Nine Gongge cipher combination counting problem

Question Description:

Android's nine Gongge password must have been seen.
A password greater than or equal to four points is connected.
So how many kinds of nine password are there?

Algorithm Description:

　　keywords: combinatorial arrangement dynamic planning

Android's password is a 3x3 lattice of a path, this path can be crossed, can "walk the word", almost omnipotent (as long as not repeating points), but there is an exception: the path does not allow to skip the way to pass the point. For example, if you connect from the upper-left point to a point in the upper-right corner, that point in the middle is automatically added to the path. But the trouble is that the rule itself has a notable place: if the middle point is already used before, then this point can be skipped.

We might as well put the nine dots in the lattice number 1 to 9 respectively. According to the above rules, 4136, 4192 are illegal, but 24136, 654192 are feasible.

Algorithmic Thinking Reference: http://www.guokr.com/article/49408/

That is: Slash there are 8, a total of 8 pairs of numbers: 1-3 ago 2 1-7 is 4 1-9 need 5 2-8 need 5

3-7 Required 5 3-9 6 4-6 required 5 7-9 required 8 otherwise failed

So: through the dynamic planning traversal arrangement and verification (first if a sequence of I number combination is a failure, then its derived from the subsequent i+1 number of combinations of a sequence is also a failure, so no longer validate, that is, no further dynamic iteration)

The code runs:

password length	1	2	3	4	5	6	7	8	9
Number of passwords	9	56	320	1624	7152	26016	72912	140704	140704	389497

Nine Gongge password is greater than or equal to four bits, so nine Gongge password a total of 389212, the password is best set 6 and more than 6.

The code is as follows:

1  Packagenine Gongge;2 3  Public classmain{4 5      Public Static voidMain (string[] argv) {6         7         //number of calculated combinations8         intK[] =New int[9];9         intP_count[] =New int[9];Ten         intSum=0; One         intOut_result=0; A          -          for(inti=0; i<9;i++){ -              the             if(i==0){ -k[i]=9; P_count[i]=1; -             } -                  +             Else{ -k[i]=k[i-1]* (9-i); +             } A             if(i>=3) atsum=sum+K[i]; -         }         - System.out.println (sum); -          -         //Calculation Arrangement -         int[] middle =New int[9] [1]; in          -          for(inti=0; i<9;i++){ to              +             intn =K[i]; -             intN =P_count[i]; the             intMid_out[][] =New intN [I+1]; *             intPai_out[][] =New intN [I+1]; $             intJ=0;Panax Notoginseng             if(i==0){ -                  the                  for(; j<n; J + + ) +Pai_out[j][0]=0; AJ=0; the                  for(; j<n; J + +) +Mid_out[j][0]=j+1; -J=0; $                 //update the last viable sequence number $K[i]=9; -             } -             Else{ the                  -                 //Get arrangementWuyi  the                  for(intt=0; t<k[i-1]; t++){ -                      Wu                     intCheck[] =New int[10]; -                      for(intm=0; m<i;m++){ About                          $Check[middle[t][m]]=1; -                          -                     } -                      for(intP=1; p<=9; p++ ){ A                          +                         if(check[p]!=1){ the                              -                         //The alignment validation succeeds in adding to the sequence $                         if(Check_success (middle[t][i-1],p,check)) { the                                                      the                              for(intm=0; m<i;m++){                             themid_out[j][m]=Middle[t][m];  the                                 } -                          inmid_out[j][i]=p; theJ + +; the                             } About                          the                         } the                          the                     } +                      -                      the                 }Bayi                 //update the last viable sequence number thek[i]=J; the             } -             if(i>2) -out_result=out_result+K[i]; theMiddle =New intN [I+1]; theMiddle =mid_out; the              for(intcount_x=0; count_x<j; count_x++){ theSystem.out.print ((count_x+1) + "th:"); -                  for(intcount_y=0; count_y<=i;count_y++) theSystem.out.print (mid_out[count_x][count_y]+ ""); the System.out.println (); the             }94System.out.println ("the" +i+ "th result as ..." +j); the              the              the         }98          AboutSYSTEM.OUT.PRINTLN ("The result is:" +out_result); -          for(inti=0; i<9; i++){101             102 System.out.println (K[i]);103         }104     } the 106      Public Static BooleanCheck_success (intAintBint[] r) {107         108         int[] k_table =New int[10] [10];109         //init ... thek_table[1][3]=2; k_table[3][1]=2; 111k_table[1][7]=4; K_table[7][1]=4;  thek_table[1][9]=5; K_table[9][1]=5;113         //init .... thek_table[2][8]=5; K_table[8][2]=5;  the         //init .... thek_table[3][7]=5; K_table[7][3]=5; 117k_table[3][9]=6; K_table[9][3]=6; 118         //init ....119k_table[4][6]=5; K_table[6][4]=5;  -         //init ....121k_table[7][9]=8; K_table[9][7]=8;122         if(k_table[a][b]>0){123             if(r[k_table[a][b]]==1)124                 return true; the             Else126                 return false;127         } -         Else129             return true; the         131          the     }133}

Nine Gongge cipher combination counting problem