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Time Complexity | Find ranks of players given their scores

I was looking through some forums for new questions to solve, and found this one:
Given an array of scores, and an integer k. Player with the same score will have the same rank, and the rank of the player is "the number of players with higher score" + 1. For instance, given scores = [10, 20, 20, 40], the corresponding rank is [4, 2, 2, 1]. Only players with a rank <= k can qualify for the next round. Return the number of player that qualify for the next round.
I have come up with a few ways to solve it, and it seems the best time complexity I can get is O(nlog(n)) with the following algorithm:
sort the array, which has time complexity O(nlogn)
then, start with rank = 1, and update it each time we pass to a lower score, so while rank <= k, keep adding in the amount of players that qualify, and this has time complexity O(n), since we may end up iterating through the whole array.
return the final count
another idea was to create some hashtable that holds the score as the key, and has the number of players as it's value:
iterate through the array and if we find someone with a certain score, then add in another player in the value for that entry in the hashtable, and also if the score we come across is larger than the smallest score in our hashtable, remove the smallest score entry, and put in the new score (so, by the end we only have the top k scores)
then add together all the values in the resulting hashtable (or, at least add together the relevant entries, as the top k scores does not necessarily mean these are the top k ranked players, but we know that only the top k scores are needed, at max, to find the amount that qualify)
This seems to have time complexity O(nk), because we need to iterate through the whole array, but each time check against the min of the current k scores we have, to ensure that we are only keeping the top k scores. This will usually take longer than O(nlogn).
However, I feel there must be an even better way then the methods I have come up with. Does anyone have any advice?
Here is the original forum post: https://leetcode.com/discuss/interview-question/1362837/goldman-sachs-new-analyst-2022-oa

Another idea is as follows:
Create a frequency table that counts the number of players for each score. This is similar to the hashtable idea you mentioned in your post. The keys are unique scores and values are the number of players for that particular score.
Using a min heap push the keys of the frequency table to the heap. As soon as the length of the heap becomes equal to k, for each new push to the heap, pop one from the heap. This guarantees that you end up with the k largest scores in the heap at the end.
Now, loop over the elements in the heap (without popping) which are keys to the freq table, and sum the number of players with those keys in the table.
Time complexity-wise we have run over the initial array in O(n) to create the freq table, we have pushed and popped the number of distinct scores from a heap and since the number of distinct scores is n in the worst case this makes it O(n * log k) operations. Notice that since the heap never grows over k it's log k and not log n. At the end we have looped over the k elements in the heap and summed their values from the freq table which is k operations.
So, this becomes n + (n * log k) + k which reduces to O(n * log k) in big O terms.

This is a small variant on the selection problem: you're looking for the kth smallest element in a list, and the answer you need to output is the number of values less than or equal to the kth element's value. There are many possible solutions, but the standard one is Quickselect, which can give an answer in linear O(n) time. Let's look at a variety of increasingly efficient approaches for the standard selection problem and see their runtime:
Sort the numbers, and count the k smallest: Runtime: O(n log n).
Keep a min-heap, with size bounded at k. Iterate over the array, pushing each value into the heap, and popping whenever the size reaches k+1. Runtime: O(n log k)
Min-Heapify the entire array, and pop k times. Sometimes called 'heapselect'. Runtime: O(n + k log n)
Quickselect. With randomized pivot selection, it has O(n) expected run-time and O(n^2) worst-case runtime, with good average performance. With median-of-medians pivot selection, it has O(n) worst-case runtime, with a higher constant factor.
If you look at the C++ standard library, in the algorithms header, this selection function is called nth element. In practice, variants of quickselect are often used, for example introselect or randomized-quickselect with heapselect fallback, which try to retain randomized QuickSelect's good average performance but without an O(n^2) worst-case.

Why is this brute force algorithm producing the incorrect result?

I'm trying to write a brute-force algorithm that minimises the number of journeys of a herd of cows, subject to the conditions in the docstring.
def brute_force_cow_transport(cows,limit=10):
"""
Finds the allocation of cows that minimizes the number of spaceship trips
via brute force. The brute force algorithm should follow the following method:
1. Enumerate all possible ways that the cows can be divided into separate trips
2. Select the allocation that minimizes the number of trips without making any trip
that does not obey the weight limitation
Does not mutate the given dictionary of cows.
Parameters:
cows - a dictionary of name (string), weight (int) pairs
limit - weight limit of the spaceship (an int)
Returns:
A list of lists, with each inner list containing the names of cows
transported on a particular trip and the overall list containing all the
trips
"""
def weight(sub):
sum = 0
for e in sub:
sum += cows[e]
return sum
valid_trips = []
for part in list(get_partitions(cows)):
if all(weight(sub) <= limit for sub in part):
valid_trips.append(part)
return min(valid_trips)
(The function get_partitions and the dictionary cows have been given in the question)
Where have I gone wrong? I've checked the weight function (that evaluates the weight of a given spaceship trip), so it must be in the last 5 lines. I've checked the code over and over, and it returns a sub-optimal answer:
[['Florence', 'Lola'],
['Maggie', 'Milkshake', 'Moo Moo'],
['Herman'],
['Oreo'],
['Millie'],
['Henrietta'],
['Betsy']]
The syntax is fine; there are no errors being produced, yet I have a sub-optimal (but valid) answer. Why is this?

The question here is:
How do I find the shortest sublist in a nested list?
To do this, change the last line to:
min(valid_trips, key=len)

Sorting points on multiple lines

Given that we have two lines on a graph (I just noticed that I inverted the numbers on the Y axis, this was a mistake, it should go from 11-1)
And we only care about whole number X axis intersections
We need to order these points from highest Y value to lowest Y value regardless of their position on the X axis (Note I did these pictures by hand so they may not line up perfectly).
I have a couple of questions:
1) I have to assume this is a known problem, but does it have a particular name?
2) Is there a known optimal solution when dealing with tens of billions (or hundreds of millions) of lines? Our current process of manually calculating each point and then comparing it to a giant list requires hours of processing. Even though we may have a hundred million lines we typically only want the top 100 or 50,000 results some of them are so far "below" other lines that calculating their points is unnecessary.

