Wednesday, August 13, 2025

If x > 3 and y < 2, then x² − 2y > 5

Suppose x > 3 and y < 2. Then x² − 2y > 5.

Let’s break this down step by step.

We’re told:

$x > 3$
$y < 2$

We want to see if these imply:

$x^{2} - 2 y > 5$

Step 1: Lower bound for $x^{2}$

If $x > 3$ , then:

$x^{2} > 9$

Step 2: Lower bound for $- 2 y$

If $y < 2$ , then multiplying both sides by $- 2$ (which reverses the inequality):

$- 2 y > - 4$

So the smallest possible value for $- 2 y$ is just greater than $- 4$ . It can be much larger if $y$ is negative (or even very large if $y$ is very negative).

Step 3: Combine

We have:

$x^{2} - 2 y > 9 - 4 = 5$

That’s already enough to conclude $x^{2} - 2 y > 5$ , because:

$x^{2}$ is strictly greater than 9,
$- 2 y$ is strictly greater than $- 4$ , so their sum is strictly greater than $9 - 4 = 5$ .

✅ Conclusion: Yes — from $x > 3$ and $y < 2$ , it follows that:

$x^{2} - 2 y > 5$

If you’d like, I can also show you the geometric interpretation of this inequality in the $(x, y)$ -plane, so you can see exactly which region these conditions describe.

Newton and Wallis

In January 1663/4 Newton dated his extracts from Wallis’s procedures for squaring the parabola and hyperbola by means of indivisibles; a little later he was annotating Wallis’s treatment of series in Arithmetica infinitorum and applying these procedures also to quadratures.⁴ All the time he was thinking and inventing, for example: “Thus Wallis doth it, but it may bee done thus . . .”⁵ The study of series brought him to the problem of how to interpolate terms. The point of interpolation – that is, finding the value of a new term, inserted between other terms in a series whose values were already established – was that by this method Newton knew he would be able to ‘square’ the general sector of the circle, for example, or the segment under an hyperbola, in terms of the two limiting ordinates. In algebra, Wallis had got so far as discovering the limits of π between two infinitely extendable series.⁶ He had also established generally that the area under any curve y = xⁿ was

$\frac{x^{n + 1}}{n + 1}$

Now the equation for the circle is y = (1 − x²)^(1/2) which can be put in Wallis’s form as y = z^(1/2). Wallis did not know how to deal with this fractional exponent, but Newton now supplied the answer by interpolation. Following Wallis, Newton wrote down a regular arithmetical progression thus:

□ · $\frac{2}{3}$ · \* · $\frac{8}{15}$ · \* · $\frac{48}{105}$ · \* · $\frac{384}{945}$

“’Twixt which termes \[he continued] if the intermediate termes □ \[and] \* can bee found the 2d □ will give the area of the line y = √aa − xx , the circle.”⁷

In effect, working (as Wallis, Pascal and others had done before) with arrays of tabulated values of coefficients, Newton arrived at a version of the bounding series used by Wallis:

$\frac{1}{2} \times \frac{2}{4} \times \frac{3}{6} \times \frac{5}{8} \times \frac{7}{1} 0 \times \frac{9}{1} 2 \times \frac{11}{14} \times \dots$

Multiplying each successive fraction into the preceding product yields another series of numbers:

1, $\frac{1}{2}$ , $\frac{1}{8}$ , $\frac{1}{16}$ , $\frac{5}{128}$ , $\frac{7}{256}$ , $\frac{21}{1024}$ , $\frac{33}{2048}$

Applying these fractions as coefficients to the series of even powers Newton produced the equivalence:

$\frac{x}{2} \sqrt{1 - x^{2}} = \frac{x}{2} - \frac{x^{3}}{4} - \frac{x^{5}}{16} - \frac{x^{7}}{32} - \frac{5 x^{9}}{256} - \dots$

From this expression the quadrature of the circle can be derived, the number of terms being extendable to the limits of the calculator’s patience. Newton worked out similar power series for sin⁻¹x and the area under the hyperbola, computing the latter at Boothby Pagnell to 51 decimal places.

Please explain Newton’s discovery in the text above.

