Develop Reference

Why is my answer wrong in Code jam 2018 "Saving the World Again"?

The problem is presented here: https://codingcompetitions.withgoogle.com/codejam/round/00000000000000cb/0000000000007966
An alien robot is threatening the universe, using a beam that will destroy all algorithms knowledge. We have to stop it!
Fortunately, we understand how the robot works. It starts off with a beam with a strength of 1, and it will run a program that is a series of instructions, which will be executed one at a time, in left to right order. Each instruction is of one of the following two types:
C (for "charge"): Double the beam's strength.
S (for "shoot"): Shoot the beam, doing damage equal to the beam's current strength.
For example, if the robot's program is SCCSSC, the robot will do the following when the program runs:
Shoot the beam, doing 1 damage.
Charge the beam, doubling the beam's strength to 2.
Charge the beam, doubling the beam's strength to 4.
Shoot the beam, doing 4 damage.
Shoot the beam, doing 4 damage.
Charge the beam, increasing the beam's strength to 8.
In that case, the program would do a total of 9 damage.
The universe's top algorithmists have developed a shield that can withstand a maximum total of D damage. But the robot's current program might do more damage than that when it runs.
The President of the Universe has volunteered to fly into space to hack the robot's program before the robot runs it. The only way the President can hack (without the robot noticing) is by swapping two adjacent instructions. For example, the President could hack the above program once by swapping the third and fourth instructions to make it SCSCSC. This would reduce the total damage to 7. Then, for example, the president could hack the program again to make it SCSSCC, reducing the damage to 5, and so on.
To prevent the robot from getting too suspicious, the President does not want to hack too many times. What is this smallest possible number of hacks which will ensure that the program does no more than D total damage, if it is possible to do so?
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each consists of one line containing an integer D and a string P: the maximum total damage our shield can withstand, and the robot's program.
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is either the minimum number of hacks needed to accomplish the goal, or IMPOSSIBLE if it is not possible.
I implemented the following logic:
- First calculate the damage of the ship.
- The S has it's greatest value when it is at the end so the swaps should start at end and continue towards the beginning of the list.
- The C at the end becomes useless so I pop it out of the list so it does not iterate over it again.
- In order to simplify the O() complexity I decided to subtract the last value of S from theSUM every time a swap is made.
The test results seem right - but the judge of the system says : Wrong Answer.
Can you help me find the mistake?
(I know only how to operate with lists and dictionaries in Python 3 and I am an absolute beginner at solving theese questions )
my code is below:
for case in range(1,T):
D, B = input().split()
D = int(D)
Blist =[]
[Blist.append(i) for i in B]
def beamDamage(Blist):
theSum=0
intS=1
Ccount = 0
for i in Blist:
if i == 'S':
theSum = theSum + intS
if i == 'C':
Ccount = Ccount +1
intS = intS*2
return theSum
def swap(Blist):
temp=''
for i in range(0,len(Blist)):
if Blist[len(Blist)- 1] == 'C':
Blist.pop()
if (Blist[len(Blist)- i - 1]) == 'C' and (Blist[len(Blist)- i] == 'S'):
temp = Blist[len(Blist)- i - 1] # C
Blist[len(Blist)- i - 1] = 'S'
Blist[len(Blist)- i] = temp
return Blist
bd = beamDamage(Blist)
y = 0
if 'C' not in B:
if beamDamage(Blist) > D:
print("Case #{}: IMPOSSIBLE".format(case))
else:
print("Case #{}: 0".format(case))
else:
while bd > D:
swap(Blist)
pwr=0
for ch in Blist:
if ch == 'C':
pwr=pwr+1
bd = bd - 2**(pwr-1)
y+=1
print("Case #{}: {}".format(case, y))

