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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995093
details
property
value
status
complete
benchmark
a.11.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n046.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
8.43214 seconds
cpu usage
8.58171
user time
4.55246
system time
4.02925
max virtual memory
735288.0
max residence set size
11988.0
stage attributes
key
value
starexec-result
YES
output
8.49/8.43 YES 8.49/8.43 8.49/8.43 DP problem for innermost termination. 8.49/8.43 P = 8.49/8.43 eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] 8.49/8.43 eval_2#(I0, I1, I2) -> eval_1#(I0, I1, I2 + 1) [I0 > I2] 8.49/8.43 eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7] 8.49/8.43 R = 8.49/8.43 eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] 8.49/8.43 eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] 8.49/8.43 eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] 8.49/8.43 8.49/8.43 We use the reverse value criterion with the projection function NU: 8.49/8.43 NU[eval_1#(z1,z2,z3)] = z1 + -1 * z3 8.49/8.43 NU[eval_2#(z1,z2,z3)] = z1 + -1 * z3 8.49/8.43 8.49/8.43 This gives the following inequalities: 8.49/8.43 z >= x ==> x + -1 * z >= x - 1 + -1 * z 8.49/8.43 I0 > I2 ==> I0 + -1 * I2 > I0 + -1 * (I2 + 1) with I0 + -1 * I2 >= 0 8.49/8.43 I3 > I5 ==> I3 + -1 * I5 >= I3 + -1 * I5 8.49/8.43 I6 > I7 ==> I6 + -1 * I8 >= I6 + -1 * I8 8.49/8.43 8.49/8.43 We remove all the strictly oriented dependency pairs. 8.49/8.43 8.49/8.43 DP problem for innermost termination. 8.49/8.43 P = 8.49/8.43 eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] 8.49/8.43 eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7] 8.49/8.43 R = 8.49/8.43 eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] 8.49/8.43 eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] 8.49/8.43 eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] 8.49/8.43 8.49/8.43 We use the reverse value criterion with the projection function NU: 8.49/8.43 NU[eval_1#(z1,z2,z3)] = z1 + -1 * z2 8.49/8.43 NU[eval_2#(z1,z2,z3)] = z1 - 1 + -1 * z2 8.49/8.43 8.49/8.43 This gives the following inequalities: 8.49/8.43 z >= x ==> x - 1 + -1 * y >= x - 1 + -1 * y 8.49/8.43 I3 > I5 ==> I3 - 1 + -1 * I4 >= I3 + -1 * (I4 + 1) 8.49/8.43 I6 > I7 ==> I6 + -1 * I7 > I6 - 1 + -1 * I7 with I6 + -1 * I7 >= 0 8.49/8.43 8.49/8.43 We remove all the strictly oriented dependency pairs. 8.49/8.43 8.49/8.43 DP problem for innermost termination. 8.49/8.43 P = 8.49/8.43 eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] 8.49/8.43 eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 R = 8.49/8.43 eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] 8.49/8.43 eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] 8.49/8.43 eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] 8.49/8.43 8.49/8.43 The dependency graph for this problem is: 8.49/8.43 0 -> 8.49/8.43 2 -> 8.49/8.43 Where: 8.49/8.43 0) eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] 8.49/8.43 2) eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] 8.49/8.43 8.49/8.43 We have the following SCCs. 8.49/8.43 8.49/11.40 EOF
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return to Integer_TRS_Innermost 2019-03-21 04.53