W20 · Mastery gates26–30 hrs

Week 20 of 28 · Mastery · due 2026-10-10 · 26–30 hrs

Advanced Combinatorics: Probabilistic Method + Algebraic Method

Advanced Combinatorics: Probabilistic Method + Algebraic Method gate complete

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This week's lesson · 10 min read

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The probabilistic and algebraic methods: proving existence without finding anything

This week's two methods share a scandalous premise: you can prove an object exists without ever exhibiting it. The probabilistic method shows a random choice works with positive probability — so SOMETHING works. The algebraic method shows a set of vectors is too big to be dependent or too constrained to be small — so a configuration is forced. These are the tools for combinatorics problems where construction is hopeless, and they mark the transition to genuinely modern contest mathematics.

Counting arguments in disguise: probability is weighted counting, rank is compressed counting — both prove existence by showing non-existence would violate arithmetic.

Key ideas — the week on one card

  1. Linearity of expectation needs NO independence — decompose into indicators, sum probabilities, and some outcome meets the average.
  2. Union bound: if the bad events' probabilities sum below 1, an object avoiding all of them exists.
  3. The algebraic method: encode objects as vectors, force independence from a diagonal-vs-off-diagonal asymmetry, and bound the count by the dimension.
  4. Match the field to the arithmetic: parity wants $\mathbb{F}_2$, mod-$p$ wants $\mathbb{F}_p$, distances want $\mathbb{R}$ with polynomials as vectors.
  5. Abundance → randomize; rigidity → linear algebra — the claim's SHAPE picks the method.

1The probabilistic method: expectation and the union bound

The two workhorses. UNION BOUND: if bad events $B_1, \dots, B_m$ have $\sum \Pr[B_i] < 1$, then with positive probability NONE occur — an object avoiding all bad properties exists. EXPECTATION: some outcome achieves at least (and some at most) the average, so computing $\mathbb{E}[X]$ by linearity proves an object with $X \ge \mathbb{E}[X]$ exists. Linearity is the entire engine: it needs NO independence, so you decompose $X$ into indicator variables, sum their individual probabilities, and never touch a joint distribution.

The canonical opener: every 2-coloring question. Color randomly, compute the expected number of monochromatic structures, and if that expectation is below 1 there is a coloring with ZERO — the classic bound that $K_n$ admits a 2-coloring with no monochromatic $K_k$ when $\binom{n}{k} 2^{1 - \binom{k}{2}} < 1$ (Erdős 1947, three lines, launched the field). The same skeleton handles tournaments, hypergraph colorings, and sum-free subsets ('every set of $n$ nonzero integers has a sum-free subset of size $> n/3$' — random dilation, expectation, done).

The deletion refinement handles expectations slightly TOO big: choose randomly, let bad events happen, then DELETE one element per bad event and count what survives — existence with a weaker but still nontrivial bound. Recognize the family: 'show there exists a subset/coloring/tournament with property P' where P resists explicit construction. Randomness is your construction.

LReach edge crosses with probability ½ — so some split cuts at least half of them
The max-cut argument: gold edges cross the random split, dim ones stay inside — linearity of expectation counts the gold.

Worked example

Show every graph with $m$ edges contains a bipartite subgraph with at least $\frac{m}{2}$ edges.

  1. Nudge 1

    Assign sides by fair coin, per vertex, independently — what is the chance one edge crosses?

    Reveal step 1

    Randomize: put each vertex in side $L$ or $R$ independently with probability $\frac12$. An edge is CUT (joins the sides) with probability exactly $\frac12$.

  2. Nudge 2

    Sum indicator expectations; note the dependence between edges never matters.

    Reveal step 2

    Linearity: $\mathbb{E}[\text{cut edges}] = \sum_{e} \Pr[e \text{ cut}] = \frac{m}{2}$ — no independence between edges needed, which is good, because the edge events are not mutually independent (a triangle's three edges can never all be cut).

  3. Nudge 3

    Quote the averaging principle: some outcome meets the expectation.

    Reveal step 3

    Some outcome meets its expectation: a partition with $\ge \frac{m}{2}$ cut edges exists, and the cut edges form the bipartite subgraph. Three sentences, no construction — and turning it INTO a construction (local switching: move any vertex with more same-side than cross-side neighbors) is the standard follow-up worth knowing.

  4. Answer

    Random bipartition cuts each edge with probability $\frac12$; linearity gives expectation $\frac{m}{2}$; some partition achieves it.

Pitfall. Concluding 'a random object HAS the property' from a positive-probability argument. The method proves EXISTENCE; the random object merely might work. Keep the conclusion existential and the write-up earns full marks — one sloppy quantifier and it doesn't.