Your data structure is a set of tuples
lines = {(y0, Δy0), (y1, Δy1), ...}
You need only the ntop points, hence build a set containing only
the top ntop yi values, with a single pass over the data
top_points = choose(lines, ntop)
EDIT --- to choose the ntop we had to keep track of the smallest
one, and this is interesting info, so let's return also this value
from choose, also we need to initialize decremented
top_points, smallest = choose(lines, ntop)
decremented = top_points
and start a loop...
while True:
Generate a set of decremented values
decremented = {(y-Δy, Δy) for y, Δy in top_points}
decremented = {(y-Δy, Δy) for y, Δy in decremented if y>smallest}
if decremented == {}: break
Generate a set of candidates
candidates = top_lines.union(decremented)
generate a new set of top points
new_top_points, smallest = choose(candidates, ntop)
The following is no more necessary
check if new_top_points == top_points
if new_top_points == top_points: break
top_points = new_top_points</strike>
of course we are in a loop...
The difficult part is the choose function, but I think that this
answer to the question
How can I sort 1 million numbers, and only print the top 10 in Python?
could help you.

It's not a really complicated thing, just a "normal" sorting problem.
Usually sorting requires a large amount of computing time. But your case is one where you don't need to use complex sorting techniques.
You on both graphs are growing or falling constantly, there are no "jumps". You can use this to your advantage. The basic algorithm:
identify if a graph is growing or falling.
write a generator, that generates the values; from left to right if raising, form right to left if falling.
get the first value from both graphs
insert the lower on into the result list
get a new value from the graph that had the lower value
repeat the last two steps until one generator is "empty"
append the leftover items from the other generator.

Given a list L labeled 1 to N, and a process that "removes" a random element from consideration, how can one efficiently keep track of min(L)?

The question is pretty much in the title, but say I have a list L
L = [1,2,3,4,5]
min(L) = 1 here. Now I remove 4. The min is still 1. Then I remove 2. The min is still 1. Then I remove 1. The min is now 3. Then I remove 3. The min is now 5, and so on.
I am wondering if there is a good way to keep track of the min of the list at all times without needing to do min(L) or scanning through the entire list, etc.
There is an efficiency cost to actually removing the items from the list because it has to move everything else over. Re-sorting the list each time is expensive, too. Is there a way around this?

To remove a random element you need to know what elements have not been removed yet.
To know the minimum element, you need to sort or scan the items.
A min heap implemented as an array neatly solves both problems. The cost to remove an item is O(log N) and the cost to find the min is O(1). The items are stored contiguously in an array, so choosing one at random is very easy, O(1).
The min heap is described on this Wikipedia page
BTW, if the data are large, you can leave them in place and store pointers or indexes in the min heap and adjust the comparison operator accordingly.

Google for self-balancing binary search trees. Building one from the initial list takes O(n lg n) time, and finding and removing an arbitrary item will take O(lg n) (instead of O(n) for finding/removing from a simple list). A smallest item will always appear in the root of the tree.
This question may be useful. It provides links to several implementation of various balanced binary search trees. The advice to use a hash table does not apply well to your case, since it does not address maintaining a minimum item.

Here's a solution that need O(N lg N) preprocessing time + O(lg N) update time and O(lg(n)*lg(n)) delete time.
Preprocessing:
step 1: sort the L
step 2: for each item L[i], map L[i]->i
step 3: Build a Binary Indexed Tree or segment tree where for every 1<=i<=length of L, BIT[i]=1 and keep the sum of the ranges.
Query type delete:
Step 1: if an item x is said to be removed, with a binary search on array L (where L is sorted) or from the mapping find its index. set BIT[index[x]] = 0 and update all the ranges. Runtime: O(lg N)
Query type findMin:
Step 1: do a binary search over array L. for every mid, find the sum on BIT from 1-mid. if BIT[mid]>0 then we know some value<=mid is still alive. So we set hi=mid-1. otherwise we set low=mid+1. Runtime: O(lg**2N)
Same can be done with Segment tree.
Edit: If I'm not wrong per query can be processed in O(1) with Linked List

If sorting isn't in your best interest, I would suggest only do comparisons where you need to do them. If you remove elements that are not the old minimum, and you aren't inserting any new elements, there isn't a re-scan necessary for a minimum value.
Can you give us some more information about the processing going on that you are trying to do?
Comment answer: You don't have to compute min(L). Just keep track of its index and then only re-run the scan for min(L) when you remove at(or below) the old index (and make sure you track it accordingly).

Your current approach of rescanning when the minimum is removed is O(1)-time in expectation for each removal (assuming every item is equally likely to be removed).
Given a list of n items, a rescan is necessary with probability 1/n, so the expected work at each step is n * 1/n = O(1).

How to approach a number guessing game (with a twist) algorithm?

