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Google foo.bar challenge "Hey I already did that", not passing all the test cases

This is the description of the problem I am trying to solve.
Hey, I Already Did That!
Commander Lambda uses an automated algorithm to assign minions randomly to tasks, in order to keep minions on their toes. But you've noticed a flaw in the algorithm -- it eventually loops back on itself, so that instead of assigning new minions as it iterates, it gets stuck in a cycle of values so that the same minions end up doing the same tasks over and over again. You think proving this to Commander Lambda will help you make a case for your next promotion.
You have worked out that the algorithm has the following process:
Start with a random minion ID n, which is a nonnegative integer of length k in base b
Define x and y as integers of length k. x has the digits of n in descending order, and y has the digits of n in ascending order
Define z = x - y. Add leading zeros to z to maintain length k if necessary
Assign n = z to get the next minion ID, and go back to step 2
For example, given minion ID n = 1211, k = 4, b = 10, then x = 2111, y = 1112 and z = 2111 - 1112 = 0999. Then the next minion ID will be n = 0999 and the algorithm iterates again: x = 9990, y = 0999 and z = 9990 - 0999 = 8991, and so on.
Depending on the values of n, k (derived from n), and b, at some point the algorithm reaches a cycle, such as by reaching a constant value. For example, starting with n = 210022, k = 6, b = 3, the algorithm will reach the cycle of values [210111, 122221, 102212] and it will stay in this cycle no matter how many times it continues iterating. Starting with n = 1211, the routine will reach the integer 6174, and since 7641 - 1467 is 6174, it will stay as that value no matter how many times it iterates.
Given a minion ID as a string n representing a nonnegative integer of length k in base b, where 2 <= k <= 9 and 2 <= b <= 10, write a function solution(n, b) which returns the length of the ending cycle of the algorithm above starting with n. For instance, in the example above, solution(210022, 3) would return 3, since iterating on 102212 would return to 210111 when done in base 3. If the algorithm reaches a constant, such as 0, then the length is 1.
My solution isn't passing 5 of the 10 test cases for the challenge. I don't understand if there's a problem with my code, as it's performing exactly as the problem asked to solve it, or if it's inefficient.
Here's my code for the problem. I have commented it for easier understanding.
def convert_to_any_base(num, b): # returns id after converting back to the original base as string
digits = []
while(num/b != 0):
digits.append(str(num % b))
num //= b
result = ''.join(digits[::-1])
return result
def solution(n, b):
minion_id_list = [] #list storing all occurrences of the minion id's
k = len(n)
while n not in minion_id_list: # until the minion id repeats
minion_id_list.append(n) # adds the id to the list
x = ''.join(sorted(n, reverse = True)) # gives x in descending order
y = x[::-1] # gives y in ascending order
if b == 10: # if number is already a decimal
n = str(int(x) - int(y)) # just calculate the difference
else:
n = int(x, b) - int(y, b) # else convert to decimal and, calculate difference
n = convert_to_any_base(n, b) # then convert it back to the given base
n = (k-len(n)) * '0' + n # adds the zeroes in front to maintain the id length
if int(n) == 0: # for the case that it reaches a constant, return 1
return 1
return len(minion_id_list[minion_id_list.index(n):]) # return length of the repeated id from
# first occurrence to the end of the list
I have been trying this problem for quite a while and still don't understand what's wrong with it. Any help will be appreciated.

Fast access to sums of pairwise ops

Given a vector of numbers v, I can access sums of sections of this vector by using cumulative sums, i.e., instead of O(n)
v = [1,2,3,4,5]
def sum_v(i,j):
return sum(v[i:j])
I can do O(1)
import itertools
v = [1,2,3,4,5]
cache = [0]+list(itertools.accumulate(v))
def sum_v(i,j):
return cache[j] - cache[i]
Now, I need something similar but for pairwise instead of sum_v:
def pairwise(i,j):
ret = 0
for p in range(i,j):
for q in range(p+1,j):
ret += f(v(p),v(q))
return ret
where f is, preferably, something relatively arbitrary (e.g., * or ^ or ...). However, something working for just product or just XOR would be good too.
PS1. I am looking for a speed-up in terms of O, not generic memoization such as functools.cache.
PS2. The question is about algorithms, not implementations, and is thus language-agnostic. I tagged it python only because my examples are in python.
PS3. Obviously, one can precompute all values of pairwise, so the solution should be o(n^2) both in time and space (preferably linear).

