Remove std::pow from reduction. (#2234)

This is a functional simplification that removes the std::pow from reduction. The resulting reduction values are within 1% of master.

This is a simplification because i believe an fp addition and multiplication is much faster than a call to std::pow() which is historically slow and performance varies widely on different architectures.

STC
LLR: 2.95 (-2.94,2.94) [-3.00,1.00]
Total: 23471 W: 5245 L: 5127 D: 13099
http://tests.stockfishchess.org/tests/view/5d27ac1b0ebc5925cf0d476b

LTC
LLR: 2.95 (-2.94,2.94) [-3.00,1.00]
Total: 51533 W: 8736 L: 8665 D: 34132
http://tests.stockfishchess.org/tests/view/5d27b74e0ebc5925cf0d493c

Bench 3765158

diff --git a/src/search.cpp b/src/search.cpp
index bd5ae75e0582145276a7f6d3f7681d7bf3e8d02f..df19108fb2db4dd28642c1ef20b34f49aa5f2a53 100644 (file)
--- a/src/search.cpp
+++ b/src/search.cpp
@@ -503,7 +503,7 @@ void Thread::search() {
 
           // If the bestMove is stable over several iterations, reduce time accordingly
           timeReduction = lastBestMoveDepth + 10 * ONE_PLY < completedDepth ? 1.95 : 1.0;
 
           // If the bestMove is stable over several iterations, reduce time accordingly
           timeReduction = lastBestMoveDepth + 10 * ONE_PLY < completedDepth ? 1.95 : 1.0;

-          double reduction = std::pow(mainThread->previousTimeReduction, 0.528) / timeReduction;
+          double reduction = (1.25 + mainThread->previousTimeReduction) / (2.25 * timeReduction);

 
           // Use part of the gained time from a previous stable move for the current move
           for (Thread* th : Threads)
 
           // Use part of the gained time from a previous stable move for the current move
           for (Thread* th : Threads)