I will not give you a complete solution, but here is one issue:
If your input is a series of "S" followed by one or more "C" (like "SSSSC"), and the calculated damage is higher than asked for, you'll clearly see that the result is wrong. It should be IMPOSSIBLE...
The reason for the failure is that the condition in if 'C' not in B: will not apply, and so the loop will kick in (when it really shouldn't). Consequently pwr remains zero and you use a calculation with 2**-1, which yields a non-integer value.
The solution is to trim the list from terminating C characters at the very start, even before doing the if test.
Secondly, I don't see the benefit of doing the damage calculation in two different ways. On the one hand you have beamDamage, and you also have the inline loop, which does roughly the same (not faster).
Finally, even if you get this right, I suspect your code might run into a timeout, because it is not doing the job efficiently. Think of keeping track of the damage incrementally, without needing to go through the whole list again.
Once you have that improvement, you may still need to tune performance furhter. In that case, think of what damage reduction you would get it you would move a "C" immediately to the very end of the list. If that reduction is still not bringing the damage below the target, you can go for that in one go (but still count the steps correctly).

Explanation on Stone Nim Game

I was doing a coding problem which I somehow passed all test cases but I did not understand exactly what was going on. The problem was a small twist on the classic nim game:
There are two players A and B. There are N piles of various stones. Each player can take any amount of stones if the pile is less than K, otherwise they must take a multiple of K stones. The last person to take stones wins.
python
# solution -> will A win the game of piles, k?
def solution(piles, k):
gn = 0 # Grundy number
for pile in piles:
if pile % 2 != 0:
gn ^= pile + 1
else:
gn ^= pile - 1
return gn != 0
I'm not sure if there was enough test cases, but k was not even used here. To be honest, I am having a difficult time even understanding what gn (Grundy number) really means. I realize there is a proof of winning the Nim game if the xor of all piles is not zero, but I don't really understand why this variation requires checking the parity of the pile.

First, the given solution is incorrect. You noticed that it does not use k, and indeed this is a big red flag. You can also look at the result it gives for a single pile game, where it seems to say that player A only wins if the size of the pile is one which you should fairly quickly be able to show is incorrect.
The structure of the answer is sort of correct, though. A lot of the power of the Grundy number is that the Grundy number of a combined game state is the nim sum (XOR in the case of finite ordinals) of the Grundy numbers of the individual game states. (This only works for a very specific way of combining game states, but this turns out to be the natural way of considering Nim piles together.) So, this problem can indeed be solved by finding the Grundy number for each pile (considering k) and XOR-ing them together to get the Grundy number for the full game state. (In Nim where you can take any number of stones from a pile and win by taking the last stone, the Grundy number of a pile is just the size of a pile. That's why the solution to that version of Nim just XOR-s the sizes of the piles.)
So, taking the theory for granted, you can solve the problem by finding the correct Grundy values for a single pile given k. You only need to consider one pile games to do this. This is actually a pretty classic problem, and IMO significantly simpler to correctly analyze than multi-pile Nim. You should give it a go.
As for how to think of Grundy numbers, there are plenty of places to read about it, but here's my approach. The thing to understand is why the combination of two game states allows the previous player (B) to win exactly when the Grundy numbers are equal.
To do this, we need only consider what effect moves have on the Grundy numbers of the two states.
By definition as the minimum excluded value of successor states, there is always a move that changes the Grundy number of a state to any lower value (ie n could become any number from 0 up to n - 1). There is never a move that leaves the Grundy number the same. There may or may not be moves that increase the Grundy number.
Then, in the case of the combination of two states with the same Grundy number, the player B can win by employing the "copycat strategy". If player A makes a move that decreases the Grundy number of one state, player B can "copy" by reducing the Grundy number of the other state to the same value. If player A makes a move that increases the Grundy number of one state, player B can "undo" it by making a move on the same state to reduce it to the same value it was before. (Our game is finite, so we don't have to worry about an infinite loop of doing and undoing.) These are the only things A can do. (Remember, importantly, there is no move that leaves a Grundy number unchanged.)
If the states don't have the same Grundy number, then the way for the first player to win is clear, then; they just reduces the number of the state with a higher value to match the state with the lower value. This reduces things to the previous scenario.
Here we should note that the minimum excluded value definition allows us to construct the Grundy number for any states recursively in terms of their successors (at least for a finite game). There are no choices, so these numbers are in fact well-defined.
The next question to address is why we can calculate the Grundy number of a combined state. I prefer not to think about XOR at all here. We can define this nim sum operation purely from the minimum excluded value property. We abstractly consider the successors of nim_sum(x, y) to be {nim_sum(k, y) for k in 0..x-1} and {nim_sum(x, k) for k in 0..y-1}; in other words, making a move on one sub-state or the other. (We can ignore successor of one of the sub-states that increase the Grundy number, as such a state would have all the successors of the original state plus nim_sum(x, y) itself as another successor, so it must then have a strictly larger Grundy number. Yes, that's a little bit hand-wavy.) This turns out to be the same as XOR. I don't have a particularly nice explanation for this, but I feel it isn't really necessary to a basic understanding. The important thing is that it is a well-defined operation.

