp>-1) {
sum3 = temp+current.value;
}
else {
posCapacity[i]--;
sum3 = solve(capacity-current.cost,players-1,posCapacity)+current.value;
posCapacity[i]++;
}
}
}
//System.out.println(sum1+ " "+ sum2+ " " + sum3 );
// Evaluate the maximum of all subproblem
int res = Math.max(Math.max(sum1,sum2), sum3);
//add current solution to Hash table
addEncode(capacity, players, posCapacity,res);
//System.out.println("Encoding: "+encode(capacity,players,posCapacity)+" Cost: "+res);
return(res);
}
return(0);
}
void getSolution(int capacity,int players,int[] posCapacity) {
if(players>=0) {
player curr = (player)playerSet.get(players);
int bestcost = isDone(capacity,players,posCapacity);
int sum1 = 0,sum2 = 0,sum3 = 0;
//System.out.println(encode(capacity,players-1,posCapacity)+" "+bestcost);
sum1 = isDone(capacity,players-1,posCapacity);
posCapacity[posCapacity.length-1]--;
sum2 = isDone(capacity-curr.cost,players-1,posCapacity) + curr.value;
posCapacity[posCapacity.length-1]++;
posCapacity[curr.pos]--;
sum3 = isDone(capacity-curr.cost,players-1,posCapacity) + curr.value;
posCapacity[curr.pos]++;
if(bestcost==0)
return;
// Check if player is not added
if(sum1==bestcost) {
getSolution(capacity,players-1,posCapacity);
}
// Check if player is added to util
else if(sum2==bestcost) {
solutionSet.add(curr);
//System.out.println(positions[posCapacity.length-1]+" added");
posCapacity[posCapacity.length-1]--;
getSolution(capacity-curr.cost,players-1,posCapacity);
posCapacity[posCapacity.length-1]++;
}
else {
solutionSet.add(curr);
//System.out.println(positions[curr.pos]+" added");
posCapacity[curr.pos]--;
getSolution(capacity-curr.cost,players-1,posCapacity);
posCapacity[curr.pos]++;
}
}
}
void getOptSet(int capacity) {
CostMatrix = new HashMap<String,Integer>();
bestCost = solve(capacity,playerSet.size()-1,posCapacity);
solutionSet = new ArrayList<player>();
getSolution(capacity, playerSet.size()-1, posCapacity);
}
public static void main(String[] args) {
KnapsackSolver ks = new KnapsackSolver();
ks.additem("david lee", 8000, 30, "PG");
ks.additem("kevin love", 12000, 50, "C");
ks.additem("kemba walker", 7300, 10, "SF");
ks.additem("jrue holiday", 12300, 30, "PF");
ks.additem("stephen curry", 10300, 80, "PG");
ks.additem("lebron james", 5300, 90, "PG");
ks.additem("kevin durant", 2300, 30, "C");
ks.additem("russell westbrook", 9300, 30, "SF");
ks.additem("kevin martin", 8300, 15, "PF");
ks.additem("steve nash", 4300, 15, "C");
ks.additem("kyle lowry", 6300, 20, "PG");
ks.additem("monta ellis", 8300, 30, "C");
ks.additem("dirk nowitzki", 7300, 25, "SF");
ks.additem("david lee", 9500, 35, "PF");
ks.additem("klay thompson", 6800, 28,"PG");
//System.out.println("Items added...");
// System.out.println(ks.playerSet);
int maxCost = 30000;
ks.getOptSet(maxCost);
System.out.println("Best Value: "+ks.bestCost);
System.out.println("Solution Set: "+ks.solutionSet);
}
}
Note: If players with certain positions are added more than its capacity then those added as util because players from any position can be added to util.
Given a Map of objects and designated proportions (let's say they add up to 100 to make it easy):
val ss : Map[String,Double] = Map("A"->42, "B"->32, "C"->26)
How can I generate a sequence such that for a subset of size n there are ~42% "A"s, ~32% "B"s and ~26% "C"s? (Obviously, small n will have larger errors).
(Work language is Scala, but I'm just asking for the algorithm.)