Here’s what’s going on, in plain terms:

Starting point (Wallis): John Wallis had shown how to “square” (i.e., find areas under) curves of the form $y = x^{n}$ for whole-number $n$ , giving the rule

$\int_{0}^{x} t^{n} d t = \frac{x^{n + 1}}{n + 1} .$

He also produced product/series tables that bounded numbers like $π$ , but he couldn’t yet handle curves with fractional powers like $y = \sqrt{1 - x^{2}}$ (the circle).
Newton’s idea (interpolation): Newton looked at Wallis’s arrays of coefficients (those strings of fractions) and asked: if we “fill in” the missing intermediate entries so the pattern works not just for integers but also for half-integers, what must the coefficients be? This “filling in” is what he called interpolation. Concretely, he used the regular product pattern

$\frac{1}{2} \times \frac{2}{4} \times \frac{3}{6} \times \frac{5}{8} \times \frac{7}{10} \times \dots$

to generate a sequence of coefficients

$1, \frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \frac{7}{256}, \dots$

which he then attached to the even powers of $x$ .
What this unlocked: By doing that, Newton effectively discovered the binomial series for fractional exponents:

$(1 + u)^{α} = 1 + α u + \frac{α (α - 1)}{2!} u^{2} + \frac{α (α - 1) (α - 2)}{3!} u^{3} + \dots$

(valid for $| u | < 1$ ). Taking $α = \frac{1}{2}$ and $u = - x^{2}$ gives a power series for the circle’s ordinate:

$\sqrt{1 - x^{2}} = 1 - \frac{x^{2}}{2} - \frac{x^{4}}{8} - \frac{x^{6}}{16} - \frac{5 x^{8}}{128} - \dots$

Multiplying by $x / 2$ (a convenient step for segment/sector areas) yields the series you quoted:

$\frac{x}{2} \sqrt{1 - x^{2}} = \frac{x}{2} - \frac{x^{3}}{4} - \frac{x^{5}}{16} - \frac{x^{7}}{32} - \frac{5 x^{9}}{256} - \dots$
Why that matters (quadrature): Once you have a power series for the curve, you can integrate term-by-term to get areas (quadratures). For the circle, integrating from $0$ to a given $x$ gives the area of a circular segment/sector; integrating all the way to $x = 1$ gives the area of a quarter circle (hence a route to $π$ ). Newton applied the same trick to other curves—e.g., the hyperbola—producing the familiar logarithm series, and to inverse trig functions like $\sin^{- 1} x$ . He pushed these computations astonishingly far (the text notes 51 decimal places for a hyperbolic area).

In one sentence: Newton took Wallis’s patterns and, by interpolating their missing terms, generalized them to fractional exponents—essentially discovering the binomial series—so he could expand curves like $\sqrt{1 - x^{2}}$ into power series and then integrate those series to compute areas with arbitrary precision. This is one of the key early steps toward his calculus of power series.

Why does this work?

Great question. The short answer is: Newton noticed the rule that generates binomial coefficients still makes sense if you let the exponent be a fraction, and the resulting power series converges to the right function on $| x | < 1$ . That’s why his “interpolated” coefficients actually work.

Here’s the clean modern view:

Extend the binomial coefficients. For integers $m$ ,

$(1 + u)^{m} = \sum_{k = 0}^{m} (\binom{m}{k}) u^{k}, \frac{(\binom{m}{k + 1})}{(\binom{m}{k})} = \frac{m - k}{k + 1} .$

Newton keeps the same recurrence law for any exponent $α$ (this is his “interpolation”):

$c_{0} = 1, c_{k + 1} = c_{k} \frac{α - k}{k + 1} .$

That uniquely gives

$c_{k} = (\binom{α}{k}) = \frac{α (α - 1) \dots (α - k + 1)}{k!} .$

For $α = \frac{1}{2}$ these $c_{k}$ are exactly the numbers he tabulates.

Show the series equals $(1 + u)^{α}$ . Define $F (u) = \sum_{k \geq 0} (\binom{α}{k}) u^{k}$ . Differentiate term-by-term (justified below):

$F^{'} (u) = \sum_{k \geq 1} k (\binom{α}{k}) u^{k - 1} = \sum_{k \geq 0} (α - k) (\binom{α}{k}) u^{k} / (1 + u) .$

Using the recurrence, this simplifies to the differential equation

$(1 + u) F^{'} (u) = α F (u), F (0) = 1,$

whose unique solution is $F (u) = (1 + u)^{α}$ . So the “interpolated” series is exactly the function with a fractional exponent.

