Don't get me started on those sort of questions, I get so mad other them. There is more than one correct answer, but they also have and winding, round about way of justifying one over another.
Sdd 2017 hsc (1 Viewer)
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It is/was a complete guess, but I assume "self checking" elements would be how an input box is static, but natively it can have spell checking or suggestions etc. As apposed to having the ability to add checking with an event listener, so you could execute a check of contents of input on the event of a "keystroke" - so the method that checks is called on when the a key is pressed.
I guess what you're saying sorta makes sense. I think I might've thought of it at the time the same way you did except I thought that sequential wasn't necessarily limited to self-checking elements as for example, you could have a sequential program which accepts two inputs in sequential order and just outputs them one after the other (a stupid program I know, but I guess that was at the time the only way I could make a decision). It wouldn't have to check the values, it would just output it. You're probably right though.
Also for that same mc question, was the option "User input does not drastically change the execution" for event-driven or sequential?
For the binary search question I changed it from high=middle to high=middle-1 (since middle and high already been checked and stuff) and from low= middle to low=middle+1, mainly because that's how it was in the textbook
I never looked over those algorithms in the textbook because I didn't think I'd need them. I guess I was wrong. I wonder if I'll get some marks anyways because it still stopped the infinite looping and, excluding the fact that I forgot to correct the lowercase 'm' in 'middle', would have given the correct output.
That option was under sequential.
Went the with same, insert + select.
Bubble can affect the position of more than just the last 2 elements in the array.
Now you've said that though, I'm questioning myself... hmmm
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Wasn't the question asking 'which sort COULD have been used', which would imply that if its possible to affect only the last 2 elements with that sort then its correct (maybe I remembered the question wrong though)? I said bubble and selection. At the time, I thought if it was insertion, on the first pass it would have checked the 2nd element and if it did need to move the 2nd element to the end of the array, it would have changed the values of not only the last two elements but also the 2nd element. Bubble could also work because it might only swap the last 2 values on the first pass.
Alright, thanks for clarifying. I must've remembered the answers wrong. I'm pretty sure I chose one of the answers where it said that for sequential "User input does not drastically change the execution". Does anyone happen to remember the options for event driven?
Answer was high = mid -1 and low = mid + 1 both lines
That way the range actually will become smaller, and it fits the scanario (just like a standard binary search)
Selection doesn't work, though?
Selection starts with the first element in the array, and then finds the max/min element and swaps the two. Insertion and bubble can work fine, selection can never.
If we were to get something like this:
1 2 3 4 6 5
Selection takes the value inside the first element: 1
It then finds the maximum, which is 6, and swaps them.
6 2 3 4 1 5
The smallest/largest value will always become the first element in the unsorted array, which means that it doesn't work with just the last two elements in an array on the first pass.
Selection starts with the first element in the array, and then finds the max/min element and swaps the two. Insertion and bubble can work fine, selection can never.
If we were to get something like this:
1 2 3 4 6 5
Selection takes the value inside the first element: 1
It then finds the maximum, which is 6, and swaps them.
6 2 3 4 1 5
The smallest/largest value will always become the first element in the unsorted array, which means that it doesn't work with just the last two elements in an array on the first pass.
Hmmm, could of been, if it was that yeah all three would be correct. Though I thought it said only affect the the last 2.
I remember narrowing it down to A and B (I think), can't remember the exact options, hopefully they put the paper up soon.
I've seen selection sorts where they swap the min/max value element with the last element in the array. The 2016 HSC Software Paper had Q13 which had the answer as a selection sort which swapped the max value element with the last element in the array.
A and B were definitely incorrect, as they said the event driven approach occurred in a fixed sequence of processes, which is definitely incorrect.
Selection can start with the last element. It doesn't have to start with the first, it can be flipped either way since it will result in the same answer in the end.
Did a quick search on selection sort, and all of the articles I've read state that the min/max value is swapped with the value at the leftmost of the unsorted section (ie first element of the unsorted).
well, shit sadlyfe: http://ee.hawaii.edu/~tep/EE160/Book/chap10/subsection2.1.2.1.html
well, shit sadlyfe: http://ee.hawaii.edu/~tep/EE160/Book/chap10/subsection2.1.2.1.html
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Mmm, it really doesn't matter which way you go, just need to find the next largest/smallest and move it to the "sorted" section of the array. Only advantage of moving from the right to left would be that decrementing loops are marginally faster than incrementing loops - even so, now I think of it you could still start from the right decremented and have the sorted part on the left.
Ehh, it'll be interesting to see the answer to this question.
NightShadeI
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For the Boolean values in a one-dimensional array I did teams winning/ losing (although drawing is an option) because different Sets of traffic lights implies a two-dimensional array and the fact that most lights are Red, yellow or green. Could of been either though.
As for the sort our software teacher told us it can start from the left or the right so I based it off that. Therefore I got D (i.e. all off the sorts)
1 2 3 5 4
Bubble sort in this case would bubble 5 past 4 on first pass (changing last 2 values)
If insertion sort starts from right it will output same result as bubble sort on first pass
Finally, if selection sort also starts from the right it will swap 5 and 4 (selecting 5 as the highest) also changing last 2 values in same way as others above.
As for the sort our software teacher told us it can start from the left or the right so I based it off that. Therefore I got D (i.e. all off the sorts)
1 2 3 5 4
Bubble sort in this case would bubble 5 past 4 on first pass (changing last 2 values)
If insertion sort starts from right it will output same result as bubble sort on first pass
Finally, if selection sort also starts from the right it will swap 5 and 4 (selecting 5 as the highest) also changing last 2 values in same way as others above.