test: add a real Lambda Feedback export as test data

.pre-commit-config.yaml

@@ -1,6 +1,9 @@
 # TODO: Add isort
 # See: https://github.com/PyCQA/isort/pull/2122
 
+# Real exports are test data: they must stay byte-for-byte as Lambda Feedback wrote them.
+exclude: ^tests/fixtures/exports/
+
 repos:
   - repo: https://github.com/pre-commit/pre-commit-hooks
     rev: v4.3.0

tests/fixtures/exports/README.md

@@ -0,0 +1,70 @@
+# Real Lambda Feedback exports
+
+Each folder here is a question set exactly as Lambda Feedback exported it. Do not edit, reformat
+or re-save these files: their value is that they are what the platform really produces. To add
+coverage, add another export as a new folder.
+
+## What an export contains
+
+```
+set_<Name>.json
+question_000_<Title_with_underscores>.json     # 000 is the question's orderNumber
+question_001_...
+media/question_000_<Title>_0001.png            # referenced from the questions by basename
+```
+
+JSON files are written on a single line.
+
+## Set
+
+`name`, `description`, `isSurvey`, `releasedAt`, `manuallyHidden`, and the visibility of final
+answers, worked solutions, structured tutorials and the chatbot.
+
+## Question
+
+| key | meaning |
+|---|---|
+| `orderNumber` | position in the set, from 0; matches the filename |
+| `title` | shown to students |
+| `skill` | difficulty (exports use 1/3 and 2/3) |
+| `guidance` | a sentence to students about the question's purpose |
+| `durationLowerBound`, `durationUpperBound` | expected minutes |
+| `masterContent` | markdown shared by every part: setup, data, figure |
+| `publish`, `displayFinalAnswer`, `displayStructuredTutorial`, `displayWorkedSolution`, `displayChatbot` | booleans |
+| `parts` | ordered from 0; students see (a), (b), ... |
+
+## Part
+
+| key | meaning |
+|---|---|
+| `content` | markdown for this part |
+| `answerContent` | the final answer shown to students, markdown |
+| `responseAreas` | the answer boxes; may be empty |
+| `workedSolution` | `{"content": ..., "children": []}`; optional. A line containing only `---` (or `***`) splits the content into the steps of the structured tutorial |
+
+## Response area
+
+Every area carries the same keys: `orderNumber`, `preResponseText` and `postResponseText` (labels
+either side of the box), `contentAfter` (markdown shown after the box, before the next one),
+`inputSymbols`, `displayInputSymbols`, `evaluationFunctionName`, `gradeParams`, `livePreview`,
+`includeInPdf`, `saveAllowed`, the feedback settings (`separateFeedback` and the colour and
+prefix fields), `tests`, `cases` and `response`.
+
+`response.responseInput` holds `responseType`, the correct `answer`, and a `config`. The pairings
+in these exports:
+
+| responseType | evaluationFunctionName | answer | notes |
+|---|---|---|---|
+| `MATH_SINGLE_LINE` | `symbolicEqual` | expression, e.g. `(pi/6)*rho*U**2*R**2` | `gradeParams` `{"strict_syntax": false}` |
+| `NUMERIC_UNITS` | `comparePhysicalQuantities` | number and unit, e.g. `0.106 kg` | `gradeParams` includes `rtol`; `config` is null |
+| `MULTIPLE_CHOICE` | `arrayEqual` | list of booleans, one per option | `config` has `single`, `options`, `randomise`; `gradeParams` null |
+
+- `inputSymbols`: `{"symbol": "\\(\\rho\\)", "code": "rho", "aliases": [...], "isVisible": true}`. `symbol` is what students see, `code` what the evaluator reads.
+- `tests`: `{"id", "payload", "expectedResponse": {"isCorrect"}}`, the author's checks of the marking.
+- `cases`: `{"id", "answer", "feedback", "isCorrect", "params"}`. A response matching `answer` is shown `feedback`, and may be marked correct.
+
+## Markdown
+
+Maths uses `$...$` inline and `$$` on its own lines for display, rendered by KaTeX: commands
+KaTeX lacks do not display (degrees written `^\circ`, for example). Images are written
+`![pictureTag](question_000_Title_0001.png){ width=60% }`.