2The algebraic method: dimension as a weapon

The template: encode combinatorial objects as vectors in a space whose dimension you know; show the vectors are linearly independent; conclude there are at most $\dim$ of them. All the content lives in ONE inner-product computation — typically $\langle v_i, v_j \rangle$ behaves differently for $i = j$ than $i \ne j$, and that asymmetry forces independence (test a dependence $\sum c_i v_i = 0$ against each $v_j$).

The signature results run on intersection parities. Oddtown: if every club has an ODD number of members and every two clubs share an EVEN number, then over $\mathbb{F}_2$ the incidence vectors satisfy $\langle v_i, v_i \rangle = 1$, $\langle v_i, v_j \rangle = 0$ — orthonormal-ish, hence independent, hence at most $n$ clubs from $n$ people. Fisher's inequality and set-pair bounds are the same three lines with different fields. The choice of FIELD is a real decision: parity conditions want $\mathbb{F}_2$; 'divisible by $p$' wants $\mathbb{F}_p$; distance conditions in geometry want $\mathbb{R}$ and polynomials as the vectors.

The polynomial-space variant is the modern upgrade: associate to each object a polynomial, show the polynomials are independent in a space of known dimension (spanned by few monomials), and the object count is bounded by the monomial count. This is how the two-distance-set bound and half of recent 'algebraic combinatorics' contest problems run — the vectors do not have to be 0-1 incidence strings; functions work.

Worked example

Oddtown, in full: $n$ residents form clubs, each of odd size, every two clubs sharing an even number of members. Show there are at most $n$ clubs.

  1. Nudge 1

    Encode clubs as 0-1 vectors and translate both hypotheses into inner products mod 2.

    Reveal step 1

    Encode clubs as incidence vectors $v_1, \dots, v_m \in \mathbb{F}_2^n$. The hypotheses become inner products: $\langle v_i, v_i \rangle = |C_i| \bmod 2 = 1$ and $\langle v_i, v_j \rangle = |C_i \cap C_j| \bmod 2 = 0$ for $i \ne j$.

  2. Nudge 2

    Test a dependence against each vector — the cross terms die by hypothesis.

    Reveal step 2

    Test independence: suppose $\sum_i c_i v_i = 0$ with $c_i \in \mathbb{F}_2$. Take the inner product with $v_j$: every cross term dies, leaving $c_j \langle v_j, v_j \rangle = c_j = 0$ — for every $j$.

  3. Nudge 3

    Independent vectors cannot outnumber the dimension.

    Reveal step 3

    Independent vectors in $\mathbb{F}_2^n$ number at most $n$: at most $n$ clubs. Note what the field did: 'odd' and 'even' became 1 and 0, and the ENTIRE combinatorial content compressed into one diagonal-vs-off-diagonal asymmetry. (Eventown — all intersections even, including self — allows $2^{\lfloor n/2 \rfloor}$ clubs: the asymmetry, not the encoding, is what bounds things.)

  4. Answer

    At most $n$: incidence vectors over $\mathbb{F}_2$ are orthogonal-with-unit-diagonal, hence independent, hence at most $\dim = n$ of them.

Pitfall. Working over the wrong field. Over $\mathbb{R}$, self-inner-products of 0-1 vectors are SIZES, not parities — the odd/even hypotheses only become clean algebra in $\mathbb{F}_2$. Match the field to the arithmetic in the hypotheses before writing anything.

3Choosing between them — and the write-up standard for both

Selection heuristics. Probabilistic: the target is an INEQUALITY ('at least half', 'more than $n/3$'), the objects are colorings/subsets/orderings, and small cases show many valid objects (abundance is what randomness detects). Algebraic: the target is an EXACT BOUND ('at most $n$'), the hypotheses are exact arithmetic conditions (parities, prescribed intersection sizes, fixed distances), and small cases show the extremal configurations are rigid. Abundance → randomize; rigidity → linear algebra.

Both methods have a write-up trap of the same shape: the magic is front-loaded, so the FRAMING must be explicit. For probability: name the sample space and the random choice precisely ('color each vertex independently, uniformly'). For algebra: name the vector space, the field, and the encoding before any computation. Graders at this level dock the setup, not the arithmetic — the arithmetic is three lines in both methods.

These are also this phase's A3-B4 recognition targets: the phrases 'show there exists', 'show at most', 'any $n$ objects satisfying [exact condition]' should now trigger a deliberate method choice from this week's pair, alongside the classical tools. Add both templates to the toolbox card with one worked instance each — the methods are short enough that ONE fully internalized example each is genuine ownership.

Worked example

Route these three: (a) 'every tournament on $n$ players has a Hamiltonian path'; (b) 'at most $n$ clubs with odd sizes and even intersections'; (c) 'some 2-coloring of $K_n$ has few monochromatic triangles'.

  1. Nudge 1

    Classify each claim's SHAPE first: exact bound or inequality? rigid or abundant?

