Function for recognize quantity of unique combinations

Hello pythonista community. Can someone adwise me formula for recognize quantity of unique combinations when you count to some max digit with sign "" (minus)? Answer can't be under zero.
Example1:
sign = ''
max = 2
22
21
11 we have only three unique combinations.
Example2:
answer can't be upper max:
max = 2
1+1 we have only one unique example.
I already have formula for two signs "+": unic_col=max**2
But i need formula for '+' and ''
And maybe someone know for:
+×
+/
+×/
/×So, i need function like that:
def uniques_col(operators, max):
if operators == '+':
return max**2
elif operators == '+':
return .....
elif operators == '':
return .....
elif operators == '+×':
return .....
......

@ccc I'm always positively surprised by the number of libraries in Python

I went into some kind of halfinsane readability/conciseness optimization/noodling loop on this. Here’s the latest version:
from itertools import product def uniques_col(ops, maximum): number_range = range(1, maximum+1) numbers = list(map(str, number_range)) return filter( lambda c: 0 <= eval(''.join(c)) <= maximum, product(numbers, ops, numbers) ) uniques = uniques_col('+/', 2) print(*[ f"{i+1:4}: {''.join(c)}" for i, c in enumerate(uniques)], sep='\n' )

numbers = [str(i + 1) for i in range(maximum)]
Read “What’s new in Python 3” for a discussion on avoiding map(), reduce(), filter().

@ccc, you take the prize for conciseness, and readability is not bad either.
I was debating the value of separating the range of numbers (problem domain issue) and the conversion to strings (a technical implementation detail).

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@ccc said:
What’s new in Python 3
For clarity, What’s new in Python 3 does not advice against using
map
andfilter
as such, but against usinglist(map(...))
when a list comprehension can be used instead.Thus, as already said, your amendment makes a lot of sense, but it does not automatically follow that we would change the filter into a comprehension, if we want to leave it up to the user of the function to decide whether to ”collapse” the iterator or not.

@cvp, ha ha. There’s a huge difference and swiftly diminishing returns between ”getting to results” and ”getting to perfect”. While this kind of noodling is fun (for me), your recent track record of real results speaks for itself.

@mikael you're too kind 😳

@cvp thank's a lot. It working for '+', '' and '+':
def uniques_col(operators, max): col = 0 if operators == '+': for i in range(1,max): for j in range(1,maxi+1): col += 1 elif operators == '': for i in range(1,max+1): for j in range(1,i+1): col += 1 elif operators == '+': col = max**2 return col while True: op = input('operator\n') max = int(input('max\n')) print(uniques_col(op,max))
And maybe you know uniques col for: +× +/ +×/ /×
??? Thank's in advance!!!

@lyubomyr83, this code:
from itertools import product def uniques_col(ops, maximum): numbers = [str(i+1) for i in range(maximum)] return filter( lambda c: 0 <= eval(''.join(c)) <= maximum, product(numbers, ops, numbers) ) ops_list = ['+*', '+/', '+/*', '/*'] to_limit = 20 for ops in ops_list: print(''*20) print('ops', ops) for maximum in range(1, to_limit+1): print(maximum, len(list( uniques_col(ops, maximum))))
... gives series up to 20 for each set of operations. For
+/
, I can deduct that the formula for possibilities is2*max*max
. For the others, not as easy, need to dig out a maths book or two.

@lyubomyr83 please try this and tell us if non integer result is allowed for division
def uniques_col(operators, max): col = 0 if len(operators) > 1: for op in operators: col += uniques_col(op, max) return col if operators == '+': for i in range(1,max): for j in range(1,maxi+1): print(i,operators,j,i+j) col += 1 elif operators == '': for i in range(1,max+1): for j in range(1,i+1): print(i,operators,j,ij) col += 1 elif operators == 'x': for i in range(1,max+1): for j in range(1,1+int(max/i)): print(i,operators,j,i*j) col += 1 return col while True: op = input('operator\n') max = int(input('max\n'))

