Remove Condition For Passed Pawns

Currently, we do not consider pawns passed if there is another pawn of
the same color in front of them. It appears that this condition is not
necessary. The idea is that the doubled pawns are likely to be weak and
one of them will be likely captured anyway. On the other hand, if we do
somehow manage to promote a pawn, then the pawn behind it becomes passed
as well. In any case, the end result is we end up with an extra
potentially passed pawn. The current evaluation for passed pawns already
handles this case by also scaling down this effect.

STC:
LLR: 2.96 (-2.94,2.94) [-3.00,1.00]
Total: 28291 W: 6287 L: 6178 D: 15826
http://tests.stockfishchess.org/tests/view/5b6c4b960ebc5902bdb9f256

LTC:
LLR: 2.96 (-2.94,2.94) [-3.00,1.00]
Total: 30717 W: 5256 L: 5151 D: 20310
http://tests.stockfishchess.org/tests/view/5b6c82980ebc5902bdb9f863

Bench: 4938285

diff --git a/src/pawns.cpp b/src/pawns.cpp
index 761d770e6772534ac15b859e210d901b47430918..212f0a2866e5f52695f6cb75d2caa78548657190 100644 (file)
--- a/src/pawns.cpp
+++ b/src/pawns.cpp
@@ -118,7 +118,6 @@ namespace {
         // which could become passed after one or two pawn pushes when are
         // not attacked more times than defended.
         if (   !(stoppers ^ lever ^ leverPush)
         // which could become passed after one or two pawn pushes when are
         // not attacked more times than defended.
         if (   !(stoppers ^ lever ^ leverPush)

-            && !(ourPawns & forward_file_bb(Us, s))

             && popcount(supported) >= popcount(lever) - 1
             && popcount(phalanx)   >= popcount(leverPush))
             e->passedPawns[Us] |= s;
             && popcount(supported) >= popcount(lever) - 1
             && popcount(phalanx)   >= popcount(leverPush))
             e->passedPawns[Us] |= s;