add data for IMO, USA_TST, USA_TSTST, USAJMO, USAMO

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IMO/segment_script/fixtures/test-case-1.md

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+## Algebra
+
+A1. Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is \(x = 2\).
+
+A2. Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+Solution 1. Subtracting 3 from both sides gives \(2y = 4\).
+
+Solution 2. Dividing both sides by 2 gives \(y = 2\).
+
+## Combinatorics
+
+C1. How many ways can 3 objects be arranged in a line?
+
+Solution. The number of permutations of 3 objects is \(3! = 6\).

IMO/segment_script/fixtures/test-case-2.md

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+## Algebra
+
+## A1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+## Combinatorics
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+## A1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is \(x = 2\).
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+Solution 1. Subtracting 3 from both sides gives \(2y = 4\).
+
+Solution 2. Dividing both sides by 2 gives \(y = 2\).
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+Solution. The number of permutations of 3 objects is \(3! = 6\).

IMO/segment_script/fixtures/test-case-3.md

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+## Algebra
+
+## A1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+## Combinatorics
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+## A 1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is \(x = 2\).
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+Solution 1. Subtracting 3 from both sides gives \(2y = 4\).
+
+Solution 2. Dividing both sides by 2 gives \(y = 2\).
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+Solution. The number of permutations of 3 objects is \(3! = 6\).

IMO/segment_script/fixtures/test-case-4.md

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+# $5^{\text {nd }}$ International Mathematical Olympiad 
+
+12 - 24 July 2011 Amsterdam The Netherlands
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_481ddb3576adc2313a2eg-01.jpg?height=2070&width=1902&top_left_y=744&top_left_x=-3)
+
+## 52nd International <br> Mathematical Olympiad <br> 12-24 July 2011 <br> Amsterdam <br> The Netherlands
+
+Problem shortlist with solutions
+
+## IMO regulation: <br> these shortlist problems have to be kept strictly confidential until IMO 2012.
+
+The problem selection committee
+
+Bart de Smit (chairman), Ilya Bogdanov, Johan Bosman,
+
+Andries Brouwer, Gabriele Dalla Torre, Géza Kós,
+
+Hendrik Lenstra, Charles Leytem, Ronald van Luijk,
+
+Christian Reiher, Eckard Specht, Hans Sterk, Lenny Taelman
+
+The committee gratefully acknowledges the receipt of 142 problem proposals by the following 46 countries:
+
+Armenia, Australia, Austria, Belarus, Belgium, Bosnia and Herzegovina, Brazil, Bulgaria, Canada, Colombia, Cyprus, Denmark, Estonia, Finland, France, Germany, Greece, Hong Kong, Hungary, India, Islamic Republic of Iran, Ireland, Israel, Japan, Kazakhstan, Republic of Korea, Luxembourg, Malaysia, Mexico, Mongolia, Montenegro, Pakistan, Poland, Romania, Russian Federation, Saudi Arabia, Serbia, Slovakia, Slovenia, Sweden, Taiwan, Thailand, Turkey, Ukraine, United Kingdom, United States of America
+
+
+## Algebra
+
+A1. Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is \(x = 2\).
+
+A2. Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+Solution 1. Subtracting 3 from both sides gives \(2y = 4\).
+
+Solution 2. Dividing both sides by 2 gives \(y = 2\).
+
+## Combinatorics
+
+C1. How many ways can 3 objects be arranged in a line?
+
+Solution. The number of permutations of 3 objects is \(3! = 6\).

IMO/segment_script/fixtures/test-case-5.md

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+## Algebra
+
+## A1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+## Combinatorics
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+## A1
+
+Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).
+
+Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is:
+
+## $\(x = 2\)$
+
+## A2
+
+Determine the value of \(y\) in the equation \(2y + 3 = 7\).
+
+Solution 1. Subtracting 3 from both sides gives \(2y = 4\).
+
+Solution 2. Dividing both sides by 2 gives \(y = 2\).
+
+## C1
+
+How many ways can 3 objects be arranged in a line?
+
+Solution. The number of permutations of 3 objects is \(3! = 6\).