Update(July 2020): Question is 9 years old but still one that I'm deeply interested in. In the time since, machine learning(RNN's, CNN's, GANS,etc), new approaches and cheap GPU's have risen that enable new approaches. I thought it would be fun to revisit this question to see if there are new approaches.
I am learning programming (Python and algorithms) and was trying to work on a project that I find interesting. I have created a few basic Python scripts, but I’m not sure how to approach a solution to a game I am trying to build.
Here’s how the game will work:
Users will be given items with a value. For example,
Apple = 1
Pears = 2
Oranges = 3
They will then get a chance to choose any combo of them they like (i.e. 100 apples, 20 pears, and one orange). The only output the computer gets is the total value (in this example, it's currently $143). The computer will try to guess what they have. Which obviously it won’t be able to get correctly the first turn.
Value quantity(day1) value(day1)
Apple 1 100 100
Pears 2 20 40
Orange 3 1 3
Total 121 143
The next turn the user can modify their numbers but no more than 5% of the total quantity (or some other percent we may chose. I’ll use 5% for example.). The prices of fruit can change(at random) so the total value may change based on that also (for simplicity I am not changing fruit prices in this example). Using the above example, on day 2 of the game, the user returns a value of $152 and $164 on day 3. Here's an example:
Quantity (day2) %change (day2) Value (day2) Quantity (day3) %change (day3) Value(day3)
104 104 106 106
21 42 23 46
2 6 4 12
127 4.96% 152 133 4.72% 164
*(I hope the tables show up right, I had to manually space them so hopefully it's not just doing it on my screen, if it doesn't work let me know and I'll try to upload a screenshot.)
I am trying to see if I can figure out what the quantities are over time (assuming the user will have the patience to keep entering numbers). I know right now my only restriction is the total value cannot be more than 5% so I cannot be within 5% accuracy right now so the user will be entering it forever.
What I have done so far
Here’s my solution so far (not much). Basically, I take all the values and figure out all the possible combinations of them (I am done this part). Then I take all the possible combos and put them in a database as a dictionary (so for example for $143, there could be a dictionary entry {apple:143, Pears:0, Oranges :0}..all the way to {apple:0, Pears:1, Oranges :47}. I do this each time I get a new number so I have a list of all possibilities.
Here’s where I’m stuck. In using the rules above, how can I figure out the best possible solution? I think I’ll need a fitness function that automatically compares the two days data and removes any possibilities that have more than 5% variance of the previous days data.
Questions:
So my question with user changing the total and me having a list of all the probabilities, how should I approach this? What do I need to learn? Is there any algorithms out there or theories that I can use that are applicable? Or, to help me understand my mistake, can you suggest what rules I can add to make this goal feasible (if it's not in its current state. I was thinking adding more fruits and saying they must pick at least 3, etc..)? Also, I only have a vague understanding of genetic algorithms, but I thought I could use them here, if is there something I can use?
I'm very very eager to learn so any advice or tips would be greatly appreciated (just please don't tell me this game is impossible).
UPDATE: Getting feedback that this is hard to solve. So I thought I'd add another condition to the game that won't interfere with what the player is doing (game stays the same for them) but everyday the value of the fruits change price (randomly). Would that make it easier to solve? Because within a 5% movement and certain fruit value changes, only a few combinations are probable over time.
Day 1, anything is possible and getting a close enough range is almost impossible, but as the prices of fruits change and the user can only choose a 5% change, then shouldn't (over time) the range be narrow and narrow. In the above example, if prices are volatile enough I think I could brute force a solution that gave me a range to guess in, but I'm trying to figure out if there's a more elegant solution or other solutions to keep narrowing this range over time.
UPDATE2: After reading and asking around, I believe this is a hidden Markov/Viterbi problem that tracks the changes in fruit prices as well as total sum (weighting the last data point the heaviest). I'm not sure how to apply the relationship though. I think this is the case and could be wrong but at the least I'm starting to suspect this is a some type of machine learning problem.
Update 3: I am created a test case (with smaller numbers) and a generator to help automate the user generated data and I am trying to create a graph from it to see what's more likely.
Here's the code, along with the total values and comments on what the users actually fruit quantities are.
#!/usr/bin/env python
import itertools
# Fruit price data
fruitPriceDay1 = {'Apple':1, 'Pears':2, 'Oranges':3}
fruitPriceDay2 = {'Apple':2, 'Pears':3, 'Oranges':4}
fruitPriceDay3 = {'Apple':2, 'Pears':4, 'Oranges':5}
# Generate possibilities for testing (warning...will not scale with large numbers)
def possibilityGenerator(target_sum, apple, pears, oranges):
allDayPossible = {}
counter = 1
apple_range = range(0, target_sum + 1, apple)
pears_range = range(0, target_sum + 1, pears)
oranges_range = range(0, target_sum + 1, oranges)
for i, j, k in itertools.product(apple_range, pears_range, oranges_range):
if i + j + k == target_sum:
currentPossible = {}
#print counter
#print 'Apple', ':', i/apple, ',', 'Pears', ':', j/pears, ',', 'Oranges', ':', k/oranges
currentPossible['apple'] = i/apple
currentPossible['pears'] = j/pears
currentPossible['oranges'] = k/oranges
#print currentPossible
allDayPossible[counter] = currentPossible
counter = counter +1
return allDayPossible
# Total sum being returned by user for value of fruits
totalSumDay1=26 # Computer does not know this but users quantities are apple: 20, pears 3, oranges 0 at the current prices of the day
totalSumDay2=51 # Computer does not know this but users quantities are apple: 21, pears 3, oranges 0 at the current prices of the day
totalSumDay3=61 # Computer does not know this but users quantities are apple: 20, pears 4, oranges 1 at the current prices of the day
graph = {}
graph['day1'] = possibilityGenerator(totalSumDay1, fruitPriceDay1['Apple'], fruitPriceDay1['Pears'], fruitPriceDay1['Oranges'] )
graph['day2'] = possibilityGenerator(totalSumDay2, fruitPriceDay2['Apple'], fruitPriceDay2['Pears'], fruitPriceDay2['Oranges'] )
graph['day3'] = possibilityGenerator(totalSumDay3, fruitPriceDay3['Apple'], fruitPriceDay3['Pears'], fruitPriceDay3['Oranges'] )
# Sample of dict = 1 : {'oranges': 0, 'apple': 0, 'pears': 0}..70 : {'oranges': 8, 'apple': 26, 'pears': 13}
print graph

We'll combine graph-theory and probability:
On the 1st day, build a set of all feasible solutions. Lets denote the solutions set as A1={a1(1), a1(2),...,a1(n)}.
On the second day you can again build the solutions set A2.
Now, for each element in A2, you'll need to check if it can be reached from each element of A1 (given x% tolerance). If so - connect A2(n) to A1(m). If it can't be reached from any node in A1(m) - you can delete this node.
Basically we are building a connected directed acyclic graph.
All paths in the graph are equally likely. You can find an exact solution only when there is a single edge from Am to Am+1 (from a node in Am to a node in Am+1).
Sure, some nodes appear in more paths than other nodes. The probability for each node can be directly deduced based on the number of paths that contains this node.
By assigning a weight to each node, which equals to the number of paths that leads to this node, there is no need to keep all history, but only the previous day.
Also, have a look at non-negative-values linear diphantine equations - A question I asked a while ago. The accepted answer is a great way to enumarte all combos in each step.

