Paging Gamers — Beyond: Two Souls Demo Now Live

I struggled to get through Heavy Rain due to its clumsy controls, so my big question about Quantic Dream’s Ellen Page “interactive drama” Beyond: Two Souls is, “How does it handle?” I can find out right now because the Beyond: Two Souls demo is live on PlayStation Network.

In a post on the official PlayStation Blog, Sony revealed the demo will place players in two “scenes” from the final game:

  • Jodie & Aiden: Join a very young Jodie as she takes part in an experiment at the Department of Paranormal Activities. As officials put her abilities to the test, you take charge of a mysterious, invisible entity.
  • Hunted: Skip forward a number of years and help Jodie escape the clutches of government agents as the mysterious entity unleashes its truly astonishing powers.

David Cage also recently revealed that you can play through all of Beyond: Two Souls using your smartphone as a controller, so I’m quite curious to see how that compares to DualShock controls. Looking forward to finding out.

Have you jumped into Ellen Page’s shoes yet? What do you think of Quantic Dream’s anticipated follow-up?

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1 Comment on Paging Gamers — Beyond: Two Souls Demo Now Live


On February 19, 2015 at 7:17 am

Menon”3,4,5 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 6 ആണല ല (3^3+4^3+5^3 = 6^3)6,8,10 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 12 ആണല ല (6^3+8^3+10^3 = 12^39,12,15 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 18 ആണല ല (9^3+12^3+15^3 = 18^3)ഈ ക ണ ന ന ബന ധത ത ന എന ത ങ ക ല ഒര ക രണ നല ക ക വ ന പറ റ മ എന ന ര സ ശയ ഉണ ട “May I approach the sioiatutn as follows:3, 4 and 5 is a Pythagorean triplet.3^3 +4^3 + 5^3 = 27 + 64 + 125=216Also [(3+4+5)/2]^3 = 6^3 =216It proves that 3^3 +4^3 + 5^3 = [(3+4+5)/2]^3We notice that each successive Pythagorean triplets that you have mentioned in your post is an integral multiple of the triplet (3,4,5) and hence a general Pythagorean triplet in your post can be expressed as (3k, 4k, 5k)where k= 1, 2, 3, 4,………Now (3k)^3 + (4k)^3 + (5k)^3 =k^3(3^3 +4^3+ 5^3)= k^3 [(3+4+5)/2]^3 by substitution from the previous relationHence, (3k)^3 + (4k)^3 + (5k)^3 =k^3 [(3+4+5)/2]^3 = [k(3+4+5)/2]^3 = [(3k+4k+5k)/2]^3i.e (3k)^3 + (4k)^3 + (5k)^3 = [(3k+4k+5k)/2]^3In the above relation put k=1, 2, 3, 4, 5….. Then we get all those relations mentioned in your post.This is a special property of Pythagorean triplets of the form (3k,4k,5k)If you examine the Pythagorean triplet (5,12,13)we see that this property namely sum of cubes equals cube of semi-perimeter does not hold.