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Integer TRS Innermost pair #487100565
details
property
value
status
complete
benchmark
poly4.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n147.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
1.09568 seconds
cpu usage
1.11459
user time
0.624795
system time
0.489795
max virtual memory
124240.0
max residence set size
8580.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] The dependency graph for this problem is: 0 -> 0 1 -> 1 2 -> 0, 2, 3 3 -> 1, 2, 3 Where: 0) eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1) eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 2) eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 3) eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We have the following SCCs. { 2, 3 } { 0 } { 1 } DP problem for innermost termination. P = eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 This gives the following inequalities: I4 >= I1 && I3 < I0 ==> I0 + -1 * I3 > I0 + -1 * (I3 + 1) with I0 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 This gives the following inequalities: j < bn && i >= an ==> bn + -1 * j > bn + -1 * (j + 1) with bn + -1 * j >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 This gives the following inequalities: I9 < I6 && I8 < I5 ==> I5 + -1 * I8 > I5 + -1 * (I8 + 1) with I5 + -1 * I8 >= 0 I14 < I11 && I13 < I10 ==> I10 + -1 * I13 >= I10 + -1 * I13 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 This gives the following inequalities: I14 < I11 && I13 < I10 ==> I11 + -1 * I14 > I11 + -1 * (I14 + 1) with I11 + -1 * I14 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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