Disclaimer: I changed my answer dramatically after temporarily deleting my answer and re-reading the question carefully as I misread some critical parts of the question. While still referencing similar topics and algorithms, the answer was greatly improved after I attempted to solve some of the problem in C# myself.
Hollywood version
The problem is a Dynamic constraint satisfaction problem (DCSP), a variation on Constraint satisfaction problems (CSP.)
Use Monte Carlo to find potential solutions for a given day if values and quantity ranges are not tiny. Otherwise, use brute force to find every potential solutions.
Use Constraint Recording (related to DCSP), applied in cascade to previous days to restrict the potential solution set.
Cross your fingers, aim and shoot (Guess), based on probability.
(Optional) Bruce Willis wins.
Original version
First, I would like to state what I see two main problems here:
The sheer number of possible solutions. Knowing only the number of items and the total value, lets say 3 and 143 for example, will yield a lot of possible solutions. Plus, it is not easy to have an algorithm picking valid solution without inevitably trying invalid solutions (total not equal to 143.)
When possible solutions are found for a given day Di, one must find a way to eliminate potential solutions with the added information given by { Di+1 .. Di+n }.
Let's lay down some bases for the upcoming examples:
Lets keep the same item values, the whole game. It can either be random or chosen by the user.
The possible item values is bound to the very limited range of [1-10], where no two items can have the same value.
No item can have a quantity greater than 100. That means: [0-100].
In order to solve this more easily I took the liberty to change one constraint, which makes the algorithm converge faster:
The "total quantity" rule is overridden by this rule: You can add or remove any number of items within the [1-10] range, total, in one day. However, you cannot add or remove the same number of items, total, more than twice. This also gives the game a maximum lifecycle of 20 days.
This rule enables us to rule out solutions more easily. And, with non-tiny ranges, renders Backtracking algorithms still useless, just like your original problem and rules.
In my humble opinion, this rule is not the essence of the game but only a facilitator, enabling the computer to solve the problem.
Problem 1: Finding potential solutions
For starters, problem 1. can be solved using a Monte Carlo algorithm to find a set of potential solutions. The technique is simple: Generate random numbers for item values and quantities (within their respective accepted range). Repeat the process for the required number of items. Verify whether or not the solution is acceptable. That means verifying if items have distinct values and the total is equal to our target total (say, 143.)
While this technique has the advantage of being easy to implement it has some drawbacks:
The user's solution is not guaranteed to appear in our results.
There is a lot of "misses". For instance, it takes more or less 3,000,000 tries to find 1,000 potential solutions given our constraints.
It takes a lot of time: around 4 to 5 seconds on my lazy laptop.
How to get around these drawback? Well...
Limit the range to smaller values and
Find an adequate number of potential solutions so there is a good chance the user's solution appears in your solution set.
Use heuristics to find solutions more easily (more on that later.)
Note that the more you restrict the ranges, the less useful while be the Monte Carlo algorithm is, since there will be few enough valid solutions to iterate on them all in reasonable time. For constraints { 3, [1-10], [0-100] } there is around 741,000,000 valid solutions (not constrained to a target total value.) Monte Carlo is usable there. For { 3, [1-5], [0-10] }, there is only around 80,000. No need to use Monte Carlo; brute force for loops will do just fine.
I believe the problem 1 is what you would call a Constraint satisfaction problem (or CSP.)
Problem 2: Restrict the set of potential solutions
Given the fact that problem 1 is a CSP, I would go ahead and call problem 2, and the problem in general, a Dynamic CSP (or DCSP.)
[DCSPs] are useful when the original formulation of a
problem is altered in some way, typically because the set of
constraints to consider evolves because of the environment. DCSPs
are viewed as a sequence of static CSPs, each one a transformation of
the previous one in which variables and constraints can be added
(restriction) or removed (relaxation).
One technique used with CSPs that might be useful to this problem is called Constraint Recording:
With each change in the environment (user entered values for Di+1), find information about the new constraint: What are the possibly "used" quantities for the add-remove constraint.
Apply the constraint to every preceding day in cascade. Rippling effects might significantly reduce possible solutions.
For this to work, you need to get a new set of possible solutions every day; Use either brute force or Monte Carlo. Then, compare solutions of Di to Di-1 and keep only solutions that can succeed to previous days' solutions without violating constraints.
You will probably have to keep an history of what solutions lead to what other solutions (probably in a directed graph.) Constraint recording enables you to remember possible add-remove quantities and rejects solutions based on that.
There is a lot of other steps that could be taken to further improve your solution. Here are some ideas:
Record constraints for item-value combinations found in previous days solutions. Reject other solutions immediately (as item values must not change.) You could even find a smaller solution sets for each existing solution using solution-specific constraints to reject invalid solutions earlier.
Generate some "mutant", full-history, solutions each day in order to "repair" the case where the D1 solution set doesn't contain the user's solution. You could use a genetic algorithm to find a mutant population based on an existing solution set.)
Use heuristics in order find solutions easily (e.g. when a valid solution is found, try and find variations of this solution by substituting quantities around.)
Use behavioral heuristics in order to predict some user actions (e.g. same quantity for every item, extreme patterns, etc.)
Keep making some computations while the user is entering new quantities.
Given all of this, try and figure out a ranking system based on occurrence of solutions and heuristics to determine a candidate solution.

