Nan-Do/CP-Ranker

SentenceTransformer based on Salesforce/SFR-Embedding-Code-400M_R

This is a sentence-transformers model finetuned from Salesforce/SFR-Embedding-Code-400M_R on around 1000000 Python samples from the Leetcode, Atcoder-ABC, Atcoder-ARC, Atcoder-AGC, Codechef, Codeforces datasets. It maps sentences & paragraphs to a 1024-dimensional dense vector space and can be used for semantic textual similarity, semantic search, paraphrase mining, text classification, clustering, and more.

Model Details

Model Description

• Model Type: Sentence Transformer
• Base model: Salesforce/SFR-Embedding-Code-400M_R
• Maximum Sequence Length: 8192 tokens
• Output Dimensionality: 1024 dimensions
• Similarity Function: Cosine Similarity
• Training Datasets:
  • Leetcode
  • Atcoder
  • Codechef
  • Codeforces
  • CodeforcesPositive

Model Sources

• Documentation: Sentence Transformers Documentation
• Repository: Sentence Transformers on GitHub
• Hugging Face: Sentence Transformers on Hugging Face

Full Model Architecture

SentenceTransformer(
  (0): Transformer({'max_seq_length': 8192, 'do_lower_case': False, 'architecture': 'NewModel'})
  (1): Pooling({'word_embedding_dimension': 1024, 'pooling_mode_cls_token': True, 'pooling_mode_mean_tokens': False, 'pooling_mode_max_tokens': False, 'pooling_mode_mean_sqrt_len_tokens': False, 'pooling_mode_weightedmean_tokens': False, 'pooling_mode_lasttoken': False, 'include_prompt': True})
)

Usage

Direct Usage (Sentence Transformers)

First install the Sentence Transformers library:

pip install -U sentence-transformers

Then you can load this model and run inference.

from sentence_transformers import SentenceTransformer

# Download from the 🤗 Hub
model = SentenceTransformer("Nan-Do/Ranker")
# Run inference
sentences = [
    'You are given integers N, M and three integer sequences of length M: X = (X_1, X_2, \\ldots, X_M), Y = (Y_1, Y_2, \\ldots, Y_M), and Z = (Z_1, Z_2, \\ldots, Z_M). It is guaranteed that all elements of X and Y are between 1 and N, inclusive.\n\n  We call a length-N sequence of non-negative integers A = (A_1, A_2, \\ldots, A_N) a good sequence if and only if it satisfies the following condition:\n\n    * For every integer i with 1 \\le i \\le M, the XOR of A_{X_i} and A_{Y_i} is Z_i.\n\n  Determine whether a good sequence A=(A_1,A_2,\\ldots,A_N) exists, and if it exists, find one good sequence that minimizes the sum of its elements \\displaystyle \\sum_{i=1}^N A_i.\n\n  Notes on XOR For non-negative integers A and B, their XOR A \\oplus B is defined as follows:\n    * In the binary representation of A \\oplus B, the digit in the place corresponding to 2^k \\,(k \\ge 0) is 1 if and only if exactly one of the digits in the same place of A and B is 1; otherwise, it is 0.\n  For example, 3 \\oplus 5 = 6 (in binary: 011 \\oplus 101 = 110).\n\n  Constraints\n\n    * 1 \\le N \\le 2\\times 10^5\n    * 0 \\le M \\le 10^5\n    * 1 \\le X_i, Y_i \\le N\n    * 0 \\le Z_i \\le 10^9\n    * All input values are integers.\n\n    Input\n\n    The input is given from Standard Input in the following format:\n\n    N M\n    X_1 Y_1 Z_1\n    X_2 Y_2 Z_2\n    \\vdots\n    X_M Y_M Z_M\n    \n\n    Output\n\n    If no good sequence exists, print -1.\n\n    If a good sequence exists, print one good sequence that minimizes the sum of its elements, separated by spaces.\n\n    If there are multiple good sequences with the same minimum sum, printing any of them is accepted.\n\n  Sample Input 1\n\n  3 2\n  1 3 4\n  1 2 3\n  \n\n  Sample Output 1\n\n  0 3 4\n  \n\n  A=(0,3,4) is a good sequence because A_1 \\oplus A_2 = 3 and A_1 \\oplus A_3 = 4.\n\n  Other good sequences include A=(1,2,5) and A=(7,4,3), but A=(0,3,4) has the smallest sum among all good sequences.\n\n  Sample Input 2\n\n  3 3\n  1 3 4\n  1 2 3\n  2 3 5\n  \n\n  Sample Output 2\n\n  -1\n  \n\n  No good sequence exists, so print -1.\n\n  Sample Input 3\n\n  5 8\n  4 2 4\n  2 3 11\n  3 4 15\n  4 5 6\n  3 2 11\n  3 3 0\n  3 1 9\n  3 4 15\n  \n\n  Sample Output 3\n\n  0 2 9 6 0',
    'import sys\r\nsys.setrecursionlimit(10**7)\r\nfrom collections import *\r\nfrom bisect import *\r\nfrom itertools import *\r\nfrom functools import cache    #関数の定義の上に  @cache\u3000をつける\r\nfrom heapq import *\r\nfrom math import *\r\n\r\n\r\nN, M = map(int,input().split())\r\nG = [list() for i in range(N+1)]\r\n\r\nfor i in range(M):\r\n    x,y,z = map(int,input().split())\r\n    G[x].append((y,z))\r\n    G[y].append((x,z))\r\n\r\n\r\n\r\nvisited = [False]*(N+1)\r\nval = [-1]*(N+1)\r\nans = [0]*(N+1)\r\n\r\ndef bfs(start):\r\n    global visited,val\r\n    comp = [start]\r\n    q = deque([start])\r\n    visited[start] = True\r\n    val[start] = 0    \r\n\r\n    while q:\r\n        pos = q.popleft()\r\n        for nxt,z in G[pos]:\r\n            if visited[nxt]:\r\n                if val[nxt]^val[pos] != z:\r\n                    print(-1)\r\n                    sys.exit()\r\n            else:\r\n                visited[nxt] = True\r\n                val[nxt] = val[pos] ^ z\r\n                q.append(nxt)\r\n                comp.append(nxt)\r\n\r\n    return comp                \r\n\r\n\r\nfor i in range(1,N+1):\r\n    if len(G[i]) == 0:\r\n        continue\r\n\r\n    if visited[i]:\r\n        continue\r\n\r\n    comp = bfs(i)\r\n\r\n    for j in range(31):\r\n        cnt = 0\r\n        for ele in comp:\r\n            if val[ele] & (1<<j):\r\n                cnt += 1\r\n\r\n        if cnt < len(comp) - cnt:\r\n            for ele in comp:\r\n                if val[ele] & (1<<j):\r\n                    ans[ele] += 1<<j\r\n        else:\r\n            for ele in comp:\r\n                if not val[ele] & (1<<j):\r\n                    ans[ele] += 1<<j\r\n\r\nprint(*ans[1:])\r\n',
    '\r\nimport sys\r\nsys.setrecursionlimit(10**7)\r\nfrom collections import *\r\nfrom bisect import *\r\nfrom itertools import *\r\nfrom functools import cache    #関数の定義の上に  @cache\u3000をつける\r\nfrom heapq import *\r\nfrom math import *\r\n\r\n\r\nN, M = map(int,input().split())\r\nG = [list() for i in range(N+1)]\r\n\r\nfor i in range(M):\r\n    x,y,z = map(int,input().split())\r\n    G[x].append((y,z))\r\n    G[y].append((x,z))\r\n\r\nans = [0]*(N+1)\r\n\r\n\r\n\r\n\r\nvisited = [False]*(N+1)\r\n\r\n\r\nfor i in range(1,N+1):\r\n    if len(G[i]) == 0:\r\n        continue\r\n\r\n    if visited[i]:\r\n        continue\r\n\r\n    answer = 0\r\n    for j in range(31):\r\n        tv = [False]*(N+1)\r\n        tmp = [0]*(N+1)\r\n        cnt1 = 0\r\n        tmp[i] = 0\r\n        q = [i]\r\n        while q:\r\n            pos = q.pop()\r\n            if tv[pos]:\r\n                continue\r\n            tv[pos] = True\r\n            for nxt,w in G[pos]:\r\n                if tv[nxt]:\r\n                    if tmp[pos]^tmp[nxt] != (w>>j) & 1:\r\n                        print(-1)\r\n                        sys.exit()\r\n                else:\r\n                    tmp[nxt] = ((w>>j) & 1)^tmp[pos]\r\n                    if tmp[nxt] == 1:\r\n                        cnt1 += 1\r\n                    q.append(nxt)\r\n\r\n        tv = [False]*(N+1)\r\n        tmp = [0]*(N+1)\r\n        cnt2 = 1\r\n        tmp[i] = 1\r\n        q = [i]\r\n        while q:\r\n            pos = q.pop()\r\n            if tv[pos]:\r\n                continue\r\n            tv[pos] = True\r\n            for nxt,w in G[pos]:\r\n                if tv[nxt]:\r\n                    if tmp[pos]^tmp[nxt] != (w>>j) & 1:\r\n                        print(-1)\r\n                        sys.exit()\r\n                else:\r\n                    tmp[nxt] = ((w>>j) & 1)^tmp[pos]\r\n \r\n                    if tmp[nxt] == 1:\r\n                        cnt2 += 1\r\n                    q.append(nxt)  \r\n        \r\n            \r\n        if cnt1 < cnt2:\r\n            ans[i] += 0\r\n        else:\r\n            ans[i] += 1<<j\r\n\r\n    \r\n    q = [i] \r\n    while q:\r\n        pos = q.pop()\r\n        if visited[pos]:\r\n            continue\r\n        visited[pos] = True\r\n\r\n        for nxt, w in G[pos]:\r\n            if visited[nxt]:\r\n                continue\r\n            \r\n            ans[nxt] = ans[pos]^w\r\n            q.append(nxt)\r\n\r\nprint(*ans[1:])\r\n',
]
embeddings = model.encode(sentences)
print(embeddings.shape)
# [3, 1024]

