details

property	value
status	complete
benchmark	random_full.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n082.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.265124082565 seconds
cpu usage	0.269873915
max memory	9355264.0

stage attributes

key	value
output-size	2503
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] rand#(I3, y) -> id_dec#(y) [0 > I3] rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0] rand#(I4, B0) -> id_inc#(B0) [I4 > 0] random#(I6) -> rand#(I6, w(0)) R = id_dec(w(x)) -> w(x - 1) id_dec(w(I0)) -> w(I0) id_inc(w(I1)) -> w(I1 + 1) id_inc(w(I2)) -> w(I2) rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0] rand(I5, B1) -> B1 [I5 = 0] random(I6) -> rand(I6, w(0)) The dependency graph for this problem is: 0 -> 0, 1 1 -> 2 -> 2, 3 3 -> 4 -> 0, 1, 2, 3 Where: 0) rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] 1) rand#(I3, y) -> id_dec#(y) [0 > I3] 2) rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0] 3) rand#(I4, B0) -> id_inc#(B0) [I4 > 0] 4) random#(I6) -> rand#(I6, w(0)) We have the following SCCs. { 2 } { 0 } DP problem for innermost termination. P = rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] R = id_dec(w(x)) -> w(x - 1) id_dec(w(I0)) -> w(I0) id_inc(w(I1)) -> w(I1 + 1) id_inc(w(I2)) -> w(I2) rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0] rand(I5, B1) -> B1 [I5 = 0] random(I6) -> rand(I6, w(0)) We use the reverse value criterion with the projection function NU: NU[rand#(z1,z2)] = 0 + -1 * z1 This gives the following inequalities: 0 > I3 ==> 0 + -1 * I3 > 0 + -1 * (I3 + 1) with 0 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0] R = id_dec(w(x)) -> w(x - 1) id_dec(w(I0)) -> w(I0) id_inc(w(I1)) -> w(I1 + 1) id_inc(w(I2)) -> w(I2) rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0] rand(I5, B1) -> B1 [I5 = 0] random(I6) -> rand(I6, w(0)) We use the reverse value criterion with the projection function NU: NU[rand#(z1,z2)] = z1 This gives the following inequalities: I4 > 0 ==> I4 > I4 - 1 with I4 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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all output return to Integ TRS Inner 43557