This problem is impossible to solve.
Let's say that you know exactly for what ratio number of items was increased, not just what is the maximum ratio for this.
A user has N fruits and you have D days of guessing.
In each day you get N new variables and then you have in total D*N variables.
For each day you can generate only two equations. One equation is the sum of n_item*price and other is based on a known ratio. In total you have at most 2*D equations if they are all independent.
2*D < N*D for all N > 2

I wrote a program to play the game. Of course, I had to automate the human side, but I believe I did it all in such a way that I shouldn't invalidate my approach when played against a real human.
I approached this from a machine learning perspective and treated the problem as a hidden markov model where the total price was the observation. My solution is to use a particle filter. This solution is written in Python 2.7 using NumPy and SciPy.
I stated any assumptions I made either explicitly in the comments or implicitly in the code. I also set some additional constraints for the sake of getting code to run in an automated fashion. It's not particularly optimized as I tried to err on the side comprehensibility rather than speed.
Each iteration outputs the current true quantities and the guess. I just pipe the output to a file so I can review it easily. An interesting extension would be to plot the output on a graph either 2D (for 2 fruits) or 3D (for 3 fruits). Then you would be able to see the particle filter hone in on the solution.
Update:
Edited the code to include updated parameters after tweaking. Included plotting calls using matplotlib (via pylab). Plotting works on Linux-Gnome, your mileage may vary. Defaulted NUM_FRUITS to 2 for plotting support. Just comment out all the pylab calls to remove plotting and be able to change NUM_FRUITS to anything.
Does a good job estimating the current fxn represented by UnknownQuantities X Prices = TotalPrice. In 2D (2 Fruits) this is a line, in 3D (3 Fruits) it'd be a plane. Seems to be too little data for the particle filter to reliably hone in on the correct quantities. Need a little more smarts on top of the particle filter to really bring together the historical information. You could try converting the particle filter to 2nd- or 3rd-order.
Update 2:
I've been playing around with my code, a lot. I tried a bunch of things and now present the final program that I'll be making (starting to burn out on this idea).
Changes:
The particles now use floating points rather than integers. Not sure if this had any meaningful effect, but it is a more general solution. Rounding to integers is done only when making a guess.
Plotting shows true quantities as green square and current guess as red square. Currently believed particles shown as blue dots (sized by how much we believe them). This makes it really easy to see how well the algorithm is working. (Plotting also tested and working on Win 7 64-bit).
Added parameters for turning off/on quantity changing and price changing. Of course, both 'off' is not interesting.
It does a pretty dang good job, but, as has been noted, it's a really tough problem, so getting the exact answer is hard. Turning off CHANGE_QUANTITIES produces the simplest case. You can get an appreciation for the difficulty of the problem by running with 2 fruits with CHANGE_QUANTITIES off. See how quickly it hones in on the correct answer then see how harder it is as you increase the number of fruit.
You can also get a perspective on the difficulty by keeping CHANGE_QUANTITIES on, but adjusting the MAX_QUANTITY_CHANGE from very small values (.001) to "large" values (.05).
One situation where it struggles is if on dimension (one fruit quantity) gets close to zero. Because it's using an average of particles to guess it will always skew away from a hard boundary like zero.
In general this makes a great particle filter tutorial.
from __future__ import division
import random
import numpy
import scipy.stats
import pylab
# Assume Guesser knows prices and total
# Guesser must determine the quantities
# All of pylab is just for graphing, comment out if undesired
# Graphing only graphs first 2 FRUITS (first 2 dimensions)
NUM_FRUITS = 3
MAX_QUANTITY_CHANGE = .01 # Maximum percentage change that total quantity of fruit can change per iteration
MAX_QUANTITY = 100 # Bound for the sake of instantiating variables
MIN_QUANTITY_TOTAL = 10 # Prevent degenerate conditions where quantities all hit 0
MAX_FRUIT_PRICE = 1000 # Bound for the sake of instantiating variables
NUM_PARTICLES = 5000
NEW_PARTICLES = 500 # Num new particles to introduce each iteration after guessing
NUM_ITERATIONS = 20 # Max iterations to run
CHANGE_QUANTITIES = True
CHANGE_PRICES = True
'''
Change individual fruit quantities for a random amount of time
Never exceed changing fruit quantity by more than MAX_QUANTITY_CHANGE
'''
def updateQuantities(quantities):
old_total = max(sum(quantities), MIN_QUANTITY_TOTAL)
new_total = old_total
max_change = int(old_total * MAX_QUANTITY_CHANGE)
while random.random() > .005: # Stop Randomly
change_index = random.randint(0, len(quantities)-1)
change_val = random.randint(-1*max_change,max_change)
if quantities[change_index] + change_val >= 0: # Prevent negative quantities
quantities[change_index] += change_val
new_total += change_val
if abs((new_total / old_total) - 1) > MAX_QUANTITY_CHANGE:
quantities[change_index] -= change_val # Reverse the change
def totalPrice(prices, quantities):
return sum(prices*quantities)
def sampleParticleSet(particles, fruit_prices, current_total, num_to_sample):
# Assign weight to each particle using observation (observation is current_total)
# Weight is the probability of that particle (guess) given the current observation
# Determined by looking up the distance from the hyperplane (line, plane, hyperplane) in a
# probability density fxn for a normal distribution centered at 0
variance = 2
distances_to_current_hyperplane = [abs(numpy.dot(particle, fruit_prices)-current_total)/numpy.linalg.norm(fruit_prices) for particle in particles]
weights = numpy.array([scipy.stats.norm.pdf(distances_to_current_hyperplane[p], 0, variance) for p in range(0,NUM_PARTICLES)])
weight_sum = sum(weights) # No need to normalize, as relative weights are fine, so just sample un-normalized
# Create new particle set weighted by weights
belief_particles = []
belief_weights = []
for p in range(0, num_to_sample):
sample = random.uniform(0, weight_sum)
# sum across weights until we exceed our sample, the weight we just summed is the index of the particle we'll use
p_sum = 0
p_i = -1
while p_sum < sample:
p_i += 1
p_sum += weights[p_i]
belief_particles.append(particles[p_i])
belief_weights.append(weights[p_i])
return belief_particles, numpy.array(belief_weights)
'''
Generates new particles around the equation of the current prices and total (better particle generation than uniformly random)
'''
def generateNewParticles(current_total, fruit_prices, num_to_generate):
new_particles = []
max_values = [int(current_total/fruit_prices[n]) for n in range(0,NUM_FRUITS)]
for p in range(0, num_to_generate):
new_particle = numpy.array([random.uniform(1,max_values[n]) for n in range(0,NUM_FRUITS)])
new_particle[-1] = (current_total - sum([new_particle[i]*fruit_prices[i] for i in range(0, NUM_FRUITS-1)])) / fruit_prices[-1]
new_particles.append(new_particle)
return new_particles
# Initialize our data structures:
# Represents users first round of quantity selection
fruit_prices = numpy.array([random.randint(1,MAX_FRUIT_PRICE) for n in range(0,NUM_FRUITS)])
fruit_quantities = numpy.array([random.randint(1,MAX_QUANTITY) for n in range(0,NUM_FRUITS)])
current_total = totalPrice(fruit_prices, fruit_quantities)
success = False
particles = generateNewParticles(current_total, fruit_prices, NUM_PARTICLES) #[numpy.array([random.randint(1,MAX_QUANTITY) for n in range(0,NUM_FRUITS)]) for p in range(0,NUM_PARTICLES)]
guess = numpy.average(particles, axis=0)
guess = numpy.array([int(round(guess[n])) for n in range(0,NUM_FRUITS)])
print "Truth:", str(fruit_quantities)
print "Guess:", str(guess)
pylab.ion()
pylab.draw()
pylab.scatter([p[0] for p in particles], [p[1] for p in particles])
pylab.scatter([fruit_quantities[0]], [fruit_quantities[1]], s=150, c='g', marker='s')
pylab.scatter([guess[0]], [guess[1]], s=150, c='r', marker='s')
pylab.xlim(0, MAX_QUANTITY)
pylab.ylim(0, MAX_QUANTITY)
pylab.draw()
if not (guess == fruit_quantities).all():
for i in range(0,NUM_ITERATIONS):
print "------------------------", i
if CHANGE_PRICES:
fruit_prices = numpy.array([random.randint(1,MAX_FRUIT_PRICE) for n in range(0,NUM_FRUITS)])
if CHANGE_QUANTITIES:
updateQuantities(fruit_quantities)
map(updateQuantities, particles) # Particle Filter Prediction
print "Truth:", str(fruit_quantities)
current_total = totalPrice(fruit_prices, fruit_quantities)
# Guesser's Turn - Particle Filter:
# Prediction done above if CHANGE_QUANTITIES is True
# Update
belief_particles, belief_weights = sampleParticleSet(particles, fruit_prices, current_total, NUM_PARTICLES-NEW_PARTICLES)
new_particles = generateNewParticles(current_total, fruit_prices, NEW_PARTICLES)
# Make a guess:
guess = numpy.average(belief_particles, axis=0, weights=belief_weights) # Could optimize here by removing outliers or try using median
guess = numpy.array([int(round(guess[n])) for n in range(0,NUM_FRUITS)]) # convert to integers
print "Guess:", str(guess)
pylab.cla()
#pylab.scatter([p[0] for p in new_particles], [p[1] for p in new_particles], c='y') # Plot new particles
pylab.scatter([p[0] for p in belief_particles], [p[1] for p in belief_particles], s=belief_weights*50) # Plot current particles
pylab.scatter([fruit_quantities[0]], [fruit_quantities[1]], s=150, c='g', marker='s') # Plot truth
pylab.scatter([guess[0]], [guess[1]], s=150, c='r', marker='s') # Plot current guess
pylab.xlim(0, MAX_QUANTITY)
pylab.ylim(0, MAX_QUANTITY)
pylab.draw()
if (guess == fruit_quantities).all():
success = True
break
# Attach new particles to existing particles for next run:
belief_particles.extend(new_particles)
particles = belief_particles
else:
success = True
if success:
print "Correct Quantities guessed"
else:
print "Unable to get correct answer within", NUM_ITERATIONS, "iterations"
pylab.ioff()
pylab.show()