For binary operations such as or, and, xor, an O(N) algorithm is possible.
Let's consider XOR for this example, but this can be easily modified for OR/AND as well.
The most important thing to note here is, the result of a binary operator on bit x of two numbers will not affect the result for bit y. (You can easily see that by trying something like 010 ^ 011 = 001. So we first count the contribution made by the leftmost bits of all numbers to the final sum, then the next least significant bit, and so on. Here's a simple algo/pseudocode for that:
Construct a simple table dp, where dp[i][j] = count of numbers in range [i,n) with jth bit set
l = [5,3,1,7,8]
n = len(l)
ans = 0
max_binary_length = max(log2(i) for i in l)+1 #maximum number of bits we need to check
for j in range(max_binary_length):
# we check the jth bits of all numbers here
for i in range(0,n):
# we need sum((l[i]^l[j]) for j in range (i+1,n))
current = l[i]
if jth bit of current == 0:
# since 0^1 = 1, we need count of numbers with jth bit 1
count = dp[i+1][j]
else:
# we need count of numbers with jth bit 0
count = (n-i)-dp[i+1][j]
# the indexing could be slightly off, you can check that once
ans += count * (2^j)
# since we're checking the jth bit, it will have a value of 2^j when set
print(ans)
In most cases, for integers, number of bits <= 32. So this should have a complexity of O(N*log2(max(A[i]))) == O(N*32) == O(N).

In principle, you can always precompute every possible output in Θ(n²) space and then answer queries in Θ(1) by just looking it up in the precomputed table. Everything else is a trade-off depending on the cost of precomputation time, space, and actual computation time; the interesting question is what can be done with o(n²) space, i.e. sub-quadratic. This will generally depend on the application, and on properties of the binary operation f.
In the particular case where f is *, we can get Θ(1) lookups with only Θ(n) space: we'll take advantage that the sum for pairs where p < q equals the sum of all pairs, minus the sum of pairs where p = q, divided by 2 to account for the pairs where p > q.
# input data
v = [1, 2, 3, 4, 5]
n = len(v)
# precomputation
partial_sums = [0] * (n + 1)
partial_sums_squares = [0] * (n + 1)
for i, x in enumerate(v):
partial_sums[i + 1] = partial_sums[i] + x
partial_sums_squares[i + 1] = partial_sums_squares[i] + x * x
# query response
def pairwise(i, j):
s = partial_sums[j] - partial_sums[i]
s2 = partial_sums_squares[j] - partial_sums_squares[i]
return (s * s - s2) / 2
More generally, this works whenever f is commutative and distributes over the accumulator operation (+ in this case). I wrote the example here without itertools, so that it is more easily translatable to other languages, since the question is meant to be language-agnostic.

Iterate in a certain base in python

I am writing a program that looks for keith numbers in different bases. Is there a way to iterate through numbers in a certain base? I have a program that can convert decimal to any base, but I'm not sure how to proceed.
This is my code so far.
def is_keith_number(n,base):
# Find sum of digits by first getting an array of all digits then adding them
c = str(n)
a = list(map(int, c))
b = sum(a)
# Now check if the number is a keith number
while b < n:
a = a[1:] + [b]
b = sum(a)
return (b == n) & (len(c) > 1)
I also have a function,
convert()
which converts a number from one base to another. Its unneccesary to show that code, but it does work. Any help?