# Get the similarity scores for the embeddings
similarities = model.similarity(embeddings, embeddings)
print(similarities)
# tensor([[1.0000, 1.0000, 1.0000],
#         [1.0000, 1.0000, 1.0000],
#         [1.0000, 1.0000, 1.0000]])

Training Details

Training Datasets

Leetcode

Leetcode

• Dataset: Leetcode
• Size: 79,862 training samples
• Columns: anchor and positive
• Approximate statistics based on the first 1000 samples:

  	anchor	positive
  type	string	string
  details	min: 83 tokens mean: 363.93 tokens max: 1064 tokens	min: 34 tokens mean: 180.39 tokens max: 1290 tokens

• Samples:

  anchor	positive
  You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the i^th type of fruit, and baskets[j] represents the capacity of the j^th basket. From left to right, place the fruits according to these rules: Each fruit type must be placed in the leftmost available basket with a capacity greater than or equal to the quantity of that fruit type. Each basket can hold only one type of fruit. If a fruit type cannot be placed in any basket, it remains unplaced. Return the number of fruit types that remain unplaced after all possible allocations are made.   Example 1: Input: fruits = [4,2,5], baskets = [3,5,4] Output: 1 Explanation: fruits[0] = 4 is placed in baskets[1] = 5. fruits[1] = 2 is placed in baskets[0] = 3. fruits[2] = 5 cannot be placed in baskets[2] = 4. Since one fruit type remains unplaced, we return 1. Example 2: Input: fruits = [3,6,1], baskets = [6,4,7] Output: 0 Explanation: fruits[0] = 3 i...	class Solution: def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int: result = 0 max_val = max(baskets) for f in fruits: if f > max_val: result += 1 continue i = 0 while i < len(baskets) and baskets[i] < f: i += 1 if i < len(baskets): baskets.pop(i) else: result += 1 #print(baskets) return result
  You are given an integer array prices representing the daily price history of a stock, where prices[i] is the stock price on the i^th day. A smooth descent period of a stock consists of one or more contiguous days such that the price on each day is lower than the price on the preceding day by exactly 1. The first day of the period is exempted from this rule. Return the number of smooth descent periods.   Example 1: Input: prices = [3,2,1,4] Output: 7 Explanation: There are 7 smooth descent periods: [3], [2], [1], [4], [3,2], [2,1], and [3,2,1] Note that a period with one day is a smooth descent period by the definition. Example 2: Input: prices = [8,6,7,7] Output: 4 Explanation: There are 4 smooth descent periods: [8], [6], [7], and [7] Note that [8,6] is not a smooth descent period as 8 - 6 ≠ 1. Example 3: Input: prices = [1] Output: 1 Explanation: There is 1 smooth descent period: [1]   Constraints: 1 <= prices.length <= 10^5 1 <= prices[i] <= 10^5	class Solution: def getDescentPeriods(self, prices: List[int]) -> int: dp = [0 for _ in range(len(prices))] result = len(prices) for i in range(1, len(prices)): if prices[i] == prices[i-1] - 1: dp[i] = dp[i-1] + 1 else: dp[i] = 0 result += dp[i] return result
  You are given a 0-indexed string array words having length n and containing 0-indexed strings. You are allowed to perform the following operation any number of times (including zero): Choose integers i, j, x, and y such that 0 <= i, j < n, 0 <= x < words[i].length, 0 <= y < words[j].length, and swap the characters words[i][x] and words[j][y]. Return an integer denoting the maximum number of palindromes words can contain, after performing some operations. Note: i and j may be equal during an operation.   Example 1: Input: words = ["abbb","ba","aa"] Output: 3 Explanation: In this example, one way to get the maximum number of palindromes is: Choose i = 0, j = 1, x = 0, y = 0, so we swap words[0][0] and words[1][0]. words becomes ["bbbb","aa","aa"]. All strings in words are now palindromes. Hence, the maximum number of palindromes achievable is 3. Example 2: Input: words = ["abc","ab"] Output: 2 Explanation: In this example, one way to get the maximum number of palindromes is: ...	class Solution: def maxPalindromesAfterOperations(self, words: List[str]) -> int: word = [0]*26 ar = list() for i in words: ar.append(len(i)) for j in range(len(i)): word[ord(i[j])-97] += 1 p,s = 0,0 for k in word: p += k//2 s += k%2 ar.sort() ans = 0 for m in ar: if m % 2 == 0: #even case if m//2 > p: break else: ans += 1 p -= m//2 else: # odd case if s == 0: #break case p -= 1 s += 2 s -= 1 if m//2 > p: break else: ans += 1 p -= m//2 return ans

• Loss: CachedMultipleNegativesRankingLoss with these parameters:

  {
      "scale": 15,
      "similarity_fct": "cos_sim",
      "mini_batch_size": 2,
      "gather_across_devices": false
  }
  

Atcoder

Atcoder

• Dataset: Atcoder
• Size: 478,734 training samples
• Columns: anchor, positive, and negative
• Approximate statistics based on the first 1000 samples:

  	anchor	positive	negative
  type	string	string	string
  details	min: 73 tokens mean: 376.31 tokens max: 1415 tokens	min: 15 tokens mean: 257.06 tokens max: 2912 tokens	min: 3 tokens mean: 260.5 tokens max: 2902 tokens

• Samples:

