So I rewrote it in a generative way (combo is a generator).# If there's only two numbers remaining in the stack#L.append( b+"+"+a) # addition and multiplication are abelian, What we'll do is take every one of these combinations, let's call them c, and make a combination out of 4, the number we took from S earlier, and c. That's what the line _combo(4,c) has also six possibilities (4+c, 4-c, 4*c, 4/c, c-4,c/4), so for each value of c we have 6 possible combinations with 4, and there are 6 values of c, so the total possibilities of _combo(4,c) is 6 * 6 = 36 possible combos : (4+(5+8)) (4+(5-8)) (4+(5*8)) (4+(5/8)) (4+(8/5)) (4+(8-5)) (4-(5+8)) (4-(5-8)) (4-(5*8)) (4-(5/8)) (4-(8/5)) (4-(8-5)) (4*(5+8)) (4*(5-8)) (4*(5*8)) (4*(5/8)) (4*(8/5)) (4*(8-5)) (4/(5+8)) (4/(5-8)) (4/(5*8)) (4/(5/8)) (4/(8/5)) (4/(8-5)) ((5+8)/4) ((5-8)/4) ((5*8)/4) ((5/8)/4) ((8/5)/4) ((8-5)/4) ((5+8)-4) ((5-8)-4) ((5*8)-4) ((5/8)-4) ((8/5)-4) ((8-5)-4) Notice how 4 is always in one side of the operation and (8,5) is on the other side ? (8+(4+5)) (8+(4-5)) (8+(4*5)) (8+(4/5)) (8+(5/4)) (8+(5-4)) (8-(4+5)) (8-(4-5)) (8-(4*5)) (8-(4/5)) (8-(5/4)) (8-(5-4)) (8*(4+5)) (8*(4-5)) (8*(4*5)) (8*(4/5)) (8*(5/4)) (8*(5-4)) (8/(4+5)) (8/(4-5)) (8/(4*5)) (8/(4/5)) (8/(5/4)) (8/(5-4)) ((4+5)/8) ((4-5)/8) ((4*5)/8) ((4/5)/8) ((5/4)/8) ((5-4)/8) ((4+5)-8) ((4-5)-8) ((4*5)-8) ((4/5)-8) ((5/4)-8) ((5-4)-8) Done ! Where : first = 6; second = +; and third = (3*2). you have cycled through all elements of your stack, and so you have all 108 possible combinations of 4,5 and 8 taken in any possible order. Each character is one byte and for 7 numbers you may have expressions like (5+(3*(8+2))/((4-2)*6) 22 characters are 22 bytes, which multiplies 117.573.120 is 2,586,608,640. avec la pédagogie PNL - Duration: 5:59.

the correct form is : 5+(6*(3+1)) which is evaluated like this : o X has one big operation inside which is 6 * (Y)It may return None when any part of the expression divides by 0.

It decerements n whenever it encounters a closing parenthesis. If the format is not respected, it may simply crash. HTTPS If all the numbers are different, all the combinations are unique. Temporibus autem quibusdam et aut officiis debitis is aut rerum necessitatibus saepes eveniet ut etes seo lage voluptates repudiandae sint et molestiae non …
If n reaches zero, that means that all opened parentheses are matched, so we're at the end of an expression contained between parentheses. That's approximatively 2.5 Gb of RAM, and that's why I got a memory error when I tried the list version of combo.

-> the recursion ends here, return all the combinations of 9 and 12 -> combine 4 with every combination of [5,8,9,12] Let's take an example with just three numbers : 4,5, and 8. If the result was correct, the TV presenter Jammot said his famous quote: "Le compte est bon!"

5:59. le son j - Duration: 4:07.

(remember that (8*5) and (8+5) are the same as (5*8) and (5+8) so they're not generated.) We can simply acheive this by putting 4 back in the stack... and make every combinations of 5 with every combinations of 4 and 8. Atout Scolaire 3,265 views. This file contains the evaluate function that computes expressions like (4+3)/(2*(5+4)).game.py feeds this function with expressions generated by the generator.combo function(file generator.py). Then, we simply retun the result of applying the operation on the evaluation of left The evalution of left or right is again the result of applying the operation contained within on the evaluation of their left and right parts respectively and so on.