Convergence (why you can integrate term-by-term). For fixed $α$ ,

$| \frac{(\binom{α}{k + 1}) u^{k + 1}}{(\binom{α}{k}) u^{k}} | = | \frac{α - k}{k + 1} | | u | ⟶ | u | (k \to \infty),$

so the radius of convergence is $1$ . On any $| u | \leq r < 1$ the series converges uniformly, hence you may integrate term-by-term. With $u = - x^{2}$ and $α = \frac{1}{2}$ ,

$\sqrt{1 - x^{2}} = \sum_{k = 0}^{\infty} (\binom{1 / 2}{k}) (- x^{2})^{k} = 1 - \frac{x^{2}}{2} - \frac{x^{4}}{8} - \frac{x^{6}}{16} - \frac{5 x^{8}}{128} - \dots,$

and multiplying by $x / 2$ or integrating gives the circular segment/sector areas (the “quadrature”).

Why Wallis’s products show up. Newton’s fractions come from the identity

$| (\binom{1 / 2}{k}) | = \frac{1 \cdot 3 \cdot 5 \dots (2 k - 1)}{2 \cdot 4 \cdot 6 \dots (2 k)} \frac{1}{2 k - 1} = \frac{(\binom{2 k}{k})}{4^{k} (2 k - 1)},$

i.e., the odd-over-even product (Wallis pattern) determines the next coefficient. That’s exactly the pattern he was “interpolating.”

So it works because Newton preserved the multiplicative law for binomial coefficients, which defines a legitimate power series; that series provably equals the target function on $| x | < 1$ and can be integrated term-by-term to get the desired areas.

Cramer’s rule

1.2 The method of determinants

Suppose that we wish to solve the following pair of equations for $x$ and $y$

$a x + b y = c$

$l x + m y = n$

where $a, b, c, l, m$ and $n$ are constants.

Multiplying the first equation by $m$ , the second equation by $b$ and subtracting eliminates $y$ leaving us with the linear equation in $x$

$(a m - l b) x = c m - n b .$

Similarly, multiplying the first equation by $l$ , the second by $a$ and subtracting would eliminate $x$ and give us a linear equation in $y$ only

$(b l - a m) y = c l - n a .$

The solution to the pair of equations is then

$x = \frac{c m - n b}{a m - l b}, y = \frac{a n - l c}{a m - l b}$

provided $a m - l b \neq 0$ . The denominator in these expressions for $x$ and $y$ is called the determinant of the coefficients of $x$ and $y$ and is defined by the notation

$| \begin{matrix} a & b \\ l & m \end{matrix} | = a m - l b .$

This is called a second order determinant because it has 2 rows and 2 columns bounded by two vertical lines, not brackets. A third order determinant would have 3 rows and 3 columns. A determinant has a single value, for example

$| \begin{matrix} 2 & 4 \\ 1 & 3 \end{matrix} | = 2 \times 3 - 4 \times 1 = 2 .$

Notice now that $x$ and $y$ can be written as fractions involving two determinants

$x = \frac{| \begin{matrix} c & b \\ n & m \end{matrix} |}{| \begin{matrix} a & b \\ l & m \end{matrix} |}, y = \frac{| \begin{matrix} a & c \\ l & n \end{matrix} |}{| \begin{matrix} a & b \\ l & m \end{matrix} |} .$

The determinants in the numerators are very similar to the denominator. We write the solution in the form

$x = \frac{Δ_{x}}{Δ}, y = \frac{Δ_{y}}{Δ}$

where

$Δ = | \begin{matrix} a & b \\ l & m \end{matrix} |, Δ_{x} = | \begin{matrix} c & b \\ n & m \end{matrix} |, Δ_{y} = | \begin{matrix} a & c \\ l & n \end{matrix} |$

The determinants $Δ_{x}$ and $Δ_{y}$ are obtained from $Δ$ by replacing the first and second columns respectively by the right-hand sides of the original linear equations.

Remember, the solution is defined only if $Δ \neq 0$ . This approach is called the method of determinants, or Cramer’s rule.

Please tell me more about Cramer’s rule.

Here’s the big picture and the useful details about Cramer’s rule.

What it is (general form)

For a square linear system $A x = b$ of size $n \times n$ with $det (A) \neq 0$ ,

$x_{i} = \frac{det (A_{i})}{det (A)} (i = 1, \dots, n),$

where $A_{i}$ is the matrix $A$ with its $i$ -th column replaced by $b$ . That’s exactly what you saw for $n = 2$ . It works over real or complex numbers.

When it applies

Unique solution: only if $det (A) \neq 0$ (i.e., $A$ is invertible).
$det (A) = 0$ : either no solution or infinitely many. (Rank test: if $rank (A) = rank ([A | b]) < n$ → infinitely many; if ranks differ → no solution.)