@lyubomyr83, here’s a version that tries to be smart where possible. Unfortunately, finding the number of unique combinations for multiplication
*
seems to be directly related to finding the factors of an integer, for which there is no straightforward formula (hence, public key cryptography works), so we still have to bruteforce it.from functools import reduce from itertools import product def unique_cols(ops, maximum): counts = { '': lambda m: m*(m+1)//2, '+': lambda m: m*(m1)//2, '/': lambda m: m**2, '*': lambda m: len(list(filter( lambda c: 0 <= eval(''.join(c)) <= m, product([str(i+1) for i in range(m)], '*', [str(i+1) for i in range(m)]) ))) } return sum([counts[op](maximum) for op in ops]) maximum = int(input('Maximum integer: ')) ops = input('Operations (one of more of +*/): ') print('Unique combinations:', unique_cols(ops, maximum))

For multiplication, it seems the number would be something like this:. Consider building the 2d multiplication table, 1..N in cold and rows. Then count how many you have in each row that are <=N. To avoid double counting, handle the diagonal separately.
For each row of the multiplication table, excluding the diagonal, you get
1*x (x!=1) ==> N1 different possiblities 2*x (x!=2) ==> (N//2)1 ... K*x (x!=K) ==> (N//k)1
Then for the diagonal, it is the number of x**2<N
Or, x<=sqrt(N)So the total is:
sum([(N//k) for k in range(1,N+1)])N+floor(sqrt (N) ) Or sum([(N//k) for k in range(2,N+1)])+floor(sqrt (N) )
Not quite brute force, but order N instead of N**2. You could also make use of the symmetry to only count through sqrt(N), which makes it order sqrt (N)..
2*sum([(N//k)k for k in range(1,1+floor(sqrt(N))])+floor(sqrt (N) )

@JonB, impressive! Need to spend some more time to really follow the logic, but the results match with the bruteforce results, so here is the updated version:
from itertools import product from math import floor, sqrt def unique_cols(ops, maximum): counts = { '': lambda m: m*(m+1)//2, '+': lambda m: m*(m1)//2, '/': lambda m: m**2, '*': lambda m: 2 * sum([(m//k)k for k in range(1, 1 + floor(sqrt(m)))]) + floor(sqrt(m)), } return sum([counts[op](maximum) for op in ops]) maximum = int(input('Maximum integer: ')) ops = input('Operations (one of more of +*/): ') print('Unique combinations:', unique_cols(ops, maximum))

@mikael shouldn't + and  be the same? In the table of allowables, in the subtraction case you are allowed everything above the diagonal that is along (1,1) to (N,N) (assuming zero is not allowed?). In the plus case, everything above the other diagonal. So both cases are (NNN)/2==N(N1)/2
I guess it depends on whether you allow 11=0 and the like. If so you'd add back N, and get N*(N+1)/2.

@JonB for multiplication, do you think it is different of mine
for i in range(1,max+1): col += int(max/i)

@cvp, oh my goodness, my apologies, it looked so simple I did not even understand it is a solution. Which also checks against the bruteforce values, hence the simplified version below.
@JonB, yes, we include 11 as per the ”rules”.
def unique_cols(ops, maximum): counts = { '': lambda m: m*(m+1)//2, '+': lambda m: m*(m1)//2, '/': lambda m: m**2, '*': lambda m: sum([m//i for i in range(1, m+1)]), } return sum([counts[op](maximum) for op in ops]) maximum = int(input('Maximum integer: ')) ops = input('Operations (one or more of +*/): ') print('Unique combinations:', unique_cols(ops, maximum))

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As terribly interesting as this has been, I am still wondering what it is for...

My last version, still without division due to an unanswered question
def uniques_col(operators, max): col = 0 if len(operators) > 1: for op in operators: col += uniques_col(op, max) return col if operators == '+': for i in range(1,max): col += max  i # comment next lines if you don't want detail for j in range(1,maxi+1): print(i,operators,j,i+j) elif operators == '': for i in range(1,max+1): col += i # comment next lines if you don't want detail for j in range(1,i+1): print(i,operators,j,ij) elif operators == 'x': for i in range(1,max+1): col += int(max/i) # comment next lines if you don't want detail for j in range(1,1+int(max/i)): print(i,operators,j,i*j) return col while True: op = input('operator\n') max = int(input('max\n')) print(uniques_col(op,max))