I am looking for an algorithmic way to solve the following problem.
Say we are given a multiple choice test with 100 questions, 4 answers per question (exactly one of those four being correct), each correctly given answer is worth one point, wrong answers are worth zero points.
If now we got a database D of lots of answer sheets and their corresponding points, e.g.
D:= { (‘ABAA…’, 80),
(‘ABAB…’, 80),
(‘ABAC…’, 80),
(‘ABAD…’, 81), … }
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How can we find out which answers are correct?
I am not looking for something probabilistic, but for answers which are definitely correct.
There are some obvious strategies like:
But what information can we get from other combinations of answers?
Viewing the answer sheets as a metric space we get
100 – S(test) = d(test, correct)
for the hamming distance d(.,.), the score S(.) and the correct sheet correct.
Maybe someone could give me a reformulation of the problem, which yields a more obvious implementation. Any contribution is appreciated.
Not considering computational complexity, couldn’t I achieve something by intersecting the balls
$$
\bigcap_i B_{d(t_i,\textrm{correct})}(t_i),
$$
with tests $t_i$ and balls $B_d(x) := \{y: d(x,y)\leq d\}$?
Today, as I was looking through some interesting math/programming problems on projecteuler, I noticed Number Mind (problem number 185), and it immediately reminded me of this math.SE question; they are practically equivalent. Searching for a solution to the projectuler problem, I found a solution (written in python) from a sister site: codereview.SE. (I haven’t actually read it though.)
What follows is what I originally posted.
Depending on the database D, we may not be able to determine the correct answers. For example, if question number 100 is very tricky and as a result everyone chooses C or D, even though the correct answer is A, then we cannot determine the correct answer.
Let’s fix some notation: Let $\mathcal{A}$ be the correct answers and $\mathcal{\hat{A}}$ be some guess of $\mathcal{A}$. (So both are strings, 100 letters long.) Let $t_i$ be a test whose actual score is 85; say $S_{\mathcal{A}}(t_i) = 85$. Further, let’s assume that if we rescore $t_i$ according to $\mathcal{\hat{A}}$, we get a new score $S_{\mathcal{\hat{A}}}(t_i) = 90$. Then we have lower and upper bounds for $d(\mathcal{A},\mathcal{\hat{A}})$ = the number of letters for which they differ:
$$5=|90-85| \leq d(\mathcal{A},\mathcal{\hat{A}}) \leq (100-85) + (100-90) = 25.$$
The first inequality holds because the test $t_i$ is graded incorrectly by $\mathcal{\hat{A}}$ for at least 5 questions. The second is the triangle inequality; $d(\mathcal{A},\mathcal{\hat{A}}) \leq d(\mathcal{A},t_i) + d(t_i,\mathcal{\hat{A}})$. (Notice that $d(\mathcal{A},t_i) = 100 – S_{\mathcal{A}}(t_i)$ etc.) The second inequality is optimal because we can imagine all 15 wrong answers of $t_i$ being counted as correct (by $\mathcal{\hat{A}}$) and 10 of its correct answers being counted as incorrect (by $\mathcal{\hat{A}}$).
The grade for each is a linear function of the selected alternatives (coefficient is 0 for a wrong alternative, 1 for the right one). Given the right mix of graded papers, you have all coefficients, and thus all right answers. Viewed this way, it is a question of linear independence of the set of papers.
Perhaps attacking this as an ANOVA (multivariable linear regression) proves useful.
Here’s what i got so far.
I gave it some thought and it seems to me, that the problem is very similar to a
trilateration problem .
So what my first approach here is, is to intersect all the spheres $S_d(x) := \{y: d(x,y)= d\}$ with $x$ being a test and $d$ being the missing points to a perfect score.
For obvious reasons it doesn’t perform well for the amount of questions I was originally going for, but for a small number of questions it seems to work.
Here is the python code I did program for the computation of
$$
\bigcap_i S_{d(t_i,\textrm{correct})}(t_i):
$$
import itertools
import random
def hammingDistance(s1, s2):
assert len(s1) == len(s2)
return sum(ch1 != ch2 for ch1, ch2 in zip(s1, s2))
def hammingSphere(word, distance, alphabet):
positions = itertools.combinations(range(len(word)), distance)
for pos in positions:
letterCombinations = itertools.product(alphabet, repeat=distance)
for letters in letterCombinations:
tmp = list(word[:])
for i,l in zip(pos, letters):
if tmp[i] == l:
break
else:
tmp[i] = l
else:
yield tuple(tmp)
def findPossiblyCorrectAnswers(tests, choices):
bestTest = max(tests, key=lambda x: x[1])
maxPoints = len(bestTest[0])
possibleAnswers = set(hammingSphere(bestTest[0], maxPoints-bestTest[1], choices))
for test in tests:
possibleAnswers = possibleAnswers.intersection(set(hammingSphere(test[0], maxPoints-test[1], choices)))
return possibleAnswers
def remainingOptions(tests, choices):
return [set(x) for x in zip(*findPossiblyCorrectAnswers(tests, choices))]
def buildTests(correct, alphabet, numTest):
tests = []
maxPoints = len(correct)
for i in range(numTest):
tests.append([random.choice(alphabet) for _ in range(len(correct))])
return [(t, maxPoints-hammingDistance(t, correct)) for t in tests]
### TESTING:
correct = "AAAAAAAA" # The correct answer
choices = "ABCD"
numTests = 8
tests = buildTests(correct, choices, numTests) # Build some tests with known correct answer sheet
print(tests) # Display the test database
print(remainingOptions(tests, choices)) # Display remaining choices
It yields the following output
[(['B', 'A', 'C', 'C', 'C', 'B', 'D', 'A'], 2),
(['C', 'D', 'C', 'B', 'C', 'B', 'A', 'C'], 1),
(['B', 'C', 'D', 'D', 'C', 'D', 'C', 'C'], 0),
(['C', 'D', 'C', 'A', 'D', 'D', 'D', 'C'], 1),
(['A', 'C', 'D', 'C', 'D', 'B', 'A', 'B'], 2),
(['C', 'C', 'D', 'B', 'A', 'B', 'A', 'C'], 2),
(['A', 'A', 'D', 'D', 'B', 'C', 'D', 'B'], 2),
(['C', 'B', 'D', 'B', 'A', 'A', 'D', 'D'], 2)]
[set(['A']), set(['A']), set(['A', 'B']), set(['A']), set(['A']), set(['A', 'B']), set(['A', 'B']), set(['A', 'D'])]
Which are the filled out tests and their scores, as well as the final knowledge about the solutions.
So for eight people randomly answering a multiple choice test with eight questions, quite a lot of information can be squeezed from it.