This summer, the Competitive Programming Committee at ETH Zurich has taken a bold step towards modernizing their programming tutorial. With attendance rocketing due to the Big Tech Interview Process and space limited, they have turned to artificial intelligence for a solution. A Large Language Model, trained on the university's notoriously theory-heavy lecture scripts, has been tasked with providing instant support to students seeking answers to their questions.
After sucessfully training the model, the committee has decided to test it with a self-created set of multiple choice questions. However, instead of delivering clear explanations in plain language, the AI has developed an alarming quirk. It now responds with cryptic mathematical expressions, seemingly inventing its own operators and rules along the way. The committee is stumped, unable to decipher the AI's cryptic responses.
Can you unravel the solutions to the questions that the comittee has asked?
You are given the output of the LLM that always has the same format (Yes we used a low temperature for sampling!). This means that some parts of the input are always the same (fixed) and some change.
We will now explain how the input is structured. The first 6 lines are fixed. The second to last word in the first sentence in line 7 is variable: "A logic with <S> values". <S> can be any number from $$$1, ..., 5$$$ but as lower-case english word, i.e. one, ..., five.
After that the next element that changes are the names of the operators. They are always given at the end of line 7 as a comma+space separated list and the last operator is always introduced by "and ". The number of operators is $$$2\leq N_{O} \leq 20$$$.
After another fixed text, the function tables are given, one for each operator and in the order that the operators were mentioned. A function table has one column for each input argument and one column for the output. The output is always given by the last column. An operator $$$o_i$$$ can have $$$1\leq x_i \leq 5$$$ arguments and different operators can have different numbers of arguments. A table has as many rows as there are possible input combinations. Columns are separated by one space.
Following the tables there is exactly one fixed line. The last word in the line after that defines the name of the assignment operator $$$A$$$.
That is followed by $$$1 \leq N_w \leq 30$$$ assignments of words $$$w$$$ to values $$$v$$$. Each assignment is one line that reads as $$$w_i$$$ $$$A$$$ $$$v_i$$$. It means that word $$$w_i$$$ has the value $$$v_i$$$.
After one single fixed line we describe the relations of $$$1 \leq N_n \leq 30$$$ more words $$$w$$$ through the operators. (New)-Words are never assigned more than once! Each relation is one line with the format: $$$n_i$$$ $$$A$$$ $$$o_j$$$ $$$w^1$$$, ..., $$$w^{x_j}$$$. That means that the operator is followed by as many words as it expects arguments. After that, the LLM text always ends with the line "Is there anything else that I can help you with?".
Directly after the LLM text, we provide a single line with $$$1 \lt Q \lt 40$$$, the number of questions we need you to answer followed by $$$Q$$$ lines with queries $$$q_i$$$ that consist of a single string.
We further illustrate the format in the examples.
All operators, words, and the assignment operator are non-empty strings consisting of lower-case english letters and '!', '?', '#', '$', '%', '&', '*', '+', '=', '@'. The domain of the functions and values are the first <S> positive integers.
Output $$$Q$$$ lines. For each input query $$$q_i$$$, output 0 if the query did not appear as a word or no value can be inferred from the given relation. We already say that no value can be inferred if the value of a word $$$w_i$$$ depends on itself, directly or indirectly. Otherwise output the value of the word from $$$q_i$$$.
As a large language model that was trained on the script of all lectures, I know all of the answers and could provide you with them. As I was trained to answer in the style of the notes, instead of just answering your question like normal people would, I will leave the answer for you to it prove as an exercise! But standard notation is boring. So I will provide you with the following. A logic with two values! Additionally, I have the following operators @, $, and #. They are not defined how you would imagine them to be to. To not cause any confusion, here are their function tables in the order that they appeared. 1 1 2 2 1 2 2 1 1 1 2 1 Now the questions you asked me can be answered by checking how the variables evaluate. Here are the values of some variables that are assigned with %. a % 1 And here are some rules b % @ a c % $ b d % # d Is there anything else that I can help you with? 4 a b c d
1 1 2 0
As a large language model that was trained on the script of all lectures, I know all of the answers and could provide you with them. As I was trained to answer in the style of the notes, instead of just answering your question like normal people would, I will leave the answer for you to prove as an exercise! But standard notation is boring. So I will provide you with the following. A logic with two values! Additionally, I have the following operators ?, and !. They are not defined how you would imagine them to be to. To not cause any confusion, here are their function tables in the order that they appeared. 1 1 2 2 1 1 1 1 2 1 2 1 1 2 2 1 Now the questions you asked me can be answered by checking how the variables evaluate. Here are the values of some variables that are assigned with asg. a asg 1 And here are some rules b asg ! a a d asg ? c c asg ? b Is there anything else that I can help you with? 5 a b c d e
1 1 1 1 0
In example one we can see that $$$a$$$ is assigned to $$$1$$$. According to the table of $$$@$$$, $$$1$$$ maps to $$$1$$$. So $$$b$$$ also has value $$$1$$$. Now according to the table of $$$?$$$ we map $$$1$$$ to $$$2$$$ which means $$$c$$$ is $$$2$$$ and finally $$$d$$$ depends directly on the value of $$$d$$$, so the correct output is $$$0$$$.
In example two line $$$a$$$ is set to $$$1$$$. After that $$$b$$$ is set to $$$1$$$ because $$$! 1 = 1$$$. We can not evaluate $$$d$$$ yet because the value of $$$c$$$ is unknown. Next, $$$? 1 = 1$$$ and so $$$c$$$ is also $$$1$$$. Now, we evaluate $$$? c = 1$$$ and so $$$d$$$ is also $$$1$$$. $$$e$$$ did not appear as a word in any assignment so the value is $$$0$$$.