Minimax Algorithm Implementation In Python3

I have been trying to build a Tic-Tac-Toe bot in Python. I tried to avoid using the Minimax algorithm, because I was QUITE daunted how to implement it. Until now.
I (finally) wrote an algorithm that sucked and could lose pretty easily, which kinda defeats the purpose of making a computer play Tic-Tac-Toe. So I finally took the courage to TRY to implement the algorithm. I stumbled upon this StackOverflow post. I tried to implement the chosen answer there, but I can't understand most of the stuff. The code in that answer follows:
def minimax(self, player, depth = 0) :
if player == "o":
best = -10
else:
best = 10
if self.complete():
if self.getWinner() == "x": # 'X' is the computer
return -10 + depth, None
elif self.getWinner() == "tie":
return 0, None
elif self.getWinner() == "o" : # 'O' is the human
return 10 - depth, None
for move in self.getAvailableMoves() :
self.makeMove(move, player)
val, _ = self.minimax(self.getEnemyPlayer(player), depth+1)
print(val)
self.makeMove(move, ".")
if player == "o" :
if val > best :
best, bestMove = val, move
else :
if val < best :
best, bestMove = val, move
return best, bestMove
First of all, why are we returning -10 + depth when the computer win and 10 -
depth when the human wins? (I get why we return 0 when it is a draw). Secondly, what is the depth parameter doing? Is there some way to omit it?
Should we omit it?
I'm probably missing something fundamental about the algorithm but, I think I understand it well enough. Please bear in mind that I'm very new to recursive algorithms...
EDIT
So, now I made myself the function:
def minimax(self, player):
won = 10
lost = -10
draw = 0
if self.has_won(HUMAN):
return lost, None
elif self.has_won(BOT):
return won, None
if not(self.board_is_empty()):
return draw, None
moves = self.get_available_moves()
for move in moves:
self.play_move(move[0], move[1], player)
make_board(self.board)
if self.board_is_empty():
val, _ = self.minimax(self.get_enemy_player(player))
self.rewind_move(move)
if val==won:
return val, move
But the problem now is I can't understand what happens when the move ends in a draw or a loss (for the computer). I think what it's doing is that it goes through a move's consequences to see if SOMEONE wins (that's probably what is happening, because I tested it) and then returns that move if SOMEONE wins. How do I modify this code to work properly?
Note:
This function is in a class, hence the self keywords.
moves is a list containing tuples. eg. moves = [(0, 1), (2, 2)] etc. So, moves contains all the empty squares. So each moves[i][j] is an integer modulo 3.
I'm using the exhaustive algorithm suggested by Jacques de Hooge in his answer below.

