tapchannel: fix split root when initiator has 0 balance

A previous bug fix was wrong... In the case where an HTLC sends all the
balance from the initiator to the recipient, then the initiator has zero
balance _and_ there is no HTLC going to the initiator. So the root split
still becomes nil and we panic.
So we re-declare the rule for the case where the initiator has zero
balance: If the initiator doesn't have any balance to carry the split
root, then the very first HTLC will carry the split root. If there is no
HTLC, it means all the balance is on the recipient's side and there is
no split (hence no split root required).

Author: guggero

tapchannel/commitment.go

@@ -667,9 +667,8 @@ func CreateAllocations(chanState *channeldb.OpenChannel, ourBalance,
 
 	// Next, we add the HTLC outputs, using this helper function to
 	// distinguish between incoming and outgoing HTLCs. The haveHtlcSplit
-	// boolean is used to determine if one of the HTLCs needs to become the
-	// split root. See below where it's used for a more in-depth
-	// explanation.
+	// boolean is used to store if one of the HTLCs has already been chosen
+	// to be the split root (only the very first HTLC might be chosen).
 	var haveHtlcSplitRoot bool
 	addHtlc := func(htlc *DecodedDescriptor, isIncoming bool) error {
 		htlcScript, err := lnwallet.GenTaprootHtlcScript(
@@ -688,34 +687,14 @@ func CreateAllocations(chanState *channeldb.OpenChannel, ourBalance,
 				"sibling: %w", err)
 		}
 
-		// The "incoming" vs. "outgoing" naming is always viewed from
-		// the perspective of the local node and independent of
-		// isOurCommit. So to find out if an HTLC is for the initiator
-		// it only depends on the direction and whether we are the
-		// initiator. The human-readable version of the below switch
-		// statement is: Either we're the initiator and the HTLC is
-		// incoming, or we're not the initiator and the HTLC is
-		// outgoing.
-		var htlcIsForInitiator bool
-		switch {
-		// If the HTLC is incoming and the local node is the channel
-		// initiator, then the HTLC is for the channel initiator.
-		case chanState.IsInitiator && isIncoming:
-			htlcIsForInitiator = true
-
-		// If the HTLC is outgoing and the local node is not the
-		// channel initiator, then the HTLC is for the channel
-		// initiator.
-		case !chanState.IsInitiator && !isIncoming:
-			htlcIsForInitiator = true
-		}
-
-		// We should always just have a single split root on the side
-		// of the initiator. So if the initiator doesn't have any normal
-		// asset output to house the split root, then we'll take the
-		// _first_ HTLC that pays to the initiator.
+		// We should always just have a single split root, which
+		// normally is the initiator's balance. However, if the
+		// initiator has no balance, then we choose the very first HTLC
+		// in the list to be the split root. If there are no HTLCs, then
+		// all the balance is on the receiver side and we don't need a
+		// split root.
 		shouldHouseSplitRoot := initiatorAssetBalance == 0 &&
-			htlcIsForInitiator && !haveHtlcSplitRoot
+			!haveHtlcSplitRoot
 
 		// Make sure we only select the very first HTLC that pays to the
 		// initiator.