recursive function in python but with strange return

I am trying to solve a primary equation with several variables. For example:11x+7y+3z=20. non-negative integer result only.
I use code below in python 3.5.1, but the result contains something like [...]. I wonder what is it?
The code I have is to test every variables from 0 to max [total value divided by corresponding variable]. Because the variables may be of a large number, I want to use recursion to solve it.
def equation (a,b,relist):
global total
if len(a)>1:
for i in range(b//a[0]+1):
corelist=relist.copy()
corelist+=[i]
testrest=equation(a[1:],b-a[0]*i,corelist)
if testrest:
total+=[testrest]
return total
else:
if b%a[0]==0:
relist+=[b//a[0]]
return relist
else:
return False
total=[]
re=equation([11,7,3],20,[])
print(re)
the result is
[[0, 2, 2], [...], [1, 0, 3], [...]]
change to a new one could get clean result, but I still need a global variable:
def equation (a,b,relist):
global total
if len(a)>1:
for i in range(b//a[0]+1):
corelist=relist.copy()
corelist+=[i]
equation(a[1:],b-a[0]*i,corelist)
return total
else:
if b%a[0]==0:
relist+=[b//a[0]]
total+=[relist]
return
else:
return
total=[]
print(equation([11,7,3],20,[]))

I see three layers of problems here.
1) There seems to be a misunderstanding about recursion.
2) There seems to be an underestimation of the complexity of the problem you are trying to solve (a modeling issue)
3) Your main question exposes some lacking skills in python itself.
I will address the questions in backward order given that your actual question is "the result contains something like [...]. I wonder what is it?"
"[]" in python designates a list.
For example:
var = [ 1, 2 ,3 ,4 ]
Creates a reference "var" to a list containing 4 integers of values 1, 2, 3 and 4 respectively.
var2 = [ "hello", ["foo", "bar"], "world" ]
var2 on the other hand is a reference to a composite list of 3 elements, a string, another list and a string. The 2nd element is a list of 2 strings.
So your results is a list of lists of integers (assuming the 2 lists with "..." are integers). If each sublists are of the same size, you could also think of it as a matrix. And the way the function is written, you could end up with a composite list of lists of integers, the value "False" (or the value "None" in the newest version)
Now to the modeling problem. The equation 11x + 7y + 3z = 20 is one equation with 3 unknowns. It is not clear at all to me what you want to acheive with this program, but unless you solve the equation by selecting 2 independent variables, you won't achieve much. It is not clear at all to me what is the relation between the program and the equation save for the list you provided as argument with the values 11, 7 and 3.
What I would do (assuming you are looking for triplets of values that solves the equation) is go for the equation: f(x,y) = (20/3) - (11/3)x - (7/3)y. Then the code I would rather write is:
def func_f(x, y):
return 20.0/3.0 - (11.0/3.0) * x - (7.0/3.0) * y
list_of_list_of_triplets = []
for (x, y) in zip(range(100),range(100)):
list_of_triplet = [x, y, func_f(x,y)]
list_of_list_of_triplets += [list_of_triplet] # or .append(list_of_triplet)
Be mindful that the number of solutions to this equation is infinite. You could think of it as a straight line in a rectangular prism if you bound the variables. If you wanted to represent the same line in an abstract number of dimensions, you could rewrite the above as:
def func_multi_f(nthc, const, coeffs, vars):
return const - sum([a*b/nth for a,b in zip(coeffs, vars)])
Where nthc is the coefficient of the Nth variable, const is an offset constant, coeffs is a list of coefficients and vars the values of the N-1 other variables. For example, we could re-write the func_f as:
def func_f(x,y):
return func_multi_f(3.0, 20.0, [11.0, 7.0], [x,y])
Now about recursion. A recursion is a formulation of a reducible input that can be called repetivitely as to achieve a final result. In pseudo code a recursive algorithm can be formulated as:
input = a reduced value or input items
if input has reached final state: return final value
operation = perform something on input and reduce it, combine with return value of this algorithm with reduced input.
For example, the fibonacci suite:
def fibonacci(val):
if val == 1:
return 1
return fibonacci(val - 1) + val
If you wanted to recusively add elements from a list:
def sum_recursive(list):
if len(list) == 1:
return list[0]
return sum_recursive(list[:-1]) + list[-1]
Hope it helps.
UPDATE
From comments and original question edits, it appears that we are rather looking for INTEGER solutions to the equation. Of non-negative values. That is quite different.
1) Step one find bounds: use the equation ax + by + cz <= 20 with a,b,c > 0 and x,y,z >= 0
2) Step two, simply do [(x, y, z) for x, y, z in zip(bounds_x, bounds_y, bounds_z) if x*11 + y*7 + z*3 - 20 == 0] and you will have a list of valid triplets.
in code:
def bounds(coeff, const):
return [val for val in range(const) if coeff * val <= const]
def combine_bounds(bounds_list):
# here you have to write your recusive function to build
# all possible combinations assuming N dimensions
def sols(coeffs, const):
bounds_lists = [bounds(a, const) for a in coeffs]
return [vals for vals in combine_bounds(bounds_lists) if sum([a*b for a,b in zip(coeff, vals)] - const == 0)

