40 + 40 x 0 + 1= ?

Short answer: 41

Long answer part 1: is this to be solved by parsing or by algebra? If it's to be solved by parsing, we need a set of parsing rules, in other words a convention. Grade school teaches things like BEDMAS/PEMDAS, but that's a fairly complex rule operating on groups. Instead let's go with one particular way of computer program parsing that is easy for a beginner programmer to write. The general algorithm goes like this:

Read the first number until the operator is found. Create a tree leaf containing the operator, with a left branch containing the first number read, and a right branch being empty. Read the next symbol: if it's a parenthesis, start over but with the right branch becoming a new "leaf" to hold the next operator. If it's another number, put it into the right branch. Now simplify by applying the leaf operator to both its branches, and storing the result inside the leaf and clipping the branches. Read the next symbol, if it's an operator create a leaf with a left branch containing the resulting value previously computed and a right branch containing nothing... repeat.

So for the above expression: we read 40, plus, 40, we stop and add those to get 80, we read times, we read 0, we stop and multiply those to get 0, we read plus, 1, stop and add to get 1, and stop with the answer 1. Update: Since this can be confusing to anyone who's never touched the subject of programming before, I decided to spend some time and make some images in Paint to further the understanding. If you don't care or think you got the gist of it already, skip to part 2 of my long answer.

Step 0: read in our expression, "40+40x0+1", and put each distinct element in its own box.


Step 1: start in the first box, read the value. "40". It's a number, so create a new "leaf" and put "40" inside.


Step 2: read the value in the second box. "+". It's an operator, so create a new, bigger "leaf" and put "+" inside, then create a child branch to the smaller leaf with 40.


Step 3: read the value in the third box. "40" again. Put it in a small leaf and add a second branch from the "+" to it.


Step 4: pause. The "+" has two child branches, so we must simplify. Perform "+" on the two leaves, so "40+40", which is "80", put "80" in the leaf where "+" was, and get cut off the old branches.


Step 5: continue and read the value in the fourth box. "x". It's an operator for multiplication, so create a big leaf and put "x" in it, and create a branch to "80".


Step 6: read the value in the fifth box. "0". Put it in a leaf that is connected by a branch to the "x".


Step 7: pause again, we have to simplify. Compute "80x0", which is 0, and put it in its own leaf.


Step 8: read the value in the sixth box. "+". Same story: big leaf, put "+" in it, create a branch that connects it with "0".


Step 9: read the last value. "1". It becomes a leaf with a branch from the "+".


Step 10: simplify for the final answer. "0+1" is 1.

Since according to rules like PEMDAS the answer should be 41, not 1, we say the above expression was "malformed". (Alternatively you might say our method of parsing is inadequate for not guessing what we mean.) You can get the right answer in this parsing system by adding parentheses in the right places, or you can use a more complicated parser that conforms to PEMDAS.


Long answer part 2: Why do we want PEMDAS's answer anyway? The answer comes from algebra. $$40 + 40*0 + 1 = x$$ is the algebraic question, which we must apply transformations to using algebraic rules. The particular rules are chosen in the same way that the particular parsing method is chosen--because we want the answer to be a particular something. In this case, we can use real-valued definitions and axioms of a vector space. The definitions describe addition and multiplication as operations on two numbers. The associativity axiom lets us rewrite as $$40 + 1 + 40*0 = x$$. Applying the definition of times, we rewrite as $$40 + 1 + 0 = x$$. Applying the definition of plus, we reduce to $$41 + 0 = x$$. Applying the definition of plus again, we reduce to $$41 = x$$.

Note that the definition of scalar multiplication implicitly takes precedence over addition. From $$40 + 1 + 40*0 = x$$, we might have said $$41 + 40*0 = 81*0 = 0 = x$$. But from the definitions, "40+1" constitute two distinct elements "40" and "1", while "40*0" does not--it constitutes one distinct element "x" in X, which in this case is "0". So the road to $$x=0$$ would be erroneous within our vector space axioms. The associative transformation above was also unnecessary since that multiplicative substitution takes precedence regardless (just like PEMDAS says), it's there just for fun.

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Posted on 2011-10-16 by Jach

Tags: fodder, math

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