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#thatlogicproblem round-up

By Christian Lawson-Perfect. Posted April 18, 2015

C: $$\begin{gathered} K_{A}m; \\ {K}_{B}d. \end{gathered}$$

A: $$\begin{gathered} \mathrm{\neg }{K}_{A}d; \\ m\models \mathrm{\neg }{K}_{B}m. \end{gathered}$$

B: $$\begin{gathered} d⊭K_{B}m; \\ (K_A(\mathrm{\neg }{K}_{B}m))\models K_B(m,d). \end{gathered}$$

A: $m\wedge K_B(m,d)\models K_A(m,d).$

Albert, Bernard and Cheryl have had a busy week. They’re the stars of #thatlogicproblem, a question from a Singapore maths test that was posted to Facebook by a TV presenter and quickly sent the internet deduction-crazy.

First of all: no, it’s not meant to be answered by an average Singaporean student. It’s a hard question from a schools Olympiad test.