For your initial rules:
From my school years, I would say that if we make an abstraction of the 5% changes, we have everyday an equation with three unknown values (sorry I don't know the maths vocabulary in English), which are the same values as previous day.
At day 3, you have three equations, three unknown values, and the solution should be direct.
I guess the 5% change each day may be forgotten if the values of the three elements are different enough, because, as you said, we will use approximations and round the numbers.
For your adapted rules:
Too many unknowns - and changing - values in this case, so there is no direct solution I know of. I would trust Lior on this; his approach looks fine! (If you have a limited range for prices and quantities.)

I realized that my answer was getting quite lengthy, so I moved the code to the top (which is probably what most people are interested in). Below it there are two things:
an explanation why (deep) neural networks are not a good approach to this problem, and
an explanation why we can't uniquely determine the human's choices with the given information.
For those of you interested in either topic, please see below. For the rest of you, here is the code.
Code that finds all possible solutions
As I explain further down in the answer, your problem is under-determined. In the average case, there are many possible solutions, and this number grows at least exponentially as the number of days increases. This is true for both, the original and the extended problem. Nevertheless, we can (sort of) efficiently find all solutions (it's NP hard, so don't expect too much).
Backtracking (from the 1960s, so not exactly modern) is the algorithm of choice here. In python, we can write it as a recursive generator, which is actually quite elegant:
def backtrack(pos, daily_total, daily_item_value, allowed_change, iterator_bounds, history=None):
if pos == len(daily_total):
yield np.array(history)
return
it = [range(start, stop, step) for start, stop, step in iterator_bounds[pos][:-1]]
for partial_basket in product(*it):
if history is None:
history = [partial_basket]
else:
history.append(partial_basket)
# ensure we only check items that match the total basket value
# for that day
partial_value = np.sum(np.array(partial_basket) * daily_item_value[pos, :-1])
if (daily_total[pos] - partial_value) % daily_item_value[pos, -1] != 0:
history.pop()
continue
last_item = (daily_total[pos] - partial_value) // daily_item_value[pos, -1]
if last_item < 0:
history.pop()
continue
basket = np.array([*partial_basket] + [int(last_item)])
basket_value = np.sum(basket * daily_item_value[pos])
history[-1] = basket
if len(history) > 1:
# ensure that today's basket stays within yesterday's range
previous_basket = history[-2]
previous_basket_count = np.sum(previous_basket)
current_basket_count = np.sum(basket)
if (np.abs(current_basket_count - previous_basket_count) > allowed_change * previous_basket_count):
history.pop()
continue
yield from backtrack(pos + 1, daily_total, daily_item_value, allowed_change, iterator_bounds, history)
history.pop()
This approach essentially structures all possible candidates into a large tree and then performs depth first search with pruning whenever a constraint is violated. Whenever a leaf node is encountered, we yield the result.
Tree search (in general) can be parallelized, but that is out of scope here. It will make the solution less readable without much additional insight. The same goes for reducing constant overhead of the code, e.g., working the constraints if ...: continue into the iterator_bounds variable and do less checks.
I put the full code example (including a simulator for the human side of the game) at the bottom of this answer.
Modern Machine Learning for this problem
Question is 9 years old but still one that I'm deeply interested in. In the time since, machine learning(RNN's, CNN's, GANS,etc), new approaches and cheap GPU's have risen that enable new approaches. I thought it would be fun to revisit this question to see if there are new approaches.
I really like your enthusiasm for the world of deep neural networks; unfortunately they simply do not apply here for a few reasons:
(Exactness) If you need an exact solution, like for your game, NNs can't provide that.
(Integer Constraint) The currently dominant NN training methods are gradient descent based, so the problem has to be differentiable or you need to be able to reformulate it in such a way that it becomes differentiable; constraining yourself to integers kills GD methods in the cradle. You could try evolutionary algorithms to search for a parameterization. This does exist, but those methods are currently a lot less established.
(Non-Convexity) In the typical formulation, training a NN is a local method, which means you will find exactly 1 (locally optimal) solution if your algorithm is converging. In the average case, your game has many possible solutions for both the original and extended version. This not only means that - on average - you can't figure out the human's choice (basket), but also that you have no control over which of the many solutions the NN will find. Current NN success stories suffer the same fate, but they tend to don't really care, because they only want some solution instead of a specific one. Some okay-ish solution beats the hell out of no solution at all.
(Expert Domain Knowledge) For this game, you have a lot of domain knowledge that can be exploited to improve the optimization/learning. Taking full advantage of arbitrary domain knowledge in NNs is not trivial and for this game building a custom ML model (not a neural network) would be easier and more efficient.
Why the game can not be uniquely solved - Part 1
Let's consider a substitute problem first and lift the integer requirement, i.e., the basket (human choice of N fruits for a given day) can have fractional fruits (0.3 oranges).
The total value constraint np.dot(basket, daily_price) == total_value limits the possible solutions for the basket; it reduces the problem by one dimension. Freely pick amounts for N-1 fruits, and you can always find a value for the N-th fruit to satisfy the constraint. So while it seems that there are N choices to make for a day, there are actually only N-1 that we can make freely, and the last one will be fully determined by our previous choices. So for each day the game goes on, we need to estimate an additional N-1 choices/variables.
We might want to enforce that all the choices are greater than 0, but that only reduces the interval from which we can choose a number; any open interval of real numbers has infinitely many numbers in it, so we will never run out of options because of this. Still N-1 choices to make.