Why it works (connection to the adjugate)

The matrix inverse has the explicit form

$A^{- 1} = \frac{1}{det (A)} adj (A),$

where $adj (A)$ is built from cofactors. Multiplying $A^{- 1} b$ yields the same “replace-a-column” determinants in each component—those are the $det (A_{i})$ .

A quick 3×3 example

Solve

${\begin{cases} 2 x - y + 3 z = 9 \\ x + 4 y - z = 1 \\ 3 x + 2 y + 2 z = 12 \end{cases} ⟺ A = [\begin{matrix} 2 & - 1 & 3 \\ 1 & 4 & - 1 \\ 3 & 2 & 2 \end{matrix}], b = [\begin{matrix} 9 \\ 1 \\ 12 \end{matrix}] .$

Compute determinants:

$Δ = det (A) = - 5, Δ_{x} = det [\begin{matrix} b & - 1 & 3 \\ 1 & 4 & - 1 \\ 12 & 2 & 2 \end{matrix}] = - 34, Δ_{y} = 10, Δ_{z} = 11.$

Hence

$x = \frac{- 34}{- 5} = \frac{34}{5}, y = \frac{10}{- 5} = - 2, z = \frac{11}{- 5} = - \frac{11}{5} .$

Geometric intuition (2×2 case)

$| det (A) |$ is the area scaling of the unit square under the linear map $A$ . Replacing a column by $b$ forms a new parallelogram; the ratio $det (A_{i}) / det (A)$ extracts the coordinate $x_{i}$ .

Pros & cons

Pros

Gives closed-form expressions for each variable.
Great for symbolic work (e.g., showing how solutions change with parameters) and small systems (2×2, occasionally 3×3).

Cons

Not computationally efficient for large $n$ : computing $n + 1$ determinants is expensive (typically $O (n^{4})$ if done via LU per determinant).
Numerically less stable than Gaussian elimination/QR for floating-point data; determinants can overflow/underflow and amplify conditioning issues.

Practical tips

For hand calculations: use it for 2×2 and 3×3 (Sarrus’ rule helps for 3×3).
For larger systems or numerics: prefer Gaussian elimination, LU/QR factorization, or direct $A^{- 1} b$ via a solver.
If $det (A)$ is (nearly) zero, the system is ill-conditioned or singular—avoid Cramer’s rule and use a stable solver and conditioning checks.

If you want, I can walk you through Sarrus’ rule for 3×3 determinants or set up a couple of practice problems with solutions.

Yes, please tell me about Sarrus’ rule.

happy to! Sarrus’ rule is a quick way to compute only 3×3 determinants.

What it is

For

$| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} |$

Sarrus’ rule says

$det = a e i + b f g + c d h - c e g - b d i - a f h .$

How to do it (the “copy columns” trick)

Rewrite the first two columns to the right:

$\begin{array}{ccccc} a & b & c & a & b \\ d & e & f & d & e \\ g & h & i & g & h \end{array}$

Multiply the three down-right diagonals and add:

$(a \cdot e \cdot i) + (b \cdot f \cdot g) + (c \cdot d \cdot h) .$

Multiply the three up-right diagonals and add:

$(c \cdot e \cdot g) + (b \cdot d \cdot i) + (a \cdot f \cdot h) .$

Subtract: (down sum) − (up sum).

Worked example

Compute

$det [\begin{matrix} 2 & - 1 & 3 \\ 1 & 4 & - 1 \\ 3 & 2 & 2 \end{matrix}] .$

Copy first two columns:

$\begin{array}{ccccc} 2 & - 1 & 3 & 2 & - 1 \\ 1 & 4 & - 1 & 1 & 4 \\ 3 & 2 & 2 & 3 & 2 \end{array}$

Down sum: $2 \cdot 4 \cdot 2 = 16$ , $(- 1) \cdot (- 1) \cdot 3 = 3$ , $3 \cdot 1 \cdot 2 = 6$ → $16 + 3 + 6 = 25$ . Up sum: $3 \cdot 4 \cdot 3 = 36$ , $2 \cdot (- 1) \cdot 2 = - 4$ , $2 \cdot 1 \cdot (- 1) = - 2$ → $36 - 4 - 2 = 30$ . Determinant $= 25 - 30 = - 5$ .

Why it works (intuition)

For 3×3 matrices the determinant is a signed sum over the six permutations of ${1, 2, 3}$ . Sarrus’ diagram just groups the three even permutations (the “down” diagonals) and subtracts the three odd permutations (the “up” diagonals).

