Test suite should have assert(isequal(...)).
It's an interesting problem. Unfortunately without any assert, the test suite does not test anything. So any syntactically correct answer succeeds even if it doesn't solve the problem.
Fixed, and (hopefully) rescored. Thanks for the heads up.
your example ([5 7 8]) is not really following the optimal bisection approach. After you choose 5, the new bounds are not 5 and 10, they are 6 and 10, so your next choice should be 8 ((6+10)/2) instead of 7...
The misplaced comment line in the first test case breaks the test suite (at least for "conventional" solutions).
Emphasized Alfonso Nieto-Castanon Comments "your example ([5 7 8]) is not really following the optimal bisection approach. After you choose 5, the new bounds are not 5 and 10, they are 6 and 10, so your next choice should be 8 ((6+10)/2) instead of 7..."
test suit is broken plz fix since its a budge problem
I used this idea to create a similar problem that requires solvers to apply the bisection method correctly (see previous comments). https://www.mathworks.com/matlabcentral/cody/problems/46603-higher-lower-correct
Times 2 - START HERE
Make the vector [1 2 3 4 5 6 7 8 9 10]
Make a simplified barcode
Return area of square
Give me Hamming on five, hold the mayo
Now 20% off!
Two fractions, one sum
Pseudo Square Root (Inspired by Project Euler 266)
I've got the power! (Inspired by Project Euler problem 29)
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