    Reveal step 1

    (a) Neither, at first read — it is exact and structural. Induction or extremal ordering wins (order by wins and swap — or note the probabilistic EXPECTATION argument counts Hamiltonian paths in random tournaments; for the plain existence claim, induction is the clean route). Method choice is not forced by topic; it is forced by the CLAIM's shape.

  2. Nudge 2

    Parity hypotheses plus 'at most n' — which method's home terrain is that?

    Reveal step 2

    (b) 'At most $n$' + parity hypotheses = rigidity: algebraic, over $\mathbb{F}_2$ — this is Oddtown verbatim.

  3. Nudge 3

    'Some coloring' with a quantity to control — randomize, and state which direction of the mean you use.

    Reveal step 3

    (c) 'Some coloring' + a quantity to keep small = abundance: probabilistic — random coloring, expected monochromatic triangles $= \binom{n}{3} \cdot \frac{1}{4}$, some coloring does at least that well... and 'at least as well as average' here means AT MOST the expected count, since some outcome is $\le \mathbb{E}$. Both directions of 'some outcome meets the mean' are available; state which one you are using.

  4. Answer

    (a) induction/extremal, (b) algebraic-$\mathbb{F}_2$, (c) probabilistic-expectation — the claim's shape (exact vs inequality, rigid vs abundant) picks the method.

Pitfall. Using $\le \mathbb{E}$ when you argued $\ge \mathbb{E}$ (or vice versa). 'Some outcome meets the average' cuts both ways, but each PROBLEM needs a specific direction — one clause in the write-up, one dropped point if missing.

Before you open the gates

  • 'There exists' + inequality target + abundant valid objects → randomize; compute the expectation by indicators and linearity.
  • Union bound when you must dodge several bad events; deletion when the expectation is slightly too big.
  • 'At most $n$' + exact parity/intersection hypotheses → vectors + dimension; pick the field to match the arithmetic ($\mathbb{F}_2$ for parity, $\mathbb{F}_p$ for mod-$p$, $\mathbb{R}$+polynomials for distances).
  • The independence test is always the same: hit the dependence with each $v_j$ and watch cross terms die.
  • Write the frame first — sample space and choice, or space/field/encoding — the methods' arithmetic is trivial once framed, and the frame is what gets graded.

Check yourself

1. Linearity of expectation requires the summed random variables to be:

2. In Oddtown, the inner product $\langle v_i, v_j \rangle$ over $\mathbb{F}_2$ computes:

3. 'Show at most $n$ objects satisfy these exact intersection conditions' most strongly suggests:

Practice ladder — three rungs, rising

Each rung: attempt cold, one hint if stuck, worked resolution only after a real try.

Rung 1 (expectation, the $\le \mathbb{E}$ direction). Prove that the edges of $K_n$ can be $2$-colored so that at most $\tfrac14\binom{n}{3}$ triangles are monochromatic.

One hint

Color each edge red or blue independently with probability $\tfrac12$. For one fixed triangle, what is the probability its three edges all match? Sum over triangles, then use that some outcome is no larger than the mean.

Worked resolution

Color each of the $\binom{n}{2}$ edges red or blue independently and uniformly. A fixed triangle is monochromatic exactly when its three edges agree, with probability $\Pr[\text{all red}] + \Pr[\text{all blue}] = (\tfrac12)^3 + (\tfrac12)^3 = \tfrac14$. Let $X$ count monochromatic triangles; by linearity $\mathbb{E}[X] = \binom{n}{3}\cdot\tfrac14$. Since a random variable takes a value $\le$ its mean for at least one outcome, some coloring has $X \le \tfrac14\binom{n}{3}$. [Source: this week's probabilistic-method section (expectation), using the $\le \mathbb{E}$ direction flagged in the method-selection section.]

Rung 2 (union bound). Show that if $\binom{n}{k}\,2^{1-\binom{k}{2}} < 1$ then the edges of $K_n$ admit a $2$-coloring with no monochromatic $K_k$; deduce the Ramsey lower bound $R(k,k) > 2^{k/2}$ for $k \ge 3$.

One hint

Color edges independently at random. Bound the probability that SOME $k$-set is monochromatic by a union bound; if that bound is below $1$, a good coloring must exist. For the deduction take $n = \lfloor 2^{k/2}\rfloor$ and estimate $\binom{n}{k} \le n^k/k!$.

