details

property	value
status	complete
benchmark	p-4.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	7.66484 seconds
cpu usage	7.79064
user time	3.99665
system time	3.79398
max virtual memory	743372.0
max residence set size	9360.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f5#(x1, x2, x3) -> f1#(x1, x2, x3) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] f1#(I12, I13, I14) -> f2#(I12, I13, I14) R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) The dependency graph for this problem is: 0 -> 5 1 -> 3, 4 2 -> 3, 4 3 -> 2 4 -> 1 5 -> 1 Where: 0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 3) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 5) f1#(I12, I13, I14) -> f2#(I12, I13, I14) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) We use the reverse value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z3 + -1 * (1 + z2) NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z2) NU[f2#(z1,z2,z3)] = z3 + -1 * (1 + z2) This gives the following inequalities: 1 + I0 <= I1 ==> I2 + -1 * (1 + I1) >= I2 + -1 * (1 + I1) ==> I5 + -1 * (1 + I4) >= I5 + -1 * (1 + I4) 1 + I7 <= I8 ==> I8 + -1 * (1 + I7) > I8 + -1 * (1 + (1 + I7)) with I8 + -1 * (1 + I7) >= 0 I11 <= I10 ==> I11 + -1 * (1 + I10) >= I11 + -1 * (1 + I10) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) The dependency graph for this problem is: 1 -> 4 2 -> 4 4 -> 1 Where: 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] We have the following SCCs. { 1, 4 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to ITS