Between two days, the total basket volume np.sum(basket) only changes by at most some_percent of the previous day, i.e. np.abs(np.sum(previous_basket) - np.sum(basket)) <= some_percent * np.sum(previous_basket). Some of the choices we could make at a given day will change the basket by more than some_percent of the previous day. To make sure we never violate this, we can freely make N-2 choices and then have to pick the N-1-th variable so that adding it and adding the N-the variable (which is fixed from our previous choices) stays within some_percent. (Note: This is an inequality constraint, so it will only reduce the number of choices if we have equality, i.e., the basket changes by exactly some_percent. In optimization theory this is known as the constraint being active.)
We can again think about the constraint that all choices should be greater 0, but the argument remains that this simply changes the interval from which we can now freely choose N-2 variables.
So after D days we are left with N-1 choices to estimate from the first day (no change constraint) and (D-1)*(N-2) choices to estimate for each following day. Unfortunately, we ran out of constraints to further reduce this number and the number of unknowns grows by at least N-2 each day. This is essentially what what Luka Rahne meant with "2*D < N*D for all N > 2". We will likely find many candidates which are all equally probable.
The exact food prices each day don't matter for this. As long as they are of some value, they will constrain one of the choices. Hence, if you extend your game in the way you specify, there is always a chance for infinitely many solutions; regardless of the number of days.
Why the game can still not be uniquely solved - Part 2
There is one constraint we didn't look at which might help fix this: only allow integer solutions for choices. The problem with integer constraints is that they are very complex to deal with. However, our main concern here is if adding this constraint will allow us to uniquely solve the problem given enough days. For this, there is a rather intuitive counter-example. Suppose you have 3 consecutive days, and for the 1st and 3d day, the total value constraint only allows one basket. In other words, we know the basket for day 1 and day 3, but not for day 2. Here, we only know it's total value, that it is within some_percent of day 1 and that day 3 is within some_percent of day 2. Is this enough information to always work out what is in the basket on day 2?
some_percent = 0.05
Day 1: basket: [3 2] prices: [10 7] total_value: 44
Day 2: basket: [x y] prices: [5 5] total_value: 25
Day 3: basket: [2 3] prices: [9 5] total_value: 33
Possible Solutions Day 2: [2 3], [3 2]
Above is one example, where we know the values for two days thanks to the total value constraint, but that still won't allow us to work out the exact composition of the basket at day 2. Thus, while it may be possible to work it out in some cases, it is not possible in general. Adding more days after day 3 doesn't help figuring out day 2 at all. It might help in narrowing the options for day 3 (which will then narrow the options for day 2), but we already have just 1 choice left for day 3, so it's no use.
Full Code
import numpy as np
from itertools import product
import tqdm
def sample_uniform(n, r):
# check out: http://compneuro.uwaterloo.ca/files/publications/voelker.2017.pdf
sample = np.random.rand(n + 2)
sample_norm = np.linalg.norm(sample)
unit_sample = (sample / sample_norm)
change = np.floor(r * unit_sample[:-2]).astype(np.int)
return change
def human(num_fruits, allowed_change=0.05, current_distribution=None):
allowed_change = 0.05
if current_distribution is None:
current_distribution = np.random.randint(1, 50, size=num_fruits)
yield current_distribution.copy()
# rejection sample a suitable change
while True:
current_total = np.sum(current_distribution)
maximum_change = np.floor(allowed_change * current_total)
change = sample_uniform(num_fruits, maximum_change)
while np.sum(change) > maximum_change:
change = sample_uniform(num_fruits, maximum_change)
current_distribution += change
yield current_distribution.copy()
def prices(num_fruits, alter_prices=False):
current_prices = np.random.randint(1, 10, size=num_fruits)
while True:
yield current_prices.copy()
if alter_prices:
current_prices = np.random.randint(1, 10, size=num_fruits)
def play_game(num_days, num_fruits=3, alter_prices=False):
human_choice = human(num_fruits)
price_development = prices(num_fruits, alter_prices=alter_prices)
history = {
"basket": list(),
"prices": list(),
"total": list()
}
for day in range(num_days):
choice = next(human_choice)
price = next(price_development)
total_price = np.sum(choice * price)
history["basket"].append(choice)
history["prices"].append(price)
history["total"].append(total_price)
return history
def backtrack(pos, daily_total, daily_item_value, allowed_change, iterator_bounds, history=None):
if pos == len(daily_total):
yield np.array(history)
return
it = [range(start, stop, step) for start, stop, step in iterator_bounds[pos][:-1]]
for partial_basket in product(*it):
if history is None:
history = [partial_basket]
else:
history.append(partial_basket)
# ensure we only check items that match the total basket value
# for that day
partial_value = np.sum(np.array(partial_basket) * daily_item_value[pos, :-1])
if (daily_total[pos] - partial_value) % daily_item_value[pos, -1] != 0:
history.pop()
continue
last_item = (daily_total[pos] - partial_value) // daily_item_value[pos, -1]
if last_item < 0:
history.pop()
continue
basket = np.array([*partial_basket] + [int(last_item)])
basket_value = np.sum(basket * daily_item_value[pos])
history[-1] = basket
if len(history) > 1:
# ensure that today's basket stays within relative tolerance
previous_basket = history[-2]
previous_basket_count = np.sum(previous_basket)
current_basket_count = np.sum(basket)
if (np.abs(current_basket_count - previous_basket_count) > allowed_change * previous_basket_count):
history.pop()
continue
yield from backtrack(pos + 1, daily_total, daily_item_value, allowed_change, iterator_bounds, history)
history.pop()
if __name__ == "__main__":
np.random.seed(1337)
num_fruits = 3
allowed_change = 0.05
alter_prices = False
history = play_game(15, num_fruits=num_fruits, alter_prices=alter_prices)
total_price = np.stack(history["total"]).astype(np.int)
daily_price = np.stack(history["prices"]).astype(np.int)
basket = np.stack(history["basket"]).astype(np.int)
maximum_fruits = np.floor(total_price[:, np.newaxis] / daily_price).astype(np.int)
iterator_bounds = [[[0, maximum_fruits[pos, fruit], 1] for fruit in range(num_fruits)] for pos in range(len(basket))]
# iterator_bounds = np.array(iterator_bounds)
# import pdb; pdb.set_trace()
pbar = tqdm.tqdm(backtrack(0, total_price,
daily_price, allowed_change, iterator_bounds), desc="Found Solutions")
for solution in pbar:
# test price guess
calculated_price = np.sum(np.stack(solution) * daily_price, axis=1)
assert np.all(calculated_price == total_price)
# test basket change constraint
change = np.sum(np.diff(solution, axis=0), axis=1)
max_change = np.sum(solution[:-1, ...], axis=1) * allowed_change
assert np.all(change <= max_change)
# indicate that we found the original solution
if not np.any(solution - basket):
pbar.set_description("Found Solutions (includes original)")

