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Integer TRS Innermost pair #487100557
details
property
value
status
complete
benchmark
csharp1.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n149.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
2.28422 seconds
cpu usage
4.52951
user time
2.3127
system time
2.21681
max virtual memory
132704.0
max residence set size
8612.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) The dependency graph for this problem is: 0 -> 1 -> 3 2 -> 1 3 -> 4, 0 4 -> 1 Where: 0) b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 4) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))] We have the following SCCs. { 1, 3, 4 } DP problem for innermost termination. P = b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) We use the extended value criterion with the projection function NU: NU[b14#(x0,x1,x2)] = x1 - 2 NU[b10#(x0,x1,x2)] = x1 - 2 NU[b15#(x0,x1,x2)] = -x0 + x1 - 2 This gives the following inequalities: ==> -I0 + I1 - 2 >= (I1 - I0) - 2 ==> I7 - 2 >= I7 - 2 sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))) ==> sv23_37 - 2 > -sv14_14 + sv23_37 - 2 with sv23_37 - 2 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) b10#(I6, I7, I8) -> b14#(I6, I7, I8) R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) The dependency graph for this problem is: 1 -> 3 3 -> Where: 1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) We have the following SCCs.
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return to Integer TRS Innermost