my task is to do alphametics in python which i did but it didnt pass time tests, any ideas please ?
#đź”’ i need to make my code faster
supple jolt
wise fossilBOT
my task is to do alphametics in python which i did but it didnt pass time tests,...
@supple jolt
Python help channel opened
supple jolt
this is the code https://paste.pythondiscord.com/QS4A
i need to make my code faster
also i cant use set.pop() or iter function
obtuse moon
I need the detailed information about what you want
supple jolt
«Word arithmetic¹» (sometimes also cryptarithm or algebrogram) is a mathematical
puzzle given as an equation with words, e.g. "SEND + MORE = MONEY".
The goal is to assign a unique² digit to each letter so that after their
substitution, the equality holds. In this case, none of the digits must start with a zero.
In this particular case (and in the decimal system), there is only one possible
solution, which is S → 9, E → 5, N → 6, D → 7, M → 1, O → 0, R → 8, Y → 2.
After this substitution with digits, ⟦9567 + 1085 = 10652⟧ is indeed true.
¹ ‹https://en.wikipedia.org/wiki/Verbal_arithmetic›
² “Unique” means that no two different letters can have the same
digit assigned to them.
The goal of this task is to write a pure function that solves similar puzzles,
in a given positional system (the basis will always be an integer between 2 and 26
inclusive). We will limit ourselves to addition only, we will not consider other arithmetic operations
. The equation at the input is given with two parameters. The left-hand side of the equation
‹lhs› is a list of (at least two) words, with each word given as a list
of letters (single-character strings). The right-hand side of the equation is then always exactly
one word (a list of letters).
The function returns a dictionary that assigns a unique
digit value to each letter of the puzzle. If there are multiple solutions, the function returns any
of them. If there is no solution, the function returns ‹None›.
Hint: Use the backtracking technique. Think back to your
elementary school years – especially the addition of numbers one after another, which always starts
from the right. Here too, a good solution is to gradually try to assign values ​​to
the digits that are as far to the right as possible in each addend, and to end the recursion
in good time when it is clear that the result cannot be achieved.
this is my assingment
letter = ""
for _ in letters:
letter = _
break
letters.remove(letter)
this for loop could be easily replaced by letter = letters.pop() and it would be much faster, but i cant use it
!close
wise fossilBOT
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