WEBVTT

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On a number line, the distance from two to five is three.

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We ignore direction and count the separation.

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A metric abstracts useful rules from examples like this, so we can

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discuss distance even when the points are not positions in a drawing.

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A metric never gives a negative distance.

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A point has distance zero from itself, and distance zero occurs only

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when the two points are the same.

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This last requirement is called identity of indiscernibles.

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A metric gives a symmetric distance: reversing the two endpoints leaves the

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number unchanged.

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The distance from two to five therefore equals the distance from five

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to two.

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Later we will deliberately relax this rule to measure directional loss.

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A direct distance cannot exceed the total distance of a detour through

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a third point.

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On the number line, travelling from zero through two to five costs

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two plus three, which equals the direct distance five.

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Replace number-line points with two-bit strings.

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To move from zero, zero to one, zero, only the first coordinate

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differs.

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Counting differing coordinates gives distance one.

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The coordinates are labelled; rearranging their order silently would change the comparison.

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From one, zero to one, one, only the second coordinate differs.

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The two-step route has total cost two.

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From zero, zero directly to one, one, both coordinates differ, so the

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direct count is also two.

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If one coordinate differs between the first and final strings, at least

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one leg of any two-leg route must change that coordinate.

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Counting all such coordinates gives the triangle inequality.

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This is an argument about every route, not just our example.

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A ball centred at v contains points whose distance from v is

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strictly less than its radius.

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A point exactly at the radius is excluded.

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The word open in this definition concerns the strict inequality, not the

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appearance of a circle.

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With our integer-valued bit distance, radius one around zero, zero contains

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only the centre.

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Points at distance one are excluded.

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Radius slightly greater than one also admits one, zero and zero, one.

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It still excludes one, one, whose distance is two.

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For the card example, X means dot and Y means stripe.

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A has only a dot, B both, C only a stripe, and

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D neither.

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Their two bits record X, then Y, with one for true and

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zero for false.

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A selected observation profile records only the answers to the chosen questions;

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we call that profile a signature.

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Return to the four valuations and observe only X.

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A and B differ in Y, yet their selected one-entry signatures

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agree.

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Counting differences in those signatures therefore gives zero between two distinct underlying

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valuations.

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This does not make the calculation wrong.

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It makes the function a pseudometric on valuations: it keeps nonnegativity, symmetry

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and the triangle inequality, but allows distinct points at distance zero.

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The measurement has forgotten a distinction present in the full state.

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With only X selected, can A and B have zero observed distance?

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Does that prove they are the same complete valuation?

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Explain the two different objects being compared.

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Yes, their selected signatures agree, so the measured distance is zero.

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No, A and B remain different valuations because Y differs.

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Identifying equal signatures later will turn this pseudometric into a metric on

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classes.

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We now have the rules needed to judge later formulas.

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Weighted disagreement will remain symmetric.

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Directed loss may be asymmetric.

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Neither formula should be inferred from how close our state cards happen

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to lie on the screen.