Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integer_Transition_Systems 2019-03-29 01.54 pair #432276252
details
property
value
status
complete
benchmark
n-15.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n091.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.47205 seconds
cpu usage
1.4997
user time
0.778183
system time
0.721514
max virtual memory
353336.0
max residence set size
8560.0
stage attributes
key
value
starexec-result
MAYBE
output
1.42/1.47 MAYBE 1.42/1.47 1.42/1.47 DP problem for innermost termination. 1.42/1.47 P = 1.42/1.47 f5#(x1) -> f1#(x1) 1.42/1.47 f4#(I0) -> f2#(I0) 1.42/1.47 f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3#(I2) -> f2#(I2) 1.42/1.47 f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 f1#(I4) -> f2#(I4) 1.42/1.47 R = 1.42/1.47 f5(x1) -> f1(x1) 1.42/1.47 f4(I0) -> f2(I0) 1.42/1.47 f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3(I2) -> f2(I2) 1.42/1.47 f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 f1(I4) -> f2(I4) 1.42/1.47 1.42/1.47 The dependency graph for this problem is: 1.42/1.47 0 -> 5 1.42/1.47 1 -> 2, 4 1.42/1.47 2 -> 1 1.42/1.47 3 -> 2, 4 1.42/1.47 4 -> 3 1.42/1.47 5 -> 2, 4 1.42/1.47 Where: 1.42/1.47 0) f5#(x1) -> f1#(x1) 1.42/1.47 1) f4#(I0) -> f2#(I0) 1.42/1.47 2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 3) f3#(I2) -> f2#(I2) 1.42/1.47 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 5) f1#(I4) -> f2#(I4) 1.42/1.47 1.42/1.47 We have the following SCCs. 1.42/1.47 { 1, 2, 3, 4 } 1.42/1.47 1.42/1.47 DP problem for innermost termination. 1.42/1.47 P = 1.42/1.47 f4#(I0) -> f2#(I0) 1.42/1.47 f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3#(I2) -> f2#(I2) 1.42/1.47 f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 R = 1.42/1.47 f5(x1) -> f1(x1) 1.42/1.47 f4(I0) -> f2(I0) 1.42/1.47 f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3(I2) -> f2(I2) 1.42/1.47 f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 f1(I4) -> f2(I4) 1.42/1.47 1.42/1.47 We use the reverse value criterion with the projection function NU: 1.42/1.47 NU[f3#(z1)] = -1 + -1 + z1 + -1 * 0 1.42/1.47 NU[f2#(z1)] = -1 + -1 + z1 + -1 * 0 1.42/1.47 NU[f4#(z1)] = -1 + -1 + z1 + -1 * 0 1.42/1.47 1.42/1.47 This gives the following inequalities: 1.42/1.47 ==> -1 + -1 + I0 + -1 * 0 >= -1 + -1 + I0 + -1 * 0 1.42/1.47 0 <= -1 + -1 + I1 ==> -1 + -1 + I1 + -1 * 0 > -1 + -1 + (-1 + I1) + -1 * 0 with -1 + -1 + I1 + -1 * 0 >= 0 1.42/1.47 ==> -1 + -1 + I2 + -1 * 0 >= -1 + -1 + I2 + -1 * 0 1.42/1.47 -1 + I3 <= 0 ==> -1 + -1 + I3 + -1 * 0 >= -1 + -1 + (-1 + I3) + -1 * 0 1.42/1.47 1.42/1.47 We remove all the strictly oriented dependency pairs. 1.42/1.47 1.42/1.47 DP problem for innermost termination. 1.42/1.47 P = 1.42/1.47 f4#(I0) -> f2#(I0) 1.42/1.47 f3#(I2) -> f2#(I2) 1.42/1.47 f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 R = 1.42/1.47 f5(x1) -> f1(x1) 1.42/1.47 f4(I0) -> f2(I0) 1.42/1.47 f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3(I2) -> f2(I2) 1.42/1.47 f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 f1(I4) -> f2(I4) 1.42/1.47 1.42/1.47 The dependency graph for this problem is: 1.42/1.47 1 -> 4 1.42/1.47 3 -> 4 1.42/1.47 4 -> 3 1.42/1.47 Where: 1.42/1.47 1) f4#(I0) -> f2#(I0) 1.42/1.47 3) f3#(I2) -> f2#(I2) 1.42/1.47 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 1.42/1.47 We have the following SCCs. 1.42/1.47 { 3, 4 } 1.42/1.47 1.42/1.47 DP problem for innermost termination. 1.42/1.47 P = 1.42/1.47 f3#(I2) -> f2#(I2) 1.42/1.47 f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 R = 1.42/1.47 f5(x1) -> f1(x1) 1.42/1.47 f4(I0) -> f2(I0) 1.42/1.47 f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] 1.42/1.47 f3(I2) -> f2(I2) 1.42/1.47 f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] 1.42/1.47 f1(I4) -> f2(I4) 1.42/1.47
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