by **Jean-Christophe** » Sat Aug 19, 2006 2:56 pm

As promised, here is my walkthrough in tiny text you may copy-paste to a text editor like note pad

I'll use an abbreviated notation for the colors : G = green, B = blue, P = pink, Y = yellow

When the sum is determined, the color is followed by the sum and number of cells eg Y 11/2 means a yellow cage of 2 cells adding up to 11

When the sum is not yet determined, the color is followed by number of cells eg Y/2 means a yellow cage of 2 cells.

When the color is not yet determined, a slash preceede the number of cells eg /2 means a cage of 2 cells.

Two possible colors are separated by a vertical bar eg P|B/3 means a cage of 3 cells either pink or blue.

Step 1

There are 4 cages of 4 cells to place in N1 and N9. Two of them are P and can't touch P 12/3 in C2.

-> P 18/4 & P 20/4 go in N9

/4 in C34 <> G = B 20/4

/4 in C1 = G 15/4

Step 2

The two /2 in R5 must be Y

There are only three Y/2, two of them are Y 16/2 = {79} and can't go both in R5

-> one of the /2 in R5 = Y 16/2 = {79}, the other = Y 11/2 = {38|56}

/2 in R4C78 = B. The only B/2 = B 5/2 = {14|23}

Step 3

The B 28/5 can't go in N7 nor in N5

-> B 28/5 go in N3

/5 in N5 & N7 = Y/5

/2 in R3C67 = G/2

/2 in R2C67 = Y/2 = the other Y 16/2 = {79}

/3 in N2 = P 13/3

/2 in R1C56 = G/2

/3 in R1C234 = Y/3

Step 4

/2 in R6C23 = P/2

/2 in R7C34 = G/2

/2 in R8C34 = P/2

/3 in N7 = B 14/3

/2 in R9C45 = G/2

/3 in R9C678 = Y/3

We now have found the colors of all cages, but only a few sums.

Step 5

In N5, /5 is either Y 20/5 or Y 29/5

See what the outies gives if it were Y 29/5

45 on N5 -> R5C37 = 20+29+12 - 45 = 16 = {79}

This can't work with the Y 16/2 either in R5C12 or R5C89

-> Y 20/5 go in N5, Y 29/5 go in N7

45 on N5 -> R5C37 = 7/2

45 on R5 -> R5C456 = 11/3

Step 6

In R5, Since either R5C12 or R5C89 = Y 16/2 = {79} -> G 20/3 in R45C4+R5C3 <> {479}

-> R5C567 = {124} = 7 (hidden triplet in R5)

45 on R5 -> R4C4 = 9, R5C34 = 11 = {38|56}

Step 7

In N3, we already found that R2C67 = Y 16/2 and B 28/5 go in N3

45 on N3 -> R2C6+R4C9 - R3C7 = 28+16+17 - 45 = 16

Max R2C6+R4C9 = 9+8 = 17 -> Max R3C7 = 17-16 = 1 = Min R3C7

-> R3C7 = 1, R2C67 = [97], R4C9 = 8

Max sum of G/2 in R3C67 = 1+8 = 9

There is only one G/2 <= 9

-> R3C67 = G 5/2, R3C6 = 4

R3C89 = {36}

Step 8

G/2 in R1C56 = 12|13 = {57|58|67} = {(5|6)...} = {(5|7)...}

-> P 13/3 in N2 <> {256|157} = {238}

-> R2C5 = 3, R3C45 = {28}

R1C56 = {57|67} = {7(5|6)}

R12C4 = {1(5|6)}

Step 9

G 12/3 in N5 = {147|246} = {4...}

-> R5C7 = 4, R5C56 = {12}, R6C6 = {67}

Since R5C37 = 7/2 -> R5C34 = [38], R3C45 = [28]

Y 11/2 in R5 = {56}

B 5/2 in R4C78 = {23}

R6C4 = 3

Step 10

P/2 in R6C23 = 7|11

Since R5C3 = 3 -> R6C23 <> {38}

-> R6C1 = 8, R6C23 = {2...} = {25|29}

R6C5 = 4

Step 11

45 on N7 -> R78C3 = 6 = {15|24}

G/2 in R7C34 = 12|13 = [48|57|49|58]

Since R45C4 = [98] -> R8C34 = [57] = G 12/2

Since R78C3 = 6 -> R8C3 = 1

P/2 in R8C34 = 7|11 -> R8C34 = 7/2, R8C4 = 6

Step 12

P/2 in R6C23 = 11/2 = {29}

Y/2 in R5C12 = 11/2 = {56}

Y/2 in R5C89 = 16/2 = {79}

R6C789 = {156}, R56C6 = [17], R5C5 = 2

R4C123 = {147}

Step 13

R12C4 = {15}, R1C56 = [76], R4C56 = [65]

G/2 in R9C45 = G 13/2 = [49]

R78C5 = [15], R7C6 = 8

R7C12 = {46}

Step 14

R2C3 = 6 (hidden single in C3)

B 20/4 in C34 = {1469} -> R34C3 = [94], R12C4 = [51]

R6C23 = [92], R19C3 = [87]

P 12/3 in R234C2 = [471]

G 15/4 in R1324C1 = [1257]

R89C1 = [93], R89C6 = [32]

Several nacked and hidden singles

Step 15

Y/3 in R1 = [385] = Y 16/3

-> Y/3 in R9 = Y 13/3 -> R9C78 = {56}

Fix all nacked and hidden singles until R17C79 = {29}

One of the P/4 in N9 = 18/4, the other = 20/4

If you read until here, I guess you can findout this yourself !

Unique solution

138576942

246139785

579284163

714965328

653821497

892347516

465718239

921653874

387492651