Worked resolution

Color the edges of $K_n$ red/blue independently and uniformly. For a fixed set $S$ of $k$ vertices, the $\binom{k}{2}$ edges inside $S$ are all equal with probability $2\cdot 2^{-\binom{k}{2}} = 2^{1-\binom{k}{2}}$. By the union bound over the $\binom{n}{k}$ such sets, $\Pr[\text{some } K_k \text{ monochromatic}] \le \binom{n}{k}\,2^{1-\binom{k}{2}}$. If this is $< 1$, then with positive probability no $K_k$ is monochromatic, so such a coloring exists. Taking $n = \lfloor 2^{k/2}\rfloor$ and $k \ge 3$: $\binom{n}{k}\,2^{1-\binom{k}{2}} \le \tfrac{n^k}{k!}\,2^{1-\binom{k}{2}} \le \tfrac{2^{k^2/2}}{k!}\,2^{1-k(k-1)/2} = \tfrac{2^{1+k/2}}{k!} < 1$, so $K_n$ has a coloring with no monochromatic $K_k$, hence $R(k,k) > n = \lfloor 2^{k/2}\rfloor$, i.e. $R(k,k) > 2^{k/2}$. [Source: this week's probabilistic-method section (the union bound); the $R(k,k) > 2^{k/2}$ bound is the Deep Study drill's assigned probabilistic-method rep.]

Rung 3 (algebraic method over $\mathbb{F}_2$). In a town of $n$ people, clubs are formed so that every club has an odd number of members and every two distinct clubs share an even number of members. Prove there are at most $n$ clubs.

One hint

Encode each club as its $0/1$ indicator vector in $\mathbb{F}_2^n$. Translate 'odd size' and 'even intersection' into inner products mod $2$, then show the vectors are linearly independent.

Worked resolution

For each club $C_i$ let $v_i \in \mathbb{F}_2^n$ be its indicator vector. Over $\mathbb{F}_2$, $\langle v_i, v_i\rangle = |C_i| \equiv 1$ (odd size) and, for $i \ne j$, $\langle v_i, v_j\rangle = |C_i \cap C_j| \equiv 0$ (even intersection). Suppose $\sum_i c_i v_i = 0$ with $c_i \in \mathbb{F}_2$. Taking the inner product with a fixed $v_j$ kills every cross term ($\langle v_i, v_j\rangle = 0$ for $i \ne j$) and leaves $c_j\langle v_j, v_j\rangle = c_j = 0$; this holds for every $j$, so the $v_i$ are linearly independent. Independent vectors in the $n$-dimensional space $\mathbb{F}_2^n$ number at most $n$, so there are at most $n$ clubs. [Source: this week's algebraic-method section — the Oddtown rank argument, assigned as the Deep Study rank-argument (Oddtown) miniature.]

Prove it — constructed response

Prove that every graph $G$ with $m$ edges contains a bipartite subgraph with at least $\frac{m}{2}$ edges. Set up the random experiment explicitly, compute the relevant expectation by linearity, and name the averaging principle that turns that expectation into an existence statement.

The gates

Technique Resources — probabilistic method + algebraic/linear-algebra method (study before the reinforcing problems)

Learning support for this week's reinforcing IMO/Putnam problems. Read/watch these before (and after) attempting the probabilistic method + algebraic/linear-algebra method problems so each attempt has a source to learn the method and to check your write-up against.

why this gate: the lesson's §1 The probabilistic method: expectation and the union bound is what it trains

sources & assignments (11)
  • Problems Recognition ledger — polynomial methodWritten, 10 minutes: state the combinatorial Nullstellensatz (Alon) informally — a nonzero polynomial of degree Σdᵢ with a nonzero coefficient on Πxᵢ^dᵢ cannot vanish on a grid S₁ × ⋯ × Sₙ with |Sᵢ| > dᵢ — and list the two recognition triggers (restricted sumsets; covering points with few hyperplanes). Recognition only, no reps; the Zhao handout in this gate is the source.Standard

Advanced Reading

Advanced Reading

- Read: Po-Shen Loh prob-comb.pdf (probabilistic method) - Read: Yufei Zhao algebraic combinatorics handout - Watch: Po-Shen Loh Extremal Combinatorics — CMU 21-738, 2 more lectures

sources & assignments (6)

Archive Block — A2/B2

Archive Block — A2/B2

- Archive: 5 Putnam A2/B2 · Combinatorics · any slot · 20 min each

sources & assignments (5)
  • Problems Archive BrowserPutnam 2002 A2, 1997 A2 · Combinatorics · 20 min eachPutnam

archive pull: 2 problems · Combinatorics · A2/B2 · Challenge · 20 min

A3/B3 Entry

A3/B3 Entry (first exposure — not nightmare pace) accelerated only

- Archive: 2 Putnam A3/B3 · Combinatorics · any slot · 30 min each - After each: write what you got, what you missed, which tool was needed

sources & assignments (5)
  • Problems Archive BrowserPutnam 2024 A3 · Combinatorics · 30 min eachNightmare

archive pull: 1 problems · Combinatorics · A3/B3 · Nightmare · 30 min

A1/B1 Maintenance

A1/B1 Maintenance

- Archive: 4 Putnam A1/B1 · mixed topics · 10 min each

sources & assignments (5)
  • Problems Archive BrowserPutnam 1990 B1 · Analysis · 10 min eachPutnam

