NEW NONUNIFORM LOWER BOUNDS FOR UNIFORM CLASSES LANCE

NEW NONUNIFORM LOWER BOUNDS FOR UNIFORM CLASSES LANCE FORTNOW GEORGIA INSTITUTE OF TECHNOLOGY RAHUL SANTHANAM UNIVERSITY OF OXFORD

SPOILERS • FOR CONSTANTS A, B WITH 1≤A<B • NTIME(NB) IS NOT CONTAINED INNTIME(NA) WITHN 1/B BITS OF ADVICE • PREVIOUSLY KNOWN WITH ONLY O(LOG N) BITS OF ADVICE • OTHER RESULTS A) WITH SUBLINEAR NONDETERMINISM • NTIME(NB) IS NOT CONTAINED IN I. ONTIME(N. K • NP DOES NOT HAVENP-UNIFORM NONDETERMINISTIC CIRCUITS OF SIZE N • RTIME(2 N) NOT INRTIME(NC)/N 1/2 C

THE POWER OF MORE • HARTMANIS AND STEARNS 1965 • A COMPUTER CAN DO MORE GIVEN MORE TIME • A COMPUTER CAN DO MORE GIVEN MORE MEMORY

DETERMINISTIC TIME HIERARCHY • FOR A>B • DTIME(NA) IS NOT CONTAINED INDTIME(NB) • ON INPUT 0 N SIMULATEMN(0 N) FORNB STEPS AND ACCEPT IF REJECT AND VICE VERSA

AN EXISTENTIAL CRISIS • FOR A>B • NTIME(NA) IS NOT CONTAINED INNTIME(NB) • ON INPUT 0 N SIMULATEMN(0 N) FORNB STEPS AND ACCEPTIF REJECT AND VICE VERSA • DOESN’T WORK OF MACHINE HAS BOTH ACCEPTING AND REJECTING PATHS

NONDETERMINISTIC TIME HIERARCHY • FOR ANY REAL R AND S, 1 ≤ R < S • NTIME(NR) IS STRICTLY CONTAINEDNTIME( IN NS) • REDUCE TO DETERMINISTIC DIAGONALIZATION • FIRST PROVED BY STEVE COOK IN 1972 • SEIFERAS-FISCHER-MEYER 1978 • IF T(N+1) = O(U(N)) THEN NTIME(T(N)) IS STRICTLY CONTAINEDNTIME( IN U(N)) • SIMPLIFIED PROOF BY 1983 • NEW PROOF BY FORTNOW AND SANTHANAM 2011

FORTNOW-SANTHANAM M(1 j 01 m 0) M(1 j 01 m 00) • DEFINE A NTM M THAT ON INPUT J 101 M 0 W • IF |W| ≤ T(J+M+2) THEN ACCEPT IF BOTH MJ(1 J 01 M 0 W 0) AND MJ(1 J 01 M 0 W 1) ACCEPT • IF |W| > T(J+M+2 ) THEN ACCEPT IF MJ(1 J 01 M 0) REJECTS ON THE PATH SPECIFIED BY W • SUPPOSE M AND MJ ACCEPT THE SAME LANGUAGE • M(1 J 01 M 0) ACCEPTS IF M(1 J 01 M 00) AND M(1 J 01 M 01) ACCEPT M(1 j 01 m 01)

M(1 j 01 m 0) FORTNOW-SANTHANAM M(1 j 01 m 00) • DEFINE A NTM M THAT ON INPUT J 101 M 0 W • IF |W| M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) ≤ T(J+M+2) THEN ACCEPT IF BOTH MJ(1 J 01 M 0 W 0) AND MJ(1 J 01 M 0 W 1) ACCEPT • M(1 j 01 m 01) IF |W| > T(J+M+2 ) THEN ACCEPT IF MJ(1 J 01 M 0) REJECTS ON THE PATH SPECIFIED BY W • SUPPOSE M AND MJ ACCEPT THE SAME LANGUAGE • M(1 J 01 M 0) ACCEPTS IF M(1 J 01 M 000) M(1 J 01 M 001) M(1 J 01 M 010) M(1 J 01 M 011) ACCEPT

M(1 j 01 m 0) FORTNOW-SANTHANAM M(1 j 01 m 00) • DEFINE A NTM M THAT ON INPUT J 101 M 0 W • IF |W| M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) ≤ T(J+M+2) THEN ACCEPT IF BOTH MJ(1 J 01 M 0 W 0) AND MJ(1 J 01 M 0 W 1) ACCEPT • M(1 j 01 m 01) 000 001 010 IF |W| > T(J+M+2 ) THEN ACCEPT IF MJ(1 J 01 M 0) REJECTS ON THE PATH SPECIFIED BY W • SUPPOSE M AND MJ ACCEPT THE SAME LANGUAGE • M(1 J 01 M 0) ACCEPTS IF M(1 J 01 M 0000) M(1 J 01 M 0001) M(1 J 01 M 0010) M(1 J 01 M 0011) M(1 J 01 M 0100) M(1 J 01 M 0101) M(1 J 01 M 0110) M(1 J 01 M 0111) ACCEPT 011 100 101 110 111

