Decision tree code

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Relational operators
If true (then part)	negation is (else part)
yes	no
no	yes
true	false
false	true
x <= y	x > y
x > y	x<= y
x >= y	x < y
x < y	x >= y
x == y	x != y
x != y	x == y

Never use "==" or "!=" on floating point approximations (double in C).

Note: To be moved to appropriate page after this lesson.

Never omit curly braces.

The curly braces enclose a blocks. Blocks are not terminated with a semicolon ";".

Statements (not blocks) are terminated with a semicolon ";". Here is the fix-it decision tree.

Note the following.

Control flow charts are not useful (bad) compared to the alternatives. This as a decision tree chart and very useful.
There are three decision nodes.
There should be three (relational) comparisons in the code.
The code should start at the root, on the left.
The code should reflect the order, top down, of the decisions (leaves on the right)

For more information, see Decision trees .

Here is the decision tree code fragment (without the other code).

if (ch1 == 'y') {
	if (ch2 == 'y') {
		printf("It moves. It should move. True-positive. No problem.\n");
		}
	else {
		printf("It moves. It should not move. False-positive. Use duct tape.\n");
		}
	}
else {
	if (ch2 == 'y') {
		printf("It does not move. It should move. False-negative. Use WD-40.\n");
		}
	else {
		printf("It does not moves. It should not move. True-negative. No problem.\n");
		}
	}

Note how the code exactly models the decision tree which reflects the specification of the logic to be used.

If ch1 is 'y' the (outer) then branch is executed. If ch1 is not 'y' then it must be (and assumed) 'n' (by assumption of the program). Thus, there is no need to check again, as the following fragment does.

if (ch1 == 'y') {
	if (ch2 == 'y') {
		printf("It moves. It should move. True-positive. No problem.\n");
		}
	else {
		printf("It moves. It should not move. False-positive. Use duct tape.\n");
		}
	}
else {
	if (ch1 == 'n') {
		if (ch2 == 'y') {
			printf("It does not move. It should move. False-negative. Use WD-40.\n");
			}
		else {
			printf("It does not moves. It should not move. True-negative. No problem.\n");
			}
		}
	}

This code also gives the impression that in the outer else part, something else might happen. It takes close inspection to see that the code has extra unneeded constraints.

One might be tempted to write the following, using conjunctions.

if (ch1 == 'y' && ch2 == 'y') {
	printf("It moves. It should move. True-positive. No problem.\n");
	}
if (ch1 == 'y' && ch2 == 'n') {
	printf("It moves. It should not move. False-positive. Use duct tape.\n");
	}
if (ch1 == 'n'  && ch2 == 'y') {
	printf("It does not move. It should move. False-negative. Use WD-40.\n");
	}
if (ch1 == 'n'  && ch2 == 'n') {
	printf("It does not moves. It should not move. True-negative. No problem.\n");
	}

This code is much less clear. Does it cover all the cases. Does it correctly model the tree. And there are 12 operators used rather than the 3 in the decision tree and in the first (and best) code presented.

Using else-if parts does not help much.

if (ch1 == 'y' && ch2 == 'y') {
	printf("It moves. It should move. True-positive. No problem.\n");
	}

else if (ch1 == 'y' && ch2 == 'n') {
	printf("It moves. It should not move. False-positive. Use duct tape.\n");
	}

else if (ch1 == 'n'  && ch2 == 'y') {
	printf("It does not move. It should move. False-negative. Use WD-40.\n");
	}

else if (ch1 == 'n'  && ch2 == 'n') {
	printf("It does not moves. It should not move. True-negative. No problem.\n");
	}

Have we covered all the cases? Is it correct? Have we missed anything. If not, the last else if could be changed to just an else, making the code even less clear.

It is often better and more clear to use if then else statements rather than logical operations on relational conditions.

If you use logical operations on relational conditions, you should be very familiar with DeMorgan's Laws for negating such expressions.

For more information, see DeMorgan's Laws .

Make it correct first. And correct in that it models closely the original specification - textual and diagram.
Make it easy to maintain - refactor when necessary.
Never try to make it faster. Rules and reasoning will be covered later.

The ways above, other than the first, are slower, require more operations to execute, and are more difficult to understand and reason about (in terms of correctness).

Here is one way to refactor the original code. Make sure you understand how the original can be converted to the following. Note that the inferior code (with 12 operators) are more difficult to refactor. If one did refactor those, one would end up with the original (best) code.

printf("It ");
if (ch1 == 'y') {
	printf("moves. It should ");
	if (ch2 == 'y') {
		printf("move. True-positive. No problem");
		}
	else {
		printf("not move. False-positive. Use duct tape");
		}
	}
else {
	printf("does not move. It should ");
	if (ch2 == 'y') {
		printf("move. False-negative. Use WD-40");
		}
	else {
		printf("not moves. It should not move. True-negative. No problem");
		}
	}
printf(".\n");

Note: The very last printf of the period also includes the newline "\n" which now appears only once.

Note: There are other ways to remove the remaining redundancy - not covered here.