archive pull: 1 problems · Analysis · A1/B1 · Standard · 10 min

Stretch

Stretch (only after required work) accelerated only

- Do: 102 Combinatorial Problems #83–102 · CMU 2024-10-Combinatorics: 2 problems

sources & assignments (6)
  • Problems 102 Combinatorial102 Combinatorial Problems #83–102 · CMU 2024-10-Combinatorics: 1 problem (displaced slot funds the Hall/probabilistic reps in Deep Study)Putnam
  • Problems Sperner / posets (stretch — accelerated lane only)Define chains and antichains in the Boolean lattice; state Dilworth's and Mirsky's theorems (one sentence each); prove Sperner's theorem — the largest antichain in 2^[n] has size C(n, ⌊n/2⌋) — via the LYM inequality, writing the normalized-chain double count in full. Recognition target: spot 'family of subsets, none containing another' instantly.Putnam

Spiral Reinforcement

Spiral Reinforcement

- Archive: 2 · Algebra A2 + 1 · Analysis A2 · any slot

sources & assignments (5)
  • Problems Archive BrowserPutnam 1987 A2 · AlgebraPutnam

archive pull: 1 problems · Algebra · A2 · Challenge · 20 min

Deep Study — Algebraic combinatorics, properly

Deep Study — Algebraic combinatorics, properly (Matousek + Stanley)

Mastery-phase learning gate: dimension arguments (Oddtown-style) are the highest-leverage new weapon for hard combinatorics slots.

Ritual: Output: 4 miniature summary cards + extremal-move sentences.

why this gate: the lesson's §2 The algebraic method: dimension as a weapon is what it trains

sources & assignments (11)
  • Problems Drill — conditioning, first-step recursion, union boundThree reps with conditioning + first-step analysis: (1) a fair coin is flipped until two consecutive heads appear; compute the expected number of flips by conditioning on the first flip(s). (2) Gambler's ruin: a simple random walk on {0, 1, …, n} starts at k and absorbs at both ends; derive P(hit n before 0) = k/n by first-step recursion — state the recurrence in one line before solving it. (3) Union bound in anger: write out the probabilistic-method lower bound R(k,k) > 2^(k/2) — random 2-coloring, union bound over monochromatic K_k's. This is the same calculation the Wootters video assigns as follow-up; doing it here replaces that separate task, so net time is unchanged. Then re-derive E[X] = E[E[X|Y]] in one paragraph and state where it silently appeared in rep (1). Probability is only 3 Kedlaya points, but conditioning is the lever in every one of them.Standard
  • Problems Required reps — Hall + the probabilistic method(1) Hall's theorem: state the marriage condition; prove the corollary that every k-regular bipartite graph (k ≥ 1) has a perfect matching (check Hall via edge counting); one line on the deficiency version's existence. (2) Probabilistic method, expectation move: prove every graph with m edges has a bipartite subgraph with at least m/2 edges (uniform random 2-partition + linearity of expectation); name the move family — union bound · expectation · alteration — this one belongs to. Time funded by trimming the stretch gate's CMU slot 2 → 1. Recognition coda, one sentence each: König's theorem (in bipartite graphs, max matching = min vertex cover) and how independent sets are vertex-cover complements. (3) Rank-argument miniature (oddtown): n clubs, each of odd size with pairwise-even intersections, over a town of n people — prove there are at most n clubs by taking 𝔽₂ characteristic vectors and showing they are linearly independent (the Gram matrix is the identity over 𝔽₂). Meta in one line: 'a set system became a vector space because incidence/parity is linear.' (Restores the Matoušek rank miniature displaced by the W22 AbA swap.)Putnam

USAMO/IMO Bridge — IMO 2011/1

USAMO/IMO Bridge — IMO 2011/1 (stretch) accelerated only

Ritual: Self-grade 0–7 against the AoPS/official solution: 7 = complete and rigorous · 5–6 = right idea, rigor gaps · 3–4 = key lemma proven · 1–2 = nontrivial progress · 0 = none. Log the score and the single biggest missing idea in the verify box.

sources & assignments (5)
  • Source IMO 2011/1 — AoPS wiki (statement + solutions + discussion link) — Sets A of 4 distinct positive integers maximizing pairs (i,j) with a_i+a_j | s_A. Extremal counting with divisibility — algebraic combinatorics under contest pressure. Time cap 60 min — do not scroll to the Solutions section until the attempt is over.
  • Source Art of Problem Solving — Lecture companion for “USAMO/IMO Bridge — IMO 2011/1 (stretch)” — full treatment; pause at each theorem statement and predict the proof's first move before it plays.
  • Source Shahriar Shahriari — Problem-session companion for “USAMO/IMO Bridge — IMO 2011/1 (stretch)” — watch one worked problem, stop, finish it yourself on paper, then compare.
  • Source Michael Penn — Interesting Integrals (playlist) — Intuition companion for “USAMO/IMO Bridge — IMO 2011/1 (stretch)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems IMO 2011/1Sets A of 4 distinct positive integers maximizing pairs (i,j) with a_i+a_j | s_A — full writeup to the Evan Chen standard, 60 min cap. Self-grade 0–7 against the AoPS/official solution: 7 = complete and rigorous · 5–6 = right idea, rigor gaps · 3–4 = key lemma proven · 1–2 = nontrivial progress · 0 = none. Log the score and the single biggest missing idea in the verify box.USAMO bridge

