Ronnie O’Sullivan aims to progress to a sixth Welsh Open final at the expense of fellow top 16 star Kyren Wilson in Cardiff on Saturday.
O’Sullivan has won through to his second ranking Semi-Final of the season and ninth in total at the Welsh Open since making his debut back in 1993.
The world number six last tasted ranking success in the Tour Championship last March but did claim the Shanghai Masters invitational title at the beginning of the season.
Runner-up to Judd Trump in the Northern Ireland Open, ‘The Rocket’ remains in hot pursuit of a record-breaking 37th career ranking title to overtake Stephen Hendry, and to do it here would see him claim a record-equalling fifth Welsh Open crown.
O’Sullivan’s run to the Last Four has seen him provisionally move into the top 16 on the one-year ranking list, bolstering his hopes of Players Championship and Tour Championship qualification.
He has so far seen off Zhang Jiankang, Stuart Carrington, Anthony Hamilton, Soheil Vahedi and rival Mark Selby the loss of just six frames.
Wilson, meanwhile, has banished his previously poor record in this prestigious event to progress beyond the Fourth Round stage for the first time in seven attempts.
The world number eight is through to his second successive ranking Semi-Final and third of the season in total as he bids to reach his first final since last year’s German Masters.
Wilson came through a last-frame decider against Jackson Page in Round One before overcoming Liam Highfield 4-2 and Martin O’Donnell 4-0.
The Kettering ace then dispatched Ding Junhui 4-2 in the Last 16 before pulling off a sensational 5-0 whitewash of man of the moment Neil Robertson on Friday.
Wilson has lost four of his five previous meetings with O’Sullivan, including an agonising 10-9 defeat in the final of the 2018 Champion of Champions.
Their latest show-down gets underway from 1pm GMT and will be contested over the best of 11 frames.
Either China Championship winner Shaun Murphy or Riga Masters champion Yan Bingtao will await the winner in Sunday’s final.