Remove dead python code #101
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JasonGross
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JasonGross
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Probably if the lists are the same lengths, then we want to compare them element-wise rather than all at once. It's way too verbose to keep expanding them, so we only do that when lists are not the same length. We now get error messages such as ``` Unable to unify: [inr [378, 381, 384]] == [inr [101, 106, 108]] Could not unify the values at index 0: [mit-plv#378, mit-plv#381, mit-plv#384] != [mit-plv#101, mit-plv#106, mit-plv#108] index 0: mit-plv#378 != mit-plv#101 (slice 0 44, [mit-plv#377]) != (slice 0 44, [mit-plv#98]) index 0: mit-plv#377 != mit-plv#98 (add 64, [mit-plv#345, mit-plv#375]) != (add 64, [mit-plv#57, mit-plv#96]) index 0: mit-plv#345 != mit-plv#57 (slice 0 44, [mit-plv#337]) != (slice 0 44, [#44]) index 0: mit-plv#337 != #44 (add 64, [#41, mit-plv#334]) != (add 64, [#25, #41]) index 1: mit-plv#334 != #25 (mul 64, [#1, mit-plv#331]) != (mul 64, [#0, #1, #22]) [(add 64, [mit-plv#329, mit-plv#329])] != [#0, (const 20, [])] [(add 64, [(mul 64, [#7, mit-plv#328]), (mul 64, [#7, mit-plv#328])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, mit-plv#327])]), (mul 64, [(const 2, []), (add 64, [#0, mit-plv#327])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, mit-plv#326])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, mit-plv#326])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])])])] != [(old 64 0, []), (const 20, [])] ``` The second to last line is generally the one to look at; the last line adds a bit more detail to it. Perhaps we should instead list out the values of indices rather than expanding one additional level?
JasonGross
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Feb 21, 2022
Probably if the lists are the same lengths, then we want to compare them element-wise rather than all at once. It's way too verbose to keep expanding them, so we only do that when lists are not the same length. We now get error messages such as ``` Unable to unify: [inr [378, 381, 384]] == [inr [101, 106, 108]] Could not unify the values at index 0: [mit-plv#378, mit-plv#381, mit-plv#384] != [mit-plv#101, mit-plv#106, mit-plv#108] index 0: mit-plv#378 != mit-plv#101 (slice 0 44, [mit-plv#377]) != (slice 0 44, [mit-plv#98]) index 0: mit-plv#377 != mit-plv#98 (add 64, [mit-plv#345, mit-plv#375]) != (add 64, [mit-plv#57, mit-plv#96]) index 0: mit-plv#345 != mit-plv#57 (slice 0 44, [mit-plv#337]) != (slice 0 44, [#44]) index 0: mit-plv#337 != #44 (add 64, [#41, mit-plv#334]) != (add 64, [#25, #41]) index 1: mit-plv#334 != #25 (mul 64, [#1, mit-plv#331]) != (mul 64, [#0, #1, #22]) [(add 64, [mit-plv#329, mit-plv#329])] != [#0, (const 20, [])] [(add 64, [(mul 64, [#7, mit-plv#328]), (mul 64, [#7, mit-plv#328])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, mit-plv#327])]), (mul 64, [(const 2, []), (add 64, [#0, mit-plv#327])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, mit-plv#326])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, mit-plv#326])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])])])] != [(old 64 0, []), (const 20, [])] ``` The second to last line is generally the one to look at; the last line adds a bit more detail to it. Perhaps we should instead list out the values of indices rather than expanding one additional level?
JasonGross
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that referenced
this pull request
Feb 22, 2022
Probably if the lists are the same lengths, then we want to compare them element-wise rather than all at once. It's way too verbose to keep expanding them, so we only do that when lists are not the same length. We now get error messages such as ``` Unable to unify: [inr [378, 381, 384]] == [inr [101, 106, 108]] Could not unify the values at index 0: [#378, #381, #384] != [#101, #106, #108] index 0: #378 != #101 (slice 0 44, [#377]) != (slice 0 44, [#98]) index 0: #377 != #98 (add 64, [#345, #375]) != (add 64, [#57, #96]) index 0: #345 != #57 (slice 0 44, [#337]) != (slice 0 44, [#44]) index 0: #337 != #44 (add 64, [#41, #334]) != (add 64, [#25, #41]) index 1: #334 != #25 (mul 64, [#1, #331]) != (mul 64, [#0, #1, #22]) [(add 64, [#329, #329])] != [#0, (const 20, [])] [(add 64, [(mul 64, [#7, #328]), (mul 64, [#7, #328])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, #327])]), (mul 64, [(const 2, []), (add 64, [#0, #327])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, #326])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, #326])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [#0, (mul 64, [#0, (const 4, [])])])])])] != [#0, (const 20, [])] [(add 64, [(mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])]), (mul 64, [(const 2, []), (add 64, [(old 64 0, []), (mul 64, [(old 64 0, []), (const 4, [])])])])])] != [(old 64 0, []), (const 20, [])] ``` The second to last line is generally the one to look at; the last line adds a bit more detail to it. Perhaps we should instead list out the values of indices rather than expanding one additional level?
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This is on top of #97, and should be merged after #97. (I don't want to update #97 because I have two PRs on top of it, and Jade mentioned working on top of it, so I don't want to rebase them.) I'm making this a PR just so that I don't forget to do this.