Half-Putnam — ONE 180-minute sitting

Half-Putnam — ONE 180-minute sitting (B1–B6, six problems)

Ritual: Time-map postmortem: reconstruct your 3 hours in 15-min blocks. Mark every block spent on a problem you didn't score on — that is the exact minute-budget leak the W25–27 full mocks must fix.

sources & assignments (5)
  • Source Kedlaya — Putnam Archive (problems + official solutions, 1985–present) — The canonical source for every real Putnam problem and official solution referenced by this gate.
  • Source Art of Problem Solving — Lecture companion for “Half-Putnam — one full B-session (B1–B6, 3 hours)” — full treatment; pause at each theorem statement and predict the proof's first move before it plays.
  • Source Shahriar Shahriari — Problem-session companion for “Half-Putnam — one full B-session (B1–B6, 3 hours)” — watch one worked problem, stop, finish it yourself on paper, then compare.
  • Source Michael Penn — Interesting Integrals (playlist) — Intuition companion for “Half-Putnam — one full B-session (B1–B6, 3 hours)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Archive BrowserOne complete B-session under exam law — the afternoon block you have never rehearsed: pull B1–B6 from a single unseen year via the archive button — ONE continuous 180-minute sitting, six problems, never split across days, no notes — write everything you would submit. Expected at this stage: B1–B2 clean, real progress on B3, honest zeros above. Log the session in the mock logger below (slot scores + total). Postmortem: time map — when did you commit to each problem, when should you have bailed? This pairs with the W18 A-session so both halves of exam day are rehearsed before the W25–26 full mocks.Putnam

archive pull: 6 problems · Mixed Putnam · B1/B2/B3/B4/B5/B6 · Putnam · 180 min

Continuation — Abstract Algebra

Continuation — Abstract Algebra: Burnside's lemma (≈2 hrs)

Ritual: Thread: Abstract Algebra. Required ~2-hr continuation swapped from the Stirling preview — Burnside is orbit-stabilizer made into a counting machine.

sources & assignments (6)
  • Source Ryan O'Donnell (CMU) — Factorial Asymptotics, Stirling's Formula — Continuation-lane companion for: Analysis: Stirling + asymptotics preview (≈2 hrs). Reused from the verified pool — watch/read with this week's lens.
  • Source MIT 18.A34 — analysis.pdf (Yufei Zhao) — Continuation-lane companion for: Analysis: Stirling + asymptotics preview (≈2 hrs). Reused from the verified pool — watch/read with this week's lens.
  • Source Shahriar Shahriari — Problem-session companion for “Continuation — Analysis: Stirling + asymptotics preview (≈2 hrs)” — watch one worked problem, stop, finish it yourself on paper, then compare.
  • Source Michael Penn — Interesting Integrals (playlist) — Intuition companion for “Continuation — Analysis: Stirling + asymptotics preview (≈2 hrs)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Source Michael Penn — Abstract Algebra (playlist) — Companion lecture series for this week's abstract-algebra rep — watch the relevant 1–2 videos, then do the written rep below. Watch-for: this playlist opens with set-theory / proof-writing videos — skip to the 'Abstract Algebra |' entries for the group/ring/field content this gate needs.
  • Problems Continuation lane — Abstract Algebra(1) State Burnside's lemma: number of orbits = (1/|G|) Σ_{g∈G} |Fix(g)|. (2) Count the distinct 2-colorings of a 4-bead necklace under cyclic rotation C₄ — compute |Fix(g)| for each rotation and average (expect 6). (3) Redo for a 6-bead necklace under C₆. (4) Extend one case to the dihedral group D₄ (add reflections) and note whether the count changes — for 2 colors and n = 4 it stays 6, because the reflections fix the same colorings the rotations already did; explain why. Displaces the Stirling-preview continuation — Stirling now lives in the W23 asymptotics ladder and keep-warm.Standard

Keep-Warm Rotation — quadratic residues · linear algebra · geometry

Keep-Warm Rotation — quadratic residues · linear algebra · geometry (30 min)

Ritual: If any micro-rep took over 10 minutes or failed, that family goes on this week's repair list — it was rusting.

