惯性聚合 高效追踪和阅读你感兴趣的博客、新闻、科技资讯
阅读原文 在惯性聚合中打开

推荐订阅源

D
DataBreaches.Net
IT之家
IT之家
博客园_首页
博客园 - 【当耐特】
V
V2EX
Apple Machine Learning Research
Apple Machine Learning Research
G
Google Developers Blog
让小产品的独立变现更简单 - ezindie.com
让小产品的独立变现更简单 - ezindie.com
Recent Announcements
Recent Announcements
F
Fortinet All Blogs
GbyAI
GbyAI
腾讯CDC
H
Hackread – Cybersecurity News, Data Breaches, AI and More
OSCHINA 社区最新新闻
OSCHINA 社区最新新闻
I
InfoQ
H
Help Net Security
T
Tailwind CSS Blog
B
Blog RSS Feed
Martin Fowler
Martin Fowler
人人都是产品经理
人人都是产品经理
The Cloudflare Blog
博客园 - 叶小钗
雷峰网
雷峰网
量子位

IBM Research

Introducing IBM and NASA’s new foundation model for the Moon Switzerland's first IBM Quantum System Two | IBM Quantum Computing Blog Cleveland Clinic, RIKEN, IBM named Gordon Bell finalists | IBM Quantum Computing Blog How llm-d makes the most of the hardware you already have Ponder This Challenge - September 2026 - Loeschian Arithmetic Progressions IBM Quantum Nighthawk r2—more circuits, faster | IBM Quantum Computing Blog What happens when information theory accounts for reasoning? Granite 4.2 brings native reasoning to enterprise agents Qiskit Fermions: a modular toolbox for fermionic systems | IBM Quantum Computing Blog IBM’s new modular architecture for cryogenic systems | IBM Quantum Computing Blog QOBLIB: tracking progress in quantum optimization | IBM Quantum Computing Blog DocLang: a markup language for LLMs From vision to reality: a unified AI solver for the grid The search for quantum advantage in differential equations Ponder This Challenge - August 2026 - The Wheel of Buttons Quantum advantage through trusted quantum computation | IBM Quantum Computing Blog All of AI benchmarking at your fingertips What are spin qubits? | IBM Quantum Computing Blog IBM to acquire HRL Laboratories IBM commits $50M in quantum access for US Genesis Mission It’s time for cryptography to get its own abstraction layer It’s time for cryptography to get its own abstraction layer This could be the largest synthetic code dataset yet How to measure the performance of a quantum computer | IBM Quantum Computing Blog Release News: Qiskit v2.5 is here! | IBM Quantum Computing Blog CoFrGeNets replace the ‘bones’ of transformer-based models How training environments can teach AI models to misbehave What’s new at IBM Quantum - Q2 2026 | IBM Quantum Computing Blog Modeling the chemistry of fusion reactor material | IBM Quantum Computing Blog Apply to IBM Quantum Developer Conference 2026 | IBM Quantum Computing Blog
Ponder This Challenge - July 2026 - Return of the Superhe...
Gadi Aleksandrowicz · 2026-07-01 · via IBM Research

Return of the Superheroes

Continuing the theme of last month, we deal with a movie franchise consisting of nn superheroes. They are joined by nn supervillains. The producers intend to pair the superheroes and the supervillains to form (hero, villain) pairs where each hero has a unique villain serving as their nemesis.

Each pairing is accepted differently by the audiences. After elaborate work, a method of assigning numerical value f(a,b)f(a,b) to each pairing (a,b)(a,b) to estimate the audiences' reaction was developed. The producers wish to find the list of pairings that maximizes the value of the pairing with the minimal value in the list. This minimal value is called the hero-villain value.

The way f(a,b)f(a,b) is computed is as follows: Let pp be some prime and define a function Ta,b(x)=x2+ax+b (mod p)T_{a,b}(x)=x^2+ax+b\ (\text{mod}\ p). By setting x0=0x_0=0 and xn+1=T(xn)x_{n+1}=T(x_n) we obtain a sequence x0,x1,x2,…x_0,x_1,x_2,\ldots which eventually repeats. Let f(a,b)f(a,b) be the number of steps until the first repeat happens. i.e. if xnx_n is the first element in the sequence such that there exists m<nm<n for which xn=xmx_n=x_m, then f(a,b)=nf(a,b)=n.

For example, for n=5n=5 and p=101p=101, one possible list of pairings is (1,3),(2,1),(3,4),(4,2),(5,5)(1,3), (2,1), (3,4), (4,2), (5,5) which yields the values 14,18,19,22,1414, 18, 19, 22, 14 for which the minimum is 14. It turns out that every list of pairings gives a value of at most 14, so 14 is hero-villain value for this case.

Your goal Find the hero-villain value for n=611n=611 and p=14411p=14411

A bonus "*" will be given for finding the optimal nn in the range 1<n<N1<n<N for N=1000N=1000 which gives the maximal hero-villain value for nn and p=17377p=17377.