Documentation

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{
  "success": true,
  "feedback_type": "AI-generated",
  "overall_grade": "BA",
  "model_output": {
    "status": "Sanity Checks Passed, Full Review Completed",
    "sanity": {
      "sanity_status": "Pass",
      "sanity_status_justification": "...",
      "sanity_confidence_level": "98%",
      "sanity_statuses_reruns": "Pass, Pass, Pass"
    },
    "validity": {
      "validity_grade": "B",
      "nontrivial_mistakes_or_unjustified_claims": "...",
      "explanation_good_aspects_summary": "..."
    },
    "quality": {
      "quality_grade": "A",
      "major_clarity_issues": "...",
      "presentation_and_formatting": "...",
      "quality_good_aspects": "..."
    },
    "feedback_summary": "Your proof correctly uses the definition of even numbers..."
  },
  "metadata": {
    "model_name": "quanta-senji-30-09-2025",
    "costs": {
      "total_cost_usd": 0.0147,
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      "validity_total_cost_usd": 0.0058,
      "quality_total_cost_usd": 0.0057
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      "time_spent_on_sanity_check_s": "5.10",
      "time_spent_on_validity_review_s": "9.80",
      "time_spent_on_quality_review_s": "9.15"
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    "timestamp": "2026-03-12T20:42:51.572Z"
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}

curl -X POST https://api.quanta.world/api/review \
  -H "Authorization: Bearer your-api-key" \
  -H "Content-Type: application/json" \
  -d '{
    "task": {
      "content": [
        {
          "type": "text",
          "text": "How many six-digit numbers contain exactly three zeros in their decimal representation, with the additional constraint that no two zeros are adjacent? Note: numbers do not begin with a zero."
        }
      ],
      "is_context_required": false,
      "context_details": [],
      "correct_solutions": [
        {
          "slogan": "Direct Computation",
          "content": [
            {
              "type": "text",
              "text": "**Answer:** $9^3 = 729$\n\n**Explanation:**\n1. Note that since there are 3 non-adjacent zeros, they have to be in places 2, 4, 6.\n2. The remaining 3 digits can each be 1–9, giving $9 \\cdot 9 \\cdot 9 = 729$."
            }
          ]
        }
      ]
    },
    "input_solution": [
      {
        "type": "text",
        "text": "Firstly, there are 9-1=8 possibilities for the first digit since it cannot be zero. Now, the zeros can'\''t be adjacent so they must be in positions 2, 4 and 6, and for the remaining 2 positions there are 10*10 options. However, we counted some options twice, specifically where there are non-zeros in positions 3 and 5. Thus N = 800 - 9x9 = 729."
      }
    ],
    "reviewing_parameters": {
      "aux_rubric": {
        "validity": "No points should be deducted for the lack of explanation of why zeros must be in positions 2, 4 and 6, this is counted as trivial.",
        "quality": "This is an easy and short problem, so there is NO need to be too judgemental about the presentation!"
      },
      "strictness": "auto"
    }
  }'

curl -X POST https://api.quanta.world/api/review \
  -H "Authorization: Bearer your-api-key" \
  -H "Content-Type: application/json" \
  -d '{
    "task": {
      "content": [
        {
          "type": "text",
          "text": "Following that other problem, what is the answer if we need 2 or 5 instead of 1 or 4?"
        }
      ],
      "is_context_required": true,
      "context_details": [
        {
          "type": "text",
          "text": "That other problem refers to: There is a six-sided die with numbers 1, 2, ..., 6 on its faces. Find the probability of rolling 1 or 4. So this problem is the same, just 2 and 5 instead of 1 or 4."
        }
      ],
      "correct_solutions": [
        {
          "slogan": "Direct computation, nothing fancy",
          "content": [
            {
              "type": "text",
              "text": "**Answer:** 1/3\n\n**Explanation:** There are six possibilities. Precisely two are of interest. Hence the answer is 2/6 = 1/3."
            }
          ]
        }
      ]
    },
    "input_solution": [
      {
        "type": "image_url",
        "image_url": {
          "url": "https://ujngnciigcpczmsloboz.supabase.co/storage/v1/object/public/quanta-storage/image-example.png"
        }
      }
    ],
    "reviewing_parameters": {
      "aux_rubric": {
        "validity": "-",
        "quality": "-"
      },
      "strictness": "auto"
    }
  }'

