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RAHUL'S ML BLOG -- notes on machine learning, worked out by hand est. 2026
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CHAPTER 19 . LEARNING BY COPYING . PART 4 OF 5
Teaching the Wind: Flow Matching by Pencil
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A machine that copies a human's game diary crashes on one kind of question: where the
diary holds TWO right answers (swing above the block, swing below -- both recorded,
both perfect), a machine graded by squared misses is forced to print the AVERAGE of
the truths, and the average of "go around this side" and "go around that side" is
"walk into it". The failure is the question shape: one printed answer, graded against
a pile of truths, lands on the pile's average.
So stop printing answers. This page teaches a machine something different: at every
point BETWEEN random noise and recorded truth, the direction toward truth -- a wind.
A machine that knows the wind everywhere never prints a destination; it can be
DROPPED anywhere and blown home. Dropped twice in different places, it blows to
different truths -- which is exactly what a pile of right answers needs.
THREE PILES AND A MIXING DIAL
noise jar mixtures truth pile
(random dust) (in between) (recorded answers)
. . . . * . * *
. . . . <------- one straight line -------> * * *
. . . *
tau = 0 tau between tau = 1
pure noise pure truth
Three characters, all living in re-ruled answer units (each answer column re-ruled to
middle 0, spread 1 -- the same food prep every machine in this chapter eats):
The TRUTH PILE: the diary's recorded answers. For pencil work, one answer here is a
one-move envelope -- two numbers, a target point -- standing in for the real
machine's eight-move, sixteen-number envelope. Same arithmetic, shorter lines.
The NOISE JAR: pure randomness. A pull from the jar is two numbers drawn from a bell
curve centred at 0 with spread 1 -- dust with no meaning at all.
The MIXING DIAL tau: a number from 0 to 1 that says how far a point sits along the
straight line from a noise pull toward a truth. At tau = 0 you hold pure noise; at
tau = 1, pure truth; at tau = 0.4, a 40/60 blend.
ONE TRAINING PAIR, EVERY STROKE WRITTEN
A training pair is manufactured, not stored. Three rolls make one:
Roll a truth from the diary: A = [0.50, 0.60]
Roll a pull from the noise jar: N = [0.10, 0.80]
Roll the dial anywhere in 0..1: tau = 0.4
Mix: the point 40% of the way from noise to truth, M = tau x A + (1 - tau) x N:
M[1] = 0.4 x 0.50 + 0.6 x 0.10 = 0.20 + 0.06 = 0.26
M[2] = 0.4 x 0.60 + 0.6 x 0.80 = 0.24 + 0.48 = 0.72
M = [0.26, 0.72]
N = [0.10, 0.80]
\
\ M = [0.26, 0.72] <- 40% of the way along
\ /
\/_____________ A = [0.50, 0.60]
And the TEACHING ANSWER -- what the machine should say when standing at M -- is the
wind: the arrow pointing from the noise end toward the truth end,
wind = A - N = [0.50 - 0.10, 0.60 - 0.80] = [0.40, -0.20].
Check the wind with feet. Standing at M with 1 - tau = 0.6 of the line still ahead,
walking that remaining fraction of the wind should land exactly on the truth:
M[1] + 0.6 x wind[1] = 0.26 + 0.6 x 0.40 = 0.26 + 0.24 = 0.50 = A[1] yes
M[2] + 0.6 x wind[2] = 0.72 + 0.6 x (-0.20) = 0.72 - 0.12 = 0.60 = A[2] yes
It lands. And because the mixing is a straight line, the SAME arrow A - N is the
right answer at every tau along this particular noise-to-truth line: the wind for
one (truth, noise) couple does not bend.
Same A = [0.50, 0.60] and N = [0.10, 0.80], but the dial rolls
tau = 0.7. What is M? And what is the teaching answer now?
CHECK: M = [0.7 x 0.50 + 0.3 x 0.10 , 0.7 x 0.60 + 0.3 x 0.80]
= [0.35 + 0.03 , 0.42 + 0.24] = [0.38, 0.66]
teaching answer: still A - N = [0.40, -0.20] -- the dial moved M along
the line; the line's arrow did not change.
WHAT THE MACHINE READS AND PRINTS
The wind machine is the same web-of-dials species as the copier -- only its mouth and
its job change. It must judge the wind FROM three things: which table situation the
answer is for, where along the mixing it is standing, and how far the dial has
turned. So its input is all three, side by side:
5 re-ruled question numbers (the table: hand x, hand y, block x, y, twist)
+ 16 mixed-envelope numbers (M, the real eight-move envelope flattened)
+ 1 the dial reading tau
= 22 wires in -> 256 -> 256 -> 256 -> 16 wires out (the wind)
Count its dials the same way as any web -- (inputs x units) + units per box:
box 1: 22 x 256 + 256 = 5,632 + 256 = 5,888
box 2: 256 x 256 + 256 = 65,792
box 3: 256 x 256 + 256 = 65,792
box 4: 256 x 16 + 16 = 4,112
total = 141,584
141,584 dials -- 4,352 more than the copier, all of them in box 1's wider mouth.