UPDATE: I resisted a random approach since, for instance, there's ~16% chance that the sequence would start with AA and ~11% chance it would start with BB and there would be very low odds that for n precisely == (sum of proportions) the distribution would be perfect. So, following #MvG's answer, I implemented as follows:
/**
Returns the key whose achieved proportions are most below desired proportions
*/
def next[T](proportions : Map[T, Double], achievedToDate : Map[T,Double]) : T = {
val proportionsSum = proportions.values.sum
val desiredPercentages = proportions.mapValues(v => v / proportionsSum)
//Initially no achieved percentages, so avoid / 0
val toDateTotal = if(achievedToDate.values.sum == 0.0){
1
}else{
achievedToDate.values.sum
}
val achievedPercentages = achievedToDate.mapValues(v => v / toDateTotal)
val gaps = achievedPercentages.map{ case (k, v) =>
val gap = desiredPercentages(k) - v
(k -> gap)
}
val maxUnder = gaps.values.toList.sortWith(_ > _).head
//println("Max gap is " + maxUnder)
val gapsForMaxUnder = gaps.mapValues{v => Math.abs(v - maxUnder) < Double.Epsilon }
val keysByHasMaxUnder = gapsForMaxUnder.map(_.swap)
keysByHasMaxUnder(true)
}
/**
Stream of most-fair next element
*/
def proportionalStream[T](proportions : Map[T, Double], toDate : Map[T, Double]) : Stream[T] = {
val nextS = next(proportions, toDate)
val tailToDate = toDate + (nextS -> (toDate(nextS) + 1.0))
Stream.cons(
nextS,
proportionalStream(proportions, tailToDate)
)
}
That when used, e.g., :
val ss : Map[String,Double] = Map("A"->42, "B"->32, "C"->26)
val none : Map[String,Double] = ss.mapValues(_ => 0.0)
val mySequence = (proportionalStream(ss, none) take 100).toList
println("Desired : " + ss)
println("Achieved : " + mySequence.groupBy(identity).mapValues(_.size))
mySequence.map(s => print(s))
println
produces :
Desired : Map(A -> 42.0, B -> 32.0, C -> 26.0)
Achieved : Map(C -> 26, A -> 42, B -> 32)
ABCABCABACBACABACBABACABCABACBACABABCABACABCABACBA
CABABCABACBACABACBABACABCABACBACABABCABACABCABACBA
For a deterministic approach, the most obvious solution would probably be this:
Keep track of the number of occurrences of each item in the sequence so far.
For the next item, choose that item for which the difference between intended and actual count (or proportion, if you prefer that) is maximal, but only if the intended count (resp. proportion) is greater than the actual one.
If there is a tie, break it in an arbitrary but deterministic way, e.g. choosing the alphabetically lowest item.
This approach would ensure an optimal adherence to the prescribed ratio for every prefix of the infinite sequence generated in this way.
Quick & dirty python proof of concept (don't expect any of the variable “names” to make any sense):
import sys
p = [0.42, 0.32, 0.26]
c = [0, 0, 0]
a = ['A', 'B', 'C']
n = 0
while n < 70*5:
n += 1
x = 0
s = n*p[0] - c[0]
for i in [1, 2]:
si = n*p[i] - c[i]
if si > s:
x = i
s = si
sys.stdout.write(a[x])
if n % 70 == 0:
sys.stdout.write('\n')
c[x] += 1
Generates
ABCABCABACABACBABCAABCABACBACABACBABCABACABACBACBAABCABCABACABACBABCAB
ACABACBACABACBABCABACABACBACBAABCABCABACABACBABCAABCABACBACABACBABCABA
CABACBACBAABCABCABACABACBABCABACABACBACBAACBABCABACABACBACBAABCABCABAC
ABACBABCABACABACBACBAACBABCABACABACBACBAABCABCABACABACBABCABACABACBACB
AACBABCABACABACBACBAABCABCABACABACBABCAABCABACBACBAACBABCABACABACBACBA
For every item of the sequence, compute a (pseudo-)random number r equidistributed between 0 (inclusive) and 100 (exclusive).
If 0 ≤ r < 42, take A
If 42 ≤ r < (42+32), take B
If (42+32) ≤ r < (42+32+26)=100, take C
The number of each entry in your subset is going to be the same as in your map, but with a scaling factor applied.
The scaling factor is n/100.
So if n was 50, you would have { Ax21, Bx16, Cx13 }.