M(1 j 01 m 0) FORTNOW-SANTHANAM M(1 j 01 m 00) • DEFINE A NTM M THAT ON INPUT 101 0 W J • IF |W| M ≤ T(J+M+2) THEN ACCEPT IF BOTH MJ(1 01 0 W 0) AND MJ(1 01 0 W 1) ACCEPT J • M J M IF |W| > T(J+M+2 ) THEN ACCEPT IF MJ(1 J 01 M 0) REJECTS ON THE PATH SPECIFIED BY W M(1 j 01 m 01) M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) 000 001 010 00… 00 011 100 101 … • SUPPOSE M AND MJ ACCEPT THE SAME LANGUAGE • M(1 01 0) ACCEPTS IF J M M(1 J 01 M 0 W) FOR ALL W M(1 J 01 M 0) REJECTS ON ALL PATHS W CONTRADICTION M(1 j 01 m 0) 110 11. . 11 111

M(1 j 01 m 0) M(1 j 01 m 01) M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) 000 001 010 00… 00 011 100 101 … 110 111 11. . 11 M(1 j 01 m 0) DEPTH IS LENGTH OF THE WITNESS. PREVIOUSLY EXPONENTIAL IN WITNESS SIZE.

OLD NONUNIFORM LOWER BOUNDS FOR UNIFORM CLASSES • GOAL: FOR CONTANTS A, B WITH 1≤A<B, • NTIME(NB) IS NOT CONTAINED INNTIME(NA) WITHN 1/B BITS OF ADVICE • HOW DO WE DO THIS FORDTIME? • DTIME(NB) IS NOT CONTAINED INDTIME(NA) WITHN BITS OF ADVICE • USE INPUT AS ADVICE • M(X) SIMULATESM|X|(X) USING ADVICE X FOR |AX|STEPS AND DO THE OPPOSITE

M(1 j 01 m 0) M(1 j 01 m 01) M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) 000 001 010 00… 00 011 100 101 … 110 111 11. . 11 M(1 j 01 m 0) • THEOREM: NTIME(NB) IS NOT CONTAINED INNTIME(NA) WITHN 1/B BITS OFADVICE FOR B>A • N IS |1 J 01 M 0| = J+M+2 • NEED TO ENCODE ADVICE FOR ALL LENGTHS UP TON+W (W = WITNESS SIZE <NA) • DO THE MATH: CAN HANDLEN 1/B BITS OF ADVICE FOR B>A

INFINITELY OFTEN HIERARCHIES • FOR B>A THERE ARE LANGUAGES IN DTIME(NB) THAT DIFFER FROM EVERY LANGUAGE IN DTIME(NA) FOR ALL BUT FINITELY MANY INPUT LENGTHS. • THERE IS AN ORACLE RELATIVE TO WHICHNEXP IS IN I. O. -NP • BUHRMAN-FORTNOW-SANTHANAM 2009

M(1 j 01 m 0) M(1 j 01 m 01) M(1 j 01 m 000) M(1 j 01 m 001) M(1 j 01 m 010) M(1 j 01 m 011) 000 001 010 00… 00 011 100 101 … 110 111 11. . 11 M(1 j 01 m 0) • PROOF REQUIRES COLLAPSE AT ALL THESE LENGTHS • GET I. O. IF WE COULD EMBED THE ENTIRE DIAGRAM IN ONE INPUT LENGTH • SIZE OF DIAGRAM IS ROUGHLY 2 W FORW = WITNESS SIZE. • WE CAN GET NTIME I. O. HIERARCHY IF WE LIMIT WITNESS SIZE TO SUBLINEAR

RE NOT IN RTIME(NC)/N 1/2 C • CASE 1: SAT IN BPP/N 1/2 C • BY TRYING ALL ADVICE AND SELF-REDUCTION, SAT IN RTIME(2 N 1/2 C POLY(N)) • NTIME(N 3 C/2) IS CONTAINED INRE • NTIME(N 3 C/2) IS NOT CONTAINED INNTIME(NC)/N 1/2 C BY MAIN THEOREM • NTIME(NC)/N 1/2 C CONTAINSRTIME(NC)/N 1/2 C

RE NOT IN RTIME(NC)/N 1/2 C • CASE 2: SAT NOT INBPP/N 1/2 C • SAT IS IN E IS IN RE • BPP/N 1/2 C CONTAINSRTIME(NC)/N 1/2 C

NP DOES NOT HAVE NP-UNIFORM NONDETERMINISTIC CIRCUITS OF SIZE NK • SUPPOSE NP DOES HAVE NP-UNIFORM NONDETERMINISTIC CIRCUITS OF SIZE NK • WE SHOW EVERY L IN NP CAN BE SIMULATED INNTIME(N 2 K+2)/N 1/4 K • PROOF USES PADDED VERSION OFL WITH ADVICE DESCRIBING CIRCUIT AND CENSUS • CONTRADICTS MAIN THEOREM FOR NTIME(N 4 K)

CONCLUSIONS • TIGHTER BOUNDS, PERHAPS SUBLINEAR ADVICE • BOUNDS FOR ADVICE IN I. O. SETTING • ONE IN A SERIES OF PAPER SHOWING SEPARATIONS FOR NONDETERMINISTIC COMPUTATION • COULD SAT NOT INLOGSPACE OR SAT NOT IN DETERMINISTIC LINEAR TIME BE FAR AWAY?