Tips & pitfalls

Only for 3×3. There’s no Sarrus shortcut for 4×4 or higher; use cofactor expansion or row-reduction.
Row/column ops help: If you can create a zero, expansion along that row/column gets easier than Sarrus.
Signs: The most common mistake is dropping a minus in the “up” group—write the two sums separately, then subtract.

Want a couple of quick practice determinants with answers?

Monday, August 11, 2025

Everyday examples of cohomology

Here’s a grab-bag of “cohomology in the wild”—ways the idea shows up in everyday-ish things. Think: local rules that don’t quite fit together globally, and the little “residue” of that mismatch is a cohomology class.

Setting one clock for a whole network (H¹ of a graph). Suppose edges of a network store measured time differences between neighboring devices. If those differences come from a single global clock, their sums around every cycle are 0. If some loop sums to +3 seconds, you can’t assign one consistent clock—your loop sum is a nonzero cohomology class.
Kirchhoff’s laws in circuits (graph cohomology). Node voltages are 0-cochains; edge voltage drops are 1-cochains. Taking differences (“gradient”) sends node voltages → edge drops. A pattern of drops is exact iff every loop sums to 0 (KVL). Batteries/induced EMFs make nonzero loop sums: that’s a nontrivial class in $H^{1}$ measuring how much “potential” fails to exist globally.
Wind around a chimney / water around a drain (de Rham $H^{1}$ ). Away from the chimney the field can be curl-free (closed), yet the circulation $\oint$ around the chimney isn’t 0. There’s no global potential function because the domain has a hole. That “circulation around the hole” lives in $H^{1}$ and counts independent loops.
Magnetic/electric induction loops. Faraday’s law $\oint E \cdot d l = - \frac{d Φ_{B}}{d t}$ says the electric field around a changing magnetic flux isn’t a pure gradient globally—even if it looks locally conservative. The cohomology class detects the “global twist” forced by the changing flux.
Panorama/map stitching (Čech $H^{1}$ ). You align overlapping photos (local matches). If the pairwise alignments are perfectly compatible, they come from one global placement. If you go around a cycle of overlaps and come back rotated/shifted, that loop defect is a 1-cocycle that isn’t a coboundary—nonzero class says “no globally consistent stitching without a cut.”
Reconstructing an image from its gradients (Poisson editing). Given desired pixel-gradients, you can recover an image iff those gradients are an exact field. If loop integrals don’t vanish, you get seams unless you cut the domain (killing $H^{1}$ ) or adjust gradients to kill the cohomology.
GPS path independence & “conservative costs.” If road “efforts” (fuel/time differentials) are integrable to a single potential, the cost from A→B is path-independent. One-way streets, toll loops, or wind currents can create loop imbalances: those are precisely the obstructions measured in $H^{1}$ .
“Counting connected pieces” ( $H^{0}$ ). $H^{0}$ is the simplest: it counts connected components. In practice: how many islands are there in your maze/map/graph? That’s the dimension of $H^{0}$ .
Global balance laws (top cohomology). Integrating a density over a region gives a number that only depends on the region’s “top” cohomology class (think total mass/charge). Change coordinates, wiggle things—integrals of exact top-forms vanish; only the cohomology class survives.

One-line motto: cohomology measures whether “locally defined stuff” (differences, gradients, patches, phases) can be glued into a single global object—and if not, how it fails.

If you want, we can pick one (say, circuits or the panorama example) and work through a tiny concrete calculation.

Ancient Ocean Sea Concept

The ancient idea of the "Ocean sea"

The ancient idea of the "Ocean Sea" (often called Mare Oceanum in Latin or Ὠκεανός in Greek) was the belief—common in many classical and medieval cultures—that all the known landmasses were surrounded by one vast, continuous body of water.

Here’s how it played out historically:

1. Origins in Ancient Greece

Homer’s world (8th century BCE): In the Iliad and Odyssey, the “Ocean” (Okeanos) is both a god and a river-like sea encircling the Earth. It wasn’t thought of as just salt water, but as a primal, life-giving boundary beyond which lay the unknown.
Geographers like Hecataeus and Herodotus: They preserved the idea that the “inhabited world” (oikoumene) was like an island surrounded by this encircling ocean.
Eratosthenes and Strabo: More scientific approaches emerged, but the encircling ocean still appeared on maps—sometimes in circular form.

2. Roman and Medieval Continuation

The Romans kept the concept, calling it Oceanus, distinguishing it from inner seas (like the Mediterranean).
Medieval European mappae mundi often show the continents (Asia, Europe, Africa) ringed by the “Ocean Sea,” with monsters, islands, and strange marvels beyond.
Norse cosmology had a similar idea—miðgarðr (the human world) surrounded by a great sea inhabited by the Midgard Serpent.

