Lower Bound for the Theory of Real Addition

1. Lower Bound for the Theory of Real Addition Lecture 23 Reporter: RICHELYN L. DAGOC MSCS – 1

2. Constructing Short Formulas for Large Integers We make use of a trick due to Fischer and Rabin [42] for constructing short formulas for full arithmetic on large integers. We illustrate the trick by constructing a formula of length O(n) that says x = 22n. We actually construct inductively a formula ψn(x, z) that says x = 22n z. Then x = 22n is expressed by ψn(x, 1).

3. Here is an inductive construction that produces formulas that are exponentially smaller than (23.1), although still too big: ψ0(x, z) ⇐de⇒f x = z + z, ψn+1(x, z) ⇐de⇒f∃y ψn(x, y) ∧ ψn(y, z). • Note that we have used only addition +, thus the ψnare syntactically correct. Under this definition, the ψnexpress the desired property: ψ0(x, z) ⇔ x = z + z ⇔ x = 220z, and by induction, ψn+1(x, z) ⇔ ∃y ψn(x, y) ∧ ψn(y, z) ⇔ ∃y x = 22n y ∧ y = 22n z ⇔ x = 22n22n z ⇔ x = 22n+1 z.

4. The problem is that ψnappears twice in the inductive definition (23.3) of ψn+1, so the length of ψn+1 is roughly twice the length of ψn. Thus the formulas ψngrow exponentially with n. Here is where the trick comes in. We use universal quantification to write ψn+1 with only one occurrence of ψn. Instead of the definition (23.3), we take ψn+1(x, z) ⇐de⇒f ∃y ∀u ∀v ((u = x ∧ v = y) ∨ (u = y ∧ v = z)) → ψn(u, v).

5. This says the same thing, but now ψn+1 contains only one occurrence of ψn, so its length is that of ψnplus a constant. Thus the formulas ψngrow only linearly with n.

6. Encoding Multiplication • Let We now show how to construct a formula of length O(n) that says, “z ∈ In and x = yz.” The basis is given by:

7. Before we give the induction step, let us prove a motivating lemma.

11. Bitstring Manipulation • Once we know how to define and manipulate large integers with short formulas, we are on our way to manipulating long bitstrings. Here are some useful formulas that will help us with this task.

14. Here is an explanation of the formula bitn(x, y). Suppose x and y are numbers satisfying bitn(x, y). Because y is a power of two, its binary representation consists of a 1 followed by a string of zeros. The formula bitn(x, y) is true precisely when x’s bit in the same position as the 1 in y is 1. We get hold of this bit in x by dividing x by y using integer division; the quotient q and remainder r are the binary numbers illustrated. The bit we are interested in is 1 iffq is odd.

15. This formula is useful for treating numbers as bit strings of exponential length and indexing into them with other numbers to extract bits.

17. And That’s AllThank You!