sources & assignments (2)
  • Problems Keep-warm rotationThree 10-minute micro-reps, no notes: (1) quadratic residues — decide one solvability question x² ≡ a (mod p) via Euler's criterion; (2) linear algebra — one rank/det/eigen rep, or re-derive rank-nullity in 5 lines; (3) geometry — one coordinate/complex geometry rep, or one Power-of-a-Point configuration proof. The dormancy rule: no tool family sleeps longer than 2 weeks — these are the three most overdue right now.Standard

archive pull: 1 problems · Number Theory/Linear Algebra/Geometry · A1/B1 · Putnam · 10 min

Track A∗ — Analysis

Track A∗ — Analysis (parallel mastery track): Continuity pathologies, functional limits

Runs EVERY mastery week so Analysis never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

sources & assignments (4)
  • Problems Analysis (depth)Putnam 1994 B2 — work it with the reading open; one clean write-up.Putnam
  • Problems Analysis (reach)Putnam 1987 A6 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track D∗ — Linear Algebra

Track D∗ — Linear Algebra (parallel mastery track): Rank, rank-nullity for existence arguments

Runs EVERY mastery week so Linear Algebra never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

sources & assignments (4)
  • Problems Linear Algebra (depth)Putnam 1988 B5 — work it with the reading open; one clean write-up.Putnam
  • Problems Linear Algebra (reach)Putnam 1990 A5 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track E∗ — Number Theory

Track E∗ — Number Theory (parallel mastery track): Pell equations & continued fractions

Runs EVERY mastery week so Number Theory never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

sources & assignments (4)
  • Problems Number Theory (depth)Putnam 1986 A2 — work it with the reading open; one clean write-up.Putnam
  • Problems Number Theory (reach)Putnam 1991 B4 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track C∗ — Combinatorics

Track C∗ — Combinatorics (parallel mastery track): Graph theory for Putnam — handshake, trees, bipartite

Continuous Combinatorics mastery spine (Track C∗). Combinatorics is the #1 Putnam genre (~20%); A1–B2 combinatorics is already drilled hard in the slots, so this spine adds the continuous theory ladder plus a weekly A3–A6 reach problem (the real gap). One depth re-solve + one fresh reach each week.

sources & assignments (4)
  • Source Combinatorics spine — GraphTheory text — This week: Graph theory for Putnam — handshake, trees, bipartite. Degree/handshake, trees, bipartite & matchings, basic colourings. Read the matching section, then do the two problems below.
  • Source Combinatorics continuous spine — Runs every Mastery week (parallel-track model) so Combinatorics — the single most frequent Putnam genre (~20%) — never goes cold. 1 depth re-solve + 1 fresh reach problem.
  • Problems Combinatorics (depth)Putnam 2000 B1 — cold re-solve with the reading open; one clean write-up.Putnam
  • Problems Combinatorics (reach)Putnam 1996 A3 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track F∗ — Geometry · Combinatorics · Probability

Track F∗ — Geometry · Combinatorics · Probability (parallel maintenance): Geometry — Power of a point & synthetic configs

Bundled parallel maintenance track so Geometry, Combinatorics, and Probability each get a fixed weekly cadence in Mastery (probability previously had multi-week gaps). Rotates focus; ~1.5 hrs.

sources & assignments (3)
  • Source Coxeter & Greitzer — Geometry Revisited — This week’s focus: Power of a point & synthetic configs. Power of a point, radical axis, Ceva/Menelaus with signed ratios.
  • Source Geometry keep-warm — Lower-cadence topic on a guaranteed weekly rotation (Toolkit Track F model). 1 problem this week; Geometry comes round again every 3rd week.
  • Problems Geometry (F∗)Putnam 1988 A4 — power of a point & synthetic configs; full attempt + one clean write-up.Putnam

Integration-Bee Micro-Spine

Integration-Bee Micro-Spine: Timed integral sprint

Integration-Bee micro-spine: fills the computational rapid-integral genre (~7% of Putnam) that the proof-heavy Analysis spine does not train for speed. Built on the local MIT Bee topics guide.

sources & assignments (3)
  • Source MIT Integration Bee — Topics Guide — This week: Timed integral sprint — speed under the clock; pick the trick in <30s, evaluate, move on.
  • Source Why this track — ~7% of Putnam problems are compute-the-integral (the "MIT BEE" genre). Proof-based integration is covered in the Analysis spine; this is the speed/trick layer.
  • Problems Integration speed10 definite integrals from the guide, timed 30 min. Goal: recognize the trick fast (symmetry / King / Feynman / series), evaluate cleanly. Mark the two you stalled on.Tutorial

Problem-Solving Reps — weekly homework

Problem-Solving Reps — weekly homework

Standing weekly homework — the problem-solving book stack, every week. Solve ALL listed; write ONE full clean solution (the rest may stay scratch). Up to 2 due re-solves from your review queue surface first. Slot reinforcement supplement: A2/B2 are deliberately overtrained so they become bankable, not occasional exposure. Slot reinforcement supplement: A3/B3 now get repeated lemma-hunt reps so slot 3 is trained as a strong reach tier.