3. The “Ocean Sea” in the Age of Discovery

By the 15th century, Europeans used “Ocean Sea” (Mar Oceana in Spanish, Mar Océano in Portuguese) to mean the Atlantic Ocean in particular.
Columbus famously set out across the “Ocean Sea” in 1492, thinking it was a relatively narrow waterway between Europe and Asia.
This reflected a lingering medieval mindset: the ocean was vast but finite, a ring to be crossed to reach the other side of the world.

4. Symbolism and Cultural Meaning

The “Ocean Sea” was both a geographical boundary and a psychological one—the edge of the known world.
It symbolized mystery, danger, and the limits of human exploration.
In maps and literature, it was often populated with sea serpents, islands of the blessed, or portals to other realms.

If you like, I can sketch you a simple diagram showing how ancient people visualized the “Ocean Sea” as a ring around the continents—it makes the concept much clearer visually.

Make an ALU using NAND gates

How do you make an ALU using nand gates?

Fun project! The clean way to “make an ALU from NAND gates” is to (1) build the basic gates from NANDs, (2) build slightly bigger blocks (XOR, full adder, multiplexers) from those, then (3) wire a 1‑bit ALU “slice” and replicate it N times.

Here’s a compact blueprint you can follow.

1) Everything from NAND

NAND is functionally complete, so first express the usual gates:

NOT: ¬A = NAND(A, A)
AND: A ∧ B = ¬NAND(A, B) = NAND(NAND(A, B), NAND(A, B))
OR (via De Morgan): A ∨ B = NAND(¬A, ¬B) = NAND(NAND(A, A), NAND(B, B))
XOR (minimal 4‑NAND version): t1 = NAND(A, B) t2 = NAND(A, t1) t3 = NAND(B, t1) A ⊕ B = NAND(t2, t3)
2:1 MUX, y = s̄·d0 + s·d1, in NANDs: s̄ = NAND(s, s) u = NAND(d0, s̄) v = NAND(d1, s) y = NAND(u, v)

These five are the only “primitives” you’ll need.

2) Full adder from NANDs

One common approach is “XOR‑XOR‑AND/OR”:

p = A ⊕ B (4 NANDs as above)
sum = p ⊕ Cin (another 4 NANDs)
c1 = A ∧ B (2 NANDs for AND)
c2 = A ∧ Cin (2 NANDs)
c3 = B ∧ Cin (2 NANDs)
Cout = c1 ∨ c2 ∨ c3 (OR = NAND(NOT,NOT); chain two ORs with NANDs)

Gate count isn’t critical unless you’re optimizing, but this maps cleanly and is easy to debug. (There’s also a well‑known 9‑NAND full‑adder topology if you’re chasing minimum count.)

3) One 1‑bit ALU slice

Give the slice two operands (Ai, Bi), a carry‑in (Ci), and control bits op[2:0]. Compute the candidate results in parallel, then use a MUX to pick one:

Common ops (examples; choose what you need):

AND: Ai ∧ Bi
OR: Ai ∨ Bi
XOR: Ai ⊕ Bi
ADD: Si = Ai ⊕ Bi ⊕ Ci, Cout from the full adder
SUB: do two’s complement: add Ai + (¬Bi) + 1 Implementation trick: feed Bi′ = (bInvert ? ¬Bi : Bi) into the adder and set Ci = bInvert for the LSB. This is exactly how the MIPS/LC‑3 textbook ALUs do it.
PASSA / PASSB: route through a MUX
SLT (set‑less‑than): in the MSB slice, take the sign of A − B and route it into bit0 via a dedicated path (others output 0). You only need this if you want comparisons.

Inside the slice:

Make Bi′ = MUX(bInvert, Bi, ¬Bi) using the NAND MUX above.
Build the full adder with (Ai, Bi′, Ci) → (Sum, Co).
Compute logic results (AND/OR/XOR) with the NAND recipes.
Select the output with a 4:1 or 8:1 MUX in NANDs, controlled by op[2:0].
- For example, use two cascaded 2:1 NAND‑MUXes to build a 4:1.

Example encoding (use any you like):

000 → AND
001 → OR
010 → XOR
011 → ADD
100 → SUB (i.e., same data path as ADD, but assert bInvert=1; for LSB, Ci=1)
101 → PASS A
110 → PASS B
111 → SLT (only meaningful in bit0; others drive 0)

Outputs per slice: result bit (Ri) and carry‑out (Ci+1). Chain Ci+1 → next slice’s Ci to form an N‑bit ripple‑carry ALU.