IMO/segment_script/fixtures/test-out-1234.jsonl

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+{"problem_type":"Algebra","problem_label":"A1","problem":"Find the roots of the quadratic equation \\(x^2 - 4x + 4 = 0\\).","solution":"The equation simplifies to \\((x - 2)^2 = 0\\), so the root is \\(x = 2\\).","tier":0}
+{"problem_type":"Algebra","problem_label":"A2","problem":"Determine the value of \\(y\\) in the equation \\(2y + 3 = 7\\).","solution":"Subtracting 3 from both sides gives \\(2y = 4\\).","tier":0}
+{"problem_type":"Algebra","problem_label":"A2","problem":"Determine the value of \\(y\\) in the equation \\(2y + 3 = 7\\).","solution":"Dividing both sides by 2 gives \\(y = 2\\).","tier":0}
+{"problem_type":"Combinatorics","problem_label":"C1","problem":"How many ways can 3 objects be arranged in a line?","solution":"The number of permutations of 3 objects is \\(3! = 6\\).","tier":0}

IMO/segment_script/fixtures/test-out-5.jsonl

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+{"problem_type":"Algebra","problem_label":"A1","problem":"Find the roots of the quadratic equation \\(x^2 - 4x + 4 = 0\\).","solution":"The equation simplifies to \\((x - 2)^2 = 0\\), so the root is:  ## $\\(x = 2\\)$","tier":0}
+{"problem_type":"Algebra","problem_label":"A2","problem":"Determine the value of \\(y\\) in the equation \\(2y + 3 = 7\\).","solution":"Subtracting 3 from both sides gives \\(2y = 4\\).","tier":0}
+{"problem_type":"Algebra","problem_label":"A2","problem":"Determine the value of \\(y\\) in the equation \\(2y + 3 = 7\\).","solution":"Dividing both sides by 2 gives \\(y = 2\\).","tier":0}
+{"problem_type":"Combinatorics","problem_label":"C1","problem":"How many ways can 3 objects be arranged in a line?","solution":"The number of permutations of 3 objects is \\(3! = 6\\).","tier":0}

