Sorting
7.1 7
Internal Base Case: For the list of one element, the double loop is not executed and
the list is not processed. Thus, the list of one element remains unaltered and
is sorted.
Induction Hypothesis: Assume that the list of n elements is sorted correctly
by Insertion Sort.
Induction Step: The list of n + 1 elements is processed by rst sorting the
top n elements. By the induction hypothesis, this is done correctly. The nal
pass of the outer for loop will process the last element (call it X ). This is
done by the inner for loop, which moves X up the list until a value smaller
than that of X is encountered. At this point, has X been properly inserted
into the sorted list, leaving the entire collection of n + 1 elements correctly
sorted. Thus, by the principle of Mathematical Induction, the theorem is
correct.
7.2 void StackSort(AStack<int>& IN) {
AStack<int> Temp1, Temp2;
while (!IN.isEmpty()) // Transfer to another stack
Temp1.push(IN.pop());
IN.push(Temp1.pop()); // Put back one element
while (!Temp1.isEmpty()) { // Process rest of elems
while (IN.top() > Temp1.top()) // Find elems place
Temp2.push(IN.pop());
IN.push(Temp1.pop());
// Put the element in
while (!Temp2.isEmpty()) // Put the rest back
IN.push(Temp2.pop());
}
}
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