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Integer TRS Innermost pair #487528892
details
property
value
status
complete
benchmark
A04.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.307931900024 seconds
cpu usage
0.268523039
max memory
9388032.0
stage attributes
key
value
output-size
1278
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = min#(I2, I3) -> if#(I2 < I3, I2, I3) cond#(y, x, y) -> minus#(x, y + 1) minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) minus#(I4, I5) -> min#(I4, I5) R = if(false, u, v) -> v if(true, I0, I1) -> I0 min(I2, I3) -> if(I2 < I3, I2, I3) cond(y, x, y) -> 1 + minus(x, y + 1) minus(I4, I5) -> cond(min(I4, I5), I4, I5) minus(I6, I6) -> 0 The dependency graph for this problem is: 0 -> 1 -> 2, 3 2 -> 1 3 -> 0 Where: 0) min#(I2, I3) -> if#(I2 < I3, I2, I3) 1) cond#(y, x, y) -> minus#(x, y + 1) 2) minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) 3) minus#(I4, I5) -> min#(I4, I5) We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = cond#(y, x, y) -> minus#(x, y + 1) minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) R = if(false, u, v) -> v if(true, I0, I1) -> I0 min(I2, I3) -> if(I2 < I3, I2, I3) cond(y, x, y) -> 1 + minus(x, y + 1) minus(I4, I5) -> cond(min(I4, I5), I4, I5) minus(I6, I6) -> 0
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