Randomize the order to your liking.
The simplest "deterministic" [in terms of #elements of each category] solution [IMO] will be: add elements in predefined order, and then shuffle the resulting list.
First, add map(x)/100 * n elements from each element x chose how you handle integer arithmetics to avoid off by one element], and then shuffle the resulting list.
Shuffling a list is simple with fisher-yates shuffle, which is implemented in most languages: for example java has Collections.shuffle(), and C++ has random_shuffle()
In java, it will be as simple as:
int N = 107;
List<String> res = new ArrayList<String>();
for (Entry<String,Integer> e : map.entrySet()) { //map is predefined Map<String,Integer> for frequencies
for (int i = 0; i < Math.round(e.getValue()/100.0 * N); i++) {
res.add(e.getKey());
}
}
Collections.shuffle(res);
This is nondeterministic, but gives a distribution of values close to MvG's. It suffers from the problem that it could give AAA right at the start. I post it here for completeness' sake given how it proves my dissent with MvG was misplaced (and I don't expect any upvotes).
Now, if someone has an idea for an expand function that is deterministic and won't just duplicate MvG's method (rendering the calc function useless), I'm all ears!
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}
return result;
}
function calc(n, proportions) {
var percents = [],
counts = [],
errors = [],
fnmap = [],
errorSum,
worstIndex;
fnmap[1] = "min";
fnmap[-1] = "max";
proportions.each(function(i, val) {
percents[i] = val / proportions.sum() * n;
counts[i] = Math.round(percents[i]);
errors[i] = counts[i] - percents[i];
});
errorSum = counts.sum() - n;
while (errorSum != 0) {
adjust = errorSum < 0 ? 1 : -1;
worstIndex = errors.indexOf(Math[fnmap[adjust]].apply(0, errors));
counts[worstIndex] += adjust;
errors[worstIndex] = counts[worstIndex] - percents[worstIndex];
errorSum += adjust;
}
return expand(counts);
}
document.body.onload = function() {
document.getElementById('output').innerHTML = calc(99, [25.1, 24.9, 25.9, 24.1]);
};
</script>
</body>
</html>
If you know what this kind of problem is called, let me know (unless you actually know the answer to the question).
If I have a set Z of objects, is there an algorithm for diving them up between a bunch of containers (each holding a certain number of objects)?
To slightly complicate the problem, let's assume the set of objects we start with has a subset X. There are X containers, and each container must hold a single element of X, in addition to other objects (if it has room).
The best way I can think of doing this currently is looking at the disjunction of Z and X, let's call it Y. Then we can generate the z choose x combinations, and then expand that out for all possible combinations of x.
Example:
The actual problem is basically generating all events in a space. Suppose we have two event triggers (X) and 2 event arguments (Y), where Z = X U Y. Each event must have a trigger, and it can have 0...N arguments (depending on the type of event, but that isn't important for now. A trigger can also be an argument. Clearly, in this situation we can have a single event with one trigger and 3 arguments (one of which is the second trigger)
Our event space is as follows (Trigger[Arguments], + indicates a new event):
X1[] + X2[]
X1[Y1] + X2[]
X1[Y2] + X2[]
X1[] + X2[Y1]
X1[] + X2[Y2]
X1[Y1] + X2[Y2]
X1[Y2] + X2[Y1]
X1[X2]
X1[X2,Y1]
X1[X2,Y2]
X1[X2,Y1,Y2]
X2[X1]
X2[X1,Y1]
X2[X1,Y2]
X2[X1,Y1,Y2]
I'm pretty sure that's all the combinations.
Update:
After thinking a bit more about the problem, I have a few thoughts on constraints and stuff: Rules for creating "events":
1) There is an event for every trigger, and every event must have a trigger
2) Event must have > 0 arguments
3) Events cannot share arguments
4) Triggers can be used as arguments
For a brute force solution, perhaps one could generate all permutations of the triggers + events and then eliminate results that don't match the above 4 rules, and treat the ordering as grouping of events?
Thanks for any problem names or ideas!
