[英]Can't figure out correct algorithm for moving within a graph
我以前發過這個問題,但我認為我會修改它並嘗試更多細節。 我在下面有這張圖片。 我想要完成的是從偶數“泡沫”開始,從零和零轉換到其他狀態。 這就是我所完成的:
1)創建具有所有狀態(偶數和奇數)的mainMap以及正確具有遠離該狀態的輸入(0和1)的另一個映射。 2)將狀態從偶數轉換為奇數轉換。
到目前為止,這是我的代碼:
public static void main(String[] args) {
//Map<Even, ArrayMap[0->even,1->odd]> for first line
Map<String, Map<String,String>> mainMap = new ArrayMap<String, Map<String,String>>();
//Map<int,even/odd>
TypedBufferReader input = new TypedBufferReader("Enter Finite Automaton Description File: ");
//read the file.
for (;;) {
try {
String line = input.readLine();
StringTokenizer st = new StringTokenizer(line, ";");
String state = st.nextToken();
Map<String,String> transitions = mainMap.get(state);
transitions = new ArrayMap<String, String>();
while (st.hasMoreTokens()) {
String intStateInput = st.nextToken();
String inputState = st.nextToken();
transitions.put(intStateInput, inputState);
}
mainMap.put(state, transitions);
} catch (EOFException e) { break;}
}
//Print in alphabetical order of states. odd/even to even/odd
List<String> mapList = new ArrayList<String>(mainMap.keys());
Collections.sort(mapList);
for (String s : mapList) {
Map<String, String> tempMap = mainMap.get(s);
System.out.println(s + " transitions = " + tempMap.toString());
}
//Process one line file.
TypedBufferReader oneLineInput = new TypedBufferReader("Enter start state/inputs file: ");
try {
String oneLine = oneLineInput.readLine();
StringTokenizer st = new StringTokenizer(oneLine,";");
String initialState = st.nextToken();
System.out.println("Initial state = " + initialState);
while (st.hasMoreTokens()) {
String inputNum = st.nextToken();
}
} catch (EOFException e) {}
}
}
我應該按如下方式讀取一行文件:“even; 1; 0; 1; 1; 0; 1”因此,它將打印偶數的初始狀態並繼續通過mainMap移動。
對於初始狀態甚至它將是這樣的:
初始狀態=偶數
input = 1 state = odd
input = 0 state = odd
input = 1 state = even
input = 1 state = odd
input = 0 state = odd
input = 1 state = even
最終狀態=偶數
請幫我解決這個問題。
我認為這很容易做到。你需要記住以前的狀態,所以你要開始
Step 1: prevState = EVEN
Step 2: Read currentState (Read them as Integer)
Step 3: Repeate Until EOF
if currentState | prevState == 0 then
state = EVEN
print EVEN
else
state = ODD
print ODD
在循環中,您將獲得鏈接到當前狀態名稱的轉換,然后獲取給定轉換的新狀態的名稱:
while (st.hasMoreTokens()) {
String inputNum = st.nextToken();
Map<String,String> mainMap transitions = mainMap.get(initialState);
initialState = transitions.get(inputNum);
}
我也會在currentState中重命名initialSate,但它只是個人偏好:)
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