There are 4 red cards and 4 blue cards, the host first take any two, and then respectively in a, B, c three people on the forehead affixed any two cards, A, B, c three people can see the other two people on the forehead of the cards, after reading let them guess what color on their forehead card, A Say do not know, B said do not know, C said do not know, then a said to know. ask how to reason, A is how to know. If you use the program, how to achieve it.
Thinking of solving problems
the first case of combination is 3x3x3 =
public void Group () { string[] arg = {"Blue-blue", "red-red", "Red-Blue"}; &nb
sp; int size = arg.length;
int index = 0; for (int i = 0; i < size; i++) { &nbs P
string A = Arg[i]; for (int j = 0; J < size; J + +) { &nbs P
string B = arg[j]; for (int k = 0; k < size; k++) { &nbs P
string C = arg[k];
index++; string group = STRING.FOrmat ("%s-%s-%s", A, B, C); string groupstr =
String.Format ("%s%s%s", A, B, C); system.out.print (
Index + ":"); system.out.println (
GROUPSTR); } &
nbsp; } } }
1: Blue-blue blue-blue-Blue 2: Blue-blue blue-blue-red- Red 3: Blue-blue-blue red-blue 4: Blue-blue red-red blue-Blue 5: Blue-blue red-red-red 6: Blue-blue-red-red-blue 7: Blue-blue-red -Blue-Blue 8: Blue-blue red-blue red-red 9: blue-blue red-blue red-blue 10: Red-red blue-blue-blue 11: Red-red blue-blue red-red 12: Red-red blue-blue red-blue 13: Red-red-red Blue-Blue 14: Red-red red-red red-red 15: Red-red red-red-blue 16: Red-red red-blue-blue 17: Red-red red-blue red-red 18: red-red red-blue red-blue 19: Red-Blue Blue-blue-blue 20: Red-blue blue-blue red-red 21: Red-blue blue-blue red-blue 22: Red-blue red-red blue-blue 23: Red-blue red-red red-red 24: Red-blue red-red red-blue 25: Red-blue red-blue-blue 26: Red-blue red-blue red-red 27: Red-blue red-blue red-blue
and then filter out some of the options in the above combinations by some conditions
Condition 1: The number of red in a combination or the number of blue cannot exceed 4
public void Group () {
string[] arg = {"Blue-blue", "red-red", "Red-Blue"};
int size = Arg.length;
int index = 0;
for (int i = 0; i < size; i++) {
String A = arg[i];
for (int j = 0; J < size; J + +) {
String B = arg[j];
for (int k = 0; k < size; k++) {
String C = arg[k];
index++;
String Group = String.Format ("%s-%s-%s", A, B, C);
String groupstr = String.Format ("%s%s%s", A, B, C);
if (Getstrcount (group)) {
System.out.print (index + ":");
System.out.println (GROUPSTR);
}}}} public boolean getstrcount (String str) {
string[] args = Str.split ("-");
int bluecount = 0;
int redcount = 0;
for (int i = 0; i < args.length; i++) {
if (args[i].equals ("Blue")) {
bluecount++;
} else if (args[i].equal S ("Red")) {
redcount++;
}
}
if (Bluecount > 4 | | redcount > 4) {
return false;
}
return true;
}Filter out the above numbered 1,3,7,14,15,17,19,23 eight cases. There are 19 of them left.
2: Blue-blue-blue red-red 4: Blue-blue-red-red-blue-blue 5: Blue-blue-red-red-red 6: Blue-blue-red-red-blue 8: Blue-blue red-blue red-red 9: blue-blue red-blue red-blue 10: red-Red Blue-blue-blue 11: Red-red-blue-blue-red 12: Red-red-blue-blue-red-blue 13: Red-red red-red blue-Blue 16: Red-red-blue-blue 18: Red-red red-blue red-blue 20: Red-Blue Blue-blue red-red 21: Red-blue blue-blue red-blue 22: Red-blue red-red blue-Blue 24: Red-blue red-red red-blue 25: Red-blue red-blue-blue 26: Red-blue red-blue red-red 27: Red-blue red-blue red-blue
condition 2. A don't know, b don't know, C don't know. Description of the three people can not each hand two cards are the same color.