  anchor	positive	negative
  We have a permutation P = P_1, P_2, \ldots, P_N of 1, 2, \ldots, N. You have to do the following N - 1 operations on P, each exactly once, in some order: * Swap P_1 and P_2. * Swap P_2 and P_3. \vdots * Swap P_{N-1} and P_N. Your task is to sort P in ascending order by configuring the order of operations. If it is impossible, print -1 instead. Constraints * All values in input are integers. * 2 \leq N \leq 2 \times 10^5 * P is a permutation of 1, 2, \ldots, N. Input Input is given from Standard Input in the following format: N
  P_1 P_2 \ldots P_N
  Output If it is impossible to sort P in ascending order by configuring the order of operations, print -1. Otherwise, print N-1 lines to represent a sequence of operations that sorts P in ascending order. The i-th line (1 \leq i \leq N - 1) should contain j, where the i-th operation swaps P_j and P_{j + 1}. If there are multiple such sequences of operations...	def main():
  import sys
  input = sys.stdin.readline
  N = int(input())
  P = list(map(int, input().split()))
  pre = 1
  res = []
  Q = list(range(1,N+1))
  for i in range(N):
  if P[i] == pre:
  P[pre-1:i+1] = P[i:i+1] + P[pre-1:i]
  for j in range(i,pre-1,-1):
  res.append(j)
  pre = i + 1
  if P == Q and len(res) == N - 1:
  for i in res:
  print(i)
  else:
  print(-1)
  main()	def main():
  import sys
  input = sys.stdin.readline
  N = int(input())
  P = list(map(int, input().split()))
  pre = 1
  res = []
  Q = list(range(1,N+1))
  for i in range(N):
  if P[i] == pre:
  P[pre-1:i+1] = P[i:i+1] + P[pre-1:i]
  if P == Q:
  print(-1)
  exit()
  for j in range(i,pre-1,-1):
  res.append(j)
  pre = i+1
  if P == Q:
  for i in res:
  print(i)
  else:
  print(-1)
  main()
  You are given an integer sequence of length N: A=(A_1,A_2,\ldots,A_N). You will perform the following consecutive operations just once: * Choose an integer x (0\leq x \leq N). If x is 0, do nothing. If x is 1 or greater, replace each of A_1,A_2,\ldots,A_x with L. * Choose an integer y (0\leq y \leq N). If y is 0, do nothing. If y is 1 or greater, replace each of A_{N},A_{N-1},\ldots,A_{N-y+1} with R. Print the minimum possible sum of the elements of A after the operations. Constraints * 1 \leq N \leq 2\times 10^5 * -10^9 \leq L, R\leq 10^9 * -10^9 \leq A_i\leq 10^9 * All values in input are integers. Input Input is given from Standard Input in the following format: N L R
  A_1 A_2 \ldots A_N
  Output Print the answer. Sample Input 1 5 4 3
  5 5 0 6 3
  Sample Output 1 14
  If you choose x=2 and y=2, you will get A = (4,4,0,3,3), for the sum of 14, which is the minimum sum achievable. Sample Input...	N,L,R = map(int,input().split())
  A = list(map(int,input().split()))
  rui1,rui2 = [],[]
  for i,v in enumerate(A):
  if i == 0:
  rui1.append(v)
  else:
  rui1.append(rui1[-1]+v)
  B = A[::-1]
  for i,v in enumerate(B):
  if i == 0:
  rui2.append(v)
  else:
  rui2.append(rui2[-1]+v)
  #print("rui1",rui1)
  #print("rui2",rui2)
  rui11,rui22 = [0],[]
  for i,v in enumerate(rui1):
  tmp = (i+1)*L
  rui11.append(rui1[i]-tmp)
  for i,v in enumerate(rui2):
  tmp = (i+1)*R
  rui22.append(rui2[i]-tmp)
  #print("rui11",rui11)
  #print("rui22",rui22[::-1])
  rui222 = [0]
  for i,v in enumerate(rui22):
  if i == 0:
  rui222.append(v)
  else:
  rui222.append(max(v,rui222[-1]))
  #print("rui222",rui222[::-1])
  rui222.reverse()
  tmpans = 0
  tmpans11 = 10*100(-1)
  for i,v in enumerate(rui11):
  tmpans11 = max(tmpans11,v)
  tmpans = max(tmpans11+rui222[i],tmpans)
  print(sum(A)-tmpans)
  	N,L,R = map(int,input().split())
  A = list(map(int,input().split()))
  rui = []
  for i,v in enumerate(A):
  if i == 0:
  rui.append(v)
  else:
  rui.append(rui[-1]+v)
  #print("rui",rui)
  rui2 = []
  for i,v in enumerate(A[::-1]):
  if i == 0:
  rui2.append(v)
  else:
  rui2.append(rui2[-1]+v)
  rui11 = []
  for i,v in enumerate(rui):
  nex = (i+1)*L
  rui11.append(v-nex)
  #print("rui11",rui11)
  rui22 = []
  for i,v in enumerate(rui2):
  nex = (i+1)*R
  rui22.append(v-nex)
  rui33 = []
  for i,v in enumerate(rui22):
  if i == 0:
  rui33.append(v)
  else:
  rui33.append(max(rui33[-1],v))
  #print("rui22",rui22[::-1])
  #print("rui33",rui33[::-1])
  rui33.reverse()
  ans = max(rui33[0],max(rui11))
  for i,v in enumerate(rui11):
  if i == len(rui11)-1:
  break
  ans = max(ans,v+rui33[i+1])
  print(sum(A)-ans) #ans
  Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. Constraints * 1 \leq A,B \leq 100 * Both A and B are integers. Input Input is given from Standard Input in the following format: A B
  Output If it is possible to give cookies so that each of the three goats can have the same number of cookies, print Possible; otherwise, print Impossible. Sample Input 1 4 5
  Sample Output 1 Possible
  If Snuke gives nine cookies, each of the three goats can have three cookies. Sample Input 2 1 1
  Sample Output 2 Impossible
  Since there are only two cookies, the three goats cannot have the same number of cookies no matte...	A,B=input().split() A=int(A) B=int(B) if A%3==0 or B%3==0 or (A+B)%3==0: print("Possible") else: print("Impossible")	x=input() x=[int(i) for i in x.split()] if (x[0]+x[1])%3==0 and (x[0]+x[1]) !=0 : print("Possible") else: print("Impossible")

• Loss: TripletLoss with these parameters:

  {
      "distance_metric": "TripletDistanceMetric.COSINE",
      "triplet_margin": 0.3
  }
  

Codechef

Codechef

• Dataset: Codechef
• Size: 68,517 training samples
• Columns: anchor, positive, and negative
• Approximate statistics based on the first 1000 samples:

  	anchor	positive	negative
  type	string	string	string
  details	min: 165 tokens mean: 575.4 tokens max: 1373 tokens	min: 16 tokens mean: 213.17 tokens max: 2489 tokens	min: 7 tokens mean: 206.87 tokens max: 2640 tokens

• Samples:

  anchor	positive	negative
  Swapping Marks Digits Alice scored $A$ marks and Bob scored $B$ marks in an exam. Both $A$ and $B$ are two-digit numbers that don't contain the digit $0$. Alice wants her marks to display higher than Bob's. For this, she can reverse her score and/or Bob's score. Is there a way for her score to display higher than Bob's? For example, if $A = 37$ and $B = 83$, Alice can reverse her score to make it $73$, and also reverse Bob's score to make it $38$, and now her score is higher. Input Format: - The first line of input will contain a single integer $T$, denoting the number of test cases. - The first and only line of each test case contains two space-separated integers $A$ and $B$ — the marks obtained by Alice and Bob, respectively. Output Format: For each test case, output on a new line the answer: "Yes" if Alice can change her score to display higher than Bob's, and "No" otherwise (without quotes). Each letter of the output may be printed in either uppercase or lowercase, i.e...	# cook your dish here
  for _ in range(int(input())):
  a,b=map(int,input().split())
  temp1=a
  temp2=b
  alice=0
  bob=0
  while(a!=0):
  alice=alice*10+a%10
  a//=10
  while(b!=0):
  bob=bob*10+b%10
  b//=10
  if(alice>bob or temp1>temp2 or temp1>bob or alice>temp2):
  print("Yes")
  else:
  print("No")	# cook your dish here
  for _ in range(int(input())):
  a,b=map(int,input().split())
  alice=0
  bob=0
  while(a!=0):
  alice=alice*10+a%10
  a//=10
  while(b!=0):
  bob=bob*10+b%10
  b//=10
  if(alice>bob):
  print("Yes")
  else:
  print("No")
  Difficulty Rating Order Our Chef has some students in his coding class who are practicing problems. Given the difficulty of the problems that the students have solved in order, help the Chef identify if they are solving them in non-decreasing order of difficulty. Non-decreasing means that the values in an array is either increasing or remaining the same, but not decreasing. That is, the students should not solve a problem with difficulty $d_1$, and then later a problem with difficulty $d_2$, where $d_1 > d_2$. Output “Yes” if the problems are attempted in non-decreasing order of difficulty rating and “No” if not. Input Format: - The first line of input will contain a single integer $T$, denoting the number of test cases. The description of the test cases follows. - Each test case consists of $2$ lines of input. - The first line contains a single integer $N$, the number of problems solved by the students - The second line contains $N$ space-separate integers, the difficulty r...	def rating_order(arr):
  for i in range(len(arr)-1):
  if (arr[i]>arr[i+1]):
  return"no"
  return"yes"
  t=int(input())
  for i in range(t):
  n=int(input())
  arr=list(map(int,input().split()))
  print(rating_order(arr))	def rating_order(arr):
  for i in range(len(arr)-1):
  if (arr[i]
  return"yes"
  return"no"
  t=int(input())
  for i in range(t):
  n=int(input())
  arr=list(map(int,input().split()))
  print(rating_order(arr))
  Read Pages Chef has started studying for the upcoming test. The textbook has $N$ pages in total. Chef wants to read at most $X$ pages a day for $Y$ days. Find out whether it is possible for Chef to complete the whole book. Input Format: - The first line of input will contain a single integer $T$, denoting the number of test cases. - The first and only line of each test case contains three space-separated integers $N, X,$ and $Y$ — the number of pages, the number of pages Chef can read in a day, and the number of days. Output Format: For each test case, output on a new line, YES, if Chef can complete the whole book in given time, and NO otherwise. You may print each character of the string in uppercase or lowercase. For example, Yes, YES, yes, and yES are all considered identical. Constraints: - $1 \leq T \leq 1000$ - $1 \leq N \leq 100$ - $1 \leq X, Y \leq 10$ Sample 1: Input: 4 5 2 3 10 3 3 7 7 1 3 2 1 Output: YES NO YES NO Explanation: Test case $1$: Chef can...	# Read the number of test cases
  T = int(input())
  # Process each test case
  for _ in range(T):
  # Read N, X, Y for the current test case
  N, X, Y = map(int, input().split())
  # Check if Chef can complete the book
  if X * Y >= N:
  print("YES")
  else:
  print("NO")	# cook your dish here
  t = int(input())
  for _ n range(t):
  n,x,y = map(int,input().split())
  if x*y>=n:
  pritn("yes")
  else:
  print("no")

