solve sudoku

Portal This question is also a very violent search ...... Because the uniqueness is empty at the beginning, and there are only a lot of size restrictions, there is no need to worry about where to start the search, and then determine whether it is legal after the brute force search. The most disgusting question is reading. Now I learned a trick to determine which part of the grid is located and use Lim [g] [I] [J] to indicate the relationship between the two grids I and j in the G Palace, the pro

/3) *3+k][int (J/3) *3+1]!=0:Try: List_a.remove (a[int (i/3) *3+k][int (J/3) *3+1]) except:Pass ifA[int (I/3) *3+k][int (J/3) *3+2]!=0:Try: List_a.remove (a[int (i/3) *3+k][int (J/3) *3+2]) except:Pass returnlist_areturnA[I][J]View CodeThe most important traversal lookup result function listDataGame_over=false #初始化未找到结果, locate the result settings game_over=truedefListData (a,posx,posy,f):GlobalGame_over#Game_over is a global variable rather than a

Import NumPy as NPImport SysSys.setrecursionlimit (#例如这里设置为一百万)def get1 (n):If nreturn 0If nReturn 3Return 6def get2 (n):If nReturn 3If nReturn 6Return 9def get3 (arr,i,j):A1=list (Arr[i,:])A2=list (Arr[:,j])A3=list (Arr[get1 (i): Get2 (i), Get1 (j): Get2 (j)].reshape (1,-1) [0])A=set (A1+A2+A3)Return b-adef SD (ARR,DEP):Count[0]+=1If Dep==len (l):Result.append (arr)ElseX,Y=L[DEP]Lac=get3 (Arr,x,y)If Len (LAC) >0 and Len (result) ==0:For I in Lac:Ac=arr.copy ()Ac[x,y]=iSD (AC,DEP+1)array= [[5,3,

This article examples for everyone to share JS Lucky Draw nine Sudoku large turntable effect, for everyone's reference, the specific content as follows Implementation code: The above is the entire content of this article, I hope to learn JavaScript program to help you.

In this paper, we introduce the corresponding method of JavaScript to realize nine Sudoku, and share them for everyone's reference, the specific contents are as follows Realize the idea:1, each lattice input value must be a number; 2, the input value can not be repeated; 3, the input value can not be less than 1 or greater than 9; 4, the value can not be empty; 5, add a total of 8, respectively, three horizontal, vertical three, two diagonal two

Sudoku concept, Idea, solution idea, C++code The overall code of the blog's thinking is very clear, it is worth learning.The above blog code DFS knowledge output an array of answers, if the number is unique to multiple solutions, simply change the DFS code simple (as follows): void dfs (int row,int col) {if (row>9)//if row exceeds 9 lines, direct output {for (int i=1; i The Overall code is as follows: #include The above code looks pretty long,

PuzzlesThird-order magic Square (nine Sudoku). Try to fill in a 3x3 table with these 9 different integers, making the sum of the numbers on each row, each column, and each diagonal 1~9.PolicyExhaustive search. Lists all the integer fill schemes, and then filters them.JavaScript Solution The code is as follows Copy Code /*** Created by Cshao on 12/28/14.*/ function Getpermutation (arr) {if (arr.length = = 1) {return [arr];}

This article describes the design code for the Nine program. A Related introduction (i) Effect display 1. The effect of the program just run: 2. Click on the page to select and slide the screen 3. Select the display after the password (b) Functional Introduction 1. After clicking on a circle, add a small circle in the center of the circle.2. The page slide appears with a following line.3. When you slide to another circle, a straight line is generated that connects.4. When the number

On the internet often see someone asked Windows Mobile under the nine Sudoku control practices, I have also many times on the internet search, but found most of the C + + version, so germination of their own to write a thought, first on the picture. When selected. The effect of the selection is actually the alternating results of two pictures, because only for the 240x320 this style, so it may be written in many places, the following to some o

(!Onceflag) { return false; } } } for(intj=0; j9; J + +) {vectorint>colnumbers; Extract_col (input,j,colnumbers); for(intm=1;mTen; m++) { BOOLonceflag=once (m,colnumbers); if(!Onceflag) { return false; } } } for(intp=0; p9; p=p+3) for(intq=0; q9; q=q+3) {vectorint>subsquarenumbers; Extract_subsquare (input,p,q,subsquarenumbers); for(intm=1;mTen; m++) { BOOLonceflag=once (m,subsqu

screen draw background Initgameview (canvas, paint); NBSP ; //Redraw Surfaceview fill new data Inflatenewnum (c Anvas,paint); }CATCH (Exception e) { e.printstacktrace (); N Bsp }finally{ if (canvas! = null) NBS P //End lock drawing and submit changes. surfaceholder.unlockcanvasandpost (canvas); //unlock canvas, update } } } nbsp; it is best to use synchronous locks, and to use try Catch finally for exception trapping. private void

