# Single Alternate Sue de Coq Trial

In the Sysudoku original Sue de Coq, two ALS in the two bent region remainders limit the cells of the box line intersection, the chute, to a required value and two alternative pairs of values. Each ALS forces an alternative in one cell. With two ALS, the Sue de Coq is an elimination method, matching ALS pairs of values with alternative values for two cells.

With one ALS, which occurs much more frequently, there is a possible trial, which may reach an evident contradiction, and gain the same removals. The box/line intersection (the chute) has five values, one of which is definitely true within the intersection.

Let’s describe the contents of the three chute cells as N(a+b)(?). Values a and b are group values in the ALS, most often a bv with values a and b. It’s a or b because of the ALS. They can’t both be in the chute, because they would see the ALS values and the ALS can give up only one.

Of the other values, two match two value groups of an ALS in a remainder. They cannot both be in the solution of the chute, because an ALS can give up only one value. If one of the ALS values is in the chute that leaves two values for the other cell, say c and d, and the contents of the three cells is

N(a+b)(c+d)

That’s not a complete description because there is another possibility. Both a and b could be missing the chute. So actually the description is

N(a+b)(c+d) + term for missing a and b.

Let’s do a couple of examples showing how s to build that term. Actually that term is what you put on trial. It specifies what the chute contains, if it is not the Sue de Coq expression. If the trial contradicts, the extra term is dropped and the Sue de Coq removals apply.

Let’s do a couple of examples showing how s to build that term. Actually that term is what you put on trial. It specifies what the chute contains, if it is not the Sue de Coq expression. If the trial contradicts, the extra term is dropped and the Sue de Coq removals apply.

Nr1 =5(3+9)(2+7) +572

Now if the test of

Nr1 = 572 fails, are there any SdC removals? Only 3r1c7. It’s in the row remainder with 39r1c2

.Maybe that’s why I chose another type of trial.

There’s another SASdC in there. Its SWc3. It’s a little harder to write the extra terms.

7(1+9)(3+8) + 783 + 873.    But graphically, you’re testing the SWc3 chute shown here:

There are a lot of early removals in the test, but only one if the trial reaches contradiction.