• Loss: TripletLoss with these parameters:

  {
      "distance_metric": "TripletDistanceMetric.COSINE",
      "triplet_margin": 0.3
  }
  

Codeforces

Codeforces

• Dataset: Codeforces
• Size: 239,311 training samples
• Columns: anchor, positive, and negative
• Approximate statistics based on the first 1000 samples:

  	anchor	positive	negative
  type	string	string	string
  details	min: 29 tokens mean: 352.67 tokens max: 1267 tokens	min: 20 tokens mean: 124.62 tokens max: 1071 tokens	min: 3 tokens mean: 137.77 tokens max: 2282 tokens

• Samples:

  anchor	positive	negative
  Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other. The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time h hours, m minutes, s seconds. Last time Misha talked with the coordinator at t1 o'clock, so now he stands on the number t1 on the clock face. The contest should be ready by t2 o'clock. In the terms of paradox it means that Misha has to go to number t2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction. Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. H...	h, m, s, t1, t2 = map(int, input().split()) m //= 5 s //= 5 a = [h, m, s, h + 12, m + 12, s + 12] up = 1 dw = 1 if t1 > t2: t1, t2 = t2, t1 t1u = t1 + 12 for i in a: if t1 <= i and i < t2: up = 0 if t2 <= i and i < t1u: dw = 0 if dw + up == 0: print('NO') else: print('YES')	h, m, s, t1, t2 = tuple(map(int, input().split())) t2 *= 5 t1 *= 5 s = float(s) m = float(m + s / 12) h = float(h * 5 + m / 12) def dt(u, v): if v < u: return (60 - u) + v return v - u def check(t1, t2, h, m, s): d = dt(t1, t2) return d < dt(t1, h) and d < dt(t1, m) and d < dt(t1, s) r = check(t1, t2, h, m, s) or check(t2, t1, h, m, s) print("YES" if r else "NO")
  A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed. To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity. Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once . Hitagi knows, however, that, apart from 0 , no integer occurs in a and b more than once in total. If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In othe...	# bsdk idhar kya dekhne ko aaya hai, khud kr!!! # import math # from itertools import * # import random # import calendar # import datetime # import webbrowser n, m = map(int, input().split()) arr = list(map(int, input().split())) discount = list(map(int, input().split())) arr[arr.index(0)] = discount[0] if m > 1 or arr != sorted(arr): print("Yes") else: print("No")	q = list(map(int, input().split())) n, k = q[0], q[1] a = list(map(int, input().split())) b = list(map(int, input().split())) if k > 1: print('Yes') else: pos = a.index(0) if b[0] > a[pos - 1] and b[0] < a[pos + 1]: print('No') else: print('Yes')
  A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input The first line contai...	n=int(input()) sx=sy=sz=0 for i in range(n): x, y, z = map(int, input().split()) sx+=x sy+=y sz+=z i+=1 if sx == 0 and sy ==0 and sz == 0: print('YES') else: print('NO')	a=int(input()) for i in range(0,a): x_i=list(input()) x=sum(int(x_i[0]) for i in range(0,a)) y=sum(int(x_i[1]) for i in range(0,a)) z=sum(int(x_i[-1]) for i in range(0,a)) if x==0 and y==0 and z==0: print('YES') else: print('NO')

• Loss: TripletLoss with these parameters:

  {
      "distance_metric": "TripletDistanceMetric.COSINE",
      "triplet_margin": 0.3
  }
  

CodeforcesPositive

CodeforcesPositive

• Dataset: CodeforcesPositive
• Size: 102,336 training samples
• Columns: anchor and positive
• Approximate statistics based on the first 1000 samples:

  	anchor	positive
  type	string	string
  details	min: 51 tokens mean: 676.52 tokens max: 1868 tokens	min: 16 tokens mean: 249.47 tokens max: 3445 tokens

• Samples:

  anchor	positive
  We will consider the numbers $$$a$$$ and $$$b$$$ as adjacent if they differ by exactly one, that is, $$$|a-b|=1$$$. We will consider cells of a square matrix $$$n \times n$$$ as adjacent if they have a common side, that is, for cell $$$(r, c)$$$ cells $$$(r, c-1)$$$, $$$(r, c+1)$$$, $$$(r-1, c)$$$ and $$$(r+1, c)$$$ are adjacent to it. For a given number $$$n$$$, construct a square matrix $$$n \times n$$$ such that: * Each integer from $$$1$$$ to $$$n^2$$$ occurs in this matrix exactly once; * If $$$(r_1, c_1)$$$ and $$$(r_2, c_2)$$$ are adjacent cells, then the numbers written in them must not be adjacent . Input The first line contains one integer $$$t$$$ ($$$1 \le t \le 100$$$). Then $$$t$$$ test cases follow. Each test case is characterized by one integer $$$n$$$ ($$$1 \le n \le 100$$$). Output For each test case, output: * -1 , if the required matrix does not exist; * the required matrix, otherwise (any such matrix if...	import sys import os.path from collections import * import math import bisect if (os.path.exists('input.txt')): sys.stdin = open("input.txt", "r") sys.stdout = open("output.txt", "w") ########################################################## t = int(input()) while t: t -= 1 n = int(input()) if n == 2: print(-1) else: arr = [[0 for i in range(n)] for i in range(n)] x = 1 for i in range(n): if(i % 2 == 0): for j in range(0,n,2): arr[i][j] = x x += 1 else: for j in range(1,n,2): arr[i][j] = x x += 1 for i in range(n): if i % 2: for j in range(0,n,2): arr[i][j] = x x += 1 else: for j in range(1,n,2): arr[i][j] = x x += 1 for i in arr: ...
  You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. Input The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct . It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. Example Input 3 2 3 1 ...	import sys input = sys.stdin.readline def pleasantPairs(n, arr): res = 0 for j in range(1, n+1): if arr[j] >= 2*j: continue tmp = 1 while (tmp * arr[j]) < (2 * j): i = tmp * arr[j] - j if i > 0 and arr[i] == tmp: res += 1 tmp += 1 return res def main(): for t in range(int(input())): n = int(input()) arr = [-1] for i in input().split(): arr.append(int(i)) print(pleasantPairs(n, arr)) if name == 'main': main()
  Let's call a positive integer $$$n$$$ ordinary if in the decimal notation all its digits are the same. For example, $$$1$$$, $$$2$$$ and $$$99$$$ are ordinary numbers, but $$$719$$$ and $$$2021$$$ are not ordinary numbers. For a given number $$$n$$$, find the number of ordinary numbers among the numbers from $$$1$$$ to $$$n$$$. Input The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$). Then $$$t$$$ test cases follow. Each test case is characterized by one integer $$$n$$$ ($$$1 \le n \le 10^9$$$). Output For each test case output the number of ordinary numbers among numbers from $$$1$$$ to $$$n$$$. Example Input 6 1 2 3 4 5 100 Output 1 2 3 4 5 18	# Ordenary numbers t = int(input()) for _ in range(t): n = int(input()) digit = 0 test = n num = 0 while test: digit += 1 test //= 10 # print(digit,n) test = 10 while test: num = 0 test -= 1 for i in range(digit): num = num * 10 + test if num <= n: break num = 0 num = digit * test num +=( (digit - 1) * (9 - test)) print(num)

• Loss: CachedMultipleNegativesRankingLoss with these parameters:

  {
      "scale": 15,
      "similarity_fct": "cos_sim",
      "mini_batch_size": 2,
      "gather_across_devices": false
  }
  

Training Hyperparameters

Non-Default Hyperparameters

• per_device_train_batch_size: 1
• per_device_eval_batch_size: 1
• num_train_epochs: 1
• lr_scheduler_type: constant
• bf16: True
• batch_sampler: no_duplicates