To disprove: if there are three people in each hand the cards are the same color. So for the first time, 3 people can guess directly at one glance.
public void Group () {string[] arg = {"Blue-blue", "red-red", "Red-blue"};
int size = Arg.length;
int index = 0;
for (int i = 0; i < size; i++) {String A = arg[i];
for (int j = 0; J < size; J + +) {String B = arg[j];
for (int k = 0; k < size; k++) {String C = arg[k];
index++;
String groupstr = String.Format ("%s%s%s", A, B, C);
Condition 1 String group = String.Format ("%s-%s-%s", A, B, C);
if (!getstrcount (group)) continue;
Condition 2 if (!iseveryonesame (A) &&!iseveryonesame (B) &&!iseveryonesame (C)) continue;
System.out.print (Index + ":");
System.out.println (GROUPSTR); }}}}//Condition 2 private boolean iseveryonesame (String one) { &nb Sp
string[] args = One.split ("-"); if (Args[0].equals (args[1])) {
return true; } return false;
}//Condition 1 Private boolean getstrcount (String str) {string[] args = Str.split ("-");
int bluecount = 0;
int redcount = 0;
for (int i = 0; i < args.length; i++) {if (Args[i].equals ("Blue")) {bluecount++;
} else if (Args[i].equals ("Red")) {redcount++;
}} if (Bluecount > 4 | | redcount > 4) {return false;
} return true; }
There are a couple of possible combinations left.
6: Blue-blue red-red red-blue 8: Blue-blue red-blue red-red 9: blue-blue red-blue red-blue 12: Red-red blue-blue red-blue 16: Red-red red-blue-blue 18: Red-red red-blue red-blue 20: Red -Blue blue-blue red-red 21: Red-blue blue-blue red-blue 22: Red-blue red-red blue-Blue 24: Red-blue red-red red-blue 25: Red-blue red-blue-blue 26: Red-blue red-blue red-red 27: Red -Blue red-blue red-blue
Number 2,4,5,10,11,13 can be excluded in six cases. There are 13 of them left.
condition 3. In the above case, to exclude C can be based on a, B, do not know, and then can be seen at a glance what the card combination.
This is the case of a hand of the card color is the same, B hands of the card color is the same. For example, 6 and 12
public void Group () {string[] arg = {"Blue-blue", "red-red", "Red-blue"};
int size = Arg.length;
int index = 0;
for (int i = 0; i < size; i++) {String A = arg[i];
for (int j = 0; J < size; J + +) {String B = arg[j];
for (int k = 0; k < size; k++) {String C = arg[k];
index++;
String groupstr = String.Format ("%s%s%s", A, B, C);
Condition 1 String group = String.Format ("%s-%s-%s", A, B, C);
if (!getstrcount (group)) continue;
Condition 2 if (Iseveryonesame (A) && iseveryonesame (B) && iseveryonesame (C)) continue;
Condition 3 if (Iseveryonesame (A) && iseveryonesame (B)) continue;
System.out.print (Index + ":");
System.out.println (GROUPSTR);
}}}}//Condition 2 Private Boolean iseveryonesame (String one) {string[] args = One.split ("-");
if (Args[0].equals (Args[1])) {return true;
} return false; }//Condition 1 Private boolean getstrcount (String str) {string[]args = Str.split ("-");
int bluecount = 0;
int redcount = 0;
for (int i = 0; i < args.length; i++) {if (Args[i].equals ("Blue")) {bluecount++;
} else if (Args[i].equals ("Red")) {redcount++;
}} if (Bluecount > 4 | | redcount > 4) {return false;
} return true;
} public static void Main (string[] args) {//TODO auto-generated method stub Problem22 problem = new Problem22 ();
Problem.group ();
}
}
There are 11 other cases left.
