Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integer_Transition_Systems 2019-03-29 01.54 pair #432274563
details
property
value
status
complete
benchmark
simple_control_on_input.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.96009 seconds
cpu usage
0.979199
user time
0.547228
system time
0.431971
max virtual memory
126024.0
max residence set size
8684.0
stage attributes
key
value
starexec-result
YES
output
0.86/0.95 YES 0.86/0.95 0.86/0.95 DP problem for innermost termination. 0.86/0.95 P = 0.86/0.95 f7#(x1, x2) -> f6#(x1, x2) 0.86/0.95 f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 0.86/0.95 f4#(I2, I3) -> f3#(I2, I3) 0.86/0.95 f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] 0.86/0.95 f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] 0.86/0.95 f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] 0.86/0.95 R = 0.86/0.95 f7(x1, x2) -> f6(x1, x2) 0.86/0.95 f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 0.86/0.95 f4(I2, I3) -> f3(I2, I3) 0.86/0.95 f5(I4, I5) -> f4(I4, I5) [1 <= I5] 0.86/0.95 f5(I6, I7) -> f1(I6, I7) [I7 <= 0] 0.86/0.95 f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 f3(I10, I11) -> f1(I10, I11) [I10 <= I11] 0.86/0.95 f1(I12, I13) -> f2(I12, I13) 0.86/0.95 0.86/0.95 The dependency graph for this problem is: 0.86/0.95 0 -> 1 0.86/0.95 1 -> 3, 4 0.86/0.95 2 -> 5, 6 0.86/0.95 3 -> 2 0.86/0.95 4 -> 0.86/0.95 5 -> 2 0.86/0.95 6 -> 0.86/0.95 Where: 0.86/0.95 0) f7#(x1, x2) -> f6#(x1, x2) 0.86/0.95 1) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 0.86/0.95 2) f4#(I2, I3) -> f3#(I2, I3) 0.86/0.95 3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] 0.86/0.95 4) f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] 0.86/0.95 5) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 6) f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] 0.86/0.95 0.86/0.95 We have the following SCCs. 0.86/0.95 { 2, 5 } 0.86/0.95 0.86/0.95 DP problem for innermost termination. 0.86/0.95 P = 0.86/0.95 f4#(I2, I3) -> f3#(I2, I3) 0.86/0.95 f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 R = 0.86/0.95 f7(x1, x2) -> f6(x1, x2) 0.86/0.95 f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 0.86/0.95 f4(I2, I3) -> f3(I2, I3) 0.86/0.95 f5(I4, I5) -> f4(I4, I5) [1 <= I5] 0.86/0.95 f5(I6, I7) -> f1(I6, I7) [I7 <= 0] 0.86/0.95 f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 f3(I10, I11) -> f1(I10, I11) [I10 <= I11] 0.86/0.95 f1(I12, I13) -> f2(I12, I13) 0.86/0.95 0.86/0.95 We use the reverse value criterion with the projection function NU: 0.86/0.95 NU[f3#(z1,z2)] = z1 + -1 * (1 + z2) 0.86/0.95 NU[f4#(z1,z2)] = z1 + -1 * (1 + z2) 0.86/0.95 0.86/0.95 This gives the following inequalities: 0.86/0.95 ==> I2 + -1 * (1 + I3) >= I2 + -1 * (1 + I3) 0.86/0.95 1 + I9 <= I8 ==> I8 + -1 * (1 + I9) > I8 + -1 * (1 + (1 + I9)) with I8 + -1 * (1 + I9) >= 0 0.86/0.95 0.86/0.95 We remove all the strictly oriented dependency pairs. 0.86/0.95 0.86/0.95 DP problem for innermost termination. 0.86/0.95 P = 0.86/0.95 f4#(I2, I3) -> f3#(I2, I3) 0.86/0.95 R = 0.86/0.95 f7(x1, x2) -> f6(x1, x2) 0.86/0.95 f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 0.86/0.95 f4(I2, I3) -> f3(I2, I3) 0.86/0.95 f5(I4, I5) -> f4(I4, I5) [1 <= I5] 0.86/0.95 f5(I6, I7) -> f1(I6, I7) [I7 <= 0] 0.86/0.95 f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] 0.86/0.95 f3(I10, I11) -> f1(I10, I11) [I10 <= I11] 0.86/0.95 f1(I12, I13) -> f2(I12, I13) 0.86/0.95 0.86/0.95 The dependency graph for this problem is: 0.86/0.95 2 -> 0.86/0.95 Where: 0.86/0.95 2) f4#(I2, I3) -> f3#(I2, I3) 0.86/0.95 0.86/0.95 We have the following SCCs. 0.86/0.95 0.86/3.93 EOF
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Integer_Transition_Systems 2019-03-29 01.54