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public class TestDictionary { public static void main(String[] args) { Map map = new HashMap(); map.put("Tag","day"); map.put("Hut","hat"); map.put("Uhr","clock"); map.put("Rad","wheel"); map.put("Ohr","ear"); map.put("Tor","gate"); System.out.println("map=" + map); System.out.println("map.size()=" + map.size()); System.out.println("map.keySet()=" + map.keySet()); System.out.println("map.values()=" + map.values()); System.out.println("map.get(\"Uhr\")=" + map.get("Uhr")); System.out.println("map.remove(\"Rad\")=" + map.remove("Rad")); System.out.println("map.get(\"Rad\")=" + map.get("Rad")); System.out.println("map=" + map); System.out.println("map.size()=" + map.size()); } }

(a) The SPICE netlist code for k 0:001 follows:

The output is:

The algorithm will know that has happened when sorted_index becomes 0 In that case, the algorithm inserts the new number as the first element in the sorted set..

Prb.1_5.CIR Vs 1 0 DC 10V R1 1 2 500ohm E 2 0 (3,0) 0.001 ; Last entry is value of k F 0 3 Vs 100 R2 3 0 100ohm RL 3 0 100ohm .DC Vs 10 10 1 .PRINT DC V(3) .END

map={Rad=wheel, Uhr=clock, Ohr=ear, Tor=gate, Hut=hat, Tag=day} map.size()=6 map.keySet()=[Rad, Uhr, Ohr, Tor, Hut, Tag] map.values()=[wheel, clock, ear, gate, hat, day] map.get("Uhr")=clock map.remove("Rad")=wheel map.get("Rad")=null map={Uhr=clock, Ohr=ear, Tor=gate, Hut=hat, Tag=day} map.size()=5 The put() method inserts key/value pairs into the table. For example, map.put("Tag","day"); inserts the key/value pair ("Tag","day"), where "Tag" is the key and "day" is the value.

(b) Edit <Prb1_5.CIR> to set k 0:05, execute the code, and poll the output le to nd vab V 3 200 V.

As we described earlier in this chapter, the familiar example is the act of moving money from a savings account to a checking account The customer thinks of the act as a single action, but the database must take two actions The database must reduce the balance in the savings account, and increase the balance in the checking account If the first action should succeed, and the second fail (perhaps the computer fails at that instant), the customer will be very unhappy, for their savings account will contain less and their checking account will contain what it did The customer would prefer that both actions be successful, but if the second action fails, the customer wants to be sure that everything will be put back as it was initially Either every change must be successful, or no change must occur.

For the circuit of Fig. 1-16, nd iL by the method of node voltages if (a) 0:9 and (b) 0.

The first call to println() invokes the HashMap.toString() method, printing the entire Map object. The second call to println() invokes the HashMap.size() method, showing that the Map object has six key/value elements. The next call to println() invokes the HashMap.keySet() method, which returns a Set object containing all the keys (the six German words). The next call to println() invokes the HashMap.values() method, which returns a Collection object containing all the values (the six English words). The next call to println() invokes the HashMap.get() method, which returns the value for a given key. This call returns the value object "clock" for the key object "Uhr". The next call to println() invokes the HashMap.remove() method, which deletes the ("Rad","wheel") pair, which is confirmed by the next call: map.get("Rad") returns null, indicating that there is no key/value pair in map whose key is "Rad". The last two lines prints the entire map again and its size, revealing that the ("Rad","wheel") pair has indeed been deleted.

Fig. 1-16 (a) With v2 and vab as unknowns and summing currents at node c, we obtain v2 vs v2 v2 vab i 0 R1 R2 R3 But i vs v2 R1 1

The order in which the key/value pairs were stored in the HashMap object in Example 8.1 seems to be random and unrelated to the order in which the pairs were inserted. The next example verifies this. EXAMPLE 8.2 Java HashMap Objects Are Hash Tables

Substituting (2) into (1) and rearranging gives 1 1 1 1 1 v v v R1 R2 R3 2 R3 ab R1 s Now, summation of currents at node a gives vab v2 v i ab 0 R3 RL Substituting (2) into (4) and rearranging yields 1 1 1 v v2 v R3 R1 R3 RL ab R1 s Substitution of given values into (3) and (5) and application of Cramer s rule nally yield 2:1 0:1vs 0:1 0:9vs 1:9vs vab 2:1 1 2:21 0:8597vs 0:1 1:1 and by Ohm s law, iL vab 0:8597vs 0:08597vs RL 10 A