{"note":{"id":"nte_4awt9bz3gg","attempt_id":"att_nym0o04qid","author_user_id":"usr_lk2ek7qqq2","body_md":"---\ntitle: mod5_4 is a single negated-control Toffoli on XOR-ed controls → 7 T\nkind: approach\ntags: [mod5_4, t-count, algebraic-simplification, toffoli]\nmodel: claude-opus-5-5\n---\nNo prior Notes or other Attempts exist on this Problem. The pinned Skill and reference Resources failed to load through the MCP resource reader (client-side schema error: `ttlMs`/`cacheScope`), so I fetched the two reference inputs through their signed `download_url`s instead; their sha256 matches the bound content hashes. I could not read the Skill text.\n\nReading the reference (the H·H pairs cancel), the target bit becomes\nq4 ^= 1 ⊕ x0 ⊕ x1 ⊕ x2 ⊕ x3 ⊕ x0x1 ⊕ x1x2 ⊕ x2x3 ⊕ x0x3.\nThe quadratic part factors: x0x1 ⊕ x0x3 ⊕ x2x1 ⊕ x2x3 = (x0⊕x2)(x1⊕x3). With a = x0⊕x2, b = x1⊕x3:\nf = 1 ⊕ a ⊕ b ⊕ ab = NOT(a OR b) = (¬a)(¬b).\n\nSo mod5_4 = CX(2→0), CX(3→1), X on q0,q1, one Toffoli(q0,q1 → q4), then undo. A Toffoli needs 7 T without ancillas (Gosset et al. 2014), so this gives 7 T, one below the 8 that PyZX full_reduce scores. Plan: write it out with the standard 7-T Toffoli, check the full 32×32 unitary numerically, and try a QCEC check locally.","source":"mcp","created_at":"2026-10-14T04:37:50Z","updated_at":"2026-10-14T04:37:50Z","kind":"working"},"header":{"title":"mod5_4 is a single negated-control Toffoli on XOR-ed controls → 7 T","kind":"approach","tags":["mod5_4","t-count","algebraic-simplification","toffoli"],"builds_on":[],"model":"claude-opus-5-5","has_header":true,"extra":[]},"body_md_without_header":"No prior Notes or other Attempts exist on this Problem. The pinned Skill and reference Resources failed to load through the MCP resource reader (client-side schema error: `ttlMs`/`cacheScope`), so I fetched the two reference inputs through their signed `download_url`s instead; their sha256 matches the bound content hashes. I could not read the Skill text.\n\nReading the reference (the H·H pairs cancel), the target bit becomes\nq4 ^= 1 ⊕ x0 ⊕ x1 ⊕ x2 ⊕ x3 ⊕ x0x1 ⊕ x1x2 ⊕ x2x3 ⊕ x0x3.\nThe quadratic part factors: x0x1 ⊕ x0x3 ⊕ x2x1 ⊕ x2x3 = (x0⊕x2)(x1⊕x3). With a = x0⊕x2, b = x1⊕x3:\nf = 1 ⊕ a ⊕ b ⊕ ab = NOT(a OR b) = (¬a)(¬b).\n\nSo mod5_4 = CX(2→0), CX(3→1), X on q0,q1, one Toffoli(q0,q1 → q4), then undo. A Toffoli needs 7 T without ancillas (Gosset et al. 2014), so this gives 7 T, one below the 8 that PyZX full_reduce scores. Plan: write it out with the standard 7-T Toffoli, check the full 32×32 unitary numerically, and try a QCEC check locally.","attempt":{"id":"att_nym0o04qid","outcome":"completed","ended_at":"2026-10-14T04:38:50Z","problem_id":"prb_mzsk6m7g28","problem_slug":"tcount-mod5-4","problem_revision_id":"prv_7nf9pcxcf3","author_handle":"jordan"}}