Here is a solution built from your second one, but without the global variable. Instead, each call passes back a list of solutions; the parent call appends each solution to the current element, making a new list to return.
def equation (a, b):
result = []
if len(a) > 1:
# For each valid value of the current coefficient,
# recur on the remainder of the list.
for i in range(b // a[0]+1):
soln = equation(a[1:], b-a[0]*i)
# prepend the current coefficient
# to each solution of the recursive call.
for item in soln:
result.append([i] + item)
else:
# Only one item left: is it a solution?
if b%a[0] == 0:
# Success: return a list of the one element
result = [[b // a[0]]]
else:
# Failure: return empty list
result = []
return result
print(equation([11, 7, 3], 20, []))

Random contiguous slice of list in Python based on a single random integer

Using a single random number and a list, how would you return a random slice of that list?
For example, given the list [0,1,2] there are seven possibilities of random contiguous slices:
[ ]
[ 0 ]
[ 0, 1 ]
[ 0, 1, 2 ]
[ 1 ]
[ 1, 2]
[ 2 ]
Rather than getting a random starting index and a random end index, there must be a way to generate a single random number and use that one value to figure out both starting index and end/length.
I need it that way, to ensure these 7 possibilities have equal probability.

Simply fix one order in which you would sort all possible slices, then work out a way to turn an index in that list of all slices back into the slice endpoints. For example, the order you used could be described by
The empty slice is before all other slices
Non-empty slices are ordered by their starting point
Slices with the same starting point are ordered by their endpoint
So the index 0 should return the empty list. Indices 1 through n should return [0:1] through [0:n]. Indices n+1 through n+(n-1)=2n-1 would be [1:2] through [1:n]; 2n through n+(n-1)+(n-2)=3n-3 would be [2:3] through [2:n] and so on. You see a pattern here: the last index for a given starting point is of the form n+(n-1)+(n-2)+(n-3)+…+(n-k), where k is the starting index of the sequence. That's an arithmetic series, so that sum is (k+1)(2n-k)/2=(2n+(2n-1)k-k²)/2. If you set that term equal to a given index, and solve that for k, you get some formula involving square roots. You could then use the ceiling function to turn that into an integral value for k corresponding to the last index for that starting point. And once you know k, computing the end point is rather easy.
But the quadratic equation in the solution above makes things really ugly. So you might be better off using some other order. Right now I can't think of a way which would avoid such a quadratic term. The order Douglas used in his answer doesn't avoid square roots, but at least his square root is a bit simpler due to the fact that he sorts by end point first. The order in your question and my answer is called lexicographical order, his would be called reverse lexicographical and is often easier to handle since it doesn't depend on n. But since most people think about normal (forward) lexicographical order first, this answer might be more intuitive to many and might even be the required way for some applications.
Here is a bit of Python code which lists all sequence elements in order, and does the conversion from index i to endpoints [k:m] the way I described above:
from math import ceil, sqrt
n = 3
print("{:3} []".format(0))
for i in range(1, n*(n+1)//2 + 1):
b = 1 - 2*n
c = 2*(i - n) - 1
# solve k^2 + b*k + c = 0
k = int(ceil((- b - sqrt(b*b - 4*c))/2.))
m = k + i - k*(2*n-k+1)//2
print("{:3} [{}:{}]".format(i, k, m))
The - 1 term in c doesn't come from the mathematical formula I presented above. It's more like subtracting 0.5 from each value of i. This ensures that even if the result of sqrt is slightly too large, you won't end up with a k which is too large. So that term accounts for numeric imprecision and should make the whole thing pretty robust.
The term k*(2*n-k+1)//2 is the last index belonging to starting point k-1, so i minus that term is the length of the subsequence under consideration.
You can simplify things further. You can perform some computation outside the loop, which might be important if you have to choose random sequences repeatedly. You can divide b by a factor of 2 and then get rid of that factor in a number of other places. The result could look like this:
from math import ceil, sqrt
n = 3
b = n - 0.5
bbc = b*b + 2*n + 1
print("{:3} []".format(0))
for i in range(1, n*(n+1)//2 + 1):
k = int(ceil(b - sqrt(bbc - 2*i)))
m = k + i - k*(2*n-k+1)//2
print("{:3} [{}:{}]".format(i, k, m))