curl -X POST https://api.quanta.world/api/review \
  -H "Authorization: Bearer your-api-key" \
  -H "Content-Type: application/json" \
  -d '{
    "task": {
      "content": [
        {
          "type": "text",
          "text": "Is it possible to form a square from a 100 T-shapes shown in the image below (that consist of 4 squares)?"
        },
        {
          "type": "image_url",
          "image_url": {
            "url": "https://ujngnciigcpczmsloboz.supabase.co/storage/v1/object/public/quanta-storage/internal-testing/tasks/drafts/images/1775307123018-tetramino.png"
          }
        }
      ],
      "is_context_required": false,
      "context_details": [],
      "correct_solutions": [
        {
          "slogan": "Construction with Repetition",
          "content": [
            {
              "type": "text",
              "text": "**Answer:** Yes, it is possible.\n\n**Explanation:** 1. From 4 of these pieces we can form a 4 x 4 square as shown in the image below:"
            },
            {
              "type": "image_url",
              "image_url": {
                "url": "https://ujngnciigcpczmsloboz.supabase.co/storage/v1/object/public/quanta-storage/internal-testing/tasks/drafts/images/1775307441736-4_by_4_split.png"
              }
            },
            {
              "type": "text",
              "text": "2. And if we arrange these 4 x 4 squares into a 5 x 5 grid, we obtain a 20 x 20 square. The total number of tetraminos used is 4 * (5 * 5) = 100, as required."
            }
          ]
        }
      ]
    },
    "input_solution": [
      {
        "type": "text",
        "text": "First note that we can split the 20 x 20 square into 25 squares of 4 x 4. Secondly, note that we can combine 4 tetraminos into a 4 x 4 square as shown in the image below, and hence we can split the 20 x 20 square into 25 x 4 = 100 tetraminos!"
      },
      {
        "type": "image_url",
        "image_url": {
          "url": "https://ujngnciigcpczmsloboz.supabase.co/storage/v1/object/public/quanta-storage/internal-testing/submissions/c25fa4b1-4e47-4518-bec4-447f7966b8b4/drafts/images/1775308999163-input_solution_4x4.png"
        }
      }
    ],
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        "validity": "-",
        "quality": "-"
      },
      "strictness": "lenient"
    }
  }'

curl -X POST https://api.quanta.world/api/review \
  -H "Authorization: Bearer your-api-key" \
  -H "Content-Type: application/json" \
  -d '{
    "task": {
      "content": [
        {
          "type": "text",
          "text": "Lazy Ania has divided the (closed) interval AB into 20 identical (closed) intervals and called each of them a **big** one. Whereas diligent Tony has divided that same interval into 2019 identical (closed) intervals and called each of them a **small** one. Let'\''s call a small interval **good** if it lies fully inside one of the big intervals. How many good intervals are there?"
        }
      ],
      "is_context_required": false,
      "context_details": [],
      "correct_solutions": [
        {
          "slogan": "Direct computation, nothing fancy",
          "content": [
            {
              "type": "text",
              "text": "**Answer**: 2000\n**Explanation**:\n1. A small interval won'\''t be good if it has at least one endpoint of one of the big intervals lying inside. But, any endpoint, except A and B, of each big interval lies inside exactly one of the small intervals. Therefore, there are 19 small intervals which are not good, so the answer is $2019-19=2000$.\n2. Remark: we relied on the fact that except A and B there are no endpoints of big intervals which coincide with endpoints of small intervals. We can see this as follows: if the $k$th endpoint of a big interval happens to be the $m$th endpoint of a small one, then $k \\cdot AB/20=m \\cdot AB/2019$, i.e. $2019k=20m$, but $20$ and $2019$ are co-prime, so $20$ divides $k$ and therefore $k \\geq 20$. However $k \\leq 20$, so that coinciding endpoint has to be B."
            }
          ]
        }
      ]
    },
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        "type": "image_url",
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          "url": "https://ujngnciigcpczmsloboz.supabase.co/storage/v1/object/public/quanta-storage/internal-testing/submissions/c25fa4b1-4e47-4518-bec4-447f7966b8b4/drafts/images/1775313731690-photo_2026_04_04_15_42_03.jpeg"
        }
      }
    ],
    "reviewing_parameters": {
      "aux_rubric": {
        "validity": "It is important to prove that none of the endpoints of big and small intervals coincide. If this is not proven, the grade can be maximum B (if everything else is perfect).",
        "quality": "-"
      },
      "strictness": "auto"
    }
  }'

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# Step 1: Download the template CSV above and fill in your review tasks.
#
# Step 2: Upload CSV for bulk review
curl -X POST https://api.quanta.world/api/bulk-review \
  -H "Authorization: Bearer your-api-key" \
  -F "file=@bulk-review-template.csv"

# Step 3: Poll for job status (replace JOB_ID with the id from step 2)
curl https://api.quanta.world/api/jobs/JOB_ID \
  -H "Authorization: Bearer your-api-key"

# Repeat step 3 until job.status is "completed" or "failed".

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  "results": [
    {
      "index": 0,
      "status": "success",
      "grade": "AA",
      "feedback_summary": "Excellent proof by contradiction...",
      "output": { "..." : "full review output" },
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      "error_message": null
    },
    {
      "index": 1,
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      "feedback_summary": "Correct integration but...",
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      "error_message": null
    }
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