tests/fixtures/exports/me2_introduction/question_000_Hydraulic_scale.json

@@ -0,0 +1 @@
+{"orderNumber":0,"title":"Hydraulic scale","skill":0.3333333333333333,"guidance":"This is a gentle warm up question to get into the swing of things after a long summer.","durationLowerBound":5,"durationUpperBound":10,"masterContent":"A piston of diameter $D = 0.1\\space\\mathrm{m}$ is fitted inside a U-shaped tube filled with liquid mercury (with density in $\\rho_{\\small \\mathrm{Hg}} = 13,540~\\mathrm{kg/m^{3}}$), as shown by the sketch below. The mercury rises by $h=1\\space\\mathrm{mm}$ under the weight of the piston.\n\n![pictureTag](question_000_Hydraulic_scale_0001.png){ width=30% }","publish":true,"displayFinalAnswer":true,"displayStructuredTutorial":true,"displayWorkedSolution":true,"displayChatbot":true,"parts":[{"orderNumber":0,"content":"What is the mass, $m$, of the piston?","answerContent":"$$\nm = \\boxed{0.106 \\space \\mathrm{kg}}\n$$\n","responseAreas":[{"orderNumber":0,"contentAfter":"","preResponseText":"$m=$","postResponseText":"","inputSymbols":[],"displayInputSymbols":false,"includeInPdf":false,"saveAllowed":false,"evaluationFunctionName":"comparePhysicalQuantities","livePreview":false,"gradeParams":{"rtol":0.05,"strict_syntax":false},"separateFeedback":true,"commonFeedbackColor":"#C4CDD5","correctFeedbackColor":"#22C55E","correctFeedbackPrefix":"Correct","incorrectFeedbackColor":"#ff5630","incorrectFeedbackPrefix":"Incorrect","tests":[{"id":"f3ab583b-87c3-44da-8f9b-b8cba1d03030","payload":"0.106 kg","expectedResponse":{"isCorrect":true}},{"id":"b1241594-b1b9-4f17-ba91-0bd329f45863","payload":"0.105 kilos","expectedResponse":{"isCorrect":true}},{"id":"ba11fe62-a4be-4003-bf1e-7e41bfff4ebf","payload":"0.23369 pound","expectedResponse":{"isCorrect":true}}],"cases":[{"id":"c9900536-fcd6-4cf4-8ec6-c8091574534d","answer":"0.106 kilos","feedback":"'Kilos' is ambiguous in general. Be explicit with units next time.","isCorrect":true,"params":null},{"id":"ac267f59-c6fe-4064-ba55-5758ca1e6b6c","answer":"0.106 Kg","feedback":"This is correct. Note that capital `K` is actually the SI unit for kelvin. In future use a lower case `k` for kilo (1,000).","isCorrect":true,"params":null},{"id":"e863f0dd-e4fa-4702-8fe3-a8d7319bba1f","answer":"1724 kg","feedback":"","isCorrect":false,"params":null},{"id":"921a8452-b8e3-4761-b0dd-f1a524e91cc1","answer":"13.54 Kg","feedback":"","isCorrect":false,"params":null}],"response":{"responseInput":{"responseType":"NUMERIC_UNITS","answer":"0.106 kg","config":null}}}],"workedSolution":{"content":"The weight of the piston applies a pressure $p = 4mg /(\\pi D^2)$. This pressure adds to the atmospheric pressure, so that the pressure on the mercury on the left-hand side of the tube is $p + p_{at}$. Applying the hydrostatic equations (or Bernoulli’s equations for steady, inviscid and irrotational fluids) we find:\n\n---\n\n$$\np+p_{\\mathrm{at}}=p_{\\mathrm{at}}+\\rho_\\mathrm{w}\\sigma_{\\mathrm{Hg}}gh\n$$\n\nwhere $\\sigma$ is 'specific gravity' (density relative to water).\n\n---\n\nHence:\n\n---\n\n$$\n\\begin{aligned}\\frac{4mg}{\\pi D^2} &= \\rho_\\mathrm{w} \\sigma_\\mathrm{\\small Hg} g h.\\\\m &= \\frac{\\pi}{4}\\rho_\\mathrm{w} \\sigma_\\mathrm{\\small Hg} h D^2 \\\\&= \\frac{\\pi}{4}* 1000 * 13.54 * 10^{-3} * 10^{-2} \\\\&= \\boxed{0.106\\,\\mathrm{kg}}\\end{aligned}\n$$","children":[]}},{"orderNumber":1,"content":"If this result is to be used to measure the piston's weight, what would you do to improve the accuracy of the measurement?\n","answerContent":"Selecting a fluid with a smaller specific gravity.\n","responseAreas":[],"workedSolution":{"content":"To improve the accuracy of the measurement, $D$, being fixed by the piston's size, it is best to make $h$ bigger (this is what is read), which translates in **selecting a fluid with a smaller specific gravity.**","children":[]}}]}