IMO/segment_script/segment.py

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+# -----------------------------------------------------------------------------
+# Author: Marina
+# Date: 2024-11-15
+# -----------------------------------------------------------------------------
+''' Script to segment IMO shortlist md files using regex. It takes as input
+files in en-shortlist and outputs en-shortlist-seg
+To run: 
+`python segment_script/segment.py`
+To debug (or see covered use cases listed in fixtures/):
+`pytest test_segment`
+'''
+
+from collections import defaultdict
+import os 
+import re
+import pandas as pd
+import json
+
+
+base = 'md'
+seg_base = 'segmented'
+
+section_re = re.compile(r'##\s+([A-Za-z]\w.*)')
+problem_re = re.compile(
+    r'^(?:##\s*)?((?:[AGNC]\s*\d+))\.*\s*(.*?)(?:\((.*?)\))?$',
+    re.MULTILINE
+)
+solution_re = re.compile(
+    r'^(?:##\s*)?(Solution(?: \d+)?\.)\s*(.*?)(?=(?:Solution|Comment|A\d+|G\d+|N\d+|C\d+|##|$))',
+    re.MULTILINE | re.DOTALL
+)
+
+def add_content(section, label, text_class, text, problems, solutions):
+        text_str = " ".join(text).strip()
+        if text_class == "problem":
+            # print(f"ADD PROBLEM {section} {label} ")
+            problems.append({"section": section, "label": label, "problem": text_str})
+        elif text_class == "solution":
+            # print(f"ADD SOLUTION {section} {label}")
+            solutions.append({"label": label, "solution": text_str})
+
+def parse(file):
+    with open(file, 'r') as file:
+        content = file.read()
+    problems, solutions = [], []
+    current_section, current_label,  current_class = None, None, None
+    current_lines = []
+    for line in content.splitlines():
+        if match := problem_re.match(line):
+            label, text, country = match.groups()
+            label = label.replace(" ", "") # clean the label
+            add_content(current_section, current_label, current_class, current_lines, problems, solutions)
+            current_class = "problem"
+            current_label = label
+            current_lines = [text]
+        elif match := solution_re.match(line):
+            label, text = match.groups()
+            add_content(current_section, current_label, current_class, current_lines, problems, solutions)
+            current_class = "solution"
+            current_lines = [text]
+        elif match := section_re.match(line):
+            add_content(current_section, current_label, current_class, current_lines, problems, solutions)
+            current_class = "section"
+            text, = match.groups()
+            current_section = text
+        else:
+            current_lines.append(line)
+    add_content(current_section, current_label, current_class, current_lines, problems, solutions)
+    problems_df = pd.DataFrame(problems).drop_duplicates(subset=["label", "problem"])
+    solutions_df = pd.DataFrame(solutions)
+    return problems_df, solutions_df
+
+def join(problems_df, solutions_df):
+    pairs_df = problems_df.merge(solutions_df, on=["label"], how="left")
+    return pairs_df
+
+def add_metadata(pairs_df):
+    pairs_df.rename(columns={"section": "problem_type", "label": "problem_label"}, inplace=True)
+    pairs_df['tier'] = 0 # according to omnimath 
+    return pairs_df
+
+def write_pairs(filename, pairs_df):
+    pairs_df.to_json(filename, orient="records", lines=True)
+
+
+os.makedirs(seg_base, exist_ok=True)
+for name in os.listdir(base):
+    if "compendium" not in name: # en-compendium is segmented in segment_compendium.py
+        print(name)
+        problems, solutions = parse(os.path.join(base, name))
+        pairs_df = join(problems, solutions)
+        pairs_df = add_metadata(pairs_df)
+        print(pairs_df)
+        basename = os.path.splitext(name)[0]
+        print(f"{seg_base}/{basename}.jsonl")
+        write_pairs(f"{seg_base}/{basename}.jsonl", pairs_df)