When the player selects a combination which will reduce the number of possibilities to 1, computer will win. Otherwise, the player can pick a combination with the constraint of the total varying within a certain percentage, that computer may never win.
import itertools
import numpy as np
def gen_possible_combination(total, prices):
"""
Generates all possible combinations of numbers of items for
given prices constraint by total
"""
nitems = [range(total//p + 1) for p in prices]
prices_arr = np.array(prices)
combo = [x for x in itertools.product(
*nitems) if np.dot(np.array(x), prices_arr) == total]
return combo
def reduce(combo1, combo2, pct):
"""
Filters impossible transitions which are greater than pct
"""
combo = {}
for x in combo1:
for y in combo2:
if abs(sum(x) - sum(y))/sum(x) <= pct:
combo[y] = 1
return list(combo.keys())
def gen_items(n, total):
"""
Generates a list of items
"""
nums = [0] * n
t = 0
i = 0
while t < total:
if i < n - 1:
n1 = np.random.randint(0, total-t)
nums[i] = n1
t += n1
i += 1
else:
nums[i] = total - t
t = total
return nums
def main():
pct = 0.05
i = 0
done = False
n = 3
total_items = 26 # np.random.randint(26)
combo = None
while not done:
prices = [np.random.randint(1, 10) for _ in range(n)]
items = gen_items(n, total_items)
total = np.dot(np.array(prices), np.array(items))
combo1 = gen_possible_combination(total, prices)
if combo:
combo = reduce(combo, combo1, pct)
else:
combo = combo1
i += 1
print(i, 'Items:', items, 'Prices:', prices, 'Total:',
total, 'No. Possibilities:', len(combo))
if len(combo) == 1:
print('Solution', combo)
break
if np.random.random() < 0.5:
total_items = int(total_items * (1 + np.random.random()*pct))
else:
total_items = int(
np.ceil(total_items * (1 - np.random.random()*pct)))
if __name__ == "__main__":
main()