4) Scale to N bits

Replicate the slice N times (4, 8, 16, 32…):

Tie all op[2:0] and bInvert lines in parallel across slices.
For ADD/SUB:
- LSB: Ci = bInvert (0 for ADD, 1 for SUB)
- Middle bits: Ci from previous slice
- MSB: exposes Cout and (optionally) the signed overflow flag: V = Cout ⊕ Cn−1
For SLT: compute A − B in the background (SUB mode), capture the MSB’s sign (or overflow‑corrected less‑than), and route it to bit0 through the 111 MUX selection. Others drive 0 for 111.

5) Practical notes

Fan‑in: stick to 2‑input NANDs. If you need wider ORs, tree them (pairwise with NAND‑based OR).
Timing: this is a ripple‑carry design; it’s simple but slowish. If you care about speed, add carry‑lookahead later (still doable in NAND‑only logic, just more gates).
Reset/flags: you can add Zero flag by NOR‑reducing all result bits (NOR = NOT(OR) using NANDs), Sign from MSB, Carry from last Cout, Overflow as above.

6) Minimal NAND‑only Verilog (illustrative)

If you want to simulate before wiring:

// 1-bit slice, NAND-only style (illustrative subset: AND/OR/XOR/ADD)
module alu1 (
  input  a, b, cin,
  input  op0, op1, op2, // select lines
  input  bInvert,       // 0:add/use b, 1:sub/use ~b
  output result,
  output cout
);
  wire nb, bsel, t1,t2,t3, xor_ab, p1,p2,p3, c1,c2,c3, sum;

  // NOT b
  nand(nb, b, b);
  // bsel = MUX(bInvert ? ~b : b)
  wire ninv; nand(ninv, bInvert, bInvert);
  wire m0, m1;
  nand(m0, nb, ninv);     // ~b & ~bInvert
  nand(m1, b,  bInvert);  //  b &  bInvert
  nand(bsel, m0, m1);     // OR via NAND-NAND → MUX output

  // XOR a ^ bsel (4-NAND)
  nand(t1, a, bsel);
  nand(t2, a, t1);
  nand(t3, bsel, t1);
  nand(xor_ab, t2, t3);

  // SUM = xor_ab ^ cin (4-NAND more)
  wire s1,s2,s3;
  nand(s1, xor_ab, cin);
  nand(s2, xor_ab, s1);
  nand(s3, cin,    s1);
  nand(sum, s2, s3);

  // Carry = (a&bsel) | (a&cin) | (bsel&cin)
  // ANDs in NAND form
  wire ab_n, ac_n, bc_n, ab, ac, bc, c12_n, c12, c_all;
  nand(ab_n, a, bsel); nand(ab, ab_n, ab_n);
  nand(ac_n, a, cin);  nand(ac, ac_n, ac_n);
  nand(bc_n, bsel, cin); nand(bc, bc_n, bc_n);
  // OR three terms: (ab | ac) | bc
  wire abac_n;                  // OR via NAND of NOTs
  // NOT ab / NOT ac
  wire nab, nac; nand(nab, ab, ab); nand(nac, ac, ac);
  nand(abac_n, nab, nac);       // ~(~ab & ~ac) = ab|ac
  wire abac; nand(abac, abac_n, abac_n);
  // (ab|ac)|bc
  wire nabac, nbc; nand(nabac, abac, abac); nand(nbc, bc, bc);
  wire carr_or_n; nand(carr_or_n, nabac, nbc);
  wire carry; nand(carry, carr_or_n, carr_or_n);

  assign cout = carry;

  // Logic ops:
  wire and_n, or_n, a_and_b, a_or_b, a_xor_b;
  // AND
  nand(and_n, a, b); nand(a_and_b, and_n, and_n);
  // OR
  wire na, nb2; nand(na, a, a); nand(nb2, b, b);
  nand(or_n, na, nb2); nand(a_or_b, or_n, or_n);
  // XOR (reuse form with original a,b)
  wire u1,u2,u3; nand(u1, a, b); nand(u2, a, u1); nand(u3, b, u1); nand(a_xor_b, u2, u3);

  // 4:1 MUX (use two 2:1 MUX stages) to choose among AND/OR/XOR/SUM
  // Here: op1:op0 select (ignore op2 in this minimal example)
  function automatic mux2 (input d0, input d1, input s);
    // synthesizable macro in real toolchains; shown behaviorally here
    mux2 = s ? d1 : d0;
  endfunction

  wire mA = mux2(a_and_b, a_or_b, op0);
  wire mB = mux2(a_xor_b, sum,    op0);
  assign result = mux2(mA, mB, op1);
endmodule

(When you build it purely from nand primitives, replace the behavioral mux2 with the 2:1 NAND‑MUX shown earlier.)

