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Figure 33 Finding all nodes reachable from a particular node procedure explore(G, v) Input: G = (V, E) is a graph; v V Output: visited(u) is set to true for all nodes u reachable from v visited(v) = true previsit(v) for each edge (v, u) E: if not visited(u): explore(u) postvisit(v)

Start with the general equation for adding two fractions: a /b + c /d = (ad + bc)/(bd ) Based on this, and on the rules you already know, prove that a /b c /d = (ad bc)/(bd ) as long as b 0 and d 0

Time period: Weekly survey The Lighter Side: The AAII Sentiment Survey, which I nicknamed The Little Guy, will keep you out of trouble when the markets get extreme A little secret: do the opposite of the little guy

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and whenever you arrive at any junction (vertex) there are a variety of passages (edges) you can follow A careless choice of passages might lead you around in circles or might cause you to overlook some accessible part of the maze Clearly, you need to record some intermediate information during exploration This classic challenge has amused people for centuries Everybody knows that all you need to explore a labyrinth is a ball of string and a piece of chalk The chalk prevents looping, by marking the junctions you have already visited The string always takes you back to the starting place, enabling you to return to passages that you previously saw but did not yet investigate How can we simulate these two primitives, chalk and string, on a computer The chalk marks are easy: for each vertex, maintain a Boolean variable indicating whether it has been visited already As for the ball of string, the correct cyberanalog is a stack After all, the exact role of the string is to offer two primitive operations unwind to get to a new junction (the stack equivalent is to push the new vertex) and rewind to return to the previous junction (pop the stack) Instead of explicitly maintaining a stack, we will do so implicitly via recursion (which is implemented using a stack of activation records) The resulting algorithm is shown in 86

Table 6-2 shows how the subtraction formula is derived from the addition formula Near the end of this proof parentheses and brackets aren t enough for grouping of an expression, so braces must be used! They look exactly the same as they braces you use to enclose lists of symbols representing set elements, but the purpose is different

Table 6-2 Derivation of a general formula for the subtraction of one fraction from another, based on the formula for addition of fractions As you read down the left-hand column, each statement is equal to all the statements above it

Figure 331 The previsit and postvisit procedures are optional, meant for performing operations on a vertex when it is rst discovered and also when it is being left for the last time We will soon see some creative uses for them More immediately, we need to con rm that explore always works correctly It certainly does not venture too far, because it only moves from nodes to their neighbors and can therefore never jump to a region that is not reachable from v But does it nd all vertices reachable from v Well, if there is some u that it misses, choose any path from v to u, and look at the last vertex on that path that the procedure actually visited Call this node z, and let w be the node immediately after it on the same path

AAII polls their members via the Internet to find out how the members feel the stock market will do over the next six months: bullish, bearish, or neutral

Statements a /b c /d a /b + [ (c /d )] a /b + [ 1(c /d )] a /b + ( 1/1)(c /d ) a /b + ( 1)c /1d a /b + ( 1)c /d [ad + b( 1)c]/bd [ad + ( 1)bc]/bd [ad + ( 1)(bc)]/bd {ad + [ (bc)]}/bd (ad bc)/bd QED Reasons Begin here Convert subtraction to addition of a negative Principle of the sign-changing element Divisive identity element : substitute 1/1 for 1 Multiplication of fractions to right of plus sign Multiplicative identity element: substitite d for 1d Formula for addition of two fractions, considering ( 1)c as a single quantity Commutative law for multiplication Group elements b and c with parentheses Principle of sign-changing element (the other way around) Convert addition of a negative to subtraction Mission accomplished

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