China DOS Union

I believe many people know the magic square.



Nine in the top, one in the bottom, three on the left, seven on the right, two and four on the shoulders, six and eight on the feet, and the sum of three numbers in horizontal, vertical, and diagonal directions is all 15.

Suddenly I thought, can we use batch processing to display the magic square? It's not just a simple echo.

We need to list the magic square pattern, with one number in each grid, and the numbers in the grids change every few seconds. For example:

Magic square           

 4  9  2

3  5  7

8  1  6



Change to

2  9  4

7  5  3

6  1  8



Then change to

6  1  8

7  5  3

2  9  4



Then change to

8  1  6

3  5  7

4  9  2



Then change to

6  7  2

1  5  9

8  3  4



Then change to

8  3  4

1  5  9

6  7  2



And so on. The requirement is to list all possibilities. Remember it's not just listing, but showing one kind each time.

[ Last edited by ccwan on 2007-1-14 at 01:31 AM ]

2 4 6 8
1 3 7 9
5
The format of the River Chart is prescribed

It's not about discussing the River Chart and Luo Book, just meet the requirements of the first floor.

Nine palaces and eight trigrams, it's obvious that there are only eight results

This is difficult.

This difficulty is indeed great, it's not easy to find the pattern

The demonstration code is as follows:

@echo off

title Nine-square grid

mode con cols=40 lines=20

set n1=0

setlocal enabledelayedexpansion



:begin

cls

set n5=5

set str= 1 2 3 4 6 7 8 9

set /a n1+=1

if %n1% gtr 9 exit

call :intercept %n1%

set /a n9=15-%n1%-%n5%

call :intercept %n9%

for %%i in (%str%) do (

    set flag=

    set n2=%%i

    set str_tmp=%str%

    call :intercept !n2!

    set /a n8=15-!n2!-!n5!

    set /a n3=15-!n1!-!n2!

    set /a n7=15-!n8!-!n9!

    set /a n4=15-!n1!-!n7!

    set /a n6=15-!n3!-!n9!

    for %%j in (!n3! !n4! !n6! !n7! !n8!) do call :intercept %%j

    if not "!str!"=="" set flag=1

    set /a sum1=!n3!+!n5!+!n7!

    set /a sum2=!n4!+!n5!+!n6!

    if !sum1! equ 15 if !sum1! equ !sum2! if not defined flag call :result

    set str=!str_tmp!

)

goto begin



:intercept

set str=!str: %1=!

goto :eof



:result

cls

echo.

echo.

echo.

echo            ┌──┬──┬──┐

echo            │ %n1%  │ %n2%  │ %n3%  │

echo            ├──┼──┼──┤

echo            │ %n4%  │ %n5%  │ %n6%  │

echo            ├──┼──┼──┤

echo            │ %n7%  │ %n8%  │ %n9%  │

echo            └──┴──┴──┘

echo.

echo ________________________________________

echo.

echo             Press any key to display the next one

pause>nul

goto :eof

There was a misspelled English word, corrected it: (By the way, the display interface was beautified.

[ Last edited by namejm on 2007-1-13 at 05:38 PM ]

Originally posted by lxmxn at 2007-1-13 04:30 PM:

　　The moderator is really amazing, PF.

I feel the same!~~

Brother namejm's code is extremely wonderful~~~! Here the nine-square grid refers to a magic square. The nine-square grid given by the original poster is a 3x3 magic square. When transforming, it just swaps the arrays on both sides with the horizontal and vertical centers as the axis. The sums of its rows, columns, and diagonals are all the same numbers added together.

You can also expand it and increase the order to 5 or 7, etc.

A magic square is a square array in which the sum of each row, each column, and the diagonal are all equal. The following is a 3x3 magic square:



8  1  6

3  5  7

4  9  2





Algorithm for printing a magic square:



1. Place 1 in the middle column of the first row;



2. For each number from 2 to n*n, place each number according to the following rule: The row of each number is 1 less than the row of the previous number, and the column number is 1 more. For example, in the above 3x3 magic square, 5 is in the row above and column to the right of 4;



3. If the row of the previous number is 1, then the row of the next number is n (referring to the bottom row). For example: 1 is in the 1st row, then 2 should be placed in the bottom row, and the column number is also increased by 1;



4. When the column number of the previous number is n, the column number of the next number should be 1, and the row number is decreased by 1. For example: 2 is in the last column of the 3rd row, then 3 should be placed in the 2nd row and 1st column;



5. If the position determined by the above rule already has a number, or the previous number is in the 1st row and nth column, then place the next number below the previous number. For example: According to the above regulations, 4 should be placed in the 1st row and 2nd column, but this position is already occupied by 1, so 4 is placed below 3. Since 6 is in the 1st row and 3rd column (i.e., the last column), 7 is placed below 6. In this way, you can get a magic square of any order.

According to the above algorithm, it is completely feasible to implement printing a magic square with batch processing. Probably need to construct a two-dimensional array.

Send a small program I wrote before to print a magic square.

[ Last edited by pengfei on 2007-1-15 at 01:29 AM ]

The level of namejm is really remarkable, I admire it a lot!

Master, KOOL````````

```
@ECHO OFF
:TOP
SET /A R=%RANDOM%%%8
SET /A R=%R%+%R%%%2
IF %R% LEQ 0 GOTO TOP
IF 0x00%R% EQU 0x00%A% (
SET Ct=1) ELSE (
SET Ct=0)
SET A=%R%
SET /A E=5
SET /A At=15-%A%
SET /A I=%At%-%E%
SET /A It=15-%I%
SET /A X=15-%A%*2-%E%
IF %X% LEQ 0 SET /A X=~%X%+1
SET /A Y=20-%A%-%X%-%I%
IF %Ct% EQU 0 (SET /A C=%X%,G=%Y%) ELSE SET /A C=%Y%,G=%X%
SET /A B=%At%-%C%,D=%At%-%G%,F=%It%-%C%,H=%It%-%G%

ECHO %A% %B% %C%
ECHO %D% %E% %F%
ECHO %G% %H% %I%
pause>nul
cls&&goto top
```