Easy Puzzle #4 - One Possible Solution Approach…
	NB:  There is only one final solution to this Sudoku puzzle…
	A	B	C	D	E	F	G	H	I
1
2	AA			BB			CC				Reference to 3x3 Blocks AA, BB & CC
3
4
5	DD			EE			FF				Reference to 3x3 Blocks DD, EE & FF
6
7
8	GG			HH			II				Reference to 3x3 Blocks GG, HH & II
9
	A	B	C	D	E	F	G	H	I
1					5			3			Easy Puzzle #4 - October 4, 2005
2		5		8	4	2	7		6		Original puzzle as published on joe-ks.com…
3	2	7			6		4		9
4		4	5	2				9
5				1		7	5
6		9			8		1	2
7	5		4	9	1			7
8	7	3			2	5	6		1
9		2	1			3	9	4
	A	B	C	D	E	F	G	H	I
1					5			3			Step 1: Refer to Blocks AA-BB-CC: #5 exists in Rows 1 & 2;
2		5		8	4	2	7		6		Since you can't repeat 5 in rows 1 or 2 OR in AA or BB,
3	2	7			6		4	5	9		5 MUST go in Block CC @ cell H3
4		4	5	2				9
5				1		7	5
6		9			8		1	2
7	5		4	9	1			7
8	7	3			2	5	6		1
9		2	1			3	9	4
	A	B	C	D	E	F	G	H	I
1					5			3			Step 2: Carry on with #5 - look @ vertical Blocks CC-FF-II;
2		5		8	4	2	7		6		You can't repeat 5 in columns G or H OR in CC or FF…
3	2	7			6		4	5	9		so the possible cells for 5 (in Block II) are I7 or I9…
4		4	5	2				9			BUT a 5 exists in Row 7 - so 5 MUST go in cell I9.
5				1		7	5
6		9			8		1	2
7	5		4	9	1			7
8	7	3			2	5	6		1
9		2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3			Step 3: Refer to middle Blocks BB-EE-HH #5's in Columns
2		5		8	4	2	7		6		E & F. Therefore, #5 must go in Block EE, and it MUST
3	2	7			6		4	5	9		go in Cell D6…
4		4	5	2				9
5				1		7	5
6		9		5	8		1	2
7	5		4	9	1			7
8	7	3			2	5	6		1
9		2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3
2		5		8	4	2	7		6
3	2	7			6		4	5	9
4	1	4	5	2				9			Step 4: #1 exists in Rows 5 & 6 - #1 MUST go in
5				1		7	5				Row 4 @ Cell A4…
6		9		5	8		1	2
7	5		4	9	1			7			Step 5: #9 is in Rows 7 & 9 - #9 MUST go in
8	7	3	9		2	5	6		1		Row 8 @ Cell C8…
9		2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3			Step 6: Look at Row 8 - we only need 2 more #s (4 & 8) to
2		5		8	4	2	7		6		complete the row; BUT since #4 is in Column H (H9),
3	2	7			6		4	5	9		#4 CAN'T go in Cell H8… Therefore, for Row #8, #8 MUST
4	1	4	5	2				9			go in Cell H8… and of course that means that the
5				1		7	5				remaining #4 must go in the only cell left in Row 8 -
6		9		5	8		1	2			@ Cell D8 (added in next step)...
7	5		4	9	1			7
8	7	3	9		2	5	6	8	1
9		2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3			Step 7: #8 is in Columns D & E, and MUST go in the only
2		5		8	4	2	7		6		Cell left in Block HH - @ Cell F7…
3	2	7			6		4	5	9
4	1	4	5	2				9
5				1		7	5
6		9		5	8		1	2
7	5		4	9	1	8		7
8	7	3	9	4	2	5	6	8	1
9		2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3			Step 8:  #8 is in Rows 7 & 8, and MUST go in
2		5		8	4	2	7		6		Row 9 @ Cell A9…
3	2	7			6		4	5	9
4	1	4	5	2				9
5				1		7	5
6		9		5	8		1	2
7	5		4	9	1	8		7
8	7	3	9	4	2	5	6	8	1
9	8	2	1			3	9	4	5
	A	B	C	D	E	F	G	H	I
1					5			3
2		5		8	4	2	7		6
3	2	7			6		4	5	9
4	1	4	5	2				9
5				1		7	5
6		9		5	8		1	2
7	5	6	4	9	1	8		7			Step 9: Look @ Block GG - Piece of Cake!
8	7	3	9	4	2	5	6	8	1		We have 8 of the 9 numbers already - only one left is #6
9	8	2	1			3	9	4	5		which MUST go in Cell #B7…
	A	B	C	D	E	F	G	H	I
1				7	5			3			Step 10: #7 is in Rows 2 & 3: where does it go in Row 1?
2