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Function for recognize quantity of unique combinations

@bennr01, with one more fancy function:
import itertools def uniques_col(operators, max_value): possibles = lambda: range(1, max_value+1) return filter( lambda c: 0 <= eval(''.join(map(str, c))) <= max_value, itertools.product(possibles(), operators, possibles()) ) print(list(uniques_col('+', 2)))

@cvp, thank you all for help!!!
With your function i receive right col for '', but not for '+':def uniques_col(operators, max):
n = 0
if operators == '+':
for i in range(1,max):
for j in range(i,maxi+1):
n += 1
print(i,operators,j,'=',i+j)
elif operators == '':
for i in range(1,max+1):
for j in range(1,i+1):
n += 1
print(i,operators,j,'=',ij)
return nwhile True:
op = input('operator\n')
max = int(input('max\n'))
print(uniques_col(op,max))What i receive:
operator
max4
1  1 = 0
2  1 = 1
2  2 = 0
3  1 = 2
3  2 = 1
3  3 = 0
4  1 = 3
4  2 = 2
4  3 = 1
4  4 = 0
10
operator+
max4
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 2 = 4
4Where is 1+3 and 2+1, so for '+' i need receive 6 unique examples.

@lyubomyr83 said:
Where is 1+3 and 2+1,
Ok, I did believe that 1+2 is the same as 2+1, thus I did not generate it.
As I told you, I was not sure to correctly understand 😀Thus it is better to use the other script (of @bennr01 and @mikael ) because it tries all combinations.


i think you would want to check 1 to max for both numbers (consider /, max/max=1)

@JonB if you do that, you have to check and skip cases where i+j>max
With « my » formula, no check is needed 

@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).


@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) )