All Hyperparameters

Click to expand

• overwrite_output_dir: False
• do_predict: False
• eval_strategy: no
• prediction_loss_only: True
• per_device_train_batch_size: 1
• per_device_eval_batch_size: 1
• per_gpu_train_batch_size: None
• per_gpu_eval_batch_size: None
• gradient_accumulation_steps: 1
• eval_accumulation_steps: None
• torch_empty_cache_steps: None
• learning_rate: 5e-05
• weight_decay: 0.0
• adam_beta1: 0.9
• adam_beta2: 0.999
• adam_epsilon: 1e-08
• max_grad_norm: 1.0
• num_train_epochs: 1
• max_steps: -1
• lr_scheduler_type: constant
• lr_scheduler_kwargs: {}
• warmup_ratio: 0.0
• warmup_steps: 0
• log_level: passive
• log_level_replica: warning
• log_on_each_node: True
• logging_nan_inf_filter: True
• save_safetensors: True
• save_on_each_node: False
• save_only_model: False
• restore_callback_states_from_checkpoint: False
• no_cuda: False
• use_cpu: False
• use_mps_device: False
• seed: 42
• data_seed: None
• jit_mode_eval: False
• use_ipex: False
• bf16: True
• fp16: False
• fp16_opt_level: O1
• half_precision_backend: auto
• bf16_full_eval: False
• fp16_full_eval: False
• tf32: None
• local_rank: 0
• ddp_backend: None
• tpu_num_cores: None
• tpu_metrics_debug: False
• debug: []
• dataloader_drop_last: False
• dataloader_num_workers: 0
• dataloader_prefetch_factor: None
• past_index: -1
• disable_tqdm: False
• remove_unused_columns: True
• label_names: None
• load_best_model_at_end: False
• ignore_data_skip: False
• fsdp: []
• fsdp_min_num_params: 0
• fsdp_config: {'min_num_params': 0, 'xla': False, 'xla_fsdp_v2': False, 'xla_fsdp_grad_ckpt': False}
• fsdp_transformer_layer_cls_to_wrap: None
• accelerator_config: {'split_batches': False, 'dispatch_batches': None, 'even_batches': True, 'use_seedable_sampler': True, 'non_blocking': False, 'gradient_accumulation_kwargs': None}
• parallelism_config: None
• deepspeed: None
• label_smoothing_factor: 0.0
• optim: adamw_torch_fused
• optim_args: None
• adafactor: False
• group_by_length: False
• length_column_name: length
• ddp_find_unused_parameters: None
• ddp_bucket_cap_mb: None
• ddp_broadcast_buffers: False
• dataloader_pin_memory: True
• dataloader_persistent_workers: False
• skip_memory_metrics: True
• use_legacy_prediction_loop: False
• push_to_hub: False
• resume_from_checkpoint: None
• hub_model_id: None
• hub_strategy: every_save
• hub_private_repo: None
• hub_always_push: False
• hub_revision: None
• gradient_checkpointing: False
• gradient_checkpointing_kwargs: None
• include_inputs_for_metrics: False
• include_for_metrics: []
• eval_do_concat_batches: True
• fp16_backend: auto
• push_to_hub_model_id: None
• push_to_hub_organization: None
• mp_parameters:
• auto_find_batch_size: False
• full_determinism: False
• torchdynamo: None
• ray_scope: last
• ddp_timeout: 1800
• torch_compile: False
• torch_compile_backend: None
• torch_compile_mode: None
• include_tokens_per_second: False
• include_num_input_tokens_seen: False
• neftune_noise_alpha: None
• optim_target_modules: None
• batch_eval_metrics: False
• eval_on_start: False
• use_liger_kernel: False
• liger_kernel_config: None
• eval_use_gather_object: False
• average_tokens_across_devices: False
• prompts: None
• batch_sampler: no_duplicates
• multi_dataset_batch_sampler: proportional
• router_mapping: {}
• learning_rate_mapping: {}