7) How to wire a 4‑bit ALU quickly

Instantiate 4 slices: bit0..bit3.
For ADD: set bInvert=0, bit0.cin=0. For SUB: set bInvert=1, bit0.cin=1.
Chain carries: bit0.cout→bit1.cin→bit2.cin→bit3.cin.
Use the same op[2:0] for all slices.
Optional flags: Z = NOR of all result bits; C = bit3.cout; V = c3_out ⊕ c2_out; N = result[3].

If you want, tell me what bit‑width and exact ops you need and I’ll map the control table and draw you a gate‑level slice with the exact NAND count.

Thursday, August 7, 2025

Rings vs semirings

Overview A semiring is an algebraic structure with two binary operations (usually called “addition” and “multiplication”) satisfying most of the familiar ring axioms except that additive inverses need not exist. A ring, by contrast, does require that every element have an additive inverse. Below is a side-by-side comparison of their defining properties, followed by examples and some remarks on when each arises in mathematics and computer science.

1. Definitions and Axioms

Property	Semiring	Ring
Underlying set	$S$	$R$
Addition $(+, 0)$	Commutative monoid: $\forall a, b, c \in S$ , $a + (b + c) = (a + b) + c$ , $a + b = b + a$ , $0 + a = a$ .	Abelian group:All semiring properties plus $\forall a \exists (- a)$ such that $a + (- a) = 0$ .
Multiplication $(\cdot, 1)$	Monoid: $\forall a, b, c \in S$ , $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ ,often with identity $1$ .	Monoid:Same as semiring. Often required to have $1$ .
Distributivity	$\forall a, b, c$ , $a \cdot (b + c) = a \cdot b + a \cdot c$ , $(b + c) \cdot a = b \cdot a + c \cdot a$ .	Same as semiring.
Zero-absorption	$\forall a,; 0 \cdot a = a \cdot 0 = 0$ .	Same as semiring.
Additive inverses	Not required	Required

Note on terminology:

Some authors require a multiplicative identity $1$ in both structures; others call a structure without $1$ a “hemiring” (for semirings without identity) or a “rng” (for rings without unity).

In this summary we assume both have $1$ , unless noted otherwise.

2. Key Differences

Additive inverses
- Ring: Every element $a$ has an inverse $- a$ so that $a + (- a) = 0$ .
- Semiring: May lack inverses; subtraction need not be defined.
Examples
- Semiring but not Ring:
  - $N = 0, 1, 2, \dots$ with usual $+$ and $\times$ .
  - The Boolean semiring $0, 1$ with $+$ = OR, $\cdot$ = AND.
- Ring:
  - $Z$ , $Q$ , $R$ , $C$ .
  - Matrix rings $M_{n} (R)$ .
Subtraction vs. cancellation
- In rings you can “cancel” addition using inverses: from $a + x = a + y$ infer $x = y$ .
- In semirings, one often works with cancellative semirings to get a weak form of cancellation, but it’s not automatic.
Applications
- Semirings:
  - Automata theory & formal languages (e.g.\ shortest-path semirings).
  - Tropical geometry (min-plus semiring).
  - Quantitative program analysis (probabilities, costs).
- Rings:
  - Core algebra and number theory.
  - Algebraic geometry, homological algebra.

3. Examples in More Detail

Natural numbers $(N, +, \times)$
- Satisfies all semiring axioms: $0$ is additive identity, $1$ is multiplicative identity.
- Fails ring axioms because there is no $n \in N$ with $3 + n = 0$ .
Tropical semiring
- Underlying set $R \cup \infty$ .
- “Addition” is $min$ , “multiplication” is $+$ .
- Used for shortest-path computations (Dijkstra, Bellman–Ford).
Boolean semiring $0, 1$
- $a + b = max (a, b)$ (logical OR), $a \cdot b = min (a, b)$ (logical AND).
- Fundamental in logic, circuit design, and automata.
Integer ring $(Z, +, \times)$
- Every integer has an additive inverse.
- Ring of central importance in algebra and number theory.

4. When to Use Which?

Use a semiring when only nonnegative (or otherwise “one-sided”) quantities make sense and subtraction would be inappropriate or undefined.
Use a ring when a full-fledged additive group structure (including negatives) is needed for constructions like modules, ideals, and homological tools.