8: Blue-blue red-blue red-red 9: blue-blue red-blue red-blue 16: Red-red red-blue-blue 18: Red-red-blue-blue 20: Red-blue-blue-red 21: Red-blue blue-blue red-blue 22: Red-blue red-red blue-Blue 24: Red-blue red-red-blue 25: Red-blue red-blue-blue 26: Red-blue red-blue red-red 27: Red-blue red-blue red-blue
The remaining 11 of the above is in accordance with the actual licensing situation, and according to the first round of a,b,c can not judge their hands of the cards these conditions filtered combinations. Then according to the last second round a will already know their cards. Description from the combination above to find, with b,c combination as the keyword, a only one case of a combination. For example, in case 8, the b,c combination is "red-blue red-red", then according to this to find out, the same is the condition of a there is no other situation. It was obvious that number 26th met. It is indicated that this combination is not in accordance with the requirements. And the 20th "blue-blue red-red" only one case, the description meets the requirements. That's how the program is implemented, we can use b+c as the keyword to build a map,value value for this keyword in the above 11 occurrences of the number of times. So we can get the keyword that appears once, so it's easy to determine what the number is.
public class Problem22 {//with BC as the keyword, number of times is value of map map<string,integer> map = new Linkedhashmap<string,intege
R> ();
Map map<string,string> map1 = new linkedhashmap<string,string> () with BC as the keyword, combined as value
public void Group () {string[] arg = {"Blue-blue", "red-red", "Red-blue"};
int size = Arg.length;
int index = 0;
for (int i = 0; i < size; i++) {String A = arg[i];
for (int j = 0; J < size; J + +) {String B = arg[j];
for (int k = 0; k < size; k++) {String C = arg[k];
index++;
String groupstr = String.Format ("%s%s%s", A, B, C);
Condition 1 String group = String.Format ("%s-%s-%s", A, B, C);
if (!getstrcount (group)) continue;
Condition 2 if (Iseveryonesame (A) && iseveryonesame (B) && Iseveryonesame (C)) continue;
Condition 3 if (Iseveryonesame (A) && iseveryonesame (B)) continue;
if (Map.containskey (B + C)) {int num = Map.get (b+c);
Map.put (B+c,++num);
}else{Map.put (b+c,1);
} map1.put (B+c,index + ":" +groupstr);
System.out.print (Index + ":");
System.out.println (GROUPSTR);
}}} set<map.entry<string, integer>> Entrys = Map.entryset ();
Iterator<map.entry<string, integer>> iter = Entrys.iterator ();
A combination of 1 found while (Iter.hasnext ()) {map.entry<string, integer> entry= iter.next (); if (Entry.getvalue () ==1) {String answer = MAP1.GET (Entry.getkey ());
SYSTEM.OUT.PRINTLN ("The qualifying answer found from the combination above is:" +answer);
}}}//Condition 2 Private Boolean iseveryonesame (String one) {string[] args = One.split ("-");
if (Args[0].equals (Args[1])) {return true;
} return false;
}//Condition 1 Private boolean getstrcount (String str) {string[] args = Str.split ("-");
int bluecount = 0;
int redcount = 0;
for (int i = 0; i < args.length; i++) {if (Args[i].equals ("Blue")) {bluecount++;
} else if (Args[i].equals ("Red")) {redcount++;
}} if (Bluecount > 4 | | redcount > 4) {return false;
} return true; } public static void Main (string[] args) {//TODO auto-generated method stub Problem22 problem = new
PROBLEM22 ();
Problem.group (); }
}
8: Blue-blue red-blue red-red 9: blue-blue red-blue red-blue 16: Red-red red-blue-blue 18: Red-red-blue-blue 20: Red-blue-blue-red 21: Red-blue blue-blue red-blue 22: Red-blue red-red blue-Blue 24: Red-blue red-red-blue 25: Red-blue red-blue-blue 26: Red-blue-red-blue-red 27: Red-blue red-blue red-Blue the answer that matches the criteria found in the combination above is: 20: Red-blue blue-blue red-red the answer to the criteria found in the combination above is: 21: Red-blue blue-blue red-Blue The qualifying answer found from the combination above is: 22: Red-blue red-red blue-Blue the answer that matches the criteria found in the combination above is: 24: Red-blue red-red red-blue
There are 4 cases in which a can guess its own cards. Of course, a can have such a strong logic.
Summary
There is no possibility that the cards in the hands are the same: red and blue. So after that you guess you're red and blue.
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