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Integ TRS Inner 43557 pair #381740780
details
property
value
status
complete
benchmark
random_full_no_wrap.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n056.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.231967926025 seconds
cpu usage
0.242806516
max memory
9187328.0
stage attributes
key
value
output-size
2416
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] rand#(I3, y) -> id_dec#(y) [0 > I3] rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0] rand#(I4, I5) -> id_inc#(I5) [I4 > 0] random#(I8) -> rand#(I8, 0) R = id_dec(x) -> x - 1 id_dec(I0) -> I0 id_inc(I1) -> I1 + 1 id_inc(I2) -> I2 rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0] rand(I6, I7) -> I7 [I6 = 0] random(I8) -> rand(I8, 0) The dependency graph for this problem is: 0 -> 0, 1 1 -> 2 -> 2, 3 3 -> 4 -> 0, 1, 2, 3 Where: 0) rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] 1) rand#(I3, y) -> id_dec#(y) [0 > I3] 2) rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0] 3) rand#(I4, I5) -> id_inc#(I5) [I4 > 0] 4) random#(I8) -> rand#(I8, 0) We have the following SCCs. { 2 } { 0 } DP problem for innermost termination. P = rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3] R = id_dec(x) -> x - 1 id_dec(I0) -> I0 id_inc(I1) -> I1 + 1 id_inc(I2) -> I2 rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0] rand(I6, I7) -> I7 [I6 = 0] random(I8) -> rand(I8, 0) We use the reverse value criterion with the projection function NU: NU[rand#(z1,z2)] = 0 + -1 * z1 This gives the following inequalities: 0 > I3 ==> 0 + -1 * I3 > 0 + -1 * (I3 + 1) with 0 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0] R = id_dec(x) -> x - 1 id_dec(I0) -> I0 id_inc(I1) -> I1 + 1 id_inc(I2) -> I2 rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3] rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0] rand(I6, I7) -> I7 [I6 = 0] random(I8) -> rand(I8, 0) We use the reverse value criterion with the projection function NU: NU[rand#(z1,z2)] = z1 This gives the following inequalities: I4 > 0 ==> I4 > I4 - 1 with I4 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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