This is a seemingly 'impossible' problem, because variously your function, "impossibleFn", must return an output of either 1 or 2 or 3 for the same input (0, the number zero). Why is this described as "seemingly impossible"? Because it does not satisfy the mathematical definition of a function. Yet there is still at least one way to pass all the tests! (One simple code has been verified to work. However, all's fair in this problem, so try whichever hacks and workarounds are at your disposal.) UPDATE: Now more than three (unrelated) approaches have been verified to pass all the tests.
Add more tests, for example:
Test1 : 0->1
Test2: 1->1
Test3: 0->2
Test4: 6->6
Test 6: 0->0
...
Hello, Jean-Marie Sainthillier. Thanks for the suggestion. If I understand you correctly, that could be done, but it would break all/most of the previous submissions, so it seems better to do that in a separate problem — if you would like to make one. —DIV
You should try the timer function.
Nice thought. I _eventually_ got that to work (Solution 1463623), but it also seems a bit clunky in the end, and it completely fails if implemented without an additional explicit instruction that ensures control is not returned to the "command prompt" (and hence the Test Suite!) until after the specified timer period has elapsed.
Would this also work if the y_correct values were 3, 2, 1 instead of 1, 2, 3?
The code _as is_ wouldn't work, but it could easily be adapted. Either with a switch statement, or by simply calculating 4 – {erstwhile result};
Determine whether a vector is monotonically increasing
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