sources & assignments (12)
  • Problems PnB/EngelPnB section 6.3 probs 1–2 · Engel Ch. 5 probs 1–6 (enumerative)Putnam
  • Problems 102 Comb102 Combinatorial Problems — Advanced probs 5–8 (probabilistic/algebraic method)Putnam
  • Problems Problems from the BookHigher Algebra in Combinatorics chapter, probs 1–2Nightmare
  • Problems Putnam ArchivePutnam 2017 A2 - 18 min clean-solve; write the final proof and fatal-slip check.Putnam
  • Problems Putnam ArchivePutnam 1989 B2 - 22 min clean-solve; write the final proof and fatal-slip check.Putnam
  • Problems Putnam ArchivePutnam 2011 A3 - 35 min lemma hunt; record one rigorous lemma even if the full proof does not close.Nightmare
  • Problems Putnam ArchivePutnam 2010 B3 - 35 min lemma hunt; compare to official solution after the attempt.Nightmare
  • Problems Putnam Slot 2 EqualizerSlot-2 Equalizer Set 7: Putnam 1989 A2, Putnam 1990 A2, Putnam 1991 A2 - clean-solve focus; write final proofs and fatal-slip checks.Putnam
  • Problems Putnam Slot 4 EqualizerSlot-4 Equalizer Set 7: Putnam 2021 B4, Putnam 2022 B4, Putnam 1992 B4 - reach/lemma focus; extract the first invariant or structure before reading solutions.Nightmare
  • Problems AMC/AIME Fluency OpenerOptional 5-minute opener: solve the three linked AMC/AIME reps before Putnam work; stop at 15 minutes total.AMC/AIME · archive: AMC geometryAMC geometryAMC geometry
  • Problems Putnam Full Past ExamFull Past Exam 17: Putnam 2011 A1, Putnam 2011 A2, Putnam 2011 A3, Putnam 2011 A4, Putnam 2011 A5, Putnam 2011 A6, Putnam 2011 B1, Putnam 2011 B2, Putnam 2011 B3, Putnam 2011 B4, Putnam 2011 B5, Putnam 2011 B6 - Primary paper protocol: A-session A1-A6 in one 180-minute block; B-session B1-B6 in one 180-minute block within 48 hours; next day score all 12 slots, tag every miss, and choose three repair pulls.Putnam
  • Problems Putnam Full Past ExamFull Past Exam 18: Putnam 2012 A1, Putnam 2012 A2, Putnam 2012 A3, Putnam 2012 A4, Putnam 2012 A5, Putnam 2012 A6, Putnam 2012 B1, Putnam 2012 B2, Putnam 2012 B3, Putnam 2012 B4, Putnam 2012 B5, Putnam 2012 B6 - Secondary paper protocol: Day 1 bankable sweep A1-A2-B1-B2 in 100 minutes; Day 2 medium/hard sweep A3-A4-B3-B4 in 100 minutes with certified-lemma goal; Day 3 reach scan A5-A6-B5-B6 in 50 minutes; Day 4 score all 12 slots and re-solve one same-family miss.Putnam

archive pull: 2 problems · Combinatorics · Nightmare · 60 min

Reflect

Reflect: reconstruct and log the pattern

- Standard ---

Ritual: Pick one combinatorics solve (probabilistic or algebraic method). Close all notes and the solution. Rewrite the full solution from memory, then compare with the original — what did you miss or reorder? Write one pattern note: "When I see [pattern], do [first move]."

sources & assignments (4)

Verify: audit one proof before closing

unlocks after: Reflect: reconstruct and log the pattern

- Standard ---

Ritual: Pick one solution. Re-check for: (1) a false claim, (2) a missing case, (3) a wrong bound. Extra check for combinatorics: does the bijection have an explicit inverse? Is the double-count counting the SAME set twice?

sources & assignments (4)

Complex numbers keep-warm

Complex numbers keep-warm (20 min)

Retention keep-warm (2026-07-02 audit): complex numbers went silent for 3-6 week stretches. ~20 minutes: two Lerma problems + one memory recall. Source: Lerma Training 2023 §5 (local PDF, hints/solutions in later parts).

Ritual: Recall (closed book): Roots-of-unity filter: extract every 3rd binomial coefficient — set up the sum.

sources & assignments (2)
  • Problems Lerma Training 2023 §5 Complex NumbersProblems 5.9–5.10 (~15 min at conversion speed). Then the recall prompt below — closed book.Putnam

Exit contract — Week 20

Verification remaining

  • reading progress…

Carry-forward repairs

  • reading queue…

Next week opens with

W21 · Advanced NT + Abstract: LTE + Orders + Burnside
the exam follows the final taper week

Train Week 20 in the trainer →