Training Logs

Click to expand

Epoch	Step	Training Loss
0.0010	1000	0.2449
0.0021	2000	0.2364
0.0031	3000	0.2453
0.0041	4000	0.2412
0.0052	5000	0.2484
0.0062	6000	0.2447
0.0072	7000	0.2475
0.0083	8000	0.2414
0.0093	9000	0.2361
0.0103	10000	0.2405
0.0114	11000	0.2419
0.0124	12000	0.2416
0.0134	13000	0.2444
0.0145	14000	0.2429
0.0155	15000	0.2412
0.0165	16000	0.2454
0.0175	17000	0.2375
0.0186	18000	0.2394
0.0196	19000	0.2407
0.0206	20000	0.2449
0.0217	21000	0.2444
0.0227	22000	0.2414
0.0237	23000	0.241
0.0248	24000	0.2417
0.0258	25000	0.2454
0.0268	26000	0.2412
0.0279	27000	0.2425
0.0289	28000	0.2483
0.0299	29000	0.2416
0.0310	30000	0.2397
0.0320	31000	0.24
0.0330	32000	0.2458
0.0341	33000	0.2479
0.0351	34000	0.2373
0.0361	35000	0.2503
0.0372	36000	0.2403
0.0382	37000	0.25
0.0392	38000	0.2387
0.0403	39000	0.2388
0.0413	40000	0.2475
0.0423	41000	0.2347
0.0434	42000	0.2412
0.0444	43000	0.2505
0.0454	44000	0.2422
0.0465	45000	0.2473
0.0475	46000	0.2401
0.0485	47000	0.241
0.0495	48000	0.2399
0.0506	49000	0.2471
0.0516	50000	0.2414
0.0526	51000	0.242
0.0537	52000	0.2413
0.0547	53000	0.2358
0.0557	54000	0.239
0.0568	55000	0.2399
0.0578	56000	0.2492
0.0588	57000	0.2402
0.0599	58000	0.2483
0.0609	59000	0.2476
0.0619	60000	0.2467
0.0630	61000	0.2393
0.0640	62000	0.2485
0.0650	63000	0.246
0.0661	64000	0.2447
0.0671	65000	0.2382
0.0681	66000	0.2472
0.0692	67000	0.2426
0.0702	68000	0.2425
0.0712	69000	0.2393
0.0723	70000	0.2452
0.0733	71000	0.2442
0.0743	72000	0.2418
0.0754	73000	0.2501
0.0764	74000	0.24
0.0774	75000	0.2448
0.0785	76000	0.2373
0.0795	77000	0.241
0.0805	78000	0.247
0.0815	79000	0.2455
0.0826	80000	0.2352
0.0836	81000	0.2404
0.0846	82000	0.2439
0.0857	83000	0.2389
0.0867	84000	0.2515
0.0877	85000	0.2425
0.0888	86000	0.246
0.0898	87000	0.2433
0.0908	88000	0.238
0.0919	89000	0.242
0.0929	90000	0.2428
0.0939	91000	0.2412
0.0950	92000	0.2489
0.0960	93000	0.2407
0.0970	94000	0.2471
0.0981	95000	0.2341
0.0991	96000	0.2389
0.1001	97000	0.2484
0.1012	98000	0.2498
0.1022	99000	0.2441
0.1032	100000	0.2463
0.1043	101000	0.2406
0.1053	102000	0.2377
0.1063	103000	0.2426
0.1074	104000	0.2406
0.1084	105000	0.2417
0.1094	106000	0.2519
0.1105	107000	0.2426
0.1115	108000	0.2384
0.1125	109000	0.2422
0.1135	110000	0.2475
0.1146	111000	0.2421
0.1156	112000	0.2425
0.1166	113000	0.2456
0.1177	114000	0.2407
0.1187	115000	0.2464
0.1197	116000	0.2406
0.1208	117000	0.2479
0.1218	118000	0.2403
0.1228	119000	0.2402
0.1239	120000	0.2422
0.1249	121000	0.2485
0.1259	122000	0.2432
0.1270	123000	0.2414
0.1280	124000	0.2495
0.1290	125000	0.2333
0.1301	126000	0.2493
0.1311	127000	0.2469
0.1321	128000	0.2365
0.1332	129000	0.2419
0.1342	130000	0.2438
0.1352	131000	0.2402
0.1363	132000	0.236
0.1373	133000	0.2431
0.1383	134000	0.2379
0.1394	135000	0.2525
0.1404	136000	0.2414
0.1414	137000	0.2361
0.1425	138000	0.2405
0.1435	139000	0.2383
0.1445	140000	0.2494
0.1455	141000	0.2425
0.1466	142000	0.2462
0.1476	143000	0.2352
0.1486	144000	0.2422
0.1497	145000	0.2458
0.1507	146000	0.2463
0.1517	147000	0.2441
0.1528	148000	0.2438
0.1538	149000	0.2444
0.1548	150000	0.2546
0.1559	151000	0.2382
0.1569	152000	0.2464
0.1579	153000	0.2369
0.1590	154000	0.2475
0.1600	155000	0.244
0.1610	156000	0.2459
0.1621	157000	0.2506
0.1631	158000	0.2412
0.1641	159000	0.2427
0.1652	160000	0.2391
0.1662	161000	0.2382
0.1672	162000	0.2412
0.1683	163000	0.241
0.1693	164000	0.2459
0.1703	165000	0.2377
0.1714	166000	0.2436
0.1724	167000	0.2429
0.1734	168000	0.2378
0.1744	169000	0.2458
0.1755	170000	0.2474
0.1765	171000	0.242
0.1775	172000	0.2433
0.1786	173000	0.2398
0.1796	174000	0.2509
0.1806	175000	0.2441
0.1817	176000	0.2443
0.1827	177000	0.2406
0.1837	178000	0.2476
0.1848	179000	0.2377
0.1858	180000	0.2413
0.1868	181000	0.2424
0.1879	182000	0.2471
0.1889	183000	0.2465
0.1899	184000	0.2469
0.1910	185000	0.2442
0.1920	186000	0.2375
0.1930	187000	0.2428
0.1941	188000	0.245
0.1951	189000	0.2454
0.1961	190000	0.2387
0.1972	191000	0.2494
0.1982	192000	0.239
0.1992	193000	0.2464
0.2003	194000	0.2468
0.2013	195000	0.2413
0.2023	196000	0.24
0.2034	197000	0.2462
0.2044	198000	0.2494
0.2054	199000	0.2436
0.2064	200000	0.2399
0.2075	201000	0.2427
0.2085	202000	0.2432
0.2095	203000	0.2388
0.2106	204000	0.2432
0.2116	205000	0.2426
0.2126	206000	0.2494
0.2137	207000	0.2457
0.2147	208000	0.2425
0.2157	209000	0.2474
0.2168	210000	0.2461
0.2178	211000	0.2464
0.2188	212000	0.2491
0.2199	213000	0.2438
0.2209	214000	0.2461
0.2219	215000	0.2398
0.2230	216000	0.24
0.2240	217000	0.2369
0.2250	218000	0.241
0.2261	219000	0.2478
0.2271	220000	0.2494
0.2281	221000	0.2485
0.2292	222000	0.2492
0.2302	223000	0.2494
0.2312	224000	0.2464
0.2323	225000	0.2436
0.2333	226000	0.2486
0.2343	227000	0.2493
0.2354	228000	0.2409
0.2364	229000	0.2453
0.2374	230000	0.2468
0.2384	231000	0.2454
0.2395	232000	0.2424
0.2405	233000	0.2437
0.2415	234000	0.2511
0.2426	235000	0.2424
0.2436	236000	0.2455
0.2446	237000	0.2405
0.2457	238000	0.2447
0.2467	239000	0.2424
0.2477	240000	0.2417
0.2488	241000	0.2429
0.2498	242000	0.2434
0.2508	243000	0.2413
0.2519	244000	0.2433
0.2529	245000	0.2401
0.2539	246000	0.2445
0.2550	247000	0.2457
0.2560	248000	0.2453
0.2570	249000	0.2439
0.2581	250000	0.2467
0.2591	251000	0.2439
0.2601	252000	0.2404
0.2612	253000	0.2426
0.2622	254000	0.2448
0.2632	255000	0.2431
0.2643	256000	0.234
0.2653	257000	0.25
0.2663	258000	0.2507
0.2674	259000	0.2501
0.2684	260000	0.2403
0.2694	261000	0.2444
0.2704	262000	0.2468
0.2715	263000	0.2407
0.2725	264000	0.2436
0.2735	265000	0.2464
0.2746	266000	0.2419
0.2756	267000	0.2493
0.2766	268000	0.2482
0.2777	269000	0.2425
0.2787	270000	0.243
0.2797	271000	0.244
0.2808	272000	0.2419
0.2818	273000	0.2448
0.2828	274000	0.2429
0.2839	275000	0.2474
0.2849	276000	0.2483
0.2859	277000	0.237
0.2870	278000	0.2463
0.2880	279000	0.2431
0.2890	280000	0.2469
0.2901	281000	0.2362
0.2911	282000	0.2419
0.2921	283000	0.2461
0.2932	284000	0.2426
0.2942	285000	0.2422
0.2952	286000	0.2458
0.2963	287000	0.249
0.2973	288000	0.243
0.2983	289000	0.2434
0.2994	290000	0.2413
0.3004	291000	0.2476
0.3014	292000	0.2446
0.3024	293000	0.2468
0.3035	294000	0.2411
0.3045	295000	0.2463
0.3055	296000	0.2464
0.3066	297000	0.2394
0.3076	298000	0.2503
0.3086	299000	0.2427
0.3097	300000	0.2515
0.3107	301000	0.2428
0.3117	302000	0.2478
0.3128	303000	0.2425
0.3138	304000	0.2474
0.3148	305000	0.2401
0.3159	306000	0.2455
0.3169	307000	0.2463
0.3179	308000	0.2441
0.3190	309000	0.2487
0.3200	310000	0.2385
0.3210	311000	0.2442
0.3221	312000	0.2454
0.3231	313000	0.2333
0.3241	314000	0.2393
0.3252	315000	0.2465
0.3262	316000	0.2349
0.3272	317000	0.2405
0.3283	318000	0.2404
0.3293	319000	0.2434
0.3303	320000	0.243
0.3314	321000	0.2409
0.3324	322000	0.2398
0.3334	323000	0.2491
0.3344	324000	0.246
0.3355	325000	0.2449
0.3365	326000	0.246
0.3375	327000	0.2445
0.3386	328000	0.2484
0.3396	329000	0.2379
0.3406	330000	0.2439
0.3417	331000	0.2385
0.3427	332000	0.2444
0.3437	333000	0.2451
0.3448	334000	0.243
0.3458	335000	0.2455
0.3468	336000	0.2413
0.3479	337000	0.2436
0.3489	338000	0.2387
0.3499	339000	0.2443
0.3510	340000	0.2396
0.3520	341000	0.2385
0.3530	342000	0.2435
0.3541	343000	0.2384
0.3551	344000	0.2386
0.3561	345000	0.2422
0.3572	346000	0.2477
0.3582	347000	0.2417
0.3592	348000	0.2466
0.3603	349000	0.2456
0.3613	350000	0.2446
0.3623	351000	0.2375
0.3634	352000	0.2358
0.3644	353000	0.2373
0.3654	354000	0.2445
0.3664	355000	0.2378
0.3675	356000	0.2459
0.3685	357000	0.2476
0.3695	358000	0.2437
0.3706	359000	0.242
0.3716	360000	0.2404
0.3726	361000	0.2473
0.3737	362000	0.2461
0.3747	363000	0.2412
0.3757	364000	0.24
0.3768	365000	0.2426
0.3778	366000	0.242
0.3788	367000	0.2428
0.3799	368000	0.2446
0.3809	369000	0.2453
0.3819	370000	0.2441
0.3830	371000	0.2529
0.3840	372000	0.246
0.3850	373000	0.2396
0.3861	374000	0.2422
0.3871	375000	0.2469
0.3881	376000	0.2436
0.3892	377000	0.2422
0.3902	378000	0.2416
0.3912	379000	0.2501
0.3923	380000	0.2443
0.3933	381000	0.2381
0.3943	382000	0.2392
0.3954	383000	0.2422
0.3964	384000	0.2496
0.3974	385000	0.243
0.3984	386000	0.2431
0.3995	387000	0.2426
0.4005	388000	0.2441
0.4015	389000	0.2434
0.4026	390000	0.2423
0.4036	391000	0.2444
0.4046	392000	0.2387
0.4057	393000	0.2386
0.4067	394000	0.2387
0.4077	395000	0.2467
0.4088	396000	0.2413
0.4098	397000	0.2471
0.4108	398000	0.2435
0.4119	399000	0.2424
0.4129	400000	0.2446
0.4139	401000	0.243
0.4150	402000	0.2436
0.4160	403000	0.2384
0.4170	404000	0.2404
0.4181	405000	0.2375
0.4191	406000	0.2455
0.4201	407000	0.2379
0.4212	408000	0.2487
0.4222	409000	0.244
0.4232	410000	0.2368
0.4243	411000	0.2415
0.4253	412000	0.2505
0.4263	413000	0.239
0.4274	414000	0.2431
0.4284	415000	0.2385
0.4294	416000	0.2515
0.4304	417000	0.2458
0.4315	418000	0.2433
0.4325	419000	0.239
0.4335	420000	0.2461
0.4346	421000	0.2474
0.4356	422000	0.2419
0.4366	423000	0.2451
0.4377	424000	0.2553
0.4387	425000	0.2486
0.4397	426000	0.2403
0.4408	427000	0.2429
0.4418	428000	0.2464
0.4428	429000	0.243
0.4439	430000	0.2446
0.4449	431000	0.2419
0.4459	432000	0.2407
0.4470	433000	0.2425
0.4480	434000	0.2429
0.4490	435000	0.2424
0.4501	436000	0.2373
0.4511	437000	0.2493
0.4521	438000	0.2403
0.4532	439000	0.2424
0.4542	440000	0.2501
0.4552	441000	0.2414
0.4563	442000	0.2472
0.4573	443000	0.2367
0.4583	444000	0.2481
0.4594	445000	0.2439
0.4604	446000	0.2466
0.4614	447000	0.241
0.4624	448000	0.2425
0.4635	449000	0.2367
0.4645	450000	0.2421
0.4655	451000	0.2386
0.4666	452000	0.2357
0.4676	453000	0.2464
0.4686	454000	0.2422
0.4697	455000	0.2443
0.4707	456000	0.2414
0.4717	457000	0.2483
0.4728	458000	0.2423
0.4738	459000	0.2386
0.4748	460000	0.2445
0.4759	461000	0.2453
0.4769	462000	0.2363
0.4779	463000	0.2432
0.4790	464000	0.2455
0.4800	465000	0.2376
0.4810	466000	0.2441
0.4821	467000	0.2466
0.4831	468000	0.2379
0.4841	469000	0.2429
0.4852	470000	0.2454
0.4862	471000	0.2439
0.4872	472000	0.2466
0.4883	473000	0.2459
0.4893	474000	0.2431
0.4903	475000	0.2429
0.4913	476000	0.2421
0.4924	477000	0.2405
0.4934	478000	0.2429
0.4944	479000	0.2435
0.4955	480000	0.2422
0.4965	481000	0.2419
0.4975	482000	0.2467
0.4986	483000	0.2409
0.4996	484000	0.2424
0.5006	485000	0.2452
