Sudoku Solver for HP-33s
Message #1 Posted by Marcel Samek on 10 May 2012, 12:35 p.m.
This is impractically slow on the HP-33s and not really useful, but it does work. It is a brute force solver for Sudoku puzzles. It will find the first solution to any solvable puzzle.
This was written for the HP 15c and is not really designed to make full use of the 33s. It could probably be a *lot* faster on the 33s with a bit of work, but my real focus was to validate the 15c code and so I have not put an effort into optimizing it on the 33s.
To run: XEQ A
It will prompt you to enter values Q through Y which represent row 1 through 9 of the unsolved sudoku puzzle.
In your input, the digit 0 represents an unsolved space. For example 100070003. You do not have to enter leading 0s.
When the program finishes (many many hours later, if you have given it a simple puzzle), You will find the solved rows in Q through Y. (There should really be a loop that spits them out).
Have fun and sorry for the lack of documentation. I will post that separately. HP 15c version is on the way.
Marcel
-----------------
/************************************************************************
getPart()
************************************************************************/
LBL F (CN=B583)
ENTER
ENTER
8
XEQ H
RCL (i)
RTN
/************************************************************************
setPart()
************************************************************************/
LBL G (CN=2942)
ENTER
8
XEQ H
STO (i)
R DOWN
RTN
LBL H (CN=F793)
+
STO i
R DOWN
RTN
/***********************************************************************
setPow2()
***********************************************************************/
LBL J (CN=8F3C)
STO Z
1
-
2
XY SWITCH
POW
X EXCH Z
RTN
/***********************************************************************
setU()
***********************************************************************/
LBL O (CN=D238)
XEQ J
RCL B
XEQ N
RCL C
XEQ N
RCL D
XEQ N
RTN
// imp subroutine that does setPart(x, getPart(x) + reg[VAR_TMP1])
LBL N (CK=ADAF)
XEQ F
RCL Z
FS? 3
CHS
+
XY SWITCH
XEQ G
XEQ M
RTN
/***********************************************************************
putA()
***********************************************************************/
LBL L (CK=7A6C)
X <> F
STO Z
RCL B
17
XEQ H
RCL (i)
RCL F
RCL Z
-
RCL E
*
+
STO (i)
RTN
/***********************************************************************
change()
***********************************************************************/
LBL K (CK=C833)
STO+ A
RCL A
9
INT /
STO B
RCL A
9
RMDR
STO C
3
INT /
RCL B
3
INT /
3
*
+
STO D
8
RCL C
-
10^X
STO E
RCL B
17
XEQ H
RCL (i)
XEQ I
STO F
STO G
R DOWN
RCL C
28
+
XEQ J
RCL B
XEQ F
RCL Z
INT /
2
RMDR
X<=0?
STO G
RTN
LBL I (CK=9AD5)
RCL E
INT /
10
RMDR
RTN
/***********************************************************************
main()
***********************************************************************/
LBL A (CK=0970)
CLRVARS
INPUT Q
INPUT R
INPUT S
INPUT T
INPUT U
INPUT V
INPUT W
INPUT X
INPUT Y
1
CHS
STO A
LBL D (CK=D600)
1
XEQ K
RCL F
X<=0?
GTO Q
XEQ O
RCL C
28
+
XEQ J
RCL B
XEQ F
RCL+ Z
XY SWITCH
XEQ G
LBL Q (CK=FEB7)
80
RCL A
X NE Y?
GTO D
1
CHS
STO A
LBL E (CK=E698)
1
XEQ K
9
RCL B
X=Y?
R/S // finished!!!!!
RCL G
x>0?
GTO E
XEQ L
LBL X (CK=0B63)
9
RCL F
X=Y?
GTO C
1
+
XEQ L
RCL F
XEQ J
CF 2
RCL B
XEQ P
FS? 2
GTO X
RCL C
XEQ P
F? 2
GTO X
RCL D
XEQ P
F? 2
GTO X
RCL F
XEQ O
GTO E
LBL C (CK=E045)
1
CHS
XEQ K
X>0?
GTO C
RCL F
SF 3
XEQ O
CF 3
GTO X
LBL P (CK=014A)
XEQ F
RCL Z
INT /
2
RMDR
X>0?
SF 2
R DOWN
R DOWN
XEQ M
RET
LBL M (CK=204F)
9
+
XEQ J
RTN
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