SOLID — Single Responsibility Principle: Visual walkthrough

This is a visual walkthrough of the central result in the parent SRP note. By the end you will see why a class serving many actors is a risk multiplier — and why splitting it to one-actor-per-class drops that risk to a flat zero.

Step 1 — What is an "actor", drawn as a person pulling a rope

WHAT. Before any maths, we need the raw ingredient. An actor is one stakeholder or role that can request a change to your class — Accounting, HR, the DBA. Picture each actor as a person, and the class as a box. If an actor can order a change, we draw a rope from that person to the box.

WHY this picture. SRP is not about how many methods a class has; it is about how many ropes are tied to the box. Counting ropes = counting reasons-to-change. That count is the single number our whole formula is built on. We will call it $A$ (for Actors).

WHAT IT LOOKS LIKE. One box, three people, three ropes. Every rope is a hand that can yank the box and possibly jolt the others.

Figure — SOLID — Single Responsibility Principle

Step 2 — What is $p$ ? One dice-roll per rope

WHAT. When one actor requests a change, the developer edits shared code. That edit might accidentally break behaviour some other actor depends on. We call the chance of one such accidental break $p$ — a probability, so a number between $0$ and $1$ .

WHY a probability and not certainty? Because a break is not guaranteed — it depends on whether the two actors happen to share a helper (recall the parent's regularHours() example). "Depends, with some chance" is exactly what a probability captures. We model it as rolling a weighted die: land in the red slice = break; land in the green = safe.

WHAT IT LOOKS LIKE. A single die where a fraction $p$ of the face is red (break) and the rest, $1 - p$ , is green (survive).

Recall

If $p = 0.2$ , what is the survival probability for one other actor? ::: $1 - 0.2 = 0.8$ .

Step 3 — Only the OTHER actors can be surprised: why $A - 1$

WHAT. One actor requests the change — they are not surprised by their own request. So the actors who might be accidentally broken are everyone except the requester. That count is $A - 1$ .

WHY subtract one? Because you cannot startle the person who pulled the rope on purpose. If $A = 3$ (Accounting requests, HR + DBA at risk), the number of bystanders is $3 - 1 = 2$ . This is the single most-missed piece of the formula, so we draw it.

WHAT IT LOOKS LIKE. Three people; one glows (the requester, safe by choice); the other two are the "at-risk" bystanders. We roll one survival-die per bystander.

Step 4 — Everybody survives: why we MULTIPLY $(1 - p)$

WHAT. We now ask a cleaner question first: what is the chance that nobody gets broken — every bystander survives? For each bystander the survival chance is $1 - p$ , and there are $A - 1$ of them.

WHY multiply, not add? Because the dice are independent — HR breaking has nothing to do with the DBA breaking. The rule for "this AND that AND that, all independent" is to multiply the probabilities: $P (all survive) = (1 - p) \times (1 - p) \times \dots = (1 - p)^{A - 1} .$ Multiplying shrinks the number: each extra bystander is another survival-die that must also come up green, and stacking green-on-green is rarer. That is why it is a power, not a product of different things — every factor is the same $1 - p$ .

WHAT IT LOOKS LIKE. A chain of green survival-dice. The joint-survival probability is the area of a shrinking green square: two dice → $(1 - p)^{2}$ , drawn as a square of side $1 - p$ .

Step 5 — Flip it: why $R = 1 - (1 - p)^{A - 1}$

WHAT. We wanted the risk: the chance that at least one bystander is broken. We just computed the opposite — the chance that none is. These two are complements, so: $R (A) = 1 - (1 - p)^{A - 1} .$

WHY compute the opposite first? "At least one broken" is messy — it could be exactly one, or two, or all of them, and you'd have to add all those cases. But "none broken" is a single clean product (Step 4). Then "at least one" is just everything else, which is $1$ minus that. This complement trick turns a hard sum into an easy subtraction.

WHAT IT LOOKS LIKE. The full bar is $1$ (certainty). The green slab is "all survive." Whatever is left — the red slab — is the risk $R$ . As you add actors, green shrinks and red grows.

Step 6 — The SRP-clean case: why $A = 1$ gives exactly $R = 0$

WHAT. Plug in the SRP-obedient class, $A = 1$ (one boss, one rope): $R (1) = 1 - (1 - p)^{1 - 1} = 1 - (1 - p)^{0} = 1 - 1 = 0.$

WHY does the exponent go to zero? With one actor there are $A - 1 = 0$ bystanders — nobody to accidentally surprise. Anything raised to the power $0$ is $1$ (an empty product of survival-dice is "fully safe"), so the green slab fills the whole bar and red vanishes.

WHAT IT LOOKS LIKE. A pure-green bar — no red at all. This is the payoff of SRP drawn as a picture: split a class until each has one actor, and the cross-actor risk is not just small — it is identically zero, for any $p$ .

Step 7 — Reading the whole curve: risk vs. number of actors

WHAT. Sweep $A$ from $1$ upward at a fixed $p = 0.2$ and plot $R (A)$ .

WHY plot it? A single number ( $R (3) = 0.36$ ) is a dot; the curve shows the trend — where SRP saves you most. It starts pinned at $0$ (Step 6), then climbs and flattens as it approaches $1$ .

WHAT IT LOOKS LIKE. A rising curve starting at $(1, 0)$ , passing through the parent's worked point $(3, 0.36)$ , bending toward the ceiling $R = 1$ . The steepest gains from splitting happen early — going from $A = 3$ to $A = 1$ erases $36%$ risk outright.

The one-picture summary

Everything above, compressed: ropes → dice → shrinking green square → red risk slab → the rising curve, with SRP pinning it back to zero.

Recall Feynman: tell the whole walkthrough to a 12-year-old

Picture a toy box with ropes tied to it — one rope per person allowed to yank it. If only you hold a rope (SRP: one boss), nobody else gets jolted, ever. But if three people hold ropes and you pull yours, the other two might get knocked over. Each of those two flips a coin that's mostly green ("I'm fine") with a little red ("ouch"). For both to stay fine, both coins must be green — and two greens in a row is rarer than one, which is why we multiply the green chances and get a smaller number. The chance that someone gets knocked over is just "everything that isn't all-green," so we take $1$ minus the all-green number. With one boss there are zero other people, all-green is guaranteed, and the risk is a clean zero. That flat zero is the reason we split classes — SRP turns a room full of tangled ropes into calm, one-rope boxes.

Connections

Cohesion and Coupling — this whole risk model is coupling made numeric: high $A$ = high cross-actor coupling.
God Object (anti-pattern) — the large- $A$ end of our curve; risk near the ceiling.
Facade Pattern — after splitting to $A = 1$ classes, a Facade restores one entry point without re-tangling ropes.
Open-Closed Principle — one-actor classes are the units you extend without editing.
Dependency Inversion Principle — separated (low- $A$ ) classes talk through abstractions.
Shotgun Surgery — over-splitting swings the other way; SRP is the balance point.
DRY Principle — a shared helper is precisely what raises $p$ between two actors.