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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995023
details
property
value
status
complete
benchmark
practical1.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n123.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
1.51003 seconds
cpu usage
1.54087
user time
0.841049
system time
0.69982
max virtual memory
687896.0
max residence set size
8688.0
stage attributes
key
value
starexec-result
YES
output
1.49/1.50 YES 1.49/1.50 1.49/1.50 DP problem for innermost termination. 1.49/1.50 P = 1.49/1.50 eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1] 1.49/1.50 eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 >= 0] 1.49/1.50 R = 1.49/1.50 eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] 1.49/1.50 eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] 1.49/1.50 1.49/1.50 We use the reverse value criterion with the projection function NU: 1.49/1.50 NU[eval_1#(z1,z2)] = z1 + -1 * 0 1.49/1.50 NU[eval_2#(z1,z2)] = z1 - 1 + -1 * 0 1.49/1.50 1.49/1.50 This gives the following inequalities: 1.49/1.50 j > i - 1 ==> i - 1 + -1 * 0 >= i - 1 + -1 * 0 1.49/1.50 I1 <= I0 - 1 ==> I0 - 1 + -1 * 0 >= I0 - 1 + -1 * 0 1.49/1.50 I2 >= 0 ==> I2 + -1 * 0 > I2 - 1 + -1 * 0 with I2 + -1 * 0 >= 0 1.49/1.50 1.49/1.50 We remove all the strictly oriented dependency pairs. 1.49/1.50 1.49/1.50 DP problem for innermost termination. 1.49/1.50 P = 1.49/1.50 eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1] 1.49/1.50 eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 R = 1.49/1.50 eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] 1.49/1.50 eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] 1.49/1.50 1.49/1.50 The dependency graph for this problem is: 1.49/1.50 0 -> 1.49/1.50 1 -> 0, 1 1.49/1.50 Where: 1.49/1.50 0) eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1] 1.49/1.50 1) eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 1.49/1.50 We have the following SCCs. 1.49/1.50 { 1 } 1.49/1.50 1.49/1.50 DP problem for innermost termination. 1.49/1.50 P = 1.49/1.50 eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 R = 1.49/1.50 eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] 1.49/1.50 eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] 1.49/1.50 eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] 1.49/1.50 1.49/1.50 We use the reverse value criterion with the projection function NU: 1.49/1.50 NU[eval_2#(z1,z2)] = z1 - 1 + -1 * z2 1.49/1.50 1.49/1.50 This gives the following inequalities: 1.49/1.50 I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 0 1.49/1.50 1.49/1.50 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.49/4.49 EOF
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