It is a little strange to give the empty list equal weight with the others. It is more natural for the empty list to be given weight 0 or n+1 times the others, if there are n elements on the list. But if you want it to have equal weight, you can do that.
There are n*(n+1)/2 nonempty contiguous sublists. You can specify these by the end point, from 0 to n-1, and the starting point, from 0 to the endpoint.
Generate a random integer x from 0 to n*(n+1)/2.
If x=0, return the empty list. Otherwise, x is unformly distributed from 1 through n(n+1)/2.
Compute e = floor(sqrt(2*x)-1/2). This takes the values 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, etc.
Compute s = (x-1) - e*(e+1)/2. This takes the values 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, ...
Return the interval starting at index s and ending at index e.
(s,e) takes the values (0,0),(0,1),(1,1),(0,2),(1,2),(2,2),...
import random
import math
n=10
x = random.randint(0,n*(n+1)/2)
if (x==0):
print(range(n)[0:0]) // empty set
exit()
e = int(math.floor(math.sqrt(2*x)-0.5))
s = int(x-1 - (e*(e+1)/2))
print(range(n)[s:e+1]) // starting at s, ending at e, inclusive

First create all possible slice indexes.
[0:0], [1:1], etc are equivalent, so we include only one of those.
Finally you pick a random index couple, and apply it.
import random
l = [0, 1, 2]
combination_couples = [(0, 0)]
length = len(l)
# Creates all index couples.
for j in range(1, length+1):
for i in range(j):
combination_couples.append((i, j))
print(combination_couples)
rand_tuple = random.sample(combination_couples, 1)[0]
final_slice = l[rand_tuple[0]:rand_tuple[1]]
print(final_slice)
To ensure we got them all:
for i in combination_couples:
print(l[i[0]:i[1]])
Alternatively, with some math...
For a length-3 list there are 0 to 3 possible index numbers, that is n=4. You have 2 of them, that is k=2. First index has to be smaller than second, therefor we need to calculate the combinations as described here.
from math import factorial as f
def total_combinations(n, k=2):
result = 1
for i in range(1, k+1):
result *= n - k + i
result /= f(k)
# We add plus 1 since we included [0:0] as well.
return result + 1
print(total_combinations(n=4)) # Prints 7 as expected.

there must be a way to generate a single random number and use that one value to figure out both starting index and end/length.
It is difficult to say what method is best but if you're only interested in binding single random number to your contiguous slice you can use modulo.
Given a list l and a single random nubmer r you can get your contiguous slice like that:
l[r % len(l) : some_sparkling_transformation(r) % len(l)]
where some_sparkling_transformation(r) is essential. It depents on your needs but since I don't see any special requirements in your question it could be for example:
l[r % len(l) : (2 * r) % len(l)]
The most important thing here is that both left and right edges of the slice are correlated to r. This makes a problem to define such contiguous slices that wont follow any observable pattern. Above example (with 2 * r) produces slices that are always empty lists or follow a pattern of [a : 2 * a].
Let's use some intuition. We know that we want to find a good random representation of the number r in a form of contiguous slice. It cames out that we need to find two numbers: a and b that are respectively left and right edges of the slice. Assuming that r is a good random number (we like it in some way) we can say that a = r % len(l) is a good approach.
Let's now try to find b. The best way to generate another nice random number will be to use random number generator (random or numpy) which supports seeding (both of them). Example with random module:
import random
def contiguous_slice(l, r):
random.seed(r)
a = int(random.uniform(0, len(l)+1))
b = int(random.uniform(0, len(l)+1))
a, b = sorted([a, b])
return l[a:b]
Good luck and have fun!