And the grading is the copier's meter, unchanged: subtract, square, average. Only
WHAT is graded changed -- printed wind against teaching wind. Say the machine, young
and half-trained, prints [0.30, -0.10] where the teaching answer is [0.40, -0.20]:
misses: 0.30 - 0.40 = -0.10 -0.10 - (-0.20) = 0.10
meter: ((-0.10)^2 + (0.10)^2) / 2 = (0.01 + 0.01) / 2 = 0.01
Every meal re-rolls everything: a fresh truth from the diary, a fresh pull from the
jar, a fresh dial reading -- the same diary envelope might train at tau = 0.13 today
and tau = 0.88 three meals later, against different noise each time. One recorded
answer therefore teaches the wind along MANY lines, one line per noise pull.
Truth A = [0.9, 0.1], noise N = [0.5, 0.5]. The teaching answer at
tau = 0.2? At tau = 0.9?
CHECK: A - N = [0.4, -0.4], both times -- tau places M on the line;
the teaching arrow belongs to the line itself.
THEN WHY TELL THE MACHINE TAU AT ALL?
The teaching answer never used tau -- so feeding tau to the machine looks like waste.
It is not, and one worked example shows why. The machine never sees which couple
(truth, noise) built its M; it sees only the POINT M and must answer for every
couple that could have built it. Different dial readings change which couples those
are.
Strip to one number. The truth pile holds two answers, +1 and -1 (the two-truths
question). The machine stands at M = 0.5 and must print a wind. Two cases:
Dial reads tau = 0. Then M IS the noise pull (at tau = 0, M = N), and noise carries
no hint about which truth this line runs to -- both are equally possible. The two
couples through this point:
couple (A = +1, N = 0.5): wind = +1 - 0.5 = +0.5
couple (A = -1, N = 0.5): wind = -1 - 0.5 = -1.5
Graded by squared misses against both, the best single print is their average
(a squared-miss
grade has one bottom and it sits at the average -- the arithmetic that doomed the
copier, here working FOR us):
best wind at (M = 0.5, tau = 0) = (0.5 + (-1.5)) / 2 = -0.5
Dial reads tau = 0.9. Now M = 0.9 x A + 0.1 x N, so the noise that explains M
depends on the truth:
if A = +1: N = (0.5 - 0.9 x 1) / 0.1 = -0.4 / 0.1 = -4.0
if A = -1: N = (0.5 + 0.9) / 0.1 = 1.4 / 0.1 = +14.0
A jar pull of -4.0 is rare (four spreads out) but happens; a pull of +14.0 is
fourteen spreads out -- beyond any real chance. So at tau = 0.9, standing at 0.5,
essentially the only couple is (A = +1, N = -4.0), and the wind commits:
wind = +1 - (-4.0) = 5.0
feet-check: 0.5 + (1 - 0.9) x 5.0 = 0.5 + 0.5 = 1.0 = the +1 truth. it lands.
Same point M = 0.5, two dial readings, two different right winds: -0.5 (undecided,
leaning toward the pile's middle) versus 5.0 (committed to the +1 truth). A machine
not told tau would have to average those too. Told tau, it can answer each case
correctly -- that is why tau gets a wire.
And this example quietly answers the mean-collapse disease. The meter still
averages -- at tau = 0 the best wind, -0.5, IS an average. But averaging ARROWS
mid-journey is survivable: an undecided early arrow just drifts the walker, and as
the dial grows the couples separate and the arrows commit to one truth. The average
moved from the DESTINATION (fatal: the hand parks inside the block) to the
DIRECTION EARLY ON (harmless: the walk still ends at a real truth). The next page
shows the ending.
AND THIS FIRST METER READING IS PREDICTABLE TOO
The copier's first meter reading was callable before training: about 1.0. Call this
one. The teaching answer is wind = A - N: truth columns re-ruled to spread 1, jar
pulls drawn with spread 1, the two rolled independently. The spread of a DIFFERENCE
of two independent spread-1 quantities obeys spreads-square-then-add:
spread^2 of (A - N) = 1^2 + 1^2 = 2
A newborn machine prints near 0, so each squared miss averages about 2, and the
first meter reading should sit near
meter ~ 2.0 before any learning.