IMO/segment_script/segment_compendium.py

@@ -0,0 +1,155 @@
+# -----------------------------------------------------------------------------
+# Author: Marina
+# Date: 2024-11-15
+# -----------------------------------------------------------------------------
+''' Script to segment IMO shortlist md files using regex. It takes as input
+the file en-compendium.md in en-shortlist and outputs the segmentation 
+(problem/solution pairs) in en-shortlist-seg
+To run: 
+`python segment_compendium.py`
+To debug (or see covered use cases by regex):
+`pytest test_segment_compendium`
+'''
+
+import os 
+import re
+import pandas as pd
+
+
+base = 'en-shortlist'
+seg_base = 'en-shortlist-seg'
+basename = 'en-compendium'
+
+
+level1_re = re.compile(r"^##\s+(Problems|Solutions|Notation and Abbreviations)$")
+year_re = re.compile(r"^[^=]*,\s+(\d{4})\s*$")
+problem_section_re = re.compile(r"^###\s+(\d+\.\d+\.\d+)\s+(.+)$")
+solution_section_re = re.compile(r"^###\s+(\d+\.\d+)\s+([\w\s]+)\s+(\d{4})$")
+problem_or_solution_re = re.compile(r"^(?:\[.*?\])?\s*(\d+)\s*\.\s*(.+)$")
+
+
+def add_content(current_dict):
+    required_keys = ["year", "category", "section_label", "label", "lines"]
+    if not all(current_dict[key] for key in required_keys):
+        return
+    text_str = " ".join(current_dict["lines"]).strip()
+    entry = {
+        "year": current_dict["year"],
+        "category": current_dict["category"],
+        "section": current_dict["section_label"],
+        "label": current_dict["label"],
+    }
+    if current_dict["class"] == "problem":
+        entry["problem"] = text_str
+        current_dict["problems"].append(entry)
+    elif current_dict["class"] == "solution":
+        entry["solution"] = text_str
+        current_dict["solutions"].append(entry)
+
+
+def get_category(s:str):
+    cat = None
+    if 'contest' in s.lower():
+        cat = 'contest'
+    elif 'shortlisted' in s.lower():
+        cat = 'shortlisted'
+    elif 'longlisted' in s.lower():
+        cat = 'longlisted'
+    return cat
+
+
+def get_matching_section_label(s:str):
+    """
+    extracts the section number to be used a a join key to pair a problem and solution
+    for problems: 3.44.1 -> 44
+    for solutions: 4.20 -> 20
+    """
+    return s.split('.')[1]
+
+
+def parse(file):
+    with open(file, 'r') as file:
+        content = file.read()
+    # problems, solutions = [], []
+    current = {
+        "year": None,
+        "category": None,
+        "section_label": None,
+        "label": None,  
+        "class": None,
+        "lines": [],
+        "problems": [],
+        "solutions": []
+    }
+    for line in content.splitlines():
+        if match := level1_re.match(line):
+            add_content(current)
+            title, = match.groups()
+            current["class"] = {
+                "Problems": "problem",
+                "Solutions": "solution",
+            }.get(title, "other")
+            current["lines"] = []
+        elif match := year_re.match(line):
+            add_content(current)
+            current["year"] = match.group(1)
+            current["lines"] = []
+        elif match := problem_section_re.match(line):
+            add_content(current)
+            number, title = match.groups()
+            current["section_label"] = get_matching_section_label(number)
+            current["category"] = get_category(title)
+            current["lines"] = []
+        elif match := solution_section_re.match(line):
+            add_content(current)
+            number, title, year = match.groups() 
+            current["section_label"] = get_matching_section_label(number)
+            current["category"] = get_category(title)
+            current["year"] = year
+            current["lines"] = []
+        elif match := problem_or_solution_re.match(line):
+            add_content(current)
+            current["label"] = match.group(1)
+            current["lines"] = [line]
+        else:
+            if current["lines"]:
+                current["lines"].append(line)
+    problems_df = pd.DataFrame(current["problems"])
+    solutions_df = pd.DataFrame(current["solutions"])
+    return problems_df, solutions_df
+
+def join(problems_df, solutions_df):
+    pairs_df = problems_df.merge(solutions_df, on=["year", "category", "section", "label"], how="outer")
+    return pairs_df
+
+def add_metadata(pairs_df):
+    problem_type_mapping = {
+    "A": "Algebra",
+    "C": "Combinatorics",
+    "G": "Geometry",
+    "N": "Number Theory"
+    }
+    pairs_df['problem_type'] = pairs_df['problem'].str.extract(r'^\d+\.\s*([ACGN])\d*')[0]
+    pairs_df['problem_type'] = pairs_df['problem_type'] .map(problem_type_mapping)
+    pairs_df['tier'] = 0 # according to omnimath 
+    pairs_df.rename(columns={"category": "problem_phase"}, inplace=True)
+    pairs_df = pairs_df.drop(columns=['section', 'label'])
+    return pairs_df
+
+def write_pairs(filename, pairs_df):
+    pairs_df.to_json(filename, orient="records", lines=True)
+
+
+problems, solutions = parse(f"{base}/{basename}.md")
+pairs_df = join(problems, solutions)
+pairs_df = pairs_df[pairs_df.notnull().all(axis=1)]
+pairs_df = add_metadata(pairs_df)
+write_pairs(f"{seg_base}/{basename}.jsonl", pairs_df)
+
+# problems contains duplicate problems (since problem in Shortlist appears in Contest, and problem in Longlist appeasr in Shortlist)
+# >>>print(len(problems))
+# 2460
+# >>>print(len(solutions))
+# 961
+# print(len(pairs_df))
+# 960