The AMD Ryzen CPU was demoed recently and what a demo it was. AMD showed that the upcoming AMD...
No Legendary Pokemon Was Sent to Any Pokemon Go Player, Says Niantic
Recently reports emerged that the first Pokemon Go legendary Pokemon has popped up in Ohio, and few of the trainers have managed to capture it. While there were red flags indicating that it was fake, but some speculated that it could be a local event.
However, Niantic has clarified the situation, as the studio has released an official response. According to spokesperson from the company, the legendary Pokemon “Articuno” was not sent by Niantic labs, it is either a spoof or a hack.
Now it remains to be seen that the legendary Pokemon was obtained through hacking or simply a fake news. Niantic themselves are not sure about it at the moment, for now they are looking in to the matter and will share the news once they get at the bottom of this.
However, the ruse has been exposed and we can expect an explanation from the developer soon. If this is a hack, then there is no telling that the hackers will further exploit the hole they have found, only time will tell.
In related news, the popularity of Pokemon Go is reaching new heights, as according to a new report that suggests the in-app time spent by players is high. According to the report, players spent 26 minutes on average in US playing Pokemon Go and almost 6 times a day the app is being opened by users.
What is more interesting is the, Pokemon Go global revenue has crossed $160 million and is showing no decline.
Pokemon Go has been officially released for only iOS and Android, while Windows mobile users were left behind. A petition was also started to get Niantic to make Pokemon Go Windows phone port, but given that how busy the developer is with already existing ports, it is unlikely we will see a Windows port. However, an unofficial Windows phone port of Pokemon Go has emerged on the internet.
Pokemon Go is an augmented reality game published by The Pokemon Company for iOS and Android.
Source: Nintendo everything.