Measured, on one real run: 2.1136 at the first meal, then 1.583 by turn 20, 1.189
by turn 40, 1.123 by turn 60. The pencil called this one too -- and called the GAP:
the wind machine's meter opens twice as high as the copier's (2 versus 1) because
its teaching answer carries two spread-1 ingredients, not one.
AN ARROW IS NOT A DESTINATION, NOISE IS NOT DIRT, TAU IS NOT A DIAL IT TUNES
"So the machine learns to print the envelope after all." In study it never prints an
envelope, not once. Its output [0.40, -0.20] is an ARROW -- a direction and a rate
-- and only a walker who follows arrows rung by rung ever arrives at an envelope.
Confuse the two and the whole design collapses back into a single-print
machine -- the very shape the two-truths question breaks.
"Tau is a dial the machine tunes." Tau is ROLLED, fresh every meal, from 0..1 -- the
machine reads it like weather, never sets it. The trainable dials are the 141,584
in the web; tau is the 22nd input wire, and its value this meal was dice.
"The noise ruins the data." The noise IS half the data. Every teaching arrow is
truth MINUS noise; without the jar there is no line, no M to stand on, and nothing
to learn between tau = 0 and tau = 1. Flour is not dirt in a bakery -- the jar is
an ingredient, not contamination.
"The teaching answer ignores tau, so the tau wire is waste." The ANSWER for one
couple ignores tau; the MACHINE answers for all couples through its M, and tau
decides which those are. At (M = 0.5, tau = 0) the honest answer is the undecided
-0.5; at (M = 0.5, tau = 0.9) it is the committed 5.0, because the only believable
noise behind that M is -4.0 (the -1 truth would need a fourteen-spread pull that
never comes). Cut the tau wire and the machine must average -0.5 with 5.0 -- one
more mean collapse, this time in the wind itself.
SEAM. Pencil ends here; below, the same numbers in Python.
No loops, no functions -- one manufactured training pair, its grade, and a second
dial roll on the same truth. Per pair: 4 multiplies + 2 adds for the mix, 2
subtracts for the wind, then the meter's 2 subtracts, 2 squares, 1 average.
# --- three rolls make one training pair ---
A = [0.50, 0.60] # a truth from the diary (re-ruled)
N = [0.10, 0.80] # a pull from the noise jar
tau = 0.4 # the mixing dial, rolled from 0..1
M1 = tau * A[0] + (1 - tau) * N[0] # 0.4*0.50 + 0.6*0.10 = 0.26
M2 = tau * A[1] + (1 - tau) * N[1] # 0.4*0.60 + 0.6*0.80 = 0.72
wind1 = A[0] - N[0] # 0.50 - 0.10 = 0.40
wind2 = A[1] - N[1] # 0.60 - 0.80 = -0.20
# feet-check: M + (1 - tau) * wind lands on A:
land1 = M1 + 0.6 * wind1 # 0.26 + 0.24 = 0.50 = A[0]
land2 = M2 + 0.6 * wind2 # 0.72 - 0.12 = 0.60 = A[1]
# machine input = [5 question numbers] + [M, flattened] + [tau] -> 22 wires
# machine output = its printed wind; grade with the same old meter:
print1, print2 = 0.30, -0.10 # a half-trained machine's print
meter = ((print1 - wind1)**2 + (print2 - wind2)**2) / 2
# = ((-0.10)**2 + (0.10)**2) / 2 = 0.01
# --- same truth, next meal: everything re-rolls ---
N_b = [-0.30, 0.20] # fresh jar pull
tau_b = 0.9 # fresh dial
M1_b = 0.9 * 0.50 + 0.1 * -0.30 # 0.45 - 0.03 = 0.42
M2_b = 0.9 * 0.60 + 0.1 * 0.20 # 0.54 + 0.02 = 0.56
wind1_b = 0.50 - -0.30 # 0.80 -- new line, new arrow
wind2_b = 0.60 - 0.20 # 0.40
# --- the scale ledger, arithmetic only ---
dials = 22*256+256 + 256*256+256 + 256*256+256 + 256*16+16 # 141584
print(M1, M2, wind1, wind2, meter) # 0.26 0.72 0.4 -0.2 0.01
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IN THIS CHAPTER (Chapter 19 -- Learning by Copying):
Part 1 -- The Diary and the Eight Envelopes: Learning by Copying
Part 2 -- The Copying Machine by Pencil
Part 3 -- One Question, Two Right Answers: Where Copying Breaks
Part 4 (this post) .
Part 5 -- Riding the Wind: From Noise to an Answer
<- Back to all posts
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