tests/fixtures/exports/me2_introduction/question_001_Friction_on_a_plate.json

@@ -0,0 +1 @@
+{"orderNumber":1,"title":"Friction on a plate","skill":0.6666666666666666,"guidance":" This question is to refresh yourself using control volumes, which were used in ME1 and will be used often in ME2.","durationLowerBound":15,"durationUpperBound":20,"masterContent":"A fluid with density $\\rho = 800\\,\\mathrm{kg/m}^3$, flows at $U_0 = 3\\,\\mathrm{m/s}$ over a flat plate of length $L = 1\\,\\mathrm{m}$ and width $W = 1\\,\\mathrm{m}$. At the trailing edge the boundary-layer thickness is $\\delta = 25\\,\\mathrm{mm}$. Assume the velocity profile at the trailing edge to be linear (in the image shown), and the flow to be two-dimensional. \n\n![pictureTag](question_001_Friction_on_a_plate_0001.png){ width=60% }","publish":true,"displayFinalAnswer":true,"displayStructuredTutorial":true,"displayWorkedSolution":true,"displayChatbot":true,"parts":[{"orderNumber":0,"content":"Compute the mass flow rate across the top surface of the control volume (noted ''ab'' in the figure).","answerContent":"$$\n\\dot{m}= \\boxed{30 \\space \\mathrm{kg/s} \\>}\n$$","responseAreas":[{"orderNumber":0,"contentAfter":"","preResponseText":"$ \\dot{m}= $","postResponseText":"","inputSymbols":[],"displayInputSymbols":false,"includeInPdf":false,"saveAllowed":false,"evaluationFunctionName":"comparePhysicalQuantities","livePreview":false,"gradeParams":{"rtol":0.05,"strict_syntax":false},"separateFeedback":true,"commonFeedbackColor":"#C4CDD5","correctFeedbackColor":"#22C55E","correctFeedbackPrefix":"Correct","incorrectFeedbackColor":"#ff5630","incorrectFeedbackPrefix":"Incorrect","tests":[{"id":"673a7103-7930-4b11-8a62-98732b27bb5e","payload":"30 kgs-1","expectedResponse":{"isCorrect":false}}],"cases":[{"id":"092d2819-a2b5-481c-b0f9-dbfecfb176fa","answer":"30 kgs-1","feedback":"To enter a negative exponent, put the exponent in parentheses (e.g. ```kg*s^(-1)``` ). The expression ```kgs-1``` is interpreted differently and is incorrect.. ","isCorrect":false,"params":null}],"response":{"responseInput":{"responseType":"NUMERIC_UNITS","answer":"30 kg/s","config":null}}}],"workedSolution":{"content":"From mass conservation:\n\n---\n\n$\\dot{m}_{ad}=\\dot{m}_{ab}+\\dot{m}_{bc}$\n\n---\n\nThis becomes:\n\n---\n\n$\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{U_0\\space \\mathrm{d}y}=\\dot{m}_{ab}+\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{u(y)\\space\\mathrm{d}y}$\n\n---\n\nSince we know that $u(y)$ is linear, we can express it in terms of $y$ and $U_0$:\n\n---\n\n$$\nu(y)=U_0\\frac{y}{\\delta},\n$$\n\n---\n\nTherefore:\n\n$$\n\\begin{aligned}\\dot{m}_{ab}&=\\rho W U_0\\left(\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{1-\\frac{y}{\\delta}\\space\\mathrm{d}y} \\right)\\\\\\dot{m}_{ab}&=\\rho W U_0\\frac{\\delta}{2},\\\\\\dot{m}_{ab}&=(800)(1)(3)\\frac{25\\times 10^{-3}}{2}.