First note that 10 - depth = - (-10 + depth).
So computer wins have opposite signs from human wins.
In this way they can be added to evaluate the value of a gameboard state.
While with tictactoe this isn't really needed, in a game like chess it is, since it is too timeconsuming to try all possible combinations until checkmate, hence gameboard states have to be evaluated somehow in terms of losses and wins (losing and winning chess pieces, each worth a certain amount of points, values drawn from experience).
Suppose now we only look at 10 - depth (so human wins).
The most attractive wins are the ones that require the least plies (moves).
Since each move or countermove results in the depth being incremented,
more moves will result in parameter depth being larger, so 10 - depth (the "amount" of advantage) being smaller. So quick wins are favored over lenghty ones. 10 is enough, since there are only in total 9 moves possible in a 3 x 3 playfield.
So in short: since tictactoe is so simple, in fact the winning combination can be found in a exhaustive recursive search. But the minimax algorithm is suitable for more complicated situations like chess, in which intermediate situations have to be evaluated in terms of the sum of losses (negative) and gains (positives).
Should the depth parameter be omitted? If you care about the quickest win: No. If you only care about a win (with tictactoe): it can indeed be omitted, since an exhaustive search is possible.
[EDIT]
Exhaustive with tictactoe just means searching 9 plies deep, since the game can never last longer.
Make a recursive function with a parameter player (o or x) and a return value win or loss, that is first decided at the deepest recursion level, and then taken upward through the recursion tree. Let it call itself with the opposite player as parameter for all free fields. A move for the machine is the right one if any sequel results in the machine winning for all branches that the human may take on each level.
Note: The assumption I made is that there IS a winning strategy. If that is not the case (ties possible), the algorithm you have may be the best option. I remember with tictactoe the one who starts the game can always enforce a win in the way described above. So the algorithm will win in at least 50% of all games.
With a non-perfect human player it may also win if the computer doesn't start, if the human player does something suboptimal,

Find the combination of variables in base of their score

I want to solve a problem with Python (like a programming riddle). It isn't an exercise or anything related with this, but something I figure and I want to find a solution for it.
Problem description:
Let's say that I have a business with 10 different working positions.
Each position has a few different sub-positions. For example (aggressive, relaxed, normal). But, the sub-positions do not be the same for every position. Other have 1, other have 2, other 3.
I have 20 people with their performance-score in every position and sub-position.
The final score of the business production will be the sum of each score in every position. I want to find the combination that will have the highest score.
A few thoughts:
Not all the people will get the job, since the positions are less than the number of people.
I cannot have two people in the same position.
My first attempt was to take the highest score of each people and trying to fill all the positions one by one. But this ended up with a problem. A problem similar to the traveling salesman.
So, what should be my next option? Any Python implementation ideas?
Edit: A few more details closer to Python
positionslist = [pos1, pos2, pos3, pos4, pos5, pos6, pos7, pos8, pos9, pos10]
subpositions = {"pos1":["A","B"], "pos2":["B","C"],"pos3":["A","B","C"],"pos4":["A"],"pos5":["A","B"],"pos6":["A","B","C"],"pos7":["B"],"pos8":["C"],"pos9":["A","C"],"pos10":["A"]
peoplelist = [{"pos1 A":15,"pos1 B": 8, "pos2 B": 2, "pos2 C": 4, "po3 A": 2, "pos3 B":5...}, {"pos1 A":1, "pos1 B":23,"pos2 B":11,.....},.........]
#run the code
print "best combination:" result
best combination:
pos1 B, person 3, score 23
pos2 C, person 5, score 11
pos3 A, person 18, score ..
pos4
pos5
pos6
pos7
pos8
pos9
pos10
Total Score: ....
As I said, my implementation in pseudo-code is:
for position in positionlist:
for every sub-position:
find the highest score from every person
find the highest sub-position from the previous step
remove that person from the peoplelist
store the position with the sub-position and the person
However, this is similar to Traveling Salesman Problem and it will end up with not the highest combination.

Problem is called maximum matching in bipartite graph or assignment problem. It is done by Hungarian algorithm.
On Wikipedia page are implementations in different languages. There is python library munkres with an implementation.

If I have read this right every person has a different score for each position, and you want to find the highest score across all the combinations of people and positions - i.e. you have to calculate everyones score for each position.
I can't think of a way of expressing this that isn't the same as the Travelling salesman problem - although this is it in reverse - effectively finding the highest score combination rather than the lowest combinations.
If that is the case, then all you will be able to do reliably in a reasonable time (for any large data set) is to get a sub-optimal score which is close to and maybe exactly the optimal, but impossible to prove it is sub-optimal or optimal.
One way to try to make this simpler would be to prioritise the positions you want to fill, and get the highest score for each position in priority order - however this may not even be a close to optimal solution when compared to the non-prioritised position list.

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()