0.5017	486000	0.2416
0.5027	487000	0.2455
0.5037	488000	0.2467
0.5048	489000	0.2467
0.5058	490000	0.243
0.5068	491000	0.2397
0.5079	492000	0.2385
0.5089	493000	0.2425
0.5099	494000	0.2501
0.5110	495000	0.2433
0.5120	496000	0.2461
0.5130	497000	0.2381
0.5141	498000	0.2464
0.5151	499000	0.248
0.5161	500000	0.2478
0.5172	501000	0.2386
0.5182	502000	0.2427
0.5192	503000	0.2484
0.5203	504000	0.2403
0.5213	505000	0.2398
0.5223	506000	0.2468
0.5233	507000	0.2444
0.5244	508000	0.2445
0.5254	509000	0.2446
0.5264	510000	0.2463
0.5275	511000	0.2495
0.5285	512000	0.2471
0.5295	513000	0.2402
0.5306	514000	0.2404
0.5316	515000	0.2442
0.5326	516000	0.2398
0.5337	517000	0.2377
0.5347	518000	0.2463
0.5357	519000	0.2418
0.5368	520000	0.2532
0.5378	521000	0.2462
0.5388	522000	0.2419
0.5399	523000	0.246
0.5409	524000	0.2365
0.5419	525000	0.2419
0.5430	526000	0.248
0.5440	527000	0.2464
0.5450	528000	0.2395
0.5461	529000	0.2405
0.5471	530000	0.24
0.5481	531000	0.2388
0.5492	532000	0.2418
0.5502	533000	0.2473
0.5512	534000	0.2448
0.5523	535000	0.2458
0.5533	536000	0.2447
0.5543	537000	0.2364
0.5553	538000	0.2483
0.5564	539000	0.2365
0.5574	540000	0.2463
0.5584	541000	0.2437
0.5595	542000	0.2387
0.5605	543000	0.2411
0.5615	544000	0.2451
0.5626	545000	0.2486
0.5636	546000	0.2374
0.5646	547000	0.2384
0.5657	548000	0.2428
0.5667	549000	0.2443
0.5677	550000	0.2404
0.5688	551000	0.2391
0.5698	552000	0.2449
0.5708	553000	0.2484
0.5719	554000	0.2422
0.5729	555000	0.2523
0.5739	556000	0.2423
0.5750	557000	0.2435
0.5760	558000	0.2373
0.5770	559000	0.2481
0.5781	560000	0.2407
0.5791	561000	0.2431
0.5801	562000	0.243
0.5812	563000	0.2457
0.5822	564000	0.2459
0.5832	565000	0.2461
0.5843	566000	0.2392
0.5853	567000	0.2416
0.5863	568000	0.245
0.5873	569000	0.2437
0.5884	570000	0.2438
0.5894	571000	0.2416
0.5904	572000	0.249
0.5915	573000	0.2409
0.5925	574000	0.2427
0.5935	575000	0.2473
0.5946	576000	0.2429
0.5956	577000	0.2429
0.5966	578000	0.242
0.5977	579000	0.2418
0.5987	580000	0.2483
0.5997	581000	0.2434
0.6008	582000	0.241
0.6018	583000	0.2393
0.6028	584000	0.2379
0.6039	585000	0.2378
0.6049	586000	0.2427
0.6059	587000	0.2417
0.6070	588000	0.2446
0.6080	589000	0.2469
0.6090	590000	0.2478
0.6101	591000	0.2489
0.6111	592000	0.2415
0.6121	593000	0.2446
0.6132	594000	0.2389
0.6142	595000	0.2381
0.6152	596000	0.2409
0.6163	597000	0.2538
0.6173	598000	0.2408
0.6183	599000	0.2436
0.6193	600000	0.2448
0.6204	601000	0.2393
0.6214	602000	0.2415
0.6224	603000	0.2404
0.6235	604000	0.2444
0.6245	605000	0.2403
0.6255	606000	0.2357
0.6266	607000	0.2502
0.6276	608000	0.2425
0.6286	609000	0.2408
0.6297	610000	0.2437
0.6307	611000	0.2416
0.6317	612000	0.2393
0.6328	613000	0.2419
0.6338	614000	0.2419
0.6348	615000	0.2395
0.6359	616000	0.2459
0.6369	617000	0.2458
0.6379	618000	0.2432
0.6390	619000	0.2474
0.6400	620000	0.235
0.6410	621000	0.2468
0.6421	622000	0.2397
0.6431	623000	0.2363
0.6441	624000	0.2442
0.6452	625000	0.2487
0.6462	626000	0.245
0.6472	627000	0.2461
0.6483	628000	0.2396
0.6493	629000	0.2429
0.6503	630000	0.2448
0.6513	631000	0.2431
0.6524	632000	0.2502
0.6534	633000	0.2525
0.6544	634000	0.2407
0.6555	635000	0.2448
0.6565	636000	0.2442
0.6575	637000	0.2417
0.6586	638000	0.2486
0.6596	639000	0.2507
0.6606	640000	0.2398
0.6617	641000	0.2471
0.6627	642000	0.2435
0.6637	643000	0.2437
0.6648	644000	0.2446
0.6658	645000	0.2467
0.6668	646000	0.2467
0.6679	647000	0.2364
0.6689	648000	0.2418
0.6699	649000	0.245
0.6710	650000	0.243
0.6720	651000	0.2387
0.6730	652000	0.2424
0.6741	653000	0.2365
0.6751	654000	0.2366
0.6761	655000	0.2425
0.6772	656000	0.2369
0.6782	657000	0.239
0.6792	658000	0.2447
0.6803	659000	0.2417
0.6813	660000	0.2457
0.6823	661000	0.2452
0.6833	662000	0.242
0.6844	663000	0.255
0.6854	664000	0.2451
0.6864	665000	0.2375
0.6875	666000	0.2441
0.6885	667000	0.2509
0.6895	668000	0.2415
0.6906	669000	0.246
0.6916	670000	0.24
0.6926	671000	0.2372
0.6937	672000	0.2504
0.6947	673000	0.2443
0.6957	674000	0.2444
0.6968	675000	0.2448
0.6978	676000	0.2411
0.6988	677000	0.2437
0.6999	678000	0.2432
0.7009	679000	0.2475
0.7019	680000	0.2352
0.7030	681000	0.246
0.7040	682000	0.2451
0.7050	683000	0.2406
0.7061	684000	0.2428
0.7071	685000	0.2433
0.7081	686000	0.2415
0.7092	687000	0.2479
0.7102	688000	0.2405
0.7112	689000	0.2379
0.7123	690000	0.2389
0.7133	691000	0.2414
0.7143	692000	0.2398
0.7153	693000	0.2369
0.7164	694000	0.2465
0.7174	695000	0.2426
0.7184	696000	0.2417
0.7195	697000	0.2422
0.7205	698000	0.2386
0.7215	699000	0.2412
0.7226	700000	0.2508
0.7236	701000	0.241
0.7246	702000	0.2427
0.7257	703000	0.2345
0.7267	704000	0.2465
0.7277	705000	0.2483
0.7288	706000	0.2395
0.7298	707000	0.2435
0.7308	708000	0.2434
0.7319	709000	0.2416
0.7329	710000	0.2421
0.7339	711000	0.2412
0.7350	712000	0.2389
0.7360	713000	0.2452
0.7370	714000	0.2428
0.7381	715000	0.2422
0.7391	716000	0.2364
0.7401	717000	0.2439
0.7412	718000	0.2435
0.7422	719000	0.2401
0.7432	720000	0.2441
0.7443	721000	0.2482
0.7453	722000	0.2417
0.7463	723000	0.2474
0.7473	724000	0.2451
0.7484	725000	0.2387
0.7494	726000	0.2422
0.7504	727000	0.2453
0.7515	728000	0.2437
0.7525	729000	0.2481
0.7535	730000	0.2451
0.7546	731000	0.2426
0.7556	732000	0.242
0.7566	733000	0.245
0.7577	734000	0.2425
0.7587	735000	0.2413
0.7597	736000	0.2501
0.7608	737000	0.2441
0.7618	738000	0.2442
0.7628	739000	0.2468
0.7639	740000	0.2419
0.7649	741000	0.2409
0.7659	742000	0.2423
0.7670	743000	0.2493
0.7680	744000	0.2469
0.7690	745000	0.249
0.7701	746000	0.2423
0.7711	747000	0.2414
0.7721	748000	0.2408
0.7732	749000	0.2407
0.7742	750000	0.2477
0.7752	751000	0.2403
0.7763	752000	0.2504
0.7773	753000	0.2358
0.7783	754000	0.2465
0.7793	755000	0.2432
0.7804	756000	0.2407
0.7814	757000	0.2407
0.7824	758000	0.2407
0.7835	759000	0.2441
0.7845	760000	0.2426
0.7855	761000	0.2395
0.7866	762000	0.2428
0.7876	763000	0.2403
0.7886	764000	0.2394
0.7897	765000	0.2424
0.7907	766000	0.2489
0.7917	767000	0.2417
0.7928	768000	0.2423
0.7938	769000	0.2366
0.7948	770000	0.2508
0.7959	771000	0.2451
0.7969	772000	0.2454
0.7979	773000	0.2419
0.7990	774000	0.2498
0.8000	775000	0.2406
0.8010	776000	0.2344
0.8021	777000	0.2518
0.8031	778000	0.249
0.8041	779000	0.2383
0.8052	780000	0.2442
0.8062	781000	0.243
0.8072	782000	0.2422
0.8082	783000	0.2476
0.8093	784000	0.2417
0.8103	785000	0.2417
0.8113	786000	0.2445
0.8124	787000	0.2414
0.8134	788000	0.2517
0.8144	789000	0.2431
0.8155	790000	0.2467
0.8165	791000	0.2483
0.8175	792000	0.2397
0.8186	793000	0.2467
0.8196	794000	0.2438
0.8206	795000	0.247
0.8217	796000	0.2446
0.8227	797000	0.2372
0.8237	798000	0.2362
0.8248	799000	0.2421
0.8258	800000	0.2404
0.8268	801000	0.2376
0.8279	802000	0.2406
0.8289	803000	0.2497
0.8299	804000	0.2475
0.8310	805000	0.2418
0.8320	806000	0.2379
0.8330	807000	0.2431
0.8341	808000	0.2472
0.8351	809000	0.2485
0.8361	810000	0.2498
0.8372	811000	0.2424
0.8382	812000	0.2434
0.8392	813000	0.2452
0.8402	814000	0.2396
0.8413	815000	0.2472
0.8423	816000	0.2529
0.8433	817000	0.2462
0.8444	818000	0.2436
0.8454	819000	0.2435
0.8464	820000	0.2417
0.8475	821000	0.2453
0.8485	822000	0.2455
0.8495	823000	0.2374
0.8506	824000	0.2416
0.8516	825000	0.2432
0.8526	826000	0.2459
0.8537	827000	0.2407
0.8547	828000	0.247
0.8557	829000	0.2399
0.8568	830000	0.2346
0.8578	831000	0.2446
0.8588	832000	0.2399
0.8599	833000	0.2433
0.8609	834000	0.2468
0.8619	835000	0.2464
0.8630	836000	0.2505
0.8640	837000	0.2437
0.8650	838000	0.2419
0.8661	839000	0.2449
0.8671	840000	0.2504
0.8681	841000	0.2415
0.8692	842000	0.2421
0.8702	843000	0.244
0.8712	844000	0.2458
0.8722	845000	0.252
0.8733	846000	0.2463
0.8743	847000	0.2492
0.8753	848000	0.2405
0.8764	849000	0.2481
0.8774	850000	0.2535
0.8784	851000	0.2379
0.8795	852000	0.2491
0.8805	853000	0.247
0.8815	854000	0.2435
0.8826	855000	0.2423
0.8836	856000	0.239
0.8846	857000	0.2439
0.8857	858000	0.24
0.8867	859000	0.2451
0.8877	860000	0.2424
0.8888	861000	0.2473
0.8898	862000	0.2402
0.8908	863000	0.2398
0.8919	864000	0.2457
0.8929	865000	0.2373
0.8939	866000	0.2426
0.8950	867000	0.2392
0.8960	868000	0.2432
0.8970	869000	0.242
0.8981	870000	0.2488
0.8991	871000	0.2429
0.9001	872000	0.2487
0.9012	873000	0.2443
0.9022	874000	0.2496
0.9032	875000	0.2497
0.9042	876000	0.245
0.9053	877000	0.233
0.9063	878000	0.2452
0.9073	879000	0.2467
0.9084	880000	0.2443
0.9094	881000	0.2477
0.9104	882000	0.249
0.9115	883000	0.2394
0.9125	884000	0.2413
0.9135	885000	0.2436
0.9146	886000	0.2451
0.9156	887000	0.2435
0.9166	888000	0.2394
0.9177	889000	0.2486
0.9187	890000	0.2502
0.9197	891000	0.2479
0.9208	892000	0.2388
0.9218	893000	0.2325
0.9228	894000	0.2395
0.9239	895000	0.2418
0.9249	896000	0.2287
0.9259	897000	0.2448
0.9270	898000	0.2418
0.9280	899000	0.2536
0.9290	900000	0.2488
0.9301	901000	0.2395
0.9311	902000	0.2347
0.9321	903000	0.2405
0.9332	904000	0.2387
0.9342	905000	0.2396
0.9352	906000	0.2456
0.9362	907000	0.2434
0.9373	908000	0.2448
0.9383	909000	0.2402
0.9393	910000	0.2363
0.9404	911000	0.2384
0.9414	912000	0.241
0.9424	913000	0.238
0.9435	914000	0.2464
0.9445	915000	0.2478
0.9455	916000	0.2452
0.9466	917000	0.2389
0.9476	918000	0.2462
0.9486	919000	0.2481
0.9497	920000	0.2444
0.9507	921000	0.2436
0.9517	922000	0.2499
0.9528	923000	0.2426
0.9538	924000	0.2472
0.9548	925000	0.2383
0.9559	926000	0.2427
0.9569	927000	0.2416
0.9579	928000	0.2415
0.9590	929000	0.2452
0.9600	930000	0.2387
0.9610	931000	0.2465
0.9621	932000	0.2452
0.9631	933000	0.2446
0.9641	934000	0.2341
0.9652	935000	0.2489
0.9662	936000	0.2437
0.9672	937000	0.2368
0.9682	938000	0.2439
0.9693	939000	0.241
0.9703	940000	0.2333
0.9713	941000	0.2456
0.9724	942000	0.2456
0.9734	943000	0.2474
0.9744	944000	0.2463
0.9755	945000	0.2488
0.9765	946000	0.2431
0.9775	947000	0.2404
0.9786	948000	0.2452
0.9796	949000	0.2397
0.9806	950000	0.2456
0.9817	951000	0.2392
0.9827	952000	0.2399
0.9837	953000	0.2416
0.9848	954000	0.2423
0.9858	955000	0.2353
0.9868	956000	0.2452
0.9879	957000	0.2404
0.9889	958000	0.2453
0.9899	959000	0.2377
0.9910	960000	0.2454
0.9920	961000	0.2412
0.9930	962000	0.2443
0.9941	963000	0.2432
0.9951	964000	0.2485
0.9961	965000	0.2412
0.9972	966000	0.249
0.9982	967000	0.2337
0.9992	968000	0.2454