In this paper, we describe the method of Python's implementation of Sudoku algorithm. Share to everyone for your reference. Specific as follows: #-*-Coding:utf-8-*-"Created on 2012-10-5@author:administrator" from collections import Defaultdictimport Itertool SA = [[0, 7, 0, 0, 0, 0, 0, 0, 0], #0 [5, 0, 3, 0, 0, 6, 0, 0, 0], #1 [0, 6, 2, 0, 8, 0, 7, 0, 0], #2 # [0, 0, 0 , 3, 0, 2, 0, 5, 0], #3 [0, 0, 4, 0, 1, 0, 3, 0, 0], #4 [0, 2, 0, 9, 0, 5, 0, 0,

Today the idea is determined, roughly for the following structure:1. Define the data in the game and initialize the lattice data.2. Define touch events.In the view class, overwrite the Ontouchevent (Motionevent event) method.3, inherit the dialog class, customize a dialog.4, do the above preparation, we can re-view class, the game layout, and to operate.5. Call the MyView object in the main class.Estimated time: 5 days, each day to complete the encoding of a class.Today's content:The class of th

{ - to if(i==Bayi) + { -n=Bayi; the return; * } $ inty=i/9, x=i%9;Panax Notoginseng if(a[y][x]==0) - { the for(intI=1; i9; i++) + { A if(check (y,x,i)) the { +a[y][x]=i; -i++; $ DFS (); $i--; - if(n==Bayi) - { theI=Bayi; - Break;Wuyi } thea[y][x]=0; - } Wu if(

canvas.drawrect (0, 0, getwidth (), getheight (), Paint );//define the Brush object and draw the grid line paint Darkpaint = new paint ();d Arkpaint.setcolor (Getresources (). GetColor (R.color.shudu_dark)); Paint Hilitepaint = new paint (); Hilitepaint.setcolor (Getresources (). GetColor (R.color.shudu_hilite)); Paint Lightpaint = new paint (); Lightpaint.setcolor (Getresources (). GetColor (R.color.shudu_light));//Draw nine lattice for (int i = 0 ; iCall the MyView object in the main class.Pa

} + } A return false; the } + intMain () { - intT, I, J; $ CharC; $scanf"%d", t); - while(t--){ - bk.clear (); thememset (Row_f,0,sizeof(Row_f)); -memset (Col_f,0,sizeof(Col_f));Wuyimemset (Block_f,0,sizeof(Block_f)); the for(i =0; I 9; ++i) { - for(j =0; J 9; ++j) { WuCIN >>C; -G[I][J] = C-'0'; About if(!G[i][j]) $ Bk.push_back (Blank (i, j)); - Else -Setnum (i, J, G[i][j],1); - } A } +

); theha[x][i]=0; -li[y][i]=0; Wuxi[(x1)/3*3+ (Y1)/3][i]=0; - } About } $ intMain () - { - for(intI=1; i9; i++) - { A for(intj=1;j9; j + +) + if(j%3) the { -scanf"%s", ch); $ if(ch[0]=='') thebh[i][j+1]=1; the Else thebh[i][j+1]=0; the } - if(i%3) in for(intj=1; j9; j + +) the { thescanf"%s", ch); About if(ch[0]=='^') thebl[i+1][j]=1; the Else t

translation数独板被部分填充，空格部分用‘.‘来填充。一个部分填充的数组是否有效只需要看其填充的部分即可。OriginalCodeThe word was inscribed for a while, wrong ... Because the input is too lazy to make, directly read someone else's writing ...classSolution { Public:inta[9];BOOLIsvalidsudoku ( vectorvectorchar>> Board) {memsetA0,sizeof(a)); for(intI=0, j=0, row=0, col=0;i9; ++j,j==9? ++i,j=0, row=col=0:0) {if(board[i][j]!='. ') {if((1 -) Row)return false;Elserow|=1 -;if((1 -) a[i/3*3+j/3])return false;Elsea[i/3*3+j/3]|=1 -;

Idea: Record the number of each row in each column that has already appeared. Data is relatively weakIn addition POJ 3074 3076 must use pruning strategies. But to achieve more trouble, or after the DLX to do it againaccepted160k0ms#include POJ 2676/2918 Sudoku (DFS)

,0,sizeof(used)); $ for(intI=1; i9; i++) - { - Charstr[Ten]; -scanf"%s", str); A for(intj=0;j9; j + +) + { thea[i][j+1]=str[j]-'0'; - if(a[i][j+1]!=0) $ { thev[i][j+1]=1; the intx=a[i][j+1]; therow[i][x]=1; thecol[j+1][x]=1; -used[(I-1)/3+1[(j+1-1)/3+1][x]=1; in } the } the } About /* the for (int i=1;i the { the for (int j=1;j + printf ("\ n"); -