Algorithm:
For all nonempty subsets Triggers of X:
For all maps from (X \ Triggers) to X:
For all maps from Y to (X union {None}):
print the combination, where an assignment of y in Y to None means y is omitted
In Python:
def assignments(xs, ys):
asgns = [[]]
for x in xs:
asgns1 = []
for y in ys:
for asgn in asgns:
asgn1 = asgn[:]
asgn1.append((x, y))
asgns1.append(asgn1)
asgns = asgns1
return asgns
def combinations(xs, ys):
xroleasgns = assignments(xs, ('argument', 'trigger'))
for xroleasgn in xroleasgns:
triggers = [x for (x, role) in xroleasgn if role == 'trigger']
if (xs or ys) and not triggers:
continue
xargs = [x for (x, role) in xroleasgn if role == 'argument']
for xargasgn in assignments(xargs, triggers):
for yargasgn in assignments(ys, [None] + triggers):
d = dict((x, []) for x in triggers)
for xarg, t in xargasgn:
d[t].append(xarg)
for yarg, t in yargasgn:
if t is not None:
d[t].append(yarg)
print ' + '.join('%s[%s]' % (t, ','.join(args)) for (t, args) in d.iteritems())
"""
>>> assign.combinations(['X1','X2'],['Y1','Y2'])
X1[X2]
X1[X2,Y1]
X1[X2,Y2]
X1[X2,Y1,Y2]
X2[X1]
X2[X1,Y1]
X2[X1,Y2]
X2[X1,Y1,Y2]
X2[] + X1[]
X2[] + X1[Y1]
X2[Y1] + X1[]
X2[] + X1[Y2]
X2[] + X1[Y1,Y2]
X2[Y1] + X1[Y2]
X2[Y2] + X1[]
X2[Y2] + X1[Y1]
X2[Y1,Y2] + X1[]
"""
Here is my java implementation over9000's solution to the original problem:
public static void main(String[] args) throws Exception {
ArrayList xs = new ArrayList();
ArrayList ys = new ArrayList();
xs.add("X1");
xs.add("X2");
ys.add("Y1");
ys.add("Y2");
combinations(xs,ys);
}
private static void combinations(ArrayList xs, ArrayList ys) {
ArrayList def = new ArrayList();
def.add("argument");
def.add("trigger");
ArrayList<ArrayList> xroleasgns = assignments(xs, def);
for(ArrayList xroleasgn:xroleasgns){
// create triggers list
ArrayList triggers = new ArrayList();
for(Object o:xroleasgn){
Pair p = (Pair)o;
if("trigger".equals(p.b.toString()))
triggers.add(p.a);
}
if((xs.size()>0 || ys.size()>0) && triggers.size()==0)
continue;
// create xargs list
ArrayList xargs = new ArrayList();
for(Object o:xroleasgn){
Pair p = (Pair)o;
if("argument".equals(p.b.toString()))
xargs.add(p.a);
}
// Get combinations!
for(ArrayList xargasgn:assignments(xargs,triggers)){
ArrayList yTriggers = new ArrayList(triggers);
yTriggers.add(null);
for(ArrayList yargasgn:assignments(ys,yTriggers)){
// d = dict((x, []) for x in triggers)
HashMap<Object,ArrayList> d = new HashMap<Object,ArrayList>();
for(Object x:triggers)
d.put(x, new ArrayList());
for(Object o:xargasgn){
Pair p = (Pair)o;
d.get(p.b).add(p.a);
}
for(Object o:yargasgn){
Pair p = (Pair)o;
if(p.b!=null){
d.get(p.b).add(p.a);
}
}
for(Entry<Object, ArrayList> e:d.entrySet()){
Object t = e.getKey();
ArrayList args = e.getValue();
System.out.print(t+"["+args.toString()+"]"+"+");
}
System.out.println();
}
}
}
}
private static ArrayList<ArrayList> assignments(ArrayList xs, ArrayList def) {
ArrayList<ArrayList> asgns = new ArrayList<ArrayList>();
asgns.add(new ArrayList()); //put an initial empty arraylist
for(Object x:xs){
ArrayList asgns1 = new ArrayList();
for(Object y:def){
for(ArrayList<Object> asgn:asgns){
ArrayList asgn1 = new ArrayList();
asgn1.addAll(asgn);
asgn1.add(new Pair(x,y));
asgns1.add(asgn1);
}
}
asgns = asgns1;
}
return asgns;
}