Framework Versions

• Python: 3.12.11
• Sentence Transformers: 5.1.0
• Transformers: 4.56.1
• PyTorch: 2.8.0+cu128
• Accelerate: 1.10.1
• Datasets: 4.1.1
• Tokenizers: 0.22.1

Citation

BibTeX

Sentence Transformers

@inproceedings{reimers-2019-sentence-bert,
    title = "Sentence-BERT: Sentence Embeddings using Siamese BERT-Networks",
    author = "Reimers, Nils and Gurevych, Iryna",
    booktitle = "Proceedings of the 2019 Conference on Empirical Methods in Natural Language Processing",
    month = "11",
    year = "2019",
    publisher = "Association for Computational Linguistics",
    url = "https://arxiv.org/abs/1908.10084",
}

CachedMultipleNegativesRankingLoss

@misc{gao2021scaling,
    title={Scaling Deep Contrastive Learning Batch Size under Memory Limited Setup},
    author={Luyu Gao and Yunyi Zhang and Jiawei Han and Jamie Callan},
    year={2021},
    eprint={2101.06983},
    archivePrefix={arXiv},
    primaryClass={cs.LG}
}

TripletLoss

@misc{hermans2017defense,
    title={In Defense of the Triplet Loss for Person Re-Identification},
    author={Alexander Hermans and Lucas Beyer and Bastian Leibe},
    year={2017},
    eprint={1703.07737},
    archivePrefix={arXiv},
    primaryClass={cs.CV}
}