Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integer_TRS_Innermost 2019-03-21 04.53 pair #429995123
details
property
value
status
complete
benchmark
Ex09.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n049.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.232244 seconds
cpu usage
0.23648
user time
0.119974
system time
0.116506
max virtual memory
113176.0
max residence set size
7764.0
stage attributes
key
value
starexec-result
MAYBE
output
0.00/0.23 MAYBE 0.00/0.23 0.00/0.23 DP problem for innermost termination. 0.00/0.23 P = 0.00/0.23 round#(x) -> if#(x % 2 = 0, x, x + 1) 0.00/0.23 minusNat#(true, I2, y) -> minus#(I2, round(I2)) 0.00/0.23 minusNat#(true, I2, y) -> round#(I2) 0.00/0.23 minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 R = 0.00/0.23 if(false, u, v) -> v 0.00/0.23 if(true, I0, I1) -> I0 0.00/0.23 round(x) -> if(x % 2 = 0, x, x + 1) 0.00/0.23 minusNat(true, I2, y) -> minus(I2, round(I2)) 0.00/0.23 minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 0.00/0.23 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.23 0.00/0.23 DP problem for innermost termination. 0.00/0.23 P = 0.00/0.23 round#(x) -> if#(x % 2 = 0, x, x + 1) 0.00/0.23 minusNat#(true, I2, y) -> minus#(I2, round(I2)) 0.00/0.23 minusNat#(true, I2, y) -> round#(I2) 0.00/0.23 minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] 0.00/0.23 minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] 0.00/0.23 R = 0.00/0.23 if(false, u, v) -> v 0.00/0.23 if(true, I0, I1) -> I0 0.00/0.23 round(x) -> if(x % 2 = 0, x, x + 1) 0.00/0.23 minusNat(true, I2, y) -> minus(I2, round(I2)) 0.00/0.23 minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 0.00/0.23 The dependency graph for this problem is: 0.00/0.23 0 -> 0.00/0.23 1 -> 5, 4, 3 0.00/0.23 2 -> 0 0.00/0.23 3 -> 0.00/0.23 4 -> 5, 4, 3 0.00/0.23 5 -> 0 0.00/0.23 Where: 0.00/0.23 0) round#(x) -> if#(x % 2 = 0, x, x + 1) 0.00/0.23 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) 0.00/0.23 2) minusNat#(true, I2, y) -> round#(I2) 0.00/0.23 3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] 0.00/0.23 5) minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] 0.00/0.23 0.00/0.23 We have the following SCCs. 0.00/0.23 { 4 } 0.00/0.23 0.00/0.23 DP problem for innermost termination. 0.00/0.23 P = 0.00/0.23 minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] 0.00/0.23 R = 0.00/0.23 if(false, u, v) -> v 0.00/0.23 if(true, I0, I1) -> I0 0.00/0.23 round(x) -> if(x % 2 = 0, x, x + 1) 0.00/0.23 minusNat(true, I2, y) -> minus(I2, round(I2)) 0.00/0.23 minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 0.00/0.23 0.00/3.21 EOF
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Integer_TRS_Innermost 2019-03-21 04.53