\\end{aligned}\n$$\n\n---\n\n$$\n\\dot{m}= \\boxed{30 \\space \\mathrm{kg/s} \\>}\n$$","children":[]}},{"orderNumber":1,"content":"Determine the drag force on the plate.","answerContent":"$$\nF_{\\mathrm{plate}}= \\boxed{30\\space \\mathrm{N}\\space}\n$$","responseAreas":[{"orderNumber":0,"contentAfter":"","preResponseText":"$F_D =$","postResponseText":"","inputSymbols":[],"displayInputSymbols":false,"includeInPdf":false,"saveAllowed":false,"evaluationFunctionName":"comparePhysicalQuantities","livePreview":false,"gradeParams":{"rtol":0.05,"strict_syntax":false},"separateFeedback":true,"commonFeedbackColor":"#C4CDD5","correctFeedbackColor":"#22C55E","correctFeedbackPrefix":"Correct","incorrectFeedbackColor":"#ff5630","incorrectFeedbackPrefix":"Incorrect","tests":[],"cases":[],"response":{"responseInput":{"responseType":"NUMERIC_UNITS","answer":"30 N","config":null}}}],"workedSolution":{"content":"From momentum conservation:\n\n***\n\n***\n\n$$\nF_{\\small \\mathrm{fluid}}=M_{ab}+M_{bc}-M_{ad}\n$$\n\n***\n\nSince we are calculating the horizontal resultant force, we need to consider the horizontal momentum in the section \"ab\"; this means that we ignore any vertical velocities. Since the velocity of the fluid at $y=\\delta$ (i.e. at \"ab\") is always $U_0$, we can say:\n\n***\n\n$$\nM_{ab}=U_0\\dot{m}_{ab},\n$$\n\n***\n\nWe then also convert the other momentum flowrate terms into their mathematical forms, and proceed to find the force of the plate on the fluid, as shown below:\n\n***\n\n$$\nF_{\\mathrm{fluid}}=M_{ab}+\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{u(y)^2\\space\\mathrm{dy}}-\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{U_0^2\\space \\mathrm{dy}}\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=U_0\\dot{m}_{ab}+\\rho WU_0^2\\left(\\LARGE{\\int}_{\\small 0}^{\\small \\delta}\\normalsize \\frac{y^2}{\\delta^2}-1\\space\\mathrm{dy} \\right)\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=(3)(30)+(800)(1)(3)^2\\left[ -\\frac{2}{3}(25\\times 10^{-3})\\right]\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=-30 \\space \\mathrm{N}\n$$\n\n***\n\nHowever, the question asks us to find the drag force **of the fluid on the plate**, hence:\n\n***\n\n$$\nF_{\\mathrm{plate}}=-F_{\\mathrm{fluid}}\n$$\n\n***\n\n$$\nF_{\\mathrm{plate}}= \\boxed{30